Chapter 6.3, Problem 1E

Let Y be a random variable with probability density function given by

$f(y)=\{\begin{array}{ll}2(1-y),\hfill & 0\le 0\le 1,\hfill \\ 0,\hfill & \text{elsewhere}.\hfill \end{array}$

a Find the density function of U₁ = 2Y − 1.

b Find the density function of U₂ = 1 − 2Y.

c Find the density function of U₃ = Y².

d Find E(U₁), E(U₂), and E(U₃) by using the derived density functions for these random variables.

e Find E(U₁), E(U₂), and E(U₃) by the methods of Chapter 4.

To determine

Find the density function for $U_1=2Y-1$.

Answer to Problem 1E

The density function for $U_1=2Y-1$ is $f_{U_1}\left(u\right)=\frac{1-u}{2}\text{, }-1\le u\le 1$.

Explanation of Solution

Calculation:

From the given information, the probability density function for Y is $f\left(y\right)=2\left(1-y\right),0\le y\le 1$.

The distribution function for Y is,

$\begin{array}{c}{F}_Y\left(y\right)={\displaystyle \underset{0}{\overset{y}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{y}{\int }}2\left(1-t\right)dt}\\ =2{\left[t-\frac{t^2}{2}\right]}_0^y\\ =2y-y^2\end{array}$

From the given information, the random variable $U_1$ is defined as $U_1=2Y-1$.

Consider the distribution function for $U_1$,

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

Limits for the random variable U₁:

The range for the random variable Y is from  0 to 1 and $U_1=2Y-1$.

For Y=  0, the value of U₁ is  ̶ 1.

For $Y=1$, the value of U₁ is 1.

Hence, the range for the random variable U₁ is from  ̶ 1 to 1.

The probability density function for $U_1$ is,

$\begin{array}{c}{f}_{U_1}\left(u\right)={F^{\prime }}_{U_1}\left(u\right)\\ =\frac{d}{du}\left(2\left(\frac{u+1}{2}\right)-{\left(\frac{u+1}{2}\right)}^2\right)\\ =\left(2\times \frac{1}{2}\right)-\left(2\times \frac{u+1}{2}\times \frac{1}{2}\right)\\ =1-\frac{u+1}{2}\end{array}$

             $=\frac{1-u}{2}\text{, }-1\le u\le 1$

To determine

Find the density function for $U_2=1-2Y$.

Answer to Problem 1E

The density function for $U_2=1-2Y$ is $f_{U_2}\left(u\right)=\frac{1+u}{2},-1\le u\le 1$.

Explanation of Solution

Calculation:

From the given information, $U_2=1-2Y$.

Consider the distribution function for $U_2$,

$\begin{array}{c}{F}_{U_2}\left(u\right)=P\left(U_2\le u\right)\\ =P\left(1-2Y\le u\right)\\ =P\left(-2Y\le u-1\right)\\ =P\left(Y>\frac{1-u}{2}\right)\end{array}$

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

Limits for the random variable U₂:

The range for the random variable Y is from  0 to 1 and $U_2=1-2Y$.

For Y= 0, the value of U₂ is 1.

For $Y=1$, the value of U₂ is  ̶ 1.

Hence, the range for the random variable U₂ is from  ̶ 1  to 1.

The probability density function for $U_2$ is,

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

To determine

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

Answer to Problem 1E

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

Explanation of Solution

Calculation:

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

Consider the distribution function for $U_3$,

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

             $\begin{array}{l}=2\sqrt{u}-{\left(\sqrt{u}\right)}^2\\ =2\sqrt{u}-u\end{array}$

The probability density function for $U_3$ is,

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

To determine

Find the value of $E\left(U_1\right)$ by using the derived density function of $U_1$.

Find the value of $E\left(U_2\right)$ by using the derived density function of $U_2$.

Find the value of $E\left(U_3\right)$ by using the derived density function of $U_3$.

Answer to Problem 1E

The value of $E\left(U_1\right)$ is $\frac{-1}{3}$.

The value of $E\left(U_2\right)$ is $\frac{1}{3}$.

The value of $E\left(U_3\right)$  is $\frac{1}{6}$.

Explanation of Solution

Calculation:

The density function for $U_1$ is $f_{U_1}\left(u\right)=\frac{1-u}{2}\text{, }-1\le u\le 1$

Consider,

$\begin{array}{c}E\left(U_1\right)={\displaystyle \underset{-1}{\overset{1}{\int }}u{f}_{U_1}}\left(u\right)du\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}u\times \frac{1-u}{2}}du\\ =\frac{1}{2}{\left[\frac{u^2}{2}-\frac{u^3}{3}\right]}_{-1}^1\\ =\frac{1}{2}\times \frac{-2}{3}\end{array}$

             $=\frac{-1}{3}$

The density function for $U_2$ is $f_{U_2}\left(u\right)=\frac{1+u}{2},-1\le u\le 1$.

Consider,

$\begin{array}{c}E\left(U_2\right)={\displaystyle \underset{-1}{\overset{1}{\int }}u{f}_{U_2}}\left(u\right)du\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}u\times \frac{1+u}{2}}du\\ =\frac{1}{2}{\left[\frac{u^2}{2}+\frac{u^3}{3}\right]}_{-1}^1\\ =\frac{1}{2}\times \frac{2}{3}\end{array}$

             $=\frac{1}{3}$

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

Consider,

$\begin{array}{c}E\left(U_3\right)={\displaystyle \underset{0}{\overset{1}{\int }}u{f}_{U_3}}\left(u\right)du\\ ={\displaystyle \underset{0}{\overset{1}{\int }}u\times \left(\frac{1}{\sqrt{u}}-1\right)}du\\ ={\left[\frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}-\frac{u^2}{2}\right]}_0^1\\ =\frac{2}{3}-\frac{1}{2}\end{array}$

              $=\frac{1}{6}$

To determine

Find the value of $E\left(U_1\right)$ by the methods of Chapter 4.

Find the value of $E\left(U_2\right)$ by the methods of Chapter 4.

Find the value of $E\left(U_3\right)$ by the methods of Chapter 4.

Answer to Problem 1E

The value of $E\left(U_1\right)$ is $\frac{-1}{3}$.

The value of $E\left(U_2\right)$ is $\frac{1}{3}$.

The value of $E\left(U_3\right)$  is $\frac{1}{6}$.

Explanation of Solution

Calculation:

Result:

Let X be the random variable, then $E\left(aX+b\right)=aE\left(X\right)+b$, where a and b are constants.

The density function for Y is $f\left(y\right)=2\left(1-y\right),0\le y\le 1$.

Consider,

$\begin{array}{c}E\left(y\right)={\displaystyle \underset{0}{\overset{1}{\int }}yf}\left(y\right)dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}y\times 2\left(1-y\right)}dy\\ =2{\left[\frac{y^2}{2}-\frac{y^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

$\begin{array}{c}E\left(U_1\right)=E\left(2Y-1\right)\\ =2E\left(Y\right)-1\\ =2\times \frac{1}{3}-1\\ =\frac{-1}{3}\end{array}$

$\begin{array}{c}E\left(U_2\right)=E\left(1-2Y\right)\\ =1-2E\left(Y\right)\\ =1-2\times \frac{1}{3}\\ =\frac{1}{3}\end{array}$

Consider,

$\begin{array}{c}E\left(y^2\right)={\displaystyle \underset{0}{\overset{1}{\int }}{y}^{2}f}\left(y\right)dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{y}^2\times 2\left(1-y\right)}dy\\ =2{\left[\frac{y^3}{3}-\frac{y^4}{4}\right]}_0^1\\ =\frac{1}{6}\end{array}$

$\begin{array}{c}E\left(U_3\right)=E\left(Y^2\right)\\ =\frac{1}{6}\end{array}$