Curry G.L., Feldman R.M. Manufacturing Systems Modeling and Analysis

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1.6 Combinations of Random Variables 33

Notice that when the random variables are pair-wise independent, i.e., X

and X

are

independent for all i and j, then E[X

]=E[X

]E[X

] and Property 1.6 is general-

ized indicating that the variance of the sum of n independent random variables is the

sum of the individual variances. In addition, when X

,··· ,X

are independent and

identically distributed (called i.i.d.), we have that

E[ S]=nE[X

] (1.28)

V [S]=nV [X

] .

1.6.2 Random Sum of Random Variables

Before discussing the random sum of random variables, we need a property of con-

ditional expectations. For this discussion we follow the development in [4] in which

these properties are developed assuming discrete random variables because the dis-

crete case is more intuitive than the continuous case. (Although the development

below only considers the discrete case, our main result — given as Property 1.8 —

is true for both discrete and continuous random variables.)

Let Y and X be two random variables. The conditional probability that the ran-

dom variable Y takes on a value b given that the random variable X takes the value

a is written as

Pr{Y = b|X = a} =

Pr{Y = b,X = a}

Pr{X = a}

, if Pr{X = a} = 0

(see Deﬁnition 1.16). Thus, the conditional expectation of Y given that X = a

changes as the value a changes so it is a function, call it g,ofa; namely,

E[Y|X = a]=

∑

bPr{Y = b|X = a} = g(a) .

Hence, the conditional expectation of Y given X is a random variable since it de-

pends on the value of X, expressed as

E[Y |X]=g(X) . (1.29)

Taking the expectation on both sides of (1.29), yields the (unconditional) expecta-

tion of Y and gives the following important property.

Property 1.8. Let Y and X be any two random variables with ﬁnite expec-

tation. The conditional expectation of Y given X is a random variable with

expectation given by

E [E[Y |X]] = E[Y ] .

34 1 Basic Probability Review

Property 1.8 can now be used to obtain the properties of a random sum of random

variables. Let S be deﬁned by

S =

∑

i=1

where X

,··· is a sequence of i.i.d. random variables, and N is a nonnegative

discrete random variable independent of each X

. (When N = 0, the random sum is

interpreted to be zero.) For a ﬁxed n,Eq.(1.28) yields



∑

i=1

|N = n



= nE[X

] , thus



∑

i=1



= NE[X

] .

The expected value of the random sum can be derived from the above result using

Property 1.8 regarding conditional expectations as follows:

E[S]=E



∑

i=1



= E[NE[X

]]

= E[N]E[X

] .

Note that the ﬁnal equality in the above arises using Property 1.6 regarding inde-

pendence and the fact that each random variable in an i.i.d. sequence has the same

mean.

We obtain the variance of the random variable S in a similar fashion, using V [S]=

E[S

] −E[S]

but we shall leave its derivation for homework with some hints (see

Problem 1.29). Thus, we have the following property:

Property 1.9. Let X

,··· be a sequence of i.i.d. random variables where

for each i, E[X

and V [X

. Let N be a nonnegative discrete random

variable independent of the i.i.d. sequence, and let S =

∑

i=1

. Then

E[ S]=

E[ N]

V [S]=

E[N]+

V [N] .

Notice that the squared coefﬁcient of variation of the random sum can also be easily

written as

[S]=C

[N]+

[X]

E[ N]

, where C

[X]=

1.6 Combinations of Random Variables 35

1.6.3 Mixtures of Random Variables

The ﬁnal type of random variable combination that we consider is a mixture of ran-

dom variables. For example, consider two products processed on the same machine,

where the two product types have different processing characteristics. Speciﬁcally,

let X

and X

denote the random processing times for types 1 and 2, respectively, and

then let T denote the processing time for an arbitrarily chosen part. The processing

sequence will be assumed to be random with p

and p

being the probability that

type 1 and type 2, respectively, are to be processed. In other words, T will equal X

with probability p

and T will equal X

with probability p

. Intuitively, we have t he

following relationship.

T =



with probability p

with probability 1 −p

Thus, T is said to be a mixture of X

and X

. In generalizing this concept, we have

the following deﬁnition.

Deﬁnition 1.21. Let X

,··· ,X

be a sequence of independent random variables and

let I be a positive discrete random variable with range 1,··· ,n independent of the

,··· ,X

sequence. The random variable T is called a mixture of random variables

with index I if it can be written as

T = X

Making use of Property 1.8, it should not be too difﬁcult to show the following

property.

Property 1.10. Let T be a mixture of X

,··· ,X

where the mean of X

and variance of X

. Then

E[ T ]=

∑

i=1

E[ T

∑

i=1





where Pr{I = i} = p

are the probabilities associated with the index.

Notice that the above property gives the ﬁrst and second moment, not the variance

directly. If the variance is desired, the equation V [T ]=E[T

]−E[T ]

must be used.

•

Suggestion: Do Problems 1.27–1.31.

36 1 Basic Probability Review

Appendix

In this appendix, two numerical problems are discussed: the computation of the

gamma function ( Eq. 1.18) and the determination of the shape and scale parameters

for the Weibull distribution. We give suggestions for those using Microsoft Excel

and those who are interested in doing the computations within a programming en-

vironment.

The gamma function: For Microsoft Excel users, the gamma function is eval-

uated by ﬁrst obtaining the natural log of the function since Excel provides an

automatic function for the log of the gamma instead of the gamma function it-

self. For example, to obtain the gamma function evaluated at 1.7, use the formula

"=EXP(GAMMALN(1.7))". This yields a value of 0.908639.

For programmers who need the gamma function, there are some good approxi-

mations are available. A polynomial approximation taken from [5, p. 155] is

(1 + x) ≈ 1 + a

x + a

+ ···+a

for 0 ≤ x ≤ 1, (1.30)

where the constants are a

= −0.5748646, a

= 0.9512363, a

= −0.6998588, a

0.4245549, and a

= −0.1010678. (Or if you need additional accuracy, an eight term

approximation is also available in [5]or[1, p. 257].) If it is necessary to evaluate

(x) for x < 1 then use the relationship

(x)=

(1 + x) . (1.31)

If it is necessary to evaluate

(n + x) for n > 1 and 0 ≤ x ≤ 1, then use the relation-

ship:

(n + x)=(n −1 + x)(n −2 + x) ···(1 + x)

(1 + x) . (1.32)

Example 1.18. Suppose we wish to compute

(0.7). The approximation given

by (1.30), yields a result of

(1.7)=0.9086. Applying (1.31) yields

(0.7)=

0.9086/0.7 = 1.298. Now suppose that we wish to obtain the gamma function eval-

uated at 5.7. From (1.32), we have

(5.7)=4.7×3.7 ×2.7×1.7×0.9086 = 72.52.



Weibull parameters: The context for this section is that we know the ﬁrst two

moments of a Weibull distribution (1.20) and would like t o determine the shape and

scale parameters. Notice that the SCV can be written as C

[X]=E[X

]/(E[X ])

−1;

thus, the shape parameter is the value of

that satisﬁes

[X]+1 =

(1 + 2/

)

(

(1 + 1/

))

, (1.33)

and the scale parameter is then determined by

E[X]

(1 + 1/

)

(1.34)

Problems 37

Example 1.19. Suppose we would like to ﬁnd the parameters of the Weibull random

variable with mean 100 and standard deviation 25. We ﬁrst note that C

[X]+1 =

1.0625. We then ﬁll in a spreadsheet with the following values and formulas.

1 mean 100

2 st.dev. 25

3 alpha-guess 1

4 first moment term =EXP(GAMMALN(1+1/B3))

5 second moment term =EXP(GAMMALN(1+2/B3))

6 ratio, Eq. (1.31) = B5/(B4

B4)

7 difference =1+B2

B2/(B1

B1)-B6

8 beta-value =B1/B4

The GoalSeek tool (found under the “Tools” menu in Excel 2003 and under the

“What-If” button on the Data Tab for Excel 2007) is ideal for solving (1.33). When

GoalSeek is clicked, a dialog box appears with three parameters. For the above

spreadsheet, the “Set cell” parameter is set to B7, the “To value” parameter is set to

0, and the “By changing cell” parameter is set to B3. The results should be that the

B3 cell is changed to 4.5 and the B8 cell is changed to 109.6. 

Problems

1.1. A manufacturing company ships (by truckload) its product to three different

distribution centers on a weekly basis. Demands vary from week to week ranging

over 0, 1, and 2 truckloads needed at each distribution center. Conceptualize an

experiment where a week is selected and then the number of truckloads demanded

at each of the three centers are recorded.

(a) Describe the sample space, i.e., list all outcomes.

(b) How many possible different events are there?

(d) If each event containing a single outcome has the same probability, what is the

probability that a total demand for three truckloads will occur?

1.2. A library has classiﬁed its books into ﬁction and nonﬁction. Furthermore, all

books can also be described as hardback and paperback. As an experiment, we shall

pick a book at random and record whether it is ﬁction or nonﬁction and whether it

is paperback or hardback.

(a) Describe the sample space, i.e., list all outcomes.

(b) Describe the event space, i.e., list all events.

paperback is 0.15, the probability of picking a nonﬁction book is 0.30, and the prob-

ability of picking a ﬁction hardback is 0.65.

(d) Using the probabilities from part (c), ﬁnd the probability of picking a ﬁction

book given that the book chosen is known to be a paperback.

38 1 Basic Probability Review

1.3. Let N be a random variable describing the number of defective items in a box

from Example 1.1. Draw the graph for the cumulative distribution function of N and

give its pmf.

1.4. Let X be a random variable with cumulative distribution function given by

G(a)=

⎧

⎨

⎩

0fora < 0,

for 0 ≤ a < 1,

1fora ≥ 1.

(a) Give the pdf for X .

(b) Find Pr{X ≥ 0.5}.

(d) Let X

and X

be independent random variables with their CDF given by G(·).

Find Pr{X

+ X

≤ 1}.

1.5. Let T be a random variable with pdf given by

f (t)=



0fort < 0.5,

−2(t−0.5)

for t ≥ 0.5.

(a) Find k.

(b) Find Pr{0.25 ≤ T ≤ 1}.

(d) Give the cumulative distribution f unction for T .

(e) Let the independent random variables T

and T

have their pdf given by f (·).

Find Pr{1 ≤ T

+ T

≤ 2}.

(f) Let Y = X + T , where X is independent of T and is deﬁned by the previous

problem. Give the pdf for Y .

1.6. Let U be a random variable with pdf given by

h(u)=

⎧

⎪

⎨

⎪

⎩

0foru < 0,

u for 0 ≤ u < 1,

2 −u for 1 ≤ u < 2,

0foru ≥ 2.

(a) Find Pr{0.5 < U < 1.5}.

(b) Find Pr{0.5 ≤U ≤ 1.5}.

read as an “and”.)

(d) Give the cumulative distribution function for U and calculate Pr{U ≤ 1.5}−

Pr{U ≤ 0.5}.

1.7. An independent rooﬁng contractor has determined that the number of jobs ob-

tained for the month of September varies. From previous experience, the probabil-

ities of obtaining 0, 1, 2, or 3 jobs have been determined to be 0.1, 0.35, 0.30, and

Problems 39

0.25, respectively. The proﬁt obtained from each job is $300. What is the expected

proﬁt and the standard deviation of proﬁt for September?

1.8. There are three investment plans for your consideration. Each plan calls for

an investment of $25,000 and the return will be one year later. Plan A will return

$27,500. Plan B will return $27,000 or $28,000 with probabilities 0.4 and 0.6, re-

spectively. Plan C will return $24,000, $27,000, or $33,000 with probabilities 0.2,

0.5, and 0.3, respectively. If your objective is to maximize the expected return, which

plan should you choose? Are there considerations that might be relevant other than

simply the expected values?

1.9. Let the random variables A,B,C denote the returns from investment plans A,

B, and C, respectively, from the previous problem. What are t he mean and standard

deviations of the three random variables?

1.10. Let N be a random variable with cumulative distribution function given by

F(x)=

⎧

⎪

⎨

⎪

⎩

0forx < 1,

0.2for1≤ x < 2,

0.5for2≤ x < 3,

0.8for3≤ x < 4,

1forx ≥ 4.

Find the mean and standard deviation of N.

1.11. Prove that the E[(X −

)

]=E[X

] −

for any random variable X whose

mean is

1.12. Find the mean and standard deviation for X as deﬁned in Problem 1.4.

1.13. Show using integration by parts that

E[ X ]=



[1 −F(x)]dx, for 0 ≤ a ≤ x ≤ b ,

where F is the CDF of a random variable with support on the interval [a,b] with

a ≥ 0. Note that the lower integration limit is 0 not a. (A random variable is zero

outside its interval of support.)

1.14. Find the mean and standard deviation for U as deﬁned in Problem 1.6.Also,

ﬁnd the mean and standard deviation using the last two properties mentioned in

Property 1.4.

Use the appropriate distribution from Sect. 1.4 to answer the questions in Prob-

lems 1.15–1.19.

1.15. A manufacturing company produces parts, 97% of which are within speciﬁ-

cations and 3% are defective (outside speciﬁcations). There is apparently no pattern

to the production of defective parts; thus, we assume that whether or not a part is

40 1 Basic Probability Review

defective is independent of other parts.

(a) What is the probability that there will be no defective parts in a box of 5?

(b) W hat is the probability that there will be exactly 2 defective parts in a box of 5?

(d) Use the Poisson distribution to approximate the probability that there will be 4

or more defective parts in a box of 40.

(e) Use the normal distribution to approximate the probability that there will be 20

or more defective parts in a box of 400.

1.16. A store sells two types of tables: plain and deluxe. When an order for a table

arrives, there is an 80% chance that the plain table will be desired.

(a) Out of 5 orders, what is the probability that no deluxe tables will be desired?

(b) Assume that each day 5 orders arrive and that today (Monday) an order came

for a deluxe table. What is the probability that the ﬁrst day in which one or more

deluxe tables are again ordered will be in three more days (Thursday)? What is the

expected number of days until a deluxe table is desired?

mean of 5. What is the probability that exactly 5 orders will arrive on a given day?

1.17. A vision system is designed to measure the angle at which the arm of a robot

deviates from the vertical; however, the vision system is not totally accurate. The

results from observations is a continuous random variable with a uniform distribu-

tion. If the measurement indicates that the range of the angle is between 9.7 and

10.5 degrees, what is the probability that the actual angle is between 9.9 and 10.1

degrees?

1.18. The dispatcher at a central ﬁre station has observed that the time between calls

is an exponential random variable with a mean of 32 minutes.

(a) A call has just arrived. What is the probability that the next call will arrive within

the next half hour.

(b) What is the probability that there will be exactly two calls during the next hour?

1.19. In an automated soldering operation, the location at which the solder is placed

is very important. The deviation from the center of the board is a normally dis-

tributed random variable with a mean of 0 inches and a standard deviation of 0.01

inches. (A positive deviation indicates a deviation t o the right of the center and a

negative deviation indicates a deviation to the left of the center.)

(a) What is the probability that on a given board the actual location of the solder

deviated by less than 0.005 inches (in absolute value) from the center?

(b) What is the probability that on a given board the actual location of the solder

deviated by more than 0.02 inches (in absolute value) from the center?

1.20. The purpose of this problem is to illustrate the dangers of statistics, espe-

cially with respect to categorical data and the use of conditional probabilities. In

this example, the data may be used to support contradicting claims, depending on

the inclinations of the person doing the reporting! The population in which we are

interested is made up of males and females, those who are sick and not sick, and

Problems 41

those who received treatment prior to becoming sick and who did not receive prior

treatment. (In the questions below, assume that the treatment has no adverse side

effects.) The population numbers are as follows.

Males

sick not sick

treated 200 300

not treated 50 50

Females

sick not sick

treated 50 100

not treated 200 370

(a) What is the conditional probability of being sick given that the treatment was

received and the patient is a male?

(b) Considering only the population of males, should the treatment be recom-

mended?

(d) Considering the entire population, should the treatment be recommended?

1.21. Let X and Y be two discrete random variables where their joint pm f

f (a, b)=Pr{X = a,Y = b}

is deﬁned by

012

10 0.01 0.06 0.03

11 0.02 0.12 0.06

12 0.02 0.18 0.10

13 0.07 0.24 0.09

with the possible values for X being 10 through 13 and the possible values for Y

being 0 through 2.

(a) Find the marginal pmf’s for X and Y and then ﬁnd the Pr{X = 11} and E[X].

(b) Find the conditional pmf for X given that Y = 1 and then ﬁnd the Pr{X = 11|Y =

1} and ﬁnd the E[X|Y = 1].

(d) Find Pr{X = 13,Y = 2} ,Pr{X = 13}, and Pr{Y = 2}. (Now make sure your

answer to part (c) was correct.)

1.22. Let S and T be two continuous random variables with joint pdf given by

f (s,t)=kst

for 0 ≤ s ≤ 1, 0 ≤t ≤ 1 ,

and zero elsewhere.

(a) Find the value of k.

42 1 Basic Probability Review

(b) Find the marginal pdf’s for S and T and then ﬁnd the Pr{S ≤ 0.5} and E[S].

0.1} and ﬁnd the E[S|T = 0.1].

(d) Are S and T independent? Why or why not?

1.23. Let U and V be two continuous random variables with joint pdf given by

g(u,v)=e

−u−v

for u ≥ 0 , v ≥ 0 ,

and zero elsewhere.

(a) Find the marginal pdf’s for U and V and then ﬁnd the Pr{U ≤ 0.5} and E[U].

(b) Find the conditional pdf for U given that V = 0.1 and then ﬁnd the Pr{U ≤

0.5|V = 0.1} and ﬁnd the E[U|V = 0.1].

1.24. This problem is to consider the importance of keeping track of history when

discussing the reliability of a machine and to emphasize the meaning of Eq. (1.16).

Let T be a random variable that indicates the time until failure for the machine.

Assume that T has a uniform distribution from zero to two years and answer the

question, “What is the probability that the machine will continue to work for at least

three more months?”

(a) Assume the machine is new.

(b) Assume the machine is one year old and has not yet failed.

answer parts (a) and (b) again.

(d) Is it important to know how old the machine is in order to answer the question,

“What is the probability that the machine will continue to work for at least three

more months?”

1.25. Determine the correlation coefﬁcient for the random variables X and Y from

Example 1.13.

1.26. A shipment containing 1,000 steel rods has just arrived. Two measurements

are of interest: the cross-sectional area and the force that each rod can support. We

conceptualize two random variables: A and B. The random variable A is the cross-

sectional area, in square centimeters, of the chosen rod, and B is the force, in kilo-

Newtons, that causes the rod to break. Both random variables can be approximated

by a normal distribution. (A generalization of the normal distribution to two random

variables is called a bivariate normal distribution.) The random variable A has a

mean of 6.05 cm

and a standard deviation of 0.1 cm

. The random variable B has a

mean of 132 kN and a standard deviation of 10 kN. The correlation coefﬁcient for A

and B is 0.8.

To answer the questions below use the fact that if X

and X

are bivariate normal

random variables with means

and

, respectively, variances

and

, respec-

tively, and a correlation coefﬁcient

, t he following hold:

• The marginal distribution of X

is normal.