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.

Short 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-by-step solution

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$R=?$

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

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

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

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

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

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

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

Explanation

Calculate the variance,

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

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

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

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

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

Final Answer

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

The variance value of $R$is localid="1647511789994" $Var\left(R\right)=\frac{\mathrm{nm}(\mathrm{n}+\mathrm{m}+1)}{12}$.