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I am working on calculating the expected value of the reciprocal of the square of a variable $X$ that follows an Inverse Gaussian distribution with parameters $\mu$ (mean) and $\lambda$ (shape). The probability density function (PDF) of the Inverse Gaussian distribution is given by:

$f(x; \mu, \lambda) = \left(\frac{\lambda}{2\pi x^3}\right)^{\frac{1}{2}} \exp\left(-\frac{\lambda (x-\mu)^2}{2\mu^2 x}\right) $

for $x > 0$. I am trying to find:

$ E\left[\frac{1}{X^2}\right] = \int_{0}^{\infty} \frac{1}{x^2} f(x; \mu, \lambda) \, dx $

which simplifies to:

$ E\left[\frac{1}{X^2}\right] = \left(\frac{\lambda}{2\pi}\right)^{\frac{1}{2}} \int_{0}^{\infty} x^{-\frac{7}{2}} \exp\left(-\frac{\lambda (x-\mu)^2}{2\mu^2 x}\right) \, dx $

I am unsure how to approach solving this integral and am wondering if there is a known closed-form solution or if it generally requires numerical methods for evaluation. Insights or references to relevant techniques or literature would be greatly appreciated.

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    $\begingroup$ X is Gamma distributed in your title and Gaussian distributed in your text... ("gaussian" must be erroneous). $\endgroup$
    – Jean Marie
    Apr 21 at 15:29

2 Answers 2

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You can have a look in the book by V. Seshadri page 53 'The inverse Gaussian ditribution' Oxford Science publications , Clarendon, 1993

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The desired definite integral is given by Eq. 3.471(9) of Gradshteyn and Ryzhik (7th Edition, 2007). It involves the modified Bessel function of the second kind.

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