Does $E[\hat{F}_n(X_1) - F(X_1) \mid \cdots]\to 0$?

40 Views Asked by At

Let $\binom{X_1}{Y_1}, \binom{X_2}{Y_2} \dots $ be iid vectors. Denote $\hat{F}^x_n(\cdot) = \frac{1}{n}\sum_{i=1}^n1[X_i\leq \cdot]$ an empirical distribution function of $X$, and similarly $\hat{F}^y_n(\cdot)$ of $Y$. Let $a_n\in(0,1)$ be konstants fulfilling $a_n\to 1^-$ as $n\to \infty$.

Does it hold that $$E[\hat{F}^x_n(X_1) - F^x(X_1) \mid \hat{F}_n^y(Y_1)>a_n]\to 0 \text{ as }n\to \infty?$$

Glivenko-Cantelli theorem says that $\sup_{x\in R}|\hat{F}_n(x)-F(x)|\to 0$ uniformly a.s., but does it imply that such conditional expected value decreases also?