Finding marginal distributions knowing the conditional distributions of an x and y Two random variables $x$ and $y$ with $x ≥ 1$ and $y ≥ 1$ are described by a probability distribution $\frac{dP}{dx dy} = p(x, y)$. The conditional probability distributions have been measured and found to be:
$$ p(x|y) = ye^{−y(x−1)} $$
$$ p(y|x) = xe^{−x(y−1)} $$
How do you then find  the normalized $p_x(x)$ and $p_y(y)$?
 A: Assuming the random variables $X$ and $Y$ take values in $[1,\infty)$, it appears the only consistent formulation is $p(x,y) = Ce^{-xy},$
where 
$$C^{-1} = \int_1^\infty \frac{e^{-x}}{x}\, dx \approx 0.218384.$$
We can verify that $p(x,y)$ is a valid joint probability density function since
$$\int_1^\infty \int_1^\infty p(x,y) \, dx \, dy = C\int_1^\infty \int_1^\infty e^{-xy} \, dx \, dy = C\int_1^\infty \left.\frac{-e^{-xy}}{y}\right|_{x=1}^\infty \, dy \\= C\int_1^\infty \frac{e^{-y}}{y} \, dy = 1.$$
Then the marginal density functions we seek are 
$$p_X(x) = \int_1^\infty Ce^{-xy} \, dy = -C \left. \frac{e^{-xy}}{x}\right|_{y=1}^\infty= C\frac{e^{-x}}{x}, \\ p_Y(y) = \int_1^\infty Ce^{-xy} \, dx = -C \left. \frac{e^{-xy}}{y}\right|_{x=1}^\infty= C\frac{e^{-y}}{y}.$$
Finally we check that the conditional density functions are as specified:
$$p(x|y) = \frac{p(x,y)}{p_Y(y)}= \frac{Ce^{-xy}}{Ce^{-y}/y} =ye^{−y(x−1)},$$
and
$$p(y|x) = \frac{p(x,y)}{p_X(x)}= \frac{Ce^{-xy}}{Ce^{-x}/x} =xe^{−x(y−1)}.$$
