I am interested in finding a monotone coupling between two random variables. Let $\alpha_1>\alpha_2$, $b<a$. Define the following two (non-normalized) densities: \begin{align*} f_{12}(x)&:=\exp \left\{ \alpha_2(a-x)+\alpha_1(x-b)-e^{a-x}-e^{x-b} \right\},\\ f_{21}(x)&:=\exp \left\{ \alpha_1(a-x)+\alpha_2(x-b)-e^{a-x}-e^{x-b} \right\}. \end{align*} Trivially, $\int_{\mathbb{R}}f_{12}(x)dx=\int_{\mathbb{R}}f_{21}(x)dx$. Moreover, one can check that for all $y$, we have \begin{equation*} \int_x^{\infty}f_{12}(y)dy \ge \int_x^{\infty}f_{21}(y)dy. \end{equation*}

Then there should exist a monotone coupling between $f_{12}$ and $f_{21}$. That is, we should be able to find a family of probability measures with densities $p_x(y)\ge0$ such that: \begin{equation*} \int_{-\infty}^x p_x(y)dy=1,\qquad \int_y^\infty f_{12}(x)p_x(y)dx=f_{21}(y) \end{equation*} for all $x$ and $y$, respectively.

When $\alpha_1-\alpha_2=1$, one such family is given by \begin{equation*} p^{(1)}_x(y)=\exp\left\{ a-y-e^{a-y}+e^{a-x} \right\}\mathbf{1}_{y\le x}. \end{equation*}

My question is: Can one find a generalization $p_x^{(\alpha_1-\alpha_2)}(y)$ of $p_x^{(1)}$ which works for all $\alpha_1>\alpha_2$?

(Of course, we can take measures with atoms for $p_x(y)$, too - it is not necessary to restrict to densities.)