# Convergence of Random Variable

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.

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.
• @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