# Expectation of the mean of the sum of random variables [closed]

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, ChristopherOct 15 '18 at 8:53

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• What have you tried? Fill the ? in $S_m^2=\sum_{i,j=1}^m ?$ so you can take $\langle S_m^2 \rangle$. – AHusain Oct 15 '18 at 0:18
• 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. – 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:
$$Var(\frac{1}{m} \sum_{i=1}^m X_i,~~ m = 1,2,\ldots,M) = m \sigma^2$$ $$Var(S_m) = \frac{m \sigma^2}{m^2}$$ $$Var(S_m) = \frac{\sigma^2}{m}$$
• These answers are wrong because you didn't divide by $m$ when considering $S_m$. – Shalop Oct 15 '18 at 1:33
• So it should be $\mathbb {E}(S_m)=\mu$ and $Var(S_m)=\frac {\sigma^2}{m}$. – Jiexiong687691 Oct 15 '18 at 1:37