Show that the Normal distribution is a member of the exponential family

Question

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?

Answer 1

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:

𝑓(𝑥|𝜇,𝜎2) = $\frac{-1}{\sqrt{2𝜋𝜎^2}}exp(\frac{−𝜇^2}{2𝜎^2})exp(\frac{−x^2}{2𝜎^2} + \frac{x𝜇}{𝜎^2})*I_{(-\infty, \infty)}(x)*I_{(-\infty, \infty)}(𝜇)*I_{(-\infty, \infty)}(𝜎)$

𝑐(𝜃) = $\frac{-1}{\sqrt{2𝜋𝜎^2}}exp(\frac{−𝜇^2}{2𝜎^2})*I_{(-\infty, \infty)}(𝜇)*I_{(-\infty, \infty)}(𝜎)$,

h(x) = $I_{(-\infty, \infty)}(x)$

Everything else you have is right!