Proving almost sure convergence

Question

Assume the sequence of random variables $X_1, X_2, \cdots$ are IID with finite mean and finite variance. Define a random variable:

\begin{align} Y_n = \frac{X_n}{n} \end{align} Show that $Y_n \to 0$ almost surely.

To converge to some value almost surely implies: \begin{align} \mathbb{P}\left( \lim_{n \to \infty}{Y_n} = 0\right) = 1 \end{align} By the way $Y_n$ is defined, this is equivalent to: \begin{align} \mathbb{P}\left( \lim_{n\to \infty}{\frac{X_n}{n}} = 0\right) =1 \end{align}

But, the limit of $\frac{1}{n}$ as $n \to \infty$ is "obviously" $0$ and intuitively I would think that if the expectation of the numerator is finite then we can expect some finite number in the numerator. As $n$ runs off to $\infty$ it would just look like some finite number (it doesn't really matter what the number is so long as it is finite) being divided by a number growing larger and larger - approaching $\infty$. So, I would think the $\lim_{n\to \infty} \frac{X_n}{n}$ would behave similarly to $\lim_{n \to \infty} \frac{1}{n}$.

EDIT: but as pointed out in the comments, the limit of $X_n$ is not necessarily $0$ for all possible definitions of $X_n$.

Thank you.

Answer 1

$$\mathbb{E}\left(\sum_{n=1}^\infty Y_n^2\right)<\infty\Rightarrow \mathbb{P}\left(\sum_{n=1}^\infty Y_n^2<\infty\right)=1\Rightarrow \mathbb{P}\left(Y_n\to 0\right)=1.$$

Answer 2

The result is still true even without the assumption of finite variance. Since $E|X|$ is finite, we have $E|X/\epsilon|$ is finite for all $\epsilon>0$. Using the layer cake formula, we get $$ \infty>E|X/\epsilon |=\int_0^\infty P(|X/\epsilon|>t)\,dt\ge\sum_{n=0}^\infty P(|X/\epsilon|>n)=\sum_n P(|X/n|>\epsilon) $$ Since the series $P(|X/n|>\epsilon)$ has a finite sum, the first Borel Cantelli lemma implies $P(|X_n/n|>\epsilon\text{ infinitely often})=0$. This holds for all $\epsilon>0$, which implies that $X_n/n\to 0$.