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.

  • 1
    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 $P(|\bar X_n|>\epsilon)\leq Var(\bar X_n)/\epsilon^2=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}Var(X_i)=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}(\log i)^2\leq \frac{(\log n)^2}{n\epsilon^2}\rightarrow 0.$

  • How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response – user593295 Dec 2 at 15:25
  • $E(X_n)=0$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.