# Expected value of 1/X^2 when X follows an inversed gamma distribution

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.

• X is Gamma distributed in your title and Gaussian distributed in your text... ("gaussian" must be erroneous). Apr 21 at 15:29