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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 ?

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$\newcommand{\eD}{\overset{\text{D}}\to}$

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.

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