Recall that we consider a subgaussian process \(\{X_t : t \in T\}\) and equip \(T\) with the distance \(d(s,t) = \sqrt{\E(X_s-X_t)^2}\).
Chaining: Instead of applying a naive union bound to control \(X_t - X_{t_0}\) for every \(t \in T\), we can break every such difference into parts and then bound each part. This potentially reduces the number of events we need to take a union bound over.
For instance, given three random variables \(X_0,X_1,X_2,X_3\), we could write
\[\begin{eqnarray*} X_3 - X_0 &=& (X_3 - X_1) + (X_1 - X_0) \\ X_2 - X_0 &=& (X_2 - X_1) + (X_1 - X_0) \\ X_1 - X_0 &=& X_1 - X_0\,. \end{eqnarray*}\]Generic chaining upper bound: For sub-gaussian processes, we can control \(\E \max_{t \in T} X_t\) using chaining.
Fix some \(t_0 \in T\) and a sequence of finite sets
\[\{t_0\} = T_0 \subseteq T_1 \subseteq T_2 \subseteq \cdots \subseteq T_n \subseteq \cdots \subseteq T\,,\]with cardinalities bounded by \(\abs{T_n} \leq 2^{2^n}\) for every \(n \geq 1\).
Then we have:
\[\begin{equation}\label{eq:gc-ub} \E \max_{t \in T} X_t \lesssim \max_{t \in T} \sum_{n \geq 0} 2^{n/2} d(t,T_n)\,, \end{equation}\]where \(d(t,T_n) = \min_{s \in T_n} d(s,t)\).
Dudley’s entropy bound: If \(\{X_t : t \in T\}\) is a subgaussian process, then
\[\begin{equation}\label{eq:dudley} \E \max_{t \in T} X_t \lesssim \sum_{n \geq 1} 2^{n/2} e_n(T)\,, \end{equation}\]where \(e_n(T)\) is the largest distance \(d\) such that there are \(2^{2^n}\) points \(y_1,y_2,\ldots,y_{2^{2^n}}\) in \(T\) with \(d(y_i,y_j) \geq d\) for all \(i \neq j\).
Easy exercise: Show that Dudley’s entropy bound follows from the generic chaining upper bound \eqref{eq:gc-ub}.
Covering numbers: Let \(N(T,d,\e)\) denote the smallest number \(N\) such that the metric space \((T,d)\) can be covered by \(N\) balls of radius \(\e\), and consider the quantity
\[C_T \seteq \sum_{h=-\infty}^{\infty} 2^h \sqrt{\log N(T,d,2^{h})}\,.\]Up to a factor of \(2\), this is equivalent to the more elegant expression \(\int_0^{\infty} \sqrt{\log N(T,d,\e)} \,d\e\). It is an exercise to show that
\[C_T \asymp \sum_{n \geq 1} 2^{n/2} e_n(T)\,.\]Exercise: Show that for any metric space \((T,d)\), it holds that
\[\begin{equation}\label{eq:CT} \sum_{n \geq 1} 2^{n/2} e_n(T) \asymp \sum_{h=-\infty}^{\infty} 2^h \sqrt{\log N(T,d,2^h)}\,. \end{equation}\]Analysis of the probability simplex
Let \(T = \{ x \in \R^n : x_1,\ldots,x_n \geq 0, x_1+\cdots+x_n = 1 \}\) denote the probability simplex.
Lemma 1: For this \(T\), let \(X_t = \langle g,t\rangle\), where \(g=(g_1,\ldots,g_n)\) is a standard \(n\)-dimensional Gaussian. Then \(\E \max_{t \in T} X_t \asymp \sqrt{\log n}\).
This is because the maximum of the linear functional \(t \mapsto \langle g,t\rangle\) is achieved at a vertex of the convex body \(T\) which is equal to the convex hull of the standard basis vectors: \(T = \mathrm{conv}(e_1,\ldots,e_n)\). Thus \(\max_{t \in T} \langle g,t\rangle = \max \{ g_i : i =1,\ldots,n\}\), and therefore
\[\E \max_{t \in T} \langle g,t\rangle = \E \max(g_1,\ldots,g_n) \asymp \sqrt{\log n}.\]Let us now see that the entropy bound \eqref{eq:dudley} is far from tight in this simple case. We will use the formulation via covering numbers coming from the equivalence \eqref{eq:CT}.
Consider vectors \(t \in T\) with \(t_i \in \{0,1/k\}\) for $i=1,\ldots,n$. There are \({n \choose k} \geq \left(\frac{n}{k}\right)^k\) such vectors. For \(k \leq \sqrt{n}\), we have \((n/k)^k \gtrsim 2^{k \log n}\), and it is possible to choose a proportional subset of those vectors (e.g., a random subset will suffice) such that
\[d(t_i,t_j) \gtrsim \left(k \cdot \frac{1}{k^2} \right)^{1/2} \asymp k^{-1/2}, \qquad i \neq j\,.\]It follows that for \(k \leq \sqrt{n}\), we have $\log N(T,d,k^{-1/2}) \gtrsim k \log n$. Consider then the \(\approx \log n\) values of \(h\) for which \(1 \leq 2^h \leq \sqrt{n}\).
This gives us \(\approx \log n\) values of \(h\) for which
\[2^{-h} \sqrt{\log N(T,d,2^{-h})} \gtrsim 2^{-h} \cdot 2^h \sqrt{\log n} = \sqrt{\log n}\,.\]Therefore Dudley’s bound \eqref{eq:CT} is \(\gtrsim (\log n) \cdot \sqrt{\log n} = (\log n)^{3/2}\). (Recall that the correct bound, given by Lemma 1, is \(\sqrt{\log n}\).)