$\bar{X}=\frac{1}{10} \sum X_i$ => $\sigma^2 = \frac{1}{10} \cdot 1$

Question

In my textbook example I have:

Let $X_1,...,X_10$ be independent $\mathcal{N}(0,1)$ distributed and let $\bar{X}=\frac{1}{10} \sum X_i$

Then $Var(\bar{X})=\sigma^2 = \frac{1}{10} \cdot 1$

Why is that? Is it a theorem or how do I show this?

Answer 1

Two things worth knowing:

• If $k$ is a finite constant then $\text{Var}(k X)=k^2\text{Var}(X)$

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

The second will tell you that here $\text{Var}(\sum X_i)=1+1+1+1+1+1+1+1+1+1=10$

and so the first will tell you $\text{Var}(\bar X)=\text{Var}(\frac{1}{10}\sum X_i)=\frac1{10^2}\times 10= \frac{1}{10}$