Errata for the ASM Study Manual for Exam P, Seventh Edition
By Dr. Krzysztof M. Ostaszewski, FSA, CERA, FSAS, CFA, MAAA
Web site: http://www.krzysio.net
E-mail: [email protected]
Effective July 5, 2013, only the latest edition of this manual will have its errata
updated. You can find the errata for all latest editions of my books at:
http://smartURL.it/errata
Posted June 26, 2013
In the solution of Problem 4 in Practice Examination 8, the fourth item in the list of
six events was mistyped as { X1 < X2 < X 3 } , while it should have been { X1 < X 3 < X2 } .
Posted April 5, 2013
The second to last formula in the solution of Problem 12 in Practice Examination 5
should be:
2 2−x
x=2
2
3
3
1 ⎞
⎛3
⎛ 3 1⎞ 1
x dy dx = ∫ x ( 2 − x ) dx = ⎜ x 2 − x 3 ⎟
= ( 3 − 2) − ⎜ − ⎟ = .
∫
⎝
⎠
⎝ 4 4⎠ 2
4
4
4
4
x=1
1 0
1
The calculation shown was correct but there was a typo in limits of integral
calculation.
Pr ( X > 1) = ∫
Posted April 2, 2013
The last formula in the solution of Problem 10 in Practice Examination 4 should be
2
+∞
2
+∞
⎛ 2.5 ⋅ 0.6 2.5 ⎞
⎛ 2.5 ⋅ 0.6 2.5 ⎞
2.5 ⋅ 0.6 2.5
5 ⋅ 0.6 2.5
E (Y ) = ∫ x ⋅ ⎜
dx
+
2
⋅
dx
=
dx
+
dx =
3.5
3.5
2.5
3.5
∫
∫
∫
⎟⎠
⎜⎝
⎟⎠
x
x
x
x
⎝
0.6
2
0.6
2
x=2
x→∞
2.5 ⋅ 0.6 2.5
2 ⋅ 0.6 2.5
2.5 ⋅ 0.6 2.5 2.5 ⋅ 0.6
0.6 2.5
=−
−
=
−
+
−
0
+
≈ 0.934273.
1.5x1.5 x=0.6
x 2.5 x=2
1.5 ⋅ 21.5
1.5
21.5
The calculation shown was correct, but a factor in the numerator of the last fraction
was mistyped as 2.5, when it should have been 2.
Posted March 6, 2013
In the solution of Problem 6 in Practice Examination 2, the expression
rectangle [ 0, 2 ] × [ 0, 3],
should be
rectangle [ 0,1] × [ 0, 2 ],
This expression appears twice in the solution.
Posted June 16, 2012
In the solution of Problem 23, Practice Examination 12, the second method of
solving the problem should be:
You can also ask yourself these two questions:
• What is the total number of possible outcomes? It is 2 n.
• What is the total number of favorable outcomes? It is equal to the number of possible
ways to get the third head on the n-th toss, which is the number of ways to put two heads
⎛ n −1 ⎞
among the first n – 1 tosses, i.e., ⎜
.
⎝ 2 ⎟⎠
This gives the probability sought as
⎛ n −1 ⎞
n+1
⎜⎝ 2 ⎟⎠
⎛ 1⎞
= ( n − 1) ⋅ ( n − 2 ) ⋅ ⎜ ⎟ .
⎝ 2⎠
2n
Posted May 9, 2012
The solution of Problem 17 in Practice Examination 9 had several typos and was not
clear. Here is an improved solution:
Let X be the random number of tosses till k successive heads occur. Note that if X = n,
then n ≥ k and the last k tosses were heads, the one just before those k was not heads (i.e.,
it was tails), and there were no k consecutive heads on the n − k −1 rolls before that.
Therefore,
k
k+1
1
⎛ 1⎞
⎛ 1⎞
Pr ( X = n ) = ⎜ ⎟ ⋅
⋅ Pr ( X > n − k − 1) = ⎜ ⎟ ⋅
Pr ( X ≥ n − k )
.
 ⎝ 2 ⎠



⎝ 2⎠
2


No k consecutive heads
=Pr( X>n−k−1) because X is discrete
Heads on the Tails on the toss
last k tosses just before that
on the first n−k−1 tosses
This is equivalent to the statement that for n ≥ k,
Pr ( X ≥ n − k ) = 2 k+1 ⋅ Pr ( X = n ).
We also have X ≥ k with probability 1 (and, notably, also X ≥ 0 with probability 1, so
that we can use the Darth Vader Rule) and
k
1
⎛ 1⎞
Pr ( X = k ) = ⎜ ⎟ = k ,
⎝ 2⎠
2
while
k
1
1
⎛ 1⎞
Pr ( X = k + 1) =
⋅
= k+1 ,
⎜⎝ ⎟⎠
2
2
2


Tails on first toss Heads on 2nd, 3rd, ..., and ( k+1)-st toss
Pr ( X = k + 2 ) =
1
⋅
Anything on the first toss
1
2

k
⋅
⎛ 1⎞
⎜⎝ ⎟⎠
2

Tails on second toss Heads on 3rd, 4th, .., and ( k+2 )-nd toss
and the same way, as long as 1 ≤ r ≤ k,
=
1
,
2 k+1
Pr ( X = k + r ) =
k
1
2

1⋅1⋅…⋅

Anything on first r−1 tosses
⎛ 1⎞
⎜⎝ ⎟⎠
2

⋅
=
1
.
2 k+1
Tails on r−th toss Heads on ( r+1)-st, .., and ( r+k )-th toss
The above gives us probabilities that X attains the values of 1, 2, 3, ..., k, k + 1, ..., 2k, and
beyond those values we can use the recursive formula we derived in the first step.
Therefore,
+∞
k−1
+∞
n=1
n=k
k−1
+∞
n=1
n=k
E ( X ) = ∑ Pr ( X ≥ n ) = ∑ Pr ( X ≥ n ) + ∑ Pr ( X ≥ n ) = ∑1 + ∑ Pr ( X ≥ n ) =
n=1
= ( k − 1) +
+∞
+∞
∑ Pr ( X ≥ n − k ) = ( k − 1) + ∑ 2
n=2 k
= ( k − 1) + 2
k+1
+∞
k+1
⋅ Pr ( X = n ) =
n=2 k
⋅ ∑ Pr ( X = n ) = ( k − 1) + 2
n=2 k
k+1
n=2 k−1
⎛ n=k−1
⎞
⋅ ⎜ 1− ∑ Pr ( X = n ) − ∑ Pr ( X = n )⎟ =

 n=k
⎜⎝
⎟⎠
n=0
=0
1
1
1 ⎞
1 k − 1⎞
⎛
⎛
= ( k − 1) + 2 k+1 ⋅ ⎜ 1− k − k+1 − ...− k+1 ⎟ = ( k − 1) + 2 k+1 ⋅ ⎜ 1− k − k+1 ⎟ = 2 k+1 − 2.
⎝ 2 2
⎝ 2
2 ⎠
2 ⎠
Answer C.
Posted April 23, 2012
The fourth formula in the solution of Problem 4 in Practice Examination 7 should
be:
2 ± 4 − 4 ⋅ 0.0005 ⋅1850 2 ± 0.5477226 ⎧⎪ 2547.72,
M=
≈
≈⎨
2 ⋅ 0.0005
0.001
⎪⎩ 1452.28.
intead of
2 ± 4 − 4 ⋅ 0.0005 ⋅1850 2 ± 0.5477226 ⎪⎧ 2547.72,
M=
≈
≈⎨
0.0001
0.001
⎩⎪ 1452.28.
Posted April 20, 2012
The first formula in the last line of the solution of Problem 9 in Practice
1
1
Examination 7 should be 2 ⋅ Pr ( X ≥ 800 ) ≤ instead of 2 ⋅ Pr ( X ≥ 800 ) ≥ .
4
4
Posted August 6, 2011
In the solution of Problem 6 in Practice Examination 9, in the first line, the words
“for and” should be “for any”.
Posted August 6, 2011
In the solution of Problem 1 in Practice Examination 9, the formula in the fifth line
should be
(
)
Pr Y(1) = 3 = Pr ( E − F ) = Pr ( E ) − Pr ( F ) .
instead of
(
)
Pr Y(1) = 3 = Pr ( F ) − Pr ( F ) .
Posted August 4, 2011
The solution of Problem 7 in Practice Examination 13 should be rephrased as
follows:
Solution.
The event of at least one color not being represented is the complement of the event of all
three colors being represented, and all colors being represented simply means that we
pick one red ball out of 3, one green ball out of 2, and one yellow ball out of 1. Thus
Pr ( At least one color not drawn ) =
⎛ 3 ⎞ ⎛
⎜⎝ 1 ⎟⎠ ⋅ ⎜⎝
= 1− Pr ( All colors drawn ) = 1−
⎛
⎜⎝
2 ⎞ ⎛ 1 ⎞
⋅
1 ⎟⎠ ⎜⎝ 1 ⎟⎠
= 0.70.
6 ⎞
3 ⎟⎠
You could also argue as follows. We are looking for the probability that all three balls are
of the same color, or of two colors only. We have
1
3! ⋅ 3!
6
1
Pr ( 3 balls of one color ) = Pr ( 3 red ) =
=
=
= ,
⎛ 6 ⎞
6!
4 ⋅ 5 ⋅ 6 20
⎜⎝ 3 ⎟⎠
and
Pr ( 3 balls of two colors only ) = Pr ( 2 red + 1 green ) + Pr ( 2 red + 1 yellow ) +
+ Pr ( 2 green + 1 red ) + Pr ( 2 green + 1 yellow ) =
⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
⎜⎝ 2 ⎟⎠ ⋅ ⎜⎝ 1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⋅ ⎜⎝ 1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⋅ ⎜⎝ 1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⋅ ⎜⎝ 1 ⎟⎠ 13
=
+
+
+
= ,
⎛ 6 ⎞
⎛ 6 ⎞
⎛ 6 ⎞
⎛ 6 ⎞
20
⎜⎝ 3 ⎟⎠
⎜⎝ 3 ⎟⎠
⎜⎝ 3 ⎟⎠
⎜⎝ 3 ⎟⎠
1 13 7
so that the total probability is
+
= .
20 20 10
Answer B.
Posted August 3, 2011
In Problem 7 in Practice Examination 14, the solution should start with the words
Because you studied this manual
instead of
Because you studied his manual
Posted July 26, 2011
In Problem 9 in Practice Examination 14, the third condition should be:
(ii) The future lifetimes follow a Weibull distribution with α = 1.5 and β = 2.0 for
smokers, and α = 2.0 and β = 2.0 for nonsmokers.
Also, the survival function of the Weibull distribution should be given as
sT ( t ) = e
⎛t⎞
−⎜ ⎟
⎝α⎠
β
.
Posted March 10, 2011
The second sentence of the solution of Problem 10 in Practice Examination 1 should
be:
As the policy has a deductible of 1 (thousand), the claim payment is
⎧ 0,
when there is no damage, with probability 0.94,
⎪
Y = ⎨ max ( 0, X − 1) , when 0 < X < 15, with probability 0.04,
⎪ 14,
in the case of total loss, with probability 0.02.
⎪⎩
Posted January 15, 2011
In Problem 16 in Practice Examination 6, the calculation of the second moment of X
should be:
1
3
3
E ( X 2 ) = ⋅ 0 2 + ⋅ (1+ 1) = .
4
4 Second
2
moment of T
instead of
E(X2 ) = E(X) =
1 2 3
3
⋅ 0 + ⋅ (1+ 1) = .

4
4 Second moment of T 2
Posted July 24, 2010
In the solution of Problem 21 in Practice Examination 6, the statement under the
first expression on the right-hand side of the third to last formula should be:
number of ways to
pick ordered samples
of size 2 from
population of size n
instead of
number of ways to
pick ordered samples
of size n − 2 from
population of size n
Posted July 17, 2010
Practice Examinations: An Introduction on page 109, the third sentence of the last
section should be:
Practice examinations 6-20 are meant to be more challenging.
Posted July 3, 2010
In the solution of Problem 1, Practice Examination 5, at the end of the first part of
the fourth sentence of the solution, 5/6 is a typo, it should be 5/36, as used in the
formula for Pr (Y = 6).
99% ⋅ 25% Knows Bayes
Posted June 9, 2010
1% ⋅ 25%
In the alternative solution of Problem
28, Practice Examination
11,Bayes
the probability
Actuarial
Knows no
tree diagram should be:
25%
98.5%⋅ 30%
Knows Bayes
Random student
30%
Mathematics
1.5% ⋅ 30%
45%
Statistics
Knows no Bayes
98% ⋅ 45%
Knows Bayes
2% ⋅ 45%
Knows no Bayes
Some numbers in the diagram were mistyped.
Posted January 5, 2010
In the description of the gamma distribution in Section 2, the condition for the
range of its MGF should be t < β , not 0 < t < β .
Posted November 20, 2009
Answers A and B in Problem 10, Practice Examination 9, have the symbol τ
mistyped as r in the numerator, and they should be:
αθ α τ yτ −1
τθ yτ −1
f
y
=
A. fY ( y ) =
B.
Y ( )
α +1
( y + θ )τ +1
yτ + θ
(
)
Posted July 1, 2009
In Section 2, the general definition of a percentile should be
the 100-p-th percentile of the distribution of X is the number x p which satisfies both of
(
)
(
)
the following inequalities: Pr X ≤ x p ≥ p and Pr X ≥ x p ≥ 1− p.
Posted June 18, 2009
The last formula in the solution of Problem 19 of Practice Examination 5 should be
1
1
1
1
instead of Pr ( X ≥ 10 ) < 2 = .
Pr ( X ≥ 10 ) ≤ 2 =
4
16
4
16
Posted June 17, 2009
In the statement of Problem 9 in Practice Examination 7, Pr ( X > 800 ) should be
Pr ( X ≥ 800 ) , and in the solution, all inequalities should be changed accordingly.
Posted April 23, 2009
In Problem 20, Practice Examination 14, the first sentence of the solution should be:
Let us define the following random variables:
X : time until death of Dwizeel by causes other than their private plane crash,
Y : time until death of Satellite Component by causes other than their private plane crash,
Z : time until death of Dwizeel and Satellite Component as a result of their private plane
crash.
The solution has the words “of” mistyped as “od.”
Posted April 23, 2009
In Problem 9, Practice Examination 14, the Greek letter τ in the statement of the
problem should be replaced by the Greek letter α .
Posted April 6, 2009
In the text of Problem 4 in Practice Examination 11, the sentence:
We put that chip aside and pick a second chip from the same contained.
should be:
We put that chip aside and pick a second chip from the same container.
Posted April 6, 2009
In the text of Problem 7 in Practice Examination 3, the word “whwther” should be
“whether”.
Posted March 19, 2009
The solution of Problem 3 in Practice Examination 8 should be:
Let E be the event that a new insured is accident-free during the second policy year, and F
be the event that a new insured is accident-free during the first policy year, and let G be
the event that this new insured was accident-free the last year, before the policy was
issued. Note that for any year only the previous year affects a given year, but not the year
before that. Therefore
(
)
Pr ( E ) = Pr ( E ∩ F ∩ G ) ∪ ( E ∩ F ∩ G C ) ∪ ( E ∩ F C ∩ G ) ∪ ( E ∩ F ∩ G C ) =
= Pr ( E ∩ F ∩ G ) + Pr ( E ∩ F ∩ G C ) + Pr ( E ∩ F C ∩ G ) + Pr ( E ∩ F C ∩ G C ) =
= Pr ( G ) ⋅ Pr ( F G ) ⋅ Pr ( E F ∩ G ) + Pr ( G C ) ⋅ Pr ( F G C ) ⋅ Pr ( E F ∩ G C ) +
(
)
(
)
+ Pr ( G ) ⋅ Pr F C G ⋅ Pr ( E F C ∩ G ) + Pr ( G C ) ⋅ Pr F C G C ⋅ Pr ( E F C ∩ G C ) =
= 0.7 ⋅ 0.8 ⋅ 0.8 + 0.3⋅ 0.6 ⋅ 0.8 + 0.7 ⋅ (1− 0.8 ) ⋅ 0.6 + 0.3⋅ (1− 0.6 ) ⋅ 0.6 = 0.748.
Answer E.
Posted March 5, 2009
The first sentence of Problem 16 in Practice Examination 6 should end with t < 1,
instead of t > 1.
Posted March 2, 2009
The properties of the cumulant moment-generating function should be:
The cumulant generating function has the following properties:
( )
d
ln E etX
dt
ψ X ( 0 ) = 0,
(
tX
d2
d E Xe
ψ X (t ) =
dt 2
dt E etX
t =0
3
( )
)
=
(
) ( ) (
( E ( e ))
=
tX
t =0
(
( )
) E ( Xe )
t =0
E X 2 etX E etX − E XetX
(
E XetX
E e
)
= E ( X ),
tX
tX
2
t =0
= Var ( X ) ,
t =0
)
d
3
ψ X ( t ) = E ( X − E ( X )) ,
3
dt
t =0
but for k > 3,
dk
k
ψ X ( t ) = ψ (Xk ) ( 0 ) < E ( X − E ( X )) .
k
dt
t =0
Also, if X and Y are independent (we will discuss this concept later),
ψ aX +b ( t ) = ψ X ( at ) + bt, and ψ X +Y ( t ) = ψ X ( t ) + ψ Y ( t ) .
(
)
Posted January 13, 2009
In Practice Examination 8, Problem 24, the calculation of the expected value had a
typo, an extra, unnecessary p in the second line, and it instead should be:
+∞
+∞
E ( X ) = ∑ k ⋅ Pr ( X = k ) = 1⋅ Pr ( X = 1) + ∑ k ⋅ Pr ( X = k ) =
k=1
k=2
+∞
+∞
+∞
k=1
k=1
k=1
= p + ∑ ( k + 1) ⋅ Pr ( X = k + 1) = p + ∑ k ⋅ Pr ( X = k + 1) + ∑ Pr ( X = k + 1) =
⎛
⎞
⎜ ⎛ +∞
⎟
⎞
= p + ∑ k ⋅ (1− p ) ⋅ Pr ( X = k ) + ⎜ ⎜ ∑ Pr ( X = k + 1)⎟ − Pr ( X = 0 + 1)⎟ =


 ⎝ k=0
⎠
k=1
⎜ 
⎟


=Pr( X=k+1)
⎝ this sum is equal to 1
⎠
+∞
+∞
= p + (1− p ) ⋅ ∑ k ⋅ Pr ( X = k ) + (1− Pr ( X = 1)) =
k=1
= p + (1− p ) ⋅ E ( X ) + (1− p ) = 1+ (1− p ) ⋅ E ( X ) .
Posted November 8, 2008
Problem 11 in Practice Examination 13 had several typos, and it should be:
Mr. Warrick Beige is gambling at the newly opened You Was Robbed casino in Nevada.
In the game he is playing, first he has to choose one of two coins: coin A or coin B. Both
coins are unfair. Coin A has the probability of heads of 0.60, and coin B has the
probability of heads of 0.40. Mr. Beige pays $20 to enter the game. He chooses a coin
randomly, but the chances of picking the coins are not equal. He has 40% chance of
picking coin A and 60% chance of picking coin B. Then he tosses the coin chosen. If the
result is heads, he is paid $250. If the result is tails, he pays $200. For an additional
payment of x dollars, Mr. Beige can test a coin chosen: he can toss it once and based on
the result, either walk away and get $20 paid initially back (he will do this if the first toss
results in tails), or toss the same coin again (he will do this if the first toss results in
heads). Assuming that Mr. Beige values all gambles based on the expected value of the
payoff (i.e., he is risk-neutral), calculate the value of x such that Mr. Beige is indifferent
between testing a coin and not testing it.
A. $6.40
B. $4.00
C. $2.50
Solution.
We begin by labelling the events:
A : Coin A is picked,
B : Coin B is picked,
H 1 : First toss results in heads,
T1 : First toss results in tails,
H 2 : Second toss results in heads,
T2 : Second toss results in tails.
D. $0.00
E. –$2.00
We know that Pr ( A ) = 0.40, Pr ( B ) = 0.60, Pr ( H 1 A ) = 0.60, and Pr ( H 1 B ) = 0.40.
Therefore, the probability of getting heads in the first toss is
Pr ( H 1 ) = Pr ( H 1 A ) ⋅ Pr ( A ) + Pr ( H 1 B ) ⋅ Pr ( B ) =
= 0.40 ⋅ 0.60 + 0.60 ⋅ 0.40 = 0.24 + 0.24 = 0.48.
Therefore, Mr. Beige’s expected gain on this game without testing first is
−$20 + 0.48 ⋅ $250 + 0.52 ⋅ ( −$200 ) = −$4.00.
Note that the two coin tosses are independent, and hence
Pr ( H 1 ∩ H 2 ) Pr ( H 1 ∩ H 2 A ) ⋅ Pr ( A ) + Pr ( H 1 ∩ H 2 B ) ⋅ Pr ( B )
Pr ( H 2 H 1 ) =
=
=
Pr ( H 1 )
Pr ( H 1 )
=
Pr ( H 1 A ) ⋅ Pr ( H 2 A ) ⋅ Pr ( A ) + Pr ( H 1 B ) ⋅ Pr ( H 2 B ) ⋅ Pr ( B )
Pr ( H 1 )
=
0.6 ⋅ 0.6 ⋅ 0.4 + 0.4 ⋅ 0.4 ⋅ 0.6
= 0.5.
0.48
For now, let us disregard the cost x of testing the coin, and calculate the expected net
payoff of the game without that additional fee. If Mr. Beige’s coin test results in tails, his
payoff for the game will be zero. Thus the expected net payoff for the case when he tests
the coin (disregarding the fee of x dollars) is
−$20 + $20 ⋅ Pr H 1C + $250 ⋅ Pr ( H 1 ) ⋅ Pr ( H 2 H 1 ) − $200 ⋅ Pr ( H 1 ) ⋅ Pr (T2 H 1 ) =
=
( )
= −$20 + $20 ⋅ 0.52 + $250 ⋅ 0.48 ⋅ 0.50 − $200 ⋅ 0.48 ⋅ 0.50 = $2.40.
This means that Mr. Beige’s expected payoff changes from –$4 to $2.40 as a result of
testing the coin. In order for him to be indifferent between the two choices, the additional
fee should be set at the difference of these two amounts, equal to his gain in the expected
payoff of the game, i.e., $6.40.
Answer A.
Posted September 1, 2008
In the solution of Problem 16 in Practice Examination 2, the left-hand side of the
second formula should be E X 2 , not E ( X ) .
( )
Posted August 5, 2008
In the solution of Problem 10 in Practice Examination 8, the phrase
Then the total number of claims among 10 less than $1,050 follows the binomial
distribution with probability of success p = 0.8413
should be
Then the total number of claims among 10 less than $1,050 follows the binomial
distribution with probability of success p = 0.5793
Posted July 24, 2008
In the solution of Problem 6 of Practice Examination 7, the phrase
Therefore, recalling that for a discrete random variable, whose only possible values are
possible integers,
should be
Therefore, recalling that for a discrete random variable, whose only possible values are
positive integers,
Posted July 23, 2008
In the solution of Problem 5 in Practice Examination 3 the expression
Var ( X2 ) = 40,000 should be Var ( X2 ) = 250,000.
Posted July 11, 2008
The formula at end of the third sentence of the solution of Problem 8 in Practice
0.26
1.26
Examination 11 should be
≈ 2.8889% instead of
≈ 2.89%.
9
9
Posted June 25, 2008
In Practice Examination 14, Problem No. 13, answer choice C should be 20762, and
the final steps of its calculation should be:
In the table of the standard normal distribution we find Φ ( 0.67 ) = 0.7486 and
Φ ( 0.68 ) = 0.7517. Using linear interpolation, we obtain the 75-th percentile of the
standard normal distribution to be
0.75 − 0.7486
z0.75 ≈ 0.67 +
⋅ ( 0.68 − 0.67 ) ≈ 0.6745.
0.7517 − 0.7486
The maximum amount paid is the 75-th percentile of the distribution of (Y X = 1200 ) ,
and that is calculated as
20280 + z0.75 ⋅ 510000 ≈ 20280 + 0.6745 ⋅ 510000 ≈ 20761.6893.
Answer C.
Posted February 7, 2008
The discussion of the lack of memory property of the geometric distribution should
have the formula
Pr ( X = n + k X ≥ n ) = Pr ( X = k )
corrected to:
Pr ( X = n + k X > n ) = Pr ( X = k ) .
Posted November 26, 2007
In the solution of Exercise 4.5, the words:
But the policy has a deductible of 2, so for N = 0 and for N = 1, actual payment is 0.
should be:
But the policy has a deductible of 2, so for N = 1 and for N = 2, actual payment is 0.
Posted November 14, 2007
The answer choice in Problem No. 20 of Practice Examination 13 should be D, not
C.
Posted November 13, 2007
In the solution of Problem No. 26 of Practice Examination 5, the sentence:
Of the five numbers, 1 can never be the median.
should be
Of the five numbers, neither 1 nor 5 can ever be the median.
Posted October 5, 2007
In the solution of Problem No. 1 in Practice Examination 3, the random variable Y
should refer to class B.
Posted September 14, 2007
In the example on page 68, the phrase:
We conclude that fZ ( z ) = 0 if z ≤ 0 or z ≥ 1,
should be
We conclude that fZ ( z ) = 0 if z ≤ 0 or z ≥ 2,
Also, in the final part of the example, showing where fZ ( z ) = 0, the condition should
be z ≤ 0 or z ≥ 2.
Posted September 13, 2007
The beginning of the third sentence in the solution of Problem No. 21 in Practice
Examination 2 should be
The mean error of the 48 rounded ages,
instead of
The mean of the 48 rounded ages.
Posted August 21, 2007
The last sentence of Problem No. 29, Practice Examination 13, should be:
1
⎛
⎞
Find the probability Pr ⎜ X + > 3⎟ .
⎝
⎠
X
It is stated so, correctly, in the text of the problem in the solutions to this Practice
1
Examination, but in the examination part, appears instead of 3.
2
Posted August 20, 2007
In the solution of Problem No. 13, Practice Examination 12, the sentence
2
.
4
But the median is given as
should be
But the mode is given as
2
.
4
Posted August 19, 2007
In the solution of Problem No. 4, Practice Examination 9, the formula
(
)
(
)
= 1− Pr ({Y ≤ 1} ∩ {Y
Pr Y( 5 ) > 1 = 1− Pr Y( 5 ) ≤ 1 =
(1)
(2)
} {
} {
} {
})
} {
})
≤ 1 ∩ Y( 3) ≤ 1 ∩ Y( 4 ) ≤ 1 ∩ Y( 5 ) ≤ 1 =
= 1− Pr ({ X1 ≤ 1} ∩ { X2 ≤ 1} ∩ { X 3 ≤ 1} ∩ { X 4 ≤ 1} ∩ {Y5 ≤ 1}) =
= 1− ( FX (1)) = 1− (1− e−1 ) ≈ 0.89907481.
5
5
should be
(
)
(
)
= 1− Pr ({Y ≤ 1} ∩ {Y
Pr Y( 5 ) > 1 = 1− Pr Y( 5 ) ≤ 1 =
(1)
(2)
} {
} {
≤ 1 ∩ Y( 3) ≤ 1 ∩ Y( 4 ) ≤ 1 ∩ Y( 5 ) ≤ 1 =
= 1− Pr ({ X1 ≤ 1} ∩ { X2 ≤ 1} ∩ { X 3 ≤ 1} ∩ { X 4 ≤ 1} ∩ { X5 ≤ 1}) =
= 1− ( FX (1)) = 1− (1− e−1 ) ≈ 0.89907481.
5
5
Posted August 15, 2007
In the solution of Problem No. 21, Practice Examination 8, the formula
f ( x, y )
f ( x, y )
10xy 2
10xy 2
x
fX ( x Y = y) =
= ∞
= y
=
= 2
y
fY ( y )
5y
2
2
∫ f ( x, y )dx ∫ 10xy dx 10y ∫ x dx
−∞
should be
fX ( x Y = y) =
f ( x, y )
=
fY ( y )
∞
f ( x, y )
=
0
10xy 2
y
∫ f ( x, y )dx ∫ 10xy
−∞
Posted August 15, 2007
On page 27, the expression
0
0
2
dx
=
10xy 2
y
10y 2 ∫ x dx
0
=
x
2x
= 2
2
0.5y
y
⎧ Pr ( min ( X,a ) > y ) , y < a, ⎫
⎪
⎪
sY ( y ) = Pr (Y > y ) = Pr ( min ( X,a ) > y ) = ⎨
⎬=
0,
y
≥
a,
⎪⎩
⎪⎭
⎧ Pr ({ X > y} ∩ { X ≤ a}) + Pr ({a > y} ∩ { X > a}) , y < a, ⎫
⎪
⎪
=⎨
⎬=
y ≥ a, ⎪
⎪⎩ 0,
⎭
⎧⎪ Pr ( y < X ≤ a ) + Pr ( X > a ) , y < 3, ⎫⎪ ⎧⎪ s ( y ) , y < a,
X
=⎨
⎬= ⎨
y ≥ 3, ⎪ ⎪ 0,
y ≥ a.
⎪⎩ 0,
⎭ ⎩
should be
⎧ Pr ( min ( X,a ) > y ) , y < a, ⎫
⎪
⎪
sY ( y ) = Pr (Y > y ) = Pr ( min ( X,a ) > y ) = ⎨
⎬=
y ≥ a, ⎪
⎪⎩ 0,
⎭
⎧ Pr ({ X > y} ∩ { X ≤ a}) + Pr ({a > y} ∩ { X > a}) , y < a, ⎫
⎪
⎪
=⎨
⎬=
y ≥ a, ⎪
⎪⎩ 0,
⎭
⎧⎪ Pr ( y < X ≤ a ) + Pr ( X > a ) , y < a, ⎫⎪ ⎧⎪ s ( y ) , y < a,
X
=⎨
⎬= ⎨
y ≥ a, ⎪ ⎪ 0,
y ≥ a.
⎪⎩ 0,
⎭ ⎩
Posted August 9, 2007
In Practice Examination 14, Problem No. 13, the expression
σ Y = Var (Y ) = 1000
should be replaced by
σ Y = Var (Y ) = 1000.
Posted August 9, 2007
In Practice Examination 14, Problem No. 3, the sentence
Let Y be the smallest integer greater than or equal to X.
should be replaced by
Let Y be the greatest integer less than or equal to X.
Also, the last formula in the solution should be:
+∞
E (Y ) = ∑ ( n + 1) e− n = 2e−1 + 3e−2 + 4e−3 + ... =
n=1
= 2 ( e−1 + e−2 + e−3 + e−4 +…) + ( e−2 + e−3 + e−4 +…) + ( e−3 + e−4 +…) +… =
2e−1
e−2
e−3
e−1
e−1
=
+
+
+… =
+
⋅ 1+ e−1 + e−2 + e−3 +…) =
−1
−1
−1
−1
−1 (
1− e
1− e
1− e
1− e
1− e
−1
−1
e
e
1
1
1
e
2e − 1
=
+
⋅
=
+
⋅
=
≈ 1.5026503.
−1
−1
−1
1− e
1− e 1− e
e − 1 e − 1 e − 1 ( e − 1)2
Posted July 25, 2007
The last sentence of Problem No. 3 in Practice Examination 12 should be:
Approximate the smallest possible number w such that the probability that the total
weight of the entire luggage on the plane is less than w is 0.95, assuming that the airplane
is full.
and the answer choice A should be: 48339.
Posted July 19, 2007
The answer choices of Problem No. 19 in Practice Examination 14 should be:
A. 769
B. 823
C. 846
D. 859
E. 893
The last part the solution should be:
Therefore,
E (Y ) = 4000 ⋅ 0.8 5 + 3000 ⋅ 0.8 4 + 2000 ⋅ 0.4 ⋅ 0.8 3 + 1000 ⋅ 0.08 ⋅ 0.8 2 = 3000.32,
E (Y 2 ) = 4000 2 ⋅ 0.8 5 + 3000 2 ⋅ 0.8 4 + 2000 2 ⋅ 0.4 ⋅ 0.8 3 + 1000 2 ⋅ 0.08 ⋅ 0.8 2
= 9799680,
Var (Y ) = E (Y 2 ) − ( E (Y )) = 9799680 − 3000.32 2 ≈ 797759.8976,
2
and finally
σ Y = Var (Y ) = 797759.8976 ≈ 893.17406.
Answer E.
Posted July 12, 2007
Problem No. 5, Practice Examination 14, should be:
November 2004 SOA Course 3 Examination, Problem No. 24
The future lifetime of a newborn follows a two-parameter Pareto distribution with θ = 50
and α = 3. The cumulative distribution function of a two-parameter Pareto random
variable T is given by the formula
α
⎛ θ ⎞
FT ( t ) = 1− ⎜
⎝ t + θ ⎟⎠
for t ≥ 0, FT ( t ) = 0 for t < 0. Calculate E (T − 20 T ≥ 20 ) .
A. 5
B. 15
C. 25
D. 35
E. 45
Solution.
Let us write
U = (T − 20 T ≥ 20 ) .
We have
sU ( u ) = Pr (T − 20 > u T ≥ 20 ) = Pr (T > u + 20 T ≥ 20 ) =
=
Pr ({T > u + 20} ∩ {T ≥ 20})
Pr ({T ≥ 20})
=
Pr (T > u + 20 ) sT ( u + 20 )
=
.
Pr (T ≥ 20 )
sT ( 20 )
Therefore
E (T − 20 T ≥ 20 ) = E (U ) =
+∞
∫
sU ( u ) du =
0
=
+∞
Pr (T > u )
∫ Pr (T ≥ 20 ) du
20
u = t + 20 u = 20 ⇒ t = 0
=
du = dt
u→∞⇒t →∞

+∞
∫
0
Pr (T > t + 20 )
dt .
Pr (T ≥ 20 )
Integration by substitution
α
α
⎛ θ ⎞
⎛ θ ⎞
Based on the CDF FT ( t ) = 1− ⎜
we get sT ( t ) = 1− FT ( t ) = ⎜
for t > 0.
⎟
⎝ t +θ ⎠
⎝ t + θ ⎟⎠
Therefore
α
θ
⎛
⎞
+∞
+∞ ⎜
Pr (T > t + 20 )
⎝ 20 + t + θ ⎟⎠
E (T − 20 T ≥ 20 ) = E (U ) = ∫
dt = ∫
dt =
α
Pr
T
≥
20
(
)
θ
⎛
⎞
0
0
⎜⎝
⎟
20 + θ ⎠
+∞
α
⎛ 20 + θ ⎞
= ∫⎜
⎟⎠ dt =
⎝
20
+
t
+
θ
0
+∞
3
t→+∞
70 3
−3+1
⎛ 70 ⎞
dt
=
⋅ ( 70 + t )
= 35.
∫0 ⎜⎝ 70 + t ⎟⎠
−3 + 1
t=0
Answer D.
Posted May 16, 2007
In the solution of Problem No. 21, Practice Examination 12, the expression:
Therefore, using Poohsticks,
fY(3) ( y ) = f X1 ( y ) + f X2 ( y ) + f X3 ( y ) − fY(1) ( y ) − fY(3) ( y ) =
= e− y + e− y + e− y − 3e−3y − 3e− y + 6e−2 y − 3e−3y = 6e−2 y − 6e−3y .
should be:
Therefore, using Poohsticks,
fY(2) ( y ) = f X1 ( y ) + f X2 ( y ) + f X3 ( y ) − fY(1) ( y ) − fY(3) ( y ) =
= e− y + e− y + e− y − 3e−3y − 3e− y + 6e−2 y − 3e−3y = 6e−2 y − 6e−3y .
There was a typo in the subscript on the left-hand side of the formula.
Posted May 15, 2007
While the solution of Problem No. 16, Practice Examination No. 10, is correct, I
decided to post a better explanation of it, and here is the whole problem:
May 1982 Course 110 Examination, Problem No. 18
Two balls are dropped in such a way that each ball is equally likely to fall into any one of
four holes. Both balls may fall into the same hole. Let X denote the number of
unoccupied holes at the end of the experiment. What is the moment generating function
of X?
A.
7 1
9
21
B. t + t 2
− t if t = 2 or 3, otherwise
4 2
4
8
t
⎞
1
1 2t
1 ⎛ 3t4
2t
3t
3t
4
C. ( 3e + e )
D. ( e + 3e )
E. ⎜ e + 3e ⎟
4
4
4⎝
⎠
Solution.
Each ball had four possible holes, so that the total number of ways for the balls to fall is
4 ⋅ 4 = 16. There can be only two or three holes unoccupied. If three holes are
unoccupied, there are four ways to put two balls in one occupied hole. Hence,
4 1
3
Pr ( X = 3) =
= ,
Pr ( X = 2 ) = 1− Pr ( X = 3) = .
16 4
4
Therefore,
3
1
1
M X ( t ) = e2t + e 3t = ( 3e2t + e 3t ) .
4
4
4
Answer C.
Posted May 15, 2007
In the solution of Problem No. 1, Practice Examination No. 7, both events were
mistakenly labeled E1 . In order to correct that, the expression
Define the following events:
E1 = { X1 = 0} ∩ { X2 = 1} ,
E1 = { X1 = 1} ∩ { X2 = 2} .
should be replaced by:
Define the following events:
E1 = { X1 = 0} ∩ { X2 = 1} ,
E2 = { X1 = 1} ∩ { X2 = 2} .
Posted May 11, 2007
The solution of Problem No. 22 in Practice Examination No. 11 is correct, but the
correct answer choice is B (not C).
Posted May 11, 2007
The first displayed formula of the solution of Problem No. 1 in Practice
Examination No. 11 should be:
FT ( t ) = Pr (T ≤ t ) = Pr ( max (T1 ,T2 ) ≤ t ) = Pr ({T1 ≤ t } ∩ {T2 ≤ t }) =
= Pr (T1 ≤ t ) ⋅ Pr (T2 ≤ t ) = (1− e−4t ) ⋅ (1− e−3t ) = 1− e−3t − e−4t + e−7t .
instead of
FT ( t ) = Pr (T ≤ t ) = Pr ( max (T1 ,T2 ) ≤ t ) = Pr ({T1 ≤ t } ∩ {T2 ≤ t }) =
= Pr (T1 ≤ t ) ⋅ Pr (T1 ≤ t ) = (1− e−4t ) ⋅ (1− e−3t ) = 1− e−3t − e−4t + e−7t .
Posted May 9, 2007
The first sentence of Problem No. 15 in Practice Examination No. 10 should be:
A fair coin is tossed until a head appears.
instead of:
A fair coin is tossed until a head appears until a head appears.
Posted May 9, 2007
Problem No. 10 in Practice Examination No. 9 should be as below (some of the
letters τ were mistyped as r):
Casualty Actuarial Society November 2005 Course 3 Examination, Problem No. 19
Claim size X follows a two-parameter Pareto distribution with parameters α and θ , for
which the density and the cumulative distribution function are given below:
αθ α
fX ( x ) =
,
( x + θ )α +1
θα
FX ( x ) = 1−
.
( x + θ )α
1
A transformed distribution Y is created by Y = X τ . Which of the following is the
probability density function of Y ?
A. fY ( y ) =
τθ yτ −1
( y + θ )τ +1
B. fY ( y ) =
αθ α τ yτ −1
C. fY ( y ) =
(y
τ
+θ )
α +1
θα θ
( y + θ )θ +1
D. fY ( y ) =
E. fY ( y ) =
⎛ y⎞
ατ ⎜ ⎟
⎝θ ⎠
τ
⎛ ⎛ y⎞τ ⎞
y ⎜ 1+ ⎜ ⎟ ⎟
⎝ ⎝θ ⎠ ⎠
(y
α +1
αθ α
τ
+θ )
α +1
Solution.
We have X = Y τ . Therefore,
dx
= τ yτ −1 .
dy
We conclude that
dx
αθ α
αθ α τ yτ −1
τ −1
fY ( y ) = f X ( x ( y )) ⋅
=
⋅
τ
y
=
α +1
α +1 .
dy
yτ + θ
yτ + θ
Answer B.
(
)
(
)
Posted May 9, 2007
The first sentence of the solution of Problem No. 2 in Practice Examination No. 9
should be:
The density of the exponential distribution with mean 1 is e− x for x > 0, while the
1
density of the exponential distribution with mean
is 2e−2 x for x > 0.
2
instead of
The density of the exponential distribution with mean 1 is e− x for x > 0, while the
1
density of the exponential distribution with mean
is 2e− x for x > 0.
2
Posted May 9, 2007
In the solution of Problem No. 15, Practice Examination No. 3, the derivative is
missing a minus sign, and it should be
2
dX
1
−
= − (8 − Y ) 3 .
dY
3
The rest of the solution is unaffected.
Posted May 9, 2007
Answer C in Problem No. 7 in Practice Examination No. 6 should be 0.2636, not
0.2626. It is the correct answer.
Posted May 9, 2007
In Problem No. 10, Practice Examination No. 7, in the first part of the Practice
Examination (not in the Solutions part), the expression E ( X ) = µY should be
E ( X ) = µX .
Posted May 8, 2007
Problem No. 24 in Practice Examination No. 5 should be (it had a typo in the list of
possible values of x)
May 1992 Course 110 Examination, Problem No. 35
Ten percent of all new businesses fail within the first year. The records of new
businesses are examined until a business that failed within the first year is found. Let X
be the total number of businesses examined prior to finding a business that failed within
the first year. What is the probability function for X?
A. 0.1⋅ 0.9 x , for x = 0,1,2, 3,...
B. 0.9 ⋅ 0.1x , for x = 0,1,2, 3,...
C. 0.1x ⋅ 0.9 x , for x = 1,2, 3,...
D. 0.9x ⋅ 0.1x , for x = 1,2, 3,...
E. 0.1⋅ ( x − 1) ⋅ 0.9 x , for x = 2, 3, 4,...
Solution.
Let a failure of a business be a success in a Bernoulli Trial, and a success of a business be
a failure in the same Bernoulli Trial. Then X has the geometric distribution with p = 0.1,
and therefore f X ( x ) = 0.1⋅ 0.9 x for x = 0, 1, 2, 3, … .
Answer A.