Uniform convergence in probability and its connection to the convergence of distribution functions

Question

I have the following question: Let $(X_i)_{i\in\mathbb{N}}$ and $(Y_i)_{i\in\mathbb{N}}$ be real stationary sequences with finite variances and further let $Z$ be a real valued random variable with finite variance, too. We denote by $F_{X_n}(u):=P(X_n\leq u)$ the distribution of $X_n$ and by $\tilde{F}_{Y_n}(u):= \frac{1}{n}\sum_{i=1}^n 1\{Y_i\leq u\}$ the empirical distribution function of $(Y_i)_{i\in\mathbb{N}}$ at $u$. If we know that \begin{align*} X_n\xrightarrow{d}Z \end{align*} ($d$ stands for convergence in distribution) and that \begin{align*} \sup_{u\in\mathbb{R}}|F_{X_n}(u)-\tilde{F}_{Y_n}(u)|\xrightarrow{p}0 \end{align*}($p$ stands for convergence in probability), does it hold then that \begin{align*} \tilde{F}_{Y_n}(u)\xrightarrow{}P(Z\leq u) \end{align*} in some sense (for example in probability) ?

Answer 1

If the distribution of $Z$ is continuous then $\sup_{u\in\mathbb{R}}|F_{X_n}(u)-\tilde{F}_Z(u)|\rightarrow0$ hence $$ \sup_{u\in\mathbb{R}}|\tilde{F}_{Y_n}(u)-P(Z\leqslant u)|\xrightarrow{p}0. $$ Otherwise one might have to delve into the specifics of the setting ($\tilde F_{Y_n}$ random Cesàro means, $(Y_i)$ stationary, etc.).