# Means of maximums

From a math point of view, it’s an interesting example of how the mean of the maximum of a set of random variables is higher than the max of the individual means – Andrew Gelman

Controlling the maximum of a set of random variables is an important problem in mathematical statistics, and it’s surprising how far a comparatively crude approach can be stretched.

Suppose you have $$m$$ random variables $$X_1$$, .., $$X_m$$, and you want to know about the mean of the maximum.  A very simple bound is that the whole is greater than the parts:
$E[\max_i |X_i|]\leq E[\sum_i |X_i|]\leq m\max E|[X_i|].$

That’s crude enough to be pretty useless, but suppose we looked at the squares of the $$X_i$$. Applying the same crude bound
$E[\max_i X_i^2]\leq E[\sum_i X_i^2]\leq m\max E[X_i^2],$
but now we can say
$\sqrt{E[\max_i X_i^2]}\leq \sqrt{m}\max \sqrt{E[X_i^2]},$
or
$\|X_i\|\leq \|X_i\|_2\leq \sqrt{m}\max_i\|X_i\|_2.$

We don’t need to to stop there:
$\|X_i\|\leq \|\max_i X_i\|_4\leq \sqrt[4]{m}\max_i\|X_i\|_4$
and in general
$\|X_i\|\leq \|\max_i X_i\|_p\leq \sqrt[p]{m}\max_i\|X_i\|_p.$

And it doesn’t even stop there. If we write $$\psi_p(x)=\exp (\|x\|^p)$$ and define a norm by
$\psi_p(\|X\|_{\psi_p})=E[\psi_p(X)],$
we get, eg,
$\|X_i\|\leq \|\max_I X_i\|_{\psi_2}\leq \sqrt{\log m}\max_i\|X_i\|_{\psi_2}.$

You still need to worry if these expectations exist, but for Normal distributions and those with lighter tails than the Normal they do.

These bounds are for finite sets, but they can be stretched to some infinite sets by a process called chaining.  Suppose you have a stochastic process $$X$$ indexed by a set $$T$$ in a metric space, and you want to control $$E\left[\sup|X({r})-X(s)|\right]$$ over all pairs with $$d(r,s)<\delta$$.  Assume we have some sort of Lipschitz condition so that $$E[|X({r})-Xs)|]<K d(r,s)$$ and (for simplicity) assume $$K=1$$. First, lay down a grid of points $$t_{1i}$$ so that for any point $$s$$ there is a $$t_{1i}$$ with  $$d(s,t_{1i})<1$$ and so also $$E[|X(s)-X(t_{1i})|]<1$$. Let $$N(1)$$ be the number of points you needed. Now, repeat with points $$t_{1/2,i}$$ so that for any point $$s$$ there is a $$t_{1i}$$ with  $$d(s,t_{1i})<1/2$$, and write $$N({1/2})$$ for the number you needed. And so on in both directions: …8,4,2,1,1/2 $$1/4, 1/8,\dots,1/{2^k},\dots$$.

For any two points $$r$$ and $$s$$ anywhere in any level of this grid, you can bound $$E[|X({r})-Xs)|]$$ by following the tree of links up to the coarsest necessary level (where the links will be of length about $$\delta$$) and then back down again. At each level, the number of links of length $$2^{-k}$$ is finite: $$N(2^{-k})$$, so the maximum over links of that length is bounded by $$2^{-k}\sqrt{\log N(2^{-k})}$$.  The total path length is bounded by a multiple of $$\sum_k 2^{-k}\sqrt{\log N(2^{-k})}$$, with the sum taken over all links shorter than $$\delta$$

And, if you were optimistic, you might hope that under reasonable conditions this bound for a dense grid might often transfer to a bound for the whole process. And, after a lot of painful details including genuine questions of measurability, it often does.

If we write $$N(\epsilon)$$ more generally for the number of points you’d need in a grid with spacing $$\epsilon$$, we can bound (up to a multiple) the sum by an integral
$\int_0^\delta \sqrt{\log N(\epsilon)}\,d\epsilon.$
Which is why integrals like that appear a lot in empirical process theory.