# On the limit of a continuous combination of sequences of random variables converging in distribution

Let $$\{X_n\}, \{Y_n\}$$ be sequences of real valued random variables converging in distribution to $$X$$ and $$Y$$ respectively. Let $$f: \mathbb R^2 \to \mathbb R$$ be a continuous function such that $$\{f(X_n,Y_n)\}$$ converges in distribution to some random variable $$Z$$.

Then is it true that $$Z$$ is identically distributed as $$f(X,Y)$$ ?

If this is not true in general, is it at least true for the function $$f(x,y)=x+y$$ ?

Not without independence assumptions (or convergence of joint distributions). Take $$\{X_n\}$$ i.i.d. with standard normal distribution, $$Y_n=-X_n$$ for all $$n$$ and $$f(x,y)=x+y$$. Then $$\{X_n\} \to X_1$$ and $$\{Y_n\} \to X_1$$ in distribution but $$X_n+Y_n \to 0$$ in distribution.

For your question to make sense, you need to mention that the vectors $$(X,Y)$$ and $$(X_n, Y_n)$$ exist (i.e. $$X_n$$ and $$Y_n$$ must live in the same probability space).

If you have the stronger assumption that $$(X_n, Y_n)$$ converges in distribution to the vector $$(X,Y)$$, then $$f(X_n, Y_n)$$ converges in distribution to $$f(X,Y)$$ by the continuous mapping theorem.

Without this stronger assumption, the claim does not hold, even for $$f(x,y) = x + y$$, as demonstrated in one of the answers in this question.