I want to show that the Normal distribution is a member of the exponential family.
I have been working under the assumption that a distribution is a member of the exponential family if its pdf/pmf can be transformed into the form:
$f(x|\theta) = h(x)c(\theta)\exp\{\sum\limits_{i=1}^{k} w_{i}(\theta)t_{i}(x)\}$
This is my approach:
$f(x|\mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}}\exp\{-\frac{(x-\mu)^2}{2 \sigma^2}\}$
Taking the logs:
$\log f(x|\mu, \sigma^2) = -\frac{1}{2}\log(2\pi\sigma^2) - \frac{(x-\mu)^2}{2 \sigma^2}$
Taking the exponential:
$f(x|\mu, \sigma^2) = \exp\{-\frac{1}{2}\log(2\pi\sigma^2)-\frac{(x-\mu)^2}{2\sigma^2}\}$
= $\exp\{-\frac{1}{2}\log(2\pi\sigma^2)-\frac{(x^2 -2\mu + \mu^2)}{2\sigma^2}\}$
= $\exp\{-\frac{1}{2}\log(2\pi\sigma^2)-\frac{x^2}{2\sigma^2} + \frac{2x\mu}{2\sigma^2} - \frac{\mu^2}{2\sigma^2}\}$
= $\exp\{-\frac{1}{2}\log(2\pi\sigma^2)\} \exp\{-\frac{x^2}{2\sigma^2} + \frac{x\mu}{\sigma^2} - \frac{\mu^2}{2\sigma^2}\}$
= $\frac{1}{\sqrt{2\pi\sigma^2}} \exp\{-\frac{x^2}{2\sigma^2} + \frac{x\mu}{\sigma^2} - \frac{\mu^2}{2\sigma^2}\}$
= $\frac{1}{\sqrt{2\pi\sigma^2}} \exp\{-\frac{\mu^2}{2\sigma^2}\} \exp\{-\frac{x^2}{2\sigma^2} + \frac{x\mu}{\sigma^2}\}$
Now I have:
$c(\theta) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\{-\frac{\mu^2}{2\sigma^2}\}$,
$w_{1}(\theta) = -\frac{1}{2\sigma^2}$,
$t_{1}(x) = x^2$,
$w_{2}(\theta) = \frac{\mu}{\sigma^2}$,
$t_{1}(x) = x$
But I am missing $h(x)$.
Am I missing something?
The exponential family distributions take into account indicator functions which help to map the domain of the function. The indicator function for x is sufficient to fill in for h(x).
The final answers with the indicator functions should be as follows:
(|,2) = $\frac{-1}{\sqrt{2^2}}exp(\frac{−^2}{2^2})exp(\frac{−x^2}{2^2} + \frac{x}{^2})*I_{(-\infty, \infty)}(x)*I_{(-\infty, \infty)}()*I_{(-\infty, \infty)}()$
() = $\frac{-1}{\sqrt{2^2}}exp(\frac{−^2}{2^2})*I_{(-\infty, \infty)}()*I_{(-\infty, \infty)}()$,
h(x) = $I_{(-\infty, \infty)}(x)$
Everything else you have is right!