Suppose $X$ and $Y$ have conditional distributions given by: \begin{align} f(x|y)&\propto ye^{-yx}\;\;\text{for}\;\;0<x<B<\infty\\ g(y|x)&\propto xe^{-xy}\;\;\text{for}\;\;0<y<B<\infty \end{align}

According to the article, the marginal distribution $g(x)$ is not easy to calculate, but it exists since $B<\infty$.

It is easy to determine that: \begin{align} f(x|y)&= \frac{e^B}{e^B-1}ye^{-yx}\;\;\text{for}\;\;0<x<B<\infty\\ g(y|x)&= \frac{e^B}{e^B-1}xe^{-xy}\;\;\text{for}\;\;0<y<B<\infty \end{align}

Now since $f(x|y)f(y)=f(x,y)=g(y|x)g(x)$, this means that $$\frac{f(y)}{g(x)}=\frac{g(y|x)}{f(x|y)}=\frac{x}{y}.$$

And thus that $$\frac{1}{g(x)}=\frac{1}{g(x)}\int_0^Bf(y)dy=\int_0^B\frac{f(y)}{g(x)}dy=\int_0^B\frac{x}{y}dy=x\ln(y)\Big|_{y=0}^{y=B}=-\infty.$$

This would imply that $g(x)=0$, and by a similar argument, that $f(y)=0$.

I'm not entirely sure what to make of this. Am I to conclude that since the ratio of marginals is an indeterminate form (zero over zero), that this method of calculating them fails in this particular case? and that they aren't in fact both zero.

Also, is this a common technique for calculating marginals given the conditionals? I came up with it while trying to see why the technique of Gibbs sampling couldn't be replaced by the (possibly numerical) evaluation of some (possibly quite difficult) integral.

Finally, like I mentioned above, the article says the marginals are not easy to calculate, which seems to imply that one could in theory calculate them, how would this be done?