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Hello, everyone. I want to calculate the expectation shown in the following formula, where $X$ follows a standard $d$-dimensional multi-variable normal distribution as $X\sim\mathbb{N}(\mathbf{0},\mathbb{I}_{d\times d})$ $$ E_X\left[\frac{X^\top\mathbf{A}XX^\top\mathbf{B}X}{\|X\|^2}\right] $$ where $\|X\|^2=X^\top X$, and $\mathbf{A},\mathbf{B}$ are both real symmetric matrices.

The expectation of the numerator is straightforward and there was result for it. However, the denominator seems to make the problem more difficult. Are there any simple approaches to deal with this problem? Thank you very much!

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    $\begingroup$ You have the spherical symmetry and the radial homogeneity of degree $2$ as opposed to degree $4$ without the denominator. So the expectation of the fraction is just $(E\|X\|^2)/E(\|X\|^4)$ times the expectation of the numerator. $\endgroup$
    – fedja
    Commented Dec 13, 2011 at 12:57
  • $\begingroup$ fedja, I have not made sense of your explanation, but I think you suggested a good way to think about this problem. I will think about it with your help. Thank you very much! $\endgroup$
    – ppyang
    Commented Dec 13, 2011 at 15:40
  • $\begingroup$ fedja, could you please explain your suggestion in more detail? Did you say that the degree is 2 just because 4(degree of numerator)-2(degree of denominator)=2? Think about the problem if I change the $\|X\|^2$ with $\|x\|^4$, and its expectation is just $1/E\|X\|^4$ times the expectation of the numerator? Thank you! $\endgroup$
    – ppyang
    Commented Dec 13, 2011 at 16:01
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    $\begingroup$ $X=RY$ where $Y$ is equidistributed over the unit sphere and $R=\|X\|$. Note that $R$ and $Y$ are independent. Now, let $W(X)$ be your numerator. Then $W(X)=R^4W(Y)$ and, by independence, $EW(X)=EW(Y)ER^4$. Also $\frac{W(X)}{R^2}=R^2W(Y)$ whence the expectation you want is $EW(Y)E(R^2)$. Now just juxtapose these two formulae and recall that $ER^k=E\|X\|^k$. And yes, if you put $\|X\|^4$, you, indeed, get pure $EW(Y)$, so you get what you wrote :) $\endgroup$
    – fedja
    Commented Dec 14, 2011 at 0:34
  • $\begingroup$ fedja, it's so kind of you to help me so much. Your idea is fantastic and I think this also provides a simple approach to solve the problem I had proposed before at mathoverflow.net/questions/81419/… Thank you very much! $\endgroup$
    – ppyang
    Commented Dec 14, 2011 at 2:25

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