The q-Hahn PushTASEP Fredholm determinant conjecture

Let $x_n(t)$ denote the position of the $n$th particle in the $q$-Hahn PushTASEP started from the step initial configuration, and let $(a;q)_\infty$ denote the $q$-Pochhammer symbol. For $\zeta\in\mathbb{C}\setminus\mathbb{R}_{>0}$, define $K_\zeta$ as the integral operator on a small positively oriented circle around $1$ with kernel

$$K_\zeta(w,w')=\frac{1}{2\pi\sqrt{-1}}\int_{-\infty\sqrt{-1}+\frac12}^{\infty\sqrt{-1}+\frac12}\frac{\pi}{\sin(-\pi s)}(-\zeta)^s\frac{g(w)}{g(q^s w)}\frac{ds}{q^s w-w'},$$

where

$$g(w)=\left(\frac{(\nu w;q)_\infty}{(w;q)_\infty}\right)^n\left(\frac{(\mu w^{-1};q)_\infty}{(\nu w^{-1};q)_\infty}\right)^t\frac{1}{(\nu w;q)_\infty}.$$

The q-Hahn PushTASEP Fredholm determinant conjecture. For the step initial configuration,

$$\mathop{\mathbb{E}}\left[\frac{1}{(\zeta q^{x_n(t)+n};q)_\infty}\right]=\det(I+K_\zeta).$$

This conjecture gives a Fredholm determinantal formula for the $e_q$-Laplace transform of the single-particle position. The identity is proved rigorously when $\nu=0$; the general case is conjectural because only finitely many of the $q$-moments are finite.