Expected value of Rayleigh quotient, non-centered Gaussian vector Let $X \sim \mathcal{N}\left(\mu, \Sigma \right)$, and let $A$ be a symmetric matrix. My understanding is that the Rayleigh quotient of vector $X$ is given by:
$$R=\frac{X^T A X}{X^T X}$$
I've been trying unsuccessfully to find an expression for the expected value of R. This seems to be a very well studied and relatively simple random variable, so I'd expect there to be some expression for $E(R)$.
I have seen these related questions(1, 2), but these ask about the generalized Rayleigh quotient, which is more complex.
I have also looked into resources discussing the distributions of quadratic forms. From his PhD thesis  beggining of section 3.2.2  (page 72), I understand that that $E(R) = \frac{E(X^T A X)}{E(X^T X)}$ (i.e. making the denominator $X^T I X$, since $I$ is idempotent as required), which seems to be off, and is in disagreement with simulations I've run. It is likely that I'm misunderstanding their claims though.
I also looked into Mathai & Provost "Quadratic forms in random variables", page 144 onwards (Ratios of Quadratic Forms), but couldn't find an answer. I may have missed the answer though, since most of the book goes over my head.
In case it's useful, this current question comes as a development of this previous question I asked here. It seems that a way to solve my previous problem is by estimating the expected value of the Rayleigh quotient.
Thanks

In general you will have to work with the approximation but, to my knowledge (which is not exhaustive), there is specific case for which the equality you mention can be analytically achieved. Specifically, in this paper you will find the following theorem:

Let $X:= (X_{1},..., X_{n})$ be normally distributed with $E(X)=0$ and
$var(X)=\Omega$ a positive definite matrix and let $A$ be any symmetric matrix.
Under this conditions $E(\frac{X^TAX}{X^T\Omega^{-1}X})$=
$\frac{E(X^TAX)}{E(X^T\Omega^{-1}X)}$ . Moreover we have that
$\frac{X^TAX}{X^T\Omega^{-1}X}$ that and $X^T\Omega^{-1}X$ are independently
distributed.

I could not open the thesis you link in the question, but I think this is probably the result they used.
