Does $X_n\xrightarrow{\text{d}}X$, $Y_n\xrightarrow{\text{d}}Y$ and $(X_n, Y_n)\xrightarrow{\text{p}}Z$ imply $(X_n,Y_n)\xrightarrow{\text{d}}(X,Y)$?

Question

Let $X_n \in \mathbb{R}^k$ and $Y_n \in \mathbb{R}^l$ be sequences of random variables and $X_n\xrightarrow{\text{d}}X$, $Y_n\xrightarrow{\text{d}}Y$. Additionally let's assume that $(X_n, Y_n)$ converges to some random variable $Z \in \mathbb{R}^{k+l}$ in probability. Does $(X_n,Y_n)\xrightarrow{\text{d}}(X,Y)$?

Let $(Z_1, Z_2) = Z$. Then I guess that $X_n \to Z_1$ and $Y_n \to Z_2$ in probability. However, this may not guarantee $Z_1 = X$ and $Z_2 = Y$. So the conclusion would be "no". But I can't find a counterexample.

score 0 · Accepted Answer · answered Nov 21 '18 at 07:11

Let $(X_n,Y_n)= (X,Y)$ be bivariate standard normal with nonzero correlation (i.e. the sequence is constant in $n$). Then let $X'$ and $Y'$ be independent standard normals. Then $X_n\to_d X'$ and $Y_n\to_d Y'$ and $(X_n,Y_n)\to_p(X,Y),$ but we don't have $(X_n,Y_n)\to_d (X',Y').$

Does $X_n\xrightarrow{\text{d}}X$, $Y_n\xrightarrow{\text{d}}Y$ and $(X_n, Y_n)\xrightarrow{\text{p}}Z$ imply $(X_n,Y_n)\xrightarrow{\text{d}}(X,Y)$?

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