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

  • 1
    $\begingroup$ I think that the result is false $\endgroup$
    – Exodd
    Commented Jun 10, 2020 at 15:58
  • 2
    $\begingroup$ There is a mistake in line 2 of your first computation: it should be $\int_{(\sqrt{2}+\epsilon)\sqrt{\log n}}^\infty \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\, dx$ (there is no $\lvert x \rvert$ in other words). Have you tried using the hint? It seems straightforward from there $\endgroup$
    – Michh
    Commented Jun 10, 2020 at 17:38
  • $\begingroup$ See my edits for proper MathJax usage. In particular either $\log n$ or $\log(n)$ (using \log, with a backslash) is correct, but $log(n)$ is not. Notice that the space to the right of $\log$ in $\log n$ is more than in $\log(n),$ i.e. you get context-dependent spacing built in to the software, and that doesn't happen if you use something like \text{log} instead. $\endgroup$ Commented Jun 10, 2020 at 17:48
  • $\begingroup$ @MichaelHardy Thanks! I am extremely new to Tex so I greatly appreciate the help! $\endgroup$ Commented Jun 10, 2020 at 17:50
  • 1
    $\begingroup$ Yes it is the same. $\mathbb{P}(X_n/\sqrt{\log n} > \epsilon \sqrt{2} \ \text{i.o.}) = 0$ for $\epsilon >1$ is equivalent to your $\mathbb{P}(X_n/\sqrt{\log n} \leq \sqrt{2}+\epsilon \ \text{eventually}) =1$ (just take the complement) and similarly for the second case. $\endgroup$
    – Michh
    Commented Jun 10, 2020 at 18:20

3 Answers 3


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:

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

  • 1
    $\begingroup$ Beautiful, thank you so much! $\endgroup$ Commented Jun 10, 2020 at 18:35

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.


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.

  • $\begingroup$ Hmm, well its a question that appeared on a previous preliminary examination, so barring an extreme overlook on the committee's part, the result has got to be correct. Perhaps I need to take an entirely different approach. $\endgroup$ Commented Jun 10, 2020 at 17:08
  • $\begingroup$ could you link it? $\endgroup$
    – Exodd
    Commented Jun 10, 2020 at 17:10
  • $\begingroup$ Unfortunately I only have a pdf document and I don't see a way to post a picture in a comment. $\endgroup$ Commented Jun 10, 2020 at 17:15
  • $\begingroup$ edit the question and put in the picture $\endgroup$
    – Exodd
    Commented Jun 10, 2020 at 17:16
  • $\begingroup$ Ok, I have inserted the picture. $\endgroup$ Commented Jun 10, 2020 at 17:23

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