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.