91Ó°ÊÓ

The given p.d.f is

\(f\left( {{x_1},{x_2}} \right) = \left\{ \begin{array}{l}{x_1} + {x_2}\;for\;0 < {x_1} < 1\;and\;0 < {x_2} < 1,\\ = 0,otherwise\end{array} \right.\)

\(\begin{array}{l}Let,\\Y = {X_1}{X_2}\\Z = \frac{{{X_1}}}{{{X_2}}}\end{array}\)

To find the range of Y and Z (both dependent and independent ranges)

Now, Clearly,

\(\)\(\begin{array}{l}0 < Y < 1\\0 < X < 1\end{array}\)

\(\begin{array}{l}{x_1} = \sqrt {yz} \\{x_2} = \sqrt {\frac{y}{z}} \\{\rm{Since,}}\\0 < {x_1} < 1\\ \Rightarrow 0 < \sqrt {yz} < 1 \ldots \left( 1 \right)\end{array}\)

\(\begin{array}{l}0 < {x_2} < 1\\ \Rightarrow 0 < \sqrt {\frac{y}{z}} < 1\\ \Rightarrow 0 < y < z \ldots \left( 2 \right)\end{array}\)

By combining \(\left( 1 \right)\;and\;\left( 2 \right)\), we get,

\(\begin{array}{l}When,\\0 < z < 1\\0 < y < z\\ [range\;of\;y\;dependent\;on\;z]\end{array}\)

\(\begin{array}{l}and\;when,\\z > 1,\\0 < y < \frac{1}{z}\\ \Rightarrow y < z < \frac{1}{y}\\ [range\;of\;z\;dependent\;on\;y]\end{array}\)

The joint distribution of Y as well as Z find,

\(\begin{array}{l}{x_1} = \sqrt {yz} \\{x_2} = \sqrt {\frac{y}{z}} \end{array}\)

Perform the Jacobian transformation

\(\begin{aligned}J &= \left| {\begin{aligned}{}{\frac{{\partial {x_1}}}{{\partial y}}}&{\frac{{\partial {x_1}}}{{\partial z}}}\\{\frac{{\partial {x_2}}}{{\partial y}}}&{\frac{{\partial {x_2}}}{{\partial z}}}\end{aligned}} \right|\\ &= \left| {\begin{aligned}{}{\frac{{\sqrt z }}{{2\sqrt y }}}&{\frac{{\sqrt y }}{{2\sqrt z }}}\\{\frac{1}{{2\sqrt y \sqrt z }}}&{ - \frac{{\sqrt y }}{{2z\sqrt z }}}\end{aligned}} \right|\\ &= \left| {\frac{1}{{2z}}} \right|\end{aligned}\)

The joint pdf of Y and Z is

\(\begin{aligned}{f_y}_z\left( {y,z} \right) &= {f_{{x_1}{x_2}}}\left( {{x_1} \cdot {x_2}} \right)\left| J \right|\\ &= \frac{{\sqrt y }}{{2z}}\left( {\frac{1}{{\sqrt z }} + \sqrt z } \right)\\ &= \frac{{\sqrt y }}{2}\left( {\frac{1}{{\sqrt z }} + \frac{1}{{{z^{\frac{3}{2}}}}}} \right),0 < y < 1,0 < z < \infty ,y < z < \frac{1}{y},0 < yz < 1,0 < \frac{y}{z} < 1\end{aligned}\)

\(\begin{array}{l}o < z < \infty \\{\rm{[Independent}}\;{\rm{range}}\;{\rm{of}}\;{\rm{z]}}\\when,\;0 < z < 1\\ \Rightarrow 0 < y < z\end{array}\)

\(\begin{array}{l}when,\;1 < z < \infty \\ \Rightarrow 0 < y < \frac{1}{z}\\{\rm{[Dependant}}\;{\rm{range}}\;{\rm{of}}\;{\rm{z]}}\end{array}\)

\(\begin{aligned}{f_Z}\left( z \right) &= \int\limits_z {{f_{yz}}\left( {y,z} \right)dy} \\ &= \left\{ \begin{aligned}\int\limits_0^z {\frac{{\sqrt y }}{2}\left( {\frac{1}{{\sqrt z }} + \frac{1}{{{z^{\frac{3}{2}}}}}} \right)dz,0 < z < 1} \\\int\limits_0^{\frac{1}{z}} {\frac{{\sqrt y }}{2}\left( {\frac{1}{{\sqrt z }} + \frac{1}{{{z^{\frac{3}{2}}}}}} \right)dz,1 < z < \infty } \end{aligned} \right.\\ &= \left\{ \begin{aligned}\frac{1}{3}\left( {z + 1} \right),0 < z \le 1,\\\frac{1}{{3{z^3}}}\left( {z + 1} \right),z > 1,\\0,z \le 0.\end{aligned} \right.\end{aligned}\)

Therefore, the required pdf is:

\(g\left( z \right) = \left\{ \begin{array}{l}\frac{1}{3}\left( {z + 1} \right),0 < z \le 1,\\\frac{1}{{3{z^3}}}\left( {z + 1} \right),z > 1,\\0,z \le 0.\end{array} \right.\)