Chapter 4.9, Problem 142E

Refer to Exercises 4.141 and 4.137. Suppose that Y is uniformly distributed on the interval (0, 1) and that a > 0 is a constant.

1. a Give the moment-generating function for Y.
2. b Derive the moment-generating function of W = aY. What is the distribution of W? Why?
3. c Derive the moment-generating function of X = −aY. What is the distribution of X? Why?
4. d If b is a fixed constant, derive the moment-generating function of V = aY + b. What is the distribution of V? Why?

To determine

Find the moment-generating function for Y.

Answer to Problem 142E

The moment-generating function of Y is $m_Y\left(t\right)=\frac{e^t-1}{t}$.

Explanation of Solution

Let Y be a random variable that has a uniform distribution on the interval $\left(0,1\right)$.

The probability density function of Y is given by

$f\left(y\right)=\{\begin{array}{l}\text{1 , 0}<y<1\\ 0,\text{ elsewhere}\text{.}\end{array}$

The moment-generating function of Y is derived below:

$\begin{array}{c}{m}_Y\left(t\right)=E\left(e^{ty}\right)\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{e}^{ty}f\left(y\right)}dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{e}^{ty}}\left(1\right)dy\\ ={\left[\frac{e^{ty}}{t}\right]}_0^1\\ m_Y\left(t\right)=\frac{e^t-1}{t}\end{array}$

Thus, the moment-generating function of Y is $m_Y\left(t\right)=\frac{e^t-1}{t}$.

To determine

Find the moment-generating function of $W=aY$

Find the distribution of W.

Answer to Problem 142E

The moment-generating function of W is $m_W\left(t\right)=\frac{e^{at}-1}{at}$.

The distribution of W is uniform distribution on the interval $\left(0,a\right)$.

Explanation of Solution

The moment-generating function of $W=aY$ is derived as follows:

$\begin{array}{c}{m}_W\left(t\right)=E\left(e^{tW}\right)\\ =E\left(e^{tay}\right)\\ =m_y\left(at\right)\\ m_W\left(t\right)=\frac{e^{at}-1}{at}\end{array}$

Thus, the moment-generating function of W is $m_W\left(t\right)=\frac{e^{at}-1}{at}$.

Hence, by the uniqueness theorem of mgf, the distribution of W is a uniform distribution on the interval $\left(0,a\right)$.

To determine

Derive the moment-generating function of $X=-aY$.

Find the distribution of X.

Answer to Problem 142E

The moment-generating function of X is $m_X\left(t\right)=\frac{e^{-at}-1}{-at}$.

The variable X is a uniform distribution on the interval $\left(-a,0\right)$.

Explanation of Solution

The moment-generating function of $X=-aY$ is obtained as follows:

$\begin{array}{c}{m}_X\left(t\right)=E\left(e^{tX}\right)\\ =E\left(e^{-tay}\right)\\ =m_y\left(-at\right)\left\{\because Y~U\left(0,1\right)\right\}\\ m_X\left(t\right)=\frac{e^{-at}-1}{-at}\end{array}$

The moment- generating function of X is $m_X\left(t\right)=\frac{e^{-at}-1}{-at}$

This indicates that the variable X is a uniform distribution on the interval $\left(-a,0\right)$.

To determine

Derive the moment-generating function of $V=aY+b$.

Find the distribution of V.

Answer to Problem 142E

The moment-generating function of V is $m_V\left(t\right)=\frac{e^{\left(b+a\right)t}-e^{bt}}{at}$.

The variable V is a uniform distribution on the interval $\left(b,b+a\right)$.

Explanation of Solution

The moment-generating function of $V=aY+b$ is obtained as follows:

$\begin{array}{c}{m}_V\left(t\right)=E\left(e^{tv}\right)\\ =E\left(e^{t\left(ay+b\right)}\right)\\ =e^{bt}E\left(e^{aty}\right)\\ =e^{bt}{m}_y\left(at\right)\left\{\because Y~U\left(0,1\right)\right\}\\ =e^{bt}\left(\frac{e^{at}-1}{at}\right)\\ m_V\left(t\right)=\frac{e^{\left(b+a\right)t}-e^{bt}}{at}\end{array}$

The moment-generating function of V is $m_V\left(t\right)=\frac{e^{\left(b+a\right)t}-e^{bt}}{at}$.

This indicates that the variable V is a uniform distribution on the interval $\left(b,b+a\right)$.