Question

Let Y be a random variable with a density function given by
f(y) = (3/2)y^2, −1≤y≤1,
0, elsewhere.
a) Find the density function of U1 = 3Y.
b) Find the density function of U2 = 3−Y.
c) Find the density function of U3 = Y^2.

Answer 1

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y<-1\\\frac{y^3+1}2&\text{for }-1\le y<1\\1&\text{for }y\ge1\end{cases}$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1<-3\\\frac{\left(\frac{{u_1}}3\right)^3+1}2&\text{for }-3\le u_1<3\\1&\text{for }u_1\ge3\end{cases}$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y<3-u_2)=1-F_Y(3-u_2)$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2<2\\1-\frac{(3-u_2)^3+1}2&\text{for }2\le u_2<4\\1&\text{for }u_2\ge4\end{cases}$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3<0\\{u_3}^{3/2}&\text{for }0\le u_3<1\\1&\text{for }u_3\ge1\end{cases}$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$