Chapter 4.9, Problem 3E

Let X₁ and X₂ be independent, each with unknown mean μ and known variance σ² = 1.

a.    Let $\widehat{{\mu }_1}=\frac{X_1+X_2}{2}$ . Find the bias, variance, and mean squared error of $\widehat{{\mu }_1}$.

b.    Let $\widehat{{\mu }_2}=\frac{X_1+2X_2}{3}$  Find the bias, variance, and mean squared error of $\widehat{{\mu }_2}$.

c.    Let $\widehat{{\mu }_3}=\frac{X_1+X_2}{4}$ Find the bias, variance, and mean squared error of $\widehat{{\mu }_3}$.

d.    For what values of µ does $\widehat{{\mu }_3}$ have smaller mean squared error than $\widehat{{\mu }_1}$?

e.    For what values of µ does $\widehat{{\mu }_3}$ have smaller mean squared error than $\widehat{{\mu }_2}$

To determine

Find the bias, variance and mean squared error of ${\widehat{\mu }}_1$, if ${\widehat{\mu }}_1=\frac{X_1+X_2}{2}$

Answer to Problem 3E

The bias of ${\widehat{\mu }}_1$ is 0.

The variance of ${\widehat{\mu }}_1$ is, $V\left({\widehat{\mu }}_1\right)=\frac{1}{2}$.

The mean square error of ${\widehat{\mu }}_1$ is, $MSE\left({\widehat{\mu }}_1\right)=\frac{1}{2}$.

Explanation of Solution

Given info:

The random variables $X_1\text{and }{X}_2$ are independent, each with unknown mean $\mu$ and known variance ${\sigma }^2=1$

Calculation:

Bias:

Assume that $\theta$ is a parameter and $\widehat{\theta }$ an estimator of $\theta$. The bias of the estimator $\widehat{\theta }$ is, ${\mu }_{\widehat{\theta }}-\theta$, that is the difference between the mean of the estimator and the true value.

Mean squared error:

The mean squared error (MSE) of $\widehat{\theta }$ is,

$MS{E}_{\widehat{\theta }}={\left({\mu }_{\widehat{\theta }}-\theta \right)}^2+{\sigma }_{\widehat{\theta }}^2$

The mean of ${\widehat{\mu }}_1$ is denoted as $E\left({\widehat{\mu }}_1\right)$.

Substitute ${\widehat{\mu }}_1=\frac{X_1+X_2}{2}$ in $E\left({\widehat{\mu }}_1\right)$.

$\begin{array}{c}E\left({\widehat{\mu }}_1\right)=E\left(\frac{X_1+X_2}{2}\right)\\ =\frac{E\left(X_1\right)+E\left(X_2\right)}{2}\\ =\frac{{\mu }_{X_1}+{\mu }_{X_2}}{2}\\ =\frac{\mu +\mu }{2}\\ =\mu \end{array}$

The bias of ${\widehat{\mu }}_1$ is,

$\begin{array}{c}E\left({\widehat{\mu }}_1\right)-\mu =\mu -\mu \\ =0\end{array}$

Thus, the bias of ${\widehat{\mu }}_1$ is 0.

The variance of ${\widehat{\mu }}_1$ is denoted as $V\left({\widehat{\mu }}_1\right)$.

Substitute ${\widehat{\mu }}_1=\frac{X_1+X_2}{2}$ in $V\left({\widehat{\mu }}_1\right)$.

$\begin{array}{c}V\left({\widehat{\mu }}_1\right)=V\left(\frac{X_1+X_2}{2}\right)\\ =\frac{V\left(X_1\right)+V\left(X_2\right)}{4}\\ =\frac{{\sigma }^2+{\sigma }^2}{4}\\ =\frac{{\sigma }^2}{2}\end{array}$

Substitute ${\sigma }^2=1$ in the above equation.

$V\left({\widehat{\mu }}_1\right)=\frac{1}{2}$

Thus, the variance of ${\widehat{\mu }}_1$ is, $V\left({\widehat{\mu }}_1\right)=\frac{1}{2}$.

The mean squared error of ${\widehat{\mu }}_1$ is the sum of the variance and the square of the bias.

Then,

$\begin{array}{c}MSE\left({\widehat{\mu }}_1\right)=\frac{1}{2}+0^2\\ =\frac{1}{2}\end{array}$

Thus, the mean square error of ${\widehat{\mu }}_1$ is, $MSE\left({\widehat{\mu }}_1\right)=\frac{1}{2}$.

To determine

Find the bias, variance and mean squared error of ${\widehat{\mu }}_2$, if ${\widehat{\mu }}_2=\frac{X_1+2X_2}{3}$

Answer to Problem 3E

The bias of ${\widehat{\mu }}_2$ is 0.

The variance of ${\widehat{\mu }}_2$ is, $V\left({\widehat{\mu }}_2\right)=\frac{5}{9}$.

The mean square error of ${\widehat{\mu }}_2$ is, $MSE\left({\widehat{\mu }}_2\right)=\frac{5}{9}$.

Explanation of Solution

Calculation:

The mean of ${\widehat{\mu }}_2$ is denoted as $E\left({\widehat{\mu }}_2\right)$.

Substitute ${\widehat{\mu }}_2=\frac{X_1+2X_2}{3}$ in $E\left({\widehat{\mu }}_2\right)$.

$\begin{array}{c}E\left({\widehat{\mu }}_2\right)=E\left(\frac{X_1+2X_2}{3}\right)\\ =\frac{E\left(X_1\right)+2E\left(X_2\right)}{3}\\ =\frac{{\mu }_{X_1}+2{\mu }_{X_2}}{3}\\ =\frac{\mu +2\mu }{3}\\ =\mu \end{array}$

The bias of ${\widehat{\mu }}_2$ is,

$\begin{array}{c}E\left({\widehat{\mu }}_2\right)-\mu =\mu -\mu \\ =0\end{array}$

Thus, the bias of ${\widehat{\mu }}_2$ is 0.

The variance of ${\widehat{\mu }}_2$ is denoted as $V\left({\widehat{\mu }}_2\right)$.

Substitute ${\widehat{\mu }}_2=\frac{X_1+2X_2}{3}$ in $V\left({\widehat{\mu }}_2\right)$.

$\begin{array}{c}V\left({\widehat{\mu }}_2\right)=V\left(\frac{X_1+2X_2}{3}\right)\\ =\frac{V\left(X_1\right)+4V\left(X_2\right)}{9}\\ =\frac{{\sigma }^2+4{\sigma }^2}{9}\\ =\frac{5{\sigma }^2}{9}\end{array}$

Substitute ${\sigma }^2=1$ in the above equation.

$V\left({\widehat{\mu }}_2\right)=\frac{5}{9}$

Thus, the variance of ${\widehat{\mu }}_2$ is, $V\left({\widehat{\mu }}_2\right)=\frac{5}{9}$.

The mean squared error of ${\widehat{\mu }}_2$ is the sum of the variance and the square of the bias.

Then,

$\begin{array}{c}MSE\left({\widehat{\mu }}_2\right)=\frac{5}{9}+0^2\\ =\frac{5}{9}\end{array}$

Thus, the mean square error of ${\widehat{\mu }}_2$ is, $MSE\left({\widehat{\mu }}_2\right)=\frac{5}{9}$.

To determine

Find the bias, variance and mean squared error of ${\widehat{\mu }}_3$, if ${\widehat{\mu }}_3=\frac{X_1+X_2}{4}$

Answer to Problem 3E

The bias of ${\widehat{\mu }}_3$ is $\frac{-\mu }{2}$.

The variance of ${\widehat{\mu }}_3$ is, $V\left({\widehat{\mu }}_3\right)=\frac{{\sigma }^2}{8}$.

The mean square error of ${\widehat{\mu }}_3$ is, $MSE\left({\widehat{\mu }}_3\right)=\frac{2{\mu }^2+1}{8}$.

Explanation of Solution

Calculation:

The mean of ${\widehat{\mu }}_3$ is denoted as $E\left({\widehat{\mu }}_3\right)$.

Substitute ${\widehat{\mu }}_3=\frac{X_1+X_2}{4}$ in $E\left({\widehat{\mu }}_3\right)$.

$\begin{array}{c}E\left({\widehat{\mu }}_3\right)=E\left(\frac{X_1+X_2}{4}\right)\\ =\frac{E\left(X_1\right)+E\left(X_2\right)}{4}\\ =\frac{{\mu }_{X_1}+{\mu }_{X_2}}{4}\\ =\frac{\mu +\mu }{4}\\ =\frac{\mu }{2}\end{array}$

The bias of ${\widehat{\mu }}_3$ is,

$\begin{array}{c}E\left({\widehat{\mu }}_3\right)-\mu =\frac{\mu }{2}-\mu \\ =-\frac{\mu }{2}\end{array}$

Thus, the bias of ${\widehat{\mu }}_3$ is $\frac{-\mu }{2}$.

The variance of ${\widehat{\mu }}_3$ is denoted as $V\left({\widehat{\mu }}_3\right)$.

Substitute ${\widehat{\mu }}_3=\frac{X_1+X_2}{4}$ in $V\left({\widehat{\mu }}_3\right)$.

$\begin{array}{c}V\left({\widehat{\mu }}_3\right)=V\left(\frac{X_1+X_2}{4}\right)\\ =\frac{V\left(X_1\right)+V\left(X_2\right)}{16}\\ =\frac{{\sigma }^2+{\sigma }^2}{16}\\ =\frac{{\sigma }^2}{8}\end{array}$

Substitute ${\sigma }^2=1$ in the above equation.

$V\left({\widehat{\mu }}_3\right)=\frac{1}{8}$

Thus, the variance of ${\widehat{\mu }}_3$ is, $V\left({\widehat{\mu }}_3\right)=\frac{1}{8}$.

The mean squared error of ${\widehat{\mu }}_3$ is the sum of the variance and the square of the bias.

Then,

$\begin{array}{c}MSE\left({\widehat{\mu }}_3\right)={\left(\frac{-\mu }{2}\right)}^2+\frac{1}{8}\\ =\frac{{\mu }^2}{4}+\frac{1}{8}\\ =\frac{2{\mu }^2+1}{8}\end{array}$

Thus, the mean square error of ${\widehat{\mu }}_3$ is, $MSE\left({\widehat{\mu }}_3\right)=\frac{2{\mu }^2+1}{8}$.

To determine

Find the values of $\mu$ so that ${\widehat{\mu }}_3$ have smaller mean squared error than ${\widehat{\mu }}_1$.

Answer to Problem 3E

The values of $\mu$ so that ${\widehat{\mu }}_3$ have smaller mean squared error than ${\widehat{\mu }}_1$ is $-1.2247<\mu <1.2247$.

Explanation of Solution

Calculation:

The mean square error of ${\widehat{\mu }}_1$ is, $MSE\left({\widehat{\mu }}_1\right)=\frac{1}{2}$ and the mean square error of ${\widehat{\mu }}_3$ is, $MSE\left({\widehat{\mu }}_3\right)=\frac{2{\mu }^2+1}{8}$.

The MSE of ${\widehat{\mu }}_3$ will be smaller than MSE of ${\widehat{\mu }}_1$, only if

$\frac{2{\mu }^2+1}{8}<\frac{1}{2}$

On simplification,

${\mu }^2<\frac{4-1}{2}$

That is,

$-\sqrt{\frac{3}{2}}<\mu <\sqrt{\frac{3}{2}}$

Implies,

$-1.2247<\mu <1.2247$

Thus, the values of $\mu$ so that ${\widehat{\mu }}_3$ have smaller mean squared error than ${\widehat{\mu }}_1$ is $-1.2247<\mu <1.2247$.

To determine

Find the values of $\mu$ so that ${\widehat{\mu }}_3$ have smaller mean squared error than ${\widehat{\mu }}_2$.

Answer to Problem 3E

The values of $\mu$ so that ${\widehat{\mu }}_3$ have smaller mean squared error than ${\widehat{\mu }}_2$ is $-1.3123<\mu <1.3123$.

Explanation of Solution

Calculation:

The mean square error of ${\widehat{\mu }}_3$ is, $MSE\left({\widehat{\mu }}_3\right)=\frac{2{\mu }^2+1}{8}$ and the mean square error of ${\widehat{\mu }}_2$ is, $MSE\left({\widehat{\mu }}_2\right)=\frac{5}{9}$.

The MSE of ${\widehat{\mu }}_3$ will be smaller than MSE of ${\widehat{\mu }}_2$, only if

$\frac{2{\mu }^2+1}{8}<\frac{5}{9}$

On simplification,

$\begin{array}{c}{\mu }^2<\frac{\frac{40}{9}-1}{2}\\ =\frac{\left(40-9\right)}{\left(9\right)\left(2\right)}\\ =\frac{31}{18}\end{array}$

That is,

$-\sqrt{\frac{31}{18}}<\mu <\sqrt{\frac{31}{18}}$

Implies,

$-1.3123<\mu <1.3123$

Thus, the values of $\mu$ so that ${\widehat{\mu }}_3$ have smaller mean squared error than ${\widehat{\mu }}_2$ is $-1.3123<\mu <1.3123$.