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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.

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    $\begingroup$ Are they independent? $\endgroup$
    – John_Wick
    Commented Dec 2, 2018 at 6:59

1 Answer 1

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If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(\log n)^2.$ Then $P(|\bar X_n|>\epsilon)\leq Var(\bar X_n)/\epsilon^2=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}Var(X_i)=\frac{1}{n^2\epsilon^2}\sum_{i=1}^{n}(\log i)^2\leq \frac{(\log n)^2}{n\epsilon^2}\rightarrow 0.$

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  • $\begingroup$ How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response $\endgroup$
    – user593295
    Commented Dec 2, 2018 at 15:25
  • $\begingroup$ $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(\log n)^2$ $\endgroup$
    – John_Wick
    Commented Dec 2, 2018 at 15:49
  • $\begingroup$ $Var(\bar X_n)=Var(1/n\sum_{i=1}^n X_i)=1/n^2\sum_i Var(X_i)$ if $X_i$'s are independent. $\endgroup$
    – John_Wick
    Commented Dec 2, 2018 at 15:50

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