Q:

What is the minimal set of expectation values I need in a statistical model?

At least if $\vec v$ is really only a one dimensional parameter, measuring all the moments $\langle v^n \rangle_f$ seems to give me all the information to compute $\langle A \rangle_f$ with $A(v)$ being a function of $v$. So integer powers of the generator of the function algebra $v$ are special functions. 
I wonder if there are other special functions whos expectation are equally useful and if so, how they depend of the system you consider. For example, let $\langle \dots \rangle_f$ be some prescrition to calculate a mean or expectation and
$$\langle A \rangle_f:=\int_\Gamma A(\vec v)\ f(\vec v)\ \text d \vec v,$$
then is there e.g. a statistical significance to $\langle f \rangle_f$?
Maybe this question has a "trivial" yes as answer, because there are more options than taylor series to give a basis for some function algebra of elements $f(v_1,v_2,\dots)$. However, I don't know how this changes if one considers noncomutative generators $[v_1,v_2]\ne0$.
The question comes in part because I'm intimidated by relations like
$$\langle T^n\rangle \equiv \int_0^\infty E^{n-1}\, e^{ - \left( {E /T } \right)^C  }{\rm d}E\ = \tfrac{\Gamma \left(\tfrac{n }{ C }\right)}{C} \cdot T^n,$$
i.e. $\langle A(E) \rangle$ expectation values are apprearently trivial to compute for weights which are polynomial in functions $f_{\ T,C}(E)=e^{ - \left( {E /T } \right)^C}$.

A:

In addition to the polynomial moments, people often consider the Fourier transform of a probability distribution. This is the expected value
$$ \int \rho(V) e^{ikV} dV $$
These exponential moments are clearly enough to reconstruct the distribution completely.
The polynomial moments are not always enough, but the countexamples are badly behaved. The moment
$$ \int x^n \rho(x) dx $$
is also the n-th derivative of the fourier transform $\rho(k)$ at k=0. If you have two distributions $\rho(1)$ and $\rho(2)$ with the same moments, their difference is a function with all moments equal to zero. In Fourier space, such a function has all derivatives zero at the origin. To make such a function, you can use
$$ f(k) = e^{-{1\over k^2}}$$
And Fourier transform, and then take the positive and negative parts of f(x) as separate probability distributions. These two distributions have all their moments equal.
As for the gamma function identity that is bothering you, you rescale the integration variable by T, and this gives the scaling (this is a form of dimensional analysis). The remaining integral is
$$ T^n \int_0^\infty u^{n-1} e^{-u^C} du $$
and a change of variables to $v=u^C$ gives the gamma function:
$$ {T^n\over C} \int_0^\infty v^{{n\over C} -1} e^{-v} dv = \Gamma({n\over C}) {T^n\over C}$$

