Suppose that $X_i$ are i.i.d. random variables, with finite absolute moment: $E|X_1| < \infty$. Then $\max(X_1, \ldots, X_n) / n \to 0$ a.s. ?

Let $M_n = \max(X_1, \ldots, X_n)$.

I know that $M_n$ has a CDF which is $F_n(x) = F_1(x)^n$.

I would like to apply Borel-Cantelli to the sets $A_n = \{ M_n > n \epsilon \}$. So I would like to show that $\Sigma P(A_n) < \infty$.

I suspect that one can obtain a bound that is something like $\Sigma P(A_n) \leq E|X_1|$. However, because one doesn't have $\Sigma 1_{A_n} \leq X_1$, this is unlikely to be literally true. At some point one will need to use the independence.

So I was trying to proceed by analyzing the product $\Pi ( 1 - P(A_n))$ instead. We would like this to converge to a nonzero limit. This means that we want $P( \cap A_n^c) > 0$, at least for small epsilon, because of the assumed independence. But the problem is that $P(A_1) = 0$ is already possible.

I would appreciate a hint to set me on the right track. :)


Hint: define $A_n :=\left\{\max_{1\leqslant j\leqslant n}\left|X_j\right| \gt n\varepsilon\right\}$. Then $$\Pr\left(A_{2^n} \right)=\Pr\left(\bigcup_{j=1}^{2^n} \left\{\left|X_j\right|\gt 2^n\varepsilon\right\}\right)\leqslant \sum_{j=1}^{2^n} \Pr\left(\left\{\left|X_j\right|\gt 2^n\varepsilon\right\}\right).$$ Now use the following facts:

  • the $X_j$'s have the same distribution, which gives $\Pr\left(A_{2^ n} \right)\leqslant 2^n\Pr\left(\left\{\left|X_1\right|\gt 2^n\varepsilon\right\}\right)$;
  • if $Y$ is an integrable non-negative random variable, then the series $\sum_{n=1}^{ +\infty}2^n\Pr\left(\left\{Y\gt 2^n\right\} \right)$ is finite. Indeed, we start from the pointwise inequality $ 2^{n} \mathbf 1\{2^n\leqslant Y \lt 2^{n +1}  \}\leqslant Y\mathbf 1\{2^n\leqslant Y \lt 2^{n +1} \}$ and we get the convergence of the series $\sum_n 2^n\Pr \{2^n\leqslant Y \lt 2^{n +1} \}$. Then note that $$\sum_{n=1}^{ +\infty}2^n\Pr\left(\left\{Y\gt 2^n\right\} \right) = \sum_{n=1}^{ +\infty}2^n\sum_{ i\geqslant n} \Pr \{2^i\leqslant Y \lt 2^{i +1} \} , $$ the switch the sums.

The independence does not seem to be needed.

  • $\begingroup$ I don't think this shows that the sum of $P(A_n)$ is finite. Are you not trying to apply Borel-Cantelli? Should I think harder? $\endgroup$ – Lorenzo Najt Nov 1 '16 at 14:55
  • $\begingroup$ I mean , we already know that A_n is bounded by 1, and I don't understand how this bounded is better. And for very small epsilon it is probably much larger than 1, like of order n if X is discretely distributed $\endgroup$ – Lorenzo Najt Nov 1 '16 at 15:05
  • $\begingroup$ @AreaMan It seems that actually, we only need to prove the convergence of the series $\sum_n P(A_{2^n} )$. $\endgroup$ – Davide Giraudo Nov 1 '16 at 19:47
  • $\begingroup$ Thanks for your patience. I'm afraid I don't see why your second bullet should hold - I guess it has something to do with the growth rate of $2^n$ (because I don't think that it follows from the first moment bound that $\Sigma_{n = 1}^{\infty} n P(Y > n) < \infty$, for that you need the second moment to be bounded... although it does follow that $\Sigma_{n = 1} P(Y > n) < \infty$...) $\endgroup$ – Lorenzo Najt Nov 2 '16 at 2:33
  • $\begingroup$ I still don't understand why your second bullet holds. Can you explain? $\endgroup$ – Lorenzo Najt Nov 30 '16 at 17:17

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