Jacobian Transformations of Two Random Variables ( probability density function) Let ${Y}_{1}$ and ${Y}_{2}$ be independent,each with density $\frac{1}{{y}^{2}}$ , $y>1$. Consider the transformation ${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$. Find probability density function of random variables $\left({U}_{1},{U}_{2}\right)$.
Here is my solution:
Since ${Y}_{1}$ and ${Y}_{2}$ are independent then $f({y}_{1},{y}_{2})=\frac{1}{{\left({y}_{1}{y}_{2}\right)}^{2}}$ , ${y}_{1}>1$ , ${y}_{2}>1$.
${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$ then ${Y}_{2}={U}_{1}{U}_{2}$ , ${Y}_{1}={U}_{2}-{U}_{1}{U}_{2}$
From here Jacobian is $J=\left|\begin{array}{cc}{u}_{2}& {u}_{1}\\ -{u}_{2}& 1-{u}_{1}\end{array}\right|={u}_{2}\Rightarrow \left|J\right|={u}_{2}$
Then I stuck here because I found the density function $f({u}_{1},{u}_{2})=\frac{{u}_{2}}{{\left({u}_{2}-{u}_{1}{u}_{2}\right)}^{2}{\left({u}_{1}{u}_{2}\right)}^{2}}$  but when I want to check it this function with integral definition I can not define the interval of integral. Any help will be appreciated.
 A: When defining joint support like this, you need to consider to specify the marginal support of which variable first. If you specify $u_1$ first, since
$$ u_1 = \frac {y_2} {y_1 + y_2} $$
then we see that
when $y_1$ fixed, $y_2 \to +\infty$, $u_1 \to 1$;
when $y_1 \to +\infty$, $y_2$ fixed, $u_1 \to 0$.
And obviously $0 < u_1 < 1$ so the support of $U_1$ is indeed $(0, 1)$
Let say we specify $u_1$ first. Then the "conditional" support of $u_2$ will be in terms of $u_1$. Since
$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_2 &> \displaystyle  \frac {1} {1 - u_1} \\  u_2 &> \displaystyle \frac {1} {u_1} \end{cases} $$
So the joint support can be specified as
$$ 0 < u_1 < 1, u_2 > \max\left\{\frac {1} {1-u_1}, \frac {1} {u_1} \right\}$$
Similarly if we want to specify $u_2$ first, obviously
$$u_2 = y_1 + y_2 > 1 + 1 = 2$$
so the marginal support of $U_2$ is $(2, +\infty)$. And
$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_1 &< \displaystyle 1 - \frac {1} {u_2} \\ u_1 &> \displaystyle \frac {1} {u_2} \end{cases} $$
So the joint support can be equivalently specified as
$$ u_2 > 2, \frac {1} {u_2} < u_1 < 1 - \frac {1} {u_2} $$
Verification:
Note
$$ \int \frac {1} {u^2(1-u)^2} du = \frac {1} {1-u} - \frac {1} {u} 
- 2\ln (1-u) + 2\ln u + C$$
Therefore
$$ \begin{align} 
&~ \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1 \\
=&~ \frac {1} {1-1+1/u_2} - \frac {1} {1-1/u_2} 
- 2\ln \left(1-1+\frac {1} {u_2}\right) 
+ 2\ln\left(1- \frac {1} {u_2} \right) \\
&~ - \frac {1} {1-1/u_2} + \frac {1} {1/u_2} 
+ 2\ln \left(1-\frac {1} {u_2}\right) 
- 2\ln\left(\frac {1} {u_2} \right) \\
=&~ 2u_2 - \frac {2u_2} {u_2-1} + 4\ln(u_2-1)\end{align} $$
As a result,
$$ \begin{align} 
&~ \int_2^{+\infty} \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1du_2 \\
=&~ \int_2^{+\infty} \left[\frac {2} {u_2^2} - \frac {2} {u_2^2(u_2-1)} + \frac {4\ln(u_2-1)} {u_2^3} \right] du_2 \\ 
=&~ \left. - \frac {2\ln(u_2-1)} {u_2^2} - \frac {2} {u_2}  \right|_2^{+\infty} \\
=&~ 1 \end{align} $$
