3
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for this---is there a place where the proof I outlined clearly went wrong? $\endgroup$ – B Merlot Mar 12 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 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 $k$. $\endgroup$ – Carlo Beenakker Mar 12 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 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 at 12:32

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.