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Proving a distribution is a member of generalised exponential family

  1. Mar 29, 2012 #1
    I've been asked to prove that the following distribution is a member of the generalised exponential family of distributions.

    f(y;β) = (ky2β(y+k))/((β+3)(y+2k)(y+1)1/2)

    I know that i have to transform the equation into the form

    f(y) = exp{(yθ-bθ)/a∅ +c(y,∅)}

    and that to do this i should take the exponential of the log


    I understand how to do this for the more simple distributions (poisson, binomial etc) however i always struggle with more complicated ones. Are there any tips or anything to look out for when answering this type of question.

    So far i have
    exp[(y+k)log(βy)-(y+2k)log(β+3)+log ky - (1/2)log(y+1)]

    but i'm pretty sure thats not correct
  2. jcsd
  3. Mar 29, 2012 #2
    what is the empty set symbol in the general formula supposed to be?

    Also, could you please demonstrate how you did it for the Poisson distribution?
  4. Mar 29, 2012 #3
    Sorry thats meant to be phi not an empty set. I typed this without my glasses! It's meant to represent the dispersion/scale parameter

    For the poisson distribution I did the following;

    f(y;θ) = λye/y!

    = exp{log(λy/y!)}

    = exp{ylogλ - λ - logy!}

    θ = logλ , a(phi) = phi = 1 , b(θ) = eθ, c(y,phi) = -logy!
  5. Mar 30, 2012 #4


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    Hey johnaphun and welcome to the forums.

    Are you aware of transformation theorems to find distributions in terms of a distribution U and a transformed distribution f(U)?
  6. Mar 30, 2012 #5
    Hi chiro, thanks for the reply.

    No i'm unaware of these theorems, would be able to explain for me?
  7. Mar 30, 2012 #6
    So, your argument in the exponential may be rewritten as:
    \left[ \log(\beta) - \log(\beta + 3) \right] y + k \log(\beta) - 2 k \log(\beta + 3) + \log(k) + (y + k + 1) \log(y) - \frac{1}{2} \log(y + 1)
    Can you read off your [itex]\theta[/itex], b, [itex]\phi[/itex], and [itex]c(y, \phi)[/itex]. Although, presonally, I don't know what you're doing and what is this generalized exponential.
  8. Mar 30, 2012 #7


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