Are matrix Fisher random variables closed under multiplication?

For those unfamiliar with the jargon, let me unpack the terms above and repose my question.

This is a question about probability distributions on rotations (i.e. on $SO(n)$). We can represent a rotation as an $n\times n$ real-valued matrix $M$. Observe that if know the first $n-1$ columns of $M$, and that the matrix is orthogonal, then the last column is determined. Therefore, to define a probability distribution on $SO(n)$, it is sufficient to consider a probability distribution on the $n \times (n-1)$ matrices whose columns are unit length and orthogonal. The set of these matrices for a fixed $n$ form an object called a Stiefel manifold, denoted $V_{n-1}(R^n)$.

The matrix Fisher distribution provides a probability distribution on $V_{k}(R^n)$ (so I'm interested in the case where $k=n-1$). In particular, consider a fixed $n \times k$ matrix $F$ (not necessarily in $V_{k}(R^n)$). Then X is a matrix Fisher random variable with parameter $F$ if its pdf $\Pr(X|F)$ is proportional to:

$$ \exp(Tr(F^TX)) $$

where $Tr$ is the trace.

Now, suppose that $X$ is a matrix Fisher random variable with parameter $F$, $Y$ is a matrix Fisher random variable with parameter $G$, and $X$ and $Y$ are independent. Consider the (matrix) product:


$Z$ is a random variable on $SO(n)$. So, to restate my original question: Does $Z$ have a matrix Fisher distribution? If so, what is its parameter as a function of $F$ and $G$? In case it makes any difference, I'm primarily interested in $n=3$.

I asked this question on Cross Validated, but didn't get any answers or comments. As I mention in that post, there's a comment in Suvorova, "Bayesian Recursive Estimation on the Rotation Group", 2013 pointing out that since rotation matrices do not, in general commute (for $n\geq 3$), then the mean of the product of matrix Fisher random variables is not necessarily the product of the mean. For (a little) more information on this subject, see Section 13.2 of Mardia and Jupp, "Directional Statistics", 2000.

  • 1
    $\begingroup$ If I understand your definitions correctly, if we choose F to be (a,0,0;0,0,0) and G to be (0,0,0;0,b,0) and a and b are very large, then essentially X is a uniformly random rotation about the x axis and Y is a uniformly random rotation about the y axis. If you compose these two, I don't think you would get a Fisher random rotation, will you? $\endgroup$ Sep 29 '14 at 21:19
  • $\begingroup$ @YoavKallus Thank you! It took me a while to work out the details of your comment (see "answer" below), but that seems to solve my problem. $\endgroup$ Oct 18 '14 at 2:17

Yoav Kallus' comment above is correct; I'm going to sketch a few of the details below for the sake of completeness.

Yoav points out that if we choose $X$ with parameter $$F=\begin{bmatrix}a&0&0\\0&0&0\end{bmatrix}$$ and $Y$ with parameter $$G=\begin{bmatrix}0&0&0\\0&a&0\end{bmatrix},$$ then for $a>0$ the probability of $X$ is maximized when it is of the form $$X=\begin{bmatrix}1&0&0\\0&\cos(\alpha)&\sin(\alpha)\end{bmatrix}$$ for any $\alpha$, and the probability of $Y$ is maximized when it is of the form $$Y=\begin{bmatrix}\cos(\beta)&0&\sin(\beta)\\0&1&0\end{bmatrix}$$ for any $\beta$.

Therefore, as $a$ grows large, the $X$ and $Y$ converge to (the Stiefel manifolds corresponding to) uniform independent rotations about the $x$ and $y$ axes, respectively.

Let $\tilde{X}$ and $\tilde{Y}$ be independent rotations around the $x$ and $y$ axes, respectively. Then we can compute the product $\tilde{Z}$: $$\tilde{Z}=\tilde{X}\tilde{Y}=\begin{bmatrix}1&0&0\\0&\cos(\alpha)&\sin(\alpha)\\0&-\sin(\alpha)&cos(\alpha) \end{bmatrix}\begin{bmatrix}\cos(\beta)&0&\sin(\beta)\\0&1&0\\-\sin(\beta)&0&cos(\beta)\end{bmatrix}$$ $$=\begin{bmatrix}\cos(\beta)&0&\sin(\beta)\\-\sin(\alpha)\sin(\beta)&\cos(\alpha)&\sin(\alpha)\cos(\beta)\\-cos(\alpha)sin(\beta)&-\sin(\alpha)&\cos(\alpha)\cos(\beta)\end{bmatrix}$$ The independence of $\tilde{X}$ and $\tilde{Y}$ and their symmetry around their respective axes imply that: $$E[Z]=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$

Suppose that $Z$ were a matrix Fisher random variable. It would then have some parameter $H$. Now, $E[Z]$ is a sufficient statistic for $H$ (since the matrix Fisher distribution is an exponential family), but then $H$ must be identically zero. In that case, $Z$ would have a uniform distribution. However, the entry in the top row, middle column is identically zero, so clearly this distribution is (highly) non-uniform. Therefore, $Z$ does not have a matrix Fisher distribution.


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.