Limit law of real-valued independent random variables

Question

Let $X_n$ and $Y_n$ be real-valued independent r.vs, each of whose limit law is $X$ and Y, resp.

i.e $X_n \overset{d}{\to} X$ and $Y_n \overset{d}{\to} Y$ for some r.vs $X$ and $Y$.

Then, are $X$ and $Y$ independent as well?

I don't think it holds, but with the lemma below, I make it, which makes me surprised.

Lemma. $X_{1}, \cdots, X_{n}$ are independent if and only if $$ \Bbb{E}[ f_{1}(X_{1})\cdots f_{n}(X_{n}) ] = \Bbb{E} f_{1}(X_{1}) \cdots \Bbb{E} f_{n}(X_{n})$$ for any bounded continuous functions $f_{1}, \cdots, f_{n}$.

Thus, I wonder if there exists such result.

Anyone, any comments would be helpful. Thanks in advance.

Don't the $f_i$'s have to be Borel-measurable as well? In any case, I don't think this lemma can directly help you. No idea for the actual question but am curious haha — BCLC, Dec 14 '14 at 10:32
each $f_i$ is mapping from $(\mathbb{R}, \mathcal{B}(\mathbb{R})) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$. Thus, I think the continuity of $f_i$ implies measurability. — inmybrain, Dec 14 '14 at 10:37
Beware! You could apply the lemma if you had convergence in distribution for the pair $(X_n,Y_n)$ to $(X,Y)$. — Siméon, Dec 14 '14 at 13:04
Oh right. http://math.stackexchange.com/questions/545159/continuity-implies-borel-measurability — BCLC, Jul 09 '15 at 19:52

score 7 · Accepted Answer · answered Dec 14 '14 at 12:04

7

Then, are $X$ and $Y$ independent as well?

Of course not, consider some nondegenerate random variable $X$, independent sequences $(X_n)$ and $(Y_n)$ i.i.d. distributed like $X$, and $Y=X$.

How you planned to apply the lemma is a mystery.

answered Dec 14 '14 at 12:04

Did

279,727

Oh, you right. Thank you so much. – inmybrain Dec 14 '14 at 12:18
Then, is the lemma true? I wonder where my proof makes a problem. – inmybrain Dec 14 '14 at 12:22
Yes the lemma is true but "how you planned to apply" it to prove a false result "is a mystery". – Did Dec 14 '14 at 12:44
Ok, after double-checking my "false" proof, I'm going to decide whether I post it or not. – inmybrain Dec 14 '14 at 12:57
I found where I made a idiotic logic. Thanks, Did. p.s. I got help from you several times, and I really appreciate every time. – inmybrain Dec 14 '14 at 13:17
You are welcome. What was the faulty step? – Did Dec 14 '14 at 13:18
I would like to know as well @inmybrain – BCLC Jul 09 '15 at 19:51
@BCLC Never mind, I made a wrong step. – inmybrain Jul 10 '15 at 04:52

user3371583 · Answer 2 · 2014-12-14T12:37:45.913

-1

~~I should leave this as a comment but I don't have enough points. I assume that $\forall (m,n) \in \mathbb{N}^2$, $X_n \perp Y_m$.~~

You can use the the properties of the characteristic function of the couple $(X_n, Y_m)$ and Levy's continuity theorem. (I suggest working with the characteristic function because it is continuous, so it's neat not to worry about treating discontinuity points).

edited Dec 14 '14 at 12:37

answered Dec 14 '14 at 11:20

user3371583

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Then, do you mean the claim holds? – inmybrain Dec 14 '14 at 11:23
1

Sorry. Yes it's a true statement. – user3371583 Dec 14 '14 at 11:28
Ok, thanks for comment! : ) – inmybrain Dec 14 '14 at 11:35
How do you plan to apply Lévy's continuity theorem? – Did Dec 14 '14 at 12:06
Nvm. Im an idiot. – user3371583 Dec 14 '14 at 12:36
He can pick independent copies of $X$ and $Y$, or maybe he can get a cvg in probability. Sunday morning is no excuse for what I wrote. – user3371583 Dec 14 '14 at 12:58

Limit law of real-valued independent random variables

2 Answers2