Expected Value and Variance of the Mean of a Random Sample

Topics: Statistics - Probability - Random Sample

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(theorem)

Let $X_1,\dots ,X_n$ be a random sample such that $\mathbb{E}[X{}_i]=\mu$ y $\mathrm{Var}(X_i)={\sigma }^2$ for every $X_i$.

Concerning the sampling distribution of the mean $\overline{X}$ of our random sample, then:

$$\begin{array}{rl}\mathbb{E}\left[\overline{X}\right]& =\mu \\ \mathrm{Var}\left(\overline{X}\right)& =\frac{{\sigma }^2}{n}\end{array}$$

Proof

See that for the expected value:

$$\begin{array}{rl}\mathbb{E}\left[\overline{X}\right]& =\mathbb{E}\left[\frac{1}{n}\sum _{i=1}^n{X}_i\right]\\ & =\mathbb{E}\left[\frac{1}{n}(X_1+\cdots +X_n)\right]\\ & =\frac{1}{n}\left(\mathbb{E}[X_1]+\cdots +\mathbb{E}[X_n]\right)\\ & =\frac{1}{n}(\mu +\cdots +\mu )\\ & =\frac{n\mu }{n}\\ & =\mu \end{array}$$

As for the variance:

$$\begin{array}{rl}\mathrm{Var}\left(\overline{X}\right)& =\mathrm{Var}\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)\\ & =\mathrm{Var}\left(\frac{1}{n}[X_1+\cdots +X_n]\right)\\ & =\frac{1}{n^2}\left[\mathrm{Var}(X_1)+\cdots +\mathrm{Var}(X_n)\right]\\ & =\frac{1}{n^2}[{\sigma }^2+\cdots +{\sigma }^2]\\ & =\frac{n{\sigma }^2}{n^2}\\ & =\frac{{\sigma }^2}{n}\end{array}$$