# Proving almost sure convergence

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.

• I don't think that limit is zero so obviously. – Giuseppe Negro Nov 15 '15 at 20:55
• If you want to use the strong law of large numbers, write ${X_n\over n}={S_n\over n}-{S_{n-1}\over n-1}({n-1\over n}).$ – user940 Nov 15 '15 at 21:10
• "the limit of $\frac{X_n}{n}$ is obviously $0$ because of the $n$ in the denominator and because the $X_n's$ have finite mean and finite variance." 1. This is not obvious. 2. That you cite every hypothesis at your disposal as reasons for this result to hold, suggests that, actually, you are not quite sure of the reasons why it should hold. Do you? – Did Nov 15 '15 at 21:22
• Then why are you writing sentences giving reasons at random? Sorry but I am not following the approach. – Did Nov 15 '15 at 21:24
• And now that you made your reasoning more explicit, one can see it is wrong: in many settings, actually as soon as the support of the common distribution of the $X_n$ is unbounded, the sequence $(X_n)$ is almost surely unbounded. So, a priori, $X_n/n$ not converging to $0$ is possible. That $X_n/n$ actually converges to $0$ requires a different argument. – Did Nov 15 '15 at 21:35

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

• So are you suggesting defining a new RV that is the sample mean of $Y_n's$? – David South Nov 15 '15 at 21:11
• I am suggesting that you consider the random variable $\sum_{n=1}^\infty Y^2_n.$ – user940 Nov 15 '15 at 21:13
• @DavidSouth The solution that I've started here has nothing to do with the strong law of large numbers. In fact, it does not need the random variables $(X_n)$ to be independent. – user940 Nov 15 '15 at 21:17
• Ahhh, is this of a more "Borel-Cantelli" flavor? This lemma is something I have not learned yet but I have seen while reading a little further in my text. So, I would prefer to use another method that does not make use of a lemma I should not know at this point. – David South Nov 15 '15 at 21:21
• @DavidSouth No, it is much easier and more direct than that. – user940 Nov 15 '15 at 21:23

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 $$X_n/n\to 0$$.