$X_1,X_2,\cdots,X_n$ are an iid sequence with standard normal distribution. Suppose that $a_n\rightarrow \infty$, find the sufficient and neccesary condition for $a_n$ such that
$$\frac{\max (X_1,X_2,\cdots,X_n)}{a_n}\rightarrow 0\quad a.e.$$
I don't know how to use law of larger number or Kolmogorov's three series theorem to prove it, so I tried this problem with Borel-Cantelli lemma but it seems to be too complicated to calculate.
Best Answer
Let $(X_n)_{n=1}^{\infty}$ be i.i.d. random variables on $\mathbb{R}$ but do not assume anything about its distribution. Write $x_+ = \max\{0, x\}$ and $M_n = \max\{X_1, \cdots, X_n\}$. Then we always have
$$ 0 \leq \limsup_{n\to\infty} \frac{M_n}{a_n} = \limsup_{n\to\infty} \frac{(X_n)_+}{a_n} $$
So $M_n/a_n \to 0$ a.s. if and only if $(X_n)_+ / a_n \to 0$ a.s. Then by the 11st & 2nd Borel-Cantelli lemmas, this holds if and only if
$$ \forall \epsilon > 0 \ : \quad \sum_{n=1}^{\infty} \mathbb{P}[X_+ \geq \epsilon a_n] < \infty $$
where $X$ has the same distribution as $X_n$.
In case of standard normal distribution, we have $\mathbb{P}[X_+ > x] \sim \frac{1}{\sqrt{2\pi}x} e^{-x^2/2} $ as $x \to \infty$. Therefore the limit in question holds if and only if
$$ \forall \epsilon > 0 \ : \quad \sum_{n=1}^{\infty} \frac{1}{a_n} e^{-\epsilon a_n^2/2} < \infty. $$