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.

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

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

  • $\begingroup$ Really very urgent. $\endgroup$
    – ABC
    Commented Sep 13, 2013 at 13:38
  • 1
    $\begingroup$ Why is it urgent? $\endgroup$ Commented Sep 13, 2013 at 14:09
  • $\begingroup$ @harry The answer is readily available at wikipedia: wiki/Inverse_Gaussian_distribution $\endgroup$
    – Sasha
    Commented Sep 13, 2013 at 14:47
  • $\begingroup$ @StefanHansen My examination is knocking at the door $\endgroup$
    – ABC
    Commented Sep 13, 2013 at 15:49
  • $\begingroup$ @Sasha I have to compute it step by step. The result is also available in the book. $\endgroup$
    – ABC
    Commented Sep 13, 2013 at 15:51

2 Answers 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


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 $f(x)$:

enter image description here

... the expectation you seek is simply:

enter image description here

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.

  • $\begingroup$ It is really excellent. But for my exam i need to do it manually. $\endgroup$
    – ABC
    Commented Sep 13, 2013 at 15:52
  • $\begingroup$ Methinks, then, perhaps, the tags that you have added (probability, statistics, prob distributions, normal distribution) are not the appropriate ones for that which you seek. What you really seek is how to integrate something step-by-step. $\endgroup$
    – wolfies
    Commented Sep 13, 2013 at 16:58
  • 2
    $\begingroup$ @wolfies: It is worth noting that you are a coauthor of mathStatica. $\endgroup$
    – robjohn
    Commented Oct 14, 2013 at 21:01

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