Let $X_1, X_2, X_3, ...$ be a sequence of independent and equally distributed random variables with $E|X_1| < \infty$

Question

Let $X_1, X_2, X_3, ...$ be a sequence of independent and equally distributed random variables with $E|X_1| < \infty$. Show that $E|\frac{S_n}{n} - EX_1| \rightarrow 0$ with $S_n = \sum_{k=1}^n X_k$

I was trying to solve this problem for a while and I feel that I'm missing something obvious. I can easily prove that $\frac{S_n}{n} - EX_1 \rightarrow 0$ almost surely, but I'm not sure how correctly make the next steps.

I was earlier recommended to work in this direction:

$E|\frac{S_n}{n} - EX_1| = E|\frac{S_n}{n} - EX_1|*I\{|X|<N\} + E|\frac{S_n}{n} - EX_1|*I\{|X|>N\}$

Basically, I'm multiplying expected value by 1 and then I work with each summand separately. However, I didn't manage to solve this problem in this way - I feel like I'm missing something.

Answer 1

All you need to do is prove that $\frac{S_{n}}{n}$ is uniformly integrable.

Then we can invoke the theorem that :-

If $\{X_{n}\}$ be a sequence of random variables , then $E|X_{n}-X|\to 0$ if and only if $\{X_{n}\}$ is uniformly integrable and $X_{n}\xrightarrow{P} X$ .

Then $\frac{S_{n}}{n}$ being uniformly integrable and convergent in Probability to $E(X_{1})$ (Due to WLLN) would imply that $E|\frac{S_{n}}{n}-E(X_{1})|\to 0$ or $\frac{S_{n}}{n}\xrightarrow{L^{1}} EX_{1}$ .

To prove that $\frac{S_{n}}{n}$ is uniformly integrable you show that $\sup_{n\in\Bbb{N}} E|\frac{S_{n}}{n}|<\infty$ and given any $\epsilon>0$ , there exists a $\delta>0$ such that for any set(in the specified sigma algebra) $A$ such that $P(A)<\delta\implies E[|\frac{S_{n}}{n}|\cdot\mathbf{1}_{A}]<\epsilon$

Both are easy enough to show due to the triangle inequality.

$E|\frac{S_{n}}{n}|\leq \frac{1}{n}\sum_{i=1}^{n}E|X_{i}|=E|X_{1}|<\infty$ and thus the first condition holds.

Secondly as $E|X_{1}|<\infty$ , you have that given any $\epsilon>0$, you can find a $\delta>0$ such that for $P(A)<\delta$ , $E|X_{1}\cdot\mathbf{1}_{A}|<\epsilon$ .

To do the above you take $E\bigg[|X_{1}|\cdot\mathbf{1}_{A}\bigg]=E|X_{1}\cdot\mathbf{1}_{A\cap \{|X_{1}|> N\}}|+E|X_{1}\cdot\mathbf{1}_{A\cap \{|X_{1}|\leq N\}}|$ . Now choose $N$ such that $E[|X_{1}|\cdot\mathbf{1}_{|X_{1}|>N}]<\frac{\epsilon}{2}$ and $\delta = \frac{\epsilon}{2N}$ .

Then you have $E[|X_{1}|\cdot\mathbf{1}_{A}]< \epsilon$ for any set $A$ with $P(A)<\delta$.

Now given $\epsilon>0$ , you choose $\delta$ just as above. That is , $\delta$ such that $P(A)<\delta$ implies $E[|X_{1}|\mathbf{1}_{A}]<\epsilon$. Then for that same choice of $\delta$ you have :-

$E[|\frac{S_{n}}{n}|\cdot\mathbf{1}_{A}]= \frac{1}{n}E[|S_{n}|\cdot\mathbf{1}_{A}]\leq \frac{1}{n}\sum_{i=1}^{n}E[|X_{i}|\cdot\mathbf{1}_{A}]< \frac{n\epsilon}{n}=\epsilon$

This proves all the conditions required for uniform integrability.

And thus $\frac{S_{n}}{n}$ is uniformly integrable and the required result follows.

For more explanations, see this well written page here. You can find all the definitions , results and theorems I have used above in the linked page. Explicitly I have used (3) and (13) from the linked page.