Question

Suppose that \({{\bf{X}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{X}}_{\bf{2}}}\)have a continuous joint distribution

for which the joint p.d.f. is as follows:

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

Find the p.d.f. of \({\bf{Y = }}{{\bf{X}}_{\bf{1}}}{{\bf{X}}_{\bf{2}}}\)

Short answer

The pdf of \(Y = {X_1}{X_2}\)

\(\begin{array}{l}g\left( y \right) = 2\left( {1 - y} \right),0 < y < 1\\{\rm{Therefore,Y}} \sim {\rm{Beta}}\left( {{\rm{1,3}}} \right)\end{array}\)\(\)

Step-by-step solution

Given information

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

Define the new variables and find their range

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

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

Now, Clearly,

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

Find the inverse of the variables and its range

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

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

Finding the dependent range

By combining 1 and 2, we get,

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

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

Finding the joint distribution

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

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

Integrating the pdf with respect to the range of z to get the pdf of y

\(\begin{aligned}{f_Y}\left( y \right) &= \int\limits_y {{f_{yz}}\left( {y,z} \right)dz} \\ &= \int\limits_y^{\frac{1}{y}} {\frac{{\sqrt y }}{2}\left( {\frac{1}{{\sqrt z }} + \frac{1}{{{z^{\frac{3}{2}}}}}} \right)dz} \\ &= 2\left( {1 - y} \right),0 < y < 1\\{\rm{Therefore,Y}} \sim {\rm{Beta}}\left( {{\rm{1,3}}} \right)\end{aligned}\)