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In one of the steps in my lecture notes, the following result was used without proof:

Given $X$ is a $p$-dimensional multivariate normal distribution, where $p\ge 3$, centred on zero, with covariance matrix equal to the $p\times p$ identity matrix, i.e.

$$X\sim N_p(0, I_p)$$ then we have $$\mathbb{E}\left(\frac{1}{\Vert X\Vert^2}\right) = \frac{1}{p-2}.$$

I have tried integrating it by brute force, but it's unwieldy. Also, I thought it might be somehow related to a $\chi^2$ distribution, but there is an inverse so I'm not sure.

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    $\begingroup$ Since $X^2$ has a $\chi^2(p)$ distribution by definition, stats.stackexchange.com/questions/198595/… is the same question and presents simple answers. My answer there explains why it's necessary that $p\gt 2.$ $\endgroup$
    – whuber
    Dec 7, 2020 at 22:48
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    $\begingroup$ Thank you. Your answer illustrates the general strategy to finding expected values well. $\endgroup$ Dec 8, 2020 at 0:30

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The question is the same as asking what is the mean of an inverse-$\chi^2$ distribution with $p$ degrees of freedom.

I could look this up in Wikipedia, but the derivation of the mean is usually via manipulation of the PDF, so I would be remiss to just accept the magical-looking PDF of a Inv-$\chi^2_\nu$:

$$\frac{2^{-\nu/2}}{\Gamma(\nu/2)}\,x^{-\nu/2-1} e^{-1/(2 x)}.$$

(Sidenote: the requirement that $p\ge 3$ is because the integrand has $x^{-\nu / 2}$, and we need the power to be less than 1 for the integral to converge, i.e. $\nu > 2$.)

However, this is easily derived from the PDF of a $\chi^2_k$:

$$\frac{1}{2^{k/2}\Gamma(k/2)}\; x^{k/2-1} e^{-x/2}.$$

How do we get this? We can derive the PDF of $\chi^2_1$ from scratch, getting $\Gamma(1/2, 2)$, and use the fact that if $X\sim \Gamma(a_1, b)$ and $Y\sim \Gamma(a_2, b)$, then $X+Y\sim\Gamma(a_1+a_2, b)$.

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  • $\begingroup$ Your last equation is ambiguous, because usually "$\Gamma(a_1,b)$" refers to a Gamma function and the sum is taken literally as a sum. There's a problem with this informal notation! $\endgroup$
    – whuber
    Dec 7, 2020 at 22:51
  • $\begingroup$ Gamma function takes one variable while the Gamma distribution takes two parameters. It just saves me typing X~G(x,b), Y~G(y,b), then X+Y~G(x+y,b). $\endgroup$ Dec 8, 2020 at 0:32
  • $\begingroup$ The Gamma function often takes two or even three parameters. Don't assume everybody uses the same convention you are familiar with: it's always best to explain your notation. $\endgroup$
    – whuber
    Dec 8, 2020 at 16:29

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