Let $\mu>0$. The probability density function of the inverse Gaussian distribution (with the shape parameter equal to 1 ) is given by

$$f(x ; \mu)=\frac{1}{\sqrt{2 \pi x^{3}}} \exp \left[-\frac{(x-\mu)^{2}}{2 \mu^{2} x}\right]$$

Show that this is a one-parameter exponential family. What is its natural parameter? Show that this distribution has mean $\mu$ and variance $\mu^{3}$.