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Central Limit Theorem Variation for Chi Square distribution?

  1. Apr 25, 2010 #1
    Central Limit Theorem Variation for Chi Square distribution???

    If this question fits into Homework Help, please move it over there. I'm not too sure.

    I encountered the following problem:

    Now, this problem seems fairly similar to a simple proof the central limit theorem. I am damn sure that this problem involves finding the mgf of D2, evaluating it and saying that it is the same as the mgf of a chi square function.

    Can you help me out with setting up the equation? that'll be a big help! Thank you!

    I've given it an attempt, but one attempt at setting up the mgf is all that i ask. Also, if there is any other way which does not involve mgf (i being wrong), please mentione that! Thank you!
     
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 25, 2010 #2
    Re: Central Limit Theorem Variation for Chi Square distribution???

    I have seen a proof of this theorem, which has been proved by assuming the value of the parameter k = 2. Essentially, it just means that the theorem has been proved using ony the asumption tht only two events may occur.

    p1 + p2 = 1 and n1 + n2 = n

    Using this and substituting in the value of D2, we arrive at a uncture where n1 is defined as sum of j from 1 to n of Yij, where Yij = 1 is A1 occurs on the jth repetition and 0 elsewhere.

    Now, central limit theorem is used over the variable n1, and if n is large, it has approximately a normal distribution.


    I distinctly remember someone mentioning the use of Principle of Mathematical Induction to solve this problem. Can anyone of you solve this problem using the PMI? Will be a big help!!!! No mgf involved till now!!!
     
  4. Apr 25, 2010 #3

    statdad

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    Homework Helper

    Re: Central Limit Theorem Variation for Chi Square distribution???

    Begin by looking at the distribution of each

    [tex]
    \frac{(n_i - np_io)}{\sqrt{np_io}}
    [/tex]

    and not that even though there are [tex] n [/tex] of them they satisfy one linear relationship.
     
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