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$X_1, X_2, \dots, X_n, \dots$ is a sequence of i.i.d random variables with $E[X_1] = 0$ and $E[X_1^2] = 1$. Show that $$ \max \left(\frac{|X_1|}{\sqrt{n}}, \dots, \frac{|X_n|}{\sqrt{n}}\right) \overset{d}{\to} 0, n \to \infty $$

I attempted to use the continuity theorem. Putting $Y_n = \max \left(\frac{|X_1|}{\sqrt{n}}, \dots, \frac{|X_n|}{\sqrt{n}}\right) $, we can show that the distribution function of $Y_n$ is $$ F_{Y_n}(y) = [F(\sqrt{n}y)]^n, $$ where $F(\cdot)$ is the distribution function of $X_1$. Then the characteristic function of $Y_n$ is $$ \varphi_{Y_n}(t) = \int_{\mathbb{R}} n^{\frac 32} [F(\sqrt{n}y)]^n e^{ity}\,\mathrm{d}F(y) $$ I can only get to this step and don't know how to proceed the proof. Is there an alternative way to prove it? Thanks for any help in advance!

probability learner
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2 Answers2

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It suffices to show that $P(Y_n\leqslant y)\to1$ for every $y\gt0$ (since convergence in probability implies convergence in distribution).

For every $y\gt0$, $P(Y_n\leqslant y)=(1-P(|X_1|^2\gt ny^2))^n$.

If $nx_n\to0$ then $(1-x_n)^n\to1$, hence it suffices to show that, if $Z$ is nonnegative and integrable then $nP(Z\gt n)\to0$ (and to apply this to $Z=|X_1|^2/y^2$).

Any idea for this last step? Note that $nP(Z\gt n)=E(Z_n)$ where $Z_n=$ $____$ hence $____$...

Did
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I think we can do this with just Chebyshev's inequality: As you pointed out, we have the exact formula $P(Y_n<\epsilon)=(1-P(X_1^2\ge n\epsilon^2))^n$. Now for fixed $\epsilon>0$, we have that $$ n\epsilon^2 P(X_1^2\ge n\epsilon^2) \le \int_{[n\epsilon^2,\infty)} x^2\, dF(x) \to 0 $$ as $n\to\infty$ by dominated convergence. Thus $P(X_1^2\ge n\epsilon^2)\le \delta/n$ eventually for arbitrary $\delta>0$ and hence $$ P(Y_n<\epsilon)\ge (1-\delta/n)^n \to e^{-\delta} . $$