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Let $X$ be standard Gaussian random variable with cdf $F(x)$. Let $\{X_i\}_{i=1}^n$ be a sequence of i.i.d. standard Gaussian random variables. And let $F_n(x)=\frac{1}{n}\sum_{i=1}^n1_{\{X_i\leq x\}}$ be the empirical distribution. We know that from Glivenko–Cantelli theorem, $\underset{x\in\mathbb{R}}{\sup}|F_n(x)-F(x)|\overset{a.s.}{\longrightarrow}0$, i.e., the empirical distribution converge uniformly (in $x$) to the true distribution for almost all $\omega\in\Omega$.

My question is: can we show that the conditional moment under empirical distribution also converges uniformly to the conditional moment under the true distribution? Namely, to show

$\underset{x\in\mathbb{R}}{\sup}\bigg|\frac{\int_{-\infty}^xtdF_n(t)}{F_n(x)}-\frac{\int_{-\infty}^xtdF(t)}{F(x)}\bigg|\overset{a.s.}{\longrightarrow}0$. Here the issue I'm worrying about is when $x\rightarrow-\infty$, the denominator goes 0, that might kill the uniform convergence.

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  • $\begingroup$ How do you define the conditional moment of the empirical distribution when $x \lt \min(X_i)$? I don't see how you can avoid coming up with some definition, since for any $x$ and no matter how large $n$ is, there is a positive chance that $x \lt \min(X_i)$. $\endgroup$
    – whuber ♦
    Commented Feb 27, 2015 at 16:25
  • $\begingroup$ @whuber Good point...Then, if I define that case you mentioned to be 0, will the uniform convergence be true? $\endgroup$
    – KevinKim
    Commented Feb 27, 2015 at 16:44
  • $\begingroup$ The supremum will always be infinite, because $\int^x t dF(t) / F(x) \lt x$ implies that for $x \lt \min(X_i)$ and $x\lt 0$, the absolute difference will exceed $|x|$. Thus the supremum cannot converge to $0$ anywhere. $\endgroup$
    – whuber ♦
    Commented Feb 27, 2015 at 16:46

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