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ASM Study Manual for Exam P, Fifth Edition
By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net)
Errata
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
Pr ( X > 1) = ∫ ∫ x dy dx = ∫ x ( 2 − x ) dx = ⎜ x 2 − x 3 ⎟
= ( 3 − 2) − ⎜ − ⎟ = .
⎝4
⎝ 4 4⎠ 2
4
4
4 ⎠ x=1
1 0
1
The calculation shown was correct but there was a typo in limits of integral
calculation.
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
+
∫2 ⎜⎝ x 3.5 ⎟⎠ 0.6∫ x 2.5
∫2 x 3.5 dx =
x 3.5 ⎟⎠
⎝
0.6
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 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 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 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).
Posted June 9, 2010
In the alternative solution of Problem 28, Practice Examination 11, the probability
tree diagram should be:
99%⋅ 25%
Actuarial
25%
Random student
30%
1%⋅ 25%
98.5%⋅ 30%
Mathematics
1.5%⋅ 30%
45%
Statistics
Knows Bayes
Knows no Bayes
Knows Bayes
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 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 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:ψ X ( 0 ) = 0,
E ( XetX )
d
tX
ln E ( e ) =
dt
E ( etX )
t=0
tX
d2
d E ( Xe )
ψ
t
=
(
)
X
dt 2
dt E ( etX )
t=0
3
(
=
= E ( X ),
t=0
E ( X 2 etX ) E ( etX ) − E ( XetX ) E ( XetX )
( E ( e ))
tX
t=0
)
3
d
ψ X ( t ) = E ( X − E ( X )) ,
3
dt
t=0
but for k > 3,
= Var ( X ) ,
2
t=0
(
)
k
dk
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, andψ 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 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 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 13, 2007
In the solution of Problem No. 26 of Practice Examination No. 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 No. 3, the random variable
Y should refer to class B.
Posted September 13, 2007
The beginning of the third sentence in the solution of Problem No. 21 in Practice
Examination No. 2 should be
The mean error of the 48 rounded ages,
instead of
The mean of the 48 rounded ages.
Posted August 19, 2007
In the solution of Problem No. 4, Practice Examination No. 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
f ( x, y )
fX ( x Y = y) =
=
fY ( y )
∞
0
f ( x, y )
=
0
10xy 2
y
∫ f ( x, y )dx ∫ 10xy
−∞
0
2
dx
=
10xy 2
y
10y 2 ∫ x dx
=
x
2x
= 2
2
0.5y
y
0
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
1 ⎛ 3t
C. ( 3e2t + e3t )
D. ( e2t + 3e3t )
E. ⎜ e 4 + 3e 4 ⎟
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
Pr ( X = 3) =
= ,
16 4
3
Pr ( X = 2 ) = 1− Pr ( X = 3) = .
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 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
FX ( x ) = 1−
θα
.
( x + θ )α
1
τ
A transformed distribution Y is created by Y = X . 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 ) =
D. fY ( y ) =
E. fY ( y ) =
(y
τ
+θ )
α +1
θα θ
( y + θ )θ +1
⎛ 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,
fY ( y ) = f X ( x ( y )) ⋅
dx
= τ yτ −1 . We conclude that
dy
dx
αθ α
αθ α τ yτ −1
τ −1
=
⋅
τ
y
=
α +1 .
τ
dy ( yτ + θ )α +1
y
+
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, and it should
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.
Posted May 5, 2007
Problem No. 10 in Practice Examination No. 3 should be (it had a typo in one of the
integrals and I decided to reproduce the whole problem to provide a better
explanation):
Sample Course 1 Examination, Problem No. 35
Suppose the remaining lifetimes of a husband and a wife are independent and uniformly
distributed on the interval (0, 40). An insurance company offers two products to married
couples:
• One which pays when the husband dies; and
• One which pays when both the husband and wife have died.
Calculate the covariance of the two payment times.
A. 0.0
B. 44.4
C. 66.7
D. 200.0
E. 466.7
Solution.
Let H be the random time to death of the husband, W be the time to death of the wife, and
X be the time to the second death of the two. Clearly,
X = max ( H ,W ) .
1
We have fH ( h ) = fW ( w ) =
for 0 ≤ h ≤ 40, and 0 ≤ w ≤ 40. Thus
40
E ( H ) = E (W ) = 20.
Furthermore,
FX ( x ) = Pr ( X ≤ x ) = Pr ( max ( H ,W ) ≤ x ) =
= Pr ({ H ≤ x} ∩ {W ≤ x}) = Pr ( H ≤ x ) ⋅ Pr (W ≤ x ) =
x x
x2
⋅
=
.
40 40 1600
This implies that
x2
sX ( x ) = 1 −
1600
for 0 ≤ x ≤ 40, and
w
40
⎛ ( h,x 2w )⎞= w
40 3 120 40 80
40
max
E ( X ) = ∫here
1
−
dx
=
40
−
=
−
=
.
⎜ 1600 ⎟⎠
4800
3
3
3
0 ⎝
In order to find covariance, we also need to find
E ( XH ) = E ( H max ( H ,W )) .
max ( h, w ) = h
We separate the double integral into two parts: one based on the region where the wife
here
lives longer and one based on the region where the husband lives longer, as illustrated in
the graph below
40
h
E ( H max ( H ,W )) =
+∞ +∞
∫ ∫ h max ( h, w ) ⋅ f ( h ) ⋅ f ( w ) ⋅ dwdh =
H
W
−∞ −∞
h = 40 w = 40
⎛ w=h 2 1 1
⎞
⎛
⎞
1 1
= ∫ ⎜ ∫ h ⋅ ⋅ dw ⎟ dh + ∫ ⎜ ∫ hw ⋅ ⋅ dw ⎟ dh =
40 40 ⎠
40 40 ⎠
h=0 ⎝ w=0
h=0 ⎝ w=h
h = 40
⎛ h2w
= ∫⎜
1600
0 ⎝
40
40 ⎛
⎞
hw 2
dh
+
⎟
∫0 ⎜⎝ 3200
w=0 ⎠
w=h
w = 40
w=h
40
40
⎞
h3
1600h − h 3
dh
=
dh
+
dh =
⎟
∫
∫
1600
3200
⎠
0
0
40
1
1
⎛1
⎞
⎛1
⎞
= ∫⎜ h+
h 3 ⎟ dh = ⎜ h 2 +
h4 ⎟
⎝2
⎝4
3200 ⎠
12800 ⎠
0
h = 40
=
h=0
1
1
⋅ 40 2 +
⋅ 40 4 = 600.
4
12800
Finally,
Cov ( X, H ) = E ( XH ) − E ( X ) E ( H ) = 600 − 20 ⋅
80
2
= 66 .
3
3
Answer C.
Posted April 30, 2007
Problem No. 3 in Practice Examination No. 10 should be:
An insurance policy has a deductible of 10. Losses follow a probability distribution with
density f X ( x ) = xe− x for x > 0 and f X ( x ) = 0 otherwise. Find the expected payment.
A. e−10
B. 2e−10
C. 10e−10
D. 12e−10
E. 100e−10
Solution.
Let X be the random variable describing losses, and let Y be the amount of payment. Then
⎧⎪ 0,
X ≤ 10,
Y =⎨
⎩⎪ X − 10, X > 10.
We can calculate the expected value of Y directly
E (Y ) =
+∞
∫ ( x − 10 ) xe
−x
+∞
dx =
10
∫
10
+∞
x e dx − ∫ 10xe− x dx =
2 −x
10
u = x2
v = −e− x
=
du = 2xdx dv = e− x dx

Integration by parts for the first integral
= ( −x 2 e− x )
=
x→∞
x=10
+∞
+
+∞
−x
−x
−10
∫ 2xe dx − ∫ 10xe dx = 100e −
10
10
−x
+∞
∫ 8xe
dx =
10
x→∞
u = 8x
v = −e
−10
−x
=
100e
+
8xe
−8
(
)
x=10
du = 8dx dv = e− x dx

Integration by parts for the first integral
−x
+∞
∫e
−x
dx
10

 

=
Survival function of
exponential distribution
with hazard rate evaluated
at x=10
= 100e−10 − 80e−10 − 8e−10 = 12e−10 .
Answer D. We could also do this problem by applying the Darth Vader Rule. We have
⎧ Pr ( X > 10 ) ,
if y = 0,
⎪
sY ( y ) = Pr (Y > y ) = ⎨
⎪⎩ Pr ( X − 10 > y ) , if y > 0,
=
u = −x
v = e− x
= −xe− x
−x
dx = −dx dv = −e


x→∞
+
x=10+y
+∞
⎫
⎪
−x
⎬ = Pr ( X > y + 10 ) = ∫ xe dx, =
10+y
⎪⎭
+∞
∫
e− x dx =


10+y
Survival function
of exponential
with hazard rate 1
evaluated at 10+y
Integration by parts
= (10 + y ) e−10−y + e−10−y = (11+ y ) e−10−y
for any nonnegative y. As the payment random variable is non-negative almost surely, the
expected payment is
+∞
∫ (11+ y ) e
−10−y
dy = 11e
−10
+∞
∫e
⋅
0
−y
dy
0

 

This equals one because
e− y for y>0 is a density
of an exponential random
variable with hazard rate 1
+e
−10
+∞
⋅
∫ ye
−y
dy = 12e−10 .
0



This equals one
because the integrand
is the density in this
problem.
Answer D.
Posted April 29, 2007
The second formula to the last in the solution of Problem No. 6 in Practice
Examination No. 3 should be:
8
t 7
⎛ ⎛ 2 + et ⎞ 8
⎞
d 2 M (t )
d t ⎛ 2 + et ⎞
2
t
t⎛2+e ⎞
t
E X =
=
3e
=
3e
+
8e
⋅
e
⎜ ⎜
⎟
⎜⎝ 3 ⎟⎠
⎟
⎜⎝ 3 ⎟⎠
dt 2 t = 0 dt
⎝ ⎝ 3 ⎠
⎠
t =0
( )
= 11.
t =0
instead of
( )
E X
2
d 2 M (t )
d t ⎛ 2 + et ⎞
=
=
3e ⎜
dt 2 t = 0 dt
⎝ 3 ⎟⎠
8
t =0
t
⎛ ⎛ 2 + et ⎞ 8
⎞
t⎛2+e ⎞
t
= ⎜ 3et ⎜
+
8e
⋅
e
⎟
⎜⎝ 3 ⎟⎠ ⎟
⎝ ⎝ 3 ⎠
⎠
= 11.
t =0
Posted April 27, 2007
In the solution of Problem No. 9 in Practice Examination No. 3, the expression
max ( X,1) should be min ( X,1) .
Posted April 20, 2007
Problem No. 16 in Practice Examination No. 1 should be:
May 2001 Course 1 Examination, Problem No. 26, also Study Note P-09-05, Problem
No. 109
A company offers earthquake insurance. Annual premiums are modeled by an
exponential random variable with mean 2. Annual claims are modeled by an exponential
random variable with mean 1. Premiums and claims are independent. Let X denote the
ratio of claims to premiums. What is the density function of X?
A.
1
2x + 1
B.
2
( 2x + 1)
C. e− x
2
D. 2e−2 x
E. xe− x
Solution.
Let U be the annual claims and let V be the annual premiums (also random), and let
fU ,V ( u, v ) be the joint density of them. Furthermore, let f X be the density of X and let FX
be its cumulative distribution function. We are given that U and V are independent, and
hence
v
−
1 −u − 2v
−u 1
2
fU ,V ( u, v ) = e ⋅ e = e e
2
2
for 0 < u < ∞, 0 < v < ∞. Also, noting the graph below,
v
u = vx or v = u/x
u
we have:
⎛U
⎞
FX ( x ) = Pr ( X ≤ x ) = Pr ⎜ ≤ x ⎟ = Pr (U ≤ Vx ) =
⎝V
⎠
∞ vx
⎛ 1
= ∫⎜− e
2
0⎝
∞
⎛ 1⎞
− v⎜ x + ⎟
⎝ 2⎠
+∞
u = vx
0
u=0
⎛ 1 −u − 2v ⎞
∫ ⎜⎝ − 2 e e ⎟⎠
1 −u − 2v
= ∫ ∫ e e dudv =
2
0 0
∞ vx
∫ ∫ f (u, v )dudv =
U ,V
0 0
∞
⎛ 1 − vx − 2v 1 − 2v ⎞
dv = ∫ ⎜ − e e + e ⎟ dv =
2
2
⎠
0⎝
⎛ 1⎞
v
⎛ 1
− v⎜ x + ⎟
− ⎞
1 − 2v ⎞
+ e ⎟ dv = ⎜
e ⎝ 2⎠ − e 2 ⎟
2
⎠
⎝ 2x + 1
⎠
+∞
=−
0
1
+ 1.
2x + 1
Finally,
f X ( x ) = FX′ ( x ) =
2
( 2x + 1)2
.
Answer B. This problem can also be done with the use of bivariate transformation. We
will now give an alternative solution using that approach. Consider the following
transformation X =
U
, Y = V. Then the inverse transformation is U = XY , V = Y . We
V
know that
1 −u − 2v
e e
2
for u > 0, v > 0. It follows that
fU ,V ( u, v ) =
∂ ( u, v ) 1 − xy − 2y
⎡ y x ⎤ 1 − xy − 2y
f X ,Y ( x, y ) = fU ,V ( u ( x, y ) , v ( x, y )) ⋅
= e e ⋅ det ⎢
⎥ = ye e
∂ ( x, y ) 2
⎣0 1 ⎦ 2
for xy > 0 and y > 0, or just x > 0 and y > 0. Therefore,
fX ( x ) =
+∞
∫
0
1 − xy − 2y
ye e dy =
2
+∞
∫
0
⎛
w=
1⎞
1 − ⎜⎝ x + 2 ⎟⎠ y
ye
dy =
2
1
y
2
z=−
1
⎛
⎜⎝ x +
e
1⎞
⎟
2⎠
⎛ 1⎞
−⎜ x+ ⎟ y
⎝ 2⎠
=
⎛ 1⎞
−⎜ x+ ⎟ y
⎝ 2⎠
1
dw = dy
dz = e
dy
2


INTEGRATION BY PARTS
⎛
⎞
⎛ 1⎞
−⎜ x+ ⎟ y ⎟
⎜ 1
1
= ⎜− y⋅
e ⎝ 2⎠ ⎟
⎜ 2 ⎛⎜ x + 1 ⎞⎟
⎟
⎝
⎝
⎠
2⎠
y→+∞
+∞
−
∫
0
y= 0
⎛
⎞
⎛ 1⎞
1⎜
1 ⎟ − ⎜⎝ x + 2 ⎟⎠ y
−
e
dy =
⎜
⎟
1⎞ ⎟
2⎜ ⎛
x+ ⎟
⎝ ⎜⎝
2⎠ ⎠
⎛
⎞
⎛
⎞
+∞
⎛ 1⎞
⎛ 1⎞ ⎟
⎜
−⎜ x+ ⎟ y
1
1 ⎜
1 ⎟ − ⎜⎝ x + 2 ⎟⎠ y
=0+ ∫
e ⎝ 2 ⎠ dy = ⎜
⎟
⎜−
⎟ ⋅e
2x + 1
⎜ 2x + 1 ⎜ ⎛ x + 1 ⎞ ⎟
⎟
0
⎟
⎜⎝
⎟⎠
⎝ ⎜⎝
2⎠ ⎠
y→+∞
=
2
( 2x + 1)2
.
y= 0
Answer B, again. The second approach is probably more complicated, but it is a good
exercise in the use of multivariate transformations.
Posted February 19, 2007
In Problem No. 20, Practice Examination Number No. 10, answer choice is D,
instead of E as stated now (typo).
Posted January 20, 2007
Exercise 2.8 should read as follows:
November 2001 Course 1 Examination, Problem No. 11
A company takes out an insurance policy to cover accidents that occur at its
manufacturing plant. The probability that one or more accidents will occur during any
3
given month is . The number of accidents that occur in any given month is independent
5
of the number of accidents that occur in all other months. Calculate the probability that
there will be at least four months in which no accidents occur before the fourth month in
which at least one accident occurs.
A. 0.01
B. 0.12
C. 0.23
D. 0.29
E. 0.41
Solution.
Consider a Bernoulli Trial with success defined as a month with an accident, and a month
3
with no accident being a failure. Then the probability of success is . Now consider a
5
negative binomial random variable, which counts the number of failures (months with no
accident) until 4 successes (months with accidents), call it X. The problem asks us to find
Pr ( X ≥ 4 ) . But
Pr ( X ≥ 4 ) = 1 − Pr ( X = 0 ) − Pr ( X = 1) − Pr ( X = 2 ) − Pr ( X = 3) =
4
4
4
2
4
3
⎛ 3⎞ ⎛ 3 ⎞
⎛ 4 ⎞ ⎛ 3 ⎞ 2 ⎛ 5⎞ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 6⎞ ⎛ 3 ⎞ ⎛ 2 ⎞
= 1− ⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟ ⋅⎜ ⎟ ⋅ − ⎜ ⎟ ⋅⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ≈
⎝ 0⎠ ⎝ 5 ⎠
⎝ 1 ⎠ ⎝ 5 ⎠ 5 ⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎝ 3⎠ ⎝ 5 ⎠ ⎝ 5 ⎠
≈ 0.289792.
Answer D.
Posted December 31, 2006
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.
It was mistyped as
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 December 28, 2006
On page 3 in the bottom paragraph, the words:
is also defined as the sent that consists of all elementary events that belong to any one of
them.
should be
is also defined as the set that consists of all elementary events that belong to any one of
them.
The word “set” was mistyped as “sent.”
Posted December 1, 2006
In Practice Examination 6, Problem 22, had inconsistencies in its assumptions, and
it should be replaced by the following:
A random variable X has the log-normal distribution with the density:
1
2
− ( ln x− µ )
1
⋅e 2
x 2π
for x > 0, and 0 otherwise, where µ is a constant. You are given that Pr ( X ≤ 2 ) = 0.4.
Find E ( X ) .
fX ( x ) =
A. 4.25
B. 4.66
C. 4.75
D. 5.00
E. Cannot be determined
Solution.
X is log-normal with parameters µ and σ 2 if W = ln X  N µ, σ 2 . The PDF of the log-
(
)
normal distribution with parameters µ and σ is
2
fX ( x ) =
1
−
1 ( ln x− µ )
2 σ2
2
e
xσ 2π
for x > 0, and 0 otherwise. From the form of the density function in this problem we see
that σ = 1. Therefore
⎛ ln X − µ ln 2 − µ ⎞
Pr ( ln X ≤ ln 2 ) = Pr ⎜
≤
⎟ = 0.4.
⎝
1
1 ⎠
Let z0.6 be the 60-th percentile of the standard normal distribution. Let Z be a standard
ln X − µ
normal random variable. Then Pr ( Z ≤ −z0.6 ) = 0.40. But Z =
is standard normal,
1
thus ln 2 − µ = −z0.6 , and µ = ln 2 + z0.6 . From the table, Φ ( 0.25 ) = 0.5987 and
Φ ( 0.26 ) = 0.6026. By linear interpolation,
0.6 − 0.5987
1
z0.6 ≈ 0.25 +
⋅ ( 0.26 − 0.25 ) = 0.25 + ⋅ 0.01 ≈ 0.2533.
0.6026 − 0.5987
3
This gives
µ = ln 2 + z0.6 ≈ 0.9464805.
The mean of the log-normal distribution is E ( X ) = e
E(X) = e
Answer A.
1
0.9464805+
2
1
µ+ σ 2
2
, so that in this case
≈ 4.2481369.
Posted December 1, 2006
In Practice Examination No. 8, Problem No. 15, answer B should be listed as
B. 0.1465
It was listed as
B. 1465,
an obvious typo.
Posted November 29, 2006
The solution of Problem No. 5 in Practice Examination No. 10 is not clear in the
explanation of the linear interpolation. Below is a more complete explanation:
May 1982 Course 110 Examination, Problem No. 1
If a normal distribution with mean µ and variance σ 2 > 0 has 46-th percentile equal to
20σ , then µ is equal to
A. 18.25σ
B. 19.90σ
C. 20.10σ
D. 21.73σ
E. Cannot be determined
Solution.
X−µ
has the standard
σ
normal distribution. Let us write x0.46 for the 46-th percentile of X. We have
⎛ X − µ x0.46 − µ ⎞
⎛ X − µ 20σ − µ ⎞
0.46 = Pr ( X < x0.46 ) = Pr ⎜
<
= Pr ⎜
<
⎟
⎟⎠ .
⎝ σ
⎝ σ
σ ⎠
σ
Let X be the random variable described in the problem. Then
From the table, Φ ( 0.1) = 0.5398 and Φ ( 0.11) = 0.5438. Therefore, Φ ( −0.1) = 0.4602
and Φ ( −0.11) = 0.4562. The probability desired is 0.46, and that is
0.4602 − 0.46
0.0002 1
=
=
0.4602 − 0.4562 0.0040 20
of the distance between 0.4562 and 0.4602, down from 0.4602. This means that the 46-th
1
percentile is approximately
of the distance between −0.1 and −0.11, down from
20
−0.1, i.e., we can get the desired percentile of the standard normal distribution via the
following linear interpolation approximation:
1
0.01
z0.46 ≈ −0.1− ⋅ (( −0.1) − ( −0.11)) = −0.1−
≈ −0.1005.
20
20
20σ − µ
Therefore,
≈ −0.1005, so that µ ≈ 20.1005σ .
σ
Answer C.
Posted November 25, 2006
The third equation from the bottom in the solution of Problem No. 10 in Practice
Examination No. 2 should be
Pr ( A ∩ E99 ) = Pr ( A E99 ) ⋅ Pr ( E99 ) = 0.03⋅ 0.20 = 0.006.
Previously it had a typo:
Pr ( A ∩ E99 ) = Pr ( A E99 ) ⋅ Pr ( E99 ) = 0.03⋅ 0.20 = 0.0036.
Posted November 24, 2006
In the description of the exponential distribution, the sentence
The exponential distribution is related to the geometric distribution in that if we have a
random variable T with exponential distribution, and we define a new random variable K
as the integer portion of T (for example, if T = 7.5, then K = 7, K is simply the value of
the greatest integer function when applied to T), then so defined K has the geometric
distribution with p = e− λ .
should be replaced by
The exponential distribution is related to the geometric distribution in that if we have a
random variable T with exponential distribution, and we define a new random variable K
as the integer portion of T (for example, if T = 7.5, then K = 7, K is simply the value of
the greatest integer function when applied to T), then so defined K has the geometric
distribution with q = e− λ .
The only difference is that the symbol in the last equation should be q, not p.
Posted November 22, 2006
The third sentence of Problem No. 12 in Practice Examination No. 7 should read:
Let X and Y be the times at which the first and second circuits fail, respectively, counting
from the moment when a circuit is activated.
Posted August 16, 2006
Problem No. 2 in Practice Examination No. 10 should be as below (it did not define
the variable Z previously)
You are given that the joint density of random variables X and Y is f X ,Y ( x, y ) = x + y for
0 < x < 1 and 0 < y < 1. What is the coefficient of variation of Z = X + Y ?
A. 0.1889
B. 1.3611
C. 1.55
D. 0.50
E. 0.3194
Solution.
Note that 0 < Z < 2 with probability 1. Using the concept of convolution
fZ ( z ) = f X+Y ( z ) =
+∞
∫ f ( x, z − x ) dx = ∫
X ,Y
−∞
⎧ z
⎪ ∫ z dx, 0 < z ≤ 1,
⎪ 0
=⎨ 1
⎪ z dx, 1 < z ≤ 2.
⎪ ∫
⎩ z−1
Therefore,
1
z dx =
2
∫
z dx =
[ 0,1]∩[ z−1,z ]
0≤x≤1
0≤z−x≤1
⎫
⎪
⎪ ⎧⎪ z 2 ,
0 < z ≤ 1,
⎬= ⎨
⎪ ⎪⎩ z ( 2 − z ) , 1 < z ≤ 2.
⎪
⎭
2
⎫⎪ ⎧⎪ z 2 ,
0 < z ≤ 1,
⎬= ⎨
2
⎪⎭ ⎩⎪ 2z − z , 1 < z ≤ 2.
1
1 ⎛2
1 ⎞
E ( Z ) = ∫ z dz + ∫ z ( 2 − z ) dz = + ∫ ( 2z 2 − z 3 ) dz = + ⎜ z 3 − z 4 ⎟
⎝
4 1
4
3
4 ⎠
0
1
3
=
z=2
=
2
z=1
1 ⎛ 16
2 1 ⎞ 1 14
3 28
31 24 7
+⎜ −4− + ⎟ = + −4 = +
−4=
−
= ,
4 ⎝ 3
3 4⎠ 2 3
6 6
6
6 6
1
2
0
1
E ( Z 2 ) = ∫ z 4 dz + ∫ z 3 ( 2 − z ) dz =
1 ⎛2
1 ⎞
= + ⎜ z4 − z5 ⎟
5 ⎝4
5 ⎠
z=2
=
z=1
2
1
+ ( 2z 3 − z 4 ) dz =
5 ∫1
1 ⎛
32 1 1 ⎞ 3
+⎜8−
− + ⎟= .
5 ⎝
5 2 5⎠ 2
Hence,
3 49 54 49 5
−
=
−
= .
2 36 36 36 36
Finally, the coefficient of variation is
5
Var ( Z )
5
= 36 =
≈ 0.31943828.
7
E ( Z)
7
6
Answer E.
Var ( Z ) =
Posted August 3, 2006
In the solution of Problem No. 4 in Practice Examination No. 1, the formula
fY ( y ) = k2 ⋅ y,
should be
fY ( y ) = k2 ⋅1.
Posted July 17, 2006
The last two lines of the solution of Problem No. 5 in Practice Examination No. 8
should be:
Alternatively,
d
1 d +∞ etn−1
1 +∞ etn−1
−1
−2
M X′ ( t ) =
− (t − 2 ) + ⋅ ∑
= (t − 2 ) + ⋅ ∑
.
dt
2 dt n=0 n!
2 n=1 ( n − 1)!
Therefore,
1
1 +∞ e−1
1 1 −1 +∞ 1 1 1 3
E ( X ) = M X′ ( 0 ) =
+
⋅
=
+ ⋅e ⋅∑ = + = .
∑
4 2 4
( −2 )2 2 n=1 ( n − 1)! 4 2
n=0 n!

(
)
=e
Posted July 3, 2006
The relationship between the survival function and the hazard rate, stated just after
the definition of the hazard rate, should be
x
sX ( x ) = e
instead of
−
∫ λ X (u ) du
−∞
x
sX ( x ) = e
∫
− λ X ( u ) du
0
.
The second equation applies in the case when the random variable X is nonnegative
almost surely.
Posted July 3, 2006
The condition concerning a continuous probability density function in its first
definition, in Section 1, should be
f X ( x ) ≥ 0 for every x ∈
instead of
0 ≤ f X ( x ) ≤ 1 for every x ∈.
Posted May 16, 2006
In the solution of Problem No. 21 in Practice Examination No. 8 the last formula
should be
x=y
y
y
⎞ 1
2x
1
1 ⎛2
E X 2 Y = y = ∫ x 2 ⋅ 2 dx = 2 ⋅ ∫ 2x 3dx = 2 ⋅ ⎜ ⋅ x 4 ⎟ = y 2 .
y
y 0
y ⎝4
2
x=0 ⎠
0
(
)
Posted May 16, 2006
The solution of Problem No. 13 in Practice Examination No. 3 should be (some of
the exponents contained typos)
Solution.
Let X be the random number of passengers that show for a flight. We want to find
Pr(X = 31) + Pr(X = 32).
We can treat each passenger arrival as a Bernoulli Trial, and then it is clear that X has
binomial distribution with n = 32, p = 0.90. Therefore,
⎛ 32 ⎞
Pr ( X = x ) = ⎜
⋅ 0.90 x ⋅ 0.10 32−x.
⎟
x
⎝
⎠
The probability desired is:
⎛ 32 ⎞
⎛ 32 ⎞
Pr ( X = 31) + Pr ( X = 32 ) = ⎜
⋅ 0.90 31 ⋅ 0.101 + ⎜
⋅ 0.90 32 ⋅ 0.10 0 ≈
⎟
⎟
31
32
⎝
⎠
⎝
⎠
≈ 0.12208654 + 0.03433684 = 0.15642337.
Answer E.
Posted May 15, 2006
The solution, but not the answer, of Problem No. 30 in Practice Examination No. 9,
should be changed to:
Solution.
First note that
sX ( x ) =
u = − (1− K + 0.01Ks ) v = e−0.01s
+∞
−0.01s
∫ 0.01⋅ (1− K + 0.01Ks ) ⋅ e ds =
x
du = −0.01Kds
dv = −0.01e−0.01s ds

INTEGRATION BY PARTS
= ( − (1− K + 0.01Ks ) ⋅ e−0.01s )
s→+∞
s=x
+∞
+ K ⋅ ∫ 0.01e−0.01s ds =
x



Survival function for
exponential with hazard
rate of 0.01
= (1− K + 0.01Kx ) ⋅ e−0.01x + Ke−0.01x = (1+ 0.01Kx ) ⋅ e−0.01x .
The payment, given that a payment is made, with a deductible of d, is
X* = ( X − d ) ( X > d ) . The survival function of that random variable is
s X* ( x ) = Pr ( X* > x ) = Pr ( X − d > x X > d ) = Pr ( X > d + x X > d ) =
=
(1+ 0.01K ( x + d )) ⋅ e
−0.01( x+d )
=
(1+ 0.01Kd ) ⋅ e
The expected value of ( X − 100 ) ( X > 100 ) is
−0.01d
+∞
1+ 0.01K ( x + d ) −0.01x
⋅e
.
1+ 0.01Kd
+∞
1+ K + 0.01Kx −0.01x
1
−0.01x
∫0 1+ K ⋅ e = 1+ K ⋅ ∫0 (1+ K + 0.01Kx ) ⋅ e =
=
u = 1+ K + 0.01Kx v = −100e−0.01x
=
du = 0.01Kdx
dv = e−0.01x dx


Integration by parts
⎛ 1+ K + 0.01Kx
⎞
= ⎜−
⋅100e−0.01x ⎟
⎝
⎠
1+ K
x→+∞
+
x=0
+∞
K
⋅
1+ K
100K
dx = 100 +
.
1+
K
0



∫e
−0.01x
Mean of exponential
with hazard rate of 0.01
1
Since this is equal to 125, we must have K = . Therefore, the expected value of
3
( X − 200 ) ( X > 200 ) is
+∞
−0.01x
∫ (1+ 0.002x ) ⋅ e .dx =
0
u = 1+ 0.002x v = −100e−0.01x
=
du = 0.002dx
dv = e−0.01x dx

Integration by parts
= ( − (1+ 0.002x ) ⋅100e
−0.01x
)
x→+∞
x=0
+∞
+ 0.2 ⋅
∫e
−0.01x
dx = 100 + 0.2 ⋅100 = 120.


0
Mean of exponential
with hazard rate of 0.01
Answer A.
=