The Method of Distribution Functions

Section 6.3 The Method of Distribution Functions

Remark 6.3.1. Summary of the Distribution Function Method.

Let $U$ be a function of the random variables $Y_1, Y_2, ..., Y_n\text{.}$

Find the region $U = u$ in the $(y_1, y_2, ..., y_n)$ space.
Find the region $U \leq u\text{.}$
Find $F_U(u) = P(U \leq u)$ by integrating $f(y_1, y_2, ..., y_n)$ over the region $U \leq u\text{.}$
Find the density function $f_U(u)$ by differentiating $F_U(u)\text{.}$ Thus, $f_U(u) = dF_U(u) / du\text{.}$

Example 6.3.2. (Example 6.5 in the book).

Let $U$ be a uniform random variable on the interval $(0, 1)\text{.}$ Find a transformation $G(U)$ such that $G(U)$ possesses an exponential distribution with mean $\beta\text{.}$

Solution.

If $U$ possesses a uniform distribution on the interval $(0, 1)\text{,}$ then the distribution function of $U$ (see [provisional cross-reference: ex-4-38]) is given by

\begin{equation*} F_U(u) = \begin{cases} 0, \amp u \lt 1,\\ u, \amp 0 \leq u \leq 1,\\ 1, \amp u \gt 1. \end{cases} \end{equation*}

Let $Y$ denote a random variable that has an exponential distribution with mean $\beta\text{.}$ Then (see [provisional cross-reference: section-4-6]) $Y$ has distribution function

\begin{equation*} F_Y(y) = \begin{cases} 0, \amp y \lt 0,\\ 1 - e^{-y / \beta}, \amp y \geq 0. \end{cases} \end{equation*}

Notice that $F_Y(y)$ is strictly increasing on the interval $[0, \infty)\text{.}$ Let $0 \lt u \lt 1$ and observe that there is a unique value $y$ such that $F_Y(y) = u\text{.}$ Thus, $F_Y^{-1}(u), 0 \lt u \lt 1\text{,}$ is well-defined. In this case, $F_Y(y) = 1 - e^{-y / \beta} = u$ if and only if $y = -\beta \ln(1 - u) = F_Y^{-1}(u)\text{.}$ Consider the random variable $F_Y^{-1}(U) = -\beta \ln(1 - U)$ and observe that, if $y \gt 0\text{,}$

\begin{align*} P(F_Y^{-1}(U) \leq y) \amp = P[-\beta \ln(1 - U) \leq y]\\ \amp = P[\ln(1 - U) \geq -y / \beta]\\ \amp = P(U \leq 1 - e^{-y / \beta})\\ \amp = 1 - e^{-y / \beta}. \end{align*}

Also, $P[F_Y^{-1}(U) \leq y] = 0$ if $y \leq 0\text{.}$ Thus, $F_Y^{-1}(U) = -\beta \ln(1 - U)$ possesses an exponential distribution with mean $\beta\text{,}$ as desired.

Computer simulations are frequently used to evaluate proposed statistical techniques. Typically, these simulations require that we obtain observed values of random variables with a prescribed distribution. As noted in [provisional cross-reference: sec-4-4], most computer systems contain a subroutine that provides observed values of a random variable $U$ that has a uniform distribution on the interval $(0, 1)\text{.}$ How can the result of Example 6.3.2 be used to generate a set of observations from an exponential distribution with mean $\beta\text{?}$ Simply use the computer’s random number generator to produce values $u_1, u_2, ..., u_n$ from a uniform $(0, 1)$ distribution and then calculate $y_i = -\beta \ln(1 - u_i), i = 1, 2, ..., n$ to obtain values of random variables with the required exponential distribution. As long as a prescribed distribution function $F(y)$ possesses a unique inverse $F^{-1}(\cdot)\text{,}$ the preceding technique can be applied.

Exercises Exercises

6.1.

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

\begin{equation*} f(y) = \begin{cases} 2(1 - y), \amp 0 \leq y \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases}\text{.} \end{equation*}

(a)

Find the density function of $U_1 = 2Y - 1\text{.}$

Solution.

First note that

\begin{equation*} F_{U_1}(u_1) = P(U_1 \leq u_1) = P(2Y - 1 \leq u_1) = P \left(Y \leq \frac{u_1 + 1}{2} \right)\text{.} \end{equation*}

Since $0 \leq y \leq 1$ is the support of $f\text{,}$ then

\begin{equation*} 0 \leq \frac{u_1 + 1}{2} \leq 1 \Leftrightarrow -1 \leq u_1 \leq 1\text{,} \end{equation*}

(i.e., the support of $f_{U_1}$ is $-1 \leq u_1 \leq 1$). Then, noting we are integrating over the support of $Y$’s density function and not $U_1$’s,

\begin{align*} F_{U_1}(u_1) \amp = \begin{cases} 0, \amp u_1 \lt -1,\\ \int_{y = 0}^{y = (u_1 + 1)/2} f(y) dy\\ \int_{y = 0}^{y = (u_1 + 1)/2} (2 - 2y) dy\\ = \left[2y - y^2 \right]_{y = 0}^{y = (u_1 + 1)/2}\\ = \left(2(\frac{u_1 + 1}{2}) - (\frac{u_1 + 1}{2})^2 \right) - \left(2(0) - (0)^2 \right)\\ = \frac{2u_1 + 2}{2} \cdot \frac{2}{2} - \frac{u_1^2 + 2u_1 + 1}{4}\\ = \frac{4u_1 + 4}{4} - \frac{u_1^2 + 2u_1 + 1}{4}\\ = -\frac{1}{4} u_1^2 + \frac{1}{2} u_1 + \frac{3}{4}, \amp -1 \leq u_1 \leq 1\\ 1, \amp 1 \lt u_1 \end{cases}\\ \implies f_{U_1}(u_1) = \frac{d}{du_1} (F_{U_1}(u_1)) \amp = \begin{cases} -\frac{1}{2} u_1 + \frac{1}{2}, \amp -1 \leq u_1 \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(b)

Find the density function of $U_2 = 1 - 2Y\text{.}$

Solution.

First note that

\begin{equation*} F_{U_2}(u_2) = P(U_2 \leq u_2) = P(1 - 2Y \leq u_2) = P \left(\frac{1 - u_2}{2} \leq Y \right) = 1 - P \left(Y \leq \frac{1 - u_2}{2} \right)\text{.} \end{equation*}

Since $0 \leq y \leq 1$ is the support of $f\text{,}$ then

\begin{equation*} 0 \leq \frac{1 - u_2}{2} \leq 1 \Leftrightarrow -1 \leq -u_2 \leq 1 \Leftrightarrow 1 \geq u_2 \geq -1, \end{equation*}

(i.e., the support of $f_{U_2}$ is $-1 \leq u_2 \leq 1$). Then

\begin{align*} F_{U_2}(u_2) \amp = \begin{cases} 0, \amp u_2 \lt -1,\\ 1 - \int_{y = 0}^{y = (1 - u_2)/2} (2 - 2y) dt\\ = \left[2y - y^2 \right]_{y = 0}^{y = (1 - u_2)/2}\\ = 1 - \left[\left(2 \left(\frac{1 - u_2}{2} \right) - \left(\frac{1 - u_2}{2} \right)^2 \right) - \left(2(0) - (0)^2 \right) \right]\\ = 1 - \left[\frac{2 - 2u_2}{2} \cdot \frac{2}{2} - \frac{1 - 2u_2 + u_2^2}{4} \right]\\ = 1 - \left[\frac{4 - 4u_2}{4} - \frac{1 - 2u_2 + u_2^2}{4} \right]\\ = \frac{1}{4} u_2^2 + \frac{1}{2} u_2 + \frac{1}{4}, \amp -1 \leq u_2 \leq 1,\\ 1, \amp 1 \lt u_2 \end{cases}\\ \implies f_{U_2}(u_2) = \frac{d}{du_2} (F_{U_2}(u_2)) \amp = \begin{cases} \frac{1}{2} u_2 + \frac{1}{2}, \amp -1 \leq u_2 \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(c)

Find the density function of $U_3 = Y^2\text{.}$

Solution.

First note that

\begin{align*} F_{U_3}(u_3) \amp = P(U_3 \leq u_3)\\ \amp = P(Y^2 \leq u_3)\\ \amp = P \left(-\sqrt{u_3} \leq Y \leq \sqrt{u_3} \right)\\ \amp = P \left(0 \leq Y \leq \sqrt{u_3} \right), \end{align*}

where this last equality is true because $Y$ only takes on nonnegative values (i.e., $0 \leq y \leq 1$). Since $0 \leq y \leq 1$ is the support of $f\text{,}$ then $0^2 \leq y^2 \leq 1^2\text{,}$ or $0 \leq y^2 \leq 1\text{.}$ By replacing $y^2$ with $u_3$ in the inequality $0 \leq y^2 \leq 1\text{,}$ we get $0 \leq u_3 \leq 1\text{,}$ and this will be the support of $f_{U_3}\text{.}$ Then

\begin{align*} F_{U_3}(u_3) \amp = \begin{cases} 0, \amp u_3 \lt 0,\\ P(U_3 \leq u_3) = P(0 \leq Y \leq \sqrt{u_3})\\ = \int_{y = 0}^{y = \sqrt{u_3}} (2 - 2y) dy\\ = \left[2y - y^2 \right]_{y = 0}^{y = \sqrt{u_3}}\\ = \left(2(\sqrt{u_3}) - (\sqrt{u_3})^2 \right) - \left(2(0) - (0)^2 \right) \\ = -u_3 + 2u_3^{1/2}, \amp 0 \leq u_3 \leq 1,\\ 1, \amp 1 \lt u_3 \end{cases}\\ \implies f_{U_3}(u_3) = \frac{d}{du_3} (F_{U_3}(u_3)) \amp = \begin{cases} -1 + u_3^{-1/2}, \amp 0 \leq u_3 \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(d)

Find $E(U_1)\text{,}$ $E(U_2)\text{,}$ and $E(U_3)$ by using the derived density functions for these random variables.

Solution.

\begin{align*} E(U_1) \amp = \int_{u_1 = -\infty}^{u_1 = \infty} u_1 f_{U_1}(u_1) du_1\\ \amp = \int_{u_1 = -1}^{u_1 = 1} u_1 \cdot \left(-\frac{1}{2} u_1 + \frac{1}{2} \right) du_1\\ \amp = \int_{u_1 = -1}^{u_1 = 1} \left(-\frac{1}{2} u_1^2 + \frac{1}{2} u_1 \right) du_1\\ \amp = \left[-\frac{1}{6} u_2^3 + \frac{1}{4} u_1^2 \right]_{u_1 = -1}^{u_1 = 1}\\ \amp = \left(-\frac{1}{6} (1)^3 + \frac{1}{4} (1)^2 \right) - \left(-\frac{1}{6} (-1)^3 + \frac{1}{4} (-1)^2 \right)\\ \amp = -\frac{1}{6} + \frac{1}{4} - \frac{1}{6} - \frac{1}{4} = -\frac{2}{6} = -\frac{1}{3}\\ E(U_2) \amp = \int_{u_2 = -1}^{u_2 = 1} u_2 \cdot \left(\frac{1}{2} u_2 + \frac{1}{2} \right) du_2\\ \amp = \int_{u_2 = -1}^{u_2 = 1} \left(\frac{1}{2} u_2^2 + \frac{1}{2} u_2 \right) du_2\\ \amp = \left[\frac{1}{6} u_2^3 + \frac{1}{4} u_2^2 \right]_{u_2 = -1}^{u_2 = 1}\\ \amp = \left(\frac{1}{6} (1)^3 + \frac{1}{4} (1)^2 \right) - \left(\frac{1}{6} (-1)^3 + \frac{1}{4} (-1)^2 \right)\\ \amp = \frac{1}{6} + \frac{1}{4} + \frac{1}{6} - \frac{1}{4} = \frac{1}{3}\\ E(U_3) \amp = \int_{u_3 = 0}^{u_3 = 1} u_3 \cdot \left(-1 + u_3^{-1/2} \right) du_3\\ \amp = \int_{u_3 = 0}^{u_3 = 1} \left(-u_3 + u_3^{1/2} \right) du_3\\ \amp = \left[-\frac{1}{2} u_3^2 + \frac{2}{3} u_3^{3/2} \right]_{u_3 = 0}^{u_3 = 1}\\ \amp = \left(-\frac{1}{2} (1)^2 + \frac{2}{3} (1)^{3/2} \right) - \left(-\frac{1}{2} (0)^2 + \frac{2}{3} (0)^{3/2} \right)\\ \amp = -\frac{1}{2} + \frac{2}{3} = -\frac{3}{6} + \frac{4}{6} = \frac{1}{6} \end{align*}

(e)

Find $E(U_1)\text{,}$ $E(U_2)\text{,}$ and $E(U_3)$ by the methods of Chapter 4.

Solution.

\begin{align*} E(Y^k) \amp = \int_{y = -\infty}^{y = \infty} y^k f(y) dy = \int_{y = 0}^{y = 1} y^k \cdot 2(1 - y) dy\\ \amp = \int_{y = 0}^{y = 1} (2y^k - 2y^{k + 1}) dy\\ \amp = \left[\frac{2}{k + 1} y^{k + 1} - \frac{2}{k + 2} y^{k + 2} \right]_{y = 0}^{y = 1}\\ \amp = \left(\frac{2}{k + 1} (1)^{k + 1} - \frac{2}{k + 2} (1)^{k + 2} \right) - \left(\frac{2}{k + 1} (0)^{k + 1} - \frac{2}{k + 2} (0)^{k + 2} \right)\\ \amp = \frac{2}{k + 1} - \frac{2}{k + 2}\\ \implies E(Y) = E(Y^1) \amp = \frac{2}{(1) + 1} - \frac{2}{(1) + 2} = \frac{2}{2} - \frac{2}{3} = 1 - \frac{2}{3} = \frac{1}{3}\\ E(Y^2) \amp = \frac{2}{(2) + 1} - \frac{2}{(2) + 2} = \frac{2}{3} - \frac{2}{4} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}\\ \implies E(U_1) \amp = E(2Y - 1) = 2E(Y) - 1 = 2 \cdot \frac{1}{3} - 1 = -\frac{1}{3}\\ E(U_2) \amp = E(1 - 2Y) = 1 - 2E(Y) = 1 - 2 \cdot \frac{1}{3} = \frac{1}{3}\\ E(U_3) \amp = E(Y^2) = \frac{1}{6} \end{align*}

6.2.

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

\begin{equation*} f(y) = \begin{cases} (3/2)y^2, \amp -1 \leq y \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{equation*}

(a)

Find the density function of $U_1 = 3Y\text{.}$

Solution.

First note that

\begin{equation*} F(y) = \begin{cases} 0, \amp y \lt -1,\\ \int_{t = -1}^{t = y} \frac{3}{2} t^2 dt\\ = \frac{1}{2} t^3 \biggr|_{t = -1}^{t = y}\\ = \frac{1}{2} (y)^3 - \frac{1}{2} (-1)^3\\ = \frac{1}{2} y^3 + \frac{1}{2}, \amp -1 \leq y \leq 1,\\ 1, \amp 1 \lt y. \end{cases} \end{equation*}

Then

\begin{equation*} F_{U_1}(u_1) = P(U_1 \leq u_1) = P(3Y \leq u_1) = P \left(Y \leq \frac{u_1}{3} \right) = F \left(\frac{u_1}{3} \right). \end{equation*}

We can see, since the support of $f$ is $-1 \leq y \leq 1\text{,}$ that the support of $f_{U_1}$ will be

\begin{equation*} -1 \leq \frac{u_1}{3} \leq 1 \Leftrightarrow -3 \leq u_1 \leq 3. \end{equation*}

Thus,

\begin{align*} F_{U_1}(u_1) \amp = F \left(\frac{u_1}{3} \right)\\ \amp = \begin{cases} 0, \amp \frac{u_1}{3} \lt -1 \Leftrightarrow u_1 \lt -3\\ \frac{1}{2} \left(\frac{u_1}{3} \right)^3 + \frac{1}{2}\\ = \frac{1}{54} u_1^3 + \frac{1}{2}, \amp -1 \leq \frac{u_1}{3} \leq 1 \Leftrightarrow -3 \leq u_1 \leq 3\\ 1, \amp 1 \lt \frac{u_1}{3} \Leftrightarrow 3 \lt u_1 \end{cases}\\ \implies f_{U_1}(u_1) \amp = \begin{cases} \frac{1}{18} u_1^2, \amp -3 \leq u_1 \leq 3,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(b)

Find the density function of $U_2 = 3 - Y\text{.}$

Solution.

First note that

\begin{equation*} F_{U_2}(u_2) = P(U_2 \leq u_2) = P(3 - Y \leq u_2) = P(3 - u_2 \leq Y) = 1 - P(Y \leq 3 - u_2) = 1 - F(3 - u_2). \end{equation*}

We can see, since the support of $f$ is $-1 \leq y \leq 1\text{,}$ that

\begin{equation*} -1 \leq 3 - u_2 \leq 1 \Leftrightarrow -4 \leq -u_2 \leq -2 \Leftrightarrow 4 \geq u_2 \geq 2. \end{equation*}

Thus,

\begin{align*} F_{U_2}(u_2) \amp = 1 - F(3 - u_2)\\ \amp = 1 - \begin{cases} 0, \amp 3 - u_2 \lt -1 \Leftrightarrow 4 \lt u_2,\\ \frac{1}{2} (3 - u_2)^3 + \frac{1}{2}, \amp -1 \leq 3 - u_2 \leq 1 \Leftrightarrow 2 \leq u_2 \leq 4,\\ 1, \amp 1 \lt 3 - u_2 \Leftrightarrow u_2 \lt 2 \end{cases}\\ \amp = \begin{cases} 0, \amp u_2 \lt 2,\\ \frac{1}{2} - \frac{1}{2} (3 - u_2)^3, \amp 2 \leq u_2 \leq 4,\\ 1, \amp 4 \lt u_2 \end{cases}\\ \implies f_{U_2}(u_2) \amp = \begin{cases} \frac{3}{2} (3 - u_2)^2, \amp 2 \leq u_2 \leq 4,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(c)

Find the density function of $U_3 = Y^2\text{.}$

Solution.

First note that

\begin{equation*} F_{U_3}(u_3) = P(U_3 \leq u_3) = P(Y^2 \leq u_3) = P(-\sqrt{u_3} \leq Y \leq \sqrt{u_3}). \end{equation*}

Notice that if $u_3 \lt 0\text{,}$ then $F_{U_3}(u_3) = P(U_3 \leq u_3) = P(Y^2 \leq u_3 \lt 0) = 0$ because $Y^2 \lt 0$ is impossible. Similarly, if $1 \lt u_3\text{,}$ then $F_{U_3}(u_3) = P(U_3 \leq u_3) = P(Y^2 \leq 1 \lt u_3) = 1$ because this covers more than even the entire support of $f\text{,}$ $-1 \leq y \leq 1\text{.}$ Thus,

\begin{align*} F_{U_3}(u_3) \amp = \begin{cases} P(Y^2 \leq u_3) = 0, \amp u_3 \lt 0,\\ P(Y^2 \leq u_3) = P(-\sqrt{u_3} \leq Y \leq \sqrt{u_3})\\ = \int_{y = -\sqrt{u_3}}^{y = \sqrt{u_3}} \frac{3}{2} y^2 dy\\ = 2 \int_{y = 0}^{y = \sqrt{u_3}} \frac{3}{2} y^2 dy\\ = 2 \left[\frac{1}{2} y^3 \right]_{y = 0}^{y = \sqrt{u_3}}\\ = 2 \left[\frac{1}{2} (\sqrt{u_3})^3 - \frac{1}{2} (0)^3 \right]\\ = u_3^{3/2}, \amp 0 \leq u_3 \leq 1,\\ P(Y^2 \leq u_3) = 1, \amp 1 \lt u_3 \end{cases}\\ \implies f_{U_3}(u_3) \amp = \begin{cases} \frac{3}{2} u_3^{1/2}, \amp 0 \leq u_3 \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

6.3.

A supplier of kerosene has a weekly demand $Y$ possessing a probability density function given by

\begin{equation*} f(y) = \begin{cases} y, \amp 0 \leq y \leq 1,\\ 1, \amp 1 \lt y \leq 1.5,\\ 0, \amp \text{elsewhere,} \end{cases} \end{equation*}

with measurements in hundreds of gallons. (This problem was introduced in [provisional cross-reference: ex-4-13].) The supplier’s profit is given by $U = 10Y - 4\text{.}$

(a)

Find the probability density function for $U\text{.}$

Solution.

First note that

\begin{equation*} F_U(u) = P(U \leq u) = P(10Y - 4 \leq u) = P \left(Y \leq \frac{u + 4}{10} \right)\text{.} \end{equation*}

Looking at the support of $f\text{,}$ when $0 \leq y \leq 1\text{,}$ we will have

\begin{equation*} 0 \leq \frac{u + 4}{10} \leq 1 \Leftrightarrow 0 \leq u + 4 \leq 10 \Leftrightarrow -4 \leq u \leq 6\text{,} \end{equation*}

and when $1 \lt y \leq 1.5\text{,}$ we will have

\begin{equation*} 1 \lt \frac{u + 4}{10} \leq 1.5 \Leftrightarrow 10 \lt u + 4 \leq 15 \Leftrightarrow 6 \lt u \leq 11\text{.} \end{equation*}

Thus,

\begin{align*} F_U(u) \amp = \begin{cases} 0, \amp u \lt -4,\\ \int_{y = 0}^{y = (u + 4)/10} y dy\\ = \frac{1}{2} y^2 \biggr|_{y = 0}^{y = (u + 4)/10}\\ = \frac{1}{2} \left(\frac{u + 4}{10} \right)^2 - \frac{1}{2} (0)^2\\ = \frac{1}{200} u^2 + \frac{1}{25} u + \frac{2}{25}, \amp -4 \leq u \leq 6,\\ \int_{y = 0}^{y = 1} y dy + \int_{y = 1}^{y = (u + 4)/10} 1 dy\\ = \frac{1}{2} y^2 \biggr|_{y = 0}^{y = 1} + y \biggr|_{y = 1}^{y = (u + 4)/10}\\ = \left[\frac{1}{2} (1)^2 - \frac{1}{2} (0)^2 \right] + \left[\left(\frac{u + 4}{10} - (1) \right) \right]\\ = \frac{1}{10} u - \frac{1}{10}, \amp 6 \lt u \leq 11,\\ 1, \amp 11 \lt u \end{cases}\\ \implies f_U(u) \amp = \begin{cases} \frac{1}{100} u + \frac{1}{25}, \amp -4 \leq u \leq 6,\\ \frac{1}{10}, \amp 6 \lt u \leq 11,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(b)

Use the answer to part (a) to find $E(U)\text{.}$

Solution.

\begin{align*} E(U) \amp = \int_{u = -4}^{u = 6} u \cdot \left(\frac{1}{100} u + \frac{1}{25} \right) du + \int_{u = 6}^{u = 11} u \cdot \frac{1}{10} du\\ \amp = \int_{u = -4}^{u = 6} \left(\frac{1}{100} u^2 + \frac{1}{25} u \right) du + \int_{u = 6}^{u = 11} \frac{1}{10} u du\\ \amp = \left[\frac{1}{300} u^3 + \frac{1}{50} u^2 \right]_{u = -4}^{u = 6} + \left[\frac{1}{20} u^2 \right]_{u = 6}^{u = 11}\\ \amp = \left[\left(\frac{1}{300} (6)^3 + \frac{1}{50} (6)^2 \right) - \left(\frac{1}{300} (-4)^3 + \frac{1}{50} (-4)^2 \right) \right] + \left[\frac{1}{20} (11)^2 - \frac{1}{20} (6)^2 \right]\\ \amp = \frac{216}{300} + \frac{36}{50} \cdot \frac{6}{6} + \frac{64}{300} - \frac{16}{50} \cdot \frac{6}{6} + \frac{121}{20} \cdot \frac{15}{15} - \frac{36}{20} \cdot \frac{15}{15}\\ \amp = \frac{216}{300} + \frac{216}{300} + \frac{64}{300} - \frac{96}{300} + \frac{1815}{300} - \frac{540}{300} = \frac{1675}{300} = \frac{67}{12} \approx 5.58333333333 \end{align*}

(c)

Find $E(U)$ by the methods of Chapter 4.

Solution.

\begin{align*} E(Y) \amp = \int_{y = 0}^{y = 1} y \cdot y dy + \int_{y = 1}^{y = 3/2} y \cdot 1 dy\\\\ \amp = \frac{1}{3} y^3 \biggr|_{y = 0}^{y = 1} + \frac{1}{2} y^2 \biggr|_{y = 1}^{y = 3/2}\\ \amp = \left(\frac{1}{3} (1)^3 - \frac{1}{3} (0)^3 \right) + \left(\frac{1}{2} \left(\frac{3}{2} \right)^2 - \frac{1}{2} (1)^2 \right)\\ \amp = \frac{1}{3} \cdot \frac{8}{8} + \frac{9}{8} \cdot \frac{3}{3} - \frac{1}{2} \cdot \frac{12}{12}\\ \amp = \frac{8}{24} + \frac{27}{24} - \frac{12}{24} = \frac{23}{24}\\ \implies E(U) \amp = E(10Y - 4) = 10E(Y) - 4\\ \amp = 10 \cdot \frac{23}{24} - 4 = \frac{230}{24} - \frac{96}{24} = \frac{134}{24} = \frac{67}{12} \end{align*}

6.4.

The amount of flour used per day by a bakery is a random variable $Y$ that has an exponential distribution with mean equal to 4 tons. The cost of the flour is proportional to $U = 3Y + 1\text{.}$

(a)

Find the probability density function for $U\text{.}$

Solution.

First note that

\begin{equation*} f(y) = \begin{cases} \frac{1}{4} e^{-y / 4}, \amp y \gt 0,\\ 0, \amp \text{elsewhere} \end{cases} \end{equation*}

and

\begin{equation*} F_U(u) = P(U \leq u) = P(3Y + 1 \leq u) = P \left(Y \leq \frac{u - 1}{3} \right)\text{,} \end{equation*}

where $0 \lt \frac{u - 1}{3} \Leftrightarrow 1 \lt u \text{.}$ Then

\begin{align*} F_U(u) \amp = \begin{cases} 0, \amp u \leq 1,\\ \int_{y = 0}^{y = (u - 1)/3} \frac{1}{4} e^{-y / 4} dy\\ = -e^{-y / 4} \biggr|_{y = 0}^{y = (u - 1)/3}\\ = -e^{-[(u - 1)/3] / 4} + e^{-(0) / 4}\\ = 1 - e^{(1 - u) / 12}, \amp 1 \lt u \end{cases}\\ \implies f_U(u) \amp = \begin{cases} 0, \amp u \leq 1,\\ \frac{1}{12} e^{(1 - u) / 12}, \amp 1 \lt u. \end{cases} \end{align*}

(b)

Use the answer in part (a) to find $E(U)\text{.}$

Solution.

\begin{align*} E(U) \amp = \int_{u = 1}^{u = \infty} u \cdot \frac{1}{12} e^{(1 - u) / 12} du\\ \amp = \int_{v = 0}^{u = \infty} \frac{(v + 1)e^{-v / 12}}{12} dv \s\s\s (\text{letting $v = u - 1$ so $dv = du$})\\ \amp = 12 \int_{v = 0}^{v = \infty} \frac{v^{2 - 1} e^{-v / 12}}{12^2 \Gamma(2)} dv + \int_{v = 0}^{v = \infty} \frac{1}{12} e^{-v / 12} dv\\ \amp = 12 \cdot 1 + 1 = 13 \end{align*}

(In the second-to-last line above, the first integrand was a gamma density with $\alpha = 2$ and $\beta = 12\text{,}$ and the second integrand was an exponential density with $\beta = 12\text{.}$)

6.5.

The waiting time $Y$ until delivery of a new component for an industrial operation is uniformly distributed over the interval from 1 to 5 days. The cost of this delay is given by $U = 2Y^2 + 3\text{.}$ Find the probability density function for $U\text{.}$

Solution.

First note that

\begin{equation*} f(y) = \begin{cases} \frac{1}{5 - 1} = \frac{1}{4}, \amp 1 \leq y \leq 5,\\ 0, \amp \text{elsewhere} \end{cases} \end{equation*}

and

\begin{align*} F_U(u) \amp = P(U \leq u)\\ \amp = P(2Y^2 + 3 \leq u)\\ \amp = P \left(Y^2 \leq \frac{u - 3}{2} \right)\\ \amp = P \left(-\sqrt{\frac{u - 3}{2}} \leq Y \leq \sqrt{\frac{u - 3}{2}} \right)\\ \amp = P \left(1 \leq Y \leq \sqrt{\frac{u - 3}{2}} \right)\text{,} \end{align*}

where this last equality is because $Y$ only takes on values in the range of $1 \leq Y \leq 5\text{.}$ By seeing that $1 = \sqrt{\frac{u - 3}{2}} \Leftrightarrow u = 5$ and $5 = \sqrt{\frac{u - 3}{2}} \Leftrightarrow u = 53\text{,}$ we know that the support of $f_U$ will be $5 \leq u \leq 53\text{.}$ Then

\begin{align*} F_U(u) \amp = \begin{cases} 0, \amp u \lt 5,\\ \int_{y = 1}^{y = \sqrt{(u - 3) / 2}} \frac{1}{4} dy\\ = \frac{1}{4} y \biggr|_{y = 1}^{y = \sqrt{(u - 3) / 2}}\\ = \frac{1}{4} \left(\sqrt{\frac{u - 3}{2}} \right) - \frac{1}{4} (1)\\ = \frac{1}{4} \left(\frac{u - 3}{2} \right)^{1/2} - \frac{1}{4}, \amp 5 \leq u \leq 53,\\ 1, \amp 53 \lt u \end{cases}\\ \implies f_U(u) \amp = \begin{cases} \frac{1}{16} \left(\frac{u - 3}{2} \right)^{-1/2}, \amp 5 \leq u \leq 53,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

6.6.

The joint distribution of amount of pollutant emitted from smokestack without a cleaning device ($Y_1$) and a similar smokestack with a cleaning device ($Y_2$) was given in [provisional cross-reference: ex-5-10] to be

\begin{equation*} f(y_1, y_2) = \begin{cases} 1, \amp 0 \leq y_1 \leq 2, 0 \leq y_2 \leq 1, 2y_2 \leq y_1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{equation*}

The reduction in amount of pollutant due to the cleaning device is given by $U = Y_1 - Y_2\text{.}$

(a)

Find the probability density function for $U\text{.}$

Solution.

First note that $P(U \leq u) = P(Y_1 - Y_2 \leq u) = P(Y_1 \leq Y_2 + u)\text{.}$ The image below shows a red triangle, which is the support of $f\text{,}$ and a green triangle, which is the intersection of the support and the region described by $y_1 \leq y_2 + u\text{.}$ The green line between the two labeled points is a segment of the line $y_1 = y_2 + u\text{.}$ The labeled points are found by 1) solving $y_1 = y_2 + u$ when $y_2 = 0$ to get $y_1 = u\text{,}$ thus the point $(u, 0)\text{,}$ and 2) by rearranging $y_1 = y_2 + u$ to $y_2 = y_1 - u$ and solving $y_1 - u = \frac{1}{2} y_1$ for $y_1 = 2u\text{,}$ and plugging that into either line equation to get $y_2 = \frac{1}{2} (2u) = (2u) - u = u\text{,}$ thus the point $(2u, u)\text{.}$ It is important to note that this is the shape of the green triangle region only when $0 \leq u \leq 1\text{,}$ because if $u = 0$ then $(u, 0) = (2u, u) = (0, 0)$ and the region degenerates, and when $u = 1$ we have $(2u, u) = (2, 1)$ and the shape of the region changes because the support of $f$ has $y_1$ only in the range $0 \leq y_1 \leq 2\text{.}$

When the green triangle region changes shape because $1 \lt u\text{,}$ it changes into a shape of the form in the following image. This time, we have a different point $(2, 2 - u)\text{,}$ which is found by solving $y_1 = y_2 + u$ when $y_1 = 2$ for $y_2 = 2 - u\text{.}$ This region shape holds when $1 \leq u \leq 2\text{,}$ because $u = 1$ is the point at which the region changed shape, and when $u = 2\text{,}$ we get $(u, 0) = (2, 2 - u) = (2, 0)$ and the region covers the entire support of $f\text{.}$ Thus, the support of $f_U$ will be $0 \leq u \leq 2\text{,}$ with different parts for $0 \leq u \leq 1$ and $1 \leq u \leq 2\text{.}$

Then

\begin{align*} F_U(u) \amp = \begin{cases} 0, \amp u \lt 0,\\ \int_{y_2 = 0}^{y_2 = u} \int_{y_1 = 2y_2}^{y_1 = y_2 + u} 1 dy_1 dy_2\\ = \int_{y_2 = 0}^{y_2 = u} y_1 \biggr|_{y_1 = 2y_2}^{y_1 = y_2 + u} dy_2\\ = \int_{y_2 = 0}^{y_2 = u} \left[(y_2 + u) - (2y_2) \right] dy_2\\ = \int_{y_2 = 0}^{y_2 = u} (u - y_2) dy_2\\ = \left[uy_2 - \frac{1}{2} y_2^2 \right]_{y_2 = 0}^{y_2 = u}\\ = \left(u(u) - \frac{1}{2} (u)^2 \right) - \left(u(0) - \frac{1}{2} (0)^2 \right)\\ = \frac{1}{2} u^2, \amp 0 \leq u \leq 1,\\ 1 - \int_{y_1 = u}^{y_1 = 2} \int_{y_2 = 0}^{y_2 = y_1 - u} 1 dy_2 dy_1\\ = 1 - \int_{y_1 = u}^{y_1 = 2} y_2 \biggr|_{y_2 = 0}^{y_2 = y_1 - u} dy_1\\ = 1 - \int_{y_1 = u}^{y_1 = 2} \left[(y_1 - u) - (0) \right] dy_1\\ = 1 - \int_{y_1 = u}^{y_1 = 2} (y_1 - u) dy_1\\ = 1 - \left[\frac{1}{2} y_1^2 - uy_1 \right]_{y_1 = u}^{y_1 = 2}\\ = 1 - \left[\left(\frac{1}{2} (2)^2 - u(2) \right) - \left(\frac{1}{2} (u)^2 - u(u) \right) \right]\\ = -\frac{1}{2} u^2 + 2u - 1, \amp 1 \leq u \leq 2,\\ 1, \amp 2 \lt u \end{cases}\\ \implies f_U(u) \amp = \begin{cases} u, \amp 0 \leq u \leq 1,\\ -u + 2, \amp 1 \leq u \leq 2,\\ 0, \amp \text{elsewhere.} \end{cases} \end{align*}

(b)

Use the answer in part (a) to find $E(U)\text{.}$ Compare your results with those of [provisional cross-reference: ex-5-78]

Solution.

\begin{align*} E(U) \amp = \int_{u = 0}^{u = 1} u \cdot u du + \int_{u = 1}^{u = 2} u \cdot (-u + 2) du\\ \amp = \left[\frac{1}{3} u^3 \right]_{u = 0}^{u = 1} + \left[-\frac{1}{3} u^3 + u^2 \right]_{u = 1}^{u = 2}\\ \amp = \left[\frac{1}{3} (1)^3 - \frac{1}{3} (0)^3 \right] + \left[\left(-\frac{1}{3} (2)^3 + (2)^2 \right) - \left(-\frac{1}{3} (1)^3 + (1)^2 \right) \right]\\\\ \amp = \frac{1}{3} - \frac{8}{3} + 4 + \frac{1}{3} - 1 = 1 \end{align*}

This is precisely what was found in [provisional cross-reference: ex-5-78].

6.7.

Suppose that $Z$ has a standard normal distribution.

(a)

Find the density function of $U = Z^2\text{.}$

(b)

Does $U$ have a gamma distribution? What are the values of $\alpha$ and $\beta\text{?}$

(c)

What is another name for the distribution of $U\text{?}$

6.8.

Assume that $Y$ has a beta distribution with parameters $\alpha$ and $\beta\text{.}$

(a)

Find the density function of $U = 1 - Y\text{.}$

(b)

Identify the density of $U$ as one of the types we studied in Chapter 4. Be sure to identify any parameter values.

(c)

How is $E(U)$ related to $E(Y)\text{?}$

(d)

How is $V(U)$ related to $V(Y)\text{?}$

6.9.

Suppose that a unit of mineral ore contains a proportion $Y_1$ of metal A and a proportion $Y_2$ of metal B. Experience has shown that the joint probability density function of $Y_1$ and $Y_2$ is uniform over the region $0 \leq y_1 \leq 1, 0 \leq y_2 \leq 1, 0 \leq y_1 + y_2 \leq 1\text{.}$ Let $U = Y_1 + Y_2\text{,}$ the proportion of either metal A or B per unit. Find

(a)

the probability density function for $U\text{.}$

(b)

$E(U)$ by using the answer to part (a).

(c)

$E(U)$ by using only the marginal densities of $Y_1$ and $Y_2\text{.}$

6.10.

The total time from arrival to completion of service at a fast food outlet, $Y_1\text{,}$ and the time spent waiting in line before arriving at the service window, $Y_2\text{,}$ were given in [provisional cross-reference: ex-5-15] with joint density function

\begin{equation*} f(y_1, y_2) = \begin{cases} e^{-y_1}, \amp 0 \leq y_2 \leq y_1 \lt \infty,\\ 0, \amp \text{elsewhere.} \end{cases} \end{equation*}

Another random variable of interest is $U = Y_1 - Y_2\text{,}$ the time spent at the service window. Find

(a)

the probability density function for $U\text{.}$

(b)

$E(U)$ and $V(U)\text{.}$ Compare your answers with the results of [provisional cross-reference: ex-5-108].

6.11.

Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the

(a)

probability density function for the average length of life of the two components.

(b)

mean and variance of this average, using the answer in part (a). Check your answer by computing the mean and variance , using [provisional cross-reference: thm-5-12].

6.12.

Suppose that $Y$ has a gamma distribution with parameters $\alpha$ and $\beta$ and that $c \gt 0$ is a constant.

(a)

Derive the density function of $U = cY\text{.}$

(b)

Identify the density of $U$ as one of the types we studied in Chapter 4. Be sure to identify any parameter values.

(c)

The parameters $\alpha$ and $\beta$ of a gamma-distributed random variable are, respectively, “shape” and “scale” parameters. How do the scale and shape parameters for $U$ compare to those for $Y\text{?}$

6.13.

If $Y_1$ and $Y_2$ are independent exponential random variables, both with mean $\beta\text{,}$ find the density function for their sum. (In [provisional cross-reference: ex-5-7], we considered two independent exponential random variables, both with mean 1 and determined $P(Y_1 + Y_2 \leq 3)\text{.}$)

6.14.

In a process of sintering (heating) two types of copper powder (see [provisional cross-reference: ex-5-152]), the density function for $Y_1\text{,}$ the volume of solid copper in a sample, was given by

\begin{equation*} f_1(y_1) = \begin{cases} 6y_1 (1 - y_1), \amp 0 \leq y_1 \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{equation*}

The density function for $Y_2\text{,}$ the proportion of type A crystals among the solid copper, was given by

\begin{equation*} f_2(y_2) = \begin{cases} 3y_2^2, \amp 0 \leq y_2 \leq 1,\\ 0, \amp \text{elsewhere.} \end{cases} \end{equation*}

The variable $U = Y_1 Y_2$ gives the proportion of the sample volume due to type A crystals. If $Y_1$ and $Y_2$ are independent, find the probability density function for $U\text{.}$

6.15.

Let $Y$ have a distribution function given by

\begin{equation*} F(y) = \begin{cases} 0, \amp y \lt 0,\\ 1 - e^{-y^2}, \amp y \geq 0. \end{cases} \end{equation*}

Find a transformation $G(U)$ such that, if $U$ has a uniform distribution on the interval $(0, 1)\text{,}$ $G(U)$ has the same distribution as $Y\text{.}$

6.16.

In [provisional cross-reference: ex-4-15], we determined that

\begin{equation*} f(y) = \begin{cases} \frac{b}{y^2}, \amp y \geq b,\\ 0, \amp \text{elsewhere,} \end{cases} \end{equation*}

is a bona fide probability density function for a random variable, $Y\text{.}$ Assuming $b$ is a known constant and $U$ has a uniform distribution on the interval $(0, 1)\text{,}$ transform $U$ to obtain a random variable with the same distribution as $Y\text{.}$

6.17.

A member of the power family of distributions has a distribution function given by

\begin{equation*} F(y) = \begin{cases} 0, \amp y \lt 0,\\ (\frac{y}{\theta})^{\alpha}, \amp 0 \leq y \leq \theta,\\ 1, \amp y \gt \theta, \end{cases} \end{equation*}

where $\alpha, \theta \gt 0\text{.}$

(a)

Find the density function.

(b)

For fixed values of $\alpha$ and $\theta\text{,}$ find a transformation $G(U)$ such that $G(U)$ has a distribution function of $F$ when $U$ possesses a uniform $(0, 1)$ distribution.

(c)

Given that a random sample of size 5 from a uniform distribution on the interval $(0, 1)$ yielded the values .2700, .6901, .1413, .1523, and .3609, use the transformation derived in part (b) to give values associated with a random variable with a power family distribution with $\alpha = 2, \theta = 4\text{.}$

6.18.

A member of the Pareto family of distributions (often used in economics to model income distributions) has a distribution function given by

\begin{equation*} F(y) = \begin{cases} 0, \amp y \lt \beta,\\ 1 - (\frac{\beta}{y})^{\alpha}, \amp y \geq \beta, \end{cases} \end{equation*}

where $\alpha, \beta \gt 0\text{.}$

(a)

Find the density function.

(b)

For fixed values of $\beta$ and $\alpha\text{,}$ find a transformation $G(U)$ so that $G(U)$ has a distribution function of $F$ when $U$ has a uniform distribution on the interval $(0, 1)\text{.}$

(c)

Given that a random sample of size 5 from a uniform distribution on the interval $(0, 1)$ yielded the values .0058, .2048, .7692, .2475, and .6078, use the transformation derived in part (b) to give values associated with a random variable with a Pareto distribution with $\alpha = 2, \beta = 3\text{.}$

6.19.

Refer to Exercises 6.17 and 6.18. If $Y$ possesses a Pareto distribution with parameters $\alpha$ and $\beta\text{,}$ prove that $X = 1/Y$ has a power family distribution with parameters $\alpha$ and $\theta = \beta^{-1}\text{.}$

6.20.

Let the random variable $Y$ possess a uniform distribution on the interval $(0, 1)\text{.}$ Derive the

(a)

distribution of the random variable $W = Y^2\text{.}$

(b)

distribution of the random variable $W = \sqrt{Y}\text{.}$

6.21.

Suppose that $Y$ is a random variable that takes on only integer values 1, 2, ... Let $F(y)$ denote the distribution function of this random variable. As discussed in [provisional cross-reference: sec-4-2], this distribution is a step function, and the magnitude of the step at each integer value is the probability that $Y$ takes on that value. Let $U$ be a continuous random variable that is uniformly distributed on the interval $(0, 1)\text{.}$ Define a variable $X$ such that $X = k$ if and only if $F(k - 1) \lt U \leq F(k), k = 1, 2, ...$ Recall that $F(0)$ because $Y$ takes on only positive integer values. Show that $P(X = i) = F(i) - F(i - 1) = P(Y = i), i = 1, 2, ...$ That is, $X$ has the same distribution as $Y\text{.}$ [Hint: Recall [provisional cross-reference: ex-4-5].]

6.22.

Use the results derived in Exercises [provisional cross-reference: ex-4-6] and 6.21 to describe how to generate values of a geometrically distributed random variable.

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