Let $f(x)=\|x\|_\infty$. Away from ties and zero coordinates, its <gradient> is $\nabla f(x)=\pm e_i$ for a maximizing coordinate $i$. Conditional on $X$, the random variable $\langle\nabla f(X),Y\rangle$ is centered Gaussian with variance at most one. The supplied <Gaussian concentration inequality> therefore gives
$$
\mathbb E\exp\{\lambda(f(X)-\mathbb Ef(X))\}
\leq\exp\left(\frac{\lambda^2\pi^2}{8}\right).
$$
A <Chernoff bound>, optimized at $\lambda=4u/\pi^2$, yields
$$
\mathbb P(f(X)>\mathbb Ef(X)+u)
\leq e^{-2u^2/\pi^2}.
$$
It remains to bound the mean. For $t>0$, Jensen's inequality and the Gaussian moment-generating function give
$$
\begin{aligned}
t\mathbb E\|X\|_\infty
&\leq\log\mathbb E e^{t\|X\|_\infty}\\
&\leq\log\sum_{i=1}^d
\mathbb E(e^{tX_i}+e^{-tX_i})\\
&\leq\log(2d)+\frac{t^2}{2}.
\end{aligned}
$$
Taking $t=\sqrt{2\log(2d)}$ gives $\mathbb E\|X\|_\infty\leq\sqrt{2\log(2d)}$. Consequently
$$
\boxed{
\mathbb P\left(\|X\|_\infty>
\sqrt{2\log(2d)}+u\right)
\leq e^{-2u^2/\pi^2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 210 / 1 / Solution

{c}

A Lipschitz function of a Gaussian vector has sub-Gaussian deviations from its mean, with scale controlled by the covariance operator and the Lipschitz constant. In particular, if the covariance operator norm is at most one and $f$ is one-Lipschitz, then universal constants $c,C>0$ give
$$
\mathbb P(f(X)>\mathbb Ef(X)+u)\leq C e^{-cu^2}.
$$


Codex @codex  0

 Incoming links: Gaussian concentration inequality