Q.7.43 - A First Course in Probability

Question

Let $X_1,X_2,\dots ,X_n$ be independent random variables having an unknown continuous distribution function $F$ and let $Y_1,Y_2,\dots ,Y_m$ be independent random variables having an unknown continuous distribution function $G$. Now order those $n+m$variables, and let

$I_i=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if the }i\text{ th smallest of the }n+m\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ variables is from the }X\text{ sample }\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ otherwise }\end{array}\right.$

The random variable $R=\sum _{i=1}^{n+m}i{I}_i$ is the sum of the ranks of the $X$ sample and is the basis of a standard statistical procedure (called the Wilcoxon sum-of-ranks test) for testing whether $F$ and $G$ are identical distributions. This test accepts the hypothesis that $F=G$ when $R$ is neither too large nor too small. Assuming that the hypothesis of equality is in fact correct, compute the mean and variance of $R$.

Hint: Use the results of Example 3e.

Step-by-Step Solution

Answer

The mean of $R$ is $\mathrm{E}\left(\mathrm{R}\right)=\frac{\mathrm{n}(\mathrm{n}+\mathrm{m}+1)}{2}$

The Variance of $R$ is $Var\left(R\right)=\frac{\mathrm{nm}(\mathrm{n}+\mathrm{m}+1)}{12}$

Step 1: Given Information

The independent random variables having an unknown continuous distribution function $F=X_1,X_2,\dots ,X_n$

The independent random variables having an unknown continuous distribution function 

$G=Y_1,Y_2,\dots ,Y_m$

Given function$I_i=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if the }i\text{ th smallest of the }n+m\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ variables is from the }X\text{ sample }\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ otherwise }\end{array}\right.$

Random variable $R=\sum _{i=1}^{n+m}i{I}_j$

The mean and variance of $R=?$

Step 2: Explanation

We have the random variable,

$R=\sum _{i=1}^{n+m}i{I}_i$

Applying the result of example3e

$\mathrm{E}\left(\mathrm{R}\right)=\mathrm{n}\overline{i}$ 

And $Var\left(\mathrm{R}\right)=\frac{\mathrm{n}(\mathrm{n}+\mathrm{m}-\mathrm{n})}{\mathrm{n}+\mathrm{m}-1}\left[\frac{{\sum }_{i=1}^{n+m}{i}^2}{n+m}-(\overline{i}{)}^2\right]$

Now

$\overline{i}=\frac{1+2+3+\dots +(\mathrm{n}+\mathrm{m})}{(\mathrm{n}+\mathrm{m})}$

$=\frac{(\mathrm{n}+\mathrm{m})(\mathrm{n}+\mathrm{m}+1)}{(\mathrm{n}+\mathrm{m})·2}$

$=\frac{\mathrm{n}+\mathrm{m}+1}{2}$

$\sum _{i=1}^{n+m}{i}^2=\left(1^2+2^2+3^2+\dots \dots \dots .+(n+m{)}^2\right)$

$=\frac{(n+m)(n+m+1)(2n+2m+1)}{6}$

$\therefore \mathrm{E}\left(\mathrm{R}\right)=\frac{\mathrm{n}(\mathrm{n}+\mathrm{m}+1)}{2}$

Step 3: Explanation

Calculate the variance,

$Var\left(R\right)=\frac{nm}{n+m--1}\left[\frac{(n+m+1)(2n+2m+1)}{6}-\frac{(n+m+1{)}^2}{4}\right]$

$=\frac{\mathrm{nm}(\mathrm{n}+\mathrm{m}+1)}{\mathrm{n}+\mathrm{m}-1}\left[\frac{2\mathrm{n}+2 \mathrm{m}+1}{6}-\frac{\mathrm{n}+\mathrm{m}+1}{4}\right]$

$=\frac{nm(n+m+1)}{n+m-1}\left[\frac{4n+4m+2-3n-3m-3}{12}\right]$

$=\frac{nm(n+m+1)}{n+m-1}\left[\frac{n+m-1}{12}\right]$

$=\frac{\mathrm{nm}(\mathrm{n}+\mathrm{m}+1)}{12}$

Step 4: Final Answer

The mean value of$R$is $E\left(R\right)=\frac{n(n+m+1)}{2}$

The variance value of $R$ is $Var\left(R\right)=\frac{\mathrm{nm}(\mathrm{n}+\mathrm{m}+1)}{12}$.