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I am interested whether Slutsky's Theorem also holds in the case of joint convergence. Let $(X_n,Y_n)$ be random variables with $(X_n,Y_n) \rightarrow (X,Y)$ in distribution for $n \to \infty$. Furthermore, let $(a_n,b_n) \rightarrow (1,1)$ a.s. for $n \to \infty$. Does also $(X_n*a_n,Y_n*b_n) \rightarrow (X,Y)$ hold in distribution for $n \to \infty$?

Many thanks in advance!

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Yes, the statement holds true. Two ways to go about this:

Proof 1: By Skorohod's representation theorem, there exists a probability space and random variables $X_n',Y_n'$, $X',Y'$ on this probability space with $$(X_n,Y_n) \sim (X_n',Y_n'), \qquad (X,Y) \sim (X',Y')$$ such that $$(X_n',Y_n') \to (X',Y') \qquad \text{almost surely.}$$ Since $a_n \to 1$ and $b_n \to 1$, we get $(a_n X_n', b_n Y_n') \to (X',Y')$ almost surely. Now, as $(a_n X_n',b_n Y_n') \sim (a_n X_n, b_n Y_n)$, it follows that $(a_n X_n,b_n Y_n) \to (X,Y)$ in distribution.

Proof 2: Convergence in distribution is equivalent to convergence of the corresponding characteristics functions (Lévy's continuity theorem). Since the convergence of the characteristic function holds even locally uniformly, it is not difficult to see that the claim holds, see this question.

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