Chapter 8.2, Problem 14E

a.

To determine

Verify that $\widehat{\theta }$ is a biased estimator for $\theta$.

Explanation of Solution

Unbiased estimator:

Consider that $\widehat{\theta }$ is point estimator of $\theta$. The $\widehat{\theta }$ is said to be unbiased estimator, if $E\left(\widehat{\theta }\right)=\theta$.

The estimator $\widehat{\theta }=\max \left(Y_1,Y_2,\dots ,Y_n\right)$.

The probability density function of $\widehat{\theta }$ is calculated as follows:

$\begin{array}{c}P\left(\widehat{\theta }\le x\right)=P\left(\max \left(Y_1,Y_2,\dots ,Y_n\right)\le x\right)\\ =P\left(Y_1\le x,Y_2\le x,\dots ,Y_n\le x\right)\\ ={\left(P\left(Y_1\le x\right)\right)}^n\\ =F_{X_1}{\left(X\right)}^n\\ ={\left({\displaystyle \underset{0}{\overset{x}{\int }}\alpha \frac{t^{\alpha -1}}{{\theta }^{\alpha }}}dt\right)}^n\\ =\frac{{\alpha }^n}{{\theta }^{n\alpha }}{\left[\frac{t^{n\alpha }}{{\alpha }^n}\right]}_0^x\\ ={\left(\frac{x}{\theta }\right)}^{n\alpha }\\ f_{\widehat{\theta }}\left(x\right)=\frac{d}{dx}{F}_{X_1}{\left(X\right)}^n\\ =\frac{n\alpha }{{\theta }^{n\alpha }}{x}^{n\alpha -1};x\in \left(0,\theta \right)\end{array}$

The expectation of $\widehat{\theta }$ is computed as follows:

$\begin{array}{c}E\left(\widehat{\theta }\right)={\displaystyle \underset{0}{\overset{\theta }{\int }}x\frac{n\alpha }{{\theta }^{n\alpha }}{x}^{n\alpha -1}}dx\\ =\frac{n\alpha }{{\theta }^{n\alpha }}{\displaystyle \underset{0}{\overset{\theta }{\int }}{x}^{n\alpha }}dx\\ =\left(\frac{n\alpha }{{\theta }^{n\alpha }}\right)\left({\frac{x^{n\alpha +1}}{n\alpha +1}|}_0^{\theta }\right)\\ =\frac{n\alpha }{{\theta }^{n\alpha }}\left(\frac{{\theta }^{n\alpha }\cdot \theta }{n\alpha +1}\right)\\ =\frac{n\alpha \left(\theta \right)}{n\alpha +1}\end{array}$

It is clear that the expectation of $\widehat{\theta }$ is not equal to $\theta$. Therefore, $\widehat{\theta }$ is a biased estimator for $\theta$.

b.

To determine

Obtain a multiple of $\widehat{\theta }$ that is an unbiased estimator of $\theta$.

Answer to Problem 14E

The multiple of $\widehat{\theta }$ that is an unbiased estimator of $\theta$ is $\frac{n\alpha +1}{n\alpha }\widehat{\theta }$.

Explanation of Solution

From Part (a), it is clear that $E\left(\widehat{\theta }\right)=\frac{n\alpha \left(\theta \right)}{n\alpha +1}$.

To be an unbiased estimator, the expected value should be $\theta$.

The multiple of $\widehat{\theta }$ that is an unbiased estimator of $\theta$ is computed as follows:

$\begin{array}{l}E\left(\widehat{\theta }\right)=\frac{n\alpha \left(\theta \right)}{n\alpha +1}\\ E\left(\frac{n\alpha +1}{n\alpha }\widehat{\theta }\right)=\theta \\ \frac{n\alpha +1}{n\alpha }E\left(\widehat{\theta }\right)=\theta \end{array}$

It is clear that $\frac{n\alpha +1}{n\alpha }\widehat{\theta }$ is an unbiased estimator of $\theta$.

c.

To determine

Obtain $\text{MSE}\left(\widehat{\theta }\right)$.

Answer to Problem 14E

The value of $\text{MSE}\left(\widehat{\theta }\right)$ is $\left(\frac{2}{\left(n\alpha +2\right)\left(n\alpha +1\right)}\right){\theta }^2$.

Explanation of Solution

From Part (a), it is clear that $E\left(\widehat{\theta }\right)=E\left(Y_{\left(n\right)}\right)=\frac{n\alpha \left(\theta \right)}{n\alpha +1}$.

The value of $\text{E}\left(\widehat{\theta }\right)$ and $\text{E}\left({\widehat{\theta }}^2\right)$ are computed as follows:

Note that $\text{E}\left(\widehat{\theta }\right)=\text{E}\left(Y_{\left(n\right)}\right)$ and $\text{E}\left({\widehat{\theta }}^2\right)=\text{E}\left(Y_{\left(n\right)}^2\right)$. The random sample $Y_{\left(n\right)}$ follows power family with parameters $n\alpha \text{and }\theta$.

$\begin{array}{c}\text{E}\left(Y\right)={\displaystyle \underset{0}{\overset{\theta }{\int }}y\alpha \frac{y^{\alpha -1}}{{\theta }^{\alpha }}}dy\\ =\frac{\alpha }{{\theta }^{\alpha }}{\displaystyle \underset{0}{\overset{\theta }{\int }}{y}^{\alpha }}dy\\ =\frac{\alpha }{{\theta }^{\alpha }}{\frac{y^{\alpha +1}}{\alpha +1}|}_0^{\theta }\\ =\frac{\alpha }{{\theta }^{\alpha }}\left(\frac{{\theta }^{{}^{\alpha +1}}}{\alpha +1}\right)\\ =\frac{\alpha }{{\theta }^{\alpha }}\left(\frac{{\theta }^{{}^{\alpha +1}}}{\alpha +1}\right)\\ E\left(Y_{\left(n\right)}\right)=\left(\frac{n\alpha }{n\alpha +1}\right)\theta \\ \text{E}\left(Y^2\right)={\displaystyle \underset{0}{\overset{\theta }{\int }}{y}^2\alpha \frac{y^{\alpha -1}}{{\theta }^{\alpha }}}dy\\ =\frac{\alpha }{{\theta }^{\alpha }}{\displaystyle \underset{0}{\overset{\theta }{\int }}{y}^{\alpha +1}}dy\\ =\frac{\alpha }{{\theta }^{\alpha }}{\frac{y^{\alpha +2}}{\alpha +2}|}_0^{\theta }\\ =\frac{\alpha }{{\theta }^{\alpha }}\left(\frac{{\theta }^{{}^{\alpha +2}}}{\alpha +2}\right)\\ E\left(Y_{\left(n\right)}^2\right)=\left(\frac{n\alpha }{n\alpha +2}\right){\theta }^{{}^2}\end{array}$

The value of $\text{MSE}\left(\widehat{\theta }\right)$ is computed as follows:

$\begin{array}{c}\text{MSE}\left(\widehat{\theta }\right)=E\left[{\left(Y_{\left(n\right)}-\theta \right)}^2\right]\\ =E\left(Y_{\left(n\right)}^2\right)-2\theta E\left(Y_{\left(n\right)}\right)+{\theta }^2\\ =\left(\frac{n\alpha }{n\alpha +2}\right){\theta }^2-2\theta \left(\frac{n\alpha }{n\alpha +1}\right)\theta +{\theta }^2\\ ={\theta }^2\left(\frac{n\alpha }{n\alpha +2}-2\frac{n\alpha }{n\alpha +1}+1\right)\\ ={\theta }^2\left(\frac{n\alpha \left(n\alpha +1\right)-2n\alpha \left(n\alpha +2\right)+\left(n\alpha +2\right)\left(n\alpha +1\right)}{\left(n\alpha +2\right)\left(n\alpha +1\right)}\right)\\ ={\theta }^2\left(\frac{n^2{\alpha }^2+n\alpha -2n^2{\alpha }^2-4n\alpha +n^2{\alpha }^2+n\alpha +2n\alpha +2}{\left(n\alpha +2\right)\left(n\alpha +1\right)}\right)\\ =\left(\frac{2}{\left(n\alpha +2\right)\left(n\alpha +1\right)}\right){\theta }^2\end{array}$

The value of $\text{MSE}\left(\widehat{\theta }\right)$ is $\left(\frac{2}{\left(n\alpha +2\right)\left(n\alpha +1\right)}\right){\theta }^2$.