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Thread: Probability

  1. #1
    Junior Member
    Aug 2009



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

    P_n \rightarrow P

    weakly, where both measures are defined in $\displaystyle R^2 $, and that $\displaystyle P_1 $ is the marginal of $\displaystyle 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:

    $\displaystyle P_n \rightarrow P$

    weakly, and if

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

    the limit $\displaystyle P $ must have the same marginals.


    Thank you very much and happy new year!
    Last edited by gustavodecastro; Dec 31st 2009 at 05:06 AM.
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

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

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

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

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

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

    (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
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  3. #3
    Junior Member
    Aug 2009
    Thank you, Moo.

    I'll read it carefully!

    Thank you very much!
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