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Standard normal limiting function chi squared

  1. Feb 18, 2012 #1
    Hi, I have a question

    If X1,X2,...,Xn are independent random variables having chi-square distribution witn v=1 and Yn=X1+X2+...+Xn, then the limiting distribution of

    (Yn/n) - 1
    Z= --------------- as n->infinity is the standard normal distribution.
    sqrt(2/n)

    I know that Yn has chi-square distribution with v=n, but how to proceed.




    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 18, 2012 #2
    I read this post on physics forum but no one had answered it. I finally found an answer and thought I should share it. The methodology might seem obvious - using the method of MGF (moment generating functions).

    The MGF of Y is that of a chi-squared distribution with n degrees of freedom:
    (1-2t)^(-n/2).

    Applying the transformation Y --> Z will give you an MGF of Z of
    e^(-√(n/2) t) (1-√(2/n) t)^(-n/2).

    Now, this is the million dollar question: you will have to use the series ln⁡(1+x)=x-1/2 x^2+1/3 x^3+⋯.

    This will help you simplify to an MGF of e^(1/2t^2) which is the mgf of a standard normal curve.
     
  4. Feb 19, 2012 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    There is nothing new to prove: this is just an application of the Central Limit Theorem. There is nothing special about chi-squared here: the result holds for any distribution having finite mean and variance, and the proof (using MGF) is almost the same.

    RGV
     
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