# 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.