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I want to show that if $X_n\to^w X$ and $Y_n\to^w Y$ which is 'weak convergence'

and the $X_n,Y_n$ are independent RV's on the same probability space,

Then we also have weak convergence of the random vector $(X_n,Y_n)\to^w (X,Y)$ Apparantly he independence condition is crucial here ..

I only know that the joint probability distribution is the product of both distributions. I'm not sure how this implies weak convergence of the random vector..

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    Are the random variables real valued?2012-11-10
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    yes sorry, forgot to mention that.2012-11-11
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    Use the Cramer Wold then compute the characteristic function using independence.2012-11-12
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    Are $X$ and $Y$ assumed to be independent?2012-11-12
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    No, so aparantly this follows...isn't it true that $X_n,Y_n$ independant implies $X,Y$ independant.?2012-11-12
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    What is $(X_n, Y_n) \xrightarrow{w} (X, Y)$ to you?2012-11-13
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    To me this means that for every continuous $f:\mathbb{R}^2\to \mathbb{R}$ we have $\mathbb{E}f(X_n,Y_n)\to \mathbb{E}f(X,Y)$2012-11-13
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    @DinkyDoe: not every continuous...2012-11-13
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    That is the definition I'm used to work with ..I remember this could be taken as an alternative definition of weak convergence.2012-11-13
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    Convergence in distribution has nothing to do with the domain, while independence does. Consider the probability space $[0,1]^2$ with uniform distirbution and let $X_n(x,y) = x$ and $Y_n(x,y) = y$. Then, $X_n \xrightarrow{w} X_1$ and $Y_n \xrightarrow{w} X_1$. So, the limit might not be independent.2012-11-14
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    oh, i meant continuous and bounded functions f :PP2012-11-21
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    @DinkyDoe: You are welcome! PS: If you use an "@" before my name, I am warned by the system about the message and then I can reply.2012-11-21

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If you do not assume that $X$ and $Y$ are independent, this is false.

Take $\Omega = [0,1]^{\mathbb{N}} \times [0,1]^{\mathbb{N}}$. And let $$ X_n(x,y) = x_n \quad\text{and}\quad Y_n(x,y) = y_n. $$ Then, $X_n \xrightarrow{w} X_1$ and $Y_n \xrightarrow{w} X_1$ (not a typo!).

But $(X_n, Y_n)$ is identically distributed, so $(X_n, Y_n) \xrightarrow{w} (X_1, Y_1)$ which has a different distribution from $(X_1, X_1)$.

For the result to be true, you will have to assume that $X$ and $Y$ are independent. That is, $(X_n, Y_n) \not \xrightarrow{w} (X_1, X_1)$.