The Krawtchouk ensemble is defined by a weight: $w(x) = \binom{K}{x}p^x q^{K-x} $ and in fact it comes from a conditioned random walk on $\mathbb{Z}^N$. It is a probability measure on the set $\{ 0, 1, \dots, K\}^N $. Each sequence has probability:

$$ \mathbb{P}[h] = \frac{1}{Z} \prod_{i < j} (h_i - h_j)^2 \prod_{j=1}^N \binom{K}{h_j} \frac{1}{2^K}$$

where $Z$ is a number such that the sum over all sequences is $\Sigma=1$.

What are the correlations of this probability distribution? Is it safe to assume this is a determinantal process? In that case it would be very easy to write down the correlations... E.g.

$$ \mathbb{P}[h_1, h_3] = \frac{1}{Z} (h_3 - h_1)^2 \times \binom{K}{h_1} \frac{1}{2^K} \times \binom{K}{h_3} \frac{1}{2^K}$$

However, I haven't shown it has a Kernel. Certainly the Gaussian Unitary Ensemble (GUE) is determinantal. The distribution of a random Hermitian matrix $N \times N$ is:

$$ \mathbb{P}(\lambda) = \prod_{i < j} (\lambda_i - \lambda_j)^2 e^{- \sum_1^N \lambda_i^2}$$

and these probabilities can all be written as determinants because there is a Kernel:

$$K_N(x,y) = \sum_{i=0}^N \phi_i (x) \, \phi_i(y) \, e^{-(x^2 + y^2)/4} $$

This looks anwful lot like the ensemble I've just written, because of the DeMoivre-Laplace limit theorem $$ \binom{n}{k} \frac{1}{2^n} \simeq \sqrt{ \frac{2}{\pi n }} \exp\left[- \frac{(k - \frac{1}{2}n)^2}{\frac{1}{2}n} \right] $$ so mainly, I am asking if there's similar Kernel for the Krawtchouk distribution above.


1 Answer 1


The short answer is "yes, it is a determinantal process". For more details, see Johansson's paper https://arxiv.org/pdf/math/9906120.pdf (which deals with the edge).

  • 2
    $\begingroup$ And to complete Ofer Zeitouni's answer the kernel is given by the same kernel than for the GUE but replacing the $\phi_i$'s by the (orthonormal) Krawtchouk polynomials, and $e^{-(x^2+y^2)/4}$ by $\sqrt{w(x)w(y)}$; the proof is exactly the same. $\endgroup$ Jan 8, 2018 at 9:21

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