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I'm trying to figure out second moment of the following quantity

$$y = \frac{<x_1, x_2>}{\|x_1\|\|x_2\|}$$

Where $x_1$, $x_2$ are sampled independently from $\mathcal{N}(0, \Sigma)$

This can be solved exactly in 2-dimensions using algebraic manipulation: suppose eigenvalues of $\Sigma$ are $a$ and $b$, then

$$E[y^2] = \frac{(a+b) \\}{\left(\sqrt{a}+\sqrt{b}\right)^2}$$

Is there a similarly elegant expression for $n$ dimensions?

(update, I extended this formula to eigenvalues $a,b,c$ "by analogy" and it seems to hold numerically)

I'm trying to figure out second moment of the following quantity

$$y = \frac{\langle x_1, x_2 \rangle}{\left\|x_1\right\|\left\|x_2\right\|}$$

Where $x_1$, $x_2$ are sampled independently from $\mathcal{N}(0, \Sigma)$

This can be solved exactly in 2-dimensions using algebraic manipulation: suppose eigenvalues of $\Sigma$ are $a$ and $b$, then

$$E[y^2] = \frac{a+b}{\left(\sqrt{a}+\sqrt{b}\right)^2}$$

Is there a similarly elegant expression for $n$ dimensions?

(update, I extended this formula to eigenvalues $a,b,c$ "by analogy" and it seems to hold numerically)

I'm trying to figure out second moment of the following quantity

$$y = \frac{<x_1, x_2>}{\|x_1\|\|x_2\|}$$

Where $x_1$, $x_2$ are sampled independently from $\mathcal{N}(0, \Sigma)$

This can be solved exactly in 2-dimensions using algebraic manipulation: suppose eigenvalues of $\Sigma$ are $a$ and $b$, then

$$E[y^2] = \frac{(a+b) \\}{\left(\sqrt{a}+\sqrt{b}\right)^2}$$

Is there a similarly elegant expression for $n$ dimensions?

I'm trying to figure out second moment of the following quantity

$$y = \frac{<x_1, x_2>}{\|x_1\|\|x_2\|}$$

Where $x_1$, $x_2$ are sampled independently from $\mathcal{N}(0, \Sigma)$

This can be solved exactly in 2-dimensions using algebraic manipulation: suppose eigenvalues of $\Sigma$ are $a$ and $b$, then

$$E[y^2] = \frac{(a+b) \\}{\left(\sqrt{a}+\sqrt{b}\right)^2}$$

Is there a similarly elegant expression for $n$ dimensions?

(update, I extended this formula to eigenvalues $a,b,c$ "by analogy" and it seems to hold numerically)