Can someone help me understand the variance part of this theorem?

Question

This is distribution of the sample mean when $\sigma$ is known

Theorem 6.1: Let $X_1, ..., X_n$ be a random sample from a distribution with mean $\mu$ and standard deviation $\sigma$

Then Var($\bar{X})$ = $\cfrac {\sigma^2}{n}$

Proof in the book: By independence, Var($\bar{X})$ = Var($\sum_{i=1}^n \cfrac{X_i}{n}$) = $\cfrac{1}{n^2}$Var($\sum_{i=1}^nX_i$) = $\cfrac {1}{n^2}(n\sigma^2) = \cfrac{\sigma^2}{n}$

Now I know sample mean is just ($\sum_{i=1}^n \cfrac{X_i}{n}$)

But where does the $\cfrac {1}{n^2}$ and $n\sigma^2$come from?

Answer 1

If you look at the basic formulas: $E(Y^k)=\int y^kdF(y)$ where $F(y)$ is the distribution function of a random variable $Y$. The variance of $Y$ is $E(Y^2)-E(Y)^2$. Now let $Y=\frac{X_i}{n}$ so $Var(\frac{X_i}{n})=\frac{E(X_i)^2}{n^2}-\frac{E(X_i)^2}{n^2}=\frac{Var(X_i)}{n^2}=\frac{\sigma^2}{n^2}$.
Since the $X_i$ are independent, the variance of the sum is $n\frac{\sigma^2}{n^2}$

Answer 2

Note that $Var(kY)=k^2Var(Y)$, hence when we move $\frac1n$ out from the variance term, we get $\frac1{n^2}$.

Also for independent random variable, the sum of the variance is equal to the variance of the sum. $$Var(\sum_{i=1}^n X_i) = \sum_{i=1}^nVar( X_i)=n\sigma^2$$