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This is the problem 3.5.1 of Shiryaev's Probability:

Let $\xi_1,\xi_2,\ldots$ be a sequence of independent identically distributed random variables with $E\xi_1=0$ and $E\xi_1^2=1$. Show that $\max\left(\frac{\mid\xi_1\mid}{\sqrt{n}},\ldots,\frac{\mid\xi_n\mid}{\sqrt{n}}\right)\overset{d}{\rightarrow} 0,\ n\to\infty$

I tried to solve it with this:

$\forall\varepsilon>0$, then $\lim_{n\to\infty}P\left\{\max\left(\frac{\mid\xi_1\mid}{\sqrt{n}},\ldots,\frac{\mid\xi_n\mid}{\sqrt{n}}\right)\leq\varepsilon\right\}=\lim_{n\to\infty}[P\{\mid\xi_1\mid\leq\sqrt{n}\varepsilon\}]^n=\lim_{n\to\infty}[1-P\{\mid\xi_1\mid>\sqrt{n}\varepsilon\}]^n=\exp[\lim_{n\to\infty}(-nP\{\mid\xi_1\mid>\sqrt{n}\varepsilon\}]$

How can I prove $\lim_{n\to\infty}(-nP\{\mid\xi_1\mid>\sqrt{n}\varepsilon\}\to 0\ ?$ Thank you!

1 Answers 1

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Actually, the problem is to show that if $X$ is an integrable random variable, we have $nP(|X|>n)\to 0$. To see that, we have that $\int_{\{n\leqslant |X| so the series $\sum_{n=1}^{+\infty}nP\{n\leqslant |X| is convergent. This implies together with $nP(|X|\geqslant n)\leqslant\sum_{k=n}^{+\infty}kP(k\leqslant |X| that $nP(|X|\geqslant n)\to 0$ as the remainder of a convergent series.

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    Replace $X$ by $|\xi_1|^2$.2012-12-25