If we assume probability density distribution functions of random variables $\alpha$, $\beta$ and $\alpha/ \beta$, we would like to devise a joint distribution of $\alpha$ and $\beta$. Although there does not exist a unique solution, we would like to just find one. Any thoughts ?

$\begingroup$ If we assume that they are positive, then we can take their logarithms and reduce the problem to the case if we know $\alpha, \beta$ and $\alpha+\beta$, which sounds maybe simpler. $\endgroup$ – domotorp Jan 24 '10 at 19:39

1$\begingroup$ I have two guesses: First, the case where alpha can be 0 is quite different from if you exclude that possibility. Second, that you can encode some hard problems this way. $\endgroup$ – Douglas Zare Jan 24 '10 at 21:06

$\begingroup$ Btw, not exactly the same thing, but related: en.wikipedia.org/wiki/Sicherman_dice $\endgroup$ – domotorp Jul 18 '10 at 11:42
I am not entirely certain what your question is. It might be
(i) Is it always possible to find a joint distribution of $(\alpha, \beta)$ for any prescribed distributions of $\alpha, \beta$ and $\alpha / \beta$ ?
(ii) Is it possible to find/calculate a joint distribution from the three distributions when you know the joint distribution exists e.g. because these are observations in an experiment?
(i) is not possible in general. Set $\alpha = \exp(X)$ and $\beta = \exp(Y)$ then $\log(\alpha / \beta) = X + Y$. Now let $X$ and $Y$ be uniform on $[0,1]$ and choose a distribution for $X+Y$ so that $P(X+Y < 0.5) = 1$. This means $P(X > 0.5) = 0$ a contradiction to uniform. A way to visualize this might be looking at mass distributions on the square $[0,1]\times[0,1]$. Prescribing the margins (here uniform) is a restriction on the projections to the axes (i.e. $0\times[0,1]$ and $[0,1]\times 0$) and the remaining freedom is distributing the mass in the square.
(ii) Looks more like statistics than probability. There are a number of ways of coming up with a joint distribution. But you would need to specify more context to find a reasonable approach.