Question

Suppose that the p.d.f. of a random variable X is as

follows:\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{2}x\,\,\,\,\,\,\,\,for\,0 < x < 2\\0\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\)

Also, suppose that \(Y = X\left( {2 - X} \right)\) Determine the cdf and the pdf of Y .

Short answer

Cdf of Y is

\(G\left( y \right) = \left\{ \begin{array}{l}0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y < 2\\\frac{1}{{36}}\,\,\,\,\,\,\,\,\,\,\,\,z \le y < 8\\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,y \ge 8\end{array} \right.\)

Pdf of Y is \(f\left( y \right) = \left\{ \begin{array}{l} = \frac{1}{{18}}\left( {y - 2} \right)\,\,if\,\,z \le y < 8\\\,\,0\,\,\,\,\,\,\,if\,y < 2\,\,or\,y \ge 8\,\,\end{array} \right.\)

Step-by-step solution

Calculating the CDF

First we have to find out the cdf of X

\(\begin{aligned}F\left( x \right) = \int\limits_{ - \infty }^x {f\left( z \right)dz} \\\left\{ \begin{aligned}if\,x < 0\,\,\,\,\,\,F\left( x \right) &= 0\\x \in \left[ {0,2} \right)\,\,\,\,\,F\left( x \right) &= \int\limits_{ - \infty }^x {F\left( z \right)dz} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \int\limits_0^x {\frac{z}{2}dz} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left. {\frac{1}{2} \times \frac{{{z^2}}}{2}} \right|_0^x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{{x^2}}}{4}\\x \ge 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F\left( x \right) &= 1\end{aligned} \right.\end{aligned}\)

Finding cdf of Y using cdf of X

\(\begin{aligned}G\left( y \right) &= {\rm P}\left( {Y \le y} \right)\\ &= {\rm P}\left( {3x + 2 \le y} \right)\\ &= {\rm P}\left( {x \le \frac{1}{3}\left( {y - 2} \right)} \right)\end{aligned}\)

\(G\left( y \right) = F\left( {\frac{1}{3}\left( {y - 2} \right)} \right)\)

If \(\frac{1}{3}\left( {y - 2} \right) < 0\,\,or\,y < 2\,\, \Rightarrow G\left( y \right) = 0\,\,or\,y < 2\)

If \(0 \le \frac{1}{3}\left( {y - z} \right)\,\,or\,\,z \le y < 8 \Rightarrow G\left( y \right) = \frac{1}{{36}}{\left( {y - z} \right)^2}\)

If \(\frac{1}{3}\left( {y - 2} \right) \ge 2\,or\,y \ge 8\, \Rightarrow G\left( y \right) = 1\)

Therefore, cdf of y is

\(G\left( y \right) = \left\{ \begin{array}{l}0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y < 2\\\frac{1}{{36}}\,\,\,\,\,\,\,\,\,\,\,\,z \le y < 8\\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,y \ge 8\end{array} \right.\)

Calculating the CDF 

Pdf of Y is

\(\begin{aligned}f\left( y \right) &= \frac{{dG\left( y \right)}}{{dy}}\\ &= \frac{1}{{36}}\left( {2y - 4} \right)\end{aligned}\)

\(\) \(f\left( y \right) = \left\{ \begin{aligned} = \frac{1}{{18}}\left( {y - 2} \right)\,\,if\,\,z \le y < 8\\\,\,0\,\,\,\,\,\,\,if\,y < 2\,\,or\,y \ge 8\,\,\end{aligned} \right.\)

Hence the required pdf is \(f\left( y \right) = \left\{ \begin{array}{l} = \frac{1}{{18}}\left( {y - 2} \right)\,\,if\,\,z \le y < 8\\\,\,0\,\,\,\,\,\,\,if\,y < 2\,\,or\,y \ge 8\,\,\end{array} \right.\)