In Example 2.2, if (X, Y) are independent and (F_{X}=F_{Y}) then (mathcal{E}(X, Y)=2 E|X-Y|-Eleft|X-X^{prime}ight|-Eleft|Y-Y^{prime}ight|=0). However, the test

Question:

In Example 2.2, if $X, Y$ are independent and $F_{X}=F_{Y}$ then $\mathcal{E}(X, Y)=2 E|X-Y|-E\left|X-X^{\prime}ight|-E\left|Y-Y^{\prime}ight|=0$. However, the test statistic $\frac{n m}{n+m} V_{n, m}=\frac{n m}{n+m} \mathcal{E}_{n, m}$ is not unbiased for $\mathcal{E}(X, Y)$. Show that under the null hypothesis, for independent random samples, the bias in the test statistic $\frac{n m}{n+m} \mathcal{E}_{n, m}$ is
\[
E\left(\frac{n m}{n+m} \mathcal{E}_{n, m}ight)=E|X-Y|
\]

Data From Exercise 2.2