I am trying to get compute at least the directional component of the following expectation, where $M$ is a symmetric, invertible, PD matrix: $$\mathbb{E}_{v \sim N(0, I)}\left[\frac{vv^T}{||Mv||_2}\right]$$

(Note that the norm is not squared). My approach so far has been to write this as \begin{align} &\mathbb{E}_{v \sim N(0, M)}\left[\frac{M^{-1/2}vv^TM^{-1/2}}{||M^{1/2}v||_2^2}\|M^{1/2}v\|_2\right] \\ &= M^{-1/2}\cdot \mathbb{E}_{v \sim N(0, M)}\left[\frac{vv^T}{||v||_M^2}\|v\|_M\right]\cdot M^{-1/2} \\ &= M^{-1/2}\cdot \mathbb{E}_{v \sim N(0, M)}\left[ \left(\frac{v}{||v||_M}\right)\left(\frac{v}{||v||_M}\right)^\top \|v\|_M\right]\cdot M^{-1/2} \end{align}

Then, it seems as though (and supported by this answer) the first two components in the expectation should be independent of the last one, and since the norm is a scalar and the expectation of the normalized outer product is $I$, the directional component of this expectation is simply M^{-1}.

---- Edit to make my approach more clear ---- Going based on this answer, we can write the aforementioned expectation as \begin{align} &= M^{-1/2}\cdot \mathbb{E}_{v \sim N(0, M)}\left[ \left(\frac{v}{||v||_M}\right)\left(\frac{v}{||v||_M}\right)^\top\right] \mathbb{E}_{v \sim N(0, M)}\left[ \|v\|_M\right]\cdot M^{-1/2} \end{align}

Since I don't care about the scaling factor, and rather only the "direction" of the resulting matrix, I am willing to omit the $\mathbb{E}_{v \sim N(0, M)}\left[ \|v\|_M\right]$ term. The other expectation term should just be identity? Leaving us with $M^{-1}\cdot C$ for some scalar $C$.

Is this correct? And if so, is there a better/more elegant way to solve this problem?


I am limiting initially to the trace of the matrix in the OP.

First of all, because of the isotropy of the distribution of the vector $v$, you may work in a basis where $M$ is diagonal, $M=\text{diag}\,(\mu_1,\mu_2,\ldots\mu_n)$. Then the expectation value you seek is $$I(\{\mu_i\})=\sum_{k=1}^n \mathbb{E}\left[\frac{v_{k}^2}{(\mu_k^2 v_k^2+Q_k)^{1/2}}\right],\;\;\text{with}\;\;Q_k=\sum_{j\neq k}\mu_{j}^2 v_{j}^2.$$ The new variable $Q_k$ is independent of $v_k$, but its distribution is cumbersome if the $\mu_j$'s are all different. An approximation by a chi-squared distribution using the Welch–Satterthwaite formula may be useful.

Even for $n=2$ an exact answer involves special functions (elliptic integrals $K$ and $E$ of the first and second kind): $$I(\mu_1,\mu_2)=\sqrt{\frac{2}{\pi }}\,\frac{1}{{\mu_1 \mu_2 \left(\mu_1^2-\mu_2^2\right)}}\left[\mu_1^3 K\left(1-\frac{\mu_1^2}{\mu_2^2}\right)-\mu_2^3 K\left(1-\frac{\mu_2^2}{\mu_1^2}\right)+\mu_1^2 \mu_2 E\left(1-\frac{\mu_2^2}{\mu_1^2}\right)-\mu_1 \mu_2^2 E\left(1-\frac{\mu_1^2}{\mu_2^2}\right)\right]$$

Instead of the trace, I might consider the individual components

$$I_k(\{\mu_i\})= \mathbb{E}\left[\frac{v_{k}^2}{(\mu_k^2 v_k^2+Q_k)^{1/2}}\right]$$

For $n=2$ I find $$\frac{I_1}{I_2}=\frac{\mu_1^2 \mu_2 E\left(1-\frac{\mu_2^2}{\mu_1^2}\right)-\mu_2^3 K\left(1-\frac{\mu_2^2}{\mu_1^2}\right)}{\mu_1^3 K\left(1-\frac{\mu_1^2}{\mu_2^2}\right)-\mu_1 \mu_2^2 E\left(1-\frac{\mu_1^2}{\mu_2^2}\right)}.$$ The direction $\arctan(I_2/I_1)$ of the vector $\mathbf{I}=(I_1,I_2)$ remains a complicated function of the ratio $\mu_1/\mu_2$, none of the simplifications suggested in the OP seem to appear.

  • $\begingroup$ Thank you for this---is there a place where the proof I outlined clearly went wrong? $\endgroup$
    – B Merlot
    Mar 12 '19 at 12:13
  • $\begingroup$ Also, in the original question I have the outer product of vv^T in the numerator, which should be a d*d matrix; has this been flipped to an inner product here? $\endgroup$
    – B Merlot
    Mar 12 '19 at 12:19
  • 1
    $\begingroup$ OK, I misread your notation; the calculation is for the inner product, so for the trace of the matrix; the average is a diagonal matrix, and for the diagonal elements you just take one of the terms in the sum over $k$. $\endgroup$ Mar 12 '19 at 12:22
  • $\begingroup$ Ah, I see. Is my logic correct for the outer product? (and if not, is the outer product potentially a simpler derivation :) ) $\endgroup$
    – B Merlot
    Mar 12 '19 at 12:24
  • $\begingroup$ I have added an edit to illustrate what I meant in my argument. In particular, I am hoping that not caring about the exact matrix allows me to simplify the result (since I really only care about having a closed form for the direction). My main fears are in (a) splitting up the expectation and (b) evaluating the outer product as Id. $\endgroup$
    – B Merlot
    Mar 12 '19 at 12:32

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