# Prove that $\liminf \frac{M_n}{\sqrt{2\log n}}\geq 1$, where $M_n=\max_{1}^n X_i$ and $(X_i)$ is a sequence of i.i.d standard normal random variables

Question

Let $$(X_n)_{n\geq 1}$$ be an i.i.d sequence of standard normals. Show that with probability one $$\liminf \frac{M_n}{\sqrt{2\log n}}\geq 1$$, where $$M_n=\max_{1}^n X_i$$.

My attempt

Given $$\varepsilon >0$$, it suffices to show that $$P(\frac{M_n}{\sqrt{2\log n}}<1-\varepsilon \quad \text{i.o.})= 0$$. To this end put $$A_n=(\frac{M_n}{\sqrt{2\log n}}<1-\varepsilon)$$ and we attempt to use the Borel-Cantelli Lemma. So $$\sum_1 ^\infty P(A_n)=\sum_{1}^\infty P(X_1 where $$c_n=(1-\varepsilon )(\sqrt{2\log n})$$ and $$\bar{\Phi}=1-\Phi$$ is the survival function. At this point since the normal distribution has no nice form for its cdf I have to use some argument involving asymptotics of this sum. To this end, I know that $$\bar{\Phi}(x)\sim\frac{\phi (x)}{x}$$ as $$x\to \infty$$ where $$\phi$$ is the density of a standard normal. More precisely, $$\left(\frac{1}{x}-\frac{1}{x^3}\right)\leq \frac{\bar{\Phi}(x)}{\phi(x)}\leq \frac{1}{x}.\tag{1}$$ We can write $$\sum_{1}^\infty P(X_1 using $$1$$, but I am not sure how to argue that this is finite.

For any $$x \geq 0$$ it holds that

$$1-x \leq e^{-x}. \tag{1}$$

For $$x:= \mathbb{P}(X_1 \geq c_n)$$ this implies that

$$\mathbb{P}(X_1 < c_n)^n = (1-\mathbb{P}(X_1 \geq c_n))^n \stackrel{(1)}{\leq} \exp \left(-n \mathbb{P}(X_1 \geq c_n) \right). \tag{2}$$ As

\begin{align*} \mathbb{P}(X_1 \geq c_n) &= \frac{1}{\sqrt{2\pi}} \int_{c_n}^{\infty} \exp \left( - \frac{y^2}{2} \right) \, dy \\ &\geq \frac{1}{\sqrt{2\pi}} \frac{c_n}{c_n^2+1} \exp \left(- \frac{c_n^2}{2} \right) \\ &\geq \frac{1}{2\sqrt{2\pi}} \frac{1}{(1-\epsilon) \sqrt{2 \log n}} \frac{1}{n^{(1-\epsilon)^2}} \end{align*}

we have

$$n \mathbb{P}(X_1 \geq c_n) \geq \frac{1}{2\sqrt{2\pi}} \frac{1}{(1-\epsilon) \sqrt{2 \log n}} n^{\epsilon}$$

for $$\epsilon \in (0,1)$$. For $$n \geq N=N(\epsilon)$$ sufficiently large this shows that

$$n \mathbb{P}(X_1 \geq c_n) \geq C n^{\epsilon/2}$$

where $$C=C(\epsilon) := 1/(2 \sqrt{4\pi} (1-\epsilon))$$. Plugging this estimate into $$(2)$$ we find that

$$\sum_{n \geq N} \mathbb{P}(X_1>c_n)^n \leq \sum_{n \geq N} \exp(- C n^{\epsilon/2})< \infty.$$