91Ó°ÊÓ

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

The independent random variables having an unknown continuous distribution function

$G=Y_1,Y_2,…,Y_m$

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

Random variable $R=\underset{i=1}{\overset{n+m}{∑}}i{I}_j$

The mean and variance of$R=?$

We have the random variable,

$R=\underset{i=1}{\overset{n+m}{∑}}i{I}_i$

Applying the result of example3e

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

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

Now

$\stackrel{¯}{i}=\frac{1+2+3+…+(\mathrm{n}+\mathrm{m})}{(\mathrm{n}+\mathrm{m})}$

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

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

$\underset{i=1}{\overset{n+m}{∑}}{i}^2=\left(1^2+2^2+3^2+……….+(n+m{)}^2\right)$

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

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

Calculate the variance,

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

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

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

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

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

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

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