Question

Suppose that the p.d.f. of X is as given in Exercise 3. Determine the p.d.f. of\(Y = 4 - {X^3}\)

Short answer

Pdf of \(Y = 4 - {X^3}\)is \(\left\{ \begin{array}{l}4 - y\,\,\,\,if\,0 \le y \le 8\\\,0\,\,\,if\,\,y < 0\,\,or\,y \ge 8\end{array} \right.\)

Step-by-step solution

Calculating the Probability.

Y depends on \({x^3}\)

Finding the cdf of \(z = {x^3}\)

\(\begin{aligned}H\left( z \right) &= {\rm P}\left( {{\rm Z} \le z} \right)\\ &= {\rm P}\left( {{X^3} \le z} \right)\end{aligned}\)

If

\(\begin{array}{l}z < 0\, \Rightarrow {\rm P}\left( {{X^3} \le z} \right) = 0\\H\left( z \right) = 0\end{array}\)

If\(\begin{array}{l}0 \le z < 8\\{X^3} \le z \Rightarrow X \le {z^{\frac{1}{3}}} \Rightarrow {\rm P}\left( {X \le {z^{\frac{1}{3}}}} \right) &= F\left( {{z^{\frac{1}{3}}}} \right)\\ &= \frac{1}{4}{z^{\frac{1}{3}}}\end{array}\)

If \(z \ge 8 \Rightarrow {\rm P}\left( {{x^3} \le z} \right) = 1\)

Therefore,

\(H\left( z \right) = \left\{ \begin{array}{l}0\,\,\,\,\,\,\,\,\,\,\,\,if\,z < 0\,\,\\\frac{1}{4}{z^{\frac{2}{3}}}\,\,\,\,\,if\,0 \le z < 8\\1\,\,\,\,\,\,\,\,\,\,\,\,\,if\,z \ge 8\end{array} \right.\)

Calculating the cdf

Now finding cdf of \(y = 4 - {x^3}\)

\(\begin{aligned}G\left( y \right) &= {\rm P}\left( {Y \le y} \right)\\ &= {\rm P}\left( {4 - z \le y} \right)\\ &= {\rm P}\left( {z \ge 4 - y} \right)\\ &= 1 - H\left( {4 - y} \right)\end{aligned}\)

Therefore,

\(G\left( y \right) = \left\{ \begin{array}{l}0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,y < 0\\4 - y\,\,\,\,\,\,\,if\,0 \le y < 8\\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,y \ge 8\end{array} \right.\)

Hence the pdf will be

\(\begin{aligned}f\left( y \right) &= \frac{{dG\left( y \right)}}{{dy}}\\f\left( y \right) &= \left\{ \begin{aligned}4 - y\,\,\,\,if\,0 \le y \le 8\\\,0\,\,\,if\,\,y < 0\,\,or\,y \ge 8\end{aligned} \right.\end{aligned}\)