# Proving a distribution is a member of generalised exponential family

• johnaphun
In summary: For the poisson distribution I did the following;f(y;θ) = λye-λ/y!withθ = logλa(phi) = phi = 1b(θ) = eθc(y,phi) = -logy!
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

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?

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) = λye/y!

= exp{log(λy/y!)}

= exp{ylogλ - λ - logy!}
with

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

johnaphun said:
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) = λye/y!

= exp{log(λy/y!)}

= exp{ylogλ - λ - logy!}
with

θ = logλ , a(phi) = phi = 1 , b(θ) = eθ, 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)?

Hi chiro, thanks for the reply.

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

johnaphun said:
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) = λye/y!

= exp{log(λy/y!)}

= exp{ylogλ - λ - logy!}
with

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

johnaphun said:
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:
$$\left[ \log(\beta) - \log(\beta + 3) \right] y + k \log(\beta) - 2 k \log(\beta + 3) + \log(k) + (y + k + 1) \log(y) - \frac{1}{2} \log(y + 1)$$
Can you read off your $\theta$, b, $\phi$, and $c(y, \phi)$. Although, presonally, I don't know what you're doing and what is this generalized exponential.

johnaphun said:
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

## 1. What is a generalised exponential family?

A generalised exponential family is a set of probability distributions that can be written in a specific mathematical form. This form includes a set of parameters that govern the shape of the distribution, and the distributions in this family have certain properties that make them useful for statistical analysis.

## 2. How can you prove that a distribution belongs to the generalised exponential family?

In order to prove that a distribution belongs to the generalised exponential family, you must show that it can be written in the form of the generalised exponential family equation. This equation includes the distribution's probability density function, the parameters that govern the distribution, and a normalising constant. If the distribution can be written in this form, then it is a member of the generalised exponential family.

## 3. What are the benefits of using the generalised exponential family in statistical analysis?

The generalised exponential family has several benefits in statistical analysis. These distributions have simple mathematical forms that make them easy to work with and they have a small number of parameters, making them more interpretable. Additionally, many commonly used distributions, such as the normal, exponential, and gamma distributions, are members of the generalised exponential family.

## 4. Can any distribution be a member of the generalised exponential family?

No, not all distributions can be members of the generalised exponential family. The distribution must have a specific mathematical form in order to belong to this family. Additionally, distributions that are not continuous, such as the binomial or Poisson distributions, cannot be members of the generalised exponential family.

## 5. How is the generalised exponential family useful in hypothesis testing?

The generalised exponential family is useful in hypothesis testing because it allows for the use of standard statistical tests and methods. Many commonly used statistical tests, such as the t-test and ANOVA, are based on distributions that belong to the generalised exponential family. This makes it easier to apply these tests and interpret their results.

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