# Tail probability of the maximum of Gaussians

Consider $$N$$ i.i.d. standard Gaussian variables, and let $$X_m$$ be the maximum. It is known that $$E[X_m] \approx \sqrt{2 \log N}$$ and that the variance goes to zero as $$N$$ grows. However, I'm looking for a good bound on the tail probability, say Prob $$[ X_m < a \sqrt{2 \log N}]$$ for some fixed $$a<1$$.

I've found the Borell-TIS Lemma, stating "inheritance of tails": essentially, the tails are not heavier than those of a unit-variance Gaussian with the correct mean. This is rather weak, as the tails do not decrease as $$N$$ grows and the variance decays. On the other hand, this lemma refers to general Gaussian vectors, not necessarily i.i.d., so I'm hoping that maybe something stronger can be said in the simpler i.i.d. case? Specifically, I'd like the lower-tail probability to decrease sharply as a function of $$N$$, for any fixed $$a<1$$.

This is an interesting question. There are probably many theorems which give you exactly what you want but I don't recall a specific one that is applicable straight away. In general, I would guess that you could obtain such bounds in the following general cases:

• maximum of IID strongly log-concave random variables.
• maximum of IID sub-gaussian random variables.
• maximum of IID random variables with sub-gaussian tails.
• relaxing the constraint to exponential tails or power-law tails would strongly impact the result.

These are variant of the same behavior: maxima of IID variables are highly regular, particularly when the tails are as sharp as that of a Gaussian.

Once you know why, you'll probably be able to derive the result you need: the CDF of the maxima of independent variables is the product of the CDF of each variable. In the Gaussian case, letting $$\Phi$$ denoting the CDF of the standard Gaussian:

$$P(X_m \geq t) = \Phi^n(t)$$

I see multiple ways to finish here:

Let me know if you need further help here (by email, preferably): I've got some unpublished results that can be brought to bear here, if you really need to push this.