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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!

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1 Answer 1

up vote 4 down vote accepted

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 $$\int_{\{n\leqslant |X|<n+1\}}|X|dP\geqslant nP\{n\leqslant |X|<n+1\},$$ so the series $$\sum_{n=1}^{+\infty}nP\{n\leqslant |X|<n+1\}$$ is convergent. This implies together with $$nP(|X|\geqslant n)\leqslant\sum_{k=n}^{+\infty}kP(k\leqslant |X|<k+1)$$ that $nP(|X|\geqslant n)\to 0$ as the remainder of a convergent series.

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Why does it suffices to show $n \cdot \mathbb{P}[|X|>n] \to 0$? I don't see why this implies $n \cdot \mathbb{P}[|X|>\sqrt{n}] \to 0$ (since $\mathbb{P}[|X|> \sqrt{n}] \geq \mathbb{P}[|X|>n]$)... –  saz Dec 25 '12 at 18:43
    
Replace $X$ by $|\xi_1|^2$. –  Davide Giraudo Dec 25 '12 at 18:51
    
Thank you very much! –  Danielsen Dec 26 '12 at 2:47

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