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The U-statistic is given by \begin{equation} \widehat{\Delta}=\frac{1}{\binom{n_1}{2}\binom{n_2}{2}}\sum_{1\leq i_1<i_2\leq n_1}\sum_{1\leq j_1<j_2\leq n_2}f(X_{i_1},X_{i_2},Y_{j_1},Y_{j_2}), \end{equation} where \begin{align*} f(X_1,X_2,Y_1,Y_2)&=\min(Y_1,\max(X_1,X_2))-\min(X_1,X_2)I(\min(X_1,X_2)>Y_1) \\&-2X_1I(Y_1<X_2,X_2>X_1). \end{align*} Since $f(X_1,X_2,Y_1,Y_2)\neq f(X_2,X_1,Y_1,Y_2)$ and $f(X_1,X_2,Y_1,Y_2)\neq f(X_1,X_2,Y_2,Y_1)$, $f$ is not symmetric in $X$ and $Y$. I have to find a symmetric kernel for the given U- statistic. I think \begin{align*} h(X_1,X_2,Y_1,Y_2)&=\frac{1}{4}\big[f(X_1,X_2,Y_1,Y_2)+f(X_2,X_1,Y_1,Y_2) \\&+f(X_1,X_2,Y_2,Y_1)+f(X_2,X_1,Y_2,Y_1)] \end{align*} is a symmetric kernel for the given u-statistic. Is this right? If my answer is wrong please let me know what the correct way is. Somebody please help me.

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