0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Have you tried converting this to a differential equation? $\endgroup$ – J. M. will be back soon Mar 25 '16 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 '16 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 '16 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 '16 at 15:36
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.