# Expectation of inverse square under multivariate standard normal

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.

• 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 $p\gt 2.$
– whuber
Dec 7, 2020 at 22:48
• Thank you. Your answer illustrates the general strategy to finding expected values well. Dec 8, 2020 at 0:30

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 $$X+Y\sim\Gamma(a_1+a_2, b)$$.

• 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!
– whuber
Dec 7, 2020 at 22:51
• 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). Dec 8, 2020 at 0:32
• 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.
– whuber
Dec 8, 2020 at 16:29