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 on the whole real line:
\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*}

UPDATE. here's how these densities look like ($f_{12}$ is the rightmost one):
[![densitites][1]][1]

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 $\alpha_1>\alpha_2$? I would like an explicit formula as an answer.**

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


  [1]: https://i.sstatic.net/PNN9i.png