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Let $(X_n)$ and $(Y_n)$ be two sequence of random variables. $(X_n)$ and $(Y_n)$ are independent to each other. If $(X_n)$ and $(Y_n)$ have limits in distribution. $(X_n)$ tends to $X$, and $(Y_n)$ tends to $Y$ in distribution. Intuitively $X$ and $Y$ are independent, is this true? How can I prove it?

Thank you very much.

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1 Answer 1

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The limits $X$ and $Y$ are not unique ... one can show that there always exist independent random variables $U$ and $V$ such that $X \stackrel{d}{=} U$ and $Y \stackrel{d}{=} V$ and $X_n \stackrel{d}{\to} U$, $Y_n \stackrel{d}{\to} V$.

Follows like this: Let $X \sim \mu$, $Y \sim \nu$ for distributions $\mu$, $\nu$. There exists a probability space $(\Omega',\mathcal{A}',\mathbb{P}')$ and independent random variables $U$ and $V$ on $\Omega'$ such that $U \sim \mu$, $V \sim \nu$ (Proof: Heinz Bauer: "Probability Theory" (De Gruyter 1996), Theorem 9.5). Since the convergence in distribution only depends on the distribution of the random variables you still have $X_n \stackrel{d}{\to} U$ and $Y_n \stackrel{d}{\to} V$.

So as far as I can see you don't even need the independence of $X_n$ and $Y_n$ (but you'll need it if you want to proof $(X_n,Y_n) \stackrel{d}{\to} (U,V)$ or $X_n + Y_n \to U+V$).

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Thanks for your answer. –  XXX11235 Nov 10 '12 at 3:30
    
@saz Can you give a reference for the existence of $U,V$ such that there are independent? I did not know this result. Thanks in advance –  math Nov 10 '12 at 8:19
    
I should have mentioned that U and V are not necessarily random variables on the same probability space as $X_n$, $Y_n$. (But this isn't a problem, since convergence in distribution ONLY depends on the distribution of the random variables.) ... You'll find the proof that there exists a probability space like that in most probability theory-books, for example "Probability Theory"-Heinz Bauer. –  saz Nov 10 '12 at 10:40

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