Let $\{X_n\}$ be i.i.d $N(0,1)$ random variables. Show that $\limsup_{n\rightarrow\infty} \frac{|X_n|}{\sqrt{\log n}}=\sqrt2$ a.s.

Question

Let $\{X_n\}_{n\ge1}$ be independent $N(0,1)$ random variables. Show that $$\limsup\limits_{n\to\infty} \frac{\left|X_n\right|}{\sqrt{\log(n)}}=\sqrt{2} \qquad \text{a.s.}$$

I aim to prove this using the fact that

$$\limsup\limits_{n\to\infty} X_n = b \quad \iff \quad \text{for all } \varepsilon>0 \ : \ \Biggl\{ \begin{array}{l} \mathbb{P}(X_n \le b+\varepsilon \text{ eventually})=1, \text{ and} \\ \mathbb{P}(X_n > b-\varepsilon\text{ i.o.})=1. \end{array}$$

I show the first of these two conditions as follows:

\begin{align*} &\hspace{-2em}\mathbb{P}\left(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\right)\\ &= \mathbb{P}\bigl(|X_1|>(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}\bigr) & \text{as $X_n$'s are identically distributed}\\ &=\int_{(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}}^{\infty} |x|\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \, dx\\ &=\int_{(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}}^{\infty} x\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \, dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{(\sqrt{2}+\varepsilon)^2\log(n)}^{\infty} e^{-u}du & \text{by making the substitution $u=\frac{x^2}{2}$}\\ &=-\frac{1}{\sqrt{2\pi}}\bigl[e^{-\infty}-e^{-\log(n)(\sqrt{2}+\varepsilon)^2}\bigr]\\ &=\frac{1}{\sqrt{2\pi}}n^{-(\sqrt{2}+\varepsilon)^2} \end{align*}

Thus, we have: \begin{align*} \sum_{n=1}^\infty \mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\Biggr)&=\sum_{n=1}^{\infty}\frac{1}{\sqrt{2\pi}}n^{-(\sqrt{2} + \varepsilon)^2}\\ &<\infty \qquad \text{ since $\sqrt{2}+\varepsilon>1$} \end{align*}

So, by the Borel–Cantelli Lemmas:

\begin{align} &\mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\text{ i.o.}\Biggr)=0\\ &\implies \mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} \le \sqrt{2} + \varepsilon\text{ eventually}\Biggr)=1 \end{align}

It remains then to show that

$$\mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} - \varepsilon \text{ i.o.}\Biggr)=1.$$

To do this, I would like to run a symmetric argument to the one above but here we need the final series to diverge which will happen if and only if $\sqrt{2}-\varepsilon \le1$, which happens if and only if $\varepsilon\ge \sqrt{2}-1=0.414\ldots$ and so $\varepsilon$ is getting away from zero, which we can't have. Unless I am making a stupid mistake or missing something obvious, I don't see a way around this issue. Is this approach doomed to fail or is there a way to fix it up? Or is there just a better approach in general for such a problem. Thanks in advance.

The problem is #1 (a) from this exam

Answer 1

Using the estimate

$$ \mathbb{P}(|X_1| > x) = 2\mathbb{P}(X_1 > x) \sim \frac{2\phi(x)}{x} \qquad \text{as } x \to +\infty $$

as in the hint, for any $a > 0$, we have

$$ \mathbb{P}(|X_n| > \sqrt{2a\log n}) \sim \Bigl(\frac{1}{\pi a \log n}\Bigr)^{1/2} \frac{1}{n^a} $$

From this, we find that

$$ \sum_{n=1}^{\infty} \mathbb{P}(|X_n| > \sqrt{2a\log n}) \ \begin{cases} =\infty, & \text{if $a < 1$} \\ <\infty, & \text{if $a > 1$} \end{cases} $$

Since the events $\{|X_n| > \sqrt{2a\log n}\}$ are independent, the Borel–Cantelli Lemmas tells:

$$ \mathbb{P}(|X_n| > \sqrt{2a\log n} \, \text{ i.o.}) = \begin{cases} 1, & \text{if $a < 1$} \\ 0, & \text{if $a > 1$} \end{cases} $$

Using the fact quoted in OP, this allows to conclude:

$$ \limsup_{n\to\infty} \frac{|X_n|}{\sqrt{\log n}} = \sqrt{2} $$

Answer 2

I can only see one mistake in your proof: you have $$ P(|X_1|>(\sqrt2+\epsilon)\sqrt{\log n})=\int_{(\sqrt2+\epsilon)\sqrt{\log n}}^\infty\color{red}{|x|}\varphi(x)\,dx $$ where $\varphi(x)$ is the pdf of $X_1$. This is not correct; the right hand side of what your wrote is equal to $E\left[|X_1|;X_1>(\sqrt2+\epsilon)\sqrt{\log n}\right]$. To correct this, you need to get rid of the $\color{red}{|x|}$. Remember, to find a probability, you integrate the pdf, and to find the expected value of a function, you integrate the function times that pdf.

The integral can no longer be written in terms of elementary functions. However, following the hint, you can get an asymptotic equivalent for the integral, which is sufficient to do the Borel-Cantelli tests.

Answer 3

Notice that your reasoning does not change if you substitute $\sqrt 2$ with $1$. It means, at least, that $$ \limsup_{n\to\infty} \frac{|X_n|}{\sqrt{\log n}}\le 1 \quad \text{a.s.} $$

My suspicion is that it goes to zero.