How do rates of convergence almost surely translate to weak rates of convergence?

Question

Suppose we have that $X_n = Y_n + Z_n$ and $Y_n$ converges to zero almost surely with a rate which we know and $Z_n$ converges in distribution with a (Berry-Esseen) rate which we also know. Then can we tell anything about the rate of convergence (in the weak sense) of $X_n$?

Answer 1

Suppose that $$ \Delta_n:=\sup_{x\in\mathbb R}\left\lvert \mathbb P\left(Z_n\leqslant x\right)-\mathbb P\left(Z\leqslant x\right)\right\rvert\to 0. $$ Let $$\Delta'_n:=\sup_{x\in\mathbb R}\left\lvert \mathbb P\left(X_n\leqslant x\right)-\mathbb P\left(Z\leqslant x\right)\right\rvert $$ and $g\colon\lambda\mapsto \sup_{x\in\mathbb R}\left\lvert \mathbb P\left(Z\leqslant x+\lambda\right)-\mathbb P\left(Z\leqslant x \right)\right\rvert$. Since for each $\lambda$, $$ \mathbb P\left(Z_n\leqslant x\right)\leqslant \mathbb P\left(Z_n\leqslant x,Y_n\leqslant \lambda\right)+\mathbb P\left(Y_n\gt\lambda\right) \leqslant \mathbb P\left(X_n\leqslant x+ \lambda\right)+\mathbb P\left(Y_n\gt\lambda\right)\\ \leqslant \Delta'_n+\mathbb P\left(Z\leqslant x+\lambda\right)+\mathbb P\left(Y_n\gt\lambda\right)\leqslant \Delta'_n+g(\lambda)+\mathbb P\left(Y_n\gt\lambda\right)+\mathbb P(Z\leqslant x), $$ we can get the bound $$ \mathbb P\left(Z_n\leqslant x\right) -\mathbb P\left(Z\leqslant x\right)\leqslant \Delta'_n+g(\lambda)+\mathbb P\left(Y_n\gt\lambda\right). $$ A lower bound can be obtained, starting from $$ \mathbb P \left(Z_n\leqslant x\right)\geqslant \mathbb P \left(Z_n\leqslant x, Y_n \leqslant -\lambda\right)\geqslant \mathbb P \left(X_n\leqslant x-\lambda, Y_n \leqslant -\lambda\right) \geqslant \mathbb P \left(X_n\leqslant x-\lambda \right) -\mathbb P \left( Y_n \leqslant -\lambda\right) , $$ where we used $\mathbb P(A\cap B)=\mathbb P(A)-\mathbb P(A\cap B^c)\geqslant \mathbb P(A)-\mathbb P( B^c)$.

We end up with a bound of the form $$ \Delta'_n\leqslant \Delta_n+\mathbb P\left(\lvert Y_n\rvert\gt\lambda\right)+\sup_{x\in\mathbb R}\lvert\mathbb P\left(Z\leqslant x+\lambda\right)-\mathbb P\left(Z\leqslant x\right)\rvert. $$ It remains to optimize this over $\lambda$, which can be done according to the information you have on the moments of $Y_n$ and the distribution of $Z$.

Note that this approach works even if $Z$ is not necessarily a normal random variable, we just need continuity of the cumulative distribution function of $Z$. When $Z$ is standard normal, then we can bound $g(\lambda)$ like in the answer by Christophe Leuridan.

Answer 2

Call $G$ le distribution function of $\mathcal{N}(0,1)$.

Berry Esseen bound yields (under adequate assumptions) $|P[X_n-x]-G(x)| \le C/\sqrt{n}$ for some constant $C$ independent of $n$ and $x$.

Convergence in probability of $(Y_n)_{n \ge 1}$ towards $0$ is sufficient

Let $n \ge 1$. Given $h>0$, you can write for every $x \in \mathbb{R}$. \begin{eqnarray*} P[X_n \le x] &\le& (P[Z_n \le x+h] \cup [Y_n < -h]) \\ &\le& P[Z_n \le x+h] + P[ Y_n < -h] \\ &\le& G(x+h) + C/\sqrt{n} + P[ Y_n < -h] \end{eqnarray*} Using that $G'$ - the density of of $\mathcal{N}(0,1)$ - is bounded above by $(2\pi)^{-1/2}$, we get \begin{eqnarray*} P[X_n \le x]-G(x) &\le& G(x+h)-G(x) + C/\sqrt{n} + P[ Y_n < -h] \\ &\le& (2\pi)^{-1/2}h + C/\sqrt{n} + P[ Y_n < -h] \end{eqnarray*} In a same way, \begin{eqnarray*} P[X_n \le x] &\ge& (P[Z_n \le x-h] \cap [Y_n \le h]) \\ &\ge& P[Z_n \le x-h] - P[ Y_n > h] \\ &\ge& G(x-h) - C/\sqrt{n} - P[ Y_n > h] \end{eqnarray*} \begin{eqnarray*} P[X_n \le x]-G(x) &\ge& G(x-h)-G(x) - C/\sqrt{n} - P[ Y_n > h] \\ &\ge& -(2\pi)^{-1/2}h - C/\sqrt{n} - P[ Y_n > h] \end{eqnarray*}

This gives an uniform upper bound of $|F_{X_n}-G|$, depending on $n$ and $h$. Then, you use the rate of convergence in probability to bound above $P[ Y_n < -h]$ and $P[ Y_n > h]$, and you choose $h$ (depending on $n$) to minimize the upper bound, or at least to approach the minimum.