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I'm trying to derive the general process of changing variables so that an exponential family has zero carrier measure. Distributions in the exponential family have cdf

$$dF(\mathbf{x}|\boldsymbol\eta) = \exp\left({\boldsymbol\eta \cdot T(\mathbf{x}) - g(\boldsymbol\eta)}\right)\, dh(\mathbf{x}).$$

I guess this is a Lebesgue integral, but I don't understand the notation that well.

I would like to find the function $z$ so that

$$dF(z(\mathbf{y})|\boldsymbol\eta) = \exp\left({\boldsymbol\eta \cdot T(z(\mathbf{y})) - g(\boldsymbol\eta)}\right)\, dh(z(\mathbf{y})).$$

and I want the function $h$ to disappear. I want the it to be the standard Lebesgue measure.

So, for the poisson distribution, $h(x) = \frac{1}{x!}$. What is $z$? Is it just $h^{-1}$?

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If $\,f(x)\in\mathbb{N}\,$ there's no problem taking $\,f(x)!\,$, otherwise you'll have to get into the function gamma and even perhaps into its analytic continuation (on the reals, to the non-integer non-positive ones, assuming your function is real)...which one is it going to be? –  DonAntonio Jun 14 '12 at 18:07
    
Yes, $f(x) \in \mathbb N$, but $x$ is over some countable subset of the reals. (It would be nice to know what that subset is as well.) –  Neil G Jun 14 '12 at 18:08
    
@NeilG But no countable set can be open, so how do you define $f'$? –  user12014 Jun 14 '12 at 18:15
    
Also, if $f$ is integer valued it must be either constant or discontinuous, hence not differentiable... –  user12014 Jun 14 '12 at 18:16
    
One moment please, I'll add my reasoning. –  Neil G Jun 14 '12 at 18:17

2 Answers 2

Mathematica says:

$f(x)\to \text{InverseFunction}\left[\int_1^{\text{$\#$1}} \frac{1}{\Gamma (K[1])} \, dK[1]\&\right]\left[c_1+x\right]$

Maybe this helps. Byt I couldn't transform the result to a better one.

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Ok, so if I understood correctly (and adding some stuff of my own to your comment), we have that $\,f:I\to\mathbb{N}\,$ , where $\,I\subset\mathbb{R}\,$ is open or at least contains some open interval within it (otherwise we won't be able to define its derivative at no point).

But if we're talking of $\,f'(x)\,$ then the function is continuos to the discrete space $\,\mathbb{N}\,$, and such functions are as boring as a friday's evening spent with an insurance agent: they are constant functions and thus their derivative is zero, so unless you want to impose some further conditions (say, different topologies for the reals and/or the naturals, or some other definition domain and range of $\,f\,$), I can't see how can we make some other sense of this question

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