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Let $X_1, X_2, ...$ be identically distributed nonnegative random variable with $EX_1<\infty$. Prove that $\frac{X_n}{n}\rightarrow 0$ almost surely.

I am trying to use Borel Cantelli Lemma (part one) to prove this.

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I think Borel-Cantelli applies. Fix $a > 0$. Let $X$ be a random variable that has the same distribution as any of the $X_n$'s. For each $m \geq 1$, define $A_m$ to be the event that $am \leq X \lt a(m+1)$. $\newcommand{\E}{\mathbf{E}}$ $$ \begin{eqnarray*} \sum\limits_{n=1}^{\infty} \Pr\left[\frac{X_n}{n} \geq a \right] &=& \sum\limits_{n=1}^{\infty} \Pr\left[X \geq an \right] \\ &=& \sum\limits_{n=1}^{\infty} \sum_{m=n}^{\infty} \Pr[am \leq X < a(m+1)] \\ &=& \sum\limits_{n=1}^{\infty} \sum_{m=n}^{\infty} \Pr[A_m] \\ &=& \sum\limits_{m=1}^{\infty} m \Pr[A_m] \ \ \ \ \ \text{switching the order of summations} \\ &=& \frac{1}{a}\sum\limits_{m=1}^{\infty} am \Pr[A_m] \\ &\leq& \frac{1}{a}\sum\limits_{m=1}^{\infty} \E[X \cdot \mathbf{1}_{A_m}] \\ &=& \frac{1}{a} \E \left[X \cdot \sum\limits_{m=1}^{\infty} \mathbf{1}_{A_m} \right] \\ &\stackrel{(1)}{\leq}& \frac{\E X}{a} < \infty. \end{eqnarray*} $$ The inequality $(1)$ follows from the fact that $A_m$ for different $m$ are disjoint.

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Thank you very much. I dont quite get how we can change the order of summation, we dont know if the series converges absolutely – bear Sep 11 '11 at 6:14
@coolio, this is bad practice to accept an answer so quickly. It prevents other people contributing (and in the present case, it makes your demand of further explanations rather bizarre...). (To avoid misundestandings, I do not intend to propose an answer.) – Did Sep 11 '11 at 7:13
@coolio Er, I think Didier is right. You shouldn't have accepted my answer so soon. I recommend that you un-accept my answer and wait for other solutions. – Srivatsan Sep 11 '11 at 7:20
@coolio: Switching the order of summation is justified because all the terms are nonnegative. It is a fact that for a double series with nonnegative terms, summation in either order gives the same result. This includes the case that the series does not converge, in which case summation in either order will give $+\infty$. You can prove this either using the monotone convergence theorem or Tonnelli's theorem. – Nate Eldredge Sep 11 '11 at 11:18
Sorry, this is my first time using this forum. So i pressed on that accidentally. – bear Sep 11 '11 at 18:25

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