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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.

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    $\begingroup$ 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 '17 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. $\endgroup$ – астон вілла олоф мэллбэрг Oct 27 '17 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 '17 at 0:10
  • $\begingroup$ Ok, I see what you are saying. I will think for some time and get back to you. $\endgroup$ – астон вілла олоф мэллбэрг Oct 27 '17 at 0:12
  • $\begingroup$ Otherwise replace $\big(P(|X_1| \leq x \sqrt n)\big)^{n}$ by $\big(P(|X_1| \leq x n)\big)^{n^2}$ and use $E(|X_1|^2) < \infty$ directly. $\endgroup$ – reuns Oct 27 '17 at 0:13
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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 $0$.

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  • $\begingroup$ Yes, thanks for the same. I knew it had to do with bounded variance, once the above comments came in. $\endgroup$ – астон вілла олоф мэллбэрг Oct 27 '17 at 11:07
  • $\begingroup$ great answer! I enjoyed. +1 $\endgroup$ – dafinguzman Nov 13 '17 at 20:02

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