(This has been inspired by a comment exchange with @guy).

Assume we have two infinite sequences of random variables, $\{X_n\}$ and $\{Y_n\}$. Assume the RVs in $\{X_n\}$ are statistically independent from the RVs in $\{Y_n\}$. Assume that $X_n \to_d X$ and $Y_n \to_d Y$.

Is it possible that $X$ and $Y$ are dependent?

If yes, an example?

@guy noted that convergence in distribution is a statement about what is the distribution of the limiting random variable, not what is the limiting random variable. So it would appear that nothing is implied about dependence/independence...
Well, intuitively, while "gradual loss of dependence:" and hence "asymptotic independence" is easy to imagine, I cannot see how dependence is absent for all finite values of the index and then suddenly it emerges at the limit... maybe cases involving discontinuities could do the trick.

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    $\begingroup$ math.stackexchange.com/questions/808884/… $\endgroup$ Commented Mar 10, 2019 at 2:03
  • $\begingroup$ @CarlosCinelli Thanks for the link. It appears that there, the assumption is that the sequence converges almost surely. $\endgroup$ Commented Mar 10, 2019 at 11:16
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    $\begingroup$ Counterexample: let $F$ be the law of a pair of independent Bernoulli$(1/2)$ variables and $G$ be the law of the variable $(Z,Z)$ where $Z$ has a Bernoulli$(1/2)$ distribution. Let every $(X_n,Y_n)$ be governed by the law $F$ and $(X,Y)$ be governed by the law $G$. Then all your assumptions hold but the conclusion does not. $\endgroup$
    – whuber
    Commented Mar 10, 2019 at 16:18
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    $\begingroup$ @whuber Ah, I think now I understand. Check: Let $W, M, Q$ be all Bernoullis$(1/2)$, totally independent from the sequence $X_n$ of other Bernoullis $(1/2)$. Still it is valid to say that $X_n \to_d W$, but also that $X_n \to_d M$, and also $X_n \to_d Q$. Am I right? $\endgroup$ Commented Mar 10, 2019 at 16:36
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    $\begingroup$ @whuber Yes, that's it, finally I got what User guy was trying to say to me all along. Thanks. $\endgroup$ Commented Mar 10, 2019 at 16:47

2 Answers 2


I think I see the disconnect between our positions. The following statement is true.

Let $X_n$ and $Y_n$ be sequences of independent random variables such that $X_n \to X^\star$ and $Y_n \to Y^\star$ in distribution. Then there exists random variables $X$ and $Y$, with $X$ independent of $Y$, such that $(X_n, Y_n) \to (X, Y)$ in distribution such that $X$ and $Y$ marginally have the same distributions as $X^\star$ and $Y^\star$.

The point I am trying to make is that the following statement is not true.

Let $X_n$ and $Y_n$ be sequences of independent random variables, and let $X$ and $Y$ be random variables such that $X_n \to X$ and $Y_n \to Y$ in distribution. Then $X$ and $Y$ are independent.

The reason is that there is no constraint imposed on $(X,Y)$ in the second statement, whereas the first statement explicitly postulates $X$ and $Y$ independent.

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    $\begingroup$ Thanks for your answer, and I totally agree with what you write. What I tried to do in my answer is to show that if in the second, not true in general statement, we add the requirement that one of the two limiting RVs has a continuous CDF, then the statement becomes true. Did I indeed prove it, or am I still missing something? $\endgroup$ Commented Mar 10, 2019 at 14:49
  • $\begingroup$ What my answer says is, that, under the additional assumption, "At the limit joint probabilities will equal the product of the limiting marginal CDFs". Is then possible to say "nevertheless, the limiting RV's may be dependent, so they will have a different joint CDF"? Namely, can we have $$\lim P(X_n \leq x, Y_n\leq y) = G(x)H(y)$$ and at the same time $$P(X \leq x, Y\leq y) \neq G(x)H(y)$$? $\endgroup$ Commented Mar 10, 2019 at 15:01
  • $\begingroup$ @AlecosPapadopoulos I think you are still missing something. WRT your last comment, the statement $\lim P(X_n \le x, Y_n \le y) = G(x) H(y)$ is equivalent to my first comment, but $P(X \le x, Y \le y) = G(x) H(y)$ is equivalent to my second comment. My initial example had $X_n \to X = Z$ and $Y_n \to Y = -Z$, even though $X_n$ and $Y_n$ are independent. The problem is that convergence in distribution is not a statement about convergence of random variables at all, but really about their distributions, so you can't constrain dependence using it. $\endgroup$
    – guy
    Commented Mar 10, 2019 at 15:45
  • $\begingroup$ Probably my confusion comes from the fact that (in)dependence is expressed through the use of the joint CDF and its relation to the marginal CDFs. My previous comment I think includes my essential question, which is "Is it possible that the joint distribution of the limiting RVs is different from the limit of the sequence of joint distributions?" Namely, when can we have $$\lim P(X_n \leq x, Y_n\leq y) \neq P(X \leq x, Y\leq y)$$, even though $X_n \to _d X,\;\; Y_n \to_d Y$? $\endgroup$ Commented Mar 10, 2019 at 16:13

We assume independence for all finite $n$, so

$$P(X_n \leq x,Y_n\leq y) = P(X_n \leq x)\cdot P(Y_n\leq y)$$


$$F_n(x,y) = G_n(x)\cdot H_n(y)$$

We want to determine

$$\lim_{n \to \infty} F_n(x,y) = \lim_{n \to \infty} \big[G_n(x)\cdot H_n(y)\big]$$

under the additional assumptions that

$$G_n(x) \to G(x),\;\;\; H_n(x) \to H(x)$$

So to preserve independence at the limit, we want the conditions under which

$$\lim_{n \to \infty} \big[G_n(x)\cdot H_n(y)\big] = G(x)H(x)$$

which is a familiar "convergence of product of function sequences" case.

Then, by standard results, to ensure the result we need at least one of the two marginal distribution function sequences to converge uniformly, and not just pointwise (so that we can say that the product converges pointwise). (we also need both of them to be bounded, and distribution functions are).

Moreover, we also know that, if the limiting distribution function has no points of discontinuity, then the convergence is uniform (see this post).

It appears then that we can state:

Let two independent sequences of random variables converge in distribution. If one of the two limiting distribution functions is continuous, then independence is preserved at the limit.

It follows that in cases where both limiting distribution functions have discontinuities, it appears that dependence can in principle arise at the limit.

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    $\begingroup$ I don’t see how the conclusion follows from your argument. You need something much stronger than convergence of the marginal distributions. Maybe we could make progress if you can explain why you don’t find the counterexample I gave convincing? You are making a big assumption that $G_n H_ n$ converted to the joint cdf of $(X,Y)$. $\endgroup$
    – guy
    Commented Mar 10, 2019 at 14:26

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