Chapter 6: Functions of Random Variables

6.3: The Method of Distribution Functions

6.3: The Method of Distribution Functions

Summary of the Distribution Function Method

Let UU be a function of the random variables Y1,Y2,...,YnY_1, Y_2, ..., Y_n.

  1. Find the the region U=uU = u in the (y1,y2,...,yn)(y_1, y_2, ..., y_n) space.

  2. Find the region UuU \leq u.

  3. Find FU(u)=P(Uu)F_U(u) = P(U \leq u) by integrating f(y1,y2,...,yn)f(y_1, y_2, ..., y_n) over the region UuU \leq u.

  4. Find the density function fU(u)f_U(u) by differentiating FU(u)F_U(u). Thus, fU(u)=dFU(u)/duf_U(u) = dF_U(u)/du.

Exercises

6.13

If Y1Y_1 and Y2Y_2 are independent exponential random variables, both with mean β\beta, find the density function for their sum. (In Exercise 5.7, we considered two independent exponential random variables, both with mean 11 and determined P(Y1+Y23)P(Y_1 + Y_2 \leq 3).)

6.14

In a process of sintering (heating) two types of copper powder (see Exercise 5.152), the density function for Y1Y_1, the volume proportion of solid copper in a sample, was given by

f1(y1)={6y1(1y1),0y11,0,elsewhere.f_1(y_1) = \begin{cases} 6y_1 (1 - y_1), & 0 \leq y_1 \leq 1,\\ 0, & \text{elsewhere.} \end{cases}

The density function for Y2Y_2, the proportion of type A crystals among the solid copper, was given as

f2(y2)={3y22,0y21,0,elsewhere.f_2(y_2) = \begin{cases} 3y_2^2, & 0 \leq y_2 \leq 1,\\ 0, & \text{elsewhere.} \end{cases}

The variable U=Y1Y2U = Y_1 Y_2 gives the proportion of the sample volume due to type A crystals. If Y1Y_1 and Y2Y_2 are independent, find the probability density function for UU.

6.15

Let YY have a distribution function given by

F(y)={0,y<0,1ey2,y0.F(y) = \begin{cases} 0, & y < 0,\\ 1 - e^{-y^2}, & y \geq 0. \end{cases}

Find a transformation G(U)G(U) such that, if UU has a uniform distribution on the interval (0,1)(0, 1), G(U)G(U) has the same distribution as YY.

6.16

In Exercise 4.15, we determined that

f(y)={by2,yb,0,elsewhere,f(y) = \begin{cases} \frac{b}{y^2}, & y \geq b,\\ 0, & \text{elsewhere,} \end{cases}

is a bona fide probability density function for a random variable, YY. Assuming bb is a known constant and UU has a uniform distribution on the interval (0,1)(0, 1), transform UU to obtain a random variable with the same distribution as YY.

6.17

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

F(y)={0,y<0,(yθ)α,0yθ,1,y>θ,F(y) = \begin{cases} 0, & y < 0,\\ \left(\frac{y}{\theta} \right)^{\alpha}, & 0 \leq y \leq \theta,\\ 1, & y > \theta, \end{cases}

where α,θ>0\alpha, \theta > 0.

a) Find the density function.

b) For fixed values of α\alpha and θ\theta, find a transformation G(U)G(U) so that G(U)G(U) has a distribution function of FF when UU possesses a uniform (0,1)(0, 1) distribution.

c) Given that a random sample of size 55 from a uniform distribution on the interval (0,1)(0, 1) yielded the values .2700.2700, .6901.6901, .1413.1413, .1523.1523, and .3609.3609, use the transformation derived in part (b) to give values associated with a random variable with a power family distribution with α=2,θ=4\alpha = 2, \theta = 4.

6.18

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

F(y)={0,y<β,1(βy)α,yβ,F(y) = \begin{cases} 0, & y < \beta,\\ 1 - \left(\frac{\beta}{y} \right)^{\alpha}, & y \geq \beta, \end{cases}

where α,β>0\alpha, \beta > 0.

a) Find the density function.

b) For fixed values of β\beta and α\alpha, find a transformation G(U)G(U) so that G(U)G(U) has a distribution function of FF when UU has a uniform distribution on the interval (0,1)(0, 1).

c) Given that a random sample of size 55 from a uniform distribution on the interval (0,1)(0, 1) yielded the values .0058.0058, .2048.2048, .7692.7692, .2475.2475 and .6078.6078, use the transformation derived in part (b) to give values associated with a random variable with a Pareto distribution with α=2,β=3.\alpha = 2, \beta = 3.

6.19

Refer to Exercises 6.17 and 6.18. If YY possesses a Pareto distribution with parameters α\alpha and β\beta, prove that X=1/YX = 1/Y has a power family distribution with parameters α\alpha and θ=β1\theta = \beta^{-1}.

6.20

Let the random variable YY possess a uniform distribution on the interval (0,1)(0, 1). Derive the

a) distribution of the random variable W=Y2W = Y^2.

b) distribution of the random variable W=YW = \sqrt{Y}.

6.21

Suppose that YY is a random variable that takes on only integer values 1,2,...1, 2, .... Let F(y)F(y) denote the distribution function of this random variable. As discussed in Section 4.2, this distribution function is a step function, and the magnitude of the step at each integer value is the probability that YY takes on that value. Let UU be a continuous random variable that is uniformly distributed on the interval (0,1)(0, 1). Define a variable XX such that X=kX = k if and only if F(k1)<UF(k),k=1,2,...F(k - 1) < U \leq F(k), k = 1, 2, .... Recall that F(0)=0F(0) = 0 because YY takes on only positive integer values. Show that P(X=i)=F(i)F(i1)=P(Y=i),i=1,2,...P(X = i) = F(i) - F(i - 1) = P(Y = i), i = 1, 2, .... That is, XX has the same distribution as YY. [Hint: Recall Exercise 4.5.]

6.22

Use the results derived in Exercises 4.6 and 6.21 to describe how to generate values of a geometrically distributed random variable.