I need to show following almost sure convergence:
$X_1$, $X_2$, …, be identically distributed non-negative random variables with $E[X_1]<\infty$. Prove that: $$\frac{X_n}{n}\to 0$$ with probability $1$.
I want to use the first Borel Cantelli lemma:
For some $\epsilon>0$:
$$A_k = \{\frac{X_k}{k}>\epsilon\}$$
$$A = \cup_{n\geq 1}\cap_{k\geq n}A_k$$
and show $P(A) = 0$. However, after apply Markov's inequality to $P(A_k)=P(\frac{X_k}{k}>\epsilon)$, I can't show $\sum_{k=1}^{\infty}P(A_k)<\infty$ …
Thank you for your help.
Best Answer
Using the fact that $(X_k)$ are identically distributed, we have
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \sum_{k=1}^{\infty} \Bbb{P}(X_1 > k \epsilon) = \sum_{k=1}^{\infty} \Bbb{E}[\mathbf{1}_{\{ X_1 > k \epsilon \}}] = \Bbb{E}\bigg[ \sum_{k=1}^{\infty} \mathbf{1}_{\{ X_1 > k \epsilon \}} \bigg]. \tag{*}$$
Now let us focus on the random variable $Y := \sum_{k=1}^{\infty} \mathbf{1}_{\{ X_1 > k \epsilon \}}$. It is not hard to see that
$$ Y = \begin{cases} k, & X_1 \in (k\epsilon, (k+1)\epsilon] \text{ for some } k = 0, 1, \cdots \\ 0, & X_1 = 0 \end{cases} $$
This proves that $Y \leq \frac{1}{\epsilon}X_1$ and hence
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \Bbb{E}[Y] \leq \frac{1}{\epsilon}\Bbb{E}[X_1] < \infty. $$
Now all you have to do is to apply the 1st Borel-Cantelli's lemma.
Remark. If we replace the assumption $\Bbb{E}[X_1] < \infty$ by $\Bbb{E}[X_1] = \infty$, then a similar argument shows that $\limsup_{n\to\infty} \frac{X_n}{n} = \infty$ a.s. Indeed, one has the inequality
$$ Y \geq \frac{1}{\epsilon}X_1 - 1.$$
Plugging this back to the computation $\text{(*)}$, we have
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \Bbb{E}[Y] \geq \frac{1}{\epsilon}\Bbb{E}[X_1] - 1 = \infty. $$
Therefore by the 2nd Borel-Cantelli lemma, $\Bbb{P}(\frac{X_n}{n} > \epsilon \text{ i.o.}) = 1$ and thus $\limsup_{n\to\infty} \frac{X_n}{n} \geq \epsilon$. Since $\epsilon > 0$ is arbitrary, the conclusion follows by letting $\epsilon \to \infty$ (along a subsequence, if you want to be rigorous).