Let $X\_0, X\_1, X\_2, \ldots$ be a sequence of i.i.d. real-valued random variables on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with continuous CDF $F(x)$ and define a sequence of empirical CDFs $F\_n(x) = \frac{1}{n} \sum\_{i = 1}^n \mathbf{1}\_{\{X\_i \leq x\}}(x)$ ($\mathbf{1}\_A$ is an indicator function). It is well known that


$$\sup\_{x \in \mathbb{R}} \left|F\_n(x) - F(x)\right| \to 0 \text{ a.s.}$$


(This is the fundamental theorem of mathematical statistics, or FTMS, supposedly.) Are we then safe to say that


$$F\_n(X\_0) \to U \text{ in law}$$


where $U$ follows a standard uniform distribution? Or could we make the even stronger claim that we can redefine the random variables into a probability space such that the convergence happens almost surely?


It seems like this would be the case. FTMS suggests we can write $F\_n(X\_0) = F(X\_0) + o(1)$ (holding almost surely) and the distribution of $F(X\_0)$ is a standard uniform distribution. While $F\_n$ is affected by the asymptotics, $X\_0$ is not. The probability $X\_0$ takes a value that is not covered by the convergence is zero thanks to FTMS being a uniform convergence result. Is it that easy? Am I overthinking this?



For a random sequence $X_0, X_1, X_2, \ldots$ and $F_n$ the empirical CDF, does $F_n(X_0)$ converge to a uniform random variable?
$\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\mathbb R}$This is indeed straightforward. Spelling out
$$\sup\_{x\in\R}|F\_n(x)-F(x)|\to0 \text{ a.s.}, $$
we see that for some subset $N$ of $\Om$ of outer probability $0$ and all $\om\in\Om\_0:=\Om\setminus N$ we have
$$\sup\_{x\in\R}\Big|\frac1n\sum\_{j=1}^n 1(X\_j(\om)\le x)-F(x)\Big|\to0$$
and hence
$$\frac1n\sum\_{j=1}^n 1(X\_j(\om)\le X\_0(\om))\to F(X\_0(\om)).$$


So, indeed we have
$$F\_n(X\_0)\to F(X\_0)$$
a.s. and hence in law. Also, if $F$ is continuous, then the random variable $F(X\_0)$ has the standard uniform distribution.


