Distribution of a sample average

As first application of the theorem, let us remind ourselves that a sample average $\overline{X}_n$ of $n$ independent variables,

$$\overline{X}_n = \sum_{i=1}^n\frac{1}{n}\,X_i,$$ (4.74)

is normally distributed, since it is a linear combination of $n$ variables $X_i$, with $c_i=1/n$. Then,

$$\overline{X}_n \sim {\cal N}(\mu_{\overline{X}_n}, \sigma_{\overline{X}_n}),$$ (4.75)

$$\mu_{\overline{X}_n} = \sum_{i=1}^n\frac{1}{n}\,\mu = \mu,$$ (4.76)

$$\sigma^2_{\overline{X}_n} = \sum_{i=1}^n \left(\frac{1}{n}\right)^2\sigma^2 = \frac{\sigma^2}{n},$$ (4.77)

$$\sigma_{\overline{X}_n} = \frac{\sigma}{\sqrt{n}}\,.$$ (4.78)

This result, we repeat, is independent of the distribution of $X$ and is already approximately valid for small values of $n$.