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.


1 Answer 1


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 $(A,B)=(X,X)$.

  • $\begingroup$ 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 $(X,X)$? $\endgroup$
    – user299158
    Dec 21, 2015 at 17:35
  • $\begingroup$ 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. $\endgroup$ Dec 21, 2015 at 17:38
  • $\begingroup$ Ok, I see, thanks. Is there any exception, for example when one of the marginals is normal? For example in the proof of Theorem 7.10 in van der Vaart "Asymptotic Statistics" it seems that the author goes from (5) to (6) but I can't understand why. $\endgroup$
    – user299158
    Dec 21, 2015 at 17:57
  • $\begingroup$ 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. $\endgroup$ Dec 22, 2015 at 12:02
  • $\begingroup$ Davide,maybe if we demand the marginals to be independent, we could also have joint convergence? $\endgroup$ Feb 16, 2018 at 18:24

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