Question

Suppose that the random variables X and Y have the following joint p.d.f.:

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ {\begin{aligned}{{}{}}{{\bf{8xy}}}&{{\bf{for}}\,{\bf{0}} \le {\bf{x}} \le {\bf{y}} \le {\bf{1,}}}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{aligned}} \right.\)

Also, let\({\bf{U = }}{\raise0.7ex\hbox{\({\bf{X}}\)} \!\mathord{\left/{\vphantom {{\bf{X}} {\bf{Y}}}}\right.}\!\lower0.7ex\hbox{\({\bf{Y}}\)}}\)and\({\bf{V = Y}}\).

a. Determine the joint p.d.f. of U and V .

b. Are X and Y independent?

c. Are U and V independent?

Short answer

a. The joint p.d.f. of U and V is,

\(f\left( {u,v} \right) = \left\{ {\begin{aligned}{{}{}}{8u{v^3}}&{for\,0 \le u \le 1,\,\,0 \le v \le 1,}\\0&{otherwise,}\end{aligned}} \right.\)

b. X and Y are not independent.

c. U and V are independent.

Step-by-step solution

Given information

The joint p.d.f. of X and Y is,

\(f\left( {x,y} \right) = \left\{ {\begin{aligned}{{}{}}{8xy}&{for\,0 \le x \le y \le 1,}\\0&{otherwise.}\end{aligned}} \right.\)

Finding the joint probability density function of U and V

Let us consider the transformation

\(u = {\raise0.7ex\hbox{$x$} \!\mathord{\left/{\vphantom {x y}}\right.}\!\lower0.7ex\hbox{$y$}}\) and \(v = y\)

So,

\(x = uv\)and\(y = v\)

The Jacobian is,

\(\begin{aligned}{}J &= \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|\\ &= \left| {\begin{aligned}{*{20}{c}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{aligned}} \right|\\ &= \left| {\begin{aligned}{{}{}}v&u\\0&1\end{aligned}} \right|\\ &= \left( {v - 0} \right)\\ &= v\, > 0\end{aligned}\)

The joint p.d.f. of U and V is,

\(\begin{aligned}{}f\left( {u,v} \right) &= f\left( {uv,v} \right) \times J\\ &= 8uv \times v \times v\\ &= 8u{v^3}\end{aligned}\)

Therefore, the joint p.d.f. of U and V is,

\(f\left( {u,v} \right) = \left\{ {\begin{aligned}{{}{}}{8u{v^3}}&{for\,0 \le u \le 1,0 \le v \le 1,}\\0&{otherwise,}\end{aligned}} \right.\)

Checking whether of X and Y are independent.

The marginal pdf of X is

\(\begin{aligned}{}f\left( x \right) &= \int_x^1 {f\left( {x,y} \right)dy} \\ &= \int_x^1 {8xydy} \\ &= 8x\int_x^1 {ydy} \\ &= 8x\left( {\frac{{{y^2}}}{2}} \right)_x^1\\ &= 4x\left( {1 - {x^2}} \right)\end{aligned}\)

The marginal pdf of Y is

\(\begin{aligned}{}f\left( y \right) &= \int_0^y {f\left( {x,y} \right)dx} \\ &= \int_0^y {8xydx} \\ &= 8y\int_0^y {xdx} \\ &= 8y\left( {\frac{{{x^2}}}{2}} \right)_0^y\\ &= 4y\left( {{y^2}} \right)\\ &= 4{y^3}\end{aligned}\)

Here,\(f\left( {x,y} \right) \ne f\left( x \right) \times f\left( y \right)\).

X and Y are not independent.

Checking whether of U and V are independent.

The marginal p.d.f. U is,

\(\begin{aligned}{}f\left( u \right) &= \int_0^1 {f\left( {u,v} \right)dv} \\ &= \int_0^1 {8u{v^3}dv} \\ &= 8u\int_0^1 {{v^3}dv} \\ &= 8u\left( {\frac{{{v^4}}}{4}} \right)_0^1\\ &= \frac{{8u}}{4}\\ &= 2u\end{aligned}\)

The marginal p.d.f. V is,

\(\begin{aligned}{}f\left( v \right) &= \int_0^1 {f\left( {u,v} \right)du} \\ &= \int_0^1 {8u{v^3}du} \\ &= 8{v^3}\int_0^1 {udu} \\ &= 8{v^3}\left( {\frac{{{u^2}}}{2}} \right)_0^1\\ &= \frac{{8{v^3}}}{2}\\ &= 4{v^3}\end{aligned}\)

\(f\left( {u,v} \right) = f\left( u \right)f\left( v \right)\)

Therefore, U and V are independent