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.
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.