# Marginal convergence in distribution implies joint convergence of a subsequence?

Consider two sequences of real-valued random variables defined on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, $X_n:\Omega\rightarrow \mathbb{R}$ and $Y_n:\Omega\rightarrow \mathbb{R}$. Suppose that $X_n\rightarrow_d A$ and $Y_n\rightarrow_d B$, where $A,B$ are real-valued random variables. This does not necessarily imply that $(X,Y)\rightarrow_d(A,B)$. However, I think it implies that there exists a subsequence $(X_{n_j}, Y_{n,j})\rightarrow_d (A,B)$ as $j\rightarrow \infty$ Here my proof:

(1) $X_n\rightarrow_d A$ $\Rightarrow$ $X_n=O_p(1)$

(2) $Y_n\rightarrow_d B$ $\Rightarrow$ $Y_n=O_p(1)$

(3) (1)+(2) $\Rightarrow$ $(X_n,Y_n)=O_p(1)$ $\Rightarrow$ $\exists \{(X_{n_j}, Y_{n,j})\}_j$ such that $(X_{n_j}, Y_{n_j})\rightarrow_d (C,D)$ as $j\rightarrow \infty$ where $C,D$ are real-valued random variables $\Rightarrow$ $X_{n_j}\rightarrow_d C$ and $Y_{n_j}\rightarrow_d D$ as $j\rightarrow \infty$

(4) $X_n\rightarrow_d A$ $\Rightarrow$ every subsequence $\{X_{n_k}\}_k \rightarrow_d A$ as $k\rightarrow \infty$ $\Rightarrow$ $C\sim A$

(5) $Y_n\rightarrow_d B$ $\Rightarrow$ every subsequence $\{Y_{n_k}\}_k \rightarrow_d B$ as $k\rightarrow \infty$ $\Rightarrow$ $D\sim B$

(6) (3)+(5) $\Rightarrow$ $(X_{n_j}, Y_{n_j})\rightarrow_d (A,B)$

Are this proof and its conclusion correct? Any hint would be really appreciated.

In order to go from (5) to (6), you need to prove that $(A,B)$ has the same distribution as $(C,D)$. But this may not be the case. For example, if $X$ is a symmetric non degenerated distribution and $X_n:=X$, $Y_n:=-X$, $A=B=X$, then $X_n\to A$ and $Y_n\to B$ in distribution. But for each $n$, $(X_n,Y_n)$ has the same distribution as $(X,-X)$, hence no subsequence of $\left(\left(X_n,Y_n\right)\right)_{n\geqslant 1}$ can converge in distribution to $(A,B)=(X,X)$.
• Thanks, may I ask you some clarifications on your example? we have that (i) $X_n:=X\rightarrow_d X$, (ii) $Y_n:=-X\rightarrow_d -X\sim X$ by symmetry, (iii) $(X_n,Y_n):=(X,-X)\sim(X,X)$ by symmetry. Why you say that $(X_n,Y_n)$ cannot converge to $(X,X)$?
• We do not have $(X,-X)\sim (X,X)$: if $\phi$ is the characteristic function of $X$, that of $(X,-X)$ is $\Psi_1(s,t):=\phi(s-t)$, that of $(X,X)$ is $\Psi_2(s,t):=\phi(s+t)$, which is in general different. Dec 21, 2015 at 17:38
• I don't know. Maybe when we can say something when the vector $(X_n,Y_n)$ is Gaussian. You could ask your question about the theorem in van der Vaart's book in an other question. Dec 22, 2015 at 12:02