# Convergence in probability of renormalized running maximum of i.i.d. square integrable random variables

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?

• 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. – Michael Hardy Sep 19 '17 at 17:02
• @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. – Lundborg Sep 19 '17 at 17:06

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 $$nP(X_1^2\geqslant n\epsilon^2)=E(Y_n)\to0$$
• Could you expand on the inequality of the $Y_n$'s? I don't quite see how you get that bound. – Lundborg Sep 19 '17 at 17:16