Is a normalized version of an exponential family distribution still an exponential family distribution? Here "normalized" means making its mean zero and variance one.
According to the following definition of an exponential family distribution from Wikipedia, since both variance and mean are functions of the parameter $\theta$ alone, after normalization wrt variance we still have an exponential family distribution, but after normalization wrt mean $\mu(\theta)$,
$$
f_{X'}(x|\theta) = h(x+\mu(\theta))\ \exp[\ \eta(\theta) \cdot T(x+\mu(\theta))\ -\ A(\theta)\ ],
$$
I am not sure.
A single-parameter exponential family is a set of probability distributions whose probability density function (or probability mass function, for the case of a discrete distribution) can be expressed in the form
$$
f_X(x|\theta) = h(x)\ \exp[\ \eta(\theta) \cdot T(x)\ -\ A(\theta)\ ]
$$
where $T(x), h(x), η(θ)$, and $A(θ)$ are known functions.An alternative, equivalent form often given is
$$
f_X(x|\theta) = h(x)\ g(\theta) \exp[\ \eta(\theta) \cdot T(x)\ ]\,
$$
or equivalently
$$
f_X(x|\theta) = \exp[\ \eta(\theta) \cdot T(x)\ -\ A(\theta) + B(x)\ ]
$$
The value $θ$ is called the parameter of the family.
Best Answer
The distributions for which this is possible are quite limited. There is a theorem of Dynkin, reported in Ferguson (1962) as Theorem 1, that states that the only one-dimensional exponential families with a location parameter have densities of the form $$f(x)=\exp\left\{ \sum_{i=1}^m e^{\alpha_ix} p_i(x)\right\}$$ against the Lebesgue measure, where the $p_i$'s are polynomials. And there is another result of Lindley, reported in Ferguson (1962) as Theorem 2, that the only one-parameter exponential family with a location parameter is any rescaled log-Gamma distribution. And a third result of Dynkin in this paper of Ferguson (1962) is that the only location-scale exponential families with $k$ parameters are those for which $k$ is even and the log density is a polynomial in $x$ of degree $k$.