Question

Let ${\mathit{x}}_{\mathbf{1}}\mathbf{,}{\mathit{x}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{x}}_{\mathbf{n}}$be independent random variables, each with density function $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$, expected value$\mathit{\mu }$ , and variance${\mathit{\sigma }}_{\mathbf{2}}$ . Define the sample meanby.$\stackrel{\mathbf{-}}{\mathbf{x}}\mathbf{=}\sum _{i=1}^n{x}_i$Showthat$\mathit{E}\mathbf{\left(}\stackrel{\mathbf{-}}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{\mu }$,and .$\mathit{v}\mathit{a}\mathit{r}\mathbf{\left(}\stackrel{\mathbf{-}}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{\sigma }}^{\mathbf{2}}}{\mathbf{n}}$ (See Problems $\mathbf{5}\mathbf{.9}\mathbf{,}\mathbf{5}\mathbf{.13}$and$\mathbf{6}\mathbf{.15}$.)

Short answer

The statement has been proven.

Step-by-step solution

Given Information

$x_1,x_2,\mathrm{...},x_n$ be the independent random variables, each with density function$f\left(x\right)$ .

Definition of the Arithmetic mean.

The arithmetic mean is the sum of all the values divided by the total number of values.

Prove the statement.

Let $x_1,x_2,\mathrm{...},x_n$be the independent random variable.

The mean is given below.

$x_1,x_2,\mathrm{...},x_n\begin{array}{c}E\left(\overline{x}\right)=E\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\\ =\frac{1}{n}\sum _{i=1}^{n}E\left(x_i\right)\\ =\frac{1}{n}\cdot n\mu \\ =\mu \end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{Var}\left(\overline{x}\right)=\mathrm{Var}\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\\ =\frac{1}{n^2}\sum _{i=1}^n\mathrm{Var}\left(x_i\right)\\ =\frac{1}{n^2}\cdot n{\sigma }^2\\ =\frac{{\sigma }^2}{n}\end{array}$

Hence the statement has been proven.