Let $(\mu\_{n})$ sequence of probability measures of $\mathbb{R}^{d}$ converging to the prob measure $\mu$.
Then by definition we know that $\int f d\mu\_{n} \longrightarrow \int f d\mu $ for f continuous and bounded function.


I was wondering if I can write the formula above as $\int\int f(x-y) d\mu\_{n}(x)d\mu\_n(y) \longrightarrow 
\int\int f(x-y) d\mu(x)d\mu(y) $ ?


Or is the latter formula implied by the definition of weak convergence ?


Do I need additional assumptions on f ?



A question on weak convergence of probability measures
$\newcommand{\eD}{\overset{\text{D}}\to}$


Let $X\_n$ and $X$ be any random vectors with distributions $\mu\_n$ and $\mu$, respectively. Let $Y\_n$ be an independent copy of $X\_n$, for each $n$, and let $Y$ be an independent copy of $X$. The questions can then be restated as follows:



> 
> Q1: Does $X\_n\eD X$ imply $X\_n-Y\_n\eD X-Y$?
> 
> 
> Q2: Vice versa, does $X\_n-Y\_n\eD X-Y$ imply $X\_n\eD X$?
> 
> 
> 


Here $\eD$ denotes the convergence in distribution.


The answer to Q1 is yes. Indeed, let $g\_Z$ denote the characteristic function (c.f.) of a random vector $Z$. Then $X\_n\eD X$ means that $g\_{X\_n}\to g\_X$ pointwise, whence $g\_{X\_n-Y\_n}=|g\_{X\_n}|^2\to|g\_X|^2=g\_{X-Y}$ pointwise, so that $X\_n-Y\_n\eD X-Y$.


The answer to Q2 is no. Indeed, let e.g. $X\_n=n=Y\_n$ and $X=Y=0$. Then $X\_n-Y\_n\eD X-Y$, but $X\_n\eD X$ is false.


