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 $$ \sum_1 ^\infty P(A_n)=\sum_{1}^\infty P(X_1<c_n)^n=\sum_{1}^\infty (1-\bar{\Phi} (c_n))^n $$ 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 $$ \sum_{1}^\infty P(X_1<c_n)^n\leq \sum_{1}^\infty(1-(c_n^{-1}-c_n^{-3})\phi(c_n))^n $$ 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

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


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.