# Proving Weak Law of large numbers by Markov's inequality

Hi I am trying to solve the problem 5.13 of the book Statistical inference by George Casella and Roger L. Berger. The problem is

Formulate and prove a version of the WLLN with a weaker assumption than a finite variance. Use the Markov inequality to prove it.

I don't what it means by $\textbf{weaker assumption than a finite variance}$.

By Markov's inequalty we have for nonnegative random variable X $$P(X\ge\epsilon)\le\frac{E(X)}{\epsilon}$$

We need to prove $$P(|\bar{X}_n-\mu|\ge\epsilon)\to 0$$

So we have $P(|\bar{X}_n-\mu|\ge\epsilon)=P((\bar{X}_n-\mu)^2\ge\epsilon^2)\le\frac{E(\bar{X}_n-\mu)^2}{\epsilon^2}=\frac{\text{Var}(\bar{X}_n)}{\epsilon^2}=\frac{\sigma^2}{n\epsilon^2}$

Since I don't know what would be the weaker version I am stuck here. Any kind of help would be highly appreciated.

• Any response to my answer below? – John Feb 13 '16 at 3:01
• @JohnZHANG Sorry for not replying any. I wanted a different one. Yours answer has also given me a different aspect. – mint Feb 13 '16 at 8:52

Let $X_n$ be sequence of r.v, $\mu_n=ES_n$, $\sigma_n^2=var(S_n)$. If $\frac{\sigma_n^2}{b_n^2}\to 0$, then $$\frac{S_n-\mu_n}{b_n}\to 0$$ in probability.
Note we don't assume each $var(X_n)\le C<\infty$ and uncorrelated $X_n$.
$P(|S_n-\mu_n|>\epsilon |b_n|)\le \frac{1}{\epsilon^2}\frac{E(S_n-\mu_n)^2}{b_n^2}=\frac{1}{\epsilon^2}\frac{\sigma_n^2}{b_n^2}\to 0$