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.
Since there are several versions of WLLN with weaker assumptions, I am not sure which one you want. So I think your question is not well-defined. But one direct generalization of WLLN is that
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 $X_n$.
The proof is easy by Markov inequality.
$P(|S_n-\mu_n|>\epsilon |b_n|)\le \frac{1}{\epsilon^2}\frac{E(S_n-\mu_n)^2}{b_n^2}=\frac{1}{\epsilon^2}\frac{\sigma_n^2}{b_n^2}\to 0$