Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression $E\left(n_i-n_j\right)$.

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the $n_i\text{'}s,i=1,2,\mathrm{...},k$ follows binomial distribution with parameters $n\text{and}{p}_i$. Further, $Cov\left(n_i,n_j\right)=-n{p}_i{p}_j\text{if}i\ne j$.

The value of the expression $E\left(n_i-n_j\right)$ is given below:

$\begin{array}{c}E\left(n_i-n_j\right)=E\left(n_i\right)-E\left(n_j\right)\\ =n{p}_i-n{p}_j\\ =n\left(p_i-p_j\right)\end{array}$

b.

To determine

Provide the unbiased estimator of $\left(p_i-p_j\right)$ with reference to Part (a).

Explanation of Solution

From Part (a), $E\left(n_i-n_j\right)=n\left(p_i-p_j\right)$.

Then,

$\begin{array}{l}E\left(\frac{n_i-n_j}{n}\right)=\frac{1}{n}E\left(n_i-n_j\right)\\ =\frac{1}{n}\left(E\left(n_i\right)-E\left(n_j\right)\right)\\ =\frac{1}{n}\left(n{p}_i-n{p}_j\right)\\ =p_i-p_j\end{array}$

An unbiased estimator for $\theta =p_i-p_j$ is given below:

$\widehat{\theta }=\frac{n_i-n_j}{n}$

c.

To determine

Verify that $V\left(n_i-n_j\right)=n\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]$.

Explanation of Solution

It is known that $Cov\left(n_i,n_j\right)=-n{p}_i{p}_j\text{if}i\ne j$. Also, the variance of a linear combination is given below:

$\begin{array}{c}V\left(aX+bY\right)=V\left(aX\right)+V\left(bY\right)+2Cov\left(aX,bY\right)\\ =a^{2}V\left(X\right)+b^{2}V\left(Y\right)+2abCov\left(X,Y\right)\end{array}$

Now,

$\begin{array}{c}V\left(n_i-n_j\right)=V\left(n_i\right)+V\left(-n_j\right)+2Cov\left(n_i,-n_j\right)\\ =V\left(n_i\right)+\left(-1^2\right)V\left(n_j\right)+2\left(-1\right)Cov\left(n_i,n_j\right)\\ =V\left(n_i\right)+V\left(n_j\right)-2Cov\left(n_i,n_j\right)\\ =n{p}_i\left(1-p_i\right)+n{p}_j\left(1-p_j\right)-2\left(-n{p}_i{p}_j\right)\\ =n{p}_i\left(1-p_i\right)+n{p}_j\left(1-p_j\right)+2n{p}_i{p}_j\\ =n\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]\end{array}$

Hence, it has been verified that $V\left(n_i-n_j\right)=n\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]$.

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

Explanation of Solution

From Part (a), $E\left(n_i-n_j\right)=n\left(p_i-p_j\right)$.

Then,

$\begin{array}{c}V\left(\frac{n_i-n_j}{n}\right)=\frac{1}{n^2}V\left(n_i-n_j\right)\\ =\frac{1}{n^2}\left[V\left(n_i\right)+V\left(-n_j\right)+2Cov\left(n_i,-n_j\right)\right]\\ =\frac{1}{n^2}\left[V\left(n_i\right)+\left(-1^2\right)V\left(n_j\right)+2\left(-1\right)Cov\left(n_i,n_j\right)\right]\\ =\frac{1}{n^2}\left[V\left(n_i\right)+V\left(n_j\right)-2Cov\left(n_i,n_j\right)\right]\\ =\frac{1}{n^2}\left[n{p}_i\left(1-p_i\right)+n{p}_j\left(1-p_j\right)+2n{p}_i{p}_j\right]\\ =\frac{1}{n^2}\left[n\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]\right]\\ =\frac{1}{n}\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]\end{array}$

e.

To determine

Obtain the consistent estimator of $n^{-1}V\left(n_i-n_j\right)$.

Explanation of Solution

A consistent estimator for $n^{-1}V\left(n_i-n_j\right)$ is ${\widehat{p}}_i-{\widehat{p}}_j$ because it is an unbiased estimator and variance reaches at 0 as n increases.

f.

To determine

Verify that a large sample $\left(1-\alpha \right)100\%$ confidence interval for $p_i-p_j$ is given below: ${\widehat{p}}_i-{\widehat{p}}_j\pm z_{\frac{\alpha }{2}}\sqrt{\frac{p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j}{n}}$.

Explanation of Solution

It is known that the variable $H_n=\frac{{\widehat{p}}_i-{\widehat{p}}_j-p_i+p_j}{\sqrt{\frac{1}{n}\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]}}$ has normal distribution.

Since $p_i\text{ and }{p}_j$ are consistent estimators, then $\frac{{\sigma }_{{\widehat{p}}_i-{\widehat{p}}_j}}{{\widehat{\sigma }}_{{\widehat{p}}_i-{\widehat{p}}_j}}$ tends to 0.

Now,

$H_n\frac{{\sigma }_{{\widehat{p}}_i-{\widehat{p}}_j}}{{\widehat{\sigma }}_{{\widehat{p}}_i-{\widehat{p}}_j}}=\frac{{\widehat{p}}_i-{\widehat{p}}_j-p_i+p_j}{\sqrt{\frac{1}{n}\left[p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j\right]}}$ has the limiting standard deviation.

In this context, it has been showed that a transformation of $H_n$ has a standard normal distribution.

Therefore, $\left(1-\alpha \right)100\%$ confidence interval for $p_i-p_j$ is given by: ${\widehat{p}}_i-{\widehat{p}}_j\pm z_{\frac{\alpha }{2}}\sqrt{\frac{p_i\left(1-p_i\right)+p_j\left(1-p_j\right)+2p_i{p}_j}{n}}$