Chapter 8.2, Problem 17E

If Y has a binomial distribution with parameters n and p, then ${\hat{p}}_1=Y/n$ is an unbiased estimator of p. Another estimator of p is ${\hat{p}}_2=(Y+1)/(n+2)$.

1. a Derive the bias of ${\hat{p}}_2$.
2. b Derive $\text{MSE}({\hat{p}}_1)$ and $\text{MSE}({\hat{p}}_2)$.
3. c For what values of p is $\text{MSE}({\hat{p}}_1)<\text{MSE}({\hat{p}}_2)$?

To determine

Compute the bias of ${\widehat{p}}_2$.

Answer to Problem 17E

The bias of ${\widehat{p}}_2$ is $\frac{1-2p}{n+2}$.

Explanation of Solution

Unbiased estimator:

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

The mean and variance of the binomial distribution for random variable Y are given as follows:

$\begin{array}{c}E\left(Y\right)=np\\ V\left(Y\right)=E\left(Y^2\right)-{\left(E\left(Y\right)\right)}^2\\ =np\left(1-p\right)\\ E\left(Y^2\right)=V\left(Y\right)+{\left(E\left(Y\right)\right)}^2\\ =np\left(1-p\right)+{\left(np\right)}^2\end{array}$

The mean square error is as follows:

$\begin{array}{c}MSE\left({\widehat{p}}_2\right)=E{\left(\widehat{\theta }-\theta \right)}^2\\ =Var\left(\widehat{\theta }\right)+{\left(b\left(\widehat{\theta }\right)\right)}^2\end{array}$

The expected value for ${\widehat{p}}_2$ is as follows:

$\begin{array}{c}E\left({\widehat{p}}_2\right)=E\left(\frac{Y+1}{n+2}\right)\\ =E\left(\frac{Y}{n+2}\right)+E\left(\frac{1}{n+2}\right)\\ =\frac{1}{n+2}E\left(Y\right)+\frac{1}{n+2}\\ =\frac{np}{n+2}+\frac{1}{n+2}\\ =\frac{np+1}{n+2}\\ b\left({\widehat{p}}_2\right)=E\left({\widehat{p}}_2\right)-p\\ =\frac{np+1}{n+2}-p\\ =\frac{np+1-\left(n+2\right)\left(p\right)}{n+2}\\ =\frac{np+1-np-2p}{n+2}\\ =\frac{1-2p}{n+2}\end{array}$

Therefore, the bias of ${\widehat{p}}_2$ is $\frac{1-2p}{n+2}$.

To determine

Compute $MSE\left({\widehat{p}}_1\right)\text{ and }MSE\left({\widehat{p}}_2\right)$.

Answer to Problem 17E

$\begin{array}{l}MSE\left({\widehat{p}}_1\right)\text{=}\frac{1}{n}p\left(1-p\right)\\ MSE\left({\widehat{p}}_2\right)=\frac{p^2\left(4-n\right)+p\left(n-4\right)+1}{{\left(n+2\right)}^2}\end{array}$.

Explanation of Solution

It is to be noted that ${\widehat{p}}_1$ is an unbiased estimator of p. $MSE\left({\widehat{p}}_1\right)$ is computed as follows:

$\begin{array}{c}MSE\left({\widehat{p}}_1\right)\text{=}Var\left({\widehat{p}}_1\right)\\ =Var\left(\frac{Y}{n}\right)\\ =\frac{1}{n^2}Var\left(Y\right)\\ =\frac{1}{n^2}np\left(1-p\right)\\ =\frac{1}{n}p\left(1-p\right)\end{array}$.

It is to be noted that ${\widehat{p}}_2$ is a biased estimator of p. $MSE\left({\widehat{p}}_2\right)$ is computed as follows:

$\begin{array}{c}MSE\left({\widehat{p}}_2\right)\text{=}Var\left({\widehat{p}}_2\right)+{\left(b\left({\widehat{p}}_2\right)\right)}^2\\ =Var\left(\frac{Y+1}{n+2}\right)+{\left(\frac{1-2p}{n+2}\right)}^2\\ =\frac{1}{{\left(n+2\right)}^2}Var\left(Y\right)+{\left(\frac{1-2p}{n+2}\right)}^2\\ =\frac{1}{{\left(n+2\right)}^2}np\left(1-p\right)+{\left(\frac{1-2p}{n+2}\right)}^2\\ =\frac{np\left(1-p\right)+1-4p+4p^2}{{\left(n+2\right)}^2}\\ =\frac{np-n{p}^2+1-4p+4p^2}{{\left(n+2\right)}^2}\\ =\frac{p^2\left(4-n\right)+p\left(n-4\right)+1}{{\left(n+2\right)}^2}\end{array}$.

$MSE\left({\widehat{p}}_1\right)\text{ and }MSE\left({\widehat{p}}_2\right)$ are obtained as $\frac{1}{n}p\left(1-p\right)$ and $\frac{p^2\left(4-n\right)+p\left(n-4\right)+1}{{\left(n+2\right)}^2}$, respectively.

To determine

Identify the values of p for which $MSE\left({\widehat{p}}_1\right)<MSE\left({\widehat{p}}_2\right)$.

Answer to Problem 17E

The values of p for which $MSE\left({\widehat{p}}_1\right)<MSE\left({\widehat{p}}_2\right)$ is $\frac{1}{2}\pm \sqrt{\frac{n+1}{8n+4}}$.

Explanation of Solution

Equate $MSE\left({\widehat{p}}_1\right)<MSE\left({\widehat{p}}_2\right)$ as given below:

$\begin{array}{c}MSE\left({\widehat{p}}_1\right)<MSE\left({\widehat{p}}_2\right)\\ \frac{1}{n}p\left(1-p\right)<\frac{np\left(1-p\right)+1-4p+4p^2}{{\left(n+2\right)}^2}\\ \frac{np\left(1-p\right)+1-4p+4p^2}{{\left(n+2\right)}^2}-\frac{1}{n}p\left(1-p\right)>0\\ \frac{n^{2}p\left(1-p\right)+n-4np+4n{p}^2-{\left(n+2\right)}^{2}p\left(1-p\right)}{n{\left(n+2\right)}^2}>0\\ \frac{n^{2}p-n^2{p}^2+n-4np+4n{p}^2-n^{2}p-4p-4np+n^2{p}^2+4p^2+4n{p}^2}{n{\left(n+2\right)}^2}>0\\ \frac{n-8np+8n{p}^2-4p+4p^2}{n{\left(n+2\right)}^2}>0\\ \left(8n+4\right)p^2-\left(8n+4\right)p+n<0\end{array}$.

Using quadratic formula, p is solved as shown below:

$\begin{array}{c}p=\frac{\left(8n+4\right)\pm \sqrt{{\left(8n+4\right)}^2-4\left(8n+4\right)n}}{2\left(8n+4\right)}\\ =\frac{\left(8n+4\right)\pm \sqrt{64n^2+16+64-32n^2+16n}}{2\left(8n+4\right)}\\ =\frac{\left(8n+4\right)}{2\left(8n+4\right)}\pm \frac{\sqrt{32n^2+16n+80}}{2\left(8n+4\right)}\\ =\frac{1}{2}\pm \sqrt{\frac{n+1}{8n+4}}\end{array}$

Therefore, the values of p for which $MSE\left({\widehat{p}}_1\right)<MSE\left({\widehat{p}}_2\right)$ is $\frac{1}{2}\pm \sqrt{\frac{n+1}{8n+4}}$.