# Probability

• Dec 31st 2009, 05:54 AM
gustavodecastro
Probability
Hello,

I'm needing some help here, please. Suppose that

$
P_n \rightarrow P
$

weakly, where both measures are defined in $R^2$, and that $P_1$ is the marginal of $P_n$.

By Portmanteau's theorem, we have

$
P(A X R) \leq \liminf P_n(A X R) = P_1(A)
$

for all A open. So far, ok. The problem arises now. I've been told that all the above implies

$
P(A X R) = P_1(A)
$

Can someone help me to prove that statement?

*******************

The real problem is the following:
If

$P_n \rightarrow P$

weakly, and if

$P_1$ and $P_2$ are the marginals of $P_n$,

the limit $P$ must have the same marginals.

***************

Thank you very much and happy new year!
• Dec 31st 2009, 08:35 AM
Moo
Hello,

What about that ? (I'm not sure at all)

We know that $P_n$ converges weakly to $P$ iff for any bounded function f, $E[f(X_{n,1},X_{n,2})] \to E[f(X_1,X_2)]$, where [tex](X_{n,1},X_{n,2}) \sim P_n[tex] and $(X_1,X_2) \sim P$

Now let's define the bounded function f : $f=g\circ \pi_1$, where g is any bounded function, and $\pi_1 ~:~ (x_1,x_2)\mapsto x_1$

Then we have (for any bouned function g) $E[g(X_{n,1})]=E[f(X_{n,1},X_{n,2})] \to E[f(X_1,X_2)]=E[g(X_1)]$

which proves that the marginal of $P_n$ converges to the marginal of $P$, so since the marginal of $P_n$ doesn't depend on n, the marginal of P is $P_1$ ?

(I hope you don't get confused with the index of P, because one designs the sequence, the other designs the marginal, but I guess you'll be able to make the difference).

I didn't use the porte-manteau theorem, because the version you have don't seem to get you anywhere...
There are many points that are available on the internet (apart from the liminf stuff).
I have something that says that if, for an open O set of R (more generally, a borelian of R), P( frontier of O )=0, then P_n(O) converges to P(O)
It deals with R, but it should work with Rē.
Then I'm not too sure... You may be able to apply the porte-manteau theorem in its "closed" version : for any closed set C, limsup P_n(C) < P(C)
and consider (R\A)\D, where D is the frontier of A (which is a closed set contained in R\A, by definition)

But I don't know if it's a valuable idea...

again, I'm not sure at all whether my method is correct or not. So just get inspired by it, and if you find something false, tell me (Worried)
• Jan 2nd 2010, 09:47 AM
gustavodecastro
Thank you, Moo.