/Exam 4 Manual (Sixteenth Edition) sorted by page

Transcription

Errata for ASM Exam C/4 Study Manual (Sixteenth Edition) Sorted by Page
1
Errata and updates for ASM Exam C/Exam 4 Manual (Sixteenth Edition)
sorted by page
Practice exam 1:9, 1:22, 1:29, 9:5, and 10:8 are defective in that none of the five answer choices is correct.
For 10:8, revised answer choices are provided. Note the change in wording to practice exam 1:1 (page 1247),
1:14 (page 1250), 1:15 (page 1251), 1:19 (page 1252), 1:24 (page 1253), 1:25 (page 1253), 1:28 (page 1254), 3:5
(page 1270), 5:15 (page 1293), 6:6 (page 1300), 8:20 (page 1326), 10:8 (page 1343), and 11:30 (page 1359).
[3/25/2014]
On page 11, in the third displayed line of Section 1.4, change e t x to e t X .
[6/12/2014]
On page 23, in the solution to exercise 1.15, the second displayed line should be
2
z 2 /2
z2
z
2
0.16 = exp − (ln 5 − ln 2) = exp
ln 2/5 =
2
2
5
[8/27/2013]
On page 30, at the end of the second line of the solution to Example 2A, add “be” after “will”.
[9/29/2013]
On page 47, in the solution to exercise
2.1, on the last line, a − 1 is missing from the numerator of the fraction.
p
4
The numerator should be (4/3) 0.35 − 1.
[1/31/2014]
On page 53, two lines above “Continuity correction”, change “call” to “called”.
[6/27/2014]
On page 97, on the first line of the answer to Example 6F, the first sentence is inaccurate; E[(X − 500)+ ] is not
calculated previously. However, it is easy to calculate: E[(X − 500)+ ] = 0.2(1000 − 500) + 0.1(2000 − 500) = 250.
[2/22/2014]
On page 97, 2 lines after the answer to Example 6H, change θ − d to θ = d .
[3/27/2014]
On page 101, in exercise 6.4, in the second bullet, change r % to 100r %.
[3/27/2014]
On page 110, in the solution to exercise 6.2, on the second displayed line, place a parenthesis around
[10/14/2013] On
20,000
.
20,000 + d 2
page 147, in Table 83, in item #3 under Tail Weight Measures, change “heavier tail” to “lighter tail”.
[2/5/2014]
On page 163, in exercise 9.13, although the exercise can be solved mechanically, the values in the table are
impossible if losses are nonnegative. If F (100) = 0.3, then E[X ∧ 100] ≥ (1 − 0.3)(100) = 70.
[4/1/2014]
On page 169, in the solution to exercise 9.4, on the first line, add “is” after “it”.
[6/2/2014]
On page 174, in the solution to exercise 9.19, on the last displayed line, there should be a 0.8 on the left hand
side, so that the part preceding the first equals sign is
0.8(E[X ∧ 5000] − E[X ∧ 100])
[6/30/2013]
On page 197, in the solution to exercise 11.1, the answer key should be (B) rather than (D).
[7/11/2013]
On page 198, in the solution to exercise 11.4, on the first line, change “a = λ and b = 0“ to “a = 0 and b = λ”.
[9/10/2014]
On page 204, in the solution to exercise 11.26, on the third line, add “1 plus” before the second “mean”: “. . . so
its mean is 1 plus the mean of an unshifted . . . ”.
[4/1/2014]
On page 237, in the warning box on the second line, change “variance distribution” to “variance formula”.
[8/27/2013]
On page 238, in exercise 14.3(v), on the second line, change the period in 10.000,000 to a comma.
[8/1/2014]
On page 273, in the solution to exercise 15.29, on the third line from the end, change 1.3965 to 1.9501.
Updated 10/12/2014
2
[6/17/2013]
On page 286, in the solution
to exercise 16.7, on the fourth displayed line, delete the first F , so that the left side
reads 50,000 1 − F (50,000) .
[7/3/2014]
On page 331, in the solution to exercise 19.1, on the last line, change 50 to 40.
[6/17/2014]
On page 332, in the solution to exercise 19.3, on the last line, change X > 3000 to S > 3000.
[4/18/2014]
On page 339, in exercise 20.11, delete the first line with “s”.
[6/23/2014]
On page 359, in exercise 21.9, on the sixth line, replace MSEβ (ψ)/ MSEβ (φ) with MSEψ (β )/ MSEφ (β ).
[6/23/2014]
On page 363, in the solution to exercise 21.9, on the first displayed line, replace MSEβ (ψ) with MSEψ (β ). On
the second displayed line, replace
M S Eβ (φ) with MSEφ (β ).
[7/29/2014]
On page 380, in exercise 23.11, on the last line, delete the hat on top of v (θ ).
[6/18/2014]
On page 384, in the solution to exercise 23.11, on the second line, change “θn is binomial” to “θn is a binomial
proportion”.
[7/29/2014]
On page 384, in the solution to exercise 23.13, on the fifth line, change q =
[6/19/2014]
On page 401, in exercise 24.29, on the fourth line, change “states” to “times”.
[4/23/2014]
On page 470, in the solution to exercise 27.9, on the third displayed line, change Fn (17) = Fn (16) to Fn (17) −
Fn (16).
[8/5/2013]
On page 480, in the second table, on the first line of the heading, change both “Age 64”’s to “Age 65”.
[8/31/2013]
On page 481, one line after equation (28.4), change the line to:
Since qˆ j = nd j /e j , this can also be written as
d
most common case), then V
ar(qˆ j ) =
[8/5/2013]
d j (e j /n − d j )
d j (e j − d j )
e j3
(e j /n )3
1
7
to q = 67 .
. In particular, if n = 1 (one-year mortality rate, the
.
On page 481, one line above Section 28.2, the displayed line should be changed to
v
t (4/9)(5/9)
54/12
nd j
= 0.23424
[8/31/2013]
On page 487, change equation (28.2) to qˆ j =
[6/11/2014]
On page 493, in the solution to exercise 28.5, on the second line, replace “toal” with “total”.
[2/2/2014]
On page 494, in the solution to exercise 28.8, replace the top 3 lines of the page with
ej
.
Total exposure in months is 24(36) + 41(34) + 24 + 12(29) + 8 + 27 + 22(25) = 3215. The estimate of 3q 25 , by
formula (28.2), is
nd j
3(1)
=
= 3/(3215/12) = 0.011198
3q 25 =
ej
3215/12
The standard deviation, by formula (28.4), is
q
v
u q (1 − q ) v
t (0.011198)(1 − 0.011198)
t j
j
d
Var(3 qˆ25 ) =
=
= 0.011135
e j /n
(3215/12)/3
Updated 10/12/2014
3
[6/11/2014]
On page 494, in the solution to exercise 28.9, on the fifth line, replace (33-5,38-5,x ) with (33.5,38-5,s ).
[8/12/2013]
On page 494, the solution to exercise 28.12 is incorrect. The correct solution is
P41 = 125 + 10 − 5 − 25 − 1 + 150 = 254. We assume uniform entry and withdrawal, so exact exposure is 254 +
0.5(8 − 10 − 2) = 252. The mortality estimate is 1 − e −2/252 = 0.0079051 .
[8/12/2013]
On pages 532–533, the solution to exercise 30.32 is incorrect. The correct solution is
The sample mean is 2000. The second moment is
5002 + 10002 + 15002 + 25002 + 45002
= 6,000,000
5
Setting mean and second moment of lognormal equal to these,
σ2
= ln 2000 = 7.6009
2
2µ + 2σ2 = ln 6,000,000 = 15.60727
µ+
Double the first equation and subtract from the second.
σ2 = 15.60727 − 2(7.6009) = 0.40547
0.40547
µ = 7.6009 −
= 7.3982
2
ln 4500 − 7.3982
Pr(X > 4500) = 1 − Φ
p
0.40547
= 1 − Φ(1.59) = 1 − 0.9441 = 0.0559
(B)
[6/24/2014]
On page 571, in exercise 32.15, in answer choice (B), the exponent 3 in the denominator should be outside the
parenthesis, so that the denominator is f (100; θ )3 .
[5/8/2014]
On page 618, in exercise 33.36, the correct solution, starting with the second line, is
ˆ=P
α
n
ln(1 + xi )
P
If the sample mean
Pn xi /n = x¯ , then at least one of the xi s is greater than or equal to x¯ . The other xi s are
ˆ → 0. (A)
nonnegative. So i =1 ln(1 + xi ) ≥ ln(1 + x¯ ). As x¯ → ∞, ln(1 + x¯ ) → ∞. Therefore α
[10/6/2013]
On page 635, replace all lines after the S (10,000; α) line of the answer to Example 34E with:
Let m be the the number of observations not exceeding 10,000. Dropping the +1 in the exponent of the denominator, which is a constant, the likelihood function is
L (α) = Qm
αm (55008α )
(8−m)α )
α
i =1 (5000 + x i ) (15,000
where the xi are the uncensored observations. We log and calculate the second derivative.

m
X
l (α) = m ln α + α 8 ln 5500 −
ln(5000 + xi ) − (8 − m) ln 15,000
i =1

m
X
dl
m
=
+ 8 ln 5500 −
ln(5000 + xi ) − (8 − m) ln 15,000
dα
α
i =1

Updated 10/12/2014
4
d2 l
m
=−
2
dα
α2
The expected value of m is 8 times the probability that an observation is less than 10,000 given that it is greater
than 500, or 1 − (5500/15,000)α , so
d2 l
8 (1 − (5500/15,000)α )
E −
=
dα2
α2
and the variance is the reciprocal of this expression. The variance can be estimated by using the estimate for
α, obtained by setting the first derivative to 0 and setting the xi s equal to the 5 observations and m = 5:
5
− (8 ln 5500 − ln 6000 − ln 7000 − ln 8000 − ln 11,000 − ln 13,000 − 3 ln 15,000) = 0
α
5
ˆ =−
α
= 0.949465
−5.26612
0.9494652
d
ˆ =
V
ar(α)
= 0.183448
8 (1 − (5500/15,000)0.949465 )
[8/4/2014]
On page 634, 2 lines from the end of the page, change e −8250/10,000 to e −10,000/8250 .
[1/7/2014]
On page 645, in exercise 34.2II, change Crarner to Cràmer.
[6/26/2014]
On page 655, in the solution to exercise 34.7, delete “negative”.
[6/27/2014]
The length of the 2-sided confidence interval for the proportion is 2 times 1.96 (the 97.5th percentile of a standard normal distribution) times the standard deviation of the estimate for the proportion. We will use the delta
method. In the previous exercise, we estimated θˆ = 125. Let g (θ ) be Pr(X > 250 | θ ). Then
g (θ ) = S (250 | θ ) = e −250/θ
By the delta method
2
Var g (θˆ ) ≈ Var(θˆ ) g 0 (θˆ )
Since θˆ is the sample mean, its variance is the variance of the distribution divided by the size of the sample.
The variance of an exponential with mean θ is θ 2 , so the variance of θˆ is
2
θ2
125
Var(θˆ ) =
≈
100
10
where as usual we approximate θ with its estimate.
250 −250/θ
e
θ2
250 −250/125
2 −2
g 0 (θˆ ) =
e
=
e
1252
125
2

125
2 −2 2
Var g (θˆ ) ≈
e
10
125
2
= 0.2e −2
g 0 (θ ) =
The standard deviation of g (θˆ ) is 0.2e −2 , and the length of the confidence interval is 2(1.96)(0.2e −2 ) = 0.1061 .
(E)
Updated 10/12/2014
5
[6/30/2014]
On page 745, in exercise 39.1, at the end of the last bullet, change “good to fit” to “goodness of fit”.
[6/30/2014]
On page 755, in exercise 39.23, after the table, add “A Pareto distribution is fit to the data.”
[8/27/2013]
On page 761, in the solution to exercise 39.6, on the sixth line, change 4041 to 4047. On the last line, change
4046 to 4047 in two places, change 990 to 900.and change the final answer to 258.
[6/19/2014]
On page 770, in the paragraph beginning “Thus if an exponential model”, on the fourth line, delete one “that”.
[4/23/2014]
On page 772, on the second-to-last line of the answer to Example 40C, replace 2(−17.8631 + 17.8625) with
2(17.8631 − 17.8625).
[8/4/2013]
On page 787, in question 41.9, none of the answer choices is correct.
[5/19/2014]
On page 790, in the solution to exercise 41.5, on the first displayed line, change F ( x ) to F ∗ (x ).
[8/4/2013]
On page 792, in the solution to exercise 41.9, the solution, starting with the fourth line, should be corrected to
the following:
3θ 3
(θ + 260)4
Multiplying these together and logging,
891
+ 3 ln θ − 4 ln(θ + 260) + constant
θ
dl
891
4
=
−
2
dθ
θ
θ + 260
l (θ ) = −3 ln θ −
Multiplying out by θ 2 (θ + 260):
891θ + 231,660 − 4θ 2 = 0
p
891 + 4,500,441
ˆ
θ=
= 376.553
8
[6/9/2014]
On page 793, in the solution to exercise 41.13, in the table, change “Transformed gamma” to “Generalized
Pareto”.
[7/22/2013]
On page 842, in the solution to exercise 44.2, on the third displayed line, the expression between the two equal
(2000)(0.052 )
signs should be
.
3
[10/12/2014] On
page 845, in the solution to exercise 44.18, replace the first line on the page with
yp2 =
2831.25(0.12 )
= 3.539
8
[9/16/2013]
On page 895, delete exercise 46.31.
[5/28/2014]
On page 963, exercise 51.26 is a duplicate of exercise 51.7.
[8/30/2014]
On pages 1034–1035, in the solution to exercise 53.17, replace the 9 lines starting with “We want to calculate”
with
We want to calculate
Pr(p < 0.1 | 1, 0) =
Pr(1, 0 and p < 0.1)
Pr(1, 0)
Updated 10/12/2014
6
=
Pr(1, 0 and p < 0.1)
Pr(1, 0 and p < 0.1) + Pr(1, 0 and p ≥ 0.1)
We will obtain the numerator by integrating, from 0 to 0.1, Pr(1, 0) = p (1−p ) over the density function of p . We
will obtain the second term of the denominator by integrating the same integrand from 0.1 to 0.2.
Replace the left side on the next line with Pr(1, 0 and p < 0.1. Replace the left side three lines after that line
with Pr(1, 0 and p ≥ 0.1). Replace the final line of the solution (page 1035) with
Pr(p < 0.1 | 1, 0) =
37
37/1200
37
=
=
37/1200 + 0.0575 37 + 69
106
[7/30/2013]
On page 1051, in exercise 54.11, on the second to last line, change 2000 to 2010.
[7/30/2013]
(B)
We will use one month as our exposure unit. Then
µ = E[θ ] = 0.05
a = Var(θ ) = 0.004
v = E[θ (1 − θ )] = E[θ ] − E[θ 2 ] = 0.05 − (0.052 + 0.004) = 0.0435
v
k = = 10.875
a
There are 28 months of exposure and 1 claim per 28 months, so
28
= 0.720257
28 + 10.875
PC = 0.720257(1/28) + (1 − 0.720257)(0.05) = 0.039711
Z=
Annualizing the estimate, we get 12(0.039711) = 0.47653 .
[5/6/2014]
On page 1069, in the solution to exercise 55.16, on the second line, change 0.2(0.5) + 0.8(3) to 0.8(0.5) + 0.2(3).
[7/31/2013]
Bayesian and Bühlmann estimation are equivalent, so we can deduce Bühlmann credibility factors using the
conditional expectation of X 2 given X 1 . Since E[X 2 | X 1 = 3] = 1.00, and 1.00 = 0.2(0.5) + 0.8(3), we see that
the credibility factor for one observation is 0.2. Then 1/(1 + k ) = 0.2 implying k = 4. The expected value of
the process variance divided by the variance of the hypothetical means is 4. Then the expected value of the
process variance is 4(6) = 24 . (D)
[6/26/2014]
On page 1144, in the solution to exercise 59.16, on the second-to-last line, change 17,0460,223 to 17,046,223.
[6/17/2014]
On page 1150, on the first displayed line, change the integrand to
5
du
u 1.5
[8/28/2013]
On page 1154, in exercise 60.1, on the second line, change “product” to “produce”. On the third line, change
y = 0.79 to u = 0.21.
[6/13/2014]
On page 1159, exercise 60.19 is a duplicate of exercise 60.2.
[8/28/2013]
On page 1164, in the solution to exercise 60.1, on the second line, change ln u to ln(1 − u ).
Updated 10/12/2014
7
p
p
2
3
0.125 to 0.125.
[7/9/2014]
On page 1167, in the solution to exercise 60.17, on the last line, change
[8/28/2013]
On page 1173, replace the last three lines of the answer to Example 61A with
The tables tell us that for a Pareto distribution, VaRp (Z ) = θ (1 − p )−1/α − 1 , so
0.488 → 800(0.512−1/3 − 1) = 200
The average is (13,986.62 + 4,437.39 + 200)/3 = 6208 .
[9/16/2013]
On page 1175, 7 lines above Example 61C, change 1.92401 to 14.92401. On the next line, change 4.47039 to
14.47039. Note that the inverse of 0.99 appears in the normal distribution table at the bottom as 2.326, so
perhaps −2.326 should be used in the calculation on that line instead of −2.33. However, the final answer is
the same whether −2.326 or −2.33 is used.
[9/9/2013]
On page 1176, in the paragraph after the table of the answer to Example 61C, on the second line, change “it
is at least equal to” to “it is less than”. On the third line, change “surrender in the first year” to “convert first
year”.
[1/14/2014]
On page 1176, two lines from the bottom, change 1 − 0.8 − 0.2 to 1 − 0.08 − 0.02.
[8/1/2013]
On page 1179, on the second line of Section 61.4, change µ + σxi to µ + σni . In Example 61H on the first line,
change “a normal distribution with µ = 5 and σ2 = 16” to “a standard normal distribution”.
[8/1/2013]
On page 1180, in Example 61I on the first line, change “a normal distribution with µ = 5 and σ2 = 16” to “a
standard normal distribution”.
[6/23/2014]
On page 1185, in exercise 61.18, on the first line, change X (l ) to X (1).
[9/24/2013]
On page 1186, in the solution to exercise 61.3, on the last line, change 3 to 4 .
[7/16/2013]
On page 1187, the solution to exercise 61.9 is incorrect. The correct solution is e −3 = 0.0498.
(0.64)(0.5) = 0.32
(0.32)(0.25) = 0.08
(0.08)(0.75) = 0.06
(0.06)(0.50) = 0.03
It took 5 numbers to get below e −3 , so we generate 4 . (C)
[6/16/2014]
On page 1187, in the solution to exercise 61.10, on the last displayed line, change 2.842434 to 2.482433.
p
On page 1189, in the solution to exercise 61.19, on the last line, change 2e to (2e ).
[6/17/2014]
On page 1213, in exercise 63.16, two lines after the enumerated list, change “longer times” to “shorter times”.
[6/2/2014]
ˆ 2 to s 2 .
On page 1226, 2 lines above Example 64F, change σ
[12/4/2013]
On page 1247, in question 1, on the last line, change “counts” to “costs”.
[6/10/2013]
On page 1250, in question 14, delete the second-to-last line beginning “The automobile collision”. On the last
line, change “payments” to “losses”.
[6/10/2013]
On page 1251(i), change x15 to x5 .
[6/11/2013]
On page 1252, in question 19, on the fourth line, add “95%“ before “log-transformed”.
[6/13/2013]
On page 1253, in question 24, after the sentence under the table, “Credibility . . . methods”, add “Annual claim
counts for each individual are assumed to follow a Poisson distribution.”
[12/3/2013]
Updated 10/12/2014
8
[9/18/2013]
On page 1253, in question 25, on the second line, change 76 to 66.
[6/11/2013]
On page 1254, in question 28(iii), change 80% to 60%.
[8/27/2013]
On page 1270, in question 5, on the second row below the heading in the table (the one starting with “2”),
change the date of policy issue from 5/1/2010 to 5/1/2000. On the last line, change q45 to q35 .
[6/11/2013]
On page 1293, in question 15, on the second line, change 720 to 240.
[6/14/2013]
On page 1300, in question 6, in the table, on the second line, change the date of policy issue from 6/1/2012 tgo
6/1/2002.
[6/11/2013]
On page 1326, in question 20, change the answer choices to
(A) 19
[1/27/2014]
(B) 20
(C) 21
(D) 22
(E) 23
On page 1343, question 8, replace 15 in the table with 19. Replace the answer choices with
(A) 43.8
(B) 44.1
(C) 44.4
(D) 44.7
(E) 45.0
[6/21/2013]
On page 1353, in question 10, on the last line, change the comma in 1.841498 to a period.
[6/11/2013]
On page 1359, in question 30, on the second line, delete “negative”.
[6/10/2013]
On page 1377, in the solution to question 7, on the second line, replace e −
P
P 1
.
line, replace θ xi with θ
xi
[6/10/2013]
On page 1377, in the solution to question 9, the last line should be
1 − p0 − p1 − p2 = 1 −
P
θ xi
with e −θ
P
1/xi
. On the third
2
2 2 2
1
1
3
2·3 1
3
−2
−
= 0.738281
4
4
4
2!
4
4
None of the five choices is correct.
[8/12/2013]
On page 1378, in the solution to question 11, on the sixth line, change e −0.175 to e 0.175 .
[8/12/2013]
On page 1379, in the solution to question 13, on the fourth line, put a negative sign before 0.203296 and change
x at the beginning of the line to a .
[6/5/2014]
On page 1379, in the solution to question 16, on the second line, in the exponent on (1 − e −1000/θ ), change 94
to 84.
[6/11/2013]
On page 1381, in the solution to question 22, change the last three lines to
60(7 − 8.375)2 + 20(12.5 − 8.375)2 − 85.8889
= 12.2620
80 − (602 + 202 )/80
85.8889
k=
= 7.0045
12.2620
60
ZA =
= 0.8955
(C)
60 + 7.0045
a=
None of the five choices is correct.
[6/11/2013]
On page 1382, in the solution to question 25, on the second line from the end, change 0.48741 to 0.23757.
[9/18/2014]
On page 1382, in the solution to question 26, on the first line, change 1000 to 1600 in two places.
[6/11/2013]
On page 1383, change the solution of question 28 to
Updated 10/12/2014
9
Annual payments are 0.6(X ∧ 10,000 − X ∧ 1,000), and we use the tables to evaluate the expected value of this.

α−1
3
θ
θ
8000
8000
E[X ∧ 1,000] =
1−
=
1−
= 793.78
α−1
θ + 1000
3
9000

3
8000
8000
1−
= 2432.56
E[X ∧ 10,000] =
3
18000
The answer is 0.6(2432.56 − 793.78) = 983.26 . (A)
[8/27/2013]
On page 1383, the solution to question 29 is incorrect. The correct solution is
Let q be the 95th percentile. Let’s calculate q .
1 − 0.6e −q /10 − 0.4e −q /20 = 0.95
0.6x 2 + 0.4x = 0.05 where x = e −q /20
p
−0.4 + 0.42 + 4(0.6)(0.05)
x=
= 0.107625
1.2
q = −20 ln x = 44.5820
TVaR0.95 (X ) = q + e (q ). We’ll calculate e (q ) using
R∞
R∞
x f (x ) dx
S (x ) dx
q
q
e (q ) =
=
0.05
0.05
The numerator is
Z∞
∞
(0.6e −x /10 + 0.4e −x /20 )dx = −10(0.6)e −x /10 − 20(0.4)e −x /20 q = 6e −q /10 + 8e −q /20 = 0.930501
q
So
TVaR0.95 (X ) = 44.5820 +
[6/11/2013]
0.930501
= 63.19
0.05
(E)
On page 1385, change the solution to question 35 to
1 plus the coefficient of variation squared can be expressed as
1 + C VX2 = 1 +
Var(X ) E[X ]2 + Var(X ) E[X 2 ]
=
=
E[X ]2
E[X ]2
E[X ]2
For a Weibull distribution with τ = 0.5,
E[X ] = θ Γ (1 + 1/τ) = θ Γ (3) = 2θ
E[X 2 ] = θ 2 Γ (1 + 2/τ) = θ 2 Γ (5) = 24θ 2
E[X 2 ] 24
=
=6
E[X ]2 22
The standard for full credibility is
2
1.96
λ =
(6) = 6403
0.06
F
(A)
The answer key on page 1375 should also be corrected.
Updated 10/12/2014
10
[6/11/2013]
On page 1392, in the solution to question 17, on the third displayed line at the right, remove the negative sign
in front of 0.9868.
[3/22/2014]
On page 1400, in the solution to question 5, change the two “Age 45”s in the heading of the table to “Age 35”.
Change the three subscripts “45” in the two lines after the table, and the subscript “40” in the displayed formula, to “35”.
[6/21/2013]
On page 1402, in question 10, on the last line, change the comma in 1.841498 to a period.
[6/11/2013]
On page 1424, in the solution to question 2, on the last line, change the equals sign before 13002 to a minus
sign.
[6/11/2013]
On page 1428, in the solution to question 15, replace the sentence starting 5 lines from the end beginning with
“For #6” with
For #6, death occurs at age 29 so the scheduled 10 months of exposure are lost.
Replace the last two sentences starting with “So the total adjustment” with
So the total adjustment to actuarial exposure in months is −6 − 6 − 10 + 6 − 6 = −22, or 1 56 years. There are 2
deaths at age 30 (#3 and #7), so the actuarial estimate of q30 is 2/(120 − 1 56 ) = 0.01693 .
[5/19/2014]
On page 1461, the solution to question 8 has typos and is unclear. Here is a clearer solution:
The function transforming X to Y is y = g (x ) = x −1 = 1/x . Notice that x = g −1 (y ) = 1/y . The derivative of
g −1 (y ) is
dg −1
1
=−
dy
y2
By the formula for transforming densities of random variables,
1
(1/(100y ))4 e −1/100y 1
f Y (y ) = f X (1/y )
=
y2
(1/y )Γ (4)
y2
−1/100y
1 e
=
1004 Γ (4)y 5
Not surprisingly, inverting X results in an inverse gamma. If you check the tables, you will see that the new
parameters are α = 4 and θ = 1/100. The mode of an inverse gamma according to the tables is θ /(α + 1) =
1/(100 · 5) = 1/500 . (B)
If you did not recognize the density as an inverse gamma, you could still find the mode by differentiating it
and set the derivative equal to 0. It is easier to differentiate the log:
1
+ ln y − ln Γ (4) − 2 ln y
100y
400
1
1
5
1
=−
+
− =− +
=0
100y 100y 2 y
y 100y 2
ln f Y = −4 ln 100y −
d ln f Y
dy
1
−5y +
=0
100
y=
[1/29/2014]
1
500
On page 1470, in the solution to question 30, on the last line, interchange 0.213061 and 0.5.
Updated 10/12/2014
[1/27/2014]
11
On page 1473, in the solution to question 5, on the second-to-last line, replace 0.03713 with 0.03488. Replace
the last line with
q
d
V
ar(5 qˆ60 ) =
v
u
t
10
= 0.03488
4002
(1 − 0.117503)2 (52 )
The 95% confidence interval is 0.117503 ± 1.96(0.03488) = (0.049, 0.186) .
None of the answer choices is correct.
[6/25/2013]
On page 1485, in the solution to question 4, on the fifth line, delete “given that it is greater than 1000”.
[1/27/2014]
On page 1486, the solution to question 8 is incorrect. The correct solution is
From the tabular values with d = 20.
E[X ] = E[X ∧ 20] + Pr(X > 20)e (20) = 19 + 0.9(200) = 199
We also have
E[X ] = E[X ∧ 50] + Pr(X > 50)e (50)
199 = E[X ∧ 50] + 180 Pr(X > 50)
(*)
Let’s get a lower bound for Pr(X > 50) by using the values of E[X ∧ 50].
Z 20
E[X ∧ 20] = 20 Pr(X > 20) +
x f (x ) dx
0
19 = 20(0.9) +
20
Z
x f (x ) dx
0
20
Z
x f (x ) dx = 1
0
E[X ∧ 50] = 50 Pr(X > 50) +
50
Z
x f (x ) dx
0
= 50 Pr(X > 50) +
20
Z
x f (x ) dx +
0
Z
50
x f (x ) dx
20
≥ 50 Pr(X > 50) + 1 + 20 Pr(20 < X ≤ 50)
= 50 Pr(X > 50) + 1 + 20 1 − 0.1 − Pr(X > 50)
= 30 Pr(X > 50) + 19
(**)
Plugging this into (*), we get
199 ≤ 30 Pr(X > 50) + 19 + 180 Pr(X > 50)
= 210 Pr(X > 50) + 19
210 Pr(X > 50) ≥ 180
Pr(X > 50) ≥
6
7
Then plugging into (**),
6
+ 19 = 44 57
7
E[X ∧ 50] ≥ 30
Updated 10/12/2014
12
To prove that this lower bound is attained, here is a discrete random variable satisfying the question’s assumptions with E[X ∧ 50] = 44 75 :
x
Pr(X = x )
10
20
230
0.1
6
7
− 0.1
6
7
[6/17/2014]
On page 1508, in the solution to question 30, on the first line of the page, change 0.109329 to 0.019329.
[6/11/2013]
On page 1522, in the solution to question 28, on the third line from the end, change the second = to −.
p
p
On page 1581, in the solution to question 1, on the fourth line, change α1 − F (u ) to α 1 − F (u ).
[7/14/2013]
Updated 10/12/2014