Question

Question:Suppose that the joint p.d.f. ofXandYis as follows:

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}2x{e^{ - y}}\;for\;0 \le x \le 1\;and\;0 < y < \infty \\0\;otherwise\end{array} \right.\)

AreXandYindependent?

Short answer

Yes. X and Y are independent.

Step-by-step solution

Given information

There are two random variables, X and Y. The joint distribution of those two random variables is,

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}2x{e^{ - y}}\;for\;0 \le x \le 1\;and\;0 < y < \infty \\0\;otherwise\end{array} \right.\)

Calculate the marginal p.d.f of X

The marginal probability distribution function of X is,

\(\begin{array}{c}f\left( x \right) = \int\limits_0^\infty {f\left( {x,y} \right)dy} \\ = \int_0^\infty {2x{e^{ - y}}dy} \\ = 2x\left( { - {e^{ - y}}} \right)_0^\infty \\ = 2x\left( {0 + 1} \right)\\ = 2x\end{array}\)

Therefore, the marginal p.d.f of X is \(f\left( x \right) = \left\{ \begin{array}{l}2x\;for\;0 \le x \le 1\\0\;otherwise\end{array} \right.\) .

Calculate the marginal p.d.f of Y

The marginal probability distribution function of Y is,

\(\begin{array}{c}f\left( y \right) = \int\limits_0^1 {f\left( {x,y} \right)dx} \\ = \int_0^1 {2x{e^{ - y}}dx} \\ = {e^{ - y}}\left( {{x^2}} \right)_0^1\\ = {e^{ - y}}\left( {1 - 0} \right)\\ = {e^{ - y}}\end{array}\)

Therefore, the marginal p.d.f of X is \(f\left( y \right) = \left\{ \begin{array}{l}{e^{ - y}}\;for\;0 \le x \le \infty \\0\;otherwise\end{array} \right.\) .

Check the independency

Hence there can be considered the marginal p.d.f as,

\(\begin{array}{c}f\left( {x,y} \right) = 2x{e^{ - y}}\\ = 2x \times {e^{ - y}}\\ = f\left( x \right) \times f\left( y \right)\end{array}\)

Therefore, there can be written the marginal p.d.f as the product of two individual marginal p.d.f of two random variable. So, X and Y are independent.