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TABLE OF CONTENTS
1.
Probability
A.
B.
C.
D.
E.
F.
2.
Discrete Probabilities
Density Histograms and Sample Percentiles
Moments of Discrete Variables
Skewness
Moment Generating Functions
Bernoulli Trials and the Binomial Distribution
Poisson Distribution
Negative Binomial Distribution
47
51
57
63
67
71
85
109
Continuous Distributions
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
K.
4.
1
7
15
21
25
29
Discrete Distributions
A.
B.
C.
D.
E.
F.
G.
H.
3.
Properties of Probability
Methods of Enumeration: Selection Without Replacement
Methods of Enumeration: Selection with Replacement
Conditional Probability
Independent Events
Bayes’ Theorem
Probability Density and Distribution Functions of Continuous Variables
Modes of Continuous Distributions
Percentiles and Medians
Moments and Expectations of Continuous Distributions
Distributions of Functions of a Random Variable
Uniform Distribution
Exponential and Gamma Distributions
Normal Distribution
Normal Approximations for Discrete Distributions
Lognormal Distribution
Beta, Pareto, and Weibull Distributions
115
123
127
131
139
145
153
169
175
181
185
Multivariate Distributions
A.
B.
C.
D.
E.
F.
G.
H.
I.
Joint Probability Distributions of Two Random Variables
Marginal Density Functions
Independent Random Variables
Conditional Distributions
Conditional Moments: Discrete Cases
Conditional Moments: Continuous Cases
Transformations of Random Variables
Order Statistics
Multinomial Distribution
189
201
207
213
223
239
247
251
255
5.
Other Topics
A.
B.
C.
D.
E.
F.
6.
Covariance
Correlation Coefficient
Bivariate Normal Distribution
Sums of Independent Random Variables
Chi-Square Distribution
Inequalities and the Central Limit Theorem
257
267
271
273
283
289
Anderson
A.
B.
Deductibles and Limits
Inflation and Coinsurance
291
303
NOTES
Questions and parts of some solutions have been taken from material copyrighted by the Casualty Actuarial
Society and the Society of Actuaries. They are reproduced in this study manual with the permission of the CAS
and SoA solely to aid students studying for the actuarial exams. Some editing of questions has been done.
Students may also request past exams directly from both societies. I am very grateful to these organizations for
their cooperation and permission to use this material. They are, of course, in no way responsible for the
structure or accuracy of the manual.
Exam questions are identified by numbers in parentheses at the end of each question. CAS questions have four
numbers separated by hyphens: the year of the exam, the number of the exam, the number of the question, and
the points assigned. SoA or joint exam questions usually lack the number for points assigned. W indicates a
written answer question; for questions of this type, the number of points assigned are also given. A indicates a
question from the afternoon part of an exam. MC indicates that a multiple choice question has been converted
into a true/false question.
Page references refer to Michael A. Bean, Probability: The Science of Uncertainty with Applications to
Investments, Insurance, and Engineering (2005); Saeed Ghahramani, Fundamentals of Probability with
Stochastic Processes, (2005); Matthew J. Hassett and Donald G. Stewart, Probability for Risk Management,
(1999); Robert V. Hogg and Elliot A. Tanis, Probability and Statistical Inference (2001); Irwin Miller and
Marylees Miller, John E. Freund's Mathematical Statistics with Applications (2004); Sheldon Ross, A First
Course in Probability (2002); and Dennis D. Wackerly, William Mendenhall III, and Richard Scheaffer,
Mathematical Statistics with Applications (2002).
Although I have made a conscientious effort to eliminate mistakes and incorrect answers, I am certain some
remain. I am very grateful to students who discovered errors in the past and encourage those of you who find
others to bring them to my attention. Please check our web site for corrections subsequent to publication. I
would also like to thank Jenny Carlson for doing initial solutions for most of the problems in the manual.
Hanover, NH 6/30/11
PJM
NOTES
Questions and parts of some solutions have been taken from material copyrighted by the Casualty
Actuarial Society and the Society of Actuaries. They are reproduced in this study manual with the
permission of the CAS and SoA solely to aid students studying for the actuarial exams. Some editing of
questions has been done. Students may also request past exams directly from both societies. I am very
grateful to these organizations for their cooperation and permission to use this material. They are, of
course, in no way responsible for the structure or accuracy of the manual.
Exam questions are identified by numbers in parentheses at the end of each question. CAS questions
have four numbers separated by hyphens: the year of the exam, the number of the exam, the number of
the question, and the points assigned. SoA or joint exam questions usually lack the number for points
assigned. W indicates a written answer question; for questions of this type, the number of points assigned
are also given. A indicates a question from the afternoon part of an exam. MC indicates that a multiple
choice question has been converted into a true/false question.
Page references refer to Michael A. Bean, Probability: The Science of Uncertainty with Applications to
Investments, Insurance, and Engineering (2005); Saeed Ghahramani, Fundamentals of Probability with
Stochastic Processes, (2005); Matthew J. Hassett and Donald G. Stewart, Probability for Risk
Management, (1999); Robert V. Hogg and Elliot A. Tanis, Probability and Statistical Inference (2001);
Irwin Miller and Marylees Miller, John E. Freund's Mathematical Statistics with Applications (2004);
Sheldon Ross, A First Course in Probability (2002); and Dennis D. Wackerly, William Mendenhall III,
and Richard Scheaffer, Mathematical Statistics with Applications (2002).
Although I have made a conscientious effort to eliminate mistakes and incorrect answers, I am certain
some remain. I am very grateful to students who discovered errors in the past and encourage those of you
who find others to bring them to my attention. Please check our web site for corrections subsequent to
publication. I would also like to thank Jenny Carlson for doing initial solutions for most of the problems
in the manual.
Hanover, NH 11/1/10
PJM
Multivariate Distributions  189
PAST CAS AND SoA EXAMINATION QUESTIONS
ON MULTIVARIATE DISTRIBUTIONS
A.
Joint Probability Distributions of Two Random Variables
A1.
Let the joint density function of X and Y be given by the following:
 x + y for 0 < x < 1 and 0 < y < 1
f(x, y) =  0 otherwise

What is P(X < 2Y)?
A. 7/32
A2.
B. 1/4
D. 19/24
E. 7/8
(792–210)
Suppose that the joint density function of X and Y is uniform over the region
R = {(x, y) | x + y < 2, x > 0, y > 0}. What is the probability that exactly one of the two events
A = {X < 1} and B = {Y > 1} occurs?
A. 1/16
A3.
C. 3/4
B. 1/4
C. 1/2
D. 5/8
E. 3/4
(792–214)
Suppose X and Y have the following joint density function:
 x + (10/3)y2 for 0 ≤ x ≤ y ≤ 1
f(x, y) =  0 otherwise

What is E[XY]?
A. 19/60
A4.
B. 1/3
C. 11/30
D. 2/5
E. 31/90
(81F–2–38)
Let the joint density function for X and Y be given by the following:
 kxy2 for 0 < x < y < 1
f(x, y) =  0 otherwise

What is the value of k?
A. 1
A5.
B. 2
C. 5
D. 6
E. 10
(83S–218)
Let X and Y be continuous random variables with the following joint density function:
 kx-3e-y/3 for 1 < x < ∞ and 1 < y < ∞
f(x, y) =  0 otherwise

Then k =
A. (1/3)e1/3
B. e-1/3
© 2010 ACTEX Publications, Inc.
C. (2/3)e1/3
D. (3/2)e-1/3
E. e1/3
(88S–110–50)
SOA Exam P and CAS Exam 1 – Peter J. Murdza
190  Multivariate Distributions
Solutions are based on Bean, pp. 29–35, 104–19, 325–40; Ghahramani, pp. 311–25, 369–75, 378–82;
Hassett, pp. 269–70, 274–78, 293–304, 320–29; Hogg, pp. 222–33; Miller, pp. 92–100; Ross, pp. 239–47; and
Wackerly pp. 210–22, 241–49, 255–61.
1 1
A1.
1
|
1
1
2
2
P(X < 2Y) = 

 x + y dy dx = 
 (xy + y /2) x/2 dx = 
 x + 1/2  5x /8 dx
0 x/2
0
P(X < 2Y) = [x2/2 + x/2 
|
1
(5/24)x3] 0
0
= 19/24
Answer: D
A2.
The region described by the inequalities is a triangle bounded by the lines x + y = 2, x = 0, and
y = 0. This can be broken up into two triangles and a square:
The upper triangle has the additional constraints X < 1 and Y > 1. The lower triangle has the additional
constraints X > 1 and Y < 1. The square has the additional constraints of X < 1 and
Y < 1. Since in the upper triangle both events occur and in the lower triangle neither events occur, neither
of these areas are included in the desired probability. In the square, which comprises 1/2 of the area, only
one of the events occurs (X < 1).
Answer: C
A3.
1 y
1
0 0
0
|
1
y
2
3
2 3
E[XY] = 
 y4/3 + (5/3)y5 dy

 xy[x + (10/3)y ] dx dy = 
 [x y/3 + (5/3)x y ] 0 dy = 
|
1
E[XY] = [y5/15 + 5y6/18] 0
0
= 1/15 + 5/18 = 31/90
Answer: E
A4.
Since probability must equal one, we get:
1 y
1
|
y
2
2 2
1 = 

 kxy dx dy = 
 kx y /2 dy 0
0 0
1
|
1
4
5
= 
 ky /2 dy = ky /10 0
0
= k/10
k = 10
0
Answer: E
A5.
Since the probabilities must integrate to unity, we get:
∞ ∞
∞
1 1
1
|

|
∞
1
-3 -y/3
-1/3
-3
-1/3 -2 1
1 = 

 3e-y/3kx-3 ∞ dx = 
 kx e dy dx = 
 3e kx dx = (3/2)e kx
1 = (3/2)e-1/3k
1
k = (2/3)e1/3
Answer: C
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza
Multivariate Distributions  191
A6.
Let X and Y be continuous random variables with the following joint density function:
 .25 for 0 ≤ x ≤ 2 and x  2 ≤ y ≤ x
f(x, y) =  0 otherwise

What is E[X3Y]?
A. 6/5
A7.
B. 4/3
C. 2
D. 4
E. 24/5
(90S–110–2)
Let X and Y be discrete random variables with the following joint probability function:
1/6 for x = 1, 2, 3; y = 2, 3
0 otherwise


f(x, y) = 
If U = X + Y. What is the probability function of U?
A.
B.
C.
D.
E.
A8.
g(u) = 1/4 for u = 3, 4, 5, 6; 0 otherwise
g(u) = 2/5 for u = 3; 1/5 for u = 4, 5, 6; 0 otherwise
g(u) = 1/6 for u = 3, 6; 1/3 for u = 4, 5; 0 otherwise
g(u) = 1/5 for u = 2, 3, 4, 5, 6; 0 otherwise
g(u) = 1/4 for u = 2, 3, 4, 5; 0 otherwise. (90S–110–9)
Let X and Y be continuous random variables with the following joint density function:
 2(x + y) for 0 < x < y < 1
f(x, y) =  0 otherwise

Then E[Y] =
A. 5/12
A9.
B. 1/2
C. 3/4
D. 1
E. 7/6
(90S–110–25) (Sample–1–37)
Let X and Y be continuous random variables with joint density function f(x, y) and marginal density
functions fX and fY, respectively, that are nonzero only on the interval (0, 1). Which one of the following
statements is always true?
A.
E[X2Y3]
(
1
1
 x2
= 
(
)(
dx
0
1
y3
1
)
dy
B.
E[X2]
 x2 f(x, y) dx
=
0
0
1
)( y
C. E[X2Y3] = 
 x2 f(x, y) dx
0
3
)
f(x, y) dy
1
D. E[X2] = 
 x2 f(x) dx
0
0
1
 y3 f(x) dx
E. E[Y3] = 
(90S–110–38)
0
A10.
Let X and Y be continuous random variables with the following joint density function:
 xy for 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1
f(x, y) =  0 otherwise

What is P(X/2 ≤ Y ≤ X)?
A. 3/32
B. 1/8
C. 1/4
© 2010 ACTEX Publications, Inc.
D. 3/8
E. 3/4
(90S–110–49)
SOA Exam P and CAS Exam 1 – Peter J. Murdza
192  Multivariate Distributions
2
A6.
E[X3Y]
x
= 
 

2
.25x3y dy dx
|
0 x2
0
2
E[X3Y]
= 

x
3 2
= 
 (1/8) x y dx x-2 dx
2
1/8[x3(x2)

x3(x

2)2]
dx = 
 x4/2  x3/2 dx
0
0
|
2
E[X3Y] = [x5/10  x4/8] 0
= (2)4(1/5  1/8) = 6/5
Answer: A
A7.
Construct a joint probability table and sum the probabilities for different values of U:
(X,Y)
U
P(X, Y)
(1, 2)
3
1/6
g(3) = g(6) = 1/6
(2, 2)
4
1/6
(2, 3)
5
1/6
(3, 1)
4
1/6
(3, 2)
5
1/6
(3, 3)
6
1/6
g(4) = g(5) = 1/3
Answer: C
1
A8.
y
1
E[Y] = 
 
 2y(x + y) dx dy = 

0
0
x2y +
0
|
y
2xy2 0
1
|
1
3
4
dy = 
 3y dy = 3y /4 0
0
= 3/4
Answer: C
1
1
A9.
E[u(x)] = 
 u(x) f(x) dx
0
E[X2]
 x2 f(x) dx
=
0
Answer: D
1 x
A10.
2 1
1
P(X/2 ≤ Y ≤ X) = 

 xy dy dx + 

 xy dy dx = 

0 x/2
1
1 x/2
|
x
xy2/2 x/2
0
2
|
1
1
|
1
dx + 
 xy2/2 x/2 dx
1
3
3
3
4
P(X/2 ≤ Y ≤ X) = 
 x /2  x /8 dx + 
 x/2  x /8 dx = 3x /32 0
0
2
|
2
+ [x2/4  x4/32] 1
P(X/2 ≤ Y ≤ X) = 3/32  0 + (1  1/2)  (1/4 1/32) = 3/8
Answer: D
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza
Multivariate Distributions  193
A11.
Let X and Y be discrete random variables with joint probability function
 y/(24x) for x = 1, 2, 4; y = 2, 4, 8; x ≤ y
f(x, y) =  0 otherwise

What is P(X + Y/2 ≤ 5)?
A. 1/8
A12.
B. 7/24
C. 3/8
D. 5/8
E. 17/24
(92S–110–6)
Let X1 and X2 be random variables with joint moment generation function
M(t1, t2) = .3 + (.1)exp(t1) + (.2)exp(t2) + (.4)exp(t1 + t2). What is E[2X1  X2]?
A. .1
B. .4
(92S–110–16)
A13.
E. .3 + (.1)exp(2t1) + (.2)exp(t2) + (.4)exp(2t1  t2)
D. .2e + .4e2
C. .8
Let X and Y be discrete random variables with joint probability function given by the following table:
(x, y)
p(x, y)
(0, 0)
0
(0, 1)
1/5
(1, 0)
2/5
(1, 1)
1/5
(2, 0)
1/5
(2, 1)
0
What is the variance of Y  X?
A. 4/25
A14.
B. 16/25
C. 26/25
D. 5/4
E. 7/5
(92S–110–40)
Let X and Y be discrete random variables with the following joint probability function:
 2(x+1-y)/9 for x = 1, 2, and y = 1, 2
p(x, y) =  0 otherwise

Calculate E[X/Y].
A. 8/9
A15.
B. 5/4
C. 4/3
D. 25/18
E. 5/3
(96W–11029)
Let X and Y be random losses with the following joint density function:
 e-(x+y) for x > 0 and y > 0
f(x, y) =  0 otherwise

An insurance policy is written to reimburse (X + Y). Calculate the probability that the reimbursement is
less than 1.
A. e-2
B. e-1
C. 1  e-1
© 2010 ACTEX Publications, Inc.
D. 1  2e-1
E. 1  2e-2
(Sample–1–4)
SOA Exam P and CAS Exam 1 – Peter J. Murdza
194  Multivariate Distributions
A11.
The specified outcome consists of the following five pairs: (1, 2), (1, 4), (1, 8), (2, 2), (2, 4).
Thus we get for a sum of their respective probabilities:
P(Outcome) = 2/24 + 4/24 + 8/24 + 2/48 + 4/48 = 17/24
Answer: E
A12.
E[X1] = MX1'(0) = (.1)exp(0) + (.4)exp(0) = .5
E[X2] = MX2'(0) = (.2)exp(0) + (.4)exp(0) = .6
E[2X1  X2] = 2E[X1]  E[X2] = (2)(.5)  (.6) = .4
Answer: B
A13.
E[Y  X] = (0)(0) + (1)(1/5) + (1)(2/5) + (0)(1/5) + (2)(1/5) + (1)(0) = 3/5
E[(Y  X)2] = (0)2(0) + (1)2(1/5) + (1)2(2/5) + (0)2(1/5) + (2)2(1/5) + (1)2(0) = 7/5
Var(Y  X) = E[(Y  X)2]  (E[Y  X])2 = 7/5  (3/5)2 = 26/25
Answer: C
A14.
E[X/Y] = (1/1)p(1, 1) + (1/2)p(1, 2) + (2/2)p(2, 2) + (2/1)p(2, 1)
E[X/Y] = (1)(2)1+1-1/9 + (1/2)(2)1+1-2/9 + (1)(2)2+1-2/9 + (2)(2)2+1-1/9 = 25/18
Answer: D
1 1x
A15.
P(X + Y < 1) = P(Y < 1  X) = 
 

0
|
1
P(X + Y < 1) = [e-x  xe-1] 0
0
1
e-(x+y)
dy dx = 

0
|
1-x
e-(x+y) 0
1
-x
-1
dx = 
 e  e dx
0
= (e-1  e-1)  (1) = 1  2e-1
Answer: D
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza
Multivariate Distributions  195
A16.
Given that future lifetimes (in months) of two components of a machine have the following joint density
function, what is the probability that both components are still functioning twenty months from now?
 (6/125,000)(50  x  y) for 0 < x < 50 – y < 50
f(x, y) =  0 otherwise

20 20
A. (6/125,000) 
(50  x  y) dy dx
 
0 0
30 50-x-y
30 50-x
B. (6/125,000) 
 (50  x  y) dy dx
 
20 20
50 50-x
C. (6/125,000) 


 (50  x  y) dy dx
D. (6/125,000) 
 (50  x  y) dy dx
 
E. (6/125,000) 


 (50  x  y) dy dx ) (00F–1–20) (Sample–P–89)
20 20
50 50-x-y
20
20
A17.
20 20
An insurance company insures a large number of drivers. Let X be the random variable representing the
company’s losses under collision insurance and let Y represent the company’s losses under liability
insurance. X and Y have the following joint density function:
(2x + 2  y)/4 for 0 < x < 1 and 0 < y < 2
 0 otherwise

f(x, y) = 
What is the probability that the total loss is at least 1?
A. .33
A18.
B. .38
C. .41
D. .71
E. .75
(00F–1–36) (Sample–P–91)
A device runs until either of two components fails, at which point the device stops running. The joint
density function of the lifetimes of the two components, both measured in hours, is:
f(x, y) = (x + y)/27
0 < x < 3 and 0 < y < 3
Calculate the probability that the device fails during its first hour of operation.
A. .04
A19.
B. .41
C. .44
D. .59
E. .96
(03S–1–16) (Sample–P–78)
Let X be the age of an insured automobile involved in an accident. Let Y be the length of time the owner
has insured the automobile at the time of the accident. X and Y have joint probability density function:
 (10  xy2)/64 for 2 ≤ x ≤ 10 and 0 ≤ y ≤ 1
f(x, y) =  0 otherwise

Calculate the expected age of an insured automobile involved in an accident.
A. 4.9
A20.
B. 5.2
C. 5.8
D. 6.0
E. 6.4
(03S–1–24) (Sample–P–121)
A device runs until either of two components fails, at which point the device stops running. If the joint
density function of the lifetimes of the two components, both measured in hours, is the following,
calculate the probability that the device fails during its first hour of operation.
f(x, y) = (x + y)/8
A. .125
B. .141
© 2010 ACTEX Publications, Inc.
C. .391
0 < x < 3 and 0 < y < 3
D. .625
E. .875
(Sample–P–77)
SOA Exam P and CAS Exam 1 – Peter J. Murdza
196  Multivariate Distributions
A16.
P(Y > 20, X > 20) = P(20 < Y < 50  X, 20 < X < 30)
30 50-x
P(Y > 20, X > 20) = 
 
 (6/125,000)(50  x  y) dy dx
20 20
30 50-x
 (50  x  y) dy dx
P(Y > 20, X > 20) = (6/125,000) 
 
20 20
Answer: B
1 2
A17.
1
|
2
P(X + Y ≥ 1) = P(Y ≥ 1  X) = 

 (2x + 2  y) /4 dy dx = 
 [(2x + 2)y  y2/2]/4 1-x dx
0 1-x
0
1
2
P(X + Y ≥ 1) = (1/4) 
 (4x + 4  2)  [(2x + 2)(1  x)  (1  x) /2] dx
0
1
|
1
 [(5/2)x2 + 3x + 1/2]/4 dx = [(5/6)x3 + (3/2)x2 + x/2]/4 0
P(X + Y ≥ 1) = 
0
= .70833
Answer: D
3 1
A18.
3
|
1
2
P(X < 1) = P(Y < 1) = 

 (x + y) /27 dx dy = 
 (1/27)(x /2 + xy) dy 0
0 0
0
3
|
3
2
P(X < 1) = 
 (1/27)(1/2 + y) dy = (1/27)(y/2 + y /2) 0
0
1 1
= (1/27)(3/2 + 9/2) = 2/9
1
|
1
2
P(X < 1) P(Y < 1) = 

 (x + y) /27 dx dy = 
 (1/27)(x /2 + xy) dy 0
0 0
1
0
|
1
2
P(X < 1) P(Y < 1) = 
 (1/27)(1/2 + y) dy = (1/27)(y/2 + y /2) 0
0
= (1/27)(1/2 + 1/2) = 1/27
P([X < 1] and [Y < 1]) = P(X < 1) + P(Y < 1)  P(X < 1) P(Y < 1)
P([X < 1] and [Y < 1]) = 2/9 + 2/9  1/27 = 11/27 = .407
Answer: B
10 1
A19.
10
E[X] = 
 
 (x/64)(10 
2
xy2)
2
|
10
E[X] = (1/64)(5x2  x3/9) 2
Answer: C
A20.
dy dx = 
 (x/64)(10y 
0
|
1
xy3/3) 0
10
dx = 
 (x/64)(10  x/3) dx
2
= (1/64)(500  1,000/9  20 + 8/9) = 52/9 = 5.78
See A18.
2
|
2
2
P(X < 1) = P(Y < 1) = 
 (1/8)(1/2 + y) dy = (1/8)(y/2 + y /2) 0
0
P(X < 1) P(Y < 1) = (1/8)(1/2 + 1/2) = 1/8
= (1/8)(2/2 + 4/2) = 3/8
P([X < 1] and [Y < 1]) = 3/8 + 3/8  1/8 = 5/8 = .625
Answer: D
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza
Multivariate Distributions  197
A21.
A device has two components. The device fails if either component fails. The joint density function of
the lifetimes of the components, measured in hours, is f(s, t), where 0 < s < 1 and 0 < t < 1. What is the
probability that the device fails during the first half hour of operation?
.5 .5
1 .5
A. 
 
 f(s, t) ds dt
1
B. 
 
 f(s, t) ds dt
0 0
.5 1
0 0
1 .5
.5 .5
.5 1
D. 
 
 f(s, t) ds dt + 
 
 f(s, t) ds dt
0
0
0
1
C. 
 
 f(s, t) ds dt
0
1 .5
E. 
 
 f(s, t) ds dt + 
 
 f(s, t) ds dt
0 .5
0
0
(Sample–P–79)
A22.
Let T1 be the time between a car accident and reporting a claim to the insurance company. The T 2 be the
time between the report of the claim and payment of the claim. The joint density function of T 1 and
T2, f(t1, t2), is constant over the region 0 < t1 < 6, 0 < t2 < 6, t1 + t2 < 10, and 0 otherwise. Determine
E[T1 + T2], the expected time between a car accident and payment of the claim.
A. 4.9
A23.
C. 5.7
D. 6.0
E. 6.7
(Sample–P–94)
Let T1 and T2 represent the lifetimes in hours of two linked components in an electronic device. The
joint density function for T1 and T2 is uniform over the region defined by 0 ≤ t1 ≤ t2 ≤ L where L is a
positive constant. Determine the expected value of the sum of the squares of T 1 and T2.
A. L2/3
A24.
B. 5.0
B. L2/2
C. 2L2/3
D. 3L2/4
E. L2
(Sample–P–97)
Let X1, X2, X3 be a random sample from a discrete distribution with probability function
p(0) = 1/3
p(1) = 2/3
Determine the moment generating function M(t), of Y = X1X2X3.
A. 19/27 + 8et/27
(Sample–P–98)
© 2010 ACTEX Publications, Inc.
B. 1 + 2et
C. (1/3 + 2et/3 )3
D. 1/27 + 8e3t/27
E. 1/3 + 2e3t/3
SOA Exam P and CAS Exam 1 – Peter J. Murdza
198  Multivariate Distributions
A21. The probability the device fails in the first half hour is the sum of the probability component S fails in the
first half hour and component T fails in the second half hour (case 1), the probability component T fails in
the first half hour and component S fails in the second half hour (case 2), and the probability both fail in
the first half hour (case 3).
A. F – This is the probability both components fail in the first half hour (case 3) but excludes the
probabilities of cases 1 and 2.
B. F – This is the probability that component S fails in the first half hour (cases 1 and 3). It excludes the
probability of case 2.
C. F – This is the probability that both components fail in the second half hour.
D. F – This is the sum of the probability that component T fails in the first half hour (cases 2 and 3) and
the probability that component S fails in the first half hour (cases 1 and 3) and thus double counts the
probability that both fail in the first half hour (case 3).
E. T – This is the sum of the probability that Component T fails in the first half hour and component S
fails in the second half hour (case 2) and the probability that component S fails in the first half hour
(cases 1 and 3).
Answer: E
4
A22.
6 10-t1
6
1 = 
 f(t1, t2) = 
 
 c dt2 dt1 + 
 
 c dt2 dt1
0
0
6
|
4
10-t1
1 = 24c + 
 ct2 dt1 0
4
0
6
[
4
1 = 24c + c[(10)(6  4)  (36  16)/2] = 34c
4
2
c = 1/34
6 10-t1
6
/ ]|6
= 
 c(10  t1) dt1 = c 10t1  (t1 2 4
4
|
6
E[t1] = 
 
 t1/34 dt2 dt1 + 
 
 t1/34 dt2 dt1 = 
 t1t2/34 dt1 0
0
4
0
4
0
0
6
|
4
E[t1] = 
 3t1/17 dt1 + 
 t1(10  t1)/34 dt1 = 3t1 34 0
0
|
4
/
2
6
4
+
10-t1
+ 
 t1t2/34 dt1 0
(15  t1)(t21 )/102|64
E[t1] = 24/17 + 162/51  88/51 = 146/51
E[t1 + t2] = 2[t1] = (2)(146)/51 = 5.72549
Answer: C
L t2
A23.
L
2
|
L
1 = 
 f(t1, t2) = 

 c dt1 dt2 = 
 ct2 dt2 = ct2 /2 0
0 0
L t2

 
(
0
[
E
0
2
t2 t1
2/L2
)(
2
t1
+
2
t2
) dt1 dt2
0
L
=

0
2
+ t2
]
(
2/L2
)(
3
t1/3
+
2
t 2t1
)
= cL2/2
|
t2
dt2 0
L
c = 2/L2
3
|
4 L
2
2
= 
 [8/(3L )]t2dt2 = [2/(3L )]t2 0
0
= 2L2/3
Answer: C
A24.
P(Y = 1) = (2/3)3 = 8/27
P(Y = 0) = 1  P(Y = 1) = 1  8/27 = 19/27
t
M(t) = P(Y = 0) + P(Y = 1)e = 19/27 + 8et/27
Answer: A
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza
Multivariate Distributions  199
A25.
An auto insurance policy will pay for damage to both the policyholder's car and the other driver's car in
the event that the policyholder is responsible for an accident. The size of the payment for damage to the
policyholder's car (X) has a marginal density function of 1 for 0 < x < 1. Given X = x, the size of the
payment for damage to the other driver's car (Y) has conditional density of 1 for x < y < x + 1. If the
policyholder is responsible for an accident, what is the probability that the payment for damage to the
other driver's car will be greater than .500?
A. 3/8
A26.
B. 1/2
C. 3/4
D. 7/8
E. 15/16
(Sample–P–119)
Let X and Y be identically distributed independent random variables such that the moment generating
function of (X + Y) is
M(t) = .09e-2t + .24e-t + .34 + .24et + .09e2t
∞ < t < ∞.
Calculate P[X ≤ 0].
A. .33
A27.
B. .34
C. .50
D. .67
E. .70
(Sample–P–137)
A machine consists of two components, whose lifetimes have the joint density function
 1/50 for x > 0, y > 0, and x + y < 10
f(x, y) =  0 otherwise

The machine operates until both components fail. Calculate the expected operational time of the
machine.
A. 1.7
A28.
B. 2.5
C. 3.3
D. 5.0
E. 6.7
(Sample–P–138)
A client spends X minutes in an insurance agent’s waiting room and Y minutes meeting the agent. The
joint function of X and Y can be modeled by
1
f(x, y) = 800 e -x/40 e -y/20, for x > 0, y > 0
f(x, y) = 0, otherwise.
Which of the following expressions represents the probability that a client spends less than 60 minutes at
the agent’s office?
40 20
A.
B.
C.
D.
E.
1

 e -x/40 e -y/20 dy dx
800 
0 0
40 20-x
1
-x/40 e -y/20 dy dx
 
e
800 
0 0
40 40-x
1
-x/40 e -y/20 dy dx
 
e
800 
0 0
60 60
1
-x/40 e -y/20 dy dx

e
800 
0 0
60 60-x
1
-x/40 e -y/20 dy dx
 
e
800 
0
0
(Sample–P–144)
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza
200  Multivariate Distributions
1 x+1
A25.
.5 x+1
1
P(Y > .5) = 
 dy dx + 
 
 dy dx = 
 y dx
 
0 .5
.5 x
.5
1
.5
|
.5
x+1
x
|
x+1
+ 
 y dx .5
0
|
.5
2
P(Y > .5) = 
 dx + 
 (x + .5) dx = .5 + (x /2 + .5x) 0
0
.5
= .875
Answer: D
A26.
Since X and Y are identical, the given m.g.f. is the square of the m.g.f. of one random variable. Thus we
get:
MX(t) = .3e-t + .4 + .3et
This is the moment generating function of a discrete random variable with probability distribution:
p(1) = .3
p(0) = .4
p(1) = .3
P[X ≤ 0] = .3 + .4 = .7
Answer: E
A27.
(0, 10)
(5, 5)
(0, 0)
(0, 10)
This diagram describes the area of f(x, y). The lower two triangles contain points where x > y and the
upper two triangles describe points where y > x. Since their distributions are uniform, and the area of
each set of two triangles equals 25, f(x) = f(y) = 1/25. The expected operational time when x > y is
calculated as follows:
5 x
10 10-x
0 0
5
5
10
2
2
E(X | X > Y) = 
 x/25 dy dx + 
 
 x/25 dy dx = 
 x /25 dx + 
 (10x – x )/25 dx

|
5
0
0
|
5
10
E(X | X > Y) = x3/75 0 + (5x2 – x3/3)/25 5
E(X | X > Y) = {1/25}{125/3 + [(5)(100) – (1,000)/3] – [(5)(25) – 125/3]} = (1/25)(125) = 5
An identical results occurs in the case where y > x.
Answer: D
A28.
The upper limit for the second integral is the greatest amount of time that can be spent in the meeting.
This equals 60 minutes less the amount of time spent in the waiting room, i.e., 60 – x.
Answer: E
© 2010 ACTEX Publications, Inc.
SOA Exam P and CAS Exam 1 – Peter J. Murdza