# A question on weak convergence of probability measures

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 ?

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

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.