# Weak LLN problem

Suppose $$(X_n)$$ is a sequence of r.v's satisfying $$P(X_n=\pm\ln (n))=\frac{1}{2}$$ for each $$n=1,2\dots$$. I am trying to show that $$(X_n)$$ satisfies the weak LLN.

The idea is to show that $$P(\overline{X_n}>\varepsilon)$$ tends to 0, but i am unsure how to do so.

• Are they independent? – John_Wick Dec 2 at 6:59

If $$X_n$$ are independent, then $$Var(X_n)=E(X_n^2)=(\log n)^2.$$ Then $$P(|\bar X_n|>\epsilon)\leq Var(\bar X_n)/\epsilon^2=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}Var(X_i)=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}(\log i)^2\leq \frac{(\log n)^2}{n\epsilon^2}\rightarrow 0.$$
• $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(\log n)^2$ – John_Wick Dec 2 at 15:49
• $Var(\bar X_n)=Var(1/n\sum_{i=1}^n X_i)=1/n^2\sum_i Var(X_i)$ if $X_i$'s are independent. – John_Wick Dec 2 at 15:50