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Math Help - Limiting Distribution

  1. #1
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    Limiting Distribution

    Hello,

    I'm stuck with this problem.

    Let Z_n be \chi^2(n) and let W_n=\frac{Z_n}{n^2}. Find the limiting distribution of W_n.

    I tried getting the mgf of W_n and then take the limit as n \to \infty to the mgf of the limiting distribution
    This is what I got, but I don't know if it's right.
    M_{W_n}(t)=\left( 1- \frac{2t}{n^2} \right)^{-\frac{n}{2}} for t < \frac{n^2}{2}.

    I don't know how to arrange the mgf so I can get an e as I take the limit....


    Thanks in advance.
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  2. #2
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    Quote Originally Posted by akolman View Post
    Hello,

    I'm stuck with this problem.

    Let Z_n be \chi^2(n) and let W_n=\frac{Z_n}{n^2}. Find the limiting distribution of W_n.

    I tried getting the mgf of W_n and then take the limit as n \to \infty to the mgf of the limiting distribution
    This is what I got, but I don't know if it's right.
    M_{W_n}(t)=\left( 1- \frac{2t}{n^2} \right)^{-\frac{n}{2}} for t < \frac{n^2}{2}.

    I don't know how to arrange the mgf so I can get an e as I take the limit....


    Thanks in advance.
    1. Note that \left( 1 - \frac{2t}{n^2}\right)^{-n/2} = \frac{1}{\left( 1 - \frac{2t}{n^2}\right)^{n/2} }.


    2. Note that \left( 1 - \frac{2t}{n^2}\right)^{n/2} = \left( 1 - \frac{\sqrt{2t}}{n}\right)^{n/2} \left( 1 + \frac{\sqrt{2t}}{n}\right)^{n/2}  = \left( 1 + \left[- \frac{\sqrt{2t}}{n}\right]\right)^{n/2} \left( 1 + \frac{\sqrt{2t}}{n}\right)^{n/2}


    3. Substitute m = \frac{n}{2}:

    = \left( 1 + \left[- \frac{\sqrt{\frac{t}{2}}}{m}\right]\right)^{m} \left( 1 + \frac{\sqrt{\frac{t}{2}}}{m}\right)^{m}.


    4. Now take the limit and find that the limiting value of the mgf is 1.

    Therefore the limiting distribution function is the Dirac Delta function (Dirac delta function - Wikipedia, the free encyclopedia).
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  3. #3
    MHF Contributor matheagle's Avatar
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    You can express Z_n=\sum_{k=1}^n Y_k where each Y_k is a \chi^2 with one degree of freedom.

    Thus {Z_n\over n}={\sum_{k=1}^n Y_k\over n}\to E(Y_1)=1 by the Strong Law of Large Numbers.

    (By the way this shows how a t_n\to N(0,1) as n\to\infty .)

    Hence {Z_n\over n^{1+\delta} }={\sum_{k=1}^n Y_k\over n^{1+\delta}}\to 0 for all \delta>0.

    MGF of one means that the rv is 0 with probability 1, same answer.

    E(e^{Xt})=1 means P(X=0)=1.
    Last edited by matheagle; March 21st 2009 at 09:32 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    I wouldn't use conjugates

    \biggl(1-{2t\over n^2}\biggr)^{n/2}= \Biggl[\biggl(1-{2t\over n^2}\biggr)^{n^2/(2t)}\Biggr]^{t/n}

    The term inside the ( ) is approaching 1/e and as the exponent t/n goes to zero,
    as n goes to infinity, you have the MGF going towards 1.
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  5. #5
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    Hello,

    MGF of one means that the rv is 0 with probability 1, same answer.
    I don't understand how you get to the MGF=1 Can you explain please ?


    And by the way, I may be a little picky, but one has to say that in the Law of large numbers, it's an almost sure convergence... (our lecturer insisted so much on it lol)

    Since it converges to 0 almost surely, then it converges to 0 in probability, and hence it converges to 0 in distribution...

    This solves the problem and looking at akolman's previous threads, I'm pretty sure he can use this link between the convergences
    (and I must say, akolman, that I love the questions you ask here, because it helps me study my lessons xD)
    Last edited by Moo; March 22nd 2009 at 12:36 AM.
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  6. #6
    MHF Contributor matheagle's Avatar
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    I (almost) certainly know what almost sure convergence is.
    LOL
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  7. #7
    Moo
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    Quote Originally Posted by matheagle View Post
    I (almost) certainly know what almost sure convergence is.
    LOL
    I don't see what's funny in it.
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  8. #8
    MHF Contributor matheagle's Avatar
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    almost certainly = almost sure
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  9. #9
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    The question is answered and moo is puzzled . A perfect state of affairs at which to close the thread.
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