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Let $(X_n)_{n \ge1}$ be a sequence of i.i.d. real-valued random variables. Define $$M_n = \max_{1\le k \leq n}X_k$$ Let $F$ be the distribution function of $X_1$. $F(x) < 1$ for all $x \in \mathbb{R}$. Assume that $EX_1^2$ is finite. Show that $$\frac{M_n}{\sqrt{n}} \overset{P}{\to} 0$$

Attempt

Let $\varepsilon > 0$ be given and note that

$$ P\left( \frac{|M_n|}{\sqrt{n}} \geq \varepsilon \right) = P(M_n^2 \geq n \varepsilon^2) = P\left(\bigcup_{k=1}^n (X_k^2 \geq n \varepsilon^2)\right)\\ \leq \sum_{k=1}^n P(X_k^2 \geq n \varepsilon^2) = nP(X_1^2 \geq n \varepsilon^2) $$

I'm not quite sure what to do from here. It seems like this final thing should converge to zero but how do I prove this?

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  • $\begingroup$ I wonder how the hypothesis that for all $x\in\mathbb R,$ $F(x)< 1$ is expected to be used. It seems obvious that if this proposition is true, then it would remain true if that hypothesis were dropped, and I would think any proof that works in general would work without that hypothesis. $\endgroup$ – Michael Hardy Sep 19 '17 at 17:02
  • $\begingroup$ @MichaelHardy This is a subproblem of a bigger problem and I have already used the assumption. The assumption of finite second moment is local to this subproblem but I don't see how that enters either. $\endgroup$ – Lundborg Sep 19 '17 at 17:06
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Indeed, if one manages to show that $$nP(X_1^2\geqslant n\epsilon^2)$$ converges to $0$ as soon as $E(X_1^2)$ is finite, your proof is complete. To do so, consider the random variables $$Y_n=n\mathbf 1_{X_1^2\geqslant n\epsilon^2}$$ and note that, for every $n$, $$Y_n\leqslant \epsilon^{-2}X_1^2$$ and that, when $n\to\infty$, $Y_n\to0$ almost surely. Hence, Lebesgue dominated convergence theorem shows that $$nP(X_1^2\geqslant n\epsilon^2)=E(Y_n)\to0$$

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    $\begingroup$ Could you expand on the inequality of the $Y_n$'s? I don't quite see how you get that bound. $\endgroup$ – Lundborg Sep 19 '17 at 17:16
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    $\begingroup$ I am sure you can get it: simply separate cases. $\endgroup$ – Did Sep 19 '17 at 17:17
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    $\begingroup$ Ah! Cheers, bud. Sometimes the simplest arguments are the trickiest :) $\endgroup$ – Lundborg Sep 19 '17 at 17:19

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