Is this proof of $\operatorname{Var}(\overline{x})=\frac{\sigma^2}{N}$ correct?

Question

Starting from $\operatorname{Var}(\overline{x})$ 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 $$\frac1N\sum_{i=1}^N \left[\overline{x}^2-2\mu \overline{x}+\mu^2\right]_i = \frac1N \sum_{i=1}^N(\overline{x}_N-\mu)_i^2=\frac{\sigma^2}{N}$$

Answer 1

Let $X,Y$ be random variables with $\mathbb{E}X^{2}<\infty$ and $\mathbb{E}Y^{2}<\infty$. Then we have the following rules.

• $\text{Var}\left(aX+b\right)=a^{2}\text{Var}X$

• If moreover $X$ and $Y$ are independent then $\text{Var}\left(X+Y\right)=\text{Var}X+\text{Var}Y$.

Applying this on iid $X_{1},\cdots,X_{N}$ with $\mathbb{E}X_{1}^{2}<\infty$ we find:$$\text{Var}\overline{X}=\text{Var}\frac{1}{N}\left(X_{1}+\cdots+X_{N}\right)=$$$$\frac{1}{N^{2}}\text{Var}\left(X_{1}+\cdots+X_{N}\right)=\frac{1}{N^{2}}\left[\text{Var}X_{1}+\cdots+\text{Var}X_{N}\right]=\frac{\text{Var}X_{1}}{N}$$