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Proof required: Sum of squared standard normal random variables is a Chi-square rv

  1. Oct 14, 2008 #1
    If Z1,Z2....Zn are standard normal random variable that are identically and independently distrubuted, then how can one prove that squaring and summing them will produce a Chi-
    squared random variable with n degrees of freedom.

    Any help on this will be greatly appreciated. I am new to this stuff and often get confused in it.

    Stattheory.
     
  2. jcsd
  3. Oct 15, 2008 #2

    statdad

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    Homework Helper

    Re: Proof required: Sum of squared standard normal random variables is a Chi-square r

    Several ways. Start with one, set up

    [tex]
    F(a) = \Pr(X^2 \le a) = \Pr(-\sqrt{a} \le X \le \sqrt{a})
    [/tex]

    and show that [tex] f(a) = F'(a) [/tex] is the density for a central chi-square with 1 degree of freedom.

    Then use the moment=generating method to show that the sum of [tex] n [/tex] chi-squares has a chi-square distribution with [tex] n [/tex] degrees of freedom.

    Or, if you haven't seen moment-generating functions, start as above for 1, then
    show that if [tex] W, Y [/tex] are two independent chi-square random variables, with [tex] n_1 [/tex] and [tex] n_2 [/tex] degrees of freedom, the sum [tex] W + Y [/tex] is chi-square with [tex] n_1 + n_2 [/tex] degrees of freedom. The use induction for your case.

    There are other ways, and I'm sure they will get proposed.
     
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