Question

Define s by the equation.${\mathit{s}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{/}\mathit{n}\mathbf{)}\sum _{i=1}^n{(x_i-\overline{x})}^2$Show that the expected valueof.${\mathit{s}}^{\mathbf{2}}\mathbf{\text{is}}\mathbf{\left[}\mathbf{\right(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{/}\mathit{n}\mathbf{]}{\mathit{\sigma }}^{\mathbf{2}}$Hints: Write

$\begin{array}{c}{\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\overline{\mathbf{x}}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{\left[}\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{-}\mathbf{(}\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{\right]}}^{\mathbf{2}}\\ \mathbf{=}{\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{\mu }\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{\mu }\mathbf{)}{\mathbf{(}\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{+}\overline{\mathbf{(}\mathbf{x}}\mathbf{-}\mathbf{\mu }\mathbf{)}}^{\mathbf{2}}\end{array}$

Find the average value of the first term from the definition of${\mathit{\sigma }}^{\mathbf{2}}$and the average value of the third term from Problem 2. To find the average value of the middle term write

$\begin{array}{c}\mathbf{(}\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{=}\mathbf{(}\frac{{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{\cdots }\mathbf{+}{\mathbf{x}}_{\mathbf{n}}}{\mathbf{n}}\mathbf{-}\mathbf{\mu }\mathbf{)}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{[}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{+}\mathbf{\cdots }\mathbf{+}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{]}\end{array}$

Show by Problemthat

$\begin{array}{c}\mathbf{E}\mathbf{\left[}\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{(}{\mathbf{x}}_{\mathbf{j}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{\right]}\mathbf{=}\mathbf{E}\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{\mu }\mathbf{)}\mathbf{E}\mathbf{(}{\mathbf{x}}_{\mathbf{j}}\mathbf{-}\mathbf{\mu }\mathbf{)}\\ \mathbf{=}\mathbf{0}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for}}\mathbf{i}\mathbf{\ne }\mathbf{j}\end{array}$

andevaluate$\mathbf{6}\mathbf{.14}$ (same as the first term). Collect terms to find

$\mathit{E}\mathbf{\left(}{\mathbf{s}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{n}}{\mathit{\sigma }}^{\mathbf{2}}$

Short answer

The expected value of$s^2$ is.$s^{2}E\left(s^2\right)=\frac{n-1}{n}{\sigma }^2$

Step-by-step solution

Given Information

The equations mentioned below:

$\begin{array}{c}{(x_i-\overline{x})}^2={\left[(x_i-\mu )-(\overline{x}-\mu )\right]}^2\\ ={(x_i-\mu )}^2-2(x_i-\mu )(\overline{x}-\mu )+{(\overline{x}-\mu )}^2\\ E{(x_i-\overline{x})}^2=E{(x_i-\mu )}^2-2E\left[(x_i-\mu )(\overline{x}-\mu )\right]+E{(\overline{x}-\mu )}^2.\end{array}$

Definition of Expectation.

The expected value (also known as expectation) is a generalisation of the weighted average in probability theory.

Step 3:Evaluate the middle term.

Solve for middle term.

$\begin{array}{c}(\overline{x}-\mu )=(\frac{x_1+x_2+\cdots +x_n}{n}-\mu )\\ =\frac{1}{n}\sum _{j=1}^n(x_j-\mu )\end{array}$

Multiply with $(x_i-\mu )$to evaluate expectation value.

$\begin{array}{c}E\left[(x_i-\mu )(\overline{x}-\mu )\right]=\frac{1}{n}\sum _{j=1}^{n}E\left[(x_i-\mu )(x_j-\mu )\right]\\ =\frac{1}{n}E\left[{(x_i-\mu )}^2\right]\\ =\frac{{\sigma }^2}{n}\end{array}$

From problem, for.$i\ne j$

$E\left[(x_i-\mu )(x_j-\mu )\right]=0$

Calculate expected value of.s2 

Solve for expected value of.${(x_i-\overline{x})}^2$

$\begin{array}{c}E{(x_i-\overline{x})}^2={\sigma }^2-2\frac{{\sigma }^2}{n}+\frac{{\sigma }^2}{n}\\ =\frac{n-1}{n}{\sigma }^2\\ E\left(s^2\right)=\frac{1}{n}\sum _{i=1}^{n}E{(x_i-\overline{x})}^2=\frac{n-1}{n}{\sigma }^2\end{array}$

Hence, the expected value of$s^2$ is$E\left(s^2\right)=\frac{n-1}{n}{\sigma }^2$ .