# Zeroing the carrier measure of an exponential family

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

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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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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