coincide.

An immediate consequence of the duality is that the $q$-moments $\mathop{\mathbb{E}}\bigl[\prod_{i=1}^{n}q^{y_i(x_i(t)+i)}\bigr]$ of the $q$-Hahn TASEP, as index by time $t$ and the vector $\vec{y}$ solves a difference equation involving the $q$-Hahn Boson operator. Using the Bethe ansatz solvability of this Boson operator allows one to write down explicit contour integral formulas for the $q$-moments, which become particularly nice when the $q$-Hahn TASEP is started from the step [@Corwin2014qmunu] or the half-stationary [@BCPS2014] initial data.

Let us recall the formulas in the step case. Encode elements of $\mathbb{Y}^N_k$ as $\vec n=(n_1\ge \ldots\ge n_k)$, $N\ge n_1$, $n_k\ge0$, where for all $m$ we have $y_m=\#\{i\colon n_i=m\}$.[^2]

\[thm:TASEP\_moments\] Fix $0<q<1$, $0\le {\boldsymbol\upnu}\le{\boldsymbol\upmu}<1$. For any $N,k\ge1$ and $\vec n$ as above, the $q$-moments of the $q$-Hahn PushTASEP with step initial data $x_i(0)=-i$, $i\ge1$, have the form $$\begin{gathered}
        \mathop{\mathbb{E}}
        \Bigl[
            \prod_{i=1}^{k}q^{x_{n_i}(t)+n_i}
        \Bigr]
        =\frac{(-1)^kq^{\frac{k(k-1)}{2}}}{(2\pi\sqrt{-1})^k}
        \oint \frac{dz_1}{z_1}\ldots
        \oint \frac{dz_k}{z_k}
        \prod_{1\le A<B\le k}\frac{z_A-z_B}{z_A-qz_B}
        \\\times\prod_{i=1}^{k}
        \Biggl[
            \left( \frac{1-{\boldsymbol\upnu}z_j}{1-z_j} \right)^{n_j}
            \left( \frac{1-{\boldsymbol\upmu}z_j}{1-{\boldsymbol\upnu}z_j} \right)^t
            \frac{1}{1-{\boldsymbol\upnu}z_j}
        \Biggr].
    \end{gathered}$$ Here all the integration contours encircle $1$ but not $0$ or ${\boldsymbol\upnu}^{-1}$, and for all $B>A$ the $z_A$ contour encircles the $qz_B$ contour.

The parameters $y_i$ are the labels of the $q$-moments. In we prove a duality result for the $q$-Hahn PushTASEP, and in 