0
$\begingroup$

Exercise: Prove that random samples from the following distributions form exponential families: Poisson, geometric, Gamma.

What I know:

A parametric family with parameter space $\Omega$ and density $f_X(x|\theta)$ with respect to measure $\nu$ on $(\mathcal{X}, \mathcal{B})$ is called an exponential family if $$f_X(x|\theta) = c(\theta)h(x)\exp\bigg(\sum\limits_{i = 1}^k\pi_i(\theta)\tau_i(x)\bigg),$$ for some measurable functions $\pi_1, ..., \pi_k, \tau_1, ..., \tau_k$ and some integer $k$. As the function $c(\theta)$ can be written as $$c(\theta) = 1\bigg/\int h(x)\exp\bigg(\sum\limits_{i = 1}^k\pi_i(\theta)\tau_i(x)\bigg)d\nu (x),$$ we see that we might as well parametrise the family by $$\pi = (\pi_1(\theta), ..., \pi_k(\theta))\in\mathbb{R}^k.$$

What I've tried: take a random sample $X$ from the Poisson distribution. I know that $$f_X(x|\theta) = e^{-\theta}\frac{\theta^x}{x\text{ ! }}, \text{ and that }c(\theta) = 1\bigg/\int h(x)\exp\bigg(\sum\limits_{i = 1}^k\pi_i(\theta)\tau_i(x)\bigg)d\nu (x). $$ I need to find $h(x), \pi_i(\theta)$ and $\tau_i(x)$ such that $$e^{-\theta}\frac{\theta^x}{x\text{ ! }} = \bigg(h(x)\bigg/\int h(x)\exp\bigg(\sum\limits_{i = 1}^k\pi_i(\theta)\tau_i(x)\bigg)d\nu (x)\bigg)\bigg(\exp\big(\sum\limits_{i = 1}^k \pi_i(\theta)\tau_i(x)\big)\bigg). $$ I don't see how I should proceed from here unfortunately..

Question: How do I prove that a random sample of for instance the Poisson distribution forms an exponential family?

Thanks in advance!

$\endgroup$
1
$\begingroup$

$$e^{-\theta}\frac{\theta^x}{x!} = e^{-\theta} \cdot \frac1{x!} \cdot \theta^x \qquad \text{where} \qquad c(\theta) = e^{-\theta}\,,~~h(x) = \frac1{x!}\mathbb{1}_{x\in\mathbb{N}}\,,~~\text{and} \\ \theta^x=\mathrm{Exp}(x\log\theta) \quad\Longleftrightarrow \quad\pi(\theta) = \log \theta\,,~~\tau(x) = x$$ There is only one "natural parameter": the vector $\pi$ is one-dimensional and it is $\pi(\theta) = \log\theta$. The summation with index $i$ for the products of $\pi_i(\theta) \cdot\tau_i(x)$ is just a single term.

The $\mathbb{1}_{x\in \mathbb{N}}$ (where $\mathbb{N}$ is understood as $0,1,2,\ldots$) is the indicator function, and it is a totally legitimate function of $x$ (and of $x$ alone). The $h(x)$ is usually where one takes care of the "domain" (in elementary terms). Note that the uniform distribution (with arbitrary domain $[a,b]$) is not part of the exponential family exactly because the domain necessarily involves the "parameters".

Is this what you want? I apologize if I misunderstood your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.