3. Math Toolkit II: Kernels, GPs & Uncertainty Bounds

RKHS, GP regression, information gain, frequentist confidence bounds and the βt problem

Before you start

This module assumes:

How to study this module

Core reading. Read the kernel and reproducing-property definitions in Section 1, then compute the one-observation posterior and noise-free error bound. Follow with information gain, the meaning of simultaneous bands, and the computable formula in Section 6. Finish with the concrete failure modes in Section 7.

Practice route. Try the three readiness checks. If a skill is rusty, use its exact review link, then return. Work through Easy P1–P4, Medium P5–P8, and Hard P9–P12; each has a separate hint and a full worked solution. Reattempt a missed problem later without opening the answer.

Optional research reading. The Fourier/Sobolev characterization, full self-normalized martingale proofs, kernel-specific information-gain rates, and newer confidence-set refinements are research extensions. Their assumptions remain attached to the results; you can learn the posterior and safety logic before proving every concentration theorem.

Readiness check — Three prerequisite skills

Try these before opening the answer. The review links lead to earlier material.

  1. Solve $(1+1/4)a=2$. Review linear solves.
  2. If variance is $0.09$, what is the standard deviation? Review variance and standard deviation.
  3. What bound does Cauchy–Schwarz give for $|\langle u,v angle|$? Review inner products.
Show readiness answers

The solution is $a=2/(5/4)=8/5$. Standard deviation is the nonnegative square root, $0.3$. Cauchy–Schwarz gives $|\langle u,v\rangle|\le\|u\|\|v\|$. GP regression repeatedly solves the first kind of equation, and the RKHS error proof repeatedly uses the third inequality.

Book contents · Apply this chapter to a real decision · Glossary

Contents
1. Kernels & the RKHS 2. GP Regression = Kernel Ridge Regression 3. The Noise-Free Error Bound 4. Maximum Information Gain 5. Frequentist Confidence Bounds & βt 6. Practical and Rigorous Bounds (Fiedler, Scherer, Trimpe) 7. When the Assumptions Fail Walkthrough: Deriving the RKHS Error Bound Interactive: GP Confidence Bands and β Application lab & chapter review Exercises Graded practice: Easy, Medium & Hard Key Papers Flashcards

Almost every safe-exploration guarantee in Modules 4–6 (the deliberate exception is LoSBO in Module 5, whose safety argument uses only a Lipschitz constant (see Primer B) and a noise bound) and the data-driven region-of-attraction certificate of Module 11 rest on one statement: a Gaussian-process model (a model in which every finite collection of function values is jointly Gaussian) of an unknown function $f$ comes with a band $\mu_t(x)\pm\beta_t\sigma_t(x)$ that contains $f(x)$ for all $x$ and all times $t$ with probability at least $1-\delta$ (see Primer C). Whether a SafeOpt run is actually safe reduces to whether that band is honest. This page builds it from scratch: kernels and their RKHS, why the GP mean is a ridge regressor, why $\sigma_t$ measures ignorance even without probability, how the information gain enters, then the three classical frequentist bounds, the Trimpe-group refinement that made them computable (and the later correction of its constant), and the ways their assumptions fail.

Notation for this page
  • $f:D\to\mathbb R$ is the unknown function on the set of allowed inputs $D\subset\mathbb R^d$, assumed compact, i.e. closed and bounded like $[0,1]^d$ (see Primer 0): a continuous function on $D$ attains its maximum and minimum, and $D$ can be covered by finitely many balls of any radius. $k:D\times D\to\mathbb R$ is a kernel with RKHS $\mathcal H_k$ and norm $\|f\|_k$ (all defined in Sections 1–2; this box is only a reference). The assumption $\|f\|_k\le B$ bounds the norm, not its square (Srinivas et al. and SafeOpt bound $\|f\|_k^2$; translated explicitly in Section 5).
  • Data $(x_i,y_i)_{i\le t}$ with $y_i=f(x_i)+\varepsilon_i$ (see Primer C); $K_t=[k(x_i,x_j)]_{i,j\le t}$, $k_t(x)=[k(x_1,x),\dots,k(x_t,x)]^\top$, $y_t$, $f_{1:t}$, $\varepsilon_{1:t}$ the stacked vectors.
  • $\lambda\gt0$ is the regulariser / nominal noise variance in $(K_t+\lambda I)^{-1}$; $\sigma_n^2$ the physical noise variance; $R$ the sub-Gaussian noise parameter (see Primer C; Gaussian noise of standard deviation $\sigma_n$ has $R=\sigma_n$). They are different objects.
  • $\mu_t,\sigma_t$: posterior mean and standard deviation after $t$ points; band $\mu_t\pm\beta_t\sigma_t$ (Chowdhury–Gopalan convention, used throughout this section); $\gamma_t$: maximum information gain (in the GP modules $\gamma$ is never the RL discount); $\delta$: failure probability.
  • Clashes: the SafeOpt threshold $h$ of Module 4 is not the barrier $h(x)$ of Module 10; $\beta_t$ is not the slope bound $\beta$ of LipSDP (Module 12); the regulariser $\lambda$ is not the Lagrange multiplier of Modules 8–9. Where a cited paper uses $\beta$, $\gamma$ or $\rho$ for something else (Lederer et al., Capone et al., Tokmak et al., Whitehouse et al.), this is flagged on the spot.

For the infinite-dimensional concentration results in Sections 5–7, assume the RKHS is separable (it has a countable dense subset) and the feature map $x\mapsto k(x,\cdot)$ is measurable, so predictable inputs give predictable Hilbert-valued features. A continuous kernel on the compact Euclidean domain above satisfies these regularity conditions. They are explicit hypotheses of Whitehouse et al.'s Hilbert-space theorem; scalar predictability alone does not state them.

1. Kernels & the RKHS

Intuition — A starting example

Think in a finite feature space first. For the linear kernel $k(x,z)=xz$, write the feature as $\varphi(x)=x$ and a function as $f(x)=ax$. The function is represented by the single coefficient $a$, its RKHS norm is $|a|$, and $\langle f,k(x,\cdot)\rangle_k=ax=f(x)$. Thus “reproducing” means that an inner product evaluates the function. General kernels use more features, possibly infinitely many, but the evaluation rule and Cauchy–Schwarz still do the work. Primer A's function-space examples give a slower bridge.

In plain words. A kernel $k(x,x')$ says how similar we expect the function values at two inputs to be: close inputs, similar values. The RKHS is the space of functions you can build by adding up "bumps" $k(z,\cdot)$ centred at different points. Its norm $\|f\|_k$ measures how much bump material you need. A function that changes sharply over short distances needs a lot of it, so the norm acts as a smoothness measure. Safe-BO guarantees assume a known bound $\|f\|_k\le B$.

Symbols in this section

A kernel is a similarity measure between inputs that behaves like an inner product. That single property is what turns the whole GP machinery into linear algebra.

Definition — Positive-definite kernel
A symmetric $k:D\times D\to\mathbb R$ is a (positive semi-definite) kernel if for every finite set $x_1,\dots,x_m\in D$ and every $c\in\mathbb R^m$ $$\sum_{i,j=1}^m c_ic_j\,k(x_i,x_j)=c^\top Kc\ge0,\qquad K=[k(x_i,x_j)]_{i,j}.$$ It is positive definite if $K$ is invertible whenever the points are distinct. With $r=\|x-x'\|$ and length-scale $\ell$: $$k_{\mathrm{SE}}=e^{-r^2/(2\ell^2)},\qquad k_{\nu=3/2}=\Big(1+\tfrac{\sqrt3r}{\ell}\Big)e^{-\sqrt3r/\ell},\qquad k_{\nu=5/2}=\Big(1+\tfrac{\sqrt5r}{\ell}+\tfrac{5r^2}{3\ell^2}\Big)e^{-\sqrt5r/\ell},\qquad k_{\mathrm{lin}}=x^\top x'.$$ The Matérn family $k_\nu$ interpolates between the rough exponential kernel ($\nu=\tfrac12$) and the SE kernel ($\nu\to\infty$).
Definition — Reproducing kernel Hilbert space
$\mathcal H_k$ is the completion (see Primer A) of all finite expansions $f=\sum_ic_i\,k(z_i,\cdot)$ under the inner product $$\Big\langle\sum_ic_ik(z_i,\cdot),\ \sum_jd_jk(w_j,\cdot)\Big\rangle_k=\sum_{i,j}c_id_j\,k(z_i,w_j),\qquad\text{so}\qquad\|f\|_k^2=c^\top K_zc .$$ Here $k(z,\cdot)$ is the whole function $x\mapsto k(z,x)$ ($z$ is held fixed, the dot marks the argument), so $f(x)=\sum_ic_ik(z_i,x)$ at every query $x$; "completion" adds all limits of such expansions in this norm. When $K_z$ is singular, different coefficient vectors can represent the same function; they give it the same norm, and a combination of norm zero is the zero function (by the bound $|f(x)|\le\|f\|_k\sqrt{k(x,x)}$ below). By the Moore–Aronszajn theorem this construction works for every positive semi-definite kernel, and $\mathcal H_k$ is the unique Hilbert space of functions on $D$ that contains every $k(x,\cdot)$ and has the following property. Its defining feature is the reproducing property: for every $f\in\mathcal H_k$ and $x\in D$, $$\langle f,\,k(x,\cdot)\rangle_k=f(x).$$ Point evaluation is a bounded linear functional represented by the feature $\varphi(x):=k(x,\cdot)\in\mathcal H_k$, with $k(x,x')=\langle\varphi(x),\varphi(x')\rangle_k$ and, by Cauchy–Schwarz (see Primer A), $|f(x)|\le\|f\|_k\sqrt{k(x,x)}$.
Background — Bounded evaluation functionals

A linear functional is a linear rule $E$ that takes a function and returns a number; it is bounded if $|E(f)|\le C\|f\|_k$ for one constant $C$ and all $f$. Point evaluation $E_x(f)=f(x)$ is bounded with $C=\sqrt{k(x,x)}$. The reproducing property represents this rule by the function $\varphi(x)=k(x,\cdot)$: $E_x(f)=\langle f,\varphi(x)\rangle_k$. The kernel pseudometric $d_k$ below is the distance between these representing functions.

For the linear kernel $k(x,z)=xz$ on $\mathbb R$, a function is $f(x)=ax$ with norm $|a|$. Then $E_2(f)=2a$, so this functional has bound $2\|f\|_k$; its representing function is $k(2,\cdot):x\mapsto2x$.

The reproducing property is the only tool needed for most of this page. Two immediate consequences:

Lemma — RKHS functions are continuous in the kernel pseudometric
For all $f\in\mathcal H_k$ and $x,x'\in D$, $$|f(x)-f(x')|\le\|f\|_k\,d_k(x,x'),\qquad d_k(x,x')^2=k(x,x)-2k(x,x')+k(x',x').$$ Proof. $|f(x)-f(x')|=|\langle f,\varphi(x)-\varphi(x')\rangle_k|\le\|f\|_k\|\varphi(x)-\varphi(x')\|_k$ by Cauchy–Schwarz, and $\|\varphi(x)-\varphi(x')\|_k^2=k(x,x)-2k(x,x')+k(x',x')$ by expanding the inner product. $\square$
For the SE kernel $d_k^2=2\big(1-e^{-r^2/(2\ell^2)}\big)\le r^2/\ell^2$ (because $1-e^{-a}\le a$), so every $f$ with $\|f\|_k\le B$ is Lipschitz with constant $L=B/\ell$. This is the "Lipschitz-like" continuity that safe-set expansion uses: SafeOpt (Module 4) assumes a separate Lipschitz constant $L$; Tokmak et al., AISTATS 2025 replace it by $\|f\|_k\,d_k$ (their Lemma 1).

In general $d_k(x,x')=\|\varphi(x)-\varphi(x')\|_k$ is only a pseudometric: it is symmetric and satisfies the triangle inequality, but distinct inputs may have distance zero (the constant kernel gives $d_k\equiv0$). It is a metric when distinct inputs have distinct features, e.g. for every positive definite kernel.

Theorem — Representer theorem
For any loss function $\mathrm{loss}(y,v)$ and $\lambda\gt0$, every minimiser of $\sum_{i=1}^t\mathrm{loss}(y_i,f(x_i))+\lambda\|f\|_k^2$ over $f\in\mathcal H_k$ has the form $f=\sum_{i=1}^t\alpha_ik(x_i,\cdot)$.
Proof. Split $f=f_\parallel+f_\perp$ with $f_\parallel\in \mathcal L_t:=\mathrm{span}\{k(x_i,\cdot)\}$ and $f_\perp\perp \mathcal L_t$. By the reproducing property $f_\perp(x_i)=\langle f_\perp,k(x_i,\cdot)\rangle_k=0$, so the loss does not see $f_\perp$, while $\|f\|_k^2=\|f_\parallel\|_k^2+\|f_\perp\|_k^2$ is strictly larger unless $f_\perp=0$. $\square$

The RKHS norm is a smoothness measure

For a translation-invariant kernel $k(x,x')=\kappa(x-x')$ on $\mathbb R^d$ with integrable, positive Fourier transform $\hat\kappa$, the norm of $f\in\mathcal H_k(\mathbb R^d)$ is (up to a normalising constant) $\|f\|_k^2=\int|\hat f(\omega)|^2/\hat\kappa(\omega)\,d\omega$: frequency content is penalised in proportion to $1/\hat\kappa(\omega)$. (On a compact $D$, the norm of $f$ in the RKHS of the restricted kernel is the smallest such norm over all extensions of $f$ to $\mathbb R^d$.) For the SE kernel $\hat\kappa(\omega)\propto\ell^d e^{-\ell^2\|\omega\|^2/2}$, so high frequencies are penalised super-exponentially and $\mathcal H_{k_{\mathrm{SE}}}$ contains only analytic functions. For Matérn-$\nu$ with the $\sqrt{2\nu}\,r/\ell$ scaling used above, $\hat\kappa(\omega)\propto(2\nu/\ell^2+\|\omega\|^2)^{-(\nu+d/2)}$ and $\mathcal H_k$ is norm-equivalent to the Sobolev space $H^{\nu+d/2}$: a statement about square-integrable weak derivatives of order $\nu+d/2$, not an exact count of classical derivatives.

Background — Fourier transforms, analytic functions and Sobolev norms

This spectral description is optional for the rest of the page. The Fourier transform $\hat f(\omega)=\int_{\mathbb R^d}f(x)e^{-i\omega^\top x}\,dx$ ($i^2=-1$, $dx=dx_1\cdots dx_d$; extended to square-integrable $f$ by limits) records how strongly $f$ oscillates with frequency vector $\omega$; for example $e^{-x^2/2}$ has transform $\sqrt{2\pi}e^{-\omega^2/2}$. The Sobolev norm of order $s$ weights frequencies polynomially, $\|f\|_{H^s}^2\propto\int(1+\|\omega\|^2)^s|\hat f(\omega)|^2\,d\omega$. For integer $s$ it measures square-integrable weak derivatives up to order $s$ (a weak derivative of $f$ is a function $v$ with $\int f\phi'=-\int v\phi$ for every smooth $\phi$ vanishing outside a bounded set, which extends differentiation to functions with corners); non-integer $s$ is defined by the same weighting. Norm-equivalent means $c\|f\|_{H^s}\le\|f\|_k\le C\|f\|_{H^s}$ for fixed constants $c,C\gt0$: the same functions with comparable sizes. Matérn gives $s=\nu+d/2$, which is not a count of classical derivatives. A function is analytic if near every point it equals a convergent power series, as $e^x=\sum_{n\ge0}x^n/n!$ does; this is stronger than having derivatives of every order. On a bounded domain all these norms are the smallest norm of an extension to $\mathbb R^d$.

Changing the length-scale moves the norm, not only the space. For the SE kernel with $\ell'\le\ell$, $\mathcal H_{k_\ell}\subseteq\mathcal H_{k_{\ell'}}$ and $\|f\|_{k_{\ell'}}\le(\ell/\ell')^{d/2}\|f\|_{k_\ell}$ (Berkenkamp, Schoellig & Krause, JMLR 2019, Lemma 2, after Bull 2011): a shorter length-scale keeps $f$ in the space, but its norm may grow by up to that factor, so $B$ has to be inflated with it. For Matérn kernels of fixed $\nu$ the space does not change with $\ell$ at all, only the norm does. This is the precise sense in which under-estimating $\ell$ is the "safe direction" (Section 7): membership survives, the old $B$ does not.

Pitfall — what "$\|f\|_k\le B$" assumes, and why no engineer can verify it
  • Membership is a hard smoothness requirement. $f\in\mathcal H_{k_{\mathrm{SE}}}$ means $f$ is analytic; a controller cost with a kink (saturation, contact) has infinite SE-RKHS norm and every bound below is void for it.
  • $B$ has no physical interpretation. Amplitude and slope bounds imply nothing about $B$: in the Fourier formula the SE weight $1/\hat\kappa(\omega)\propto e^{\ell^2\|\omega\|^2/2}$ is, at frequency $\|\omega\|=10/\ell$, already $e^{50}$ times its value at $\omega=0$. A wiggle of amplitude $10^{-3}$ at that frequency changes $\sup|f|$ (see Primer 0) by $10^{-3}$ and $\sup|f'|$ by about $10^{-2}/\ell$, yet its contribution to $\|f\|_k^2$ carries this enormous weight (the exact value also depends on the wiggle's shape and frequency spread). Small $\sup|f|$ and $\sup|f'|$ therefore supply no RKHS norm bound.
  • Bayesian and frequentist models exclude each other. If $\mathcal H_k$ is infinite-dimensional (SE, Matérn) and $f\sim\mathcal{GP}(0,k)$, then $f\notin\mathcal H_k$ almost surely, i.e. $\|f\|_k=\infty$: typical sample paths lie outside the RKHS of their own covariance; for Matérn kernels they are about $d/2$ orders of Sobolev smoothness rougher (Srinivas et al., ICML 2010, Section 4, citing Wahba; Kanagawa et al. 2018, Section 4). For a finite-rank kernel such as the linear one the samples do lie in $\mathcal H_k$. So one cannot "sample $f$ from the prior" of an SE or Matérn GP and then invoke an RKHS bound for that same kernel.
  • This is the practical obstacle identified by Fiedler, Menn, Kreisköther & Trimpe, TMLR 2024: with a correct $\beta_t$, safety stands or falls with the appropriateness of $B$. Section 7 and Module 5 show what happens when it is under-estimated and what the group did about it.
Worked example — a Gram matrix, and what the RKHS norm measures

A Gram matrix. For the SE kernel with $\ell=1$ and the points $0$, $1$, $2$:

$$K=\begin{bmatrix}1&0.6065&0.1353\\0.6065&1&0.6065\\0.1353&0.6065&1\end{bmatrix},\qquad\text{eigenvalues }0.207,\ 0.865,\ 1.928 .$$

All eigenvalues are positive, so $c^{\top}Kc\gt0$ for every $c\ne0$. For example, the "wiggly" combination $c=(1,-2,1)$ gives $c^{\top}Kc=1.418$, which is $\|k(0,\cdot)-2k(1,\cdot)+k(2,\cdot)\|_k^2$.

Steep changes cost norm. What is the smallest RKHS norm of a function with value $+1$ at $x=0$ and $-1$ at $x=d$? The minimum-norm interpolant (Section 3) has $\|f\|_k^2=y^{\top}K^{-1}y$ with $y=(1,-1)$, which here equals $2/(1-k(0,d))$:

  • $d=1$: $k(0,1)=0.6065$, so $\|f\|_k=\sqrt{2/0.3935}=2.25$;
  • $d=0.1$: $k(0,0.1)=0.9950$, so $\|f\|_k=\sqrt{2/0.0050}=20.0$.

The same jump of size 2, made ten times faster, needs about nine times the norm. The Lipschitz lemma above gives the same number from the other side: $|f(0)-f(d)|=2\le\|f\|_k\,d_k(0,d)$ forces $\|f\|_k\ge2/d_k(0,0.1)=20.0$.

2. GP Regression = Kernel Ridge Regression

In plain words. GP regression predicts an unknown function from noisy samples and says how unsure it is. Near the data the prediction follows the measurements and the uncertainty is small. For kernels whose correlations decay with distance, such as SE and Matérn, the prediction far from all data falls back to the prior mean, zero, and the uncertainty returns to its prior level. Other kernels, such as the linear kernel, extrapolate differently. The same formula has two readings: Bayesian (the posterior of a random function) and frequentist (kernel ridge regression of a fixed unknown function). Safe BO uses the second.

Symbols in this section

A Gaussian process $f\sim\mathcal{GP}(0,k)$ is the statement that the values of $f$ at any finite set of inputs are jointly Gaussian with covariance $K$. With the observation model $y_i=f(x_i)+\varepsilon_i$, $\varepsilon_i\sim\mathcal N(0,\sigma_n^2)$ i.i.d. and independent of $f$, the vector $(y_t,f(x))$ is jointly Gaussian too (see Primer C): $$\begin{pmatrix}y_t\\ f(x)\end{pmatrix}\sim\mathcal N\!\left(0,\begin{pmatrix}K_t+\sigma_n^2I & k_t(x)\\ k_t(x)^\top & k(x,x)\end{pmatrix}\right).$$ Conditioning on $y_t$ is a Schur-complement computation (Module 2).

The displayed joint law uses the physical noise variance $\sigma_n^2$, so the exact Bayesian posterior has $\lambda=\sigma_n^2$ below. Any other $\lambda$ gives the posterior of a working likelihood (equivalently, the ridge estimator of the next subsection), which need not be the physical conditional law. Also, $\sigma_t^2(x)$ is the variance of the latent value $f(x)$; a new noisy observation at $x$ has predictive variance $\sigma_t^2(x)+\sigma_n^2$ under the correctly specified model.

Key equation — GP posterior
Writing $\lambda$ for the noise variance used in the likelihood, $$\mu_t(x)=k_t(x)^\top(K_t+\lambda I)^{-1}y_t,\qquad\sigma_t^2(x)=k(x,x)-k_t(x)^\top(K_t+\lambda I)^{-1}k_t(x),$$ and under that working likelihood $f(x)\mid y_t\sim\mathcal N(\mu_t(x),\sigma_t^2(x))$. The posterior covariance of two points is $k_t(x,x')=k(x,x')-k_t(x)^\top(K_t+\lambda I)^{-1}k_t(x')$.
Derivation — Gaussian conditioning via the Schur complement

Let $(a,b)\sim\mathcal N(0,\Sigma)$ with $\Sigma=\begin{pmatrix}A&C\\C^\top&B\end{pmatrix}$, $\Sigma\succ0$ (hence $A\succ0$), and let $S=B-C^\top A^{-1}C$ be the Schur complement of $A$. The block inverse is (see Primer A) $$\Sigma^{-1}=\begin{pmatrix}A^{-1}+A^{-1}CS^{-1}C^\top A^{-1} & -A^{-1}CS^{-1}\\ -S^{-1}C^\top A^{-1} & S^{-1}\end{pmatrix},$$ which one checks by multiplying out. The joint density is $\propto\exp(-\tfrac12[a;b]^\top\Sigma^{-1}[a;b])$; collecting the terms that involve $b$, $$b^\top S^{-1}b-2\,b^\top S^{-1}C^\top A^{-1}a=(b-C^\top A^{-1}a)^\top S^{-1}(b-C^\top A^{-1}a)-(\text{terms in }a\text{ only}),$$ so as a function of $b$ the conditional density is Gaussian with mean $C^\top A^{-1}a$ and covariance $S$. With $a=y_t$, $A=K_t+\lambda I$, $C=k_t(x)$, $B=k(x,x)$ this is the posterior above: the mean is linear in the data, the variance is the Schur complement and does not depend on the observed values when the design and kernel are fixed; an adaptive design can depend on past observations.

$\Sigma\succ0$ also gives $S\succ0$ (the Schur complement of a positive definite matrix is positive definite), so $S^{-1}$ exists. If the joint Gaussian is degenerate ($\Sigma$ singular), the mean and covariance formulas still hold as long as $A\succ0$, which is the case here whenever $\lambda\gt0$; only the density argument must be replaced (by an affine-transformation argument or a limit). A zero conditional variance means the conditioned quantity is determined exactly.

The same formula from an optimisation problem

Theorem — Kernel ridge regression equals the GP posterior mean
The minimiser of $\sum_{i=1}^t(y_i-f(x_i))^2+\lambda\|f\|_k^2$ over $f\in\mathcal H_k$ is $f^\star(x)=k_t(x)^\top(K_t+\lambda I)^{-1}y_t=\mu_t(x)$. (Kanagawa, Hennig, Sejdinovic & Sriperumbudur (2018) give the full dictionary between the GP and the kernel view, including the sample-path facts used above.)
Derivation — representer theorem plus normal equations

By the representer theorem $f=\sum_i\alpha_ik(x_i,\cdot)$, so $f(x_j)=(K_t\alpha)_j$ and $\|f\|_k^2=\alpha^\top K_t\alpha$. The objective becomes the convex quadratic (see Primer B) $J(\alpha)=\|y_t-K_t\alpha\|^2+\lambda\alpha^\top K_t\alpha$ with gradient $$\nabla J=-2K_t(y_t-K_t\alpha)+2\lambda K_t\alpha=-2K_t\big(y_t-(K_t+\lambda I)\alpha\big),$$ which vanishes for $\alpha=(K_t+\lambda I)^{-1}y_t$ (unique when $K_t\succ0$; otherwise one minimiser among several that define the same function). Then $f^\star(x)=\sum_i\alpha_ik(x_i,x)=k_t(x)^\top\alpha$. Nothing probabilistic was used: $\mu_t$ is a deterministic function of the data and $\lambda$ a regularisation weight.

Why several minimisers define the same function: $J$ is convex, so $\alpha$ and $\alpha'$ both minimise it exactly when both make $\nabla J$ vanish, i.e. $K_t(K_t+\lambda I)(\alpha-\alpha')=0$. The two matrices commute and $K_t+\lambda I$ is invertible, so $d=\alpha-\alpha'$ satisfies $K_td=0$. The function $h=\sum_id_ik(x_i,\cdot)$ then has $\|h\|_k^2=d^\top K_td=0$, hence $h=0$: the coefficients may differ, the fitted function does not.

Feature-space form and the variance identity

Let $\Phi_t:\mathcal H_k\to\mathbb R^t$, $\Phi_tg=[\langle\varphi(x_1),g\rangle_k,\dots,\langle\varphi(x_t),g\rangle_k]^\top$, be the design operator. Then $K_t=\Phi_t\Phi_t^*$, $k_t(x)=\Phi_t\varphi(x)$, $f_{1:t}=\Phi_tf$, and two identities carry the whole theory:

Background — Adjoints and finite-rank operators

The star is an adjoint, the function-space counterpart of transpose. Here it has the concrete formula $\Phi_t^*v=\sum_{i=1}^t v_i\varphi(x_i)$, characterised by $v^\top\Phi_t g=\langle\Phi_t^*v,g\rangle_k$. Thus $(\Phi_t\Phi_t^*)_{ij}=\langle\varphi(x_i),\varphi(x_j)\rangle_k=k(x_i,x_j)$. For a single feature, $\varphi(x)^*g=\langle\varphi(x),g\rangle_k$ (the analogue of a row vector), and $\varphi(x_i)\varphi(x_i)^*$ is the rank-one operator $g\mapsto\varphi(x_i)\langle\varphi(x_i),g\rangle_k$. Write $A_t=\Phi_t^*\Phi_t=\sum_i\varphi(x_i)\varphi(x_i)^*$, so $A_tg=\sum_i\varphi(x_i)\langle\varphi(x_i),g\rangle_k$.

All nonzero action of $A_t$ lies in the finite span of the observed features. On that span use an orthonormal basis and ordinary symmetric-matrix algebra; on its orthogonal complement $A_t=0$. Consequently $A_t+\lambda I$ acts there as $\lambda I$ and is invertible for $\lambda>0$.

With a linear feature $\varphi(x)=(1,x)^\top$ and one observation at $x=2$, the design matrix is $\Phi=(1,2)$, its adjoint is the column $(1,2)^\top$, and $A=\Phi^*\Phi=\begin{pmatrix}1&2\\2&4\end{pmatrix}$. The same multiplication works in an RKHS; only the feature vectors may be functions.

Key equation — push-through and variance identities
$$\Phi_t^*(\Phi_t\Phi_t^*+\lambda I)^{-1}=(\Phi_t^*\Phi_t+\lambda I)^{-1}\Phi_t^*,\qquad\qquad\sigma_t^2(x)=\lambda\,\big\langle\varphi(x),(\Phi_t^*\Phi_t+\lambda I)^{-1}\varphi(x)\big\rangle_k .$$ Hence $\mu_t(x)=\langle\varphi(x),(\Phi_t^*\Phi_t+\lambda I)^{-1}\Phi_t^*y_t\rangle_k$ is ridge regression in feature space, and $\sigma_t^2/\lambda$ is the squared $V_t^{-1}$-norm (see Primer A) of the feature at $x$, with $V_t:=\Phi_t^*\Phi_t+\lambda I$.
Background — weighted norms and $V_t^{-1/2}$

For a self-adjoint operator $M$ with positive eigenvalues, $\|g\|_M^2:=\langle g,Mg\rangle_k$. For $V_t=A_t+\lambda I$ the eigen-directions are explicit: on the span of the observed features take an orthonormal eigenbasis of $A_t$ (eigenvalues $a_j\ge0$, so $V_t$ has eigenvalues $a_j+\lambda$), and on the orthogonal complement $V_t=\lambda I$. $V_t^{-1}$ and $V_t^{-1/2}$ replace each eigenvalue $m$ by $m^{-1}$ and $m^{-1/2}$, so $\|V_t^{-1/2}g\|_k^2=\langle g,V_t^{-1}g\rangle_k=\|g\|_{V_t^{-1}}^2$. The identity operator is written $I$ or $\mathrm{id}$.

Derivation — the two identities

Push-through. $(\Phi^*\Phi+\lambda I)\Phi^*=\Phi^*\Phi\Phi^*+\lambda\Phi^*=\Phi^*(\Phi\Phi^*+\lambda I)$; both bracketed operators are invertible for $\lambda\gt0$, so multiply by their inverses on the appropriate sides.

Variance. By definition of $\Phi^*$, $(\Phi^*\Phi+\lambda I)\varphi(x)=\Phi^*k_t(x)+\lambda\varphi(x)$. Apply $(\Phi^*\Phi+\lambda I)^{-1}$, then push-through: $$\varphi(x)=\Phi^*(K_t+\lambda I)^{-1}k_t(x)+\lambda(\Phi^*\Phi+\lambda I)^{-1}\varphi(x).$$ Take the inner product with $\varphi(x)$ and use $\langle\varphi(x),\Phi^*v\rangle_k=k_t(x)^\top v$: $$k(x,x)=k_t(x)^\top(K_t+\lambda I)^{-1}k_t(x)+\lambda\langle\varphi(x),(\Phi^*\Phi+\lambda I)^{-1}\varphi(x)\rangle_k ,$$ and the first term on the right is $k(x,x)-\sigma_t^2(x)$. This is equation (13) in the proof of Chowdhury & Gopalan, ICML 2017.

Key insight — the double role of $\lambda$
The same formulas admit two readings, and the confidence bounds of Sections 5–6 live entirely in the second.
Bayesian readingFrequentist reading
$f$a random draw from $\mathcal{GP}(0,k)$a fixed unknown element of $\mathcal H_k$, $\|f\|_k\le B$
$\lambda$the physical noise variance $\sigma_n^2$a free regulariser; the noise enters only through $R$
$\mu_t$posterior meankernel ridge regressor
$\sigma_t^2$posterior variancea data-location-dependent "power function"; never sees $y_t$
"with probability $1-\delta$"over the prior and the noiseover the noise only, for every fixed $f$ (worst case over the ball)
Chowdhury & Gopalan (2017) prove their bound for a posterior computed with $\lambda=1+\eta$, $\eta=2/T$, whatever the physical noise is; Fiedler, Scherer & Trimpe (AAAI 2021) note that the nominal $\lambda$ is independent of the actual sub-Gaussian noise and their bound holds for every $\lambda\gt0$. A control engineer who sets $\lambda$ to the measured sensor variance is making a modelling choice, not satisfying an assumption.
Worked example — a GP posterior from two measurements

Take the SE kernel with $\ell=1$, two measurements, $y=1$ at $x=0$ and $y=0.5$ at $x=1$, and $\lambda=0.01$. Then

$$K_t+\lambda I=\begin{bmatrix}1.01&0.6065\\0.6065&1.01\end{bmatrix},\qquad \mu_t(x)=k_t(x)^{\top}(K_t+\lambda I)^{-1}\begin{bmatrix}1\\0.5\end{bmatrix} .$$
Query $x$$k_t(x)$mean $\mu_t(x)$std. dev. $\sigma_t(x)$Reading
0(1, 0.6065)0.9890.099at a data point: near the measurement, small uncertainty
0.5(0.8825, 0.8825)0.8190.191between the data: in between, more uncertain
3(0.0111, 0.1353)−0.0090.987far away: back to the prior, mean near 0, std. dev. near 1

At $x=0$ the mean is $0.989$ rather than $1$. With $\lambda\gt0$ the regression does not interpolate exactly, because it allows for measurement noise; as $\lambda\to0$ it becomes the interpolant of Section 3. The standard deviation never looks at the measured values $y$, only at where the data are.

3. The Noise-Free Error Bound

In plain words. With exact, noise-free measurements, how far can the true function be from the interpolant through the data? The answer of this section is short: at most the function's RKHS norm times the posterior standard deviation, $\|f\|_k\,P_t(x)$. Near the data $P_t$ is small and the interpolant can be trusted. For a decaying kernel such as SE, far from the data $P_t$ is close to its prior value, and the norm bound is all you have. (In general $P_t(x)$ measures how much of "the value at $x$" cannot be reproduced from the measured values, which need not grow with distance.)

Symbols in this section

Before any probability enters, here is the cleanest reason why $\sigma_t$ measures ignorance. Take noise-free observations $y_i=f(x_i)$ at distinct inputs with $K_t\succ0$ and let $\lambda\to0$: the GP mean becomes the interpolant $s_t(x)=k_t(x)^\top K_t^{-1}f_{1:t}$ and $\sigma_t^2$ becomes the power function $P_t(x)^2=k(x,x)-k_t(x)^\top K_t^{-1}k_t(x)$.

Theorem — Noise-free (interpolation) error bound
For every $f\in\mathcal H_k$ and every $x\in D$: $\ |f(x)-s_t(x)|\le\|f\|_k\,P_t(x)$. Moreover $s_t$ is the minimum-norm interpolant, $s_t=\operatorname*{arg\,min}\{\|g\|_k:\ g(x_i)=f(x_i),\ i\le t\}$, and the bound is sharp.
In words: the interpolant is off by at most the norm of $f$ times the posterior standard deviation, deterministically, for every function in the ball.
Derivation — the power-function argument (orthogonal projection)

Step 1: $s_t$ is the orthogonal projection of $f$ onto $\mathcal L_t=\mathrm{span}\{\varphi(x_1),\dots,\varphi(x_t)\}$. Let $\Pi_t$ denote that projection. $\Pi_tf=\sum_i\alpha_i\varphi(x_i)$ is characterised by $\langle f-\Pi_tf,\varphi(x_j)\rangle_k=0$ for all $j$, i.e. (reproducing property) $f(x_j)=\sum_i\alpha_ik(x_i,x_j)$, i.e. $K_t\alpha=f_{1:t}$; so $\alpha=K_t^{-1}f_{1:t}$ and $(\Pi_tf)(x)=k_t(x)^\top K_t^{-1}f_{1:t}=s_t(x)$.

Step 2: write the error as an inner product with a residual feature. Since $f-\Pi_tf\perp \mathcal L_t$ and $\Pi_t\varphi(x)\in \mathcal L_t$, $$f(x)-s_t(x)=\langle f-\Pi_tf,\ \varphi(x)\rangle_k=\langle f-\Pi_tf,\ \varphi(x)-\Pi_t\varphi(x)\rangle_k .$$

Step 3: Cauchy–Schwarz and Pythagoras. $|f(x)-s_t(x)|\le\|f-\Pi_tf\|_k\,\|\varphi(x)-\Pi_t\varphi(x)\|_k\le\|f\|_k\,\|\varphi(x)-\Pi_t\varphi(x)\|_k$, because $\|f\|_k^2=\|\Pi_tf\|_k^2+\|f-\Pi_tf\|_k^2$.

Step 4: the residual feature has norm $P_t(x)$. Step 1 applied to $\varphi(x)$, whose data vector is $[\varphi(x)(x_i)]_i=k_t(x)$, gives $\Pi_t\varphi(x)=\sum_i(K_t^{-1}k_t(x))_i\varphi(x_i)$, so $\|\Pi_t\varphi(x)\|_k^2=k_t(x)^\top K_t^{-1}K_tK_t^{-1}k_t(x)=k_t(x)^\top K_t^{-1}k_t(x)$ and $$\|\varphi(x)-\Pi_t\varphi(x)\|_k^2=k(x,x)-k_t(x)^\top K_t^{-1}k_t(x)=P_t(x)^2 .\qquad\square$$

Minimum norm. Any interpolant $g$ satisfies $\langle g-f,\varphi(x_i)\rangle_k=g(x_i)-f(x_i)=0$, so $g-f\perp \mathcal L_t$, hence $\Pi_tg=\Pi_tf=s_t$ and $\|g\|_k^2=\|s_t\|_k^2+\|g-s_t\|_k^2\ge\|s_t\|_k^2$.

Sharpness. Take $f=B\,(\varphi(x)-\Pi_t\varphi(x))/P_t(x)$. It vanishes at every $x_i$ (its data vector is $k_t(x)-K_tK_t^{-1}k_t(x)=0$), so $s_t\equiv0$, while $f(x)=B\,P_t(x)^2/P_t(x)=B\,P_t(x)$: the worst case is the residual feature itself.

This construction needs $P_t(x)\gt0$. If $P_t(x)=0$ the residual feature vanishes, the bound says that every $f\in\mathcal H_k$ is reproduced exactly at $x$, and there is nothing to sharpen.

Why it matters
$P_t(x)$ is the distance from the evaluation functional at $x$ to the span of the evaluation functionals at the data: how much of "the value at $x$" cannot be reproduced from "the values at $x_1,\dots,x_t$", for any function of bounded norm. No prior, no noise, no probability. Everything in Section 5 is this argument plus a separate bound for the noise, and the term $B\sigma_t(x)$ in every frequentist $\beta_t$ is exactly this theorem with $\lambda\gt0$ (Step 2 of the walkthrough).
Worked example — how big the error can be, near and far from the data

Take data at $x=0$ and $x=1$ and the SE kernel with $\ell=1$. The power function is

$$P_t(x)^2=1-k_t(x)^{\top}K_t^{-1}k_t(x),\qquad P_t(0.5)=0.175,\qquad P_t(3)=0.987 .$$

Suppose you know $\|f\|_k\le B=2$. Then the theorem guarantees $|f(0.5)-s_t(0.5)|\le2\times0.175=0.35$ between the data, but only $|f(3)-s_t(3)|\le2\times0.987=1.97$ far away.

The bound is sharp. Open Problem T1 builds the function that attains it at $x=0.5$: it vanishes at both data points and equals $0.0305$ at the midpoint, with norm $0.1745$. Scaled to norm $B$, it reaches exactly $B\,P_t(0.5)$ (Open Problems, T1). That function is invisible in the data, so no amount of noise-free data at $0$ and $1$ can tell you whether you are in the good case or the bad one. Only $B$ can.

4. Maximum Information Gain

In plain words. Each measurement teaches the GP something about $f$, but a measurement close to earlier ones teaches little that is new. The information gain counts how much is learned in total, in units of information (nats). The maximum information gain $\gamma_T$ is the most that $T$ measurements could ever reveal, and it is what regret and sample-complexity bounds depend on. If it grows slowly, the uncertainty must shrink fast wherever the algorithm keeps sampling.

Symbols in this section

Regret and sample-complexity bounds (see Primer E) must control how fast the posterior variances shrink along whatever sequence the algorithm chooses. The right quantity is the amount of information that $t$ noisy observations can reveal about $f$ under the GP model.

Definition — Information gain and maximum information gain
Under the GP model with noise variance $\lambda$, the mutual information (see Primer C) between the observations $y_A$ at a fixed design $A=(x_1,\dots,x_T)$ (repetitions allowed) and the values $f_A$ is $$I(y_A;f_A)=H(y_A)-H(y_A\mid f_A)=\tfrac12\log\det\big(I+\lambda^{-1}K_A\big),$$ and the maximum information gain after $T$ rounds is $\gamma_T:=\sup_{x_1,\dots,x_T\in D}\tfrac12\log\det\big(I_T+\lambda^{-1}K_T\big)$ (Srinivas, Krause, Kakade & Seeger, ICML 2010, who write the maximum over sets $A\subset D$ with $|A|=T$; the analysis needs repeated inputs, so the supremum runs over designs with repetition). It depends on $k$, $D$, $\lambda$ and $T$ only, not on the data.

Closed form. $y_A\sim\mathcal N(0,K_A+\lambda I)$ and $y_A\mid f_A\sim\mathcal N(f_A,\lambda I)$; with $H(\mathcal N(m,\Sigma))=\tfrac12\log\det(2\pi e\Sigma)$ the $2\pi e$ factors cancel and $I=\tfrac12\log\det(K_A+\lambda I)-\tfrac12\log\det(\lambda I)=\tfrac12\log\det(I+\lambda^{-1}K_A)$.

Lemma — Sequential form of the information gain
For any sequence $x_1,\dots,x_T$, fixed in advance or chosen adaptively from past data, the realised information gain $G_T:=\tfrac12\log\det(I+\lambda^{-1}K_T)$ satisfies, pathwise, $$G_T=\tfrac12\sum_{t=1}^T\log\big(1+\lambda^{-1}\sigma_{t-1}^2(x_t)\big)\le\gamma_T .$$ Proof. Partition $K_t+\lambda I$ with $x_t$ in the last row and take the Schur complement of $K_{t-1}+\lambda I$: $\det(K_t+\lambda I)=\det(K_{t-1}+\lambda I)\,\big(k(x_t,x_t)+\lambda-k_{t-1}(x_t)^\top(K_{t-1}+\lambda I)^{-1}k_{t-1}(x_t)\big)=\det(K_{t-1}+\lambda I)\,\big(\lambda+\sigma_{t-1}^2(x_t)\big)$. Divide by $\lambda^t$, take logarithms and telescope; $G_T\le\gamma_T$ is the definition of $\gamma_T$ applied to the realised design. $\square$
For a fixed design $G_T=I(y_{1:T};f_{1:T})$, and the identity is the chain rule of entropy $H(y_{1:T})=\sum_{t=1}^TH(y_t\mid y_{1:t-1})$, written with the scalar observations $y_t$ (and $y_{1:t-1}$ the vector of earlier ones): under the model $y_t\mid y_{1:t-1}\sim\mathcal N(\mu_{t-1}(x_t),\sigma_{t-1}^2(x_t)+\lambda)$, and one subtracts $H(y_t\mid f(x_t))=\tfrac12\log(2\pi e\lambda)$ term by term (Lemma 5.3 of Srinivas et al.; Lemma 3 of Chowdhury & Gopalan). For an adaptive design the identity is used pathwise, as proved above; the realised $G_T$ is then a random quantity, not the mutual information of a fixed design. In feature space, with $A_t:=\Phi_t^*\Phi_t$ and the (Fredholm) determinant $\det(I+A_t/\lambda)=\det(I_t+K_t/\lambda)$, each step multiplies the determinant by $1+\langle\varphi(x_t),V_{t-1}^{-1}\varphi(x_t)\rangle_k=1+\lambda^{-1}\sigma_{t-1}^2(x_t)$, where $V_{t-1}=A_{t-1}+\lambda I$ (matrix determinant lemma, see Primer A): the "elliptical potential" of linear bandits.
Background — Finite-rank determinants

Only a finite-rank special case is needed here. If the nonzero eigenvalues of $A_t$ are $a_1,\ldots,a_r$, define $\det(I+A_t/\lambda)=\prod_{j=1}^r(1+a_j/\lambda)$; the infinitely many remaining identity directions contribute factors of one. The nonzero eigenvalues of $\Phi_t^*\Phi_t$ and $\Phi_t\Phi_t^*=K_t$ coincide, with multiplicities (if $K_tv=av$ with $a\ne0$, then $A_t\Phi_t^*v=\Phi_t^*K_tv=a\,\Phi_t^*v$, and conversely), so this product equals $\det(I_t+K_t/\lambda)$. We never take an ordinary infinite-dimensional determinant of $V_t$ itself.

For $A=uu^*$ with $\|u\|_k^2=3$ and $\lambda=2$, the only nonzero eigenvalue is $3$ ($Au=u\|u\|_k^2$), so the determinant is $1+3/2=2.5$.

Lemma — Sum of posterior variances
If $k(x,x)\le1$ on $D$, then for every sequence $$\sum_{t=1}^T\sigma_{t-1}^2(x_t)\le\frac{2}{\log(1+\lambda^{-1})}\,\gamma_T,\qquad\text{hence}\qquad\sum_{t=1}^T\sigma_{t-1}(x_t)\le\sqrt{\frac{2T\gamma_T}{\log(1+\lambda^{-1})}} .$$
Derivation — sum of variances from the sequential form

$g(s)=s/\log(1+s)$ is non-decreasing on $(0,\infty)$: $g'(s)$ has the sign of $h(s)=\log(1+s)-s/(1+s)$, and $h\ge0$ because $h(0)=0$ and $h'(s)=s/(1+s)^2\ge0$. Since $0\le\sigma_{t-1}^2(x_t)\le k(x_t,x_t)\le1$, the number $s=\lambda^{-1}\sigma_{t-1}^2(x_t)$ lies in $[0,\lambda^{-1}]$, so $g(s)\le g(\lambda^{-1})$ (for $s=0$ the inequality below reads $0\le0$): $$\lambda^{-1}\sigma_{t-1}^2(x_t)\le\frac{\lambda^{-1}}{\log(1+\lambda^{-1})}\log\big(1+\lambda^{-1}\sigma_{t-1}^2(x_t)\big)\ \Longrightarrow\ \sigma_{t-1}^2(x_t)\le\frac{\log\big(1+\lambda^{-1}\sigma_{t-1}^2(x_t)\big)}{\log(1+\lambda^{-1})} .$$ Sum over $t$ and use the sequential lemma: $\sum_t\sigma_{t-1}^2(x_t)\le\frac{2}{\log(1+\lambda^{-1})}G_T\le\frac{2}{\log(1+\lambda^{-1})}\gamma_T$. The second form is Cauchy–Schwarz for the vectors $(\sigma_{t-1}(x_t))_{t\le T}$ and $(1,\dots,1)$: $\big(\sum_t\sigma_{t-1}(x_t)\big)^2\le T\sum_t\sigma_{t-1}^2(x_t)$.

Where the constant goes. In the bandit problem (see Primer E) an algorithm queries $x_1,x_2,\dots$ to maximise $f$; with $x^*$ a maximiser, the instantaneous regret is $r_t=f(x^*)-f(x_t)$ and the cumulative regret $R_T=\sum_{t\le T}r_t$. GP-UCB queries a maximiser of $U_t(x)=\mu_{t-1}(x)+c_t\sigma_{t-1}(x)$, where $c_t\sigma_{t-1}(x)$ is the confidence half-width. On the event that all bands hold, $f(x^*)\le U_t(x^*)\le U_t(x_t)$ and $f(x_t)\ge\mu_{t-1}(x_t)-c_t\sigma_{t-1}(x_t)$, so $r_t\le2c_t\sigma_{t-1}(x_t)$. Srinivas et al. write $c_t=\beta_t^{1/2}$, i.e. $r_t^2\le4\beta_t\sigma_{t-1}^2(x_t)$; this page writes $c_t=\beta_t$. For non-decreasing $c_t$ the lemma gives $R_T\le2c_T\sqrt{2T\gamma_T/\log(1+\lambda^{-1})}$, which for $\lambda=\sigma_n^2$ and $c_T=\beta_T^{1/2}$ is Srinivas et al.'s $R_T\le\sqrt{C_1T\beta_T\gamma_T}$ with $C_1=8/\log(1+\sigma_n^{-2})$. SafeOpt's sample-complexity bound (Module 4), which also depends on the size of the region reachable safely from the initial observations, inherits the same constant. Chowdhury & Gopalan's Lemma 4 is the same computation with $\lambda=1+\eta$ and $\log(1+\alpha)\ge\alpha/2$ on $[0,1]$, giving $\sum_t\sigma_{t-1}(x_t)\le\sqrt{4(T+2)\gamma_T}$.

How fast does $\gamma_T$ grow?

Growth rates use asymptotic notation (see Primer 0): as $T\to\infty$, $a_T=O(b_T)$ means $|a_T|\le Cb_T$ for all large $T$ with a constant $C$ independent of $T$; $a_T=o(b_T)$ means $a_T/b_T\to0$; $a_T=\Omega(b_T)$ means $a_T\ge cb_T$ eventually for some $c\gt0$; $\widetilde O$ also hides logarithmic factors. Kernel, dimension, domain and regulariser are fixed, and the hidden constants may depend on them. A regret bound is sublinear if $R_T/T\to0$. An $O(\cdot)$ rate is not a computable finite-sample certificate.

$\gamma_T$ is governed by the eigenvalue decay of the kernel operator: if the first few eigenfunctions carry most of the mass, $T$ observations cannot learn much more than a finite-dimensional model. Srinivas et al. bound it through this spectrum (their Theorems 4–5); Vakili, Khezeli & Picheny, AISTATS 2021 sharpen the Matérn case by projecting onto the span of the first $M$ Mercer eigenfunctions $e_m$ and controlling the tail. Under their Assumption 1 (a bounded kernel whose Mercer eigenfunctions are uniformly bounded, $|e_m(x)|\le\psi$), $\gamma_T=O(M\log T+\Delta_MT)$ for every $M$, with tail mass $\Delta_M=\psi^2\sum_{m\gt M}\lambda_m$ (their Theorem 3, stated precisely below; here $\lambda_m$ are the Mercer eigenvalues, not the regulariser). Choosing $M$ optimally gives the rates below, which for Matérn kernels are tight up to logarithms against the known lower bounds.

Background — Mercer eigenfunctions

Choose a fixed reference probability measure $q$ on $D$. The kernel integral operator acts on functions by $(T_kg)(x)=\int_D k(x,z)g(z)\,q(dz)$. An eigenfunction satisfies $T_ke_m=\lambda_m e_m$, the function-space analogue of $Av=\lambda v$. For a continuous PSD kernel on compact $D$ and a Borel probability measure $q$ with support $D$ (every nonempty open part of $D$ receives positive weight), Mercer's theorem gives $k(x,z)=\sum_m\lambda_m e_m(x)e_m(z)$ with $\lambda_m\ge0$ and the $e_m$ orthonormal for the inner product $\int gh\,dq$. These $\lambda_m$ describe the whole kernel and reference measure; they are not the eigenvalues of a particular observed Gram matrix and are unrelated to the regulariser $\lambda$.

For $D=[0,1]$, uniform $q$ and $k(x,z)=1$, $(T_kg)(x)=\int_0^1g(z)\,dz$: the output is the constant average of $g$. Here $e_1(x)=1$, $\lambda_1=1$, and every zero-average function is mapped to zero.

Theorem — Spectral truncation bound (Vakili, Khezeli & Picheny 2021)

Let $D$ be compact, $q$ a Borel probability measure with support $D$, and $k$ continuous and PSD with $k(x,x)\le\kappa$. Suppose its Mercer eigenfunctions satisfy $|e_m(x)|\le\psi$ for all $m$ and $x$. Fix $\lambda\gt0$. For all integers $T,M\ge1$, with $\Delta_M=\psi^2\sum_{m\gt M}\lambda_m$,

$$\gamma_T\le\frac M2\log\left(1+\frac{T\kappa}{M\lambda}\right)+\frac{T\Delta_M}{2\lambda}.$$

Proof idea. Split $k$ into its first $M$ Mercer terms and the tail. The first part gives a Gram matrix of rank at most $M$ and trace at most $T\kappa$; Jensen's inequality for the concave logarithm (see Primer B) bounds its $\log\det$ by the first term. The tail gives a PSD Gram matrix with trace at most $T\Delta_M$, and $\log\det(I+C)\le\operatorname{tr}C$ for PSD $C$ bounds its extra contribution. In words: the retained directions cost logarithmically each, the discarded tail costs its total mass.

Vakili et al., 2021, Theorem 3

Derivation — choosing $M$: from eigendecay to the rates in the table

For $\lambda_m\le C_pm^{-\beta_p}$ with $\beta_p\gt1$, comparison with an integral gives $\sum_{m\gt M}\lambda_m\le C_pM^{1-\beta_p}/(\beta_p-1)$, so the bound is a constant times $M\log T$ plus a constant times $TM^{1-\beta_p}$. Balance the two terms by choosing $M$ of order $(T/\log T)^{1/\beta_p}$, rounded to an integer; substitution gives $T^{1/\beta_p}(\log T)^{1-1/\beta_p}$, the polynomial-decay row (the constants hidden by $O$ are not identified). For Matérn, $\beta_p=(2\nu+d)/d$ turns the exponents into $1/\beta_p=d/(2\nu+d)$ and $1-1/\beta_p=2\nu/(2\nu+d)$. For a finite-rank kernel of rank $d$, take $M=d$, so $\Delta_M=0$. For exponential decay $\lambda_m\le C e^{-c m^{1/d}}$, comparison with an integral and the substitution $v=u^{1/d}$ bound the tail by a constant times $M^{(d-1)/d}e^{-cM^{1/d}}$; choosing $M=\lceil(2\log T/c)^d\rceil$ makes $e^{-cM^{1/d}}\le T^{-2}$, so the tail term is $O((\log T)^{d-1}/T)$ while the retained term is $O((\log T)^{d+1})$: the SE row.

KernelEigendecaySrinivas et al. 2010 (Thm 5)Vakili et al. 2021 (Cor. 1, Remark 2)
Linear, $d$ dimsfinite rank$O(d\log T)$—
SE$\lambda_m=O(e^{-m^{1/d}})$$O((\log T)^{d+1})$$O(\log^{d+1}T)$
Matérn-$\nu$$\lambda_m=O(m^{-(2\nu+d)/d})$$O\big(T^{\frac{d(d+1)}{2\nu+d(d+1)}}\log T\big)$, $\nu\gt1$$O\big(T^{\frac{d}{2\nu+d}}\log^{\frac{2\nu}{2\nu+d}}T\big)$, $\nu\gt\tfrac12$ (Remark 2, under Assumption 1)
general, $\lambda_m\le C_pm^{-\beta_p}$, $\beta_p\gt1$ (their Definition 1, under Assumption 1)polynomial—$O\big(T^{1/\beta_p}\log^{1-1/\beta_p}T\big)$
Caveat — Matérn kernels and sublinearity
Plugging a $\gamma_T$ rate into an RKHS regret bound $O(\beta_T\sqrt{T\gamma_T})$ with $\beta_T=O(\sqrt{\gamma_T})$ gives $O(\sqrt T\gamma_T)$, sublinear only if $\gamma_T=o(\sqrt T)$: with the Matérn rate $T^{d/(2\nu+d)}$ this needs $d\lt2\nu$, and with the original Srinivas rate it fails for many $(\nu,d)$. Vakili et al. fixed $\gamma_T$; Whitehouse, Ramdas & Wu, NeurIPS 2023 fixed the regret by regularising in proportion to kernel smoothness (Section 6). These rates are what every safe-BO sample-complexity statement in Modules 4–6 depends on.
Worked example — repeated measurements are worth less

Take the SE kernel with $\ell=1$ and $\lambda=0.01$, and compare designs of one or two measurements:

Design$\tfrac12\log\det(I+\lambda^{-1}K_A)$
one point, $x=0$$\tfrac12\log(1+100)=2.31$
the same point twice, $x=0,0$$\tfrac12\log(1+200)=2.65$
two close points, $x=0,\ 0.1$$2.85$
two far-apart points, $x=0,\ 5$$\log(101)=4.62$

For the repeated point, $K_A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ has eigenvalues $2$ and $0$, so the determinant is $(1+200)(1+0)=201$. The second measurement only averages out noise and is worth about $0.34$ nats. Two far-apart points have $K_A\approx I$ and give $\tfrac12\log(101^2)$, twice a single point's gain.

The sequential form in the lemma above gives the same total. For the repeated point it is $\tfrac12\big[\log(1+100)+\log(1+\sigma_1^2/\lambda)\big]$ with $\sigma_1^2=1-1/1.01=0.0099$, again $2.65$.

5. Frequentist Confidence Bounds & $\beta_t$

In plain words. Safe BO needs error bars that hold at every input and every round at once, with probability at least $1-\delta$, even though the algorithm chooses where to measure based on earlier noisy data. The answer is a band $\mu_{t-1}(x)\pm\beta_t\sigma_{t-1}(x)$. The width multiplier $\beta_t$ collects four ingredients: the norm bound $B$, the noise level $R$, the failure probability $\delta$, and the information gain. Only the information gain grows with data.

Symbols in this section

The goal is: with probability at least $1-\delta$, for all $t\ge1$ and all $x\in D$ simultaneously, $|\mu_{t-1}(x)-f(x)|\le\beta_t\,\sigma_{t-1}(x)$. "Simultaneously" is the hard part: a safe-exploration algorithm chooses $x_t$ from the past, the noise is only a martingale difference sequence (its conditional law may depend on the history; see Primer C), and a naive union bound assigning the same failure budget to every point and time is insufficient. Three generations of results achieve it.

A union bound does work over countably many times if the failure budgets are summable, e.g. $\delta_t=6\delta/(\pi^2t^2)$, which add up to $\delta$ because $\sum_t1/t^2=\pi^2/6$; Srinivas et al. use such a union bound over time. It cannot give a useful bound over the uncountably many points of $D$ by granting each point the same positive failure probability. The RKHS argument handles all $x$ at once through one norm bound, and the self-normalised bounds below handle all $t$ at once through a maximal inequality instead of a time budget.

Theorem — Srinivas, Krause, Kakade & Seeger (ICML 2010), Theorem 6 (the concentration result behind their RKHS regret bound, Theorem 3)
Assumptions. $f\in\mathcal H_k$ with $\|f\|_k^2\le B$ (bound on the squared norm); $k(x,x)\le1$ on $D$ (their standing assumption); the inputs are predictable, $0\lt\delta\lt1$, $\sigma\gt0$, and noise $\varepsilon_t$ is a martingale difference sequence uniformly bounded by $\sigma$; the posterior is computed with prior $\mathcal{GP}(0,k)$ and noise variance $\lambda=\sigma^2$ (a deliberately misspecified likelihood).
Statement. With $\beta_t=2B+300\,\gamma_t\log^3(t/\delta)$ (Theorem 6 itself is printed with $2\|f\|_k^2$; since $\beta_t$ is increasing in it, any $B\ge\|f\|_k^2$ may be substituted, as in their Theorem 3 and in SafeOpt), $$\Pr\Big\{\forall T\ge1,\ \forall x\in D:\ |\mu_T(x)-f(x)|\le\beta_{T+1}^{1/2}\,\sigma_T(x)\Big\}\ge1-\delta .$$ In words. Band $\mu\pm\beta^{1/2}\sigma$ with $\beta$ growing like $\gamma_t\log^3t$. The proof does not discretise $D$: it controls the error in the norm of the posterior RKHS, $\|\mu_T-f\|_{k_T}^2\le\beta_{T+1}$, by an inductive martingale argument with Freedman's (Bernstein-type) inequality and a union bound over $T$, and then gets uniformity in $x$ for free from Cauchy–Schwarz in $\mathcal H_{k_T}$, $|\mu_T(x)-f(x)|\le\sigma_T(x)\,\|\mu_T-f\|_{k_T}$. Freedman's inequality needs bounded martingale increments, which is where the bounded-noise assumption comes from; the union bound over $T$ and the sub-exponential regime of Freedman's tail produce the enormous constant and the $\log^3$ factor. This is the bound behind SafeOpt's Theorem 1 (Sui et al., ICML 2015) and SafeMDP.
Going deeper — the two tools behind Srinivas et al.'s proof

Freedman's inequality (Freedman 1975, as used in Appendix B.1 of Srinivas et al.). Let $Z_1,Z_2,\dots$ be a martingale difference sequence for a filtration $(F_s)$, i.e. $\mathbb E[Z_s\mid F_{s-1}]=0$ (see Primer C), with steps bounded above by a constant, $Z_s\le b$, and let $W_n=\sum_{s\le n}\mathbb E[Z_s^2\mid F_{s-1}]$ be the accumulated conditional variance. Then for all $u,v\gt0$, $$\Pr\Big\{\sum_{s\le n}Z_s\ge u\ \text{ and }\ W_n\le v\Big\}\le\exp\Big(-\frac{u^2}{2v+2bu/3}\Big).$$ For $u$ small compared with $v/b$ this is a Gaussian tail with variance $v$; for larger $u$ it decays only exponentially in $u$, and this sub-exponential regime is what produces Srinivas et al.'s large constant and $\log^3$ factor. The step bound $b$ is why their theorem needs bounded noise; a two-sided statement applies the inequality to $-Z_s$ as well, with its own failure budget.

The posterior RKHS (Srinivas et al., Appendix B.2). For fixed inputs $x_1,\dots,x_t$ (repetitions allowed) and $\lambda\gt0$, the posterior covariance $k_t(x,x')$ of Section 2 is itself a kernel. Its RKHS contains the same functions as $\mathcal H_k$, with the larger norm $$\|g\|_{k_t}^2=\|g\|_k^2+\lambda^{-1}\sum_{i=1}^tg(x_i)^2,$$ which charges extra for values at the measured inputs. Since $k_t(x,x)=\sigma_t^2(x)$, the bound $|g(x)|\le\|g\|_{k_t}\sigma_t(x)$ holds for every $x$ at once. Applied to $g=\mu_t-f$ (in $\mathcal H_k$, because $\mu_t$ is a finite kernel expansion), a single bound on $\|\mu_t-f\|_{k_t}$ yields the whole band.

Theorem — Abbasi-Yadkori, Pál & Szepesvári (NeurIPS 2011), Theorem 1: self-normalised bound
Assumptions. $\{F_t\}$ a filtration; $X_t\in\mathbb R^d$ is $F_{t-1}$-measurable; $\varepsilon_t$ is $F_t$-measurable and conditionally $R$-sub-Gaussian, $\mathbb E[e^{\alpha\varepsilon_t}\mid F_{t-1}]\le e^{\alpha^2R^2/2}$ for all $\alpha\in\mathbb R$ (they write $\eta_t$ for the noise); $V\succ0$ fixed; $S_t=\sum_{s\le t}\varepsilon_sX_s$, $\bar V_t=V+\sum_{s\le t}X_sX_s^\top$.
Statement. With probability at least $1-\delta$, for all $t\ge0$ simultaneously, $$\|S_t\|_{\bar V_t^{-1}}^2\le2R^2\log\!\Big(\frac{\det(\bar V_t)^{1/2}\det(V)^{-1/2}}{\delta}\Big).$$ In words. The noise-weighted sum of features, measured in the norm of its own growing covariance, grows only like a log-determinant. Their Theorem 2 turns it into the ridge-regression ellipsoid $\|\hat\theta_t-\theta_*\|_{\bar V_t}\le R\sqrt{2\log(\det(\bar V_t)^{1/2}\det(\lambda I)^{-1/2}/\delta)}+\lambda^{1/2}B$ for $\|\theta_*\|\le B$ (they write $S$ for this bound): the linear-kernel case of everything below, where $B=\|f\|_k$ for $f(x)=\theta_*^\top x$ (Abbasi-Yadkori et al., NeurIPS 2011).
Derivation — from Theorem 1 to the ellipsoid of Theorem 2

For the linear model $y_s=X_s^\top\theta_*+\varepsilon_s$, take $V=\lambda I$ and the ridge estimate $\hat\theta_t=\bar V_t^{-1}\sum_{s\le t}X_sy_s$. Substituting $y_s$ gives $\bar V_t\hat\theta_t=(\bar V_t-\lambda I)\theta_*+S_t$, i.e. $\bar V_t(\hat\theta_t-\theta_*)=S_t-\lambda\theta_*$. Since $\|v\|_{\bar V_t}=\|\bar V_tv\|_{\bar V_t^{-1}}$, the triangle inequality and $\bar V_t\succeq\lambda I$ (so $\|\theta\|_{\bar V_t^{-1}}\le\lambda^{-1/2}\|\theta\|$) give $\|\hat\theta_t-\theta_*\|_{\bar V_t}\le\|S_t\|_{\bar V_t^{-1}}+\lambda\|\theta_*\|_{\bar V_t^{-1}}\le\|S_t\|_{\bar V_t^{-1}}+\sqrt\lambda\|\theta_*\|$. Inserting Theorem 1 with $\det V=\det(\lambda I)$ gives the displayed ellipsoid.

Proof sketch — the method of mixtures in four lines

1. One supermartingale per direction. For fixed $\xi\in\mathbb R^d$ put $M_t^\xi=\exp\big(\xi^\top S_t/R-\tfrac12\sum_{s\le t}(\xi^\top X_s)^2\big)$. Sub-Gaussianity gives $\mathbb E[\exp(\varepsilon_t\xi^\top X_t/R)\mid F_{t-1}]\le\exp(\tfrac12(\xi^\top X_t)^2)$ ($X_t$ is known at time $t-1$), so $\mathbb E[M_t^\xi\mid F_{t-1}]\le M_{t-1}^\xi$ (a supermartingale, see Primer C) and $\mathbb EM_t^\xi\le1$.

2. Mix over $\xi$. Integrate against $\xi\sim\mathcal N(0,V^{-1})$; completing the square in the Gaussian integral gives the closed form $$M_t:=\int M_t^\xi\,\mathcal N(\xi;0,V^{-1})\,d\xi=\Big(\frac{\det V}{\det\bar V_t}\Big)^{1/2}\exp\Big(\frac{\|S_t\|^2_{\bar V_t^{-1}}}{2R^2}\Big),$$ still a non-negative supermartingale with $\mathbb EM_t\le1$. Why mix: the direction that makes $\|S_t\|_{\bar V_t^{-1}}$ large is random and unknown in advance; the Gaussian mixture covers all directions at once.

Completing the square (for $R\gt0$; if $R=0$ the noise vanishes and there is nothing to bound). The mixing density is $(2\pi)^{-d/2}\det(V)^{1/2}e^{-\xi^\top V\xi/2}$, so the integrand's exponent is $\xi^\top S_t/R-\tfrac12\xi^\top\bar V_t\xi=-\tfrac12(\xi-m)^\top\bar V_t(\xi-m)+\|S_t\|_{\bar V_t^{-1}}^2/(2R^2)$ with $m=\bar V_t^{-1}S_t/R$. The remaining Gaussian integral is $(2\pi)^{d/2}\det(\bar V_t)^{-1/2}$; multiplying by the density's constant gives the displayed $M_t$. Integrating the inequalities $\mathbb E[M_t^\xi\mid F_{t-1}]\le M_{t-1}^\xi$ over $\xi$ keeps them, so $M_t$ is again a supermartingale.

3. Ville's inequality. For a non-negative supermartingale, $\Pr\{\sup_tM_t\ge1/\delta\}\le\delta\,\mathbb EM_0\le\delta$: a maximal inequality, which is what makes the result simultaneous over all $t$ (Abbasi-Yadkori et al. phrase it with a stopping time).

4. Rearrange. $M_t\lt1/\delta$ for all $t$ is exactly $\|S_t\|^2_{\bar V_t^{-1}}\lt2R^2\log(\det(\bar V_t)^{1/2}\det(V)^{-1/2}/\delta)$. $\square$

In an RKHS, $X_s=\varphi(x_s)$ is infinite-dimensional and the Gaussian mixture becomes a Gaussian process; with $A_t:=\Phi_t^*\Phi_t=\sum_{s\le t}\varphi(x_s)\varphi(x_s)^*$ (finite rank), the Fredholm determinant $\det(I+\lambda^{-1}A_t)$ equals $\det(I_t+\lambda^{-1}K_t)$ (Weinstein–Aronszajn), so the infinite dimension never shows up in the bound. Chowdhury & Gopalan do this with a "double mixture" over functions and sequences; Whitehouse, Ramdas & Wu (2023) give a projection-and-limit proof and note the result is essentially Corollary 3.5 of Abbasi-Yadkori's PhD thesis (2013).

Theorem — RKHS self-normalised bound (Chowdhury & Gopalan 2017, Thm 1; Abbasi-Yadkori 2013 / Whitehouse et al. 2023, Cor. 1)
Assumptions. $x_t$ is $F_{t-1}$-measurable, $\varepsilon_t$ is $F_t$-measurable and conditionally $R$-sub-Gaussian; $k$ a kernel with Gram matrices $K_t$; $S_t=\sum_{s\le t}\varepsilon_s\varphi(x_s)$, $A_t=\Phi_t^*\Phi_t=\sum_{s\le t}\varphi(x_s)\varphi(x_s)^*$ (Whitehouse et al. call it $V_t$; on this page $V_t=A_t+\lambda I$ includes the regulariser).
Statement (Whitehouse et al. form). For any $\rho\gt0$, with probability at least $1-\delta$, simultaneously for all $t\ge0$, $$\big\|(A_t+\rho\,\mathrm{id})^{-1/2}S_t\big\|_k\le R\sqrt{2\log\Big(\tfrac1\delta\sqrt{\det(I_t+\rho^{-1}K_t)}\Big)}=R\sqrt{\log\det(I_t+\rho^{-1}K_t)-2\log\delta}\,;$$ in kernel coordinates (for $K_t\succ0$) the left side equals $\|(I_t+\rho K_t^{-1})^{-1/2}\varepsilon_{1:t}\|_2=\sqrt{\varepsilon_{1:t}^\top K_t(K_t+\rho I)^{-1}\varepsilon_{1:t}}$.
Statement (Chowdhury & Gopalan form). For any $\eta\gt0$ (or $\eta=0$ if $K_t\succ0$ a.s.), with probability at least $1-\delta$, for all $t\ge0$, $$\|\varepsilon_{1:t}\|^2_{((K_t+\eta I)^{-1}+I)^{-1}}\le2R^2\log\Big(\tfrac1\delta\sqrt{\det((1+\eta)I+K_t)}\Big).$$ The two coincide at $\rho=1$, $\eta\downarrow0$ and are otherwise not equivalent; qualitatively, increasing $\rho$ corresponds to decreasing $\eta$ (Whitehouse et al., Section 3).
Theorem — Chowdhury & Gopalan (ICML 2017), Theorem 2
Assumptions. $f\in\mathcal H_k$ with $\|f\|_k\le B$ (bound on the norm); $x_t$ predictable, $\varepsilon_t$ conditionally $R$-sub-Gaussian (no boundedness); $\mu_{t-1},\sigma_{t-1}$ computed with $\lambda=1+\eta$, $\eta=2/T$, $T\ge1$ a fixed integer horizon, a fixed kernel and $0\lt\delta\lt1$; $\gamma$ uses this same $\lambda$.
Statement. With probability at least $1-\delta$, for all $x\in D$ and all $1\le t\le T$, $$|\mu_{t-1}(x)-f(x)|\le\Big(B+R\sqrt{2\big(\gamma_{t-1}+1+\ln(1/\delta)\big)}\Big)\sigma_{t-1}(x)=:\beta_t\,\sigma_{t-1}(x).$$ In words. Band $\mu\pm\beta\sigma$ (no square root): the deterministic bias term $B$ of Section 3 plus a noise term growing like the square root of the information gain. Why each assumption matters: $\|f\|_k\le B$ feeds the bias bound; sub-Gaussianity feeds the supermartingale; $\lambda=1+\eta\ge1$ is needed because their Theorem 1 normalises with a unit regulariser in feature space; the horizon enters only through $\eta(t-1)\le2$, so the "$+1$" in $\beta_t$ is justified for rounds within the horizon $T$ (the paper prints "for all $t\ge1$"; the next theorem shows that this is nevertheless true). The derivation is the walkthrough below.
Theorem — All-time version of the band (Whitehouse, Ramdas & Wu 2023, Corollary 1, plus the bias–noise split)
Assumptions. $f\in\mathcal H_k$ with $\|f\|_k\le B$; $x_t$ predictable; $\varepsilon_t$ conditionally $R$-sub-Gaussian; one fixed kernel and one fixed $\lambda\gt0$, used for $\mu_{t-1},\sigma_{t-1}$ and $\gamma_{t-1}$ alike; $0\lt\delta\lt1$.
Statement. With probability at least $1-\delta$, for all $x\in D$ and all $t\ge1$ simultaneously (at $t=1$: $\mu_0=0$, $\sigma_0^2(x)=k(x,x)$, $\gamma_0=0$), $$|\mu_{t-1}(x)-f(x)|\le\Big(B+\frac{R}{\sqrt\lambda}\sqrt{2\gamma_{t-1}+2\ln(1/\delta)}\Big)\sigma_{t-1}(x).$$ In words. This is Step 4 of the walkthrough with the realised $\ln\det(I+K_{t-1}/\lambda)=2G_{t-1}$ bounded by $2\gamma_{t-1}$; no horizon enters. For $\lambda=1+2/T\ge1$ the multiplier is at most the Chowdhury–Gopalan one, because $R/\sqrt\lambda\le R$ and $2\gamma+2\ln(1/\delta)\le2(\gamma+1+\ln(1/\delta))$; so their band holds for every $t\ge1$, only by a different proof.
Common mistake — four convention traps
  1. $\beta$ versus $\beta^{1/2}$. Srinivas et al., SafeOpt (Sui et al. 2015) and SafeMDP (Turchetta et al. 2016) write $Q_t=\mu\pm\beta_t^{1/2}\sigma$; Chowdhury–Gopalan-style papers (StageOpt, CONFIG, ISE, Real-$\beta$-SafeOpt, Berkenkamp et al. 2017, Fiedler et al. 2021) write $\mu\pm\beta_t\sigma$. The interval-width recursion is $w_t^2\le4\beta_t\sigma^2$ in the first convention and $4\beta_t^2\sigma^2$ in the second, which is why StageOpt's sample complexity carries $\beta^2$ where SafeOpt's carries $\beta$.
  2. $B$ on the norm versus the squared norm. Srinivas/SafeOpt assume $\|f\|_k^2\le B$ and use $2B$; Chowdhury–Gopalan assume $\|f\|_k\le B$ and use $B$ linearly. StageOpt states $\|g_i\|_k^2\le B$ but plugs $B$ linearly into the Chowdhury–Gopalan formula; read such statements with $B\to\sqrt B$ in mind.
  3. Which $\lambda$. Srinivas et al. compute the posterior with $\lambda=\sigma^2$; Chowdhury–Gopalan with $\lambda=1+2/T$; Fiedler et al. with any $\lambda\gt0$. A $\beta$ is not transferable between posteriors computed with different $\lambda$, because $\sigma_t$ itself changes.
  4. Index shifts. $\beta_{T+1}\sigma_T$ (Srinivas), $\beta_t\sigma_{t-1}$ with $\gamma_{t-1}$ (Chowdhury–Gopalan), $\beta_N\sigma_N$ (Fiedler). Harmless unless a proof uses monotonicity of $\beta_t$.

How large are these $\beta$ really?

Take an ordinary case: $\|f\|_k=2$ (Srinivas' $B=4$), $t=100$, $\gamma_t=10$ (the same value for both theorems, a simplification: $\gamma_t$ depends on $\lambda$), $\delta=0.05$, and zero-mean noise bounded by $0.1$, so that Srinivas' $\sigma=0.1$ and, by Hoeffding's lemma, $R=0.1$ as well. (Gaussian noise would not do here: it is not bounded, so Srinivas' theorem would not apply to it.) Srinivas et al.: $\beta_t=8+300\cdot10\cdot\ln^3(2000)\approx8+300\cdot10\cdot439\approx1.32\times10^6$, a band of $\beta_t^{1/2}\approx1150$ posterior standard deviations. Chowdhury–Gopalan: $\beta_t=2+0.1\sqrt{2(10+1+3.0)}\approx2.53$, but for a posterior computed with $\lambda\approx1$, a hundred times $\sigma^2=0.01$, whose $\sigma_{t-1}$ is correspondingly inflated. The data-dependent bounds of Section 6 land at roughly $B$ plus a few units to a few tens for the $\lambda$ a practitioner would use (the explorer prints them). Against this, the constants used in practice are $\beta\equiv2$ (Berkenkamp et al. 2016, Turchetta et al. 2016), $\beta\equiv3$ (GoSafe) and $\beta\equiv4$ (GoSafeOpt), and Srinivas et al. already reported scaling their own $\beta_t$ down by a factor of five in experiments. Fiedler et al. (TMLR 2024) quantify the consequence: with $\beta\equiv2$, on average $2727\pm3882$ of $10^4$ sampled data sets per test function violated the uncertainty tube (100 test functions), and SafeOpt with $\beta\equiv2$ produced $2862$ unsafe runs out of $10^4$ on one test function; Real-$\beta$-SafeOpt with the true RKHS norm produced none, and with the norm under-estimated as $2.5$ instead of $10$ it produced $1338$ (Module 5).

Key insight
A confidence bound has four ingredients that the user must supply or choose: the norm bound $B$ (unknowable, Section 1), the noise level $R$ (measurable), the failure probability $\delta$ (it enters only through $\log(1/\delta)$) and the regulariser $\lambda$ (a free choice that trades the size of $\beta$ against the size of $\sigma_t$). The information term is the only one that grows with data, and for the SE kernel only polylogarithmically in $t$. In the modern bounds $\beta$ is dominated either by $B$ or by the noise term $(R/\sqrt\lambda)\sqrt{\log\det}$, and the explorer shows both regimes. Of all the ingredients only $B$ cannot be measured: it is the price of admitting that one does not know how wiggly $f$ is.
Worked example — computing $\beta$ and the band it gives

Use the all-time bound above,

$$\beta_t=B+\frac{R}{\sqrt\lambda}\sqrt{2\gamma_{t-1}+2\ln(1/\delta)},$$

with $B=1$, noise level $R=0.1$, $\lambda=0.01$ (so $R/\sqrt\lambda=1$), $\delta=0.01$, and the SE kernel, for which $k(x,x)=1$. At round $t=2$ we need $\gamma_1$, the most one measurement can reveal: $\gamma_1=\tfrac12\log(1+1/\lambda)=2.31$ (Section 4). Then

$$\beta_2=1+\sqrt{2(2.31)+2\ln100}=1+\sqrt{4.62+9.21}=4.72 .$$
  • Where $\sigma_1(x)=0.1$, the band is $\mu_1(x)\pm0.47$. The common heuristic $\beta=2$ would give $\pm0.2$: less than half as wide, and not backed by the theorem.
  • Tightening to $\delta=0.001$ raises $\beta_2$ only to $5.29$, because the failure probability enters through a logarithm.
  • The norm bound enters additively: doubling $B$ to $2$ adds $1$ to $\beta$, a full extra standard deviation.

Underestimating $B$ or $R$ makes the band too narrow, and the guarantee silently fails. That is the gap between theory and practice that Module 5 opens with (Module 5 §1).

6. Practical and Rigorous Bounds (Fiedler, Scherer, Trimpe)

For learning-based control one rarely needs an a-priori bound: the data set is fixed when the controller is synthesised, so the bound may depend on it. Fiedler, Scherer & Trimpe (AAAI 2021) exploit exactly this: stop the Chowdhury–Gopalan proof at the log-determinant instead of bounding it by $\gamma_t$.

Theorem — Fiedler, Scherer & Trimpe (AAAI 2021), Theorem 1, as corrected in arXiv v2
Assumptions. $k$ positive definite on $D\ne\emptyset$; $f\in\mathcal H_k$ with $\|f\|_k\le B$; $(x_n)$ predictable w.r.t. a filtration, $\varepsilon_n$ conditionally $R$-sub-Gaussian; $\mu_N,\sigma_N$ the GP posterior with likelihood noise variance $\lambda\gt0$ (any value); $\bar\lambda:=\max\{1,\lambda\}$.
Statement. For any $\delta\in(0,1)$, with $$\beta_N=B+\frac{R}{\sqrt\lambda}\sqrt{\log\det\Big(\frac{\bar\lambda}{\lambda}K_N+\bar\lambda I_N\Big)-2\log\delta}\,,\qquad\Pr\big[\,|\mu_N(x)-f(x)|\le\beta_N\sigma_N(x)\ \ \forall N\in\mathbb N,\ \forall x\in D\,\big]\ge1-\delta .$$ In words. Same structure as Chowdhury–Gopalan, but the information term is the realised $\log\det$ of the actual Gram matrix: computable after the fact, no $\gamma_t$, any kernel (custom or linearly constrained GPs), any $\lambda$.
Caveat — what the correction changed and why
The AAAI 2021 print version states $\beta_N=B+R\sqrt{\log\det(K_N+\bar\lambda I_N)-2\log\delta}$; arXiv v2 (August 2023) says in footnote 2 that the denominator $\sqrt\lambda$ and the factor $\bar\lambda/\lambda$ in the determinant were missing. The mechanism: Chowdhury–Gopalan's Theorem 1 needs a regulariser of at least one in feature space. For $\lambda\ge1$ one applies it directly with $\eta=\lambda-1$: then $\bar\lambda=\lambda$, the determinant is $\det(K_N+\lambda I)$ in both versions, and the corrected noise term is the printed one divided by $\sqrt\lambda$. So for $\lambda\ge1$ the printed bound remains valid, merely conservative, and at $\lambda=1$ the two coincide (the authors' remark that for $\lambda\ge1$ the original results hold without correction). For $0\lt\lambda\lt1$ one must rescale the kernel to $\bar k=k/\lambda$ with unit regulariser: the posterior mean is unchanged, $\bar\sigma_N^2=\sigma_N^2/\lambda$, $\|f\|_{\bar k}=\sqrt\lambda\|f\|_k$ and the determinant becomes $\det(K_N/\lambda+I)$, which is where both missing factors come from. For $\lambda\lt1$ the printed formula is strictly smaller than the corrected one, so no theorem covers it any more; Exercise 3.6 shows that it cannot be valid for all $\lambda$, and the explorer shows it being violated at small $\lambda$. The CDC 2021 robust-synthesis paper by the same authors (Module 11) prints a pre-correction form in its Proposition 1, $\beta_D=B+2R\sqrt{\log\det(K_D+\bar\lambda I)-2\log\delta}$ (no $1/\sqrt\lambda$, no $\bar\lambda/\lambda$); re-derive with the corrected constant before using it.
Fact — Scaling the kernel rescales the norm
For a PSD kernel $k$ and $c\gt0$, the kernel $ck$ has the same RKHS as $k$, with $\|f\|_{ck}=\|f\|_k/\sqrt c$. Proof. $f=\sum_ia_ik(z_i,\cdot)=\sum_i(a_i/c)(ck)(z_i,\cdot)$ has squared $ck$-norm $(a/c)^\top(cK_z)(a/c)=a^\top K_za/c=\|f\|_k^2/c$, and limits of such expansions inherit the identity. $\square$ With $c=1/\lambda$ (so $\bar k=k/\lambda$ and $\|f\|_{\bar k}=\sqrt\lambda\|f\|_k$) and regulariser $1$ instead of $\lambda$, substituting into the formulas of Section 2 leaves the posterior mean unchanged and divides the posterior variance by $\lambda$.
Derivation — why the correction is exactly $\sqrt\lambda$ and $\bar\lambda/\lambda$

From Steps 1–3 of the walkthrough, for every $x$ and $N$, $$|f(x)-\mu_N(x)|\le\|f\|_k\sigma_N(x)+\frac{1}{\sqrt\lambda}\sigma_N(x)\sqrt{\varepsilon_N^\top K_N(K_N+\lambda I)^{-1}\varepsilon_N}\,.$$ Write $\bar K_N=(\bar\lambda/\lambda)K_N$ and observe $K_N(K_N+\lambda I)^{-1}=\bar K_N(\bar K_N+\bar\lambda I)^{-1}$ (multiply inside and outside the inverse by $\bar\lambda/\lambda$). Since $\bar\lambda\ge1$, Chowdhury–Gopalan's Theorem 1 applies to the kernel $(\bar\lambda/\lambda)k$ with $\eta=\bar\lambda-1\ge0$ (the case $\eta=0$, i.e. $\lambda\le1$, needs $\bar K_N\succ0$, which Fiedler et al. take from the positive definiteness of $k$ and which holds for distinct inputs; the $\rho$-form of Whitehouse et al. needs no such condition) and gives, with probability at least $1-\delta$ for all $N$, $$\varepsilon_N^\top\bar K_N(\bar K_N+\bar\lambda I)^{-1}\varepsilon_N\le\varepsilon_N^\top\big((\bar K_N+\eta I)^{-1}+I\big)^{-1}\varepsilon_N\le R^2\big(\log\det(\bar K_N+\bar\lambda I)-2\log\delta\big),$$ where the first inequality is not a general monotonicity rule for matrix functions but holds because both matrices have the eigenvectors of $\bar K_N=Q\operatorname{diag}(a_j)Q^\top$ (see Primer A): with $\bar\lambda=1+\eta$ their eigenvalues are $a_j/(a_j+1+\eta)$ and $(a_j+\eta)/(a_j+\eta+1)$, whose difference $\eta/(a_j+1+\eta)$ is $\ge0$ (Chowdhury–Gopalan's proof phrases this as $\bar K_N\preceq\bar K_N+\eta I$). Substituting yields the corrected $\beta_N$: the $1/\sqrt\lambda$ is the Cauchy–Schwarz factor of Step 3 (simply dropped originally), the $\bar\lambda/\lambda$ is the rescaling needed for "regulariser $\ge1$". The bound is uniform in $x$ because the one random factor controlled by the martingale bound, $\varepsilon_N^\top K_N(K_N+\lambda I)^{-1}\varepsilon_N=\|S_N\|^2_{V_N^{-1}}$, does not involve $x$: the same high-probability event covers every query point, and $x$ enters only through $\sigma_N(x)$. (Under adaptive sampling $\sigma_N$ and $K_N$ do depend on past noise through the inputs; the self-normalised bound allows this because the inputs are predictable.)

Repeated inputs. They make $K_N$ singular even for a positive definite kernel, so the $\eta=0$ route ($\lambda\le1$) does not apply as written. The $\rho$-form bound of Section 5 with $\rho=\lambda$ needs no inverse of $K_N$ and gives the Abbasi-Yadkori band listed below; the corrected Fiedler band equals it for $\lambda\le1$ and is wider for $\lambda\gt1$, so it stays valid for the explorer's designs, which may repeat inputs.

Variants: log-det-free, Abbasi-Yadkori form, and tighter

The Bayesian counterpart: Lederer, Umlauft & Hirche (NeurIPS 2019)

Theorem — Uniform error bound for GP sample paths (Lederer et al. 2019, Theorem 3.1)
Assumptions. $f$ is a sample from $\mathcal{GP}(0,k)$ on a compact $X$; observations $y=f(x)+\epsilon$ with i.i.d. $\mathcal N(0,\sigma_n^2)$ noise; $k$ continuous with Lipschitz constant $L_k$ (their eq. 3); $f$ Lipschitz with constant $L_f$; $M(\tau,X)$ the $\tau$-covering number of $X$ (Euclidean balls of radius $\tau$). Their $\nu_N$, $\beta(\tau)$, $\gamma(\tau)$ are the posterior mean and two constants of this theorem, not the $\mu_t$, $\beta_t$, $\gamma_t$ of this page.
Statement. The posterior mean $\nu_N$ and standard deviation $\sigma_N$ are continuous, with Lipschitz constant $L_{\nu_N}$ and modulus of continuity $\omega_{\sigma_N}$ (see Primer B) bounded by $$L_{\nu_N}\le L_k\sqrt N\,\|(K+\sigma_n^2I)^{-1}y\|,\qquad\omega_{\sigma_N}(\tau)\le\sqrt{2\tau L_k\Big(1+N\,\|(K+\sigma_n^2I)^{-1}\|\max_{x,x'\in X}k(x,x')\Big)}\,.$$ For any $\delta\in(0,1)$ and $\tau\gt0$, with $$\beta(\tau)=2\log\frac{M(\tau,X)}{\delta},\qquad\gamma(\tau)=(L_{\nu_N}+L_f)\tau+\sqrt{\beta(\tau)}\,\omega_{\sigma_N}(\tau),$$ $$\Pr\Big[\,|f(x)-\nu_N(x)|\le\sqrt{\beta(\tau)}\,\sigma_N(x)+\gamma(\tau)\ \ \forall x\in X\Big]\ge1-\delta,\qquad M(\tau,[0,r]^d)\le\big(1+r\sqrt d/\tau\big)^d .$$ The covering bound is written with the diameter $r\sqrt d$ of the cube, as in their appendix (eq. 53, $r=\max_{x,x'}\|x-x'\|$); their main-text eq. (9) prints $(1+r/\tau)^d$ with $r$ the edge length, which in the Euclidean metric is too small in high dimension. Ingredients. On a $\tau$-grid, $f(x)-\nu_N(x)$ is Gaussian with variance $\sigma_N^2(x)$, so a Gaussian tail bound plus a union bound over the $M(\tau,X)$ grid points gives $\sqrt{\beta(\tau)}\sigma_N$; between grid points, Lipschitz continuity of $f$ and $\nu_N$ and the modulus of $\sigma_N$ give the additive $\gamma(\tau)$, which shrinking $\tau$ makes negligible at only logarithmic cost in $\beta(\tau)$. Their Theorem 3.2 supplies $L_f$ probabilistically from the kernel.

The statement is for one fixed $N$ and a $\tau$ chosen in advance. If $L_f$ comes from their Theorem 3.2 and is valid with probability at least $1-\delta_L$, a union bound gives the band with probability at least $1-\delta-\delta_L$; several values of $N$ or $\tau$ need their own failure budgets (or a result that is uniform over them).

Background — Finite nets and covering numbers

A $\tau$-net is a finite set $Z\subseteq X$ such that every $x\in X$ has some $z\in Z$ with $\|x-z\|\le\tau$; $M(\tau,X)$ is the smallest possible number of centres. For such a pair, $|f(x)-\nu_N(x)|\le|f(z)-\nu_N(z)|+(L_f+L_{\nu_N})\tau$, and $\sigma_N(z)\le\sigma_N(x)+\omega_{\sigma_N}(\tau)$. Substituting the simultaneous grid bound gives exactly the additive $\gamma(\tau)$. For a cube, partition each coordinate into intervals of length at most $\tau/\sqrt d$ and use cell centres to obtain the displayed conservative covering count.

For $X=[0,1]$ and $\tau=1/4$, the centres $1/4,3/4$ cover the interval with two closed balls. Compactness ensures some finite net exists at every positive radius (see Primer 0); its size controls the union-bound cost.

This bound needs no $B$ and gives modest $\sqrt{\beta(\tau)}$ (about $4.3$–$7.4$ for $\delta=0.01$, $\tau\in[10^{-3},10^{-2}]$ on the unit cube in $d=1,2,3$), which is why it is popular in GP-based control. But it is a statement about a random $f$ under the prior with Gaussian noise: it does not certify a fixed unknown plant, and the Bayesian-versus-RKHS incompatibility of Section 1 means it cannot be combined with the frequentist results. Capone, Lederer & Hirche (ICML 2022) extend it to unknown length-scales. They assume stationary kernels $k_\vartheta(x,x')=\kappa\big((x_1-x'_1)/\vartheta_1,\dots,(x_d-x'_d)/\vartheta_d\big)$ with $\kappa(0)=1$ and a Fourier transform that is a non-negative, non-increasing function of $\|\omega\|_2$ (their Assumption 2.1; SE and Matérn qualify), and a known scaling function $\beta(\vartheta)$ that yields a valid uniform bound with probability $1-\rho$ for each fixed length-scale vector $\vartheta$ (Assumption 3.2, e.g. from Lederer et al.). Then: if $\vartheta'\le\vartheta\le\vartheta''$ componentwise, $\sigma_\vartheta(x)\le\gamma\,\sigma_{\vartheta'}(x)$ with $\gamma^2=\prod_i\vartheta''_i/\vartheta'_i$ (Lemma 3.3), and with the inflated $\bar\beta=\gamma^2\big(\max_{\vartheta'\le\vartheta\le\vartheta''}\beta^{1/2}(\vartheta)+2\|y\|_2/\sigma_n\big)^2$ one has $\Pr[|f(x)-\mu_{\vartheta_0}(x)|\le\bar\beta^{1/2}\sigma_{\vartheta'}(x)\ \forall x]\ge1-\rho$ for a sample $f$ of the GP with any $\vartheta$ in the box and any working length-scale $\vartheta_0$ in the box (Theorem 3.5); with a hyperprior on $\vartheta$ (see Primer C) and the box chosen as a $1-\delta$ posterior credible region, the bound holds with probability $(1-\delta)(1-\rho)$ (Theorem 3.7). Their $\gamma$ is an inflation factor and their $\rho$ a probability, not the information gain and the Whitehouse regulariser.

Here the box describes plausible true length-scales, while $\vartheta_0$ is the one actually used to fit the mean. The probability $(1-\delta)(1-\rho)$ is not an independence calculation: the box contains the true $\vartheta$ with posterior probability at least $1-\delta$, and for every $\vartheta$ in the box the band holds with conditional probability at least $1-\rho$, so conditioning (the tower rule) multiplies the two. Like Lederer et al.'s bound, this is a statement under the GP model and the hyperprior, not a certificate for a fixed function.

BoundModel of $f$NoisePosterior $\lambda$NeedsA posteriori?Typical size
Srinivas et al. 2010, Thm 6$\|f\|_k^2\le B$bounded by $\sigma$$\sigma^2$$\gamma_t$no$\beta^{1/2}\sim10^3$
Chowdhury & Gopalan 2017, Thm 2$\|f\|_k\le B$$R$-sub-Gaussian$1+2/T$$\gamma_t$, horizon $T$no$B+O(R\sqrt{\gamma_t})$
Fiedler et al. 2021, Thm 1 (corrected)$\|f\|_k\le B$$R$-sub-Gaussianany$\log\det$ of the datayes$B+O(R\sqrt{\log\det})$
Abbasi-Yadkori 2013 / Fiedler 2024 eq. 7$\|f\|_k\le B$$R$-sub-Gaussianany$\log\det$ of the datayes$\le$ the row above
Flynn & Reeb 2025$\|f\|_k\le B$$R$-sub-GaussianoptimisedSOCP or 1-D dualyesbelow the Abbasi-Yadkori and Chowdhury–Gopalan rows for matched $c$, $\alpha$ (Thm 6.1)
Lederer et al. 2019, Thm 3.1$f\sim\mathcal{GP}(0,k)$Gaussian$\sigma_n^2$$L_f$, $L_k$, covering numberyes$\sqrt{\beta(\tau)}\approx4$–$7.5$ plus $\gamma(\tau)$

7. When the Assumptions Fail

Each rigorous bound is only as good as its assumptions, and four of them fail routinely in control practice.

Unknown RKHS norm

Under-estimating $B$ shrinks the bias part $B\sigma_t$ of every band and the guarantee silently disappears (the explorer's "under-estimated $B$" option shows it once the bias term dominates, e.g. for $\lambda\approx1$; Fiedler et al. 2024 measured $1338/10^4$ unsafe runs with $B=2.5$ against a true norm of $10$). Tokmak, Krishnan, Schön & Baumann (AISTATS 2025) estimate an over-estimate $B_t\ge\|f\|_k$ from data: sample $m$ random RKHS functions consistent with the observations, sort their norms and discard the largest few (a sampling-and-discarding scenario programme, see Module 15 for a later treatment of scenario methods). Their Theorem 1: under their Assumption 2, that the norms of the random functions and of $f$ are i.i.d. draws from one unknown distribution, and for $m$ large enough, with confidence at least $1-\kappa$ the estimate satisfies $B_t\ge\|f\|_k$ with probability at least $1-\gamma$ (their $\gamma$ and $\kappa$ are probability levels, not the information gain); the resulting confidence intervals (Theorem 2, using the Abbasi-Yadkori form of Section 6) then hold with probability $(1-\gamma)(1-\delta)$, and safety follows as in SafeOpt with $d_k$ in place of the Lipschitz metric (Theorem 3). The i.i.d. assumption is the honest price: the norm of the true function is treated as one more sample. The same group's PACSBO (Tokmak, Schön & Baumann, SysDO 2024) estimates the norm bound locally, for $f$ restricted to a neighbourhood of the explored region rather than globally, which makes it less conservative, again with a probably-approximately-correct safety guarantee.

Theorem — Sampling and discarding for a norm bound (the order-statistic core of Tokmak et al. 2025, Theorem 1)
Assumptions. The norms $Z_1,\dots,Z_m$ of the sampled functions and the norm $Z=\|f\|_k$ of the true function are i.i.d. draws from one (unknown) distribution (their Assumption 2). Sort the samples, $Z_{(1)}\le\dots\le Z_{(m)}$, discard the $r$ largest ($0\le r\lt m$) and set $B=Z_{(m-r)}$. Let $\gamma,\kappa\in(0,1)$ satisfy $\sum_{j=0}^{r}\binom mj\gamma^j(1-\gamma)^{m-j}\le\kappa$.
Statement. With probability at least $1-\kappa$ over the samples, $\Pr_Z(Z\le B)\ge1-\gamma$: a probability about a probability.
In words. $B$ falls below the upper $\gamma$-tail of the distribution only if at most $r$ of the $m$ samples land in that tail, a binomial event with the displayed probability (distributions with atoms only make this more conservative). For $r=0$ the condition is $(1-\gamma)^m\le\kappa$: $\gamma=0.05$ and $\kappa=0.01$ need $m\ge\log\kappa/\log(1-\gamma)\approx89.8$, i.e. $m=90$. Tokmak et al.'s Algorithm 3 discards the largest admissible $r$, their Theorem 1 asks for $(1-\gamma)^{m-1}(1+\gamma(m-1))\le\kappa$ (the $r=1$ case of the sum), and their Corollary 1 extends the statement to all iterations jointly. Everything rests on the i.i.d. modelling assumption.
Fact — What a local norm bound certifies
For a nonempty $U\subseteq D$, the restriction $f|_U$ lies in the RKHS of $k_U=k|_{U\times U}$, with $\|f|_U\|_{k_U}=\min\{\|g\|_k:\ g\in\mathcal H_k,\ g|_U=f|_U\}\le\|f\|_k$: explaining fewer locations can require a smaller norm. If $B_U\ge\|f|_U\|_{k_U}$ and all observations lie in $U$, the Section 6 bands computed with $B_U$ hold on $U$ (there the posterior computed with $k_U$ equals the one computed with $k$) and say nothing outside $U$. If $B_U$ is itself a sampling-based estimate as in the theorem above, its failure probability adds: under that i.i.d. model the band holds with probability at least $1-\kappa-\gamma-\delta$ by a union bound. This is the mechanism behind PACSBO's local norm bounds, not a statement of PACSBO's own guarantees.

Hyperparameters chosen from the data

Every RKHS bound is for a fixed kernel. Maximising the marginal likelihood over the length-scale (see Primer C) after each observation changes $\mathcal H_k$ (or at least its norm) at every step and makes the kernel depend on the noise, so neither $\|f\|_k\le B$ nor the fixed-kernel martingale argument is automatically justified: the theorems do not apply to "ML-II, then Theorem 1". Partial remedies: Fiedler et al. 2021, Propositions 3–4: if the GP uses a kernel whose RKHS contains the true one (for the isotropic SE kernel, a length-scale $\ell\le\tilde\ell$), Theorem 1 holds unchanged provided $B$ bounds the norm of $f$ in the RKHS of the kernel actually used. That norm is finite because $\mathcal H_{\tilde k}\subseteq\mathcal H_k$, but it can exceed $\|f\|_{\tilde k}$ (by up to the factor $(\tilde\ell/\ell)^{d/2}$ of Section 1), since the inclusion map has norm greater than one in general; this is the clarification stated in the Corrections section of v2 (the footnote it adds to Proposition 4 is worded "valid for $\tilde k$"; by the argument just given, what Theorem 1 needs is a bound on the norm for the kernel $k$ used in the GP). Under-estimating $\ell$ is the safe direction for membership, over-estimating it is not. Their Theorem 5 covers a true kernel $\tilde k$ that differs from the working kernel $k$ by at most a known $\tilde\epsilon$ everywhere: the band is widened and shifted by explicit terms that do not vanish even where $\sigma_N=0$, because $f\notin\mathcal H_k$ is possible while $\mu_N\in\mathcal H_k$ always (a simplified version with a complete proof follows below). Capone et al. 2022 handle a box of length-scales in the Bayesian setting (Section 6). On the regret side, Berkenkamp, Schoellig & Krause (JMLR 2019) let the function class grow over time, shrinking the length-scales and inflating the norm bound with monotonically increasing scaling functions $g(t)$, $b(t)$ that may also depend on the data (A-GP-UCB, with $B_t=b(t)g(t)^dB_0$ in their eq. (9)). Their Section 2, Lemma 1 and Algorithm 1 use $B$ as a bound on the norm and insert it linearly into their confidence multiplier $\beta_t^{1/2}$ (Srinivas-style notation), while their Theorem 1 prints the squared-norm assumption $\|f\|_{k_\theta}^2\le B$, inconsistently with those definitions; read the algorithmic $B$ as a norm bound (the factor $g(t)^d$ is then conservative, since by their Lemma 2 the norm itself grows by at most $g(t)^{d/2}$). Their confidence bounds become valid from the (unknown) round on which the class contains $f$, which is enough for no-regret but gives no safety from the first query.

Going deeper — what ML-II maximises, and why a fixed-kernel theorem does not cover it

With a zero-mean working GP and fixed $\lambda$, the log marginal likelihood of the data under kernel parameters $\vartheta$ (e.g. length-scales) is, up to an additive constant, $L(\vartheta)=-\tfrac12y_t^\top(K_\vartheta+\lambda I)^{-1}y_t-\tfrac12\log\det(K_\vartheta+\lambda I)$, where $K_\vartheta$ is the Gram matrix of the kernel with parameters $\vartheta$; ML-II (empirical Bayes) picks the maximiser. A theorem that holds for each fixed kernel separately says nothing about a kernel selected with the same observations. For a finite family of $J$ kernels declared in advance, one repair is to demand all $J$ events at once with failure budgets $\delta/J$ (a union bound), given a valid norm bound for each kernel; this does not justify unrestricted continuous hyperparameter fitting.

Theorem — Robust band under a kernel mismatch (simplified version of Fiedler et al. 2021, Theorem 5)
Assumptions. $k,\tilde k$ PSD kernels on $D$ with $|k(x,z)-\tilde k(x,z)|\le\epsilon_k$ for all $x,z$; $f\in\mathcal H_{\tilde k}$ with $\|f\|_{\tilde k}\le B$ (the norm bound must hold for the true kernel); predictable inputs; conditionally $R$-sub-Gaussian noise; $\lambda\gt0$, $\delta\in(0,1)$; $K_N$, $k_N(x)$, $\mu_N$, $\sigma_N$ computed with the working kernel $k$ (repeated inputs allowed). Put $u_N=\sqrt N\epsilon_k$, $w_N=N\epsilon_k/\lambda^2$ and $$\begin{aligned}C_N(x)&=u_N/\lambda+w_N\|k_N(x)\|_2,\\ S_N^2(x)&=\epsilon_k+u_N\big(2\|k_N(x)\|_2+u_N\big)/\lambda+w_N\|k_N(x)\|_2^2,\\ \bar\beta_N&=B+\frac R{\sqrt\lambda}\sqrt{\ln\det\Big(I+\frac{K_N+N\epsilon_kI}{\lambda}\Big)-2\ln\delta}.\end{aligned}$$ Statement. With probability at least $1-\delta$, for every $N\ge1$ and $x\in D$, $\ |f(x)-\mu_N(x)|\le\bar\beta_N\sqrt{\sigma_N^2(x)+S_N^2(x)}+C_N(x)\|y_N\|_2$.
In words. The kernel error moves the fitted mean (the shift $C_N\|y_N\|_2$) and the uncertainty (the inflation $S_N$), so both need margins; for $\epsilon_k=0$ this is the Abbasi-Yadkori band of Section 6. Fiedler et al.'s Theorem 5 has the same structure with their own, different expressions for $S_N$, $C_N$ and $\bar\beta_N$; the cruder ones here are fully derived below.
Proof — perturbing the posterior of the true kernel

Let $\tilde\mu_N,\tilde\sigma_N$ be the posterior computed with the true kernel $\tilde k$, and $A=K_N+\lambda I$, $\tilde A=\tilde K_N+\lambda I$. The Abbasi-Yadkori band of Section 6 for $\tilde k$ gives, with probability at least $1-\delta$ for all $N$ and $x$, $|f(x)-\tilde\mu_N(x)|\le\tilde\beta_N\tilde\sigma_N(x)$, where $\tilde\beta_N$ contains $\ln\det(I+\tilde K_N/\lambda)$; since $\tilde K_N\preceq K_N+\|\tilde K_N-K_N\|_2I\preceq K_N+N\epsilon_kI$, $\tilde\beta_N\le\bar\beta_N$. Entrywise errors give $\|\tilde K_N-K_N\|_2\le N\epsilon_k$ and $\|\tilde k_N(x)-k_N(x)\|_2\le u_N$; both inverses have norm at most $1/\lambda$, and $\tilde A^{-1}-A^{-1}=\tilde A^{-1}(A-\tilde A)A^{-1}$ has norm at most $w_N$. Hence $\|\tilde A^{-1}\tilde k_N-A^{-1}k_N\|_2\le\|\tilde A^{-1}(\tilde k_N-k_N)\|_2+\|(\tilde A^{-1}-A^{-1})k_N\|_2\le C_N(x)$, so $|\tilde\mu_N(x)-\mu_N(x)|\le C_N(x)\|y_N\|_2$. For the variances, $\tilde\sigma_N^2-\sigma_N^2=[\tilde k(x,x)-k(x,x)]-(\tilde k_N-k_N)^\top\tilde A^{-1}(\tilde k_N+k_N)-k_N^\top(\tilde A^{-1}-A^{-1})k_N\le S_N^2(x)$, so $\tilde\sigma_N\le\sqrt{\sigma_N^2+S_N^2}$. The triangle inequality $|f-\mu_N|\le|f-\tilde\mu_N|+|\tilde\mu_N-\mu_N|$ finishes the proof. $\square$

Fact — When the growing class of A-GP-UCB contains $f$
Take the isotropic schedule $\ell_t=\ell_0/g(t)$, $B_t=b(t)g(t)^dB_0$ with non-decreasing $g,b\ge1$ tending to infinity, and let $f$ lie in the SE RKHS with length-scale $\ell_*$ and norm at most $B_*$ (same amplitude and dimension). Once $\ell_t\le\ell_*$, the inclusion of Section 1 gives $\|f\|_{k_{\ell_t}}\le(\ell_*/\ell_t)^{d/2}B_*=(\ell_*g(t)/\ell_0)^{d/2}B_*$, which is at most $B_t$ as soon as $b(t)g(t)^{d/2}B_0\ge(\ell_*/\ell_0)^{d/2}B_*$. Both conditions hold for all large $t$, so the working class eventually contains $f$ with a valid norm bound. The crossing time depends on the unknown $\ell_*$ and $B_*$, so nothing is certified before it: enough for no-regret, not for safety from the first query.

Misspecified function class: $f$ only close to the RKHS

Bogunovic & Krause (NeurIPS 2021) formalise $\min_{f\in\mathcal F_k(D;B)}\|f-f^*\|_\infty\le\epsilon$, where $\mathcal F_k(D;B)=\{g\in\mathcal H_k:\|g\|_k\le B\}$ is the norm ball and $\|g\|_\infty=\sup_{x\in D}|g(x)|$: the observed $f^*$ differs from some bounded-norm $\tilde f$ by an unknown $m(x)=f^*(x)-\tilde f(x)\in[-\epsilon,\epsilon]$.

Lemma — Bias of the posterior mean under $\epsilon$-misspecification (Bogunovic & Krause 2021, Lemma 2)
Let $\mu_t^*$ be the posterior mean computed from the actual observations $y_i^*=f^*(x_i)+\varepsilon_i$ and $\mu_t$ the one computed from the hypothetical (unobservable) noisy observations $y_i=\tilde f(x_i)+\varepsilon_i=y_i^*-m(x_i)$ of $\tilde f$: same inputs, same noise realisation $\varepsilon_i$, same $\lambda\gt0$, so that $y^*_{1:t}-y_{1:t}=m_{1:t}$ (noise-free values of $\tilde f$ would leave $\varepsilon_{1:t}$ in this difference, and the bound would fail even for $\epsilon=0$). Then for all $x$ and $t$: $\ |\mu_t(x)-\mu_t^*(x)|\le\dfrac{\epsilon\sqrt t}{\sqrt\lambda}\,\sigma_t(x)$.
Derivation — the misspecification term has the shape of the noise term

$\mu_t^*(x)-\mu_t(x)=k_t(x)^\top(K_t+\lambda I)^{-1}m_{1:t}$ with $\|m_{1:t}\|_\infty\le\epsilon$, so $\|m_{1:t}\|_2\le\epsilon\sqrt t$. Exactly as in Step 3 of the walkthrough, $$|k_t(x)^\top(K_t+\lambda I)^{-1}m_{1:t}|=|\langle V_t^{-1/2}\varphi(x),V_t^{-1/2}\Phi_t^*m_{1:t}\rangle_k|\le\frac{\sigma_t(x)}{\sqrt\lambda}\sqrt{m_{1:t}^\top K_t(K_t+\lambda I)^{-1}m_{1:t}}\le\frac{\sigma_t(x)}{\sqrt\lambda}\|m_{1:t}\|_2,$$ using $K_t(K_t+\lambda I)^{-1}\preceq I$ (its eigenvalues $a_j/(a_j+\lambda)$, with $a_j\ge0$ the eigenvalues of $K_t$, lie in $[0,1)$). $\square$ The difference from the noise term: $m$ is not a martingale difference, nothing averages out, and the bound grows like $\sqrt t$ instead of $\sqrt{\log\det}$.

Consequently, on the event where the well-specified bound $|\tilde f-\mu_{t-1}|\le\beta_t\sigma_{t-1}$ holds (with $\beta_t$ computed for $\|\tilde f\|_k\le B$), the triangle inequality gives

$$|f^*(x)-\mu_{t-1}^*(x)|\le\underbrace{|f^*(x)-\tilde f(x)|}_{\le\epsilon}+\underbrace{|\tilde f(x)-\mu_{t-1}(x)|}_{\le\beta_t\sigma_{t-1}(x)}+\underbrace{|\mu_{t-1}(x)-\mu_{t-1}^*(x)|}_{\le\epsilon\sqrt{t}\,\sigma_{t-1}(x)/\sqrt\lambda}\le\epsilon+\Big(\beta_t+\frac{\epsilon\sqrt t}{\sqrt\lambda}\Big)\sigma_{t-1}(x):$$

(the lemma with $t-1$ observations gives the factor $\epsilon\sqrt{t-1}/\sqrt\lambda$; the display uses the slightly larger $\epsilon\sqrt t/\sqrt\lambda$): the enlarged width of their EC-GP-UCB rule (eq. 15) plus the pointwise misfit $\epsilon$, which matters for safety because the constraint is on $f^*$, not on $\tilde f$. In the regret this costs an extra $O(\epsilon T\sqrt{\gamma_T})$ term in the proved upper bound (Theorem 1), and they show that an $\Omega(\epsilon T)$ dependence is unavoidable for any algorithm (Appendix C). For safety the point is that the enlargement does not vanish with data: the band never gets narrower than $\epsilon$ and its multiplier $\epsilon\sqrt t/\sqrt\lambda$ grows with $t$, so a safe set built from the unenlarged band is over-confident by an amount that more data cannot remove. Whether $\epsilon$ is known is the same problem as whether $B$ is known.

Noise that is not sub-Gaussian

All martingale bounds above need $\mathbb E[e^{\alpha\varepsilon_t}\mid F_{t-1}]\le e^{\alpha^2R^2/2}$. A common $R$ already covers heteroscedastic noise (noise whose scale depends on the input or the history; see Primer C) as long as every conditional law is $R$-sub-Gaussian (at the price of using the worst noise level everywhere). Polynomial-tailed disturbances without finite exponential moments violate it (occasional large observations alone do not show this: Gaussian noise produces outliers too), and then nothing above applies; then either bounded noise (deterministic statements, Module 5) or estimators with their own concentration results are needed. Fiedler et al. 2021 stress that their construction is flexible in the underlying concentration result (their Proposition 2 already swaps in a $\chi^2$-type bound), so the architecture survives whenever a suitable inequality exists for the noise class at hand. A scenario-based route is taken by Tokmak, Schön & Baumann (IEEE Control Systems Letters, accepted; arXiv 2025): they bound the measurement noise itself with high probability via the scenario approach, for noise models up to heteroscedastic heavy-tailed ones, insert that bound into the confidence intervals, and prove safety and optimality of the resulting safe BO algorithm.

Theorem — Scenario bound on the noise, scalar case (Tokmak, Schön & Baumann 2025, Lemma 2 and Theorem 1)
Assumptions. One can draw i.i.d. samples $\varepsilon^{(1)},\dots,\varepsilon^{(m)}$ from the noise distribution (their Assumption 1), independent of the noise value $\varepsilon$ that actually occurs; $\nu,\kappa\in(0,1)$ with $(1-\nu)^m\le\kappa$. Let $E=\max_{j\le m}|\varepsilon^{(j)}|$.
Statement. With probability at least $1-\kappa$ over the samples, $\Pr_\varepsilon(|\varepsilon|\gt E)\le\nu$. Repeating this at every round $t$ with $\kappa_t=6\kappa/(\pi^2t^2)$ makes it hold for all rounds simultaneously with probability at least $1-\kappa$ (their Theorem 1, since $\sum_t\kappa_t=\kappa$).
In words. No moment or tail assumption is needed, only samples from the right distribution (for heteroscedastic noise, the one at the current input). The observed maximum is an estimated upper $\nu$-quantile, not a hard bound: each round's noise may still exceed its $E_t$ with probability up to $\nu$, so a guarantee that no noise value exceeds its bound over $H$ rounds costs an extra $H\nu$ (or summable $\nu_t$).
Connection to Module 5 — the Trimpe-group conclusion
Put together: the only computable rigorous $\beta_t$ needs $B$, which is unknowable; adapting hyperparameters or under-estimating $B$ voids the guarantee; and practitioners replaced $\beta_t$ by 2, 3 or 4 anyway. Fiedler et al. (TMLR 2024) draw the consequence in two directions: Real-$\beta$-SafeOpt uses the honest bound (Section 6), which in the numerical experiments of Fiedler et al. 2021 was often not much larger than common heuristics (a performance comparison with tuned heuristics is, as the TMLR paper notes in its Section 5.2, not meaningful, because the two address different problems); LoSBO drops the GP from the safety argument entirely and certifies safety from a Lipschitz constant and a noise bound alone (Module 5), keeping the GP only to decide where to sample. The GP bounds of this page reappear in Module 11 as the uncertainty tubes fed into robust synthesis (Fiedler, Scherer & Trimpe, CDC 2021) and in Module 6 as the reason GoSafeOpt's experiments carry a $\beta$ caveat (Module 6).

Open problems

Walkthrough: Deriving the RKHS Error Bound

The central derivation of this page: from the kernel-ridge representation to $|\mu_t(x)-f(x)|\le\beta_t\sigma_t(x)$, following the proof of Theorem 2 of Chowdhury & Gopalan with the regulariser $\lambda$ kept general, so that the data-dependent bound of Fiedler et al. drops out at Step 4 and the a-priori bound at Step 5.

Derivation — the inequality of Step 2, spelled out

With $u=(A+\lambda I)^{-1}\varphi(x)$, $A\succeq0$ gives $\lambda\|u\|_k^2\le\langle u,(A+\lambda I)u\rangle_k=\langle\varphi(x),(A+\lambda I)^{-1}\varphi(x)\rangle_k=\sigma_t^2(x)/\lambda$ by the variance identity of Section 2; multiplying by $\lambda$ gives $\|\lambda(A+\lambda I)^{-1}\varphi(x)\|_k^2\le\sigma_t^2(x)$. Steps 2 and 3 bound the absolute values of the bias and noise terms, and the triangle inequality adds the two bounds.

Interactive: GP Confidence Bands and $\beta$

A 1-D GP regression whose truth is an exact RKHS element $f=\sum_{i=1}^8c_i\,k_{\mathrm{SE}}(\cdot,z_i)$ with $\|f\|_k=\sqrt{c^\top K_zc}$ normalised to $3$ (so $|f|\le3$ by the reproducing property), so every assumption of the theorems can be met or broken on purpose.

The domain $D$ is the 201-point grid on $[0,1]$. The value $3$ is the norm of $f$ on $[0,1]$; the centres $z_i$ lie off the grid, and restricting $f$ to $D$ can only lower its norm (the minimum over all extensions, Section 1). Thus $B=3$ is a valid upper bound on the grid norm; equality is not needed for the certificate.

The $t$ inputs are drawn uniformly from $D$ (seeded and nested, so increasing $t$ adds points; repetitions are possible), with Gaussian noise of standard deviation $\sigma_n$, hence $R=\sigma_n$. The posterior is a Cholesky solve (see Primer A) of $K_t+\lambda I$.

The a-priori "Chowdhury–Gopalan type" rule is their proof with general $\lambda$, i.e. the corrected Fiedler et al. formula with the realised $\log\det(I+K_t/\lambda)$ replaced by $2\hat\gamma_t$, where $\hat\gamma_t=\tfrac{e}{e-1}I_{\mathrm{greedy}}\ge\gamma_t$ is a certified upper bound on this finite $D$ (the greedy design of Srinivas et al., which repeatedly observes the point of largest posterior variance, is within a factor $1-1/e$ of optimal by submodularity). The data-dependent rules use the realised $\log\det$. Red dots mark grid points where the true $f$ leaves the band.

The percentage of red points describes this one simulated data set; it is not an estimate of $\delta$, which bounds the probability, over repeated experiments, that the band fails anywhere at any time. Moving the length-scale slider rebuilds the true $f$ for the new $\ell$ (same centres and relative weights, norm renormalised to $3$), so it changes the experiment rather than refitting one fixed plant; each theorem concerns one fixed kernel and norm bound, not settings chosen after looking at the plot.

Going deeper — how the explorer computes the posterior

The implementation never forms an inverse. Factor $K_t+\lambda I=LL^\top$, solve $Lv=y_t$ and $L^\top\alpha=v$, and evaluate $\mu_t(x)=k_t(x)^\top\alpha$. If $Lw=k_t(x)$, then $\sigma_t^2(x)=k(x,x)-\|w\|_2^2$. Also $\log\det(I+K_t/\lambda)=2\sum_i\log L_{ii}-t\log\lambda$. Small numerical floors in the code prevent roundoff from producing negative variances; the display illustrates the exact-arithmetic theorem rather than providing a formally verified floating-point certificate.

Theorem — The greedy design certifies $\gamma_t$ on a finite domain (Srinivas et al. 2010, Section 5)
Setting. A finite ground set of possible observations: every point of $D$ in $t$ copies (so that repeats are allowed), each copy with its own independent $\mathcal N(0,\lambda)$ noise; $F(S)=\tfrac12\log\det(I+K_S/\lambda)$ is the information gain of observing the set $S$. $F(\emptyset)=0$, $F$ is monotone (more observations never lose information) and submodular, i.e. has diminishing returns, $F(A\cup\{v\})-F(A)\ge F(B\cup\{v\})-F(B)$ for $A\subseteq B$, $v\notin B$: the increment is $\tfrac12\log(1+\sigma_S^2(v)/\lambda)$ (Section 4), and conditioning on more data only shrinks the posterior variance $\sigma_S^2(v)$.
Statement. The greedy design, which repeatedly adds the observation with the largest increment, i.e. the largest posterior variance, reaches after $t$ steps a value $G_t^{\rm greedy}\ge(1-1/e)\max_{|S|=t}F(S)=(1-1/e)\gamma_t$ (the classical greedy guarantee of Nemhauser, Wolsey & Fisher 1978, used by Srinivas et al.). Hence $G_t^{\rm greedy}\le\gamma_t\le\tfrac{e}{e-1}G_t^{\rm greedy}=\hat\gamma_t$, a certified upper bound on this finite $D$, not on the whole interval.
Proof idea. With $O$ the optimal value and $G_j$ the greedy value after $j$ steps, submodularity gives $O\le G_j+t\cdot(\text{largest single increment})$, so the greedy step gains at least $(O-G_j)/t$. Hence $O-G_{j+1}\le(1-1/t)(O-G_j)$ and $O-G_t\le(1-1/t)^tO\le O/e$.
GP confidence bands and β

What to try. (i) Set $\sigma_n$ to $0$: with a valid norm bound for the working kernel, the correctly specified rules collapse to $\beta=B$, i.e. to the deterministic bias bound $B\sigma_t$ (Step 2 of the walkthrough; Section 3's bound when also $\lambda\to0$), and the band holds with certainty (the "B under-estimated" rule gives $(B/4)\sigma_t$, deliberately not certified). (ii) Lower $\lambda$ far below $\sigma_n^2$ (e.g. $0.001$): the posterior mean over-fits the noise and $\beta\equiv2$ is violated at many points, while the rigorous rules widen through $R/\sqrt\lambda$ and stay valid. (iii) Compare the a-priori and data-dependent rows of the table: the gap is $2\hat\gamma_t$ versus the realised $\log\det$ (last row). (iv) Select "B under-estimated ×0.25", raise $\lambda$ to about $1$ and keep $t\le20$: once the bias term $B\sigma_t$ dominates, the band is violated; at small $\lambda$ the noise term happens to cover the error, which is luck, not a guarantee. (v) Switch on the misspecification, set $\sigma_n=0$ and $t\ge50$: the norm bound for the original kernel does not certify the new kernel. The band may fail; whether it fails for a particular simulation is a separate numerical question (Section 7). (vi) Set $\lambda=0.001$ and compare the corrected and uncorrected AAAI rows: the uncorrected band is violated.

From the mathematics to a real decision

Learning objectives

A commissioning decision

A thermal controller uses a nominal model to predict the next enclosure temperature. A learned residual $f(z)$ corrects that prediction at sensor location $z$, where location is measured in metres. The residual and its observations have units degrees Celsius. The controller will accept a proposed heating action only when its nominal next temperature plus an upper residual bound is at most $60\,{}^\circ\mathrm C$.

Two measurements are available: $y=(2,0)^\top$ degrees at two locations. A fixed kernel gives the matrix $K=\left[\begin{smallmatrix}4&2\\2&4\end{smallmatrix}\right]$ in squared degrees, and the regularization parameter is $\lambda=1$ squared degree. At a midpoint $z_*$, the kernel vector is $k_*=(3,3)^\top$ and the prior variance is $k(z_*,z_*)=4$. Assume the resulting three-point kernel matrix is valid and the measurement model and kernel have been fixed before observing these data.

For this decision calculation, explicitly assume an event on which $|f(z)-\mu(z)|\le2\sigma(z)$ simultaneously at the candidate locations. The number 2 is supplied as part of that hypothetical event; it is not claimed to be a theorem-valid multiplier for arbitrary kernels, functions, or adaptive experiments. A real confidence statement would still require the relevant norm, noise, and simultaneous-coverage assumptions from this chapter.

Worked decision, with its limits

Solve once. Rather than form an inverse in software, solve $(K+\lambda I)\alpha=y$. Here the exact arithmetic is small enough to display:

$$K+\lambda I=\begin{bmatrix}5&2\\2&5\end{bmatrix},\qquad \alpha=\begin{bmatrix}10/21\\-4/21\end{bmatrix}.$$

Predict the correction. At the midpoint, $\mu_*=k_*^\top\alpha=6/7\approx0.857143$ degrees. Solve $(K+\lambda I)v=k_*$, obtaining $v=(3/7,3/7)^\top$. Consequently the posterior variance is $\sigma_*^2=4-k_*^\top v=10/7$ squared degrees and the standard deviation is $\sqrt{10/7}\approx1.195229$ degrees.

Build the decision margin. The upper residual envelope is $b_* =6/7+2\sqrt{10/7}\approx3.247600$ degrees. A proposed action whose nominal next temperature is 57 degrees is rejected: its certified upper temperature would be $60.247600$. Reducing the nominal prediction to 56.5 gives the upper temperature $59.747600$, so that revised action passes on the assumed confidence event.

Interpret the remaining uncertainty. The positive correction mean is not the whole correction. Substituting only the mean would accept 57 degrees and report $57.857143$, silently discarding uncertainty. Nor does the upper bound describe a noisy future sensor reading: the displayed posterior variance concerns the latent function, and a prediction interval for a new observation would include its observation noise as well.

The algebra certifies a conditional implication: if the error envelope holds and the nominal update plus residual really describes the plant, the revised one-step temperature stays below the limit. It does not prove an all-time closed-loop guarantee. Such a guarantee additionally needs a confidence event covering future uses and a controller that preserves the conditions required for the next decision.

A tempting wrong approach

Common mistake — Squared uncertainty in a temperature sum

Adding $2\sigma_*^2=20/7$ to the mean mixes squared degrees with degrees. It also gives a different decision. The uncertainty term in this envelope is the multiplier times the standard deviation. A second error is choosing a more convenient kernel after seeing these two readings and reusing a fixed-kernel confidence theorem without accounting for that selection.

Transfer the argument

Exercise 3.B1 — Medium: Repeat a measurement before heating

At one location, use prior variance 4, regularization 1, and three identical readings of 2 degrees. With the same explicitly assumed multiplier 2, compute the latent posterior and decide whether nominal temperature 57 degrees passes.

Review: Posterior mean and variance.

Show hint

The three-by-three kernel matrix has every entry equal to 4. Solve along its all-ones direction.

Show worked solution

The posterior mean is $\mu=4(2+2+2)/(1+3\cdot4)=24/13\approx1.846154$. Its variance is $4/(1+3\cdot4)=4/13$, so the upper residual is $24/13+2\sqrt{4/13}\approx2.955554$. The total $59.955554$ passes. Repetition reduces observation uncertainty under the assumed model; the posterior mean approaches 2 rather than zero. This is a different dataset from the midpoint calculation and should not be mixed into it without recomputing the joint posterior.

Exercise 3.B2 — Hard: A calibration discrepancy survives the posterior

Return to the two-location midpoint. The actual physical residual is $f(z)+b(z)$, where the modeled $f$ obeys the assumed envelope but the unknown discrepancy obeys only $|b(z)|\le0.4$ degrees. Does nominal temperature 56.5 still pass? Find its largest permitted replacement.

Review: Model misspecification.

Show hint

Bound the sum of two unknown terms separately. The discrepancy need not be zero-mean noise.

Show worked solution

The physical residual upper bound is $3.247600+0.4=3.647600$ degrees. The old proposal gives $60.147600$ and fails. Its largest permitted nominal prediction is $60-3.647600\approx56.352400$ degrees. More repeated observations do not by themselves eliminate a discrepancy excluded from the statistical model. Treating this bounded, possibly systematic term as independent measurement noise would change the assumptions and requires a fresh justification.

Synthesis and bridge

The useful output of regression is an envelope with an explicit scope, not a visually narrow curve. Units identify several mistakes immediately, while the confidence-event sentence identifies the deeper ones. Kernel choice, noise assumptions, and unmodeled discrepancy determine whether a number computed from a posterior can support a physical decision.

Safe Bayesian optimization will use these envelopes to choose the next experiment. Its extra challenge is that the learner controls where data arrive: a candidate must be certified before it can be measured. This makes the safe seed and the geometry of transferring bounds as important as the regression calculation.

Exercises

Graded practice — Build the argument yourself

There are 12 new problems: four Easy, four Medium, and four Hard. Start with the level that lets you make progress without opening the solution. Easy problems rebuild individual operations; Medium problems combine them; Hard problems ask for proofs, boundary cases, and the limits of a guarantee. The required concepts are explained on this page or in the linked earlier material.

Easy — Warm up one skill at a time.

Exercise 3.P1 — Easy: A two-point kernel matrix

A kernel has $k(a,a)=k(b,b)=1$ and $k(a,b)=k(b,a)=1/2$. Write the Gram matrix and find its eigenvalues. For $f=k(a,\cdot)-k(b,\cdot)$, compute $f(a)$, $f(b)$ and $\|f\|_k^2$.

Review if needed: Primer A: eigenvalues and PSD matrices. Apply here: this module's explanation.

Show hint

Use coefficient vector $c=(1,-1)^\top$ and $\|f\|_k^2=c^\top Kc$.

Show worked solution

The Gram matrix is $K=\begin{bmatrix}1&1/2\\1/2&1\end{bmatrix}$. Its eigenvalues are $3/2$ on $(1,1)^\top$ and $1/2$ on $(1,-1)^\top$, so it is positive definite.

Evaluation gives $f(a)=1-1/2=1/2$ and $f(b)=1/2-1=-1/2$. The squared norm is $c^\top Kc=1-1/2-1/2+1=1$, hence $\|f\|_k=1$. The coefficient norm squared is 2; it differs from the RKHS norm squared because the two kernel sections are not orthogonal.

Exercise 3.P2 — Easy: One observation, one posterior

Use a zero-mean GP with $k(a,a)=k(b,b)=1$, $k(a,b)=1/2$, nominal noise variance $\lambda=1/4$, and one observation $y=2$ at $a$. Compute the posterior means and latent variances at $a$ and $b$.

Review if needed: Primer C: Gaussian conditioning. Apply here: this module's explanation.

Show hint

The matrix inverse is just the scalar $1/(1+\lambda)=4/5$.

Show worked solution

At $a$, $\mu(a)=1(4/5)2=8/5=1.6$ and $\sigma^2(a)=1-1(4/5)1=1/5=0.2$. At $b$, $\mu(b)=(1/2)(4/5)2=4/5=0.8$ and $\sigma^2(b)=1-(1/2)(4/5)(1/2)=4/5=0.8$.

The observed point remains uncertain because the observation is noisy. Correlation transfers some information to $b$, but less than to $a$. These are variances of the latent function value; under the correctly specified likelihood, a fresh noisy observation has an additional variance $1/4$.

Exercise 3.P3 — Easy: Standard deviation is not variance

At one input, a posterior has mean 1, latent variance $0.04$, and half-width multiplier $\beta=3$. Compute the band $\mu\pm\beta\sigma$. Does it certify the threshold $f(x)\ge0.5$ on the valid-band event?

Review if needed: Primer C: variance and standard deviation. Apply here: this module's explanation.

Show hint

Take the square root before applying the multiplier.

Show worked solution

The standard deviation is $\sigma=\sqrt{0.04}=0.2$. The half-width is $3(0.2)=0.6$, giving the interval $[0.4,1.6]$. Its lower endpoint is below 0.5, so it does not certify safety at threshold 0.5.

Using the variance as though it were the standard deviation would give the incorrect narrower band $[0.88,1.12]$ and a false certificate. The true value might still exceed 0.5; “uncertified” means that the available lower bound is insufficient, not that the input is known unsafe.

Exercise 3.P4 — Easy: Translate the square-root convention

Paper A writes its band as $\mu\pm\sqrt{b_t}\sigma$, while this module writes $\mu\pm\beta_t\sigma$. If $b_t=9$, what is this module's multiplier? If Paper A bounds $\|f\|_k^2\le16$, what norm bound $B$ should be used here?

Review if needed: Primer 0: functions and notation. Apply here: this module's explanation.

Show hint

Match the actual coefficient multiplying $\sigma$, and take the nonnegative square root of the squared norm bound.

Show worked solution

The half-widths agree when $\beta_t=\sqrt{b_t}$, so $b_t=9$ translates to $\beta_t=3$. The squared norm bound $\|f\|_k^2\le16$ translates to $\|f\|_k\le4$, hence $B=4$ in this module.

These are two separate conversions: one concerns a confidence multiplier, the other a function-class assumption. Reusing the symbol $\beta$ or $B$ from another paper without checking its defining formula can inflate a constant, or dangerously shrink it. Match equations before comparing numbers.

Medium — Combine definitions and compute a certificate.

Exercise 3.P5 — Medium: Noise-free uncertainty at an unseen point

One exact observation is available at $a$, with $k(a,a)=k(b,b)=1$ and $k(a,b)=0.6$. Compute the power function $P_1(b)$ and the guaranteed interpolation error when $\|f\|_k\le2$. What happens at $a$?

Review if needed: Primer A: projection and norm bounds. Apply here: this module's explanation.

Show hint

Use $P_1(b)^2=k(b,b)-k(a,b)^2/k(a,a)$.

Show worked solution

The power-function squared is $1-0.6^2=0.64$, so $P_1(b)=0.8$. The deterministic error bound is $|f(b)-s_1(b)|\le2(0.8)=1.6$. The interpolant is $s_1(b)=0.6f(a)$, so the unknown value lies within 1.6 of that prediction.

At $a$, $P_1(a)^2=1-1=0$ and interpolation is exact. The norm assumption controls the part of the function not determined by the exact data. No failure probability is involved in this noise-free bound; its uncertainty is a worst-case function-class statement.

Exercise 3.P6 — Medium: Repeated noisy measurements still help

At a single input $a$ with prior variance 1, take $n$ independent observations with nominal noise variance $\lambda=1$. Show that the posterior latent variance is $1/(n+1)$. Compare $n=1$ and $n=3$, and distinguish this from a noiseless repeated measurement.

Review if needed: Primer A: matrix solves. Apply here: this module's explanation.

Show hint

The Gram matrix is $\mathbf1\mathbf1^\top$; solve $(I+\mathbf1\mathbf1^\top)v=\mathbf1$.

Show worked solution

The vector $v=\mathbf1/(n+1)$ solves that system, because $\mathbf1\mathbf1^\top\mathbf1=n\mathbf1$. Hence $\sigma_n^2(a)=1-\mathbf1^\top v=1-n/(n+1)=1/(n+1)$.

For $n=1$ the variance is $1/2$, and for $n=3$ it is $1/4$. Independent noisy repeats reduce uncertainty, with diminishing gains. Repeating an exact noiseless observation adds no new information once the value is known; a direct noiseless inverse formula would also need care because repeated inputs make the Gram matrix singular.

Exercise 3.P7 — Medium: Information gain from eigenvalues

A two-observation Gram matrix has eigenvalues $3/2$ and $1/2$, and $\lambda=1$. Compute the realized information gain $I=\tfrac12\log\det(I_2+K)$ in nats. Is this automatically the maximum information gain $\gamma_2$?

Review if needed: Primer A: determinants and log determinants. Apply here: this module's explanation.

Show hint

The eigenvalues of $I_2+K$ are $1$ plus the eigenvalues of $K$.

Show worked solution

The determinant is $(1+3/2)(1+1/2)=(5/2)(3/2)=15/4$. Thus $I=\tfrac12\log(15/4)\approx0.660878$ nats, where the logarithm is natural.

This is the information gain of this particular two-point design. The maximum $\gamma_2$ optimizes over all allowed two-query designs, including repeats in the convention used here, so $I\le\gamma_2$. Equality needs evidence that this design is optimal. A realized log determinant and a worst-case maximum serve different roles in confidence bounds.

Exercise 3.P8 — Medium: Evaluate a data-dependent confidence multiplier

Use the valid-bound formula $\beta=B+(R/\sqrt\lambda)\sqrt{\log\det(I+K/\lambda)+2\log(1/\delta)}$. Take $B=2$, $R=0.1$, $\lambda=0.04$, $\delta=0.05$, and realized log determinant 3. Compute $\beta$ and the half-width at $\sigma(x)=0.2$.

Review if needed: Primer C: confidence levels. Apply here: this module's explanation.

Show hint

First compute $R/\sqrt\lambda=1/2$; then keep the logarithms inside the square root.

Show worked solution

The multiplier is $2+0.5\sqrt{3+2\log20}\approx3.4992885$, so the half-width is $0.2\beta\approx0.6998577$. The norm contribution alone is $B\sigma=0.4$, and the remainder comes from the noise/concentration term.

The arithmetic does not establish the assumptions. This bound requires the fixed function to have RKHS norm at most $B$, the stated conditional sub-Gaussian noise bound, predictable query choices, and the matching regularizer. A plausible-looking interval produced from an invalid $B$ does not inherit the advertised probability.

Hard — Explain why the argument works and where it stops.

Exercise 3.P9 — Hard: Minimum norm required by two conflicting values

Let $k(a,a)=k(b,b)=1$ and $k(a,b)=r$ with $0\le r\lt1$. Among RKHS functions with $f(a)=1$ and $f(b)=-1$, derive the minimum squared norm and evaluate it at $r=0.5$ and $r=0.99$.

Review if needed: Primer A: Cauchy–Schwarz. Apply here: this module's explanation.

Show hint

Use the minimum-norm interpolant with $y=(1,-1)^\top$, or Cauchy–Schwarz applied to $k(a,\cdot)-k(b,\cdot)$.

Show worked solution

The Gram matrix satisfies $Ky=(1-r)y$, so the interpolation coefficient is $K^{-1}y=y/(1-r)$. Its squared norm is $y^\top K^{-1}y=2/(1-r)$. Projection onto the data span preserves the two values and can only decrease the norm, so this interpolant is the minimum.

At $r=0.5$, the squared norm is 4 and the norm is 2. At $r=0.99$, the squared norm is 200 and the norm is $\sqrt{200}\approx14.1421$. Highly correlated inputs are expected to have similar values; forcing opposite values needs a much larger norm. Independently, $2=|f(a)-f(b)|\le\|f\|_k\sqrt{2(1-r)}$ yields the same lower bound.

Exercise 3.P10 — Hard: A misspecification floor remains

Suppose $|f^*(x)-\tilde f(x)|\le\epsilon$, $|\tilde f(x)-\mu(x)|\le\beta\sigma(x)$, and $|\mu(x)-\mu^*(x)|\le\epsilon\sqrt{n/\lambda}\,\sigma(x)$. Combine these bounds. Evaluate at $\epsilon=0.05$, $n=16$, $\lambda=0.25$, $\beta=2$, $\sigma(x)=0.1$.

Review if needed: Primer A: triangle inequality. Apply here: this module's explanation.

Show hint

Insert $\tilde f$ and $\mu$ between $f^*$ and $\mu^*$, then apply the triangle inequality.

Show worked solution
$$|f^*(x)-\mu^*(x)|\le\epsilon+\left(\beta+\epsilon\sqrt{n/\lambda}\right)\sigma(x).$$

Here $\sqrt{16/0.25}=8$, so the error bound is $0.05+(2+0.05\cdot8)(0.1)=0.29$. Keeping only the original $\beta\sigma=0.2$ would omit both the direct model mismatch and its effect on the fitted mean.

For fixed $n$, even when $\sigma(x)$ tends to zero this certificate retains the floor $\epsilon=0.05$. More measurements alone do not justify pretending that the unknown function belongs to the smaller assumed class. The three inequalities are assumptions of this calculation; their validity must be established by the corresponding model-mismatch argument.

Exercise 3.P11 — Hard: Spend a confidence budget over infinitely many rounds

For round $t\ge1$, suppose failure has probability at most $\delta_t=\delta/[t(t+1)]$. Prove that failure at any round has probability at most $\delta$. Contrast this with using $\delta_t=0.01$ forever.

Review if needed: Primer 0: telescoping series. Apply here: this module's explanation.

Show hint

Use $1/[t(t+1)]=1/t-1/(t+1)$, then take a limit of finite unions.

Show worked solution

The first $T$ budgets sum to $\delta\sum_{t=1}^T(1/t-1/(t+1))=\delta(1-1/(T+1))\le\delta$. The union bound therefore controls failure within the first $T$ rounds by that number. As $T\to\infty$, the finite failure unions increase to “failure at some round,” and continuity of probability gives the bound $\delta$.

A constant budget 0.01 sums to $0.01T$, which becomes uninformative once it reaches 1. Separate 99% marginal statements are therefore insufficient for a 99% infinite-run guarantee. Genuine confidence-sequence theorems establish a simultaneous event directly rather than requiring independent pointwise statements.

Exercise 3.P12 — Hard: Verify a posterior without forming an inverse

Take $K=\begin{bmatrix}1&1/2\\1/2&1\end{bmatrix}$, $\lambda=1/2$, observations $y=(1,-1)^\top$, and query vector $k_x=(1/2,0)^\top$ with $k(x,x)=1$. Solve $(K+\lambda I)\alpha=y$ and $(K+\lambda I)v=k_x$. Use the solves to obtain the posterior mean and variance.

Review if needed: Primer A: linear systems and block algebra. Apply here: this module's explanation.

Show hint

The first solve uses an eigenvector. For the second, multiply the two scalar equations by 2.

Show worked solution

The matrix is $M=\begin{bmatrix}3/2&1/2\\1/2&3/2\end{bmatrix}$. Since $My=y$, $\alpha=(1,-1)^\top$. The second solve is $3v_1+v_2=1$, $v_1+3v_2=0$, giving $v=(3/8,-1/8)^\top$.

Therefore $\mu(x)=k_x^\top\alpha=1/2$ and $\sigma^2(x)=1-k_x^\top v=1-3/16=13/16=0.8125$. The query covariance is consistent: the full latent covariance has positive Schur complement $1-k_x^\top K^{-1}k_x=2/3$. Numerical GP implementations use factorization and solves for these quantities; a displayed inverse is a mathematical formula, not a requirement to compute an explicit inverse.

Further practice — Original problems and research connections

The original exercises below retain their numbering. Some compare later methods or ask for longer research derivations; use the graded set above first, and return to a research-connection problem after reading the relevant linked module.

Exercise 3.1 — RKHS continuity, Lipschitz constants and sharpness

(a) Prove $|f(x)-f(x')|\le\|f\|_k\,d_k(x,x')$ with $d_k^2=k(x,x)-2k(x,x')+k(x',x')$ from the reproducing property. (b) For the SE kernel with length-scale $\ell$ show that every $f$ with $\|f\|_k\le B$ is $B/\ell$-Lipschitz. (c) For which $f$ is (a) an equality?

Show answer

(a) By the reproducing property $f(x)-f(x')=\langle f,\varphi(x)\rangle_k-\langle f,\varphi(x')\rangle_k=\langle f,\varphi(x)-\varphi(x')\rangle_k$. Cauchy–Schwarz gives $|f(x)-f(x')|\le\|f\|_k\|\varphi(x)-\varphi(x')\|_k$, and $\|\varphi(x)-\varphi(x')\|_k^2=\langle\varphi(x),\varphi(x)\rangle_k-2\langle\varphi(x),\varphi(x')\rangle_k+\langle\varphi(x'),\varphi(x')\rangle_k=k(x,x)-2k(x,x')+k(x',x')$.

(b) For the SE kernel $k(x,x)=1$ and $d_k^2=2\big(1-e^{-r^2/(2\ell^2)}\big)$ with $r=\|x-x'\|$. Since $1-e^{-a}\le a$ for $a\ge0$, $d_k^2\le r^2/\ell^2$, hence $|f(x)-f(x')|\le(\|f\|_k/\ell)\,r\le(B/\ell)\,r$. (Note the constant depends only on $B$ and $\ell$, never on the data, and that $d_k\le\sqrt2$ always: for far-apart points the RKHS bound is better than the Lipschitz bound.)

(c) If $\varphi(x)=\varphi(x')$, both sides are zero for every $f$. Otherwise Cauchy–Schwarz is tight iff $f$ is a multiple of $\varphi(x)-\varphi(x')=k(x,\cdot)-k(x',\cdot)$; for that $f$ the two evaluations differ by exactly $\|f\|_kd_k(x,x')$. So no continuity argument can improve the lemma without extra assumptions.

Exercise 3.2 — The GP posterior via the Schur complement, and the rank-one update

(a) Starting from the joint Gaussian of $(y_t,f(x))$, derive $\mu_t(x)$ and $\sigma_t^2(x)$ with the block-inverse formula. (b) Show that conditioning on one more observation $(x_t,y_t)$ updates the posterior covariance as $k_t(x,x')=k_{t-1}(x,x')-\dfrac{k_{t-1}(x,x_t)\,k_{t-1}(x_t,x')}{k_{t-1}(x_t,x_t)+\lambda}$, and conclude that $\sigma_t^2(x)\le\sigma_{t-1}^2(x)$ for every $x$.

Show answer

(a) With $A=K_t+\lambda I$, $C=k_t(x)$, $B=k(x,x)$ and $S=B-C^\top A^{-1}C$, the derivation in Section 2 gives $f(x)\mid y_t\sim\mathcal N(C^\top A^{-1}y_t,\ S)$, i.e. $\mu_t(x)=k_t(x)^\top(K_t+\lambda I)^{-1}y_t$ and $\sigma_t^2(x)=k(x,x)-k_t(x)^\top(K_t+\lambda I)^{-1}k_t(x)$. The key facts used: the Schur complement of the $y$-block is the conditional covariance, and the cross term of the block inverse gives the conditional mean.

(b) The posterior after $t-1$ points is again a GP, with mean $\mu_{t-1}$ and covariance $k_{t-1}(\cdot,\cdot)$; use it as the prior for the single observation $y_t=f(x_t)+\varepsilon_t$. The joint covariance of $(y_t,f(x),f(x'))$ under this prior has entries $k_{t-1}(x_t,x_t)+\lambda$, $k_{t-1}(x,x_t)$, $k_{t-1}(x',x_t)$, and the $1\times1$ Schur complement gives $k_t(x,x')=k_{t-1}(x,x')-k_{t-1}(x,x_t)k_{t-1}(x_t,x')/(k_{t-1}(x_t,x_t)+\lambda)$. (Sequential conditioning equals joint conditioning for Gaussians.) Setting $x'=x$, the subtracted term is $k_{t-1}(x,x_t)^2/(\sigma_{t-1}^2(x_t)+\lambda)\ge0$, so $\sigma_t^2(x)\le\sigma_{t-1}^2(x)$: data never increase the posterior variance anywhere, whatever the observed values. This monotonicity is why the GP intervals $Q_t=[\mu_t\pm\beta\sigma_t]$ narrow as data arrive, apart from the slow growth of $\beta_t$; SafeOpt's $C_t=C_{t-1}\cap Q_t$ is nested by construction, so that growth cannot widen it. The same rank-one update is what the explorer uses for the greedy information-gain bound.

Exercise 3.3 — Kernel ridge regression is the GP mean; the two limits of $\lambda$

(a) Show that the minimiser of $\sum_i(y_i-f(x_i))^2+\lambda\|f\|_k^2$ over $\mathcal H_k$ is $\mu_t$. (b) Show that as $\lambda\to0$ (with $K_t\succ0$) it converges to the minimum-norm interpolant of Section 3, and that as $\lambda\to\infty$, $\mu_t(x)=\lambda^{-1}k_t(x)^\top y_t+O(\lambda^{-2})$ while $\sigma_t^2(x)\to k(x,x)$.

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(a) By the representer theorem write $f=\sum_i\alpha_ik(x_i,\cdot)$; then $[f(x_i)]_i=K_t\alpha$ and $\|f\|_k^2=\alpha^\top K_t\alpha$, so the objective is $J(\alpha)=\|y_t-K_t\alpha\|^2+\lambda\alpha^\top K_t\alpha$. Expanding, $J=y_t^\top y_t-2\alpha^\top K_ty_t+\alpha^\top K_t(K_t+\lambda I)\alpha$; the stationarity condition $K_t\big((K_t+\lambda I)\alpha-y_t\big)=0$ is solved by $\alpha=(K_t+\lambda I)^{-1}y_t$, and $J$ is convex ($K_t(K_t+\lambda I)\succeq0$), so this is a minimiser. Evaluating, $f(x)=k_t(x)^\top\alpha=\mu_t(x)$.

(b) $\lambda\to0$: $(K_t+\lambda I)^{-1}\to K_t^{-1}$, so $\mu_t\to k_t(x)^\top K_t^{-1}y_t=s_t(x)$, the interpolant, which Section 3 showed is the norm-minimising interpolant; its squared norm is $y_t^\top K_t^{-1}y_t$. $\lambda\to\infty$: $(K_t+\lambda I)^{-1}=\lambda^{-1}(I+\lambda^{-1}K_t)^{-1}=\lambda^{-1}I-\lambda^{-2}K_t+O(\lambda^{-3})$, so $\mu_t(x)=\lambda^{-1}k_t(x)^\top y_t+O(\lambda^{-2})\to0$ (the prior mean) and $\sigma_t^2(x)=k(x,x)-O(\lambda^{-1})\to k(x,x)$ (the prior variance): heavy regularisation means "trust the data less", which is exactly why the Chowdhury–Gopalan choice $\lambda\approx1$ produces a small $\beta$ with a wide $\sigma_t$.

Background — Matrix inverse remainders

Diagonalise the fixed matrix $K_t=Q\operatorname{diag}(a_j)Q^\top$, $a_j\ge0$. For each eigenvalue, $1/(\lambda+a_j)=\lambda^{-1}-a_j\lambda^{-2}+a_j^2/[\lambda^2(\lambda+a_j)]$ (put the right side over the common denominator to check). Reassembling in the eigenbasis gives $(K_t+\lambda I)^{-1}=\lambda^{-1}I-\lambda^{-2}K_t+E$ with remainder $E=Q\operatorname{diag}\big(a_j^2/[\lambda^2(\lambda+a_j)]\big)Q^\top$ of spectral norm at most $\max_ja_j^2/\lambda^3$: this is the meaning of $O(\lambda^{-3})$, and it gives the stated limits of the mean and variance. Example: for $a=2$, $\lambda=10$ the first two terms give $0.1-0.02=0.08$, and the remainder $4/(100\cdot12)=0.00\overline{3}$ completes the exact $1/12$.

Exercise 3.4 — Information gain of the linear kernel

Let $d\ge1$ and $T\ge1$ be integers, and let $k(x,x')=x^\top x'$ on $D=\{x\in\mathbb R^d:\|x\|\le1\}$ with noise variance $\lambda$. Show that $\gamma_T\le\frac d2\log\big(1+\frac{T}{\lambda d}\big)=O(d\log T)$, exhibit a design that attains it when $d$ divides $T$, and show that it cannot be attained when $T\lt d$.

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Stack the design into $X\in\mathbb R^{T\times d}$; then $K_A=XX^\top$ and, by the Weinstein–Aronszajn identity $\det(I_T+\lambda^{-1}XX^\top)=\det(I_d+\lambda^{-1}X^\top X)$, $$I(y_A;f_A)=\tfrac12\log\det(I_d+\lambda^{-1}X^\top X)=\tfrac12\sum_{i=1}^d\log(1+\mu_i/\lambda),$$ where $\mu_1,\dots,\mu_d\ge0$ are the eigenvalues of $X^\top X$ with $\sum_i\mu_i=\operatorname{tr}(X^\top X)=\sum_t\|x_t\|^2\le T$. By concavity of $\log$ (Jensen), $\sum_i\log(1+\mu_i/\lambda)\le d\log\big(1+\frac{1}{d}\sum_i\mu_i/\lambda\big)\le d\log(1+T/(\lambda d))$. Maximising over designs gives $\gamma_T\le\frac d2\log(1+T/(\lambda d))$.

Equality needs all $\mu_i$ equal to $T/d$ and all $\|x_t\|=1$: cycling through the unit vectors $e_1,\dots,e_d$ gives $X^\top X=(T/d)I$ when $d$ divides $T$, so the bound is attained. (For other $T\ge d$ a unit-norm tight frame does the same: unit vectors with $\sum_tx_tx_t^\top=(T/d)I$, e.g. the three unit vectors at angles $0$, $2\pi/3$, $4\pi/3$ for $d=2$, $T=3$, whose outer products sum to $\tfrac32I_2$.) For $T\lt d$ it cannot be attained: $\operatorname{rank}(X^\top X)\le T\lt d$ forces $d-T$ eigenvalues to vanish, so they cannot all equal $T/d$. The maximum is then $\frac T2\log(1+1/\lambda)$: apply Jensen in the $T\times T$ form, where $XX^\top$ has $T$ eigenvalues with sum $\sum_t\|x_t\|^2\le T$, so $\log\det(I_T+XX^\top/\lambda)\le T\log(1+1/\lambda)$, with equality for orthonormal inputs ($XX^\top=I_T$), e.g. $\tfrac12\log2\approx0.347\lt\log1.5\approx0.405$ for $T=1$, $d=2$, $\lambda=1$. Interpretation: a $d$-parameter model can only ever reveal $O(d\log T)$ nats; the SE and Matérn rates of Section 4 replace $d$ by an effective dimension that grows slowly with $T$.

Exercise 3.5 — Translating Srinivas' $\beta_t$ into the Chowdhury–Gopalan convention

A paper states "$\|f\|_k\le2$" and computes SafeOpt's confidence intervals with the Srinivas–Sui formula. Write both bands in the form $\mu\pm c\,\sigma$ for $t=100$ (also the Chowdhury–Gopalan horizon $T$), $\gamma_{t}=10$ (for simplicity the same value for both theorems, although their $\gamma$ use different regularisers), $\delta=0.05$, noise bounded by $\sigma=0.1$ (so also $R=0.1$-sub-Gaussian). Then explain why the two $c$ are still not comparable.

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Srinivas / Sui. Their $B$ bounds the squared norm, so $B=2^2=4$ and $\beta_t=2\cdot4+300\cdot10\cdot\ln^3(100/0.05)$. With $\ln2000\approx7.60$, $\ln^32000\approx439$, so $\beta_t\approx8+1.32\times10^6$ and the band is $\mu\pm\beta_t^{1/2}\sigma$ with $c=\beta_t^{1/2}\approx1.15\times10^3$.

Chowdhury–Gopalan. Their $B$ bounds the norm, so $B=2$ and $c=\beta_t=2+0.1\sqrt{2(10+1+\ln20)}\approx2+0.1\sqrt{2\cdot13.996}\approx2+0.53=2.53$. The ratio of the two multipliers $c$ (not yet of the widths) is about $450$.

Why still not comparable. The two theorems require different posteriors: Srinivas' with $\lambda=\sigma^2=0.01$, Chowdhury–Gopalan's with $\lambda=1+2/T=1.02$. The $\sigma_{t-1}(x)$ multiplying $c$ is therefore a different function in the two cases. The posterior variance is non-decreasing in $\lambda$, so for $\lambda=1.02$ it is never smaller, and near the data, where the small $\lambda$ lets it collapse, it can be much larger (Exercise 3.3b); there the ratio of the actual widths $c\,\sigma_{t-1}(x)$ drops below $450$. Far from the data, where $k_{t-1}(x)\approx0$, both variances are $\approx k(x,x)$ whatever $\lambda$ is, and the width ratio is essentially the multiplier ratio. So the width ratio is at most about $450$; how much smaller it is depends on the inputs, the kernel and the query, and cannot be computed from the quantities given. A fair comparison fixes $\lambda$ and uses the data-dependent formula of Section 6 for both. Also note that Srinivas assumes bounded noise and Chowdhury–Gopalan only sub-Gaussian noise, so the second theorem applies to Gaussian noise and the first does not.

Derivation — why the posterior variance grows with $\lambda$

For fixed inputs and kernel, write $K_t=Q\operatorname{diag}(a_j)Q^\top$ and $b=Q^\top k_t(x)$. Then $\sigma_t^2(x)=k(x,x)-k_t(x)^\top Q\operatorname{diag}\big(1/(a_j+\lambda)\big)Q^\top k_t(x)=k(x,x)-\sum_jb_j^2/(a_j+\lambda)$, whose derivative in $\lambda$ is $\sum_jb_j^2/(a_j+\lambda)^2\ge0$: a larger $\lambda$ never gives a smaller posterior variance.

Exercise 3.6 — What the uncorrected AAAI 2021 constant does for $\lambda\lt1$

Compare $\beta^{\mathrm{orig}}_N=B+R\sqrt{\log\det(K_N+\bar\lambda I)-2\log\delta}$ with the corrected $\beta_N$ for $0\lt\lambda\lt1$. (a) Show $\beta^{\mathrm{orig}}_N\lt\beta_N$ whenever $R\gt0$, and evaluate both noise terms for $N=1$, $k(x_1,x_1)=1$, $\lambda=0.01$, $R=0.1$, $\delta=0.05$. (b) Use the rescaling $k\to ck$, $\lambda\to c\lambda$ to show that the original formula cannot be a valid bound for all $\lambda$.

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(a) For $\lambda\lt1$, $\bar\lambda=1$, so the corrected noise term is $\frac{R}{\sqrt\lambda}\sqrt{\log\det(I+K_N/\lambda)-2\log\delta}$ and the original is $R\sqrt{\log\det(I+K_N)-2\log\delta}$. Since $K_N/\lambda\succeq K_N$, $\log\det(I+K_N/\lambda)\ge\log\det(I+K_N)$, and $1/\sqrt\lambda\gt1$; both factors go the wrong way for the original, so $\beta^{\mathrm{orig}}_N\lt\beta_N$. Numbers: $-2\log0.05\approx5.99$ and $\ln101\approx4.62$; corrected: $\frac{0.1}{0.1}\sqrt{\ln101-2\ln0.05}\approx3.26$; original: $0.1\sqrt{\ln2-2\ln0.05}\approx0.26$. The printed constant under-states the noise contribution by a factor of about $12.6$ at a single data point, and by at least $1/\sqrt\lambda=10$ always.

(b) Replace $k$ by $ck$ and $\lambda$ by $c\lambda$ with $c\gt0$ (keep $c\lambda\le1$). The posterior mean is unchanged, $(ck_t)^\top(cK_t+c\lambda I)^{-1}y_t=k_t^\top(K_t+\lambda I)^{-1}y_t$; the posterior variance scales as $c\sigma_t^2$; and $\|f\|_{ck}=\|f\|_k/\sqrt c$, so $B\to B/\sqrt c$. The error $|\mu-f|$ is thus the same random variable for every $c$. The corrected formula scales as $\beta\to\beta/\sqrt c$ exactly, $\frac{B}{\sqrt c}+\frac{R}{\sqrt{c\lambda}}\sqrt{\log\det(I+cK/(c\lambda))-2\log\delta}=\frac1{\sqrt c}\beta_N$, so its band $\beta\sqrt c\,\sigma$ does not depend on $c$. The original gives $\frac{B}{\sqrt c}+R\sqrt{\log\det(I+cK)-2\log\delta}$, whose band $\beta\sqrt c\,\sigma$ shrinks to $B\sigma$ as $c\to0$. Exact scale equivariance is not necessary for validity (a conservative bound may change its width), so this alone proves nothing; a concrete failure does. Take $f=0$, $B=0$, one observation with $k(x_1,x_1)=1$ and Gaussian noise $\varepsilon_1\sim\mathcal N(0,R^2)$, $R\gt0$ (which is $R$-sub-Gaussian). After rescaling, $\mu(x_1)=\varepsilon_1/(1+\lambda)$ for every $c$, while $\sigma^2(x_1)=c\lambda/(1+\lambda)$, and the original formula claims the half-width $R\sqrt{\log(1+c)-2\log\delta}\,\sqrt{c\lambda/(1+\lambda)}\to0$ as $c\to0$. The probability that the band covers $f(x_1)=0$ therefore tends to $0$: about $0.017$ for $\lambda=0.5$, $R=1$, $\delta=0.05$, $c=10^{-4}$, instead of at least $0.95$. The corrected half-width here is $R\sqrt{\log(1+1/\lambda)-2\log\delta}/\sqrt{1+\lambda}\approx2.17$ for every $c$ (coverage $\approx0.999$). So the original formula cannot hold for all $\lambda$; the corrected one does. For $\lambda\ge1$ the two differ only by the factor $\sqrt\lambda$ on the noise term, the original being the larger, so it remains valid there (conservatively), and at $\lambda=1$ they coincide.

Derivation — the two coverage probabilities

At the observed point, $\mu=\varepsilon/(1+\lambda)$ with $\varepsilon\sim\mathcal N(0,R^2)$. For any proposed half-width $w$, coverage of the true value zero is $\Pr(|\mu|\le w)=2\Phi((1+\lambda)w/R)-1$. Substituting the original and corrected half-widths gives the two reported probabilities. Here $\Phi$ is the standard-normal CDF, not the design operator $\Phi_t$.

Key Papers

PaperVenueContributionWhy read it
Chowdhury & Gopalan, On Kernelized Multi-armed BanditsICML 2017RKHS self-normalised inequality (double mixture); Theorem 2 confidence bound $\beta_t=B+R\sqrt{2(\gamma_{t-1}+1+\ln(1/\delta))}$; IGP-UCB, GP-TS.The default $\beta_t$ of modern safe-BO analyses; the proof is the walkthrough.
Fiedler, Scherer & Trimpe, Practical and Rigorous Uncertainty Bounds for Gaussian Process RegressionAAAI 2021 (arXiv v2 corrected)Data-dependent, computable $\beta_N$ with $\log\det$ instead of $\gamma_t$; log-det-free bound for independent noise; robustness to kernel misspecification.The bound that makes RKHS guarantees usable in control; read the Corrections section.
Srinivas, Krause, Kakade & Seeger, Gaussian Process Optimization in the Bandit Setting: No Regret and Experimental DesignICML 2010 (IEEE TIT 2012)GP-UCB, maximum information gain $\gamma_T$, spectral bounds for linear/SE/Matérn kernels, first RKHS bound (Theorem 6).Origin of the $\beta_t/\gamma_T$ machinery and of the $\beta^{1/2}$, squared-norm conventions.
Abbasi-Yadkori, Pál & Szepesvári, Improved Algorithms for Linear Stochastic BanditsNeurIPS 2011Self-normalised martingale bound by the method of mixtures; anytime ridge-regression confidence ellipsoids.The engine behind every RKHS confidence bound.
Kanagawa, Hennig, Sejdinovic & Sriperumbudur, Gaussian Processes and Kernel Methods: A Review on Connections and EquivalencesarXiv 2018GP regression versus kernel ridge regression, posterior variance as a worst-case RKHS error, GP sample paths versus the RKHS.The rigorous dictionary behind Sections 1–3 and the Bayesian-versus-frequentist pitfall.
Fiedler, Menn, Kreisköther & Trimpe, On Safety in Safe Bayesian OptimizationTMLR 2024Shows heuristic $\beta$ voids safety guarantees; Real-$\beta$-SafeOpt with the Abbasi-Yadkori-form bound (eq. 7); identifies the RKHS-norm bound as the obstacle; LoSBO.The corrective for anyone implementing safe BO; bridge to Module 5.
Vakili, Khezeli & Picheny, On Information Gain and Regret Bounds in Gaussian Process BanditsAISTATS 2021Tight $\gamma_T$ bounds from eigendecay (Mercer kernel with uniformly bounded eigenfunctions): Matérn $O(T^{d/(2\nu+d)}\log^{2\nu/(2\nu+d)}T)$ for $\nu\gt\tfrac12$, SE $O(\log^{d+1}T)$.The rates plugged into every sample-complexity bound.
Whitehouse, Ramdas & Wu, On the Sublinear Regret of GP-UCBNeurIPS 2023Simplified Hilbert-space self-normalised bound with explicit regulariser $\rho$; sublinear GP-UCB regret for Matérn kernels with $\nu\gt\tfrac12$ in every dimension (bounded-eigenfunction assumption).Cleanest statement of the $\rho$-form bound; settles the Matérn question.
Lederer, Umlauft & Hirche, Uniform Error Bounds for Gaussian Process Regression with Application to Safe ControlNeurIPS 2019Bayesian uniform bound via covering plus Lipschitz continuity; probabilistic Lipschitz constants of GP samples.The standard Bayesian-setting bound in GP-based control, and its assumptions.
Capone, Lederer & Hirche, Gaussian Process Uniform Error Bounds with Unknown Hyperparameters for Safety-Critical ApplicationsICML 2022Robust Bayesian bound over a box of length-scales with an explicit inflation factor $\gamma^2=\prod\vartheta''_i/\vartheta'_i$.Hyperparameter misspecification is the main practical failure mode.
Berkenkamp, Schoellig & Krause, No-Regret Bayesian Optimization with Unknown HyperparametersJMLR 2019A-GP-UCB: shrink the length-scales and inflate the norm bound over time with monotone scaling functions; norm-inflation lemma, in the isotropic case $\|f\|_{k_{\ell'}}\le(\ell/\ell')^{d/2}\|f\|_{k_\ell}$ for $\ell'\le\ell$.The frequentist answer to "which kernel and which $B$?" on the regret side, and why a shorter length-scale needs a larger $B$.
Bogunovic & Krause, Misspecified Gaussian Process Bandit OptimizationNeurIPS 2021$\epsilon$-misspecified kernel bandits; enlarged confidence $\beta_t+\epsilon\sqrt t/\sqrt\lambda$; unavoidable $\epsilon T$ regret term.Quantifies how much a wrong function class inflates the band.
Tokmak, Krishnan, Schön & Baumann, Safe exploration in reproducing kernel Hilbert spacesAISTATS 2025Data-driven PAC over-estimate of $\|f\|_k$ by a scenario approach; $d_k$-based safe-set expansion.Attacks the unknown-$B$ problem directly.
Flynn & Reeb, Tighter Confidence Bounds for Sequential Kernel RegressionAISTATS 2025Two-ellipsoid confidence sequences; exact UCB as an SOCP with a one-dimensional dual; tighter than Abbasi-Yadkori and Chowdhury–Gopalan when the mixture scale and regulariser are matched to theirs (Theorem 6.1).The current tightest practical RKHS bounds.
Sui, Gotovos, Burdick & Krause, Safe Exploration for Optimization with Gaussian ProcessesICML 2015SafeOpt; its Theorem 1 is stated with the Srinivas-type $\beta_t=2B+300\gamma_t\log^3(t/\delta)$ and $\|f\|_k^2\le B$.Where the convention traps enter safe BO (Module 4).

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