Given $X_i, i \geq 1$ iid random variables, with mean zero and variance one, I would like to show that $$ M_n = \max_{1 \leq k \leq n}\left\{\frac{|X_k|}{\sqrt n}\right\} \xrightarrow{d} 0$$ as $n \to \infty$.

I know that $P(M_n \leq x) = \big(P(|X_1| \leq x \sqrt n)\big)^{n}$. However, as $n \to \infty$, it may not be necessary that this goes to one for positive $x$.

I tried using Chebyshev's inequality after finding the distribution functions, but the factors coming from the variance and the scaling factor cancel out, so the right hand side is not meaningful there.

I would like to see if I can use characteristic functions to do this (find the characteristic function of $M_n$, and show that it goes to $1$ everywhere). But I was unable to proceed this way as well.

  • 1
    $\begingroup$ Did you try something like this : $E(|X_1|^2) = -\int_0^\infty x^2 dP(|X_1| > x)=\int_0^\infty 2x P(|X_1| > x)dx=\int_0^\infty 2 a t^{1/2} P(X_1 > a\sqrt{t})\frac{a\, dt}{\sqrt{t}}$. That it converges, by comparison with $\int_1^{t_k} \frac{1}{t}dt$ implies there is a sequence $t_k \to \infty$ such that $t_k P(X_1 > a\sqrt{t_k}) \to 0$ and your claim follows from $1-(1-\epsilon)^n \sim n\epsilon$ $\endgroup$
    – reuns
    Oct 27, 2017 at 0:03
  • $\begingroup$ I have seen the second equality, I tried playing around with it. How do you see the third one converging? I don't see how it compares with $\int \frac 1t dt$, since $\sqrt t$ on top and bottom cancel in that expression. $\endgroup$ Oct 27, 2017 at 0:09
  • $\begingroup$ The variance converges that's in the assumption. What is the contradiction of there is a sequence $t_k \to \infty$ such that $t_k P(X_1 > a\sqrt{t_k}) \to 0$ ? $\endgroup$
    – reuns
    Oct 27, 2017 at 0:10
  • $\begingroup$ Ok, I see what you are saying. I will think for some time and get back to you. $\endgroup$ Oct 27, 2017 at 0:12
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    $\begingroup$ @runr I see! The quantity $P(|X|^2 \leq \epsilon^2 n)$ is what should be attempt to be controlled. I will think more about this, but yeah, certainly the convergence to zero part should be correct, while the other part, I expect it to go through as well. $\endgroup$ Apr 29, 2021 at 16:11

1 Answer 1


Note that $$ \begin{align} P\left(\frac{\max_{1\leq i\leq n}|X_i|}{\sqrt n} > \epsilon \right)&= 1-P\left(\max_{1\leq i\leq n}|X_i| \leq \epsilon \sqrt n \right)\\ &= 1-P\left(|X_1| \leq \epsilon \sqrt n \right)^n \\ &= 1-P\left(|X_1|^2 \leq \epsilon ^2 n \right)^n \end{align}$$

For any integrable $X$, $\lim_{x\to \infty} xP(X> x )=0$. Indeed $$\begin{align} xP(X> x ) = x\int \mathbb 1_{X> x}(w) dP(w) &\leq x\int \mathbb 1_{X> x}(w)\frac{X(w)}{x} dP(w)\\ &=\int \mathbb 1_{X> x}(w) X(w) dP(w) \end{align}$$ and $\displaystyle \lim_{x\to \infty} \int \mathbb 1_{X> x}(w) X(w) dP(w)=0$ by dominated convergence.

Here, this implies $P\left(|X_1|^2 > \epsilon ^2 n \right)=o\left( \frac{1}{n}\right)$, hence $$ \begin{align} P\left(\frac{\max_{1\leq i\leq n}|X_i|}{\sqrt n} > \epsilon \right)&= 1-\left(1+o\left( \frac{1}{n}\right) \right)^n\\ &= 1-\exp\left(n\ln \left(1+o\left( \frac{1}{n}\right) \right) \right)\\ &= 1-\exp\left(o\left( 1\right) \right)\\ &= o\left(1\right) \end{align}$$

This proves convergence in probability to $0$, hence convergence in distribution to $0$.

  • $\begingroup$ Yes, thanks for the same. I knew it had to do with bounded variance, once the above comments came in. $\endgroup$ Oct 27, 2017 at 11:07
  • 2
    $\begingroup$ great answer! I enjoyed. +1 $\endgroup$ Nov 13, 2017 at 20:02

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