Difference between Empirical Function and Real Distribution when Conditioning on Filtration

Question

Let $X_1,\dots, X_n$ are i.i.d (F), and the empirical distribution is given by $\hat{F_n}(x) = \sum\limits_{i=1}^n \frac{1}{n} 1_{\{X_i\leq x\}}$. By Kolmogorov-Smirnov theorem, $\hat{F_n}(x)$ converges to the real c.d.f $F(x)$ with probability 1 (uniformly over all $x$). I am wondering whether such convergence will somehow hold if I am conditioning on filtration?

Like will it be true that for any constant $\epsilon>0$, $P(\sup\limits_{x}|\hat{F_n}(x)-F(x)|>\epsilon| X_1,\dots, X_n)$ goes to zero when $n\to\infty$?

Please guide me if there are existing materials on this kind of topic. Thanks!

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02/13 Update: Many Thanks to @Botnakov for giving the statements and guiding the proofs!

I have a follow-up question: I see in the article "A Uniform Bound for the Tail Probability of Kolmogorov-Smirnov Statistics" that the tail probability of Kolmogorov-Smirnov test statistic (for $n$ samples) can be bounded by something like $c_1 e^{-c_2 n}$, where $c_1,c_2$ are constants. The thing I am looking for here is whether we can show $P(\sup\limits_{x}|\hat{F_n}(x)-F(x)|>\epsilon| X_1,\dots, X_n)$ goes to $0$ uniformly. Now we know $P(\sup\limits_{x}|\hat{F_n}(x)-F(x)|>\epsilon| X_1,\dots, X_n)\to 0$ in the sense of $L^p$ for all $p\geq1$, by which I guess we can show that $\lVert P(\sup\limits_{x}|\hat{F_n}(x)-F(x)|>\epsilon| X_1,\dots, X_n)\rVert_{\infty} := \lVert D_n \rVert_{\infty}\to 0$ (limiting $L^p$ norm). I am wondering whether there exists some bounds like "$c_1 e^{-c_2 n}$" for $\lVert D_n \rVert_{\infty}$?

Please guide me if there are existing/related materials about this question. Thanks!

Answer 1

Yes.

Statement: If $\mathcal{G}$ is an arbitrary $\sigma$-algebra then for any fixed $\varepsilon > 0$ we have $\mathbf{P}( \sup\limits_{x}|\hat{F_n}(x)-F(x)| > \varepsilon| \mathcal{G}) \overset{a.s}{\to} 0$. Moreover, if $p \ge 1$ is fixed then $\mathbf{P}( \sup\limits_{x}|\hat{F_n}(x)-F(x)| > \varepsilon| \mathcal{G}) \overset{L_p}{\to} 0$.

Proof.

1. Put $\xi_n = \sup\limits_{x}|\hat{F_n}(x)-F(x)|$. As $-1 \le \hat{F_n}(x)-F(x) \le 1$ we have that $0 \le \xi_n \le 1$. By Glivenko–Cantelli theorem $\xi_n \overset{a.s.}\to 0$.

2. Let $\mathcal{G}$ be an arbitrary $\sigma$-algebra. As $0 \le \xi_n \le 1$ and $\xi_n \overset{a.s}\to 0$ we get that $$\mathbf{E}(\xi_n | \mathcal{G}) \overset{a.s.}\to \mathbf{E}(0| \mathcal{G}) = 0$$ by dominated convergence theorem for condit.expectations.

3. Fix $\varepsilon > 0$. As $0 \le \xi_n \le 1$ we have $0 \le \mathbf{E}(\xi_n | \mathcal{G}) \le 1$. Hence $$\mathbf{E}(\xi_n | \mathcal{G}) = \mathbf{E}(\xi_n I_{\xi_n > \varepsilon}| \mathcal{G}) + \mathbf{E}(\xi_n I_{0 \le \xi_n \le \varepsilon}| \mathcal{G}) \ge \mathbf{E}(\xi_n I_{\xi_n > \varepsilon}| \mathcal{G}) \ge$$ $$\ge \mathbf{E}(\varepsilon I_{\xi_n > \varepsilon}| \mathcal{G}) = \varepsilon \mathbf{E}( I_{\xi_n > \varepsilon}| \mathcal{G}) \ge 0.$$ So $\mathbf{E}(\xi_n | \mathcal{G}) \ge \varepsilon \mathbf{E}( I_{\xi_n > \varepsilon}| \mathcal{G}) \ge 0$ and hence $\mathbf{E}( I_{\xi_n > \varepsilon}| \mathcal{G}) \overset{a.s}{\to} 0$ by sandwich theorem. So $$\mathbf{P}( \sup\limits_{x}|\hat{F_n}(x)-F(x)| > \varepsilon| \mathcal{G}) = \mathbf{E}( I_{\xi_n > \varepsilon}| \mathcal{G}) \overset{a.s}{\to} 0.$$

4. Put $\eta_n = \mathbf{P}( \sup\limits_{x}|\hat{F_n}(x)-F(x)| > \varepsilon| \mathcal{G}) = \mathbf{E}( I_{\xi_n > \varepsilon}| \mathcal{G}) $. As $0 \le I_{\xi_n > \varepsilon} \le 1$ it follows that $0 \le \eta_n \le 1$. We have already proved that $\eta_n \overset{a.s}{\to} 0.$

5. We have $0 \le \eta_n \le 1$ and $\eta_n \overset{a.s}{\to} 0$. It follows from Dominated convergence theorem that for all $p \ge 1$ we have $\eta_n \overset{L_p}{\to} 0$ - see Dominated convergence in Lp-spaces (corollary) https://en.wikipedia.org/wiki/Dominated_convergence_theorem

Answer to the updated question may be obtained via the previous technique: conditional probability $\mathbf{E}( I_{\xi_n > \varepsilon}| \mathcal{G})$ may be estimated by $\varepsilon^{-1} \mathbf{E}(\xi_n | \mathcal{G})$. Further, $||\xi_n|| \to 0$ and hence $||\mathbf{E}(\xi_n | \mathcal{G})|| \le ||\xi_n|| \to 0$. We know rate of convergence of $||\xi_n|| \to 0$. Hence we know rate of convergence of $||\mathbf{E}(\xi_n | \mathcal{G})|| \to 0$.