Question

Let X and Y be random variables for which the jointp.d.f. is as follows:

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ \begin{array}{l}{\bf{2}}\left( {{\bf{x + y}}} \right)\;\;\;\;\;\;\;\;\;\;{\bf{for}}\;{\bf{0}} \le {\bf{x}} \le {\bf{y}} \le {\bf{1,}}\\{\bf{0}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\bf{otherwise}}\end{array} \right.\)

Find the p.d.f. of Z = X + Y.

Short answer

The pdf is \({f_Z}\left( z \right) = \left\{ \begin{array}{l}{z^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{for}}\;0 \le z \le 1\\z\left( {2 - z} \right)\;\;\;\;\;\;\;\;\;{\rm{for}}\;1 < z < 2\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Step-by-step solution

Given information

The given p.d.f is

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}2\left( {x + y} \right)\;\;\;\;\;\;\;{\rm{for}}\;0 \le x \le y \le 1,\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{array} \right.\)

Define convolution rule

Convolution rule:Let \({X_1}\;and\;{X_2}\) be two independent random variables with their respective p.d.f. defined by \({f_X}\left( x \right)\)and \({f_Y}\left( y \right)\).

Let \({\bf{Z = X + Y}}\). The newly defined variable Z is convolution of distribution of X and Y.

The p.d.f of Z is

\(\begin{aligned}{f_Z}\left( z \right) &= \int_{ - \infty }^\infty {{f_x}} \left( {z - y} \right){f_Y}\left( y \right)dy\\ &= \int_{ - \infty }^\infty {{f_x}\left( {z - y,y} \right)} dy\end{aligned}\)

Apply Convolution rule evaluate the pdf

The positive region for the given interval is for \(0 \le z - t \le z \le 1\). So,for \(0 \le z \le 1\) it is positive only when \(\frac{z}{2} \le t \le z\),

\(\begin{aligned}{f_Z}\left( z \right) &= \int\limits_y^{} {{f_y}\left( z \right)} dt\\ &= \int\limits_{\frac{z}{2}}^z {2z} dt\,\\ &= 2z\left( {z - \frac{z}{2}} \right)\\ &= {z^2}\end{aligned}\)

For\(1 < z < 2\), the positive integrand lies in \(\frac{z}{2} \le t \le 1\). Thus,

\(\begin{aligned}{f_Z}\left( z \right) &= \int\limits_t^{} {{f_y}\left( z \right)} dyt\\ &= \int\limits_{\frac{z}{2}}^1 {2z} dt\,\,\,\\ &= 2z\left( {1 - \frac{z}{2}} \right)\\ &= z\left( {2 - z} \right)\end{aligned}\)

Therefore, the pdf of \({\bf{Z}} = {\bf{X}} + {\bf{Y}}\)

\({f_Z}\left( z \right) = \left\{ \begin{array}{l}{z^2},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0 \le z \le 1\\z\left( {2 - z} \right),\;\;\;\;\;\;\;1 < z < 2\\0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)