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```							ASM Study Manual for Exam P, Second Edition
By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net)
Errata

Posted June 18, 2009
The last formula in the solution of Problem 19 of Practice Examination 5 should be
1    1                            1   1
Pr ( X ≥ 10 ) ≤ 2 =    instead of Pr ( X ≥ 10 ) < 2 = .
4    16                           4   16

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 5, 2009
The first sentence of Problem 16 in Practice Examination 6 should end with t < 1,

Posted March 2, 2009
The properties of the cumulant moment-generating function should be:
The cumulant generating function has the following properties:
(
E XetX     )
ψ X ( 0 ) = 0,
d
( )
ln E etX            =                         = E ( X ),
dt              t =0    ( )E e   tX
t =0

d2
ψ X (t ) =
d E Xe (tX
)          =
(
E X 2 etX   ) E ( e ) − E ( Xe ) E ( Xe )
tX               tX              tX

= Var ( X ) ,
dt 2         t =0
dt E etX   ( )       t =0                         ( E ( e ))
tX
2

t =0

(                        )
3
d
ψ X ( t ) = E ( X − E ( X )) ,
3
3
dt            t =0
but for k > 3,
dk
dt k
t =0
(
ψ X ( t ) = ψ (Xk ) ( 0 ) < E ( X − E ( X )) .
k
)
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 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 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,
The mean of the 48 rounded ages.

Posted May 9, 2007
In the solution of Problem No. 15, Practice Examination No. 3, the derivative is
missing a minus, and it should
dX      1          2
= − (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 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, … .

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 ) . We have
1
fH ( h ) = fW ( w ) =       for 0 ≤ h ≤ 40, and 0 ≤ w ≤ 40. Thus E ( H ) = E (W ) = 20.
40
Furthermore,
FX ( x ) = Pr ( X ≤ x ) = Pr ( max ( H ,W ) ≤ x ) =
x x   x2
= Pr ({ H ≤ x} ∩ {W ≤ x}) = Pr ( H ≤ x ) ⋅ Pr (W ≤ x ) =     ⋅  =     .
40 40 1600
This implies that
x2
sX ( x ) = 1 −
1600
for 0 ≤ x ≤ 40, and
40
⎛     x2 ⎞            40 3 120 40 80
E ( X ) = ∫ ⎜1 −         dx = 40 −      =   −    =     .
0 ⎝
1600 ⎟
⎠           4800   3   3     3
In order to find covariance, we also need to find
E ( XH ) = E ( H max ( H ,W )) .
We separate the double integral into two parts: one over the region where the wife lives
longer and one over the region where the husband lives longer, as illustrated below:

w
40      max ( h, w ) = w
here

max ( h, w ) = h
here

40                        h
+∞ +∞

E ( H max ( H ,W )) =    ∫ ∫ h max ( h, w ) ⋅ f ( h ) ⋅ f ( w ) ⋅ dwdh =
H             W
−∞ −∞

⎛ w=h 2 1 1
h = 40
⎞        ⎛
h = 40 w = 40
1 1   ⎞
= ∫ ⎜ ∫ h ⋅ ⋅ dw ⎟ dh + ∫ ⎜ ∫ hw ⋅ ⋅ dw ⎟ dh =
h=0 ⎝ w=0
40 40 ⎠   h=0 ⎝ w=h
40 40 ⎠
⎛ h2w
40          w=h
⎞      40 ⎛
hw 2
w = 40
⎞      40
h3
40
1600h − h 3
= ∫⎜               ⎟ dh + ∫ ⎜                  ⎟ dh = ∫       dh + ∫              dh =
0 ⎝
1600     w=0 ⎠      0 ⎝
3200   w=h      ⎠      0
1600      0
3200
40                                                                 h = 40
⎛1   1      ⎞      ⎛1        1     ⎞                                           1            1
= ∫⎜ h+     h 3 ⎟ dh = ⎜ h 2 +      h4 ⎟                                       =     ⋅ 40 2 +       ⋅ 40 4 = 600.
0
⎝2  3200 ⎠         ⎝4      12800 ⎠                                h=0
4          12800
Finally,
80     2
Cov ( X, H ) = E ( XH ) − E ( X ) E ( H ) = 600 − 20 ⋅                       = 66 .
3     3

Posted April 29, 2007
The second formula to the last in the solution of Problem No. 6 in Practice
Examination No. 3 should be:
d 2 M (t )
8
⎛ 2 + et ⎞
( )
E X   2
=
d
= 3et ⎜
dt 2 t = 0 dt   ⎝ 3 ⎟    ⎠
=
t =0

⎛ ⎛ 2 + et ⎞ 8          t 7      ⎞
t⎛2+e ⎞
= ⎜ 3e ⎜
t
⎟   + 8e ⎜       ⋅ et ⎟                          = 11.
⎝ ⎝ 3 ⎠             ⎝ 3 ⎟⎠       ⎠                   t =0
d 2 M (t )
8
d t ⎛ 2 + et ⎞
( )
E X   2
=              = 3e ⎜
dt 2 t = 0 dt  ⎝ 3 ⎟    ⎠
=
t =0

⎛ ⎛ 2 + et ⎞  8
⎛ 2 + et ⎞ t ⎞
= ⎜ 3et ⎜    ⎟ + 8et ⎜          ⋅e ⎟                      = 11.
⎝ ⎝ 3 ⎠            ⎝ 3 ⎟ ⎠  ⎠
t =0

Posted April 27, 2007
In the solution of Problem No. 9 in Practice Examination 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?

1                                 2
A.                            B.                              C. e− x                   D. 2e−2 x          E. xe− x
2x + 1                          ( 2x + 1)   2

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
1 −v 1       −
v
hence fU ,V ( u, v ) = e−u ⋅ e 2 = e−u e 2 for 0 < u < ∞, 0 < v < ∞. Also, noting the graph
2       2
below, we have:

v

u = vx or v = u/x

u

∞ vx
⎛U    ⎞
FX ( x ) = Pr ( X ≤ x ) = Pr ⎜ ≤ x ⎟ = Pr (U ≤ Vx ) =                 ∫ ∫ f (u, v )dudv =
⎝V    ⎠                                  0 0
U ,V

∞ vx                           +∞              u = vx        ∞
1    −
v
⎛ 1 −u − 2 ⎞
v
⎛ 1        −
v
1 −v ⎞
= ∫ ∫ e−u e 2 dudv =              ∫⎜ 2
− e e ⎟               dv = ∫ ⎜ − e− vx e 2 + e 2 ⎟ dv =
0 0
2                           0⎝          ⎠   u=0           0⎝
2           2    ⎠
+∞
⎛ 1 − v⎛ x + 1 ⎞ 1 − v ⎞
∞
⎜       ⎟            ⎛ 1           ⎛ 1⎞
− v⎜ x + ⎟   − ⎞
v
1
= ∫⎜− e   ⎝ 2⎠
+ e ⎟ dv = ⎜
2
e    ⎝ 2⎠
−e ⎟
2
=−          + 1.
0⎝
2               2    ⎠    ⎝ 2x + 1                   ⎠                  0
2x + 1
Finally,
2
f X ( x ) = FX ( x ) =
′                        .
( 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
U
X=       ,          Y = V.
V
Then the inverse transformation is
U = XY ,              V = Y.
v
1 −u − 2
We know that fU ,V ( u, v ) = e e for u > 0, v > 0. It follows that
2
∂ ( u, v ) 1 − xy − 2
y
⎡ y x ⎤ 1 − xy − 2
y
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,
⎛ 1⎞
1                    1            −⎜ x+ ⎟ y
⎝ 2⎠
w= y        z=−                 e
+∞                     +∞        ⎛ 1⎞                   2                ⎛  1⎞
⎜x+ ⎟
y               −⎜ x+ ⎟ y
1        −             1
fX ( x ) =   ∫ 2 ye
− xy
e dy =
2
∫ 2 ye
⎝ 2⎠
dy =                         ⎝  2⎠                        =
0                      0                                                         ⎛   1⎞
1          −⎜ x+ ⎟ y
dw = dy  dz = e ⎝ 2 ⎠ dy
2   
            
INTEGRATION BY PARTS
y→+∞
⎛                       ⎞                                 ⎛         ⎞
⎜ 1     1     −⎜ x+ ⎟ y ⎟
⎛ 1⎞                                +∞
1 ⎜     1 ⎟ −⎜ x+ 2⎟ y
⎛ 1⎞

e ⎝ 2⎠ ⎟                              ∫
⎝   ⎠
= ⎜− y⋅                                          −          ⎜−        ⎟e         dy =
⎜ 2 ⎛     1⎞            ⎟                               2⎜ ⎛     1⎞ ⎟
⎜x+ ⎟
⎝
0
⎝ ⎜
x+ ⎟
⎝         2⎠            ⎠            y= 0
⎝   2⎠ ⎠
y→+∞
⎛        ⎛           ⎞            ⎞
+∞
1
⎛ 1⎞
−⎜ x+ ⎟ y   ⎜ 1 ⎜          1     ⎟ −⎛ x+ 1⎞ y ⎟
⎜   ⎟                                           2
=0+ ∫         e ⎝ 2 ⎠ dy = ⎜        ⎜−          ⎟ ⋅e
⎝ 2⎠
⎟                            =                    .
2x + 1              ⎜ 2x + 1 ⎜ ⎛ x + 1 ⎞ ⎟            ⎟                                 ( 2x + 1)2
⎝ ⎜       ⎟
0
⎜
⎝           ⎝    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 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) =
⎛ 3⎞ ⎛ 3 ⎞   ⎛ 4 ⎞ ⎛ 3 ⎞ 2 ⎛ 5⎞ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 6⎞ ⎛ 3 ⎞ ⎛ 2 ⎞
4            4             4     2          4      3

= 1− ⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟ ⋅⎜ ⎟ ⋅ − ⎜ ⎟ ⋅⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ≈
⎝ 0⎠ ⎝ 5 ⎠   ⎝ 1 ⎠ ⎝ 5 ⎠ 5 ⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎝ 3⎠ ⎝ 5 ⎠ ⎝ 5 ⎠
≈ 0.289792.

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 No. 6, Problem No. 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
1    − ( ln x − µ )
2

fX ( x ) =      ⋅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 ) .
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

1 ( ln x − µ )
2

1            −
fX ( x ) =           e           2      σ2
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
1
normal, 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.
1
µ+ σ2
The mean of the log-normal distribution is E ( X ) = e                      2
, so that in this case
1
0.9464805 +
E(X) = e                     2
≈ 4.2481369.

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.
Pr ( A ∩ E99 ) = Pr ( A E99 ) ⋅ Pr ( E99 ) = 0.03 ⋅ 0.20 = 0.0036.

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 3, 2006
The relationship between the survival function and the hazard rate, stated just after
the definition of the hazard rate, should be
x
−   ∫ λX (u ) du
sX ( x ) = e       −∞

x

∫
− λ X ( u ) du
sX ( x ) = e 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 ∈
0 ≤ f X ( x ) ≤ 1 for every x ∈.

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.

Posted March 1, 2006
The formula for the variance of a linear combination of random variables should
be:
Var ( a1 X1 + a2 X2 + … + an Xn ) =

⎣      (
= [ a1 a2 … an ] ⋅ ⎡Cov Xi , X j ⎤  (           )             )
⎦i, j =1,2,…,n ⋅ [ a1 a2 … an ] =
T

⎡ a1 ⎤
⎢a ⎥
(
= [ a1 a2 … an ] ⋅ ⎡Cov Xi , X j ⎤
⎣                (
⎦i, j =1,2,…,n )             )   ⋅⎢ ⎥ =
⎢⎥
2

⎢ ⎥
⎣ an ⎦

(     )
n                 n       n
= ∑ ai2Var ( Xi ) + ∑     ∑ 2a a    i   j   ⋅ Cov Xi , X j .
i =1              i =1 j =i +1

The only incorrect piece was in the last double sum, where a1a2 appeared instead of
ai a j .

Posted March 1, 2006
The moment generating function of the chi-square distribution should be:
n
⎛ 1 ⎞2
M X (t ) = ⎜
⎝ 1 − 2t ⎟
.
⎠
(its argument was mistakenly written as x instead of t).

Posted February 11, 2006
The solution of Problem No. 18 of the Practice Examination No. 4 should be:
Let X be the claim for employee who incurred a loss in excess of 2000, and Y be the
claim for the other employee. The joint distribution of X and Y , given that each loss
occurs, is uniform on
[1000, 5000 ]2 .
This means that we can calculate all probabilities by comparing areas. The probability
that total losses exceed reimbursement, i.e.,
X + Y > 8000,
given that X > 2000 is the ratio of the area of the triangle in the right upper corner of the
square
[1000, 5000 ]2
in the figure below to the area of the rectangle
[ 2000, 5000 ] × [1000, 5000 ],
i.e.,
1
⋅ 2000 ⋅ 2000 1
2                = .
3000 ⋅ 4000      6
Y

5000

X + Y = 8000
However, this is conditional on the occurrence of the loss of the employee whose loss is
Y (but no longer conditional on the occurrence of the loss for the employee whose loss is
X, because the event considered is already conditional on X > 2000, so the loss has
occurred). The probability of the loss occurring is 40% = 0.4. Therefore the probability
1 2 1 1
sought is 0.40 ⋅ = ⋅ = .
6 5 6 15

Posted January 31, 2006
In the solution of Problem No. 21 in Practice Examination No. 2, May 2000 Course 1
Examination, Problem No. 19, the fourth sentence should begin
1 48
The mean of the 48 rounded ages, E =     ∑ Ei
48 i =1

Posted January 26, 2006
In the solution of Problem No. 15 in Practice Examination No. 2, May 2000 Course 1
Examination, Problem No. 7, the survival function should be written as:
s X ( x ) = (1 + x ) .
−3

Posted January 18, 2006
In the description of the geometric distribution:
pet
M X (t ) =         . Some texts use a different version of this distribution, which only
1 − qet
counts failures until the first success. That’s just too pessimistic for this author. The two
distributions differ by one. For this other form of random variable (similar in its design to
q               q
the negative binomial distribution, just below) Y = X – 1, and E (Y ) =     , Var ( X ) = 2 ,
p              p
p
M X (t ) =          .
1 − qet
should be
pet
M X (t ) =         . Some texts use a different version of this distribution, which only
1 − qet
counts failures until the first success. That’s just too pessimistic for this author. The two
distributions differ by one. For this other form of random variable (similar in its design to
q              q
the negative binomial distribution, just below) Y = X – 1, and E (Y ) = , Var (Y ) = 2 ,
p             p
p
M Y (t ) =         .
1 − qet

Posted September 8, 2005
In the solution of Exercise 2.2, the first sentence should be:
It is important to realize that there is a point-mass at 1 -- this can be seen by analyzing the
limit of CDF at 1 from the left (equal to 0) and the right-hand side limit of CDF at 1
(equal to 0.5).

Posted August 22, 2005
In the solution of Problem No. 11 in Practice Examination No. 6 these words
(         )
Similarly, Pr X( 7 ) < m is the probability that either none of eight elements of the
random sample X1 , X1 ,…, X8 are greater than or equal to m
should be replaced by
(         )
Similarly, Pr X( 7 ) < m is the probability that either none or one of eight elements of the
random sample X1 , X1 ,…, X8 are greater than or equal to m

Posted August 21, 2005
In Exercise 3.11, the second to last formula should be:
∞              ∞
1 −3    1 −2      1
f X ( 2 ) = ∫ y dy = − y     = .
1
2       4    1    4
It was missing y −2 .

Posted August 20, 2005
In Exercise 3.15, November 2001 Course 1 Examination, Problem No. 40, in the
solution, first two displayed formulas should be:
E ( X + Y ) = E ( X ) + E (Y ) = 50 + 20 = 70,
Var ( X + Y ) = Var ( X ) + Var (Y ) + 2Cov ( X,Y ) = 50 + 30 + 20 = 100.
The problem with existing formulas was that Y was improperly replaced by X.

Posted August 8, 2005
In Problem No. 7 of Practice Examination No. 4, the five choices of answers should
be:
20 20
6
( 50 − x − y )dydx
125, 000 ∫ ∫
A.
0  0
30       50 − x
6
B.            ∫
125, 000 20           ∫ ( 50 − x − y )dydx
20
30       50 − x − y
6
C.            ∫
125, 000 20             ∫ ( 50 − x − y )dydx
20
50       50 − x
6
D.            ∫
125, 000 20           ∫ ( 50 − x − y )dydx
20
50       50 − x − y
6
E.          ∫
125, 000 20              ∫ ( 50 − x − y )dydx
20

Posted August 5, 2005
The solution of Problem No. 13, Practice Examination No. 5 should be:
Note the region where the density is positive and the region describing the probabilities
we are calculating in the figure below

y

1

xy = w

1                    x

Therefore,
⎛    w⎞
FW ( w ) = Pr (W ≤ w ) = Pr ( XY ≤ w ) = Pr ⎜ Y ≤ ⎟ =
⎝    X⎠
⎛w        ⎞
∫(                    )
w
⎛y       ⎞                     1
⎜
y
⎟
w
x=y
1
⎛ 2 x= w ⎞
=   ∫       ⎜∫ 8xydx ⎟ dy +                ∫ ⎜ ∫ 8xydx ⎟ dy =                  4x y x = 0 dy +       ∫ ⎜ 4x y x = 0y ⎟ dy =
2

⎝0       ⎠                     w⎜ 0
⎝             ⎠
0
⎝         ⎟
⎠
0                             w

w                      1
4w 2
( )                (                  )
y= w                                y=1
=   ∫       4y dy + ∫      dy = y 4                               + 4w 2 ln y                    =
3

0               w
y                              y= 0                                y= w

(                  )               (
= w 2 − 0 + 4w 2 ln1 − ln w = w 2 − 4w 2 ln w.        )
Hence,
1   1
fw ( w ) = FW ( w ) = 2w − 8w ln w − 4w 2
′                                                                      ⋅   =
w 2 w
= 2w − 8w ln w − 2w = −8w ln w.

Posted June 12, 2005
In Section 1, the statement of the Binomial Theorem should be:
n
⎛ n⎞
( a + b )n = ∑ ⎜ ⎟ a k b n − k ,
k=0 ⎝ k⎠

i.e., the sum should start from k = 0, not k = 1.

Posted May 20, 2005
On the second page of Section 4: Risk and Insurance, page 77 of the manual, the
formula for the variance should be:
( )                          ( )
Var ( X ) = E X 2 − ( E ( X )) = q ⋅ E B 2 − ( q ⋅ E ( B )) = q ⋅ E B 2 − q ⋅ (1 − p ) ⋅ ( E ( B )) =
2                                           2
( )                               2

= q ⋅ ( E ( B ) − ( E ( B )) ) + pq ⋅ ( E ( B ))                             = q ⋅ Var ( B ) + pq ⋅ ( E ( B )) .
2                          2                      2                                              2

( i.e., + pq ⋅ ( E ( B))               2
instead of − pq ⋅ ( E ( B ))
2
)

Posted May 18, 2005
In Exercise 2.13, the five answer choices should be:
−0.2
A. 10y 0.8 e−8 y
−0.8
B. 8y −0.2 e−10 y
C. 8y −0.2 e− ( 0.1y)
1.25

D. ( 0.1y )         e−0.125⋅( 0.1y)
1.25                          0.25
E. 0.125 ⋅ ( 0.1y ) e− ( 0.1y)
0.25        1.25

(parentheses were missing in answers D and E)

Posted May 18, 2005
The last line of the solution of Problem No. 11, Practice Examination No. 3, should
be:
⎛ 4   5⎞
(                               )
E max (Y + Z, 2 ) − 2 X = 0 = 1⋅ ⎜   + ⎟ + 2⋅
⎝ 27 27 ⎠
6 7
= .
27 9
6            5
(2 is multiplied by     , not by     ).
27           27

Posted May 18, 2005
The last line of the solution of Problem No. 26, Practice Examination No. 3, should
be:

( )
E X = 2 M X (t ) = 2 e
2    d2
dt        t =0
d 2 3t +t 2
dt          t =0
=
d
dt
(
( 3 + 2t ) e3t +t
2

t =0
)=

(
= 2e3t +t + ( 3 + 2t ) e3t +t
2            2           2

)   t =0
= 2 + 9 = 11.
(the superscript 2 was missing in the symbol of the second derivative)

Posted May 15, 2005
In Problem No. 30, Practice Examination No. 5, the figure illustrating the solution
should be:

y
y=x

1                                      x
Posted May 14, 2005
In Problem No. 25, Practice Examination No. 5, in the solution, this formula
Y − X1 Y2 − X2       Y − X100
Y −X= 1        +        + … + 100       ,
100       100           100
should replace
X − Y X − Y2          X −Y
Y −X= 1 1+ 2            + … + 100 100 .
100       100           100

Posted May 14, 2005
In Problem No. 9, Practice Examination No. 5, the solution is correct but the answer
choice should be E.

Posted May 14, 2005
In Problem No. 14, Practice Examination No. 4, the solution is correct but the

Posted May 14, 2005
The density of the log-normal distribution should be:
1 ( ln x − µ )
2

1         −
fX ( x ) =           e       2      σ2
.
xσ 2π

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