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Relationship between the Chi_squared and Gamma Distributions ?

  1. Jan 28, 2014 #1

    It has been a long time since I have worked with pdfs so perhaps someone can help. According to Wikipedia (http://en.wikipedia.org/w/index.php?title=Chi-squared_distribution#Additivity) the pdf of the addition of n independend Chi_squared distributed R.V.s is also Chi_squared distributed but with n*k degrees of freedom where the original R.V.s each had k DofF. It goes on though to say that the mean of n such Chi_squared R.V.s has a Gamma Distribution (http://en.wikipedia.org/w/index.php?title=Chi-squared_distribution#Sample_mean)
    But the mean is just the sum scaled by 1/n, does this imply that the Gamma distribution is essentially the same as the Chi_squared distribution (just compressed along the x-axis) ? Or is the Wikipedia entry wrong ? I just find it odd that there would be two 'standard' distributions that are just transforms of one-another, can anyone anyone confirm this ?


  2. jcsd
  3. Jan 29, 2014 #2

    Stephen Tashi

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  4. Jan 30, 2014 #3
    Thanks to the above reference I have been able to confirm that the sum of n IID Chi_squared(k) RVs will be distributed as:
    Chi_squared(n*k) or equivalently Gamma(nk/2,2)
    I still see a problem with the Wikipedia entry though, it asserts that the mean of n IID Chi_squared(k) RVs will be:
    Gamma(n*k/2,2/n) (see http://en.wikipedia.org/wiki/Chi-squared_distribution#Sample_mean)
    My problem with this is that the mean is just the sum scaled by 1/n => by the standard result for scaling of an RV:
    If X has pdf p(x) then Y = aX has pdf 1/a*p(y/a)
    So the sum of n Chi_squared(k) should be Chi_squared(n*k) and the mean should be n*Chi_squared(nx;nk) or eqivalently n*Gamma(nx;nk/2,2)
    The Wikipedia article lists the distribution of the sample mean of IID Chi_squared RVs as:
    Gamma(x;nk/2,2/n) which is not the same.
    So it seems to me that the Wikipedia entry is incorrect, can anyone confirm/disprove this, I'd rather not change it without someone else confirming as I am clearly not an expert in this subject.
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