Characteristic Function of Inverse Gaussian Distribution

Question

The pdf of Inverse Gaussian distribution, IG$(\mu,\lambda)$, is :

$$p_X(x)=\sqrt\frac{\lambda}{2\pi x^3}\exp\left[\frac{-\lambda}{2\mu^2x}(x-\mu)^2\right];\quad x>0,\lambda,\mu>0$$

I have to compute the Characteristic Function, $\phi_X(t)$.

$$\phi_X(t)=\mathbb E(e^{itX})=\int_0^\infty e^{itx}\sqrt\frac{\lambda}{2\pi x^3}\exp\left[\frac{-\lambda}{2\mu^2x}(x-\mu)^2\right] \, dx$$

I tried to fall it under Gamma function.

$$\phi_X(t)=\sqrt\frac{\lambda}{2\pi}e^{\lambda/\mu}\int_0^\infty x^{\frac{-3}{2}}\exp\left[-\left(\frac{\lambda}{2\mu^2}+\frac{\lambda}{2x^2}-it\right)x\right]dx$$

$$ = \sqrt\frac{\lambda}{2\pi}e^{\lambda/\mu}\int_0^\infty x^{\frac{-3}{2}}\exp\left[\left(it-\frac{\lambda}{2\mu^2}\right)x-\frac{\lambda}{2x}\right]dx $$

Answer 1

Why not use a computer algebra system to do the manual work for you? Given random variable $X\sim \operatorname{InverseGaussian}(\mu, \lambda)$ with pdf $f(x)$:

... the expectation you seek is simply:

where Expect is a function from the mathStatica add-on to Mathematica (I am one of the authors of the former). There are other packages for Maple etc (I am not an author of that package) that can solve these sorts of problems too, and then you can solve them yourself just as easily.

Answer 2

So what you actually have is called a generalised inverse Gaussian distribution (http://en.wikipedia.org/wiki/Generalized_inverse_Gaussian_distribution)

$$ \sqrt\frac{\lambda}{2\pi}e^{\lambda/\mu}\int_0^\infty x^{\frac{-3}{2}}\exp\left[\left(it-\frac{\lambda}{2\mu^2}\right)x-\frac{\lambda}{2x}\right]dx \\ =C\int_0^\infty x^{-\frac{1}{2}-1}\exp\left(-\frac{1}{2}\left(\left(\frac{\lambda}{\mu^2}-2it\right)x+\frac{\lambda}{x}\right)\right)dx $$

Using the normalising constant of a generalised inverse Gaussian and the constants, $C=\sqrt\frac{\lambda}{2\pi}e^{\lambda/\mu}$, $~a=\frac{\lambda}{\mu^2}-2it$, $~b=\lambda$, $~p=-\frac{1}{2}$ above is:

$$C\frac{2\mathcal{K}_p(\sqrt{ab})}{(a/b)^{p/2}}\int_0^\infty \frac{(a/b)^{p/2}}{2\mathcal{K}_p(\sqrt{ab})}x^{p-1}\exp\left(-\frac{1}{2}\left(ax+\frac{b}{x}\right)\right)dx\\ =C\frac{2\mathcal{K}_p(\sqrt{ab})}{(a/b)^{p/2}}=\sqrt\frac{\lambda}{2\pi}e^{\lambda/\mu}\frac{2\mathcal{K}_{1/2}(\sqrt{ab})}{(a/b)^{1/4}}$$

Where $\mathcal{K}_p$ is the modified besssel function of the second kind, and is invariant to positive or negative indeces, furthermore $\mathcal{K}_{1/2}(u)=\sqrt{\frac{\pi}{2u}}\exp(-u)$

I will let you complete the rest