Proving a distribution is a member of generalised exponential family

[johnaphun]

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

exp{log(f(y;β))}

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 that's not correct

 

[Dickfore]

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?

 

[johnaphun]

Sorry that's 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) = λ^ye^-λ/y!

= exp{log(λ^y/y!)}

= exp{ylogλ - λ - logy!}
with

θ = logλ , a(phi) = phi = 1 , b(θ) = e^θ, c(y,phi) = -logy!

 

[chiro]

Sorry that's 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) = λ^ye^-λ/y!

= exp{log(λ^y/y!)}

= exp{ylogλ - λ - logy!}
with

θ = logλ , a(phi) = phi = 1 , b(θ) = e^θ, c(y,phi) = -logy!

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)?

 

[johnaphun]

Hi chiro, thanks for the reply.

No I'm unaware of these theorems, would be able to explain for me?

 

[Dickfore]

Sorry that's 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) = λ^ye^-λ/y!

= exp{log(λ^y/y!)}

= exp{ylogλ - λ - logy!}
with

θ = logλ , a(phi) = phi = 1 , b(θ) = e^θ, c(y,phi) = -logy!

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

exp{log(f(y;β))}

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 that's not correct

So, your argument in the exponential may be rewritten as:
<br /> \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)<br />
Can you read off your \theta, b, \phi, and c(y, \phi). Although, presonally, I don't know what you're doing and what is this generalized exponential.

 

[chiro]

Hi chiro, thanks for the reply.

No I'm unaware of these theorems, would be able to explain for me?

Transformations allow you to find the form of a distribution given one known PDF and a transformed random variable involving the PDF. A quick google search gave us this:

http://www.ebyte.it/library/docs/math04a/PdfChangeOfCoordinates04.html