# Effect of the measure on exponential families

This might be a very naive question.

Wikipedia describes an exponential family as a distribution

$$f(x \mid \theta) = h(x) \exp( - \theta x - A(\theta)),$$

where $$A(\theta) = \log\left(\int h(x) \exp( - \theta x) dx\right).$$

I want to understand the role of the underlying measure or basically the $h(x)$ in the above equation.

For instance, can any function of $\theta$ be written as $A(\theta)$ for some choice of $h(x)$.

• $A$ is the Laplace transform of $h$. Depending on what you mean by "any function," the Laplace transform can be inverted.
– whuber
Oct 20 '15 at 16:31
• $A(x)$ would actually be $\log\left(\int h(x) \exp(-\theta x)\right)$ Dec 20 '17 at 17:19

## 1 Answer

This is neither a complete nor thorough answer, but an illustration instead.

Consider the product space $\{0, 1\}^n$ that serves as sample space for the extraction of $n$ i.i.d. Bernoulli r.v.s with parameter $p\in(0, 1)$ $$P(X_1=x_1, \ldots, X_n=x_n) = p^{x_1+\ldots+x_n}(1-p)^{n-x_1-\ldots-x_n}$$ and their transformation into the binomial r.v. $X:=X_1+\ldots+X_n \sim\text{Binom}(n, p)$ with $$P(X=x) = \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x},$$ which can be written as member of the exponential family in the form $$P(X=x) = \exp\left\{ x \cdot \log\left(\frac{p}{1-p}\right) + n \log(1-p) + \log\left(\frac{n!}{x!(n-x)!}\right) \right\},$$ which would give $A(p)=n \log(1-p)$ and $h(x)=\log\left(\frac{n!}{x!(n-x)!}\right)$.

In that example, $h(x)$ accounts for the multiplicity of elements $(x_1, \ldots, x_n)$ of $\{0, 1\}^n$ which yield $x_1+\ldots+x_n=x$ $$h(x) = \log\left( \#\{ (x_1, \ldots, x_n)\in\{0, 1\}^n : x_1+\ldots+x_n=x \} \right) = \log\left(\frac{n!}{x!(n-x)!}\right).$$