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When we have two independent sequences or random variables $\{X_{n}\}$ and $\{Y_{n}\}$ for which $X_{n}$ converges weakly to $X$ ( $X_n \overset{w}{\rightarrow}X$) and $Y_n \overset{w}{\rightarrow}Y$ I want to show that holds $$(X_n,Y_n) \overset{w}{\rightarrow}(X,Y)$$

I know that holds for $f\in C_{b}$: $\mathbb{E}_{X}f(X_n)\rightarrow \mathbb{E}f(X)$ and the same for $Y$. I want to show that $\mathbb{E}_{X\times Y}f(X_n,Y_n)\rightarrow \mathbb{E}_{X\times Y}f(X,Y)$

As $X_n$ and $Y_n$ are independent I can write $\mathbb{E}_{X\times Y} = \mathbb{E}_X \times \mathbb{E}_Y$, the product measure.

So $$\mathbb{E}_{X\times Y}f(X_n,Y_n)=\mathbb{E}_X (X_n,Y_n)\times \mathbb{E}_Y (X_n,Y_n)=\mathbb{E}_{X}(X,Y_{n})\times \mathbb{E}_Y (X_n,Y)$$

I now need some help making the last step to get $\mathbb{E}_{X\times Y}(X,Y)$. Could anyone help me with this? I'd really prefer to do it without using characteristic functions if possible.

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Another approach is to look at characteristic functions along with Lévy's continuity theorem. –  Stefan Hansen Nov 14 '12 at 16:54
    
As I've not had characteristic functions yet I would really prefer not to.. –  Math Girl Nov 14 '12 at 16:56
    
What would work is the following: $$ \lim_{n\rightarrow \infty} P(X_n \leq x, Y_n \leq y) = \lim_{n\rightarrow \infty} P(X_n \leq x)P(Y_n \leq y) = P(X \leq x)P(Y \leq y)=P(X \leq x,Y \leq y)$$ For all points(x,y) of continuity of the cdf. We just need to show that the discontinuity set has measure 0. –  Gautam Shenoy Nov 14 '12 at 17:12
    
The following might help: $\partial(A\times B) = (\partial B\times \overline{A}) \cup (\partial A\times \overline{B})$. $\partial$ means boundary. I am urging u to use the portmanteau theorem. –  Gautam Shenoy Nov 14 '12 at 17:26
    
I'm sorry I've never heard of this theorem.. I can't see how this is relevant. I am trying to keep it simple. So again I'd prefer to do it my way if that is possible without using theorems I havent encountered yet. –  Math Girl Nov 14 '12 at 17:30

1 Answer 1

In general your assertion isn't true (see this post). But you can prove that there always exist independent random variables $U,V$ such that $X \sim U$, $Y \sim V$ and $(X_n,Y_n) \stackrel{w}{\to} (U,V)$ (see also this post).

Follows like this: There exist independent random variables $U$, $V$ such that $X \sim U$, $Y \sim V$, $X_n \stackrel{w}{\to} U$, $Y_n \stackrel{w}{\to} V$ (proof). We have $$\mathbb{E}e^{\imath \, (\xi,\eta) \cdot (X_n,Y_n)} = \mathbb{E}e^{\imath \, \xi \cdot X_n} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y_n} \to \mathbb{E}e^{\imath \, \xi \cdot U} \cdot \mathbb{E}e^{\imath \, \eta \cdot V} = \mathbb{E}e^{\imath \, (\xi,\eta) \cdot (U,V)}$$ By Lévy's theorem: $(X_n,Y_n) \stackrel{w}{\to} (U,V)$.

Remark: Your last equation above isn't correct: Since $f: \mathbb{R}^2 \to \mathbb{R}$ you can't simply write the expectation value as a product:

$$\mathbb{E}(f(X_n,Y_n)) = \int f(x,y) \, d\mathbb{P}_{X_n,Y_n}(x,y) = \int f(x,y) \, d\mathbb{P}_{X_n}(x) \, d\mathbb{P}_{Y_n}(y) \\ \not= \int f(x,y) \, d\mathbb{P}_{X_n}(x) \cdot \int f(x,y) \, d\mathbb{P}_{Y_n}(y)$$

This works only if $f(x,y)=g_1(x) \cdot g_2(y)$ where $g_1,g_2 \in C_b$.

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Is there any way of proving this without using characteristic functions? I'd really prefer to do it without. –  Math Girl Nov 14 '12 at 18:06
    
As I already wrote: The problem is that you can only write the integral as a product of integrals if $f$ is a product itself. Probably it suffices to check the equality for these functions $f(x,y) = g_1(x) \cdot g_2(y)$ (since Lévy's theorem is even stronger). But I don't know a proof ... –  saz Nov 14 '12 at 18:15

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