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Let $\{X_n\}$ be a sequence of identically distributed random variables with $\text{E}(|X_1|)< \infty$. Show that $$\dfrac{\max(X_1,\ldots,X_n)}{n}\xrightarrow{\text{a.s}} 0$$

I need to show $P\left[\left|\dfrac{\max(X_1,\cdots,X_n)}{n}\right|> \epsilon\right]\to 0$.But unable to do that. Please help.

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Without loss of generality assume $X_n \ge 0$. Then $$\int_0^\infty P[X_n \ge x] \, dx = \int_0^\infty \int_{\{X_n \ge x\}} 1 \, dP \, dx = \int_\Omega \int_{0}^{X_n(\omega)} 1 \, dx \, dP = E[X_n] < \infty.$$ In particular $$\sum_{n=1}^\infty P[X_n \ge \varepsilon n] \le \int_0^\infty P[\frac{X_n}{\varepsilon} \ge x] \, dx = E[X_n]/\varepsilon < \infty.$$ Notice now that $$P\left[\frac{\max(X_1,\dots,X_N)}{N} \ge \varepsilon\right] \le \sum_{n=1}^N P[X_n \ge \varepsilon N].$$ Now $$\sum_{n=1}^N P[X_n \ge \varepsilon N] \le \sum_{n=1}^\infty P[X_n \ge \varepsilon n],$$ so by dominated convergence theorem as $N \to \infty$ we have $$P\left[\frac{\max(X_1,\dots,X_N)}{N} \ge \varepsilon\right] \to 0.$$

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  • $\begingroup$ How you consider Without loss of generality assume $X_n \ge 0$ $\endgroup$
    – A.D
    Apr 8, 2014 at 15:21
  • $\begingroup$ @A.D: If the claim did not hold for some $X_n$, we could consider $Y_n = |X_n|$. The claim would not then hold for $Y_n$ either since $P[\max Y_i / n \ge \varepsilon] \ge P[\max X_i / n \ge \varepsilon]$, which would be a contradiction assuming we have proved the result for non-negative random variables. $\endgroup$
    – J. J.
    Apr 8, 2014 at 15:26

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