Is this proof of $\operatorname{Var}(\overline{x})=\frac{\sigma^2}{N}$ correct? 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
$$\frac1N\sum_{i=1}^N \left[\overline{x}^2-2\mu \overline{x}+\mu^2\right]_i = \frac1N \sum_{i=1}^N(\overline{x}-\mu)_i^2=\frac{\sigma^2}{N}$$
 A: 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:
$$\begin{align*}
&= \dfrac{1}{n^2}\sum_{i=1}^n  E\Big((X_i - \mu)^2\Big)  \\
&= \dfrac{1}{n^2}\sum_{i=1}^n  Var(X_i)  \\
&= \dfrac{1}{n^2}\sum_{i=1}^n  \sigma^2  \\
&= \dfrac{1}{n^2}  n\sigma^2  \\
&= \dfrac{\sigma^2}{n}
\end{align*}$$
A: $$
\begin{align*}
Var(\bar{X}) &= Var\left(\dfrac{1}{n}\sum_{i=1}^n X_i\right)\\
&= \dfrac{1}{n^2}Var\left(\sum_{i=1}^n X_i\right)
\end{align*}$$
variance of sum is equal to sum of variances because $X_i$ are independent
$$
\begin{align*}
&= \dfrac{1}{n^2}\sum_{i=1}^n Var(X_i) \\
&= \dfrac{n\sigma^2}{n^2}\\
&= \dfrac{\sigma^2}{n}
\end{align*}
$$
