Determinantal point processes 2

So, to try to understand these determinantal processes better, let’s take a look at the case when the set \(E\) contains only two points, say \(E=\{x,y\}\).  Then the kernel \(K\) that defines the process is a symmetric \(2\times 2\) matrix, say

\[ K = \left(\begin{array}{cc} a & b \\ b& d \end{array}\right) , \]

and the definition of a determinantal process gives us

\[ {\mathbb P}( x \in {\mathfrak S}) = a, \quad {\mathbb P}( y \in {\mathfrak S}) = d, \quad {\mathbb P}( x,y \in {\mathfrak S}) = ad-b^2 \]

for a random subset \(\mathfrak S\).   From these we can obtain by the inclusion-exclusion principle the exact probabilities of all four subsets of \(E\): the probability of \(E=\{x,y\}\) is \(ad-b^2\), that of \(\{x\}\) is \( a – (ad-b^2)\), that of \(\{y\}\) is \(d-(ad-b^2) \), and finally that of \(\emptyset\) is \( 1-a-d+(ad-b^2) = (1-a)(1-d)-b^2 \).

By construction these four quantities add up to 1, but in order to have a valid probability measure we also require that they all be  non-negative.  What condition will bring this about?

Lemma If the matrix \(K\) is positive and contractive (which is to say that its eigenvalues are \(\le 1\), or equivalently that \(\|K\|\le 1\) then the four quantities above are non-negative (and thus we have a determinantal probability measure).

Proof Let the eigenvalues be \(\lambda,\mu \in [0,1]\).  The probabilities defined above are, respectively, \(\lambda \mu, a-\lambda\mu, d-\lambda\mu, (1-\lambda)(1-\mu)\).  The first and fourth quantities are obviously non-negative, and the second and third are as well since \(a,d\) lie in the closed interval whose endpoints are \(\lambda\) and \(\mu\).

Conversely suppose that all the named quantities are strictly positive.  Since \(a,d>0\) are convex combinations of \(\lambda,\mu\), at least one eigenvalue is positive.  The determinant is positive, so they have the same sign, hence they are both positive.  Since also \((1-\lambda)(1-\mu)> 0 \) the eigenvalues are either both less than 1 or both greater than 1.  They can’t both be greater than one or else their sum \(a+d\) would be greater than 2 and therefore one of \(a,d\) would be greater than 1.

Thus at least in the \(2\times 2\) case we have established that determinantal measures are defined by positive contractions.  Of course this is very ad hoc.  To proceed more generally we need to use exterior algebra – topic for the next post.

 

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>