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Math Help - Probability

  1. #1
    Junior Member
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    Probability

    Hello,

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

    <br />
P_n \rightarrow P<br />

    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

    <br />
P(A X R) \leq \liminf P_n(A X R) =  P_1(A)<br />

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

    <br />
P(A X R) = P_1(A)<br />

    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!
    Last edited by gustavodecastro; December 31st 2009 at 05:06 AM.
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  2. #2
    Moo
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    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
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  3. #3
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    Thank you, Moo.

    I'll read it carefully!

    Thank you very much!
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