1
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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).

$\endgroup$
1
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.