Natural parametrization of pdfs

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SUMMARY

The discussion centers on the natural parametrization of probability density functions (pdfs) within the exponential family. The original pdf is expressed as f(x;θ)=exp[∑_{j=1}^k A_j(θ)B_j(x)+C(x)+D(θ)], where A and D are functions of θ, while B and C depend solely on x. The natural parametrization reformulates this as f(x;φ)=exp[∑_{j=1}^k φ_jB_j(x)+C(x)+D(φ)], with φ_j=A_j(θ). Key questions raised include the method for determining D(φ) and the applicability of natural parametrization across all pdfs in the exponential family.

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WantToBeSmart
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I am struggling to understand the concept of natural parametrization of pdf of exponential family. Say that we have a function with the following pdf:

[tex]f(x;\theta)=exp\left[\sum_{j=1}^k A_j(\theta)B_j(x)+C(x)+D(\theta)\right][/tex]

where A and D are functions of [itex]\theta[/itex] alone and B and C are functions of x alone.

Natural parametrization.

[tex]f(x;\phi)=exp\left[\sum_{j=1}^k \phi_jB_j(x)+C(x)+D(\phi)\right][/tex]

where [tex]\phi_j=A_j(\theta)[/tex]

My two questions are:

1 How to I find [itex]D(\phi)?[/itex]
2 Can we perform natural parametrization on all pdfs belonging to the exponential family? If not why is that the case?

Thank you in advance!
 
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WantToBeSmart said:
1 How to I find [itex]D(\phi)?[/itex]
It isn't clear from your notation whether [itex]\phi[/itex] is a scalar or a vector.

2 Can we perform natural parametrization on all pdfs belonging to the exponential family? If not why is that the case?

I had to look up "natural parameterization" on the web and from that glance, I think what you are asking entails such questions as: How do we write an arbitrary function of two variables as a sum of two functions, each involving one variable? I don't know a procedure for doing that. If you allowed to define new variables then for f(x,y) = xy, one could certainly set p = xy and write f(x,y) as a function of one variable p. For a more complicated function, I think it is a complicated question. Perhaps some other forum member will know the answer.
 

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