Can someone write in details how i can derive the Laplace transform of the Inverse Gaussian distribution? I think i am missing something during the calculation of the interval which gives the Laplace transform $L(s)=E(e^{sX})$, so can anyone help by providing the exact calculations?

$$\sqrt{\frac{\lambda}{2\pi x^3}}\text{exp}\bigg(\frac{\lambda (x\mu)^2 }{2\mu ^2 x}\bigg)$$ Is this it? – Gappy Hilmore May 30 '15 at 16:04

Yes, i am trying to use this form of the Inverse Gaussian distribution. – john May 30 '15 at 17:36

$$e^{\frac{\lambda }{\mu }\frac{\sqrt{\frac{\lambda }{\mu ^2}+2 s}}{\sqrt{\frac{1}{\lambda }}}}$$ I would like to see someone do it by hand. – Gappy Hilmore May 30 '15 at 17:52
1 Answers
This is not an easy problem. Lets get rid of some of the parameters so that we will not fill the screen with them. $\lambda=2$ and $\mu=1$. We need to find:
$$\frac{1}{\sqrt \pi}\int _0 ^{\infty}t^{3/2} e^ {\frac{(t1)^2}{t}st} \,dt$$
Expand the exponent:
$$\frac{1}{\sqrt \pi}\int _0 ^{\infty}t^{3/2} e^ {t+2\frac1tst} \,dt$$
$e^{t}$ corresponds to shift in laplace domain, we will take this into account at the last step. We take $e^2$ out:
$$\frac{1}{\sqrt \pi}e^2\int _0 ^{\infty}t^{3/2} e^ {\frac1t}e^{st} \,dt$$
It turns out the laplace transform of $t^{3/2} e^ {\frac1t}$ is $\sqrt{\pi } e^{2 \sqrt{s}}$. EDIT: For completeness, you can check this solution to see how you can obtain this result yourself: enter link description here Or you can look here for a very detailed inverse transformation by Ron Gordon: enter link description here
So we get
$$e^{22 \sqrt{s+1}}$$
In more general:
$$e^{\frac{\lambda }{\mu }\frac{\sqrt{\frac{\lambda }{\mu ^2}+2 s}}{\sqrt{\frac{1}{\lambda }}}}$$
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