0
$\begingroup$

By definition a RV is $\nu$-sub-gaussian, if the log of the moment generating function is bounded such that

$$ \log(\mathbb{E}[\exp(\lambda X)]) \leq \frac{\lambda^2\nu}{2} $$ Also, it can be shown that all the sub-gaussian RVs are centered i.e. $\mathbb{E}[X] = 0$.

Just now, I saw a result which proves, for a sub-gaussian RV $$ \mathbb{E}[\max\limits_{i=1\dots N} X_i] \leq \sqrt(2\nu\log N) $$

We already know that expectation is zero, then what is the point of bounding the expected maximum of the RV ?. Moreover, expectation is defined over a sample or population right ?. What is the intuition behind this bound and what exactly are we bounding ?

$\endgroup$
0
$\begingroup$

Looking at the random variable $M_N=\max_{1\le i\le N}X_i$ is an example of extreme value theory, an area of probability theory which is very active at the moment. For $\{X_i\}$ mean zero, the fact that $E[X_i]=0$ implies that, roughly speaking, $X_i>0$ approximately half of the time, so for large $N$ we would expect $M_N>0$ with high probability. If $X_i$ satisfies $P(X_i\le x)<1$ for all $x$ (which is the case for Gaussian random variables, for example), then $P(M_N\le x)=P(X_1\le x)^N\to0$ as $N\to\infty$, so in fact one can show we have $M_N\to+\infty$ almost surely. The question then becomes: how quickly? If $X_i\sim\mathcal N(0,1)$, then one can show $E[M_N]\approx\sqrt{2\log N}$. The purpose of this question is to show that the maximum sub-Gaussian random variables grows no faster than this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.