# From univariate to joint convergence in distribution

Let $$X_n \rightarrow_d X$$ and $$Y_n \rightarrow_d Y$$ where $$X$$ and $$Y$$ are i.i.d standard exponential random variables. However, I do not have that for any $$n$$, $$X_n$$ and $$Y_n$$ are independent.

Can I still have the result that for any continuous sets $$A$$ and $$B$$:

$$lim_{n \rightarrow \infty} P(X_n \in A \cap Y_n \in B) = P(X \in A)P(Y \in B)?$$

Which conditions on the dependence of $$X_n$$ and $$Y_n$$ do I need to claim this?

• What does "they become independent only when $n$ tends to infinity" mean? Commented Aug 12, 2022 at 11:01
• Can you even say $\lim\limits_{n \rightarrow \infty} P(X_n \in A) = P(X \in A)$ here for all measurable $A$? That is not what I would call convergence in distribution, and I would expect $A$ need only be a continuity set or something similar Commented Aug 12, 2022 at 11:06
• You're right. $A$ is a continuous set. I updated the question. Commented Aug 12, 2022 at 11:11
• (1) What's a "continuous set"?? (2) The question is strange, because it only assumes something about the distributions of the sequences of random variables. That doesn't give you any information at all about independence. Thus, your result has no basis. As a simple counterexample, let $X_n=Y_n$ for all $n.$
– whuber
Commented Aug 12, 2022 at 13:17

## 1 Answer

The comment of @whuber below the OP is crucial, and it brings into the surface a confusion that established notation can create to less experienced users like me.

When we write $$X_n \to_d X$$ we only mean $$F_n(x) \to F(x)$$ (these being distribution functions). Namely we only mean that the leading term of the sequence of $$X_n$$'s, viewed on their own, acquires at the limit a distribution that is the same as the marginal distribution of some $$X$$ random variable. Same comments go for $$Y_n$$ and $$Y$$. But then, the information "$$X$$ and $$Y$$ are independent" does not play a role. We only used the symbols $$X$$ and $$Y$$ to invoke their marginal distributions, and not their possible statistical association. The fact that $$X$$ and $$Y$$ are independent has no bearing on what happens to the limits of $$X_n$$ and $$Y_n$$, because the limiting rv's for $$X_n$$ and $$Y_n$$ are not necessarily $$X$$ and $$Y$$. They may be the random variables $$X_n \to_p W_x$$ and $$Y_n \to_p Z_y$$, which may have the marginal distributions of $$X$$ and $$Y$$, but they are not $$X$$ or $$Y$$. They just have the same marginal distributions. But then, $$X$$ and $$Y$$ may be independent while $$W_x$$ and $$Z_y$$ are not.

Only if we can claim, making additional assumptions (see https://stats.stackexchange.com/a/379971/28746), that convergence in distribution does imply convergence in probability, then the statistical relation of $$X$$ and $$Y$$ becomes relevant.

In such a more narrow case, if we have that $$X_n \to_p X$$ and $$Y_n \to_p Y$$, we can say the following:

Assume that $$\{X_n\}$$ and $$\{Y_n\}$$ are comprised of continuous random variables. Then their joint distribution function $$H_n(x,y)$$ has a unique Copula representation

$$H_n(x,y) = C_n\big[F_n(x), G_n(y)\big].$$

Evidently, the Copula is a proper distribution function itself w.r.t $$F_n(x), G_n(y)$$.

Moreover the limit $$C$$ of $$C_n$$ itself will depend on the assumed dependence relation between the limiting random variables. These are assumed independent, so it follows that $$C_n$$ converges to the Independence Copula $$\Pi = F(x)G(y)$$. Now, the limiting distribution functions $$F(x)$$ and $$G(y)$$ are continuous, so the convergence to them of $$F_n(x)$$ and $$G_n(x)$$ is uniform, and distribution functions are bounded. Also the product $$F(x)G(y)$$ is continuous so the Copula converges uniformly. So we can look at $$C_n\big[F_n(x), G_n(y)\big]$$ and see a composition of functions that each converges uniformly to its respective limit. It follows that $$\lim C_n\big[F_n(x), G_n(y)\big] = C\big[F(x), G(y)\big] = \Pi[F(x), G(y)] = F(x)G(y).$$

In other words, under convergence in probability, some "regularity/smooth behavior" assumptions are sufficient, together with limiting independence, to obtain

$$H_n(x,y) \to F(x)G(y).$$

• Re "it follows that $C_n$ converges to the Independence Copula:" why? As far as I can tell, because the assumptions about the $X_n$ and $Y_n$ are only about convergence in distribution, you shouldn't be able to deduce anything at all about the dependence structures of the joint variables $(X_n,Y_n).$ In fact, let $X_n=X=Y_n$ for all $n.$ Don't all the stated assumptions hold (trivially)?
– whuber
Commented Aug 30, 2022 at 18:54
• @whuber The OP writes that the limits $X$ and $Y$ "are i.i.d. standard exponential". So it appears that it is part of the premises here, that the limits are independent rv's . Commented Aug 30, 2022 at 20:43
• @whuber Probably I am missing something here. I understand the distinction that may arise since we do not assume convergence in probability, but I have trouble describing it. Still thinking about it. Commented Aug 30, 2022 at 20:52
• @whuber ok, I think I got it. If convergence is only in distribution, then the information "$X$ nd $Y$ are independent" is irrelevant and misleading -because with convergence in distribution, we only mean the convergence of marginal distribution functions, not convergence to some specific random variables. The shorthand symbols $X_n \to_d X$ is what does the damage. I will change my post accordingly. Commented Aug 30, 2022 at 20:57
• Okay, let's return to my initial query: I still do not fathom how you deduce that $C_n$ converges to anything at all. Exactly what additional assumptions are you making that would imply that?
– whuber
Commented Aug 30, 2022 at 21:14