If $X_i$'s are independent and identified random variables, each with mean $\mu$ and variance $\sigma^2$. Let's say $S_m = \frac{1}{m} \sum_{i=1}^m X_i,~~ m = 1,2,\ldots,M.$ What are the values of $\mathbb {E}[S_m]$ and $Var(S_m)$?


closed as off-topic by Shalop, StubbornAtom, Leucippus, José Carlos Santos, Christopher Oct 15 '18 at 8:53

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  • $\begingroup$ What have you tried? Fill the ? in $S_m^2=\sum_{i,j=1}^m ?$ so you can take $\langle S_m^2 \rangle$. $\endgroup$ – AHusain Oct 15 '18 at 0:18
  • $\begingroup$ I got $\mathbb {E}[S_m]=\mathbb {E}[\frac{1}{m} \sum_{i=1}^m X_i]=\frac{1}{m} \sum_{i=1}^m \mathbb {E}[X_i]=\mu$. But it's too routine to believe it's true. $\endgroup$ – Jiexiong687691 Oct 15 '18 at 0:23

Recall that $E(X+Y) = E(X) + E(Y)$ hence we have:
$$E(S_m) = \frac{m*\mu}{m}$$ $$E(S_m) = \mu $$ Recall that $Var(X+Y) = Var(X) + Var(Y)$. Also recall that $Var(cX) - c^2 Var(x)$ hence we have:
$$Var(\frac{1}{m} \sum_{i=1}^m X_i,~~ m = 1,2,\ldots,M) = m \sigma^2$$ $$Var(S_m) = \frac{m \sigma^2}{m^2}$$ $$Var(S_m) = \frac{\sigma^2}{m}$$

I hope that helps.


  • $\begingroup$ These answers are wrong because you didn't divide by $m$ when considering $S_m$. $\endgroup$ – Shalop Oct 15 '18 at 1:33
  • $\begingroup$ @Shalop You are right. I will fix me post. $\endgroup$ – Bob Oct 15 '18 at 1:36
  • $\begingroup$ So it should be $\mathbb {E}(S_m)=\mu$ and $Var(S_m)=\frac {\sigma^2}{m}$. $\endgroup$ – Jiexiong687691 Oct 15 '18 at 1:37

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