Consider $n$ points in the euclidean plane, $p_i = (x_i,y_i)_{1\leq i \leq n}$. Now consider a $2 \times 2$ matrix $M = \left(\begin{array}{cc}a & b\\c& d\end{array}\right)$ a vector $r = \left(\begin{array}{c}e\\f\end{array}\right)$, and the function $\Phi$ representing the c.d.f of a standard normal.

Let $$f:v \rightarrow \Phi(Mv+r)$$ be a mapping between the plane and the unit square. $\Phi$ is taken component by component. We are interested in the product of the jacobian of $f$ across all points $p$.

The higher this product, the more spread out the mapping of the points over the unit square. Intuitively, the points will be most spread out when $r$ and $M$ respectively demean and perform a PCA decomposition of the set of points $p$. I think I may have proved it at some point but I forgot, it definitely sounds truthy.

Beyond the optimal value of $M$ and $r$, one can interpret the product of the jacobians as a probability density, and we would like to sample values of $M$ and $r$ with respect to this density.

Let $$A = \sum_{i=1}^n \left(\begin{array}{ccc}x_i^2& x_i y_i & x_i\\ x_i y_i & y_i^2 & y_i\\ x_i & y_i & 1\end{array}\right)$$

I think that the log product of the jacobians can be written as

$$\sum_{i=1}^n \log |J(f)(p_i)| = C(n) -\frac{1}{2}(a~b~e~c~d~f)\left(\begin{array}{cc}A & 0\\0 & A\end{array}\right)(a~b~e~c~d~f)^{\top}+n \log |a d - b c|$$

where $C(n)$ is a constant. The first term represents the contribution of $\Phi$, and it tries to make the coefficients small to stay in the region where $\Phi$ is most expansive. The second represents the contribution of the multiplication by $M$, and it tries to make the determinant large to spread out the points.

Here's where I'm stuck...

As it is, I have the product of a multivariate gaussian density in $(a~b~e~c~d~f)$ (which is great and easy), and a function of the determinant of $M$, (which isn't).

1) Is there any trick to sample from this density, or to get rid of this determinant?

2) Am I even correct that this is maximized with PCA?

  • $\begingroup$ Since M is 2*2, you could just analytically find its eigenvalues (it will be some simple quadratic equation) and then use the fact that the determinant is the product of the eigenvalues $\endgroup$ – Sid May 13 '15 at 18:51
  • $\begingroup$ how would that help me exactly ? $\endgroup$ – Arthur B. May 13 '15 at 19:14

Actually performing demeaning and PCA first doesn't change the density up to a constant but makes things way simpler! Write the density in terms of the SVD of the matrix, and it depends only on the singular values! Furthermore, it is separable in those singular values, and I can sample them individually from

$f(x) = e^{-\frac{n}{2} x^2} x^n$

pick a random sign for each of them, pick two random rotation matrices, and that's it!

For $n>10$ or so, $f$ is very closely approximated by a gaussian distribution with mean $$\sqrt{\frac{n}{2}} \frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)} ~= 1+\frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$ and variance $$1+\frac{1}{n}-\frac{\Gamma\left(1+\frac{n}{2}\right)\Gamma\left(\frac{n}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)^2}~= \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$

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