I need some help on the following problem:

Let $X_1$ and $X_2$ be two random variables with the following joint distribution pdf $$f_{X_1,X_2}(x_1,x_2)= \begin{cases} 4x_1x_2, & \text{if } 0<x_1<1,0<x_2<1 \\ 0, & \text{otherwise} \\ \end{cases}.$$ Find the joint pdf of random variables $U=X_1/X_2$ and $V=X_1X_2$.

By using the transformation of the formula, I got the result $f_{U,V}(u,v)=2u^{-1}v$. I hope this is right. However, I may got stuck on the domain of $U$ and $V$. I may feel that the domain for $U$ is $(0,+\infty)$ and for $V$ is $(0,1)$. However, if I integrate the pdf I got, the result is NOT 1.

I really do not know where went wrong. Any help, please.

  • $\begingroup$ I would expect the pdf to have a different form for $u \lt 1$ and $u \gt 1$ $\endgroup$ – Henry Oct 18 '13 at 6:18

The general technique to deduce the domain of the new variables $(u,v)$ from the domain of the old ones $(x_1,x_2)$ is almost the same as the one used to compute the Jacobian:

Express $(x_1,x_2)$ in terms of $(u,v)$ and apply the constraints on $(x_1,x_2)$ to these formulas.

In the present case, $(x_1,x_2)=(\sqrt{uv},\sqrt{vu^{-1}})$ hence the $(u,v)$-domain is defined by $$ u\gt0,\quad v\gt0,\quad \sqrt{uv}\lt1,\quad\sqrt{vu^{-1}}\lt1. $$ or, equivalently, $$ u\gt0,\quad v\gt0,\quad v\lt u\lt v^{-1}, $$ or, still equivalently and perhaps more conveniently, $$ u\gt0,\quad 0\lt v\lt\min\{u,u^{-1}\}. $$ Another general principle is to include the domains in the formulas for the densities, here one gets $$ f_{X_1,X_2}(x_1,x_2)=4x_1x_2\mathbf 1_{0\lt x_1\lt1}\mathbf 1_{0\lt x_2\lt1}, $$ and $$ f_{U,V}(u,v)=2vu^{-1}\mathbf 1_{u\gt0}\mathbf 1_{0\lt v\lt\min\{u,u^{-1}\}}. $$ Addendum: To check the multiplicative constant in the density of $(U,V)$, one can note that $$ f_U(u)=\int_\mathbb R f_{U,V}(u,v)\mathrm dv=u^{-1}\mathbf 1_{u\gt0}\int_0^{\min\{u,u^{-1}\}}2v\mathrm dv=u^{-1}\min\{u,u^{-1}\}^2\mathbf 1_{u\gt0}, $$ hence $$ \int_\mathbb R f_{U}(u)\mathrm du=\int_0^1u\mathrm du+\int_1^\infty u^{-3}\mathrm du=\left.\frac12u^2\right|_0^1-\left.\frac12u^{-2}\right|_1^\infty=1, $$ as this should be.


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