# How to calculate E(1/Y) when Y is Inverse Gaussian distributed?

The Inverse Gaussian Distribution density is : $$\frac{\phi^{\frac{1}{2}}}{\sqrt{2\pi y^3}} exp[\frac{-\phi(y - \mu)^2}{2\mu^2y}]$$

Got to this integral: $$\int_0^\infty \frac{1}{y} \frac{\phi^{\frac{1}{2}}}{\sqrt{2\pi y^3}} exp[\frac{-\phi(y - \mu)^2}{2\mu^2y}]dy$$

What kind of trick do I use to get the value here? Since the score expected value must be 0 the result should be $\frac{1}{\phi}$, but the point is to show that.

• Hint: Evaluate the moment generating function as described at stats.stackexchange.com/a/176814/919 in the case of a Normal ("Gaussian") distribution. Exactly the same technique works: complete the square and compute the integral from your knowledge that the PDF has unit integral. – whuber Sep 7 '18 at 17:25