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I don't know where to begin to calculate the expectation value of the random variable $1/V$, where $V$ is a random variable with chi-square distribution $\chi^2(\nu)$.

Could somebody help me?

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1 Answer 1

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The pdf of a chi-square distribution is $$\frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-1} e^{-x/2}.$$

So you want to calculate $$\int_0^{\infty} \frac{1}{x} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-1} e^{-x/2} dx = \int_0^{\infty} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-2} e^{-x/2} dx.$$

Rewrite the integrand so that it is the pdf of a $\chi^2(\nu-2)$ random variable, which will then integrate to 1. The leftover constant factor will be the expected value you're looking for.

If you want a more detailed hint, just ask.

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