# On the expectation of the reciprocal of a quadratic form of a standard normal variable

$\newcommand{\Tr}{\operatorname{Tr}}$We know that if $x$ is a p-variate standard normal random variable, then $x^T A x$ converges to the $\Tr(A)$. Reference.

Suppose that $p>2$.

Question: What if instead we wanted to find the distribution of $(x^TAx)^{-1}$?

Attempt: The obvious answer is that it should behave like $$\chi_2^{-1}(p) \Tr(A)^{-1}\,\,,$$ which has expectation: $$\frac{1}{(p-2)Tr(A)}\,.$$

Is this true?

• Taking the cases $p=1$ or $p=2$ makes it obvious this result cannot generally be true. – whuber Dec 15 '15 at 19:20
• It's a good nitpick, but your observation is true about the expectation of the inverse chi-square distribution in general. I've added clarity in an edit. – Lepidopterist Dec 15 '15 at 22:28
• It's not a nitpick: it indicates your speculation is very wrong. Simply barring the obviously incorrect cases does not make the other ones any more correct! BTW, in what sense do you mean "converge"? Where is there any sequence? Since your speculated answer involves both $p$ and $A$, they are cannot be varying. Therefore the distribution of $(x^\prime A x)^{-1}$ is whatever it is--and it's certainly not a constant. – whuber Dec 15 '15 at 22:30
• You are right that I was very sloppy in my question. The real question is about the expectation. I meant to say that the random variable clusters around this point, but spoke awkardly because in fact I am working in a setting where p grows. But I am not interested in varying p for the sake of this question. But I don't understand your point about the nitpick. I didn't claim that your observation about p=1 or p=2 proves that what I say is true for p>2. It was just my speculation. – Lepidopterist Dec 16 '15 at 4:13