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Maximum of Chi-square RV's

  1. Jan 14, 2010 #1
    Hello,
    I am trying to find the distribution of the maximum of a set of independed Chi-square RV's with 2-degrees of freedom. Actually I only want to find the MEAN value.
    I am using the following formula to find the PDF.

    [tex]f_{X_{\mathsf{max}}}(x) = NF_X(x)^{N-1}f_X(x)[/tex]

    Following PDF and CDF is used:
    [tex]f_X(x)=\frac{1}{2}e^{-\frac{x}{2}}[/tex]
    [tex]F_X(x) = 1-e^{-\frac{x}{2}}[/tex]

    So what I want to find is: (Assuming N variables)
    [tex]E[f_{X_{MAX}}(x)] =N \int_0^R xF_X(x)^{N-1}f_X(x)[/tex]

    I am stuck in a neverending integration by-parts.
    If anybody know any solution to this or any method to find this, please let me know.

    Cheer.s
     
    Last edited: Jan 14, 2010
  2. jcsd
  3. Jan 15, 2010 #2
    Did you mean to say [tex]E[X_{MAX}][/tex].

    Mathematica gives the answer 2*HarmonicNumber[n] but I'm not sure how to derive it.
     
  4. Jan 15, 2010 #3
    Thank you bpet.
    Yes I want to find [tex]E[X_{MAX}][/tex]. Sorry for the mistake.
    I will search for Harmonicanumber. Havent heard of it before.


    Cheers.
     
  5. Jan 24, 2010 #4
    You need to calculate the mean of the maximum of independent Chi-square RVs, is it right? So:

    [tex]E_{X^*}[X]=\,\int_0^{\infty}x\,f_{X^*}(x)\,dx[/tex]

    where [tex]f_{X^*}(x)[/tex] is the first equation you wrote, but the PDF and CDF must be of Chi-square not of exponentials. Then substiute these data into the integration and evaluate the integral.
     
  6. Jan 24, 2010 #5
    Thank you very much David.
    I'm not sure what you meant by but the PDF and CDF must be of Chi-square not of exponentials?

    Because of the exponentials I could not find a closed-from expression there is a pattern and I could find a recursive solution. Could you please explain a bit what you meant earlier?

    Cheers.
     
  7. Jan 24, 2010 #6
    I mean the PDF and CDF in the first equation must be of a Chi-square random variable. To elaborate, suppose that we have n independent and identically distributed Chi-square random variables: [tex]X_1,\,X_2,\ldots,\,X_n[/tex] with [tex]f_X(x)[/tex] and [tex]F_X(x)[/tex] as the PDF and CDF, respectively. Arrange them in ascending order as: [tex]X_{(1)}\leq X_{(2)}\leq X_{(3)} \leq \cdots\leq X_{(n)}[/tex]. Then the PDF of [tex]X_{(n)}[/tex] (the maximum RV) is:

    [tex]f_{X_{(n)}}(x)=\,n\left[F_X(x)]^{n-1}\,f_X(x)[/tex]

    You can find the distributions of a Chi-square RV from any probability book.
     
  8. Jan 24, 2010 #7
    m26k9's expression for the distribution is correct.
     
  9. Jan 24, 2010 #8
    Thank you very much David and Bpet.

    The problem I had was to evaluate the integral and was thinking if there is a commonly known closed-form expression for [tex]f_{X_{(n)}}(x)[/tex]. I guess I have to do with the recursive one.

    Thank you again guys.
     
  10. Jan 25, 2010 #9
    Yes, bpet is right, I am sorry, where Chi-square distributions with two degrees of freedom become as m26k9 wrote with the assumption that the variance is unity. So, the PDF of the maximum RV becomes:

    [tex]f_{X_{(n)}}(x)=\,n\left[F_X(x)\right]^{n-1}\,f_X(x)=\,\frac{n}{2}\left[1-\text{e}^{-x/2}\right]^{n-1}\,\text{e}^{-x/2}[/tex].

    Where using binomial expansion:

    [tex]\left[1-\text{e}^{-x/2}\right]^{n-1}=\sum_{k=0}^{n-1}(-1)^k{n-1\choose k}\text{e}^{-kx/2}[/tex]

    After simple manipulation, use the table of integral to solve the resulting integral.

    Good luck
     
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