Starting from $Var(\overline{x})$ I am trying to algebraically show that it is equal to $\frac{\sigma^2}{N}$ using the fact that the variance of the sum equals to the sum of variances. I start by $$\operatorname{Var}(\overline{x})=\operatorname{Var}\left(\frac1N\sum_{i=1}^Nx_i\right)$$ then $$\frac1N\sum_{i=1}^N \left(\frac1N \sum_{i=1}^Nx_i-\mu\right)_i^2 = \frac1N \sum_{i=1}^N \left[\frac1{N^2} \left(\sum_{i=1}^N x_i\right)^2-\frac{2\mu}{N}\sum_{i=1}^N x_i+\mu^2 \right]_i$$ which becomes $$\frac1N\sum_{i=1}^N \left[\overline{x}^2-2\mu \overline{x}+\mu^2\right]_i = \frac1N \sum_{i=1}^N(\overline{x}-\mu)_i^2=\frac{\sigma^2}{N}$$
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1$\begingroup$ what is $M$? What is the first expression on 2nd line? $\endgroup$– bdeonovicCommented Aug 13, 2016 at 13:15
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$\begingroup$ what is $\bar{x}_N$? $\endgroup$– David CCommented Aug 13, 2016 at 13:29
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$\begingroup$ Sample mean where a sample has $N$ values of the random variable. $\overline{x_N}=\frac1N\sum_{i=1}^Nx_i$ $\endgroup$– boneheadCommented Aug 13, 2016 at 13:34
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1$\begingroup$ How is $\bar{x}$ different than $\bar{x}_N$? $\endgroup$– David CCommented Aug 13, 2016 at 13:42
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1$\begingroup$ How do you go from the right hand side of the 1st line to the left-hand side of the 2nd line? And how do you go from the middle of the 3rd line to the right-hand side of the 3rd line? Those don't make any sense at all, thereby rendering your proof incorrect. You made one error with the first of these, and you made an offsetting error with the second of these. $\endgroup$– Mark L. StoneCommented Aug 13, 2016 at 14:37
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2 Answers
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It doesn't really seem like your proof makes sense, but it might just be that you are skipping steps, which is making it difficult to understand. Here is a more complete proof of $Var(\bar{X}) = \sigma^2/n$
$$ \begin{align*} Var(\bar{X}) &= E\Big( (\bar{X} - \mu)^2 \Big) \\ &= E\Big( \big(\dfrac{1}{n}\sum_{i=1}^n X_i - \mu\big)^2 \Big) \\ &= E\Big( \big(\dfrac{1}{n}(X_1+\ ...\ + X_n) - \mu\big)^2 \Big) \\ &= E\Big( \big(\dfrac{1}{n}\big((X_1-\mu)+\ ...\ + (X_n-\mu)\big)\big)^2 \Big) \\ &= \dfrac{1}{n^2}E\Big( \big((X_1-\mu)+\ ...\ + (X_n-\mu)\big)^2 \Big) \\ &= \dfrac{1}{n^2}E\Big( \sum_{i=1}^n \sum_{j=1}^n(X_i - \mu)(X_j-\mu) \Big) \\ &= \dfrac{1}{n^2}\sum_{i=1}^n \sum_{j=1}^n E\Big((X_i - \mu)(X_j-\mu)\Big) \\ \end{align*}$$ Note, however, that since the $X_i$s are independent, then if $i\neq j$, then $E((X_i - \mu)(X_j - \mu)) =Cov(X_i,X_j) = 0$. Thus: $$\begin{align*} &= \dfrac{1}{n^2}\sum_{i=1}^n E\Big((X_i - \mu)^2\Big) \\ &= \dfrac{1}{n^2}\sum_{i=1}^n Var(X_i) \\ &= \dfrac{1}{n^2}\sum_{i=1}^n \sigma^2 \\ &= \dfrac{1}{n^2} n\sigma^2 \\ &= \dfrac{\sigma^2}{n} \end{align*}$$
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$$ \begin{align*} Var(\bar{X}) &= Var\left(\dfrac{1}{n}\sum_{i=1}^n X_i\right)\\ &= \dfrac{1}{n^2}Var\left(\sum_{i=1}^n X_i\right) \end{align*}$$ variance of sum is equal to sum of variances because $X_i$ are independent $$ \begin{align*} &= \dfrac{1}{n^2}\sum_{i=1}^n Var(X_i) \\ &= \dfrac{n\sigma^2}{n^2}\\ &= \dfrac{\sigma^2}{n} \end{align*} $$
answered Aug 13, 2016 at 12:48
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$\begingroup$ I think the proof I posted is perhaps more fundamental because it doesn't use any properties of variance like Bienaymé's formula. $\endgroup$– boneheadCommented Aug 13, 2016 at 12:57
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$\begingroup$ @ahra But you have. Otherwise where are all the $\sum_i\sum_j$ terms in the second step of your derivation? $\endgroup$ Commented Aug 13, 2016 at 13:12
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$\begingroup$ Well I did use the fact that $X_i$ since I'm studying this in context of experimental errors, but I don't see where I used any variance properties. Regarding the sum multiplication terms I'm not sure exactly where should they appear? $\endgroup$– boneheadCommented Aug 13, 2016 at 13:26