Dr. Krzysztof Ostaszewski, FSA, FSAS, CERA, CFA, MAAA
http://www.krzysio.net
E-mail: krzysio@krzysio.net
Errata for the Solutions Manual for Bowers' et al Actuarial Mathematics, 2007 Edition
Posted April 7, 2012
While the solution of Exercise 2.15, part b, is not wrong, it is confusing because of
the way it uses the claim count random variable. This is a better, hopefully clear,
solution:
b. Let us write I for a random variable that is equal to 1 if there is a claim, and 0 if there is no claim. We have E ( S ) = E E ( S I ) = 0.96 ! E ( S I = 0 ) + 0.04 ! E ( S I = 1) = 0 + 0.04 ! E ( S I = 1) =
20000
30000
50000
10000 %
" 10000
= 0.04 ! $
! 80 +
! 35 +
! 25 +
!15 +
! 5 ' = 70000.
# 2
&
2
2
2
2
For each of the 80 contracts for amount 10,000, the variance is Var ( S ) = E Var ( S I ) + Var E ( S I ) =
(
)
(
)
(
)
10000 2
" 10000 %
= 0.04 !
+ 0.04 ! 0.96 ! $
(
1,293,
333.33,
# 2 '&
12
for each contract for amount 20,000, Var ( S ) = E Var ( S I ) + Var E ( S I ) =
2
(
)
(
)
20000 2
" 20000 %
+ 0.04 ! 0.96 ! $
(
5,173,
333.33,
# 2 '&
12
for each contract for amount 30,000, Var ( S ) = E Var ( S I ) + Var E ( S I ) =
2
= 0.04 !
(
)
(
)
30000 2
" 30000 %
+ 0.04 ! 0.96 ! $
(
11,640,000,
# 2 '&
12
for contracts for amount 50,000 Var ( S ) = E Var ( S N ) + Var E ( S N ) =
2
= 0.04 !
(
)
(
)
50000 2
" 50000 %
= 0.04 !
+ 0.04 ! 0.96 ! $
( 32, 333, 333.33,
'
# 2 &
12
and for contracts for amount 100,000 Var ( S ) = E Var ( S I ) + Var E ( S I ) =
100000 2
100000
= 0.04 !
+ 0.04 ! 0.96 !
" 129, 333, 333,
12
2
so that the total variance is 80 !1, 293, 333.33 + 35 ! 5,173, 333.33 + 25 !11, 640, 000 +
+ 15 ! 32, 333, 333.33 + 5 !129, 333, 333 " 1707200000.
2
(
)
(
)
Posted April 7, 2012
The value of probability calculated in the last step of the solution of Exercise 2.10
should be: 0.07477068, not 0.7477068, i.e., the solution has a typo, missing a zero
after the decimal point.
Posted April 7, 2012
The ending part of the solution of Exercise 2.13, part b, should be:
Therefore, for retained business E ( S ) = 8000 !1! 0.02 + 3500 ! 2 ! 0.02 + 2500 ! 3 ! 0.02 + 2000 ! 5 ! 0.02 = 650, and Var ( S ) = 8000 !12 ! 0.02 ! (1" 0.02 ) + 3500 ! 2 2 ! 0.02 ! (1" 0.02 ) +
+ 2500 ! 32 ! 0.02 ! (1" 0.02 ) + 2000 ! 5 2 ! 0.02 ! (1" 0.02 ) = 1852.20.
The total coverage in the plan is calculated in the book as 35,000 units, while the retained coverage is 8000 !1 + 3500 ! 2 + 2500 ! 3 + 2000 ! 5 = 32, 500. The total amount reinsured is 35, 000 ! 32, 500 = 2, 500. The reinsurance cost is 2, 500 ! 0.025 = 62.50. The total cost is S + 62.50. The probability we are seeking is " S ! 650 762.50 ! 650 %
Pr ( S + 62.50 > 825 ) = Pr ( S > 762.50 ) = $
>
'=
# 1890
1852.20 &
" 762.50 ! 650 %
= 1! ( $
) 0.004447422.
# 1852.20 '&
In the printed solution, the variance calculation omits multiplication by 0.98, which is, of course, a standard part of Bernoulli Trial variance calculation. Posted April 24, 2011
In the solution of Exercise 11.7, the value of 10,000 should be replaced by 15,000,
and the value of 25,000 should be replaced by 40,000. The formula displayed should
be:
1
" #36
k+
)
S
(% ) &
0.10 ! $ v 2 ! 1 p35
! ( 40, 000 ! 35+k #15, 000 !1.05k + .
k+
S35
'
*
k=0
2
Posted April 3, 2011
In the solution of Exercise 11.9, the value of 8000 should be replaced everywhere by
2% of 450,000, i.e., 9000, and the value of 720 should be replaced everywhere by 2%
of 42,000, i.e., 840.
Posted November 2, 2010
Solutions to 6.4b and 6.4c should be:
b. Let P be the premium rate sought. We have L = e!" T ! PaT and this is a decreasing
function of T. The 50-th percentile of L is found at the value corresponding to the 50-th
ln 2 ln 2
=
! 34.66 years. Based on this
percentile of T , which is ln2 times the mean, or
µ
0.02
0=e
!0.06"
ln 2
0.02
!0.06"
!P"
1! e
0.06
ln 2
0.02
so that
P=
e
!0.06"
1
2 !3
8 = 6 # 0.008571.
=
!3 =
7 100 7 "100
1! 2
"
8 6
0.06
ln 2
0.02
!0.06"
ln 2
0.02
1! e
0.06
c. Let P be the premium rate sought. With zero force of interest e!" T = 1 and the
prospective loss function at policy duration 0 becomes
T
L = 1 ! " P dt = 1 ! PT .
0
Then
"1
%
"
1%
0.50 = Pr L > 0 = Pr 1 ! PT > 0 = Pr $ > T ' = Pr $ T < ' .
P&
#P
&
#
(
This implies that
)
(
)
1 ln 2
1
µ
0.02
, or P =
=
! 0.028854.
is the median of T, i.e., =
P
µ
P
ln 2 ln 2
Posted September 19, 2009
In the solution of Exercise 3.4b the last formula should be:
11 11 7
11 15
44 45 1
s! (1) = " + " = " +
="
+
=
> 0.
6 4 8
6 8
24 24 24
instead of
11 11 7
11 37
176 111
65
" 1%
s! $ ' = ( + (
=( +
=(
+
=(
# 2&
6 8 32
6 32
96 96
96.