# Uniform convergence in probability

Let $X_i$ and $Y_i$ $(i\in\mathbb{N})$ be two strictly stationary sequences of real valued random variables with finite variance. We have two empirical distribution functions $F_{1,n}:=\frac{1}{n} \sum_{i=1}^n 1\{X_i\leq u\}$ and $F_{2,n}:=\frac{1}{n} \sum_{i=1}^n 1\{Y_i\leq u\}$. Assume that \begin{align*} \sup_{u\in\mathbb{R}} |F_{1,n}(u)-F_{2,n}(u)|\xrightarrow{p} 0 \end{align*}in probability and that $F_{1,n}(u)\xrightarrow{n\rightarrow\infty}c(u) Z$ in distribution where $Z$ is a real valued random variable with continuous distribution and $c: \mathbb{R}\rightarrow \mathbb{R_{>0}}$ continuous and increasing. Follows now that $F_{2,n}(u)\xrightarrow{n\rightarrow\infty}c(u) Z$?

Yes. To see that, show and use the fact that if $(X_n,n\geqslant 1)$ is a sequence of random variables converging in distribution to $X$, and $(Y_n,n\geqslant 1)$ converges in probability to $0$, then $X_n+Y_n\to X$ in distribution.
We have to prove that $P(X_n+Y_n\leqslant t)\to P(X\leqslant t)$ at each point where $s\mapsto P(X\leqslant s)$ is continuous. Fix $t_0$ such a point. Then use the fact that the term $Y_n$ does not perturb too much.
• See edit.${}{}$ – Davide Giraudo Jul 5 '13 at 8:09
• Ok, we have now \begin{align*}F_{2,n}(u)\xrightarrow{n\rightarrow\infty}c(u) Z.\end{align*} for all $u\in\mathbb{R}$. Can we now argue that $F_{2,n}(u)\xrightarrow{n\rightarrow\infty}c(u) Z$ uniform in $u$? – stroem Jul 10 '13 at 8:59