Chapter 15, Problem 76SE

a.

To determine

Show that $E\left(U\right)=\frac{n_1{n}_2}{2}\text{ when }{H}_0\text{ is true}$.

Explanation of Solution

Consider

$\begin{array}{c}{W}_A={\displaystyle \sum _{i=1}^{n_1}R\left(A_i\right)}\\ ={\displaystyle \sum _{i=1}^{n_1}{X}_i}\end{array}$

Where, $\begin{array}{c}{X}_i=\{\begin{array}{l}R\left(z_i\right)\text{ if }{z}_i\text{ is from sample }A\\ 0\text{ if }{z}_i\text{ is from sample B}\end{array}\\ \end{array}$

If $H_0$ is true, then $E\left(X_i\right)$ is given below:

$\begin{array}{c}E\left(X_i\right)=R\left(z_i\right)P\left(X_i=R\left(z_i\right)\right)+0P\left(X_i=0\right)\\ =R\left(z_i\right)\frac{n_1}{N}+0\\ =R\left(z_i\right)\frac{n_1}{N}\end{array}$

The value of $E\left(W_A\right)$ is given below:

$\begin{array}{c}E\left(W_A\right)={\displaystyle \sum _{i=1}^{N}E\left(X_1\right)}\\ ={\displaystyle \sum _{i=1}^{N}R\left(z_i\right)\frac{n_1}{N}}\\ =\frac{n_1}{N}{\displaystyle \sum _{i=1}^{N}R\left(z_i\right)}\\ =\frac{n_1}{N}\left(\frac{N\left(N+1\right)}{2}\right)\\ =\frac{n_1\left(N+1\right)}{2}\end{array}$

From Exercise 15.75, $U=n_1{n}_2+\left(\frac{1}{2}\right)n_1\left(n_1+1\right)-W_A$.

The value of $E\left(U\right)$ is given below:

$\begin{array}{c}E\left(U\right)=n_1{n}_2+\left(\frac{1}{2}\right)n_1\left(n_1+1\right)-E\left(W_A\right)\\ =n_1{n}_2+\left(\frac{1}{2}\right)n_1\left(n_1+1\right)-\frac{n_1\left(N+1\right)}{2}\\ =\frac{2n_1{n}_2+n_1\left(n_1+1\right)-n_1\left(N+1\right)}{2}\\ =\frac{n_1{n}_2}{2}\end{array}$

Hence, $E\left(U\right)=\frac{n_1{n}_2}{2}\text{ when }{H}_0\text{ is true}$.

b.

To determine

Show that $V\left(U\right)=\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}\text{ when }{H}_0\text{ is true}$.

Explanation of Solution

The value of $E\left(X_i^2\right)$ is given below:

$E\left(X_i^2\right)={\left[R\left(z_i\right)\right]}^2\frac{n_1}{N}$

The value of $V\left(X_i\right)$ is given below:

$\begin{array}{c}V\left(X_i\right)=E\left(X_i^2\right)-{\left[E\left(X_i\right)\right]}^2\\ ={\left[R\left(z_i\right)\right]}^2\frac{n_1}{N}-{\left[R\left(z_i\right)\frac{n_1}{N}\right]}^2\\ ={\left[R\left(z_i\right)\right]}^2\left(\frac{n_1}{N}-\frac{n_1^2}{N^2}\right)\\ ={\left[R\left(z_i\right)\right]}^2\left(\frac{n_{1}N-n_1^2}{N^2}\right)\\ ={\left[R\left(z_i\right)\right]}^2\left(\frac{n_1\left(N-n_1\right)}{N^2}\right)\end{array}$

The value of $E\left(X_i,X_j\right)$ and $Cov\left(X_i,X_j\right)$ is given below:

$\begin{array}{c}E\left(X_i,X_j\right)=R\left(z_i\right)R\left(z_j\right)P\left(X_i=R\left(z_i\right),X_j=R\left(z_j\right)\right)\\ =R\left(z_i\right)R\left(z_j\right)\left(\frac{n_1}{N}\right)\left(\frac{n_1-1}{N-1}\right)\\ Cov\left(X_i,X_j\right)=R\left(z_i\right)R\left(z_j\right)\left(\frac{-n_1\left(N-n_1\right)}{N^2\left(N-1\right)}\right)\end{array}$

The value of $V\left(W_A\right)$ is given below:

$\begin{array}{c}V\left(W_A\right)={\displaystyle \sum _{i=1}^{N}V\left(X_i\right)}+{\displaystyle \sum _{i\ne j}Cov\left(X_i,X_j\right)}\\ =\left(\frac{n_1\left(N-n_1\right)}{N^2}\right){\displaystyle \sum _{i=1}^N{\left[R\left(z_i\right)\right]}^2}-\left(\frac{n_1\left(N-n_1\right)}{N^2\left(N-1\right)}\right)\left[{\displaystyle \sum _{i=1}^N{\displaystyle \sum _{j=1}^{N}R\left(z_i\right)R\left(z_j\right)}-{\displaystyle \sum _{i=1}^{N}R{\left(z_i\right)}^2}}\right]\\ =\left(\frac{n_1\left(N-n_1\right)}{N^2}\right)\left(\frac{N\left(N+1\right)\left(2N+1\right)}{6}\right)-\left(\frac{n_1\left(N-n_1\right)}{N^2\left(N-1\right)}\right)\left\{{\left({\displaystyle \sum _{i=1}^{N}R\left(z_i\right)}\right)}^2-{\displaystyle \sum _{i=1}^{N}R{\left(z_i\right)}^2}\right\}\\ =\left[\begin{array}{l}\left(\frac{n_1\left(N-n_1\right)}{N^2}\right)\left(\frac{N\left(N+1\right)\left(2N+1\right)}{6}\right)\\ -\left(\frac{n_1\left(N-n_1\right)}{N^2\left(N-1\right)}\right)\left\{\frac{N^2{\left(N+1\right)}^2}{4}-\left(\frac{N\left(N+1\right)\left(2N+1\right)}{6}\right)\right\}\end{array}\right]\\ =\left[\begin{array}{l}\left(\frac{2n_1\left(N-n_1\right)\left(N+1\right)\left(2N+1\right)}{12N}\right)\\ -\left(\frac{n_1\left(N-n_1\right)}{N^2\left(N-1\right)}\right)\left\{\frac{N^2{\left(N+1\right)}^2}{4}-\left(\frac{N\left(N+1\right)\left(2N+1\right)}{6}\right)\right\}\end{array}\right]\\ =\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}\left[\frac{4N+2}{N}-\frac{\left(3N+2\right)\left(N-1\right)}{n\left(N-1\right)}\right]\\ =\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}\end{array}$

The value of $V\left(U\right)$ is given below:

$\begin{array}{c}V\left(U\right)=V\left(W_A\right)\\ =\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}\end{array}$

Hence, $V\left(U\right)=\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}\text{ when }{H}_0\text{ is true}$.