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Math Help - Limiting distribution involving chi square distribution

  1. #1
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    Limiting distribution involving chi square distribution

    Let Zn be a chi square distribution with n degrees of freedom. Let Wn = Zn/n^2. Find the limiting distribution of Wn.

    I know the MGF of a chi square distribution is (1 - 2t)^{\frac{-n}{2}}

    E(e^{tW_n}) = E(e^{t \{Z_n}{n^2})= (1-\frac{2t}{n^2})^{\frac{-n}{2}}

    I'm trying to use the taylor expansion (1+b/n)^c = e^bc as n approaces infinity, but can't since the denominator is n^2, and not n, and I'm stuck here. I'm assuming I made a mistake somewhere, but can't figure out where.

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by statmajor View Post
    Let Zn be a chi square distribution with n degrees of freedom. Let Wn = Zn/n^2. Find the limiting distribution of Wn.

    I know the MGF of a chi square distribution is (1 - 2t)^{\frac{-n}{2}}

    E(e^{tW_n}) = E(e^{t \{Z_n}{n^2})= (1-\frac{2t}{n^2})^{\frac{-n}{2}}

    I'm trying to use the taylor expansion (1+b/n)^c = e^bc as n approaces infinity, but can't since the denominator is n^2, and not n, and I'm stuck here. I'm assuming I made a mistake somewhere, but can't figure out where.

    Any help would be appreciated.
    See here: - Wolfram|Alpha

    So now you need to think about a distribution that has a moment generating function equal to 1 .... (see here Dirac delta function - Wikipedia, the free encyclopedia)
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  3. #3
    MHF Contributor matheagle's Avatar
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    na, use the limit of exp(x).

    By that I meant \biggl(1+{a\over x}\biggr)^x\to e^a as x goes towards infinity.

    However, there's nothing to this problem

    {Z_n\over n}\to 1 by the STONG law of large numbers

    Since Z_n can be represented as a sum of i.i.d. chi-squares with 1 df.

    So {Z_n\over n^2}\to 0
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