Question

Suppose that X and Y are i.i.d. random variables, and that each has the following p.d.f.:

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\left\{ {\begin{aligned}{{}{}}{{{\bf{e}}^{{\bf{ - x}}}}}&{{\bf{for}}\,{\bf{x > 0,}}}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{aligned}} \right.\)

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

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

b. Are U and V independent?

Short answer

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

\(f\left( {u,v} \right) = \left\{ {\begin{aligned}{{}{}}{v{e^{ - v}}}&{for\,0 < u < 1,v > 0}\\0&{otherwise.}\end{aligned}} \right.\)

b. Yes, U and V are independent.

Step-by-step solution

Given information

The p.d.f. of X is given by,

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{{e^{ - x}}}&{for\,x > 0,}\\0&{otherwise.}\end{aligned}} \right.\)

The p.d.f. of Y is given by,

\(f\left( y \right) = \left\{ {\begin{aligned}{{}{}}{{e^{ - y}}}&{for\,y > 0,}\\0&{otherwise.}\end{aligned}} \right.\)

Finding the joint probability density function U and V

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

\(\begin{aligned}{}f\left( {x,y} \right) &= f\left( x \right) \times f\left( y \right)\\ &= {e^{ - x}}{e^{ - y}}\\ &= {e^{ - \left( {x + y} \right)}};\,for\,x > 0,y > 0\end{aligned}\)

Let, consider the given transformation

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

Now,

\(x = uv\)and\(y = v\left( {1 - u} \right)\)

The Jacobian is,

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

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

\(\begin{aligned}{}f\left( {u,v} \right) &= f\left( {uv,v\left( {1 - u} \right)} \right) \times J\\ &= {e^{ - \left( {uv + v\left( {1 - u} \right)} \right)}} \times v\\ &= v{e^{ - \left( {uv + v - uv} \right)}}\\ &= v{e^{ - v}}\,;0 < u < 1,v > 0,\end{aligned}\)

Therefore, the joint p.d.f.ofU and V is,

\(f\left( {u,v} \right) = \left\{ {\begin{aligned}{{}{}}{v{e^{ - v}}}&{for\,0 < u < 1,v > 0}\\0&{otherwise.}\end{aligned}} \right.\)

Checking whether U and V are independent

The marginal p.d.f. of U is,

\(\begin{aligned}{}f\left( u \right) &= \int_0^\infty {f\left( {u,v} \right)dv} \\ &= \int_0^\infty {v{e^{ - v}}dv} \\ &= v\int_0^\infty {{e^{ - v}}dv} - \int_0^\infty {\left( {\left( {\frac{{dv}}{{dv}}} \right)\int_0^\infty {{e^{ - v}}dv} } \right)dv} \\ &= v\int_0^\infty {{e^{ - v}}dv} + \int_0^\infty {\left( {{e^{ - v}}} \right)dv} \\ &= \left( {v{e^{ - v}} - {e^{ - v}}} \right)_0^\infty \\ &= 1\end{aligned}\)

The marginal p.d.f. of V is,

\(\begin{aligned}{}f\left( v \right) &= \int_0^1 {f\left( {u,v} \right)du} \\ &= \int_0^1 {v{e^{ - v}}du} \\ &= v{e^{ - v}}\left( u \right)_0^1\\ &= v{e^{ - v}}\left( {1 - 0} \right)\\ &= v{e^{ - v}}\end{aligned}\)

So,

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

Therefore, U and V are independent.