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

Transcribed Image Text:

Example 2.2 (A two-sample energy statistic). A two-sample energy statistic for testing if two independent random samples, 1,..., n and yi,..., ym are drawn from equal distributions can be defined with the following two-sample kernel. Let h(x, Tj; Yk , Ye) = ( [xi Yk | + |; Ye| + |xi - Ye| + |xj - Yk| - -|T - Tj|- |j xi|- |yk - ye|- |ye Yj). Then his symmetric in its arguments (ri, Tj) and (yk, ye). The corresponding energy V-statistic for testing H: Fx = Fy is nm Lijke h(xi, j; Yk , Ye). To simplify the equations, introduce distance matrices D = (dik) = (- Yk), A = (aj) = (x - 2), and B = (bke) = (yk - yel). Let d. denote the sum of matrix D, and similarly define aand b... The V-statistic can be written En.m = = = 1 27h (Fi, Tj; Yk, ye) nm 1 2nm (4 nmd -2ma - 2nb.) TL m -| - | i=1 k=1 nm 7 = 72 72 |-2;| i=1 j=1 m m m n+m (2.4) |yw-yel. k=1 l=1 (2.5) En.m. (The factor is half the har- The test statistic is m V monic mean of n and m.) n+m n, m Alternately, one could apply the corresponding U-statistic with the same kernel (2.4). n+m