Answer

We first calculate the inverse transforms

\[ \begin{aligned} X_{1} &= Y_{1}X_{2} = Y_{1}Y_{2}\newline X_{2} &= Y_{2}\newline \implies J &= \begin{vmatrix} \frac{\partial x_{1}}{y_{1}} &\frac{\partial x_{1}}{y_{2}}\newline \frac{\partial x_{2}}{y_{1}} &\frac{\partial x_{2}}{y_{2}} \end{vmatrix} = \begin{vmatrix} y_{2} &y_{1}\newline 0 &1 \end{vmatrix} = \detm{y_{2}} = y_{2}\newline f_{Y}(y) &= f_{X}(y_{1}y_{2}, y_{2}) \detm{J}\newline &= 10(y_{1}y_{2})(y_{2}^{2})y_{2} = 10y_{1}y_{2}^{4} \end{aligned} \]

since \(y_{2} = x_{2}\), it is always positive.

We can also find the support by considering the inequalities

\[ \begin{aligned} 0 < x_{1} < x_{2} < 1\newline \implies 0 < y_{1}y_{2} < y_{2} < 1\implies 0 < y_{1}, y_{2} < 1 \end{aligned} \]

Marginal distributions can be easily found out from the joint distribution

\[ \begin{aligned} f_{Y_{1}}(y_{1}) &= \int_{0}^{1} 10y_{1}y_{2}^{4} dy_{2} = 2y_{1}\newline f_{Y_{2}}(y_{2}) &= \int_{0}^{1} 10y_{1}y_{2}^{4} dy_{1} = 5y_{2}^{4}\newline \end{aligned} \]

Clearly, \(f(y_{1}, y_{2}) = f(y_{1}) f(y_{2})\) implying that the two are independent.