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I wish to find a joint PDF $h(p,b) $or CDF $H(p,b)$, given the following expression:

$(1-L) r \int _0^r\int _0^{\frac{p}{1-L}}h(p,b)dbdp- r L \int _0^r\int _0^{\frac{p-r}{1-L}+\frac{r}{L}}h(p,b)dbdp=0,$

where $0<L<1/2$ and $1/2<L<1$ and $r>0$ is some constaant. I naively tried the simple 

Solve[{Integrate[h[p, b], {p, 0, r}, {b, 0, (p/(1 - L))}] (1 - L) r - Integrate[h[p, b], {p, 0, r}, {b, 0, (1/(1 - L)) (p - r) + (r/L)}] L r == 0 }, h[p, b]]

and of course it "does not work" since I receive the following answer

{{h[p, b] -> 0}}.

This answer is surely incorrect. Ideally, I should get some joint probability distribution function of the two variables $p$ and $b$. Also note that probably there are two solutions: one for $0<L1/2$ and the other for $1/2<L<1$ (so $l=1/2$ is excluded).

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  • $\begingroup$ Have you tried converting this to a differential equation? $\endgroup$
    – J. M.'s slightly less busy ♦
    Mar 25, 2016 at 12:16
  • $\begingroup$ @J.M. If you mean to take partial derivatives with respect o $p$ and $b$, technically I can do that. But for the problem I am trying to solve, I cannot do that. $\endgroup$
    – Beck
    Mar 25, 2016 at 12:49
  • $\begingroup$ I suspect that there must be more than one solution depending on if you need a joint distribution that satisfies the condition for all $L$ in $0<L<1$ and for all $r>0$ or just particular values of $L$ and $r$. For example, suppose $L=1/2$. Then any value of $r>0$ satisfies your condition for any legitimate $h[p,b]$. $\endgroup$
    – JimB
    Mar 25, 2016 at 15:30
  • $\begingroup$ @JimBaldwin Thanks for the answer. My apologies, for I forgot to add additional information on $L$. I will make necessary changes in the original question $\endgroup$
    – Beck
    Mar 25, 2016 at 15:36

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This is more of an extended comment. I wonder if rather than first going after a general solution that you try some specific joint probability density functions and observe what values of $L$ and $r$ satisfy your condition.

For example, consider $p$ and $b$ having independent exponential distributions:

f = (1 - L)*r*Integrate[Exp[-p]*Exp[-b], {p, 0, r}, {b, 0, p/(1 - L)}] – 
  L*r*Integrate[Exp[-p]*Exp[-b], {p, 0, r}, {b, 0, (p - r)/(1 - L) + r/L}]

A contour plot shows the solutions for f==0:

ContourPlot[f == 0, {L, 0.000001, 0.4}, {r, 0.000001, 10}, PlotPoints -> 200, 
  Frame -> True, FrameLabel -> Map[Style[#, Bold, Large, Italic] &, {"L", "r"}]]

Contour plot

I don’t think there are any solutions for $0.5 < L < 1$ for this particular joint density function. In fact other than the solution where $L=0.5$, there doesn’t appear to be any combinations of $L$ and $r$ with $L>0.38198$ that can satisfy the condition.

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