Chapter 6.3, Problem 2E

a.

To determine

Find the density function for $U_1=3Y$.

Answer to Problem 2E

The density function for $U_1=3Y$ is $f_{U_1}\left(u\right)=\frac{u^2}{18}\text{, }-3\le u\le 3$.

Explanation of Solution

Calculation:

From the given information, the probability density function for Y is $f\left(y\right)=\frac{3}{2}{y}^2,-1\le y\le 1$.

The distribution function for Y is,

$\begin{array}{c}{F}_Y\left(y\right)={\displaystyle \underset{-1}{\overset{y}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{-1}{\overset{y}{\int }}\frac{3}{2}{t}^{2}dt}\\ =\frac{3}{2}{\left[\frac{t^3}{3}\right]}_{-1}^y\\ =\frac{1}{2}\left(y^3-1\right)\end{array}$

From the given information, the random variable $U_1$ is defined as $U_1=3Y$.

Consider the distribution function for $U_1$,

$\begin{array}{c}{F}_{U_1}\left(u_1\right)=P\left(U_1\le u\right)\\ =P\left(3Y\le u\right)\\ =P\left(Y\le \frac{u}{3}\right)\\ =\frac{1}{2}\left[{\left(\frac{u}{3}\right)}^3-1\right]\end{array}$

Limits for the random variable U₁:

The range for the random variable Y is from  ̶ 1 to 1 and $U_1=3Y$.

For Y=  ̶ 1, the value of U₁ is  ̶ 3.

For $Y=1$, the value of U₁ is 3.

Hence, the range for the random variable U₁ is from  ̶ 3 to 3.

The probability density function for $U_1$ is,

$\begin{array}{c}{f}_{U_1}\left(u\right)={F^{\prime }}_{U_1}\left(u\right)\\ =\frac{d}{du}\left(\frac{1}{2}\left[{\left(\frac{u}{3}\right)}^3-1\right]\right)\\ =\frac{1}{2}\times 3{\left(\frac{u}{3}\right)}^2\times \frac{1}{3}\\ =\frac{u^2}{18}\text{, }-3\le u\le 3\end{array}$

Thus, the density function for $U_1=3Y$ is $f_{U_1}\left(u\right)=\frac{u^2}{18}\text{, }-3\le u\le 3$.

b.

To determine

Find the density function for $U_2=3-Y$.

Answer to Problem 2E

The density function for $U_2=3-Y$ is $f_{U_2}\left(u\right)=\frac{3}{2}\times {\left(3-u\right)}^2,2\le u\le 4$.

Explanation of Solution

Calculation:

From the given information, $U_2=3-Y$.

Consider the distribution function for $U_2$,

$\begin{array}{c}{F}_{U_2}\left(u\right)=P\left(U_2\le u\right)\\ =P\left(3-Y\le u\right)\\ =P\left(-Y\le u-3\right)\\ =P\left(Y>3-u\right)\end{array}$

             $\begin{array}{c}=1-F\left(3-u\right)\\ =1-\frac{1}{2}\left[{\left(3-u\right)}^3-1\right]\\ =-\frac{1}{2}{\left(3-u\right)}^3+\frac{3}{2}\end{array}$

Limits for the random variable U₂:

The range for the random variable Y is from  ̶ 1 to 1 and $U_2=3-Y$.

For Y=  ̶ 1, the value of U₂ is 4.

For $Y=1$, the value of U₂ is 2.

Hence, the range for the random variable U₂ is from  2 to 4.

The probability density function for $U_2$ is,

$\begin{array}{c}{f}_{U_2}\left(u\right)={F^{\prime }}_{U_2}\left(u\right)\\ =\frac{d}{du}\left(-\frac{1}{2}{\left(3-u\right)}^3+\frac{3}{2}\right)\\ =\frac{3}{2}\times {\left(3-u\right)}^2,2\le u\le 4\end{array}$

Thus, the density function for $U_2=3-Y$ is $f_{U_2}\left(u\right)=\frac{3}{2}\times {\left(3-u\right)}^2,2\le u\le 4$.

c.

To determine

Find the density function for $U_3=Y^2$.

Answer to Problem 2E

The density function for $U_3=Y^2$ is $f_{U_3}\left(u\right)=\frac{3}{2}\sqrt{u},0\le u\le 1$.

Explanation of Solution

Calculation:

From the given information, $U_3=Y^2$.

Consider the distribution function for $U_3$,

$\begin{array}{c}{F}_{U_3}\left(u\right)=P\left(U_3\le u\right)\\ =P\left(Y^2\le u\right)\\ =P\left(-\sqrt{u}\le Y\le \sqrt{u}\right)\\ =F\left(\sqrt{u}\right)-F\left(-\sqrt{u}\right)\end{array}$

             $\begin{array}{c}=\frac{1}{2}\left({\left(\sqrt{u}\right)}^3-1\right)-\frac{1}{2}\left({\left(-\sqrt{u}\right)}^3-1\right)\\ =\frac{1}{2}{\left(\sqrt{u}\right)}^3-\frac{1}{2}+\frac{1}{2}{\left(\sqrt{u}\right)}^3+\frac{1}{2}\\ =u^{\frac{3}{2}}\end{array}$

The probability density function for $U_3$ is,

$\begin{array}{c}{f}_{U_3}\left(u\right)={F^{\prime }}_{U_3}\left(u\right)\\ =\frac{d}{du}\left(u^{\frac{3}{2}}\right)\\ =\frac{3}{2}\times \left(u^{\frac{1}{2}}\right),0\le u\le 1\end{array}$

Thus, the density function for $U_3=Y^2$ is $f_{U_3}\left(u\right)=\frac{3}{2}\sqrt{u},0\le u\le 1$.