How to show that the inverse Gaussian density integrates to 1?

Question

How to prove $\int_{0}^{\infty}\left[\frac{\lambda}{2\pi x^3}\right]^{1/2}\exp\left\{\frac{-\lambda(x-\mu)^2}{2\mu^2 x}\right\}dx=1$?

Answer 1

You can use the closed form formula for the more general integral

$$\int_{ 0 }^{\infty} (ax^m)^{s} e^{\frac{-b(x-\mu)^2}{x}} dx = 2\,{a}^{s}{\mu}^{ms+1}{{\rm e}^{2b\mu}} {{\rm K_{m s+1}}\left(2\,b\mu\right)},$$

where $\rm{K}_{\nu}(x)$ is the modified Bessel function of the second kind.

Answer 2

Define$$\begin{align}f(x)&:=\sqrt{\frac{\lambda}{2\pi x^3}}\exp-\frac12\left(\frac{\sqrt{\lambda}}{\mu}x^{1/2}-\sqrt{\lambda}x^{-1/2}\right)^2\\&:=\sqrt{\frac{\lambda}{2\pi x^3}}\exp\left(\frac{-2\lambda}{\mu}\right)\exp-\frac12\left(\frac{\sqrt{\lambda}}{\mu}x^{1/2}+\sqrt{\lambda}x^{-1/2}\right)^2,\\F(x)&:=\Phi\left(\frac{\sqrt{\lambda}}{\mu}x^{1/2}-\sqrt{\lambda}x^{-1/2}\right)+\exp\left(\frac{2\lambda}{\mu}\right)\Phi\left(-\frac{\sqrt{\lambda}}{\mu}x^{1/2}-\sqrt{\lambda}x^{-1/2}\right),\end{align}$$where $\Phi$ is the $N(0,\,1)$ CDF. Now verify$$F(0)=0,\,F(\infty)=1,\,F^\prime=f.$$I'll do everything apart from $F^\prime=f$, but you can probably see how to do that now from linear combinations of my two expressions for $f$. As $x\to0^+$, each $\Phi$'s argument $\to-\infty$; as $x\to\infty$, the first $\Phi$ approaches $\Phi(\infty)=1$, while the second approaches $\Phi(-\infty)=0$.

Answer 3

Remember that google is your friend. Here are some sources where you can find the proof

Youtube:

http://www.youtube.com/watch?v=g_bCDcNWcgU

Math StackExchange:

How to show the normal density integrates to 1?

Wikipedia:

http://en.wikipedia.org/wiki/Gaussian_integral