- #1

johnaphun

- 14

- 0

f(y;β) = (ky

^{2}β

^{(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