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Let $(X_1, Y_1), ...., (X_n, Y_n)$ be iid random vectors with marginal distributions functions $F(x)$ and $G(x)$ (both are continuous distributions) respectively such that $F(0)=G(0)=\frac{1}{2}$. Then find a $U$-statistic estimator of $\int_0^{\infty}G(x)dF(x)$. Also find the limiting distribution of this $U$-statistic when $F(x)=G(x)$ and $X, Y$ are independent. $$$$We can write $$\int_0^{\infty}G(x)dF(x)=\int_{-\infty}^{\infty}\mathbb{P}(Y_1 \leq x)dF(x)=\int_{-\infty}^{\infty}\mathbb{P}(Y_1 \leq X_2|X_2=x)dF(x)=\mathbb{P}(Y_1 \leq X_2).$$ Then a symmetric unbiased estimator of $\mathbb{P}(Y_1 \leq X_2)$ is given by $$h((x_1, y_1), (x_2, y_2))=\frac{\mathbf{1}(y_2 \leq x_1)+\mathbf{1}(y_1 \leq x_2)}{2}$$, where $\mathbf{1}(.)$ denotes the indicator function. So the corresponding $U$-statistic is given by $$U_n((X_1, Y_1),....,(X_n, Y_n))=\frac{1}{n\choose 2}\sum_{i<j}\frac{\mathbf{1}(Y_j \leq X_i)+\mathbf{1}(Y_i \leq X_j)}{2}$$. Now we know that $\sqrt{n}(U-\theta(F))$ converges in distribution to $N(0, k^2\sigma^2_1)$, where $\theta(F)$ is the parametric function to be estimated, $k$ is the degree of the unbiased estimator (in our case $k=2$) and $\sigma^2_1=\text{Var}(h_1((X_1, Y_1))$ and $$h_1((x_1, y_1))=\mathbb{E}[h((x_1, y_1), (X_2, Y_2))]=\frac{\mathbb{P}(Y_2 \leq x_1)+\mathbb{P}(X_2 \geq y_1)}{2}$$. Now under the assumption thet $F(x)=G(x)$ and $X, y$ are independent we get $$h_1((X_1, Y_1))=\frac{1-F(Y_1)+F(X_1)}{2}.$$ As $X, Y$ are independent so $$\sigma^2_1=\text{Var}(h_1((X_1, Y_1))=\frac{1}{2}(\frac{1}{12}+\frac{1}{12})=\frac{1}{12}.$$ Also under these assumptions $$\theta(F)=\mathbb{P}(Y_1 \leq X_2)=\mathbb{P}(Y_1 \leq X_1)=\frac{1}{2}.$$ And so $\sqrt{n}(U-\frac{1}{2})$ converges in distribution to $N(0, \frac{1}{3})$. $$$$Is the solution correct?

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  • $\begingroup$ In your last line you write, $$\mathbb{P}(Y_1 \leq X_2)=\mathbb{P}(Y_1 \leq X_1).$$ For this to hold we need to assume that $Y_1$ and $ X_1$ are independent, something that is not declared in the premises, and something that you cleverly sidestepped earlier by exploiting the i.i.d. assumption over the index $i$, when you transformed a marginal probability into a conditional between $Y_1$ and $X_2$ (which are independent). $\endgroup$ Commented Mar 12 at 23:44
  • $\begingroup$ We have to find the limiting distribution under the assumptions that $X$ and $Y$ are independent and $F(x)=G(x)$. And under these assumptions $\mathbf{P}(Y_1 \leq X_2)=\mathbf{P}(Y_1 \leq X_1)=\frac{1}{2}$. $\endgroup$
    – user771946
    Commented Mar 15 at 6:01
  • $\begingroup$ Ok, but writing, "Let $(X_1, Y_1), ...., (X_n, Y_n)$ be i.i.d. random vectors", means only that $(X_1, Y_1)$ is independent of $(X_2, Y_2)$ etc, not that inside each pair there exists independence. $\endgroup$ Commented Mar 15 at 20:24

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