Solutions to the
Spring 2016
CAS Exam S
There were 45 questions in total, of equal value, on this 4 hour exam.
There was a 15 minute reading period in addition to the 4 hours.
The Exam S is copyright 2016 by the Casualty Actuarial Society.
The exam is available from the CAS.
The solutions and comments are solely the responsibility of the author.
CAS Exam S
prepared by
Howard C. Mahler, FCAS
Copyright 2016 by Howard C. Mahler.
Howard Mahler
[email protected]
www.howardmahler.com/Teaching
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 1
1. Alice, Bob and Chris are hired by a firm to drive three vehicles - a bus, a taxi and a train (only one
driver is needed for each). Each employee has different skills and requires different amounts of
training on each vehicle. The cost to train each employee i for vehicle j, Ci,j, is an independent
exponential random variable with mean of 200.
To minimize the total cost of training, the firm uses the following procedure to assign the employees
to their vehicle:
• Alice is assigned to the vehicle which minimizes her training cost, CAlice,j.
• Bob is then assigned one of the two remaining vehicles which minimizes CBob,j.
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€
€
• Chris is then assigned the remaining vehicle.
Calculate the firm's expected total cost of training these three employees, if it uses this assignment
algorithm.
A. Less than 200
B. At least 200, but less than 300
C. At least 300, but less than 400
D. At least 400, but less than 500
E. At least 500
1. C. E[Training costs of Alice] = E[minimum of 3 exponentials] = θ/3.
E[Training costs of Bob] = E[minimum of 2 exponentials] = θ/2.
E[Training costs of Chris] = E[exponential] = θ.
Expected total training costs: θ(1/3 + 1/2 + 1) = (200)(11/6) = 366.67.
Comment: An example of a “Greedy Algorithm for Assignment” as discussed in Example 5.7 of
An Introduction to Probability Models, by Ross.
Solutions Spring 2016 CAS Exam S,
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Page 2
2. You are given the following information:
• Vehicles arrive at the drive-through bank at a Poisson rate of 30 per hour.
• On average, 10% of these vehicles require cash transactions.
Calculate the probability that at least 3 vehicles require cash transactions between 10 AM and noon.
A. Less than 0.915
B. At least 0.915, but less than 0.925
C. At least 0.925, but less than 0.935
D. At least 0.935, but less than 0.945
E. At least 0.945
2. D. The number of cash transactions over two hours is Poisson with mean: (2)(10%)(30) = 6.
Prob[at least 3] = 1 - f(0) - f(1) - f(2) = 1 - e-6 (1 + 6 + 62 /2) = 93.80%.
Solutions Spring 2016 CAS Exam S,
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3. You are given:
• The number of claims, N(t), follows a Poisson process with intensity:
λ(t) = t/2, 0 < t < 5
λ(t) = t/4, t ≥ 5
• By time t = 4, 15 claims have occurred.
Calculate the probability that exactly 16 claims will have occurred by time t = 6.
A. Less than 0.075
B. At least 0.075, but less than 0.125
C. At least 0.125, but less than 0.175
D. At least 0.175, but less than 0.225
E. At least 0225
3. B. The number of claims from time 4 to time 6 is Poison with mean:
5
6
∫ t / 2 dt + ∫ t / 4 dt = (52 - 42)/4 + (62 - 52)/8 = 3.625.
4
5
Prob[exactly one claim between time 4 and 6] = 3.625 e-3.625 = 9.66%.
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Solutions Spring 2016 CAS Exam S,
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4. You are given the following information about the rate at which calls come into
a customer service center:
• The customer service center is open 24 hours a day.
• Calls are received from four regions.
• The rate at which calls arrive is 10 calls per hour and follows a Poisson distribution.
• The probability that a call will be from a given region is independent of other calls and
is shown in the table below.
Region
Probability
1
0.2
2
0.3
3
0.1
4
0.4
Calculate the probability that in a one-hour time period at least one call will have been received from
each region.
A. Less than 0.25
B. At least 0.25, but less than 0.30
C. At least 0.30, but less than 0.35
D. At least 0.35, but less than 0.40
E. At least 0.40
4. E. The number of calls from each region are independent Poissons with means: 2, 3, 1, and 4.
Prob[at least one call from each region] =
Prob[> 0 from Region 1] Prob[> 0 from Reg. 2] Prob[> 0 from Reg. 3] Prob[> 0 from Reg. 4] =
(1 - e-2) (1 - e-3) (1 - e-1) (1 - e-4) = 50.98%.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 5
5. You are given the following information on a health insurance policy:
• Claims occur according to a Poisson process with mean λ = 10 per week.
• Claim amounts follow the Pareto distribution with probability density function
f(x) =
€
€
(4) (10004 )
,0<x
(x + 1000)5
Calculate the probability that the total claim amount in 52 weeks exceeds 200,000
using the Normal approximation.
€ than 0.02
A. Less
B. At least 0.02, but less than 0.03
C. At least 0.03, but less than 0.04
D. At least 0.04, but less than 0.05
E. At least 0.05
5. B. Looking in the tables attached to the exam, the Pareto Distribution has α = 4 and θ = 1000.
The Pareto has mean: θ/(α-1) = 1000/(4-1) = 333.333.
The Pareto has second moment:
2 θ2
(2) (10002)
=
= 1/3 million.
(α - 1)(α - 2) (4 -1) (4 - 2)
Thus the mean aggregate loss is: (52)(10)(333.333) = 173,333.
The variance of aggregate loss is: (52)(10)(1/3 million) = 173.333 million.
€ 52 weeks exceeds 200,000 is approximately:
Probability that the total €
claim amount in
200,000 - 173,333
1 - Φ[
] = 1 - Φ[2.026] = 2.14%.
173.333 million
€
Solutions Spring 2016 CAS Exam S,
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Page 6
6. You are given the following information:
• The lifetimes of all light bulbs follow the exponential distribution.
• A new incandescent light bulb has a hazard rate of 0.20 per year.
• The expected lifetime of an LED light bulb is twice the expected lifetime of a 2-year-old
incandescent light bulb.
Calculate the hazard rate per year for an LED light bulb.
A. Less than 0.05
B. At least 0.05, but less than 0.07
C. At least 0.07, but less than 0.09
D. At least 0.09, but less than 0.11
E. At least 0.11
6. D. The hazard rate for an Exponential Distribution is constant and equal to 1/mean.
The expected lifetime of a 2-year old incandescent light bulb is: 1/0.2 = 5,
same as that of a new one.
Thus LED light bulbs have mean: (2)(5) = 10, and hazard rate: 1/10.
Solutions Spring 2016 CAS Exam S,
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Page 7
7. You are given the following information:
• Consider a system of three independent components, each of which functions for an amount of
time (in months) uniformly distributed over (0, 1).
• Under current design, the system will fail if any of the 3 components fail.
• A new design was made such that the system will fail if 2 or more components fail.
Calculate the increase in expected system life in months under the new system design.
A. Less than 0.195
B. At least 0.195, but less than 0.215
C. At least 0.215, but less than 0.235
D. At least 0.235, but less than 0.255
E. At least 0.255
7. D. Under the current design, the expected lifetime is the minimum of three uniform distributions
over (0, 1): 1/(3+1) = 1/4.
Under the new design, the expected lifetime is the 2nd order statistic from three uniform distributions
over (0, 1): 2/(3+1) = 1/2.
The increase in expected system life is: 1/2 - 1/4 = 1/4.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 8
8. In an airport, you are given the following information:
• Taxis arrive in accordance with a Poisson process with rate of λT = 2 per minute.
• Customers arrive in accordance with a Poisson process with rate λC = 3 per minute.
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• A taxi will wait no matter how many taxis are present.
• An arriving customer that does not find a taxi waiting will leave.
Calculate the proportion of arriving customers that find taxis.
A. Less than 0.2
B. At least 0.2, but less than 0.4
C. At least 0.4, but less than 0.6
D. At least 0.6, but less than 0.8
E. At least 0.8
8. D. Let Pi = proportion of the time there are i taxis present.
The balance equations are:
2P0 = 3P1 . ⇒ P1 = (2/3) P0 .
2P1 = 3P2 . ⇒ P2 = (2/3) P1 = (2/3)2 P0 .
2P€
2 = 3P3 . ⇒ P3 = (2/3) P2 = (2/3)3 P0 .
In general,
Pi = (2/3)i P0 .
€
Since
€ the proportions add to one: 1 = P0 (1 + 2/3 + (2/3)2 + (2/3)3 + ...} = 3P0 .
⇒ P0 = 1/3.
The proportion of arriving customers that find taxis is: 1 - P0 = 2/3.
€
Comment: This is an example of a birth and death process. The rate at which the process goes from
state i to state i+1 must be the same as the rate at which the process goes from state i+1 to state i.
The proportions follow a Geometric Distribution with β = 2.
For a Birth-Death Process that has the rate of births λ and rate of deaths µ each constant,
P0 = 1 - λ/µ. In this case: 1 - P0 = λ/µ = 2/3.
Solutions Spring 2016 CAS Exam S,
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9. You are given the following information:
• A firm has modeled the annual movement of its workers using a Markov chain with
the following three states:
0. Fully Employed
1. Temporarily Disabled
2. Retired
• The associated transition matrix is:
⎛0.7 0.1 0.2⎞
⎜
⎟
P = ⎜0.4 0.3 0.3⎟
⎜
⎟
⎝0.0 0.0 1.0⎠
• Tom is Fully Employed with the firm this year.
Calculate the probability that Tom will be Fully Employed in two years.
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A. Less than 0.52
B. At least 0.52, but less than 0.54
C. At least 0.54, but less than 0.56
D. At least 0.56, but less than 0.58
E. At least 0.58
9. B. (1, 0, 0) P = (0.7, 0.1, 0.2). (0.7, 0.1, 0.2) P = (0.53, 0.10, 0.37).
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Solutions Spring 2016 CAS Exam S,
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10. You are given the following information:
• A taxi driver provides service for two cities, C1 and C2, only.
• If the taxi driver is in the city C1, the probability that he has to drive the customer to city C2
is 0.35.
• If the taxi driver is in the city C2, the probability that he has to drive the customer to city C1
is 0.75.
• The expected profit for each trip is:
• Within C1: $6
• Within C2: $5
• Between C1 and C2: $10
€ Calculate the long-term expected profit per trip.
€ A. Less than $7.50
€ B. At least $7.50, but less than $8.00
C. At least $8.00, but less than $8.50
D. At least $8.50, but less than $9.00
E. At least $9.00
10. B. Let the states of a Markov chain be the city in which the taxi starts a trip.
1 ⎛0.65 0.35 ⎞
The transition matrix for this Markov chain is: ⎜
⎟ .
2 ⎝0.75 0.25 ⎠
The equations for the stationary distribution are:
0.65π1 + 0.75π2 = π1 . 0.35π1 + 0.25π2 = π2 . π1 + π2 = 1. ⇒ π1 = 15/22, and π2 = 7/22.
€€
The average profit when the taxi starts in city 1 is: (0.65)(6) + (0.35)(10) = 7.4.
The average profit when the taxi starts in city 2 is: (0.25)(5) + (0.75)(10) = 8.75.
The long term average profit per trip is: (15/22)(7.4) + (7/22)(8.75) = 7.83.
Comment: Similar to 3, 1/01, Q.22.
With only two states: π1 = P21/(P21 + P12) and π2 = P12/(P21 + P12).
π 1 = 0.75/(0.75 + 0.35) = 15/22, and π2 = 0.35/(0.75 + 0.35) = 7/22.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 11
11. You are given the following information:
A three-state Markov Chain, with the following transition probability matrix, is used to model the
movement of policyholders between three States:
States
0
1
2
0
0.5
0.5
0.0
1
0.4
0.4
0.2
2
0.3
0.4
0.3
Calculate the stationary percentage of policyholders in State 2.
A. Less than 13%
B. At least 13%, but less than 23%
C. At least 23%, but less than 33%
D. At least 33%, but less than 43%
E. At least 43%
11. A. The balance equations are:
0.5π0 + 0.4π1 + 0.3π2 = π0.
0.5π0 + 0.4π1 + 0.4π2 = π1.
0.2π1 + 0.3π2 = π2. ⇒ π1 = 3.5 π2.
From the first equation: 0.5π0 = 0.4π1 + 0.3π2 = 1.7 π2. ⇒ π0 = 3.4 π2.
Also π0 + π1 + π2 = 1.
Thus, 3.4 π€2 + 3.5 π2 + π2 = 1. ⇒ π2 = 1/(3.4 + 3.5 + 1) = 1/7.9 = 0.1266.
€
⇒ π0 = 3.4/7.9 = 0.4304. π1 = 3.5/7.9 = 0.4430.
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Solutions Spring 2016 CAS Exam S,
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12. You are given the following information:
• A 3-year term insurance policy on a life age (80) provides for a death benefit,
payable at the end of the year of death.
• The death benefit is 1000t, t ∈ {1, 2, 3}, where t is the year of death.
• This policy is purchased by a single premium, P, at time 0.
• If (80) lives to age 83, the single premium is returned without interest.
• Mortality rates follow€the Illustrative Life Table.
• i = 0.10
Calculate P using the equivalence principle.
A. Less than 830
B. At least 830, but less than 850
C. At least 850, but less than 870
D. At least 870, but less than 890
E. At least 890
12. E. APV of future benefits is: 1000 v q80 + 2000 v2 2 q80 + 3000 v3 3 q80 =
(1000/1.1)
3,914,365 - 3,600,038
3,600,038 - 3,284,542
+ (2000/1.12 )
3,914,365
3,914,365
3,284,542 - 2,970,496
= 387.06.
3,914,365
€
€
APV of future premiums including the possible return of the single premium is:
2,970,496
P - P v3 3 q80€= P (1 /1.13 ) = 0.42985 P.
3,914,365
+ (3000/1.13 )
Thus, 0.42985 P = 387.06. ⇒ P = 900.45.
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Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 13
13. Let Y1 , ... ,Yn denote a random sample from the uniform distribution on the interval: (θ, θ + 1).
You are given:
• There are two unbiased estimators for θ, θ^ 1 and θ^ 2
θ^ 1 = Y - 1/2
€
θ^ 2 = Y(n) - n/(n+1), where€Y(n) = max(Y
1 , ... ,Yn )
€
€
€
• The efficiency of θ^ 1 relative to θ^ 2 is given by eff( θ^ 1 , θ^ 2 ) =
€
€
€
€
12 n2
(n + 2) (n +1)2
• s is the minimum sample size required to make θ^ 2 more efficient than θ^ 1
€
€
€ €
• s is less than 10
€
Calculate Var( θ^ 2 | n = s).
A. Less than 0.007
B. At least 0.007, but less than 0.009
C. At€least 0.009, but less than 0.011
D. At least 0.011, but less than 0.013
E. At least 0.013
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€
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Page 14
13. C. In order for θ^ 2 to be more efficient than θ^ 1 , the efficiency of θ^ 1 relative to θ^ 2 has to be less
12 n2
than one. We try values of n, eff( θ^ 1 , θ^ 2 ) =
.
(n + 2) (n +1)2
€
€
€
€
For n = 7, eff( θ^ 1 , θ^ 2 ) = 1.02083. For n = 8, eff( θ^ 1 , θ^ 2 ) = 0.948148.
€ €
So the minimum sample size that works
is 8.
€
Var[€
θ^ 1 ] €= Var[ Y - 1/2] = Var[ Y ] = Var[Y]
€ /€n.
However, Y is uniformly distributed over an interval of width one. ⇒ Var[Y] = 1/12.
^
^
⇒ Var[
€ θ 1 ] = 1/(12n).
€ For n = 8, Var[ θ 1 ] = 1/96.
€
Now for n = 8, Var[ θ^ 2 ] / Var[ θ^ 1 ] = eff( θ^ 1 , θ^ 2 ) = 0.948148.
€
€ ⇒ Var[ θ^ ] = 0.948148/96 =€0.009877.
2
€
€ Var[ θ^ €
€ €
Alternately,
2 ] = Var[Y(n) - n/(n+1)] = Var[Y(n)].
€ Var[Y(n)] is the same as the variance of the maximum of a sample of size n from a uniform on (0, 1).
€
This maximum follows a Beta Distribution with a = n = 8 and b = 1.
€
a(a +1)
(8)(9)
Mean = a/(a+b) = 8/9. Second Moment =
=
= 0.8.
(a + b)(a + b + 1) (9)(10)
Variance = 0.8 - (8/9)2 = 0.009877.
Comment: Var[ θ^ 1 ] =
€
€
Var[ θ^ 2 ] is the variance of a Beta Distribution with a = n and b = 1:
€
⎛ n ⎞2
⎛ n ⎞2
n(n +1) €
n
n
- ⎜
⎟ =
- ⎜
⎟ =
.
(n +1)(n + 1+1) ⎝ n + 1⎠
n + 2 ⎝ n + 1⎠
(n + 2) (n +1)2
€
12 n2
⇒ eff( θ^ 1 , θ^ 2 ) = Var[ θ^ 2 ] / Var[ θ^ 1 ] =
, as given in the question.
(n + 2) (n +1)2
€
€
€
€
€
€
1
.
12n
€
€
€
€
€
Solutions Spring 2016 CAS Exam S,
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Page 15
14. You are given the following information:
• Loss severity follows an exponential distribution with mean θ.
• Small claims are handled directly by the insured, and no information about losses less than 10,000
is available to you.
• You observe the following four claim payments net of the 10,000 deductible:
11,000
10,500
12,000
15,000
Calculate the Maximum Likelihood Estimate of θ.
A. Less than 2,000
B. At least 2,000, but less than 2,050
C. At least 2,050, but less than 2,100
D. At least 2,100, but less than 2,150
E. At least 2,150
14. E. For the Exponential Distribution,
θ^ = (sum of the payments) / (number of uncensored values)
€
= (11,000 + 10,500 + 12,000 + 15,000) / 4 = 12,125.
Comment: Based on the given choices, I suspect the writer of this question at some point intended
the data to be loss amounts prior to the effect of the deductible rather than claim payments net of the
10,000 deductible. In the case of the given data being loss amounts rather than payments:
θ^ = (1000 + 500 + 2000 + 5000)/4 = 2,125, corresponding to choice D.
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Page 16
15. You are given three normal distributions with the same mean and varying variances,
1 1 1
,
,
.
θ 2θ 4θ
You select one random draw from each of the distributions respectively:
0, 4, 8
€ The
€ probability
€
density of the normal distribution is:
f(x I µ, σ2) =
1
(x - µ)2
exp[].
2σ2
σ 2π
Determine the Maximum Likelihood Estimate of θ.
A. Less than 0.04
€
B. At least€0.04, but less than 0.06
C. At least 0.06, but less than 0.08
D. At least 0.08, but less than 0.10
E. At least 0.10
15. B. ln[f(x)] = -(x- µ)2 /(2σ2) - lnσ + constants.
Ignoring constants, the loglikelihood is:
-µ2θ/2 + lnθ /2 - (4- µ)2 θ + ln[2θ] /2 - (8- µ)2 2θ + ln[4θ] /2 =
-θ {µ2/2 + (µ - 4)2 + 2(µ - 8)2 } + 1.5 ln[θ] + ln[2]/2 + ln[4]/2.
Set the partial derivative with respect mu equal to zero:
0 = -θ {µ + 2(µ - 4) + 4(µ - 8)}. ⇒ µ = 40/7.
Set the partial derivative with respect theta equal to zero:
0 = -{µ2/2 + (µ - 4)2 + 2(µ - 8)2 } + 1.5/θ.
€
⇒ θ = 1.5 / {(40/7)2 /2 + (40/7 - 4)2 + 2(40/7 - 8)2 } = 0.0505.
€
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Page 17
16. You are given:
• A density function, f(x) = 5θ x3 exp[-θx4 ]
• A random sample of size n from this distribution
Calculate the Rao-Cramer lower bound for the variance of an unbiased estimator of θ.
€
A.
€
€
€
θ2
n
B.
1
θn
C.
D.
16. A. ln[f(x)] = ln5 + lnθ + 3lnx -θx4 .
€
€
€
∂ ln[f(x)]
= 1/θ - x4 .
∂θ
5θ2
n- 1
n
E. 2
θ
€
∂2 ln[f(x)]
= -1/θ2.
∂θ2
Rao-Cramer lower bound is:
€
θ2
4n
-1
∂2 ln[f(x)]
n E[
]
∂θ2
= θ2/n.
Comment: The Rao-Cramer lower bound goes down as 1/n, eliminating choices D and E.
This is a Weibull Distribution, parameterized differently than in the tables attached to the exam.
€
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Page 18
17. Determine the Fisher Information of n independent samples from a geometric distribution with
mean β.
A. 1/β
B. n/β
C. n/β - n/(β + 1)
D. nβ/(β + 1)
E. β(β+1)/n
17. C. The method of moments is equal to maximum likelihood for the Geometric.
^
^
β = X . Var[ β ] = Var[ X ] = Var[X] /n = β(1+β)/n.
^
Fisherʼs Information is: 1/Var[ β ] = n/ {β(1+β)} = n/β - n/(β + 1).
€
€
Alternately,
€
f(x) = βx/(1+β)x+1.
€
ln[f(x)] = x lnβ - (x+1) ln[1+β].
€
∂ ln[f(x)]
= x/β - (x+1)/(1+β).
∂β
∂2 ln[f(x)]
= -x/β2 + (x+1)/(1+β)2 .
∂β2
€
E[
€
∂2 ln[f(x)]
] = -β/β2 + (β+1) /(1+β)2 = -1/β + 1/(1+β).
∂β2
Fisherʼs Information is: -n E[
€
∂2 ln[f(x)]
] = n/β - n/(β + 1).
∂β2
Comment: Fisherʼs Information is proportional to n, eliminating choices A and E.
€
Solutions Spring 2016 CAS Exam S,
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Page 19
18. You are testing whether the number of claims in year 2 is independent of the number of claims in
year 1.
You are given the following contingency table:
Claims in Year 2
Claims in Year 1
0
1+
0
20
21
1+
30
21
Calculate the Chi-Square test statistic.
A. Less than 0.5
B. At least 0.5, but less than 1.0
C. At least 1.0, but less than 1.5
D. At least 1.5, but less than 2.0
E. At least 2.0
18. B. Here is the calculation of the contributions to the Chi-Square Statistic of 0.924:
Yr Two 0
Yr Two 1+
Sum
Yr One 0
Yr One 1+
20
30
21
21
41
51
Sum
50
42
92
Yr One 0
Yr One 1+
Sum
Yr One 0
Yr One 1+
Sum
Expected Number
22.2826
18.7174
27.7174
23.2826
Sum
41.0000
51.0000
50.0000
42.0000
92.0000
Contribution to Chi-Square
0.2338
0.2784
0.1880
0.2238
Sum
0.5122
0.4118
0.4218
0.9240
0.5022
For example, in the first row and first column, the expected number is: (50)(41)/92 = 22.2826.
The contribution from the first row and first column is: (20 - 22.2826)2 / 22.2826 = 0.2338
Comment: There is: (2 - 1)(2 - 1) = 1 degree of freedom.
Using a computer, the p-value is 33.6%.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 20
19. To test whether taking a driver's education course improves driving skills, you set up an
experiment. Participants are given a pretest of their driving skills and are tested on their skills again
after the course. The data is contained in the table below.
User ID
Pretest
Posttest
Difference
1
62
60
-2
2
93
94
1
3
97
100
3
4
77
80
3
5
69
75
6
6
72
78
6
7
99
100
1
8
55
57
2
9
90
91
1
10
66
63
-3
11
76
83
7
12
51
49
-2
13
74
84
10
14
52
61
9
15
97
99
2
16
54
52
-2
Sum
1184
1226
42
Sum of Squares
91,940
98,616
352
Assume the scores follow a Normal distribution.
Calculate the statistic to test if there is a significant difference in the two scores.
A. Less than -3
B. At least -3, but less than -1
C. At least -1, but less than 1
D. At least 1, but less than 3
E. At least 3
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 21
19. D. The differences Wi are Normal with unknown variance. We apply a t-test.
W = 42/16 = 2.625.
S W2 = (16/15) (352/16 - 2.6252 ) = 16.1167.
€
t=
€
W
SW 2 / n
=
2.625
16.1167 / 16
= 2.615.
Comment: The t-statistic has 16 - 1 = 15 degrees of freedom.
For a two-sided test, the p-value is 1.95%.
€
In general, the sign of the t-statistic depends on the order in which one takes the differences; as most
people would, I have used the differences given in the question.
Solutions Spring 2016 CAS Exam S,
€
€
HCM 6/28/16,
Page 22
20. You are given the following information:
• The Underwriting Department tells you that 40% of homeowners who stated they have
alarm systems have cancelled their alarm service.
• A survey of 1,000 homeowners who stated they have alarm systems shows that
428 homeowners have cancelled their service.
• You test the following hypotheses:
H0 ; The Underwriting Department's claim is accurate.
H1 : The Underwriting Department's claim is inaccurate.
€
Calculate the minimum significance level at which you would reject the null hypothesis.
A. Less than 0.005
B. At least 0.005, but less than 0.010
C. At least 0.010, but less than 0.025
D. At least 0.025, but less than 0.050
E. At least 0.050
20. E. If H0 is true, then the mean is 400, and the variance is: (0.4)(0.6)(1000) = 240.
Using the continuity correction, for the two-sided test the p-value is:
427.5 - 400
2{1 - Φ[
]} = (2){1 - Φ[1.775]} = (2)(1 - 0.9621) = 7.58%.
240
Alternately, Z =
428 - 400
240
= 1.807.
€
For the two-sided test, the p-value is: (2){1 - Φ[1.807]} = (2)(1 - 0.9646) = 7.08%.
€
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 23
21. Given
• X1 , X2 , ... , X10 are random samples from a Normal distribution with mean µx and variance 7.
• Y1 , Y2 , ... , Y8 are random samples from a Normal distribution with mean µy and variance 9.
€
• X and Y are independent
• W = X - Y is a Normal distribution with mean µx - µy
€
• H0 : µx = µy
€
€
€
• H1 : µx > µy
For W = 2, what is the result of the hypothesis test using the Normal distribution?
A. Reject H0 at the 0.025 level.
€
B. Reject H0 at the 0.050 level, but not at the 0.025 level.
C. Reject H0 at the 0.075 level, but not at the 0.050 level.
D. Reject H0 at the 0.100 level, but not at the 0.075 level.
E. Do not reject H0 at the 0.100 level.
21. C. Var[W] = 7/10 + 9/8 = 1.825.
Z = 2/ 1.825 = 1.480.
The p-value of a one-sided test is: 1 - Φ[1.480] = 6.9%.
€
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 24
22. You are given the following:
• The time it takes an insurance company to adjust a claim follows a Normal distribution with µ = 90
and σ = 30. (Time is in minutes)
• The insurance company is considering purchasing software from a vendor that claims the software
will reduce adjusting time by ten minutes per claim but without any impact on variation.
Before purchasing the software, the insurance company will test the software on a sample of
100 claims and measure the time to adjust claims.
• The insurance company will purchase the software only if the improvement is shown to be at least
ten minutes and the sample result is subject to no more than a 5% Type I error.
Calculate the probability of a Type II error if the vendor's claim is true.
A. At least 1%, but less than 2%
B. At least 2%, but less than 3%
C. At least 3%, but less than 4%
D. At least 4%, but less than 5%
E. At least 5%
€
€
€
22. D. Let H0 : µ = 90.
Let H1 : µ ≤ 80.
5% = Prob[Type I error] = Prob[ X < c | µ = 90] = Φ[
c - 90
302 / 100
].
c - 90
= -1.645. ⇒ c = 90 - (3)(1.645) = 85.065.
€
3
€
85.065 - 80
Prob[Type II error] = 1 - Prob[ X < 85.065 | µ = 80] = 1 - Φ[
] = 1 - Φ[1.688] = 4.57%.
€
302 / 100
⇒
€
€
Comment: Personally, I found it a little unclear exactly how to translate the given language into
hypothesis testing. “if€
the vendor's claim is true” is redundant and/or confusing.
€
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 25
23. To test if the distribution of the grades of two classes which have the same teacher are the
same, 7 students from Class A and 9 from Class B are randomly selected.
The sum of the ranks for Class A is 75.
Calculate the Mann-Whitney U test statistic.
A. Less than 17
B. 17
C. 18
D. 19
E. At least 20
23. A. U = n1 n2 + n1 (n1 +1)/2 - (sum of the ranks for the first sample of size n1 ) =
(7)(9) + (7)(8)/2 - 75 = 16.
Comment: Looking in the tables attached to the exam: F(16) = 5.7%.
Thus for a two-sided test, the p-value is: (2)(5.7%) = 11.4%.
It is somewhat arbitrary which one takes as the first sample; here it makes sense to take A as the first
sample since it is both mentioned first and is the smaller sample. While the value of U does, the
result of the hypothesis test does not depend on this choice.
The total ranks are: (16)(17)/2 = 136. Thus the sum of the ranks for Class B is: 136 - 75 = 61.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 26
24. You are given:
• Y1 , Y2 , and Y3 denotes a random sample from an exponential distribution
f(y) = e-y/5 / 5,
0<y<∞
• The second-order statistic is Y(2)
Calculate E[Y(2)].
€
A. Less than 3.0
B. At least 3.0, but less than 4.0
C. At least 4.0, but less than 5.0
D. At least 5.0, but less than 6.0
E. At least 6.0
24. C. For a sample of size N from an Exponential Distribution with mean θ:
⎧ 1
1
1 ⎫
E[X(r)] = ⎨ +
+ ... +
⎬ θ.
⎩N
N -1
N + 1-r ⎭
Here θ = 5 and E[Y(2)] = (1/3 + 1/2) (5) = 4.1667.
Alternately, Prob[ E[Y(2)] > y] = Prob[exactly two values > y] + Prob[3 values > y] =
€
3F(y)S(y)2 + S(y)3 = 3(1 - e-y/5)(e-y/5)2 + (e-y/5)3 = 3 e-y2/5 - 2e-y3/5.
Integrating from 0 to infinity: E[Y(2)] = (3)(5/2) - (2)(5/3) = 4.1667.
Solutions Spring 2016 CAS Exam S,
€
€
€
HCM 6/28/16,
Page 27
25. Two models are used to determine annual losses for a group of policies.
• Ten policies are selected at random
Policy
Model Y
Model Z
A
500
400
B
500
525
C
500
650
D
750
925
E
750
950
F
1000
1225
G
1250
1125
H
1250
1300
I
1250
1500
J
1500
1575
• H0 : MedianY = MedianZ
• H1 : MedianY < MedianZ
Calculate the smallest value for significance level for which H0 can be rejected using
the Signed-Rank Wilcoxon test for a matched pairs experiment.
A. Less than 0.010
B. At least 0.010, but less than 0.025
C. At least 0.025, but less than 0.050
D. At least 0.050, but less than 0.100
E. At least 0.100
25. C. The differences of Z minus Y: -100, 25, 150, 175, 200, 225, -125, 50, 250, 75.
Rank these by absolute value: 25, 50, 75, -100, -125, 150, 175, 200, 225, 250.
The sum of the negative ranks is: 4 + 5 = 9.
Consulting the table attached to the exam for N = 10 for a one-sided test:
8 < 9 ≤ 10. Therefore, reject at 5% and not at 2.5%.
Comment: If the alternative were true, we would expect lots of large absolute value positive
differences in Z - Y, and thus we would reject if the sum of the ranks for the negative differences is
small (and there the sum of the positive ranks is large).
For a one-sided test, using the tables attached to the exam, one can not reject at 10%; thus the
choices in the question could have been better.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 28
26. You are given:
• X1 , X2 , ... , Xn , is a random sample from a N(µ, σ2)
• σ2 = 200
• The prior distribution is a normal N(µ0, σ02) distribution
• µ0 = 70 and σ02 = 10
• There are 5 observations from this distribution, Xt:
53
22
61
86
70
Calculate the lower bound of the 95% Bayesian credibility interval for µ.
A. Less than 55
B. At least 55, but less than 60
C. At least 60, but less than 65
D. At least 65, but less than 70
E. At least 70
26. C. The sum of the observed loss is 292.
Using greek letters for parameters of the prior Normal, the posterior distribution is also Normal,
L σ 2 + µ s2
(292)(10) + (70)(200)
with mean =
=
= 67.68,
2
2
Cσ + s
(5)(10) + 200
σ2 s2
(10)(200)
and variance =
=
= 8.
C σ2 + s 2 (5)(10) + 200
€
€
Thus a posterior 95% confidence interval for µ is:
67.68€± 1.960 8 = (62.14 , 73.22).
€
Alternately, K = 200/10 = 20. Thus, the Buhlmann Credibility is: Z = 5/(5+20) = 1/5.
The posterior distribution is also Normal,
with mean
€ = (1/5)(292/5) + (1 - 1/5)(70) = 67.68,
and variance = (1 - Z)(σ02) = (4/5)(10) = 8. Proceed as before.
Comment: The third bullet would have been better if it had said:
“The prior distribution of µ is a normal N(µ0, σ02) distribution.”
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 29
27. For ABC Insurance Company, you are given the following information:
• All insureds come from one of three population classes.
• The claim frequency of each insured follows a Poisson process.
• The claims information is as follows:
€
€
€
Population
Expected time
Probability of being
Claim
Class
between claims
in class
cost
I
12 months
0.25
$5,000
II
18 months
0.25
$5,000
III
36 months
0.50
$5,000
Calculate the expected loss in year 2 for an insured that had no claims in year 1.
A. Less than 2,400
B. At least 2,400, but less than 2,500
C. At least 2,500, but less than 2,600
D. At least 2,600, but less than 2,700
E. At least 2,700
27. C. For the three classes, λ = 1, 1/ 1.5 = 2/3, and 1/3.
The chance of the observation is: e− λ.
We apply Bayes Analysis:
A
B
A Priori
Population Chance of
Class
This Class
4-sided
6-sided
8-sided
Overall
0.250
0.250
0.500
C
Chance
of the
Observation
0.36788
0.51342
0.71653
D
E
F
Prob. Weight = Posterior
Product
Chance of
Mean
of Columns
This Type Frequency
B&C
of Class
0.09197
0.12835
0.35827
15.90%
22.18%
61.92%
1.0000
0.6667
0.3333
0.57859
1.000
0.5133
For example, e-1 = 0.36788. (0.25)(0.36788) = 0.09197. 0.09197/0.57859 = 15.90%.
(15.90%)(1) + (22.18%)(2/3) + (61.92%)(1/3) = 0.5133.
Thus the estimated future loss is: ($5000)(0.5133) = $2567.
Solutions Spring 2016 CAS Exam S,
€
€
€
HCM 6/28/16,
Page 30
28. You are given the following information:
• There are five observations, y1 , y2 , y3 , y4 , y5 , from a gamma distribution:
6.85
10.3
20.89
15.06
6.26
• Var(Y) = 37.67
• The gamma distribution G(µ, ν) is a member of the exponential family with:
a(θ) = -In(-θ)
E(Y) = µ
Var(Y) = µ2/ν
Calculate the maximum likelihood estimate of the dispersion parameter, ϕ, for this distribution.
A. Less than 0.00
B. At least 0.00, but less than 1.00
C. At least 1.00, but less than 2.00
€
D. At least 2.00, but less than 3.00
E. At least 3.00
28. B. The dispersion parameter for the Gamma Distribution is 1/α.
They intend for us to find the maximum Iikelihood conditional on Var(Y) = 37.67.
Therefore, we want: αθ2 = 37.67.
€
When fitting an exponential family via maximum Iikelihood, the fitted mean is equal to the sample
mean. Therefore we want: αθ = (6.85 + 10.3 + 20.89 + 15.06 + 6.26) / 5 = 11.872.
Dividing the two equations in two unknowns: θ = 37.67 / 11.872 = 3.173.
⇒ α = 11.872/3.173 = 3.742. ⇒ 1/α = 1/3.742 = 0.267.
Comment: The dispersion parameter is always positive, eliminating choice A.
de Jong and Heller use ν for what is α in the Tables attached to the exam.
€ a Gamma via maximum likelihood conditional on the variance;
One does not usually fit
this is not covered in the syllabus readings.
This differs from fitting via maximum likelihood with both parameters unknown.
(Using a computer, the maximum likelihood fit is α = 4.886 and θ = 2.430.)
37.67 is the sample variance. Thus this weird method of fitting a Gamma is equivalent to setting the
sample mean equal to the theoretical mean and the sample variance equal to the theoretical
variance. The method of moments would also set the sample mean equal to the theoretical mean
but would instead set the biased estimate of the variance equal to the theoretical variance.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 31
29. You are given the following information for a fitted GLM:
Response variable
Occurrence of Accidents
Response distribution
Binomial
Link
Logit
Parameter
df
^
β
Intercept
1
x
Driver's Age
2
1
1 €
0.288
2
1
0.064
3
0
0
Area
2
A
1
-0.036
B
1
0.053
C
0
0
Vehicle Body
2
Bus
1
1.136
Other
1
-0.371
Sedan
0
0
The probability of a driver in age group 2, from area C and with vehicle body type Other,
having an accident is 0.22.
Calculate the odds ratio of the driver in age group 3, from area C and with vehicle body type Sedan
having an accident.
A. Less than 0.200
B. At least 0.200, but less than 0.250
C. At least 0.250, but less than 0.300
D. At least 0.300, but less than 0.350
E. At least 0.350
€
€
29. E. In order to solve for the unknown intercept, we use the given probability of accident for a
driver in age group 2, from area C and with vehicle body type Other.
0.22 = exp[x + 0.064 - 0.371] / {exp[x + 0.064 - 0.371] + 1}.
⇒ exp[x + 0.064 - 0.371] = 0.22 / (1 - 0.22) = 0.28205.
⇒ x + 0.064 - 0.371 = ln[0.28205] = -1.2657. ⇒ x = -0.9587.
Thus for a driver in age group 3, from area C and with vehicle body type Sedan, the odds (ratio) is:
π / (1 - π) = exp[-0.9857 + 0 + 0 + 0] = 0.3834.
Comment: The probability of having an€accident for a driver in age group 3, from area C and with
vehicle body type Sedan is: 0.3834 / (1 + 0.3834) = 0.277. Note that 0.277 / (1 - 0.277) = 0.383.
π / (1 - π) is called the odds, but is also sometimes called the odds ratio.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
30. You are given the following information for a fitted GLM:
Response variable
Occurrence of Accidents
Response distribution
Binomial
Link
Logit
Parameter
df
^
β
se
Intercept
1
-2.358
0.048
Area
2
Suburban € 0
0.000
Urban
1
0.905
0.062
Rural
1
-1.129
0.151
Calculate the modeled probability of an Urban driver having an accident.
A. Less than 0.01
B. At least 0.01, but less than 0.05
C. At least 0.05, but less than 0.10
D. At least 0.10, but less than 0.20
E. At least 0.20
30. D.
€
Exp[-2.358 + 0.905]
= 0.190.
1 + Exp[-2.358 + 0.905]
Page 32
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
31. You are given the following information for a fitted GLM:
Response variable
Claim size
Response distribution
Gamma
Link
Log
Dispersion Parameter
1
^
Parameter
df
β
Intercept
1
2.100
€4
1
1
1
0
1
7.678
4.227
1.336
0.000
1.734
Vehicle Class
Convertible
Coupe
Sedan
Truck
Minivan
Station wagon
Utility
6
1
1
0
1
1
1
1
1.200
1.300
0.000
1.406
1.875
2.000
2.500
Driver Age
Youth
Middle age
Old
2
1
0
1
2.000
0.000
1.800
Zone
1
2
3
4
5
Calculate the predicted claim size for an observation from Zone 3,
with Vehicle Class Truck and Driver Age Old.
A. Less than 650
B. At least 650, but less than 700
C. At least 700, but less than 750
D. At least 750, but less than 800
E. At least 800
31. D. exp[2.100 + 1.336 + 1.406 + 1.800] = 766.63.
Page 33
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
32. You are given the following information for a fitted GLM:
Response variable
Claim size
Response distribution
Gamma
Link
Log
Dispersion parameter
1
^
Parameter
df
β
Intercept
1
2.100
€4
1
1
1
0
1
7.678
4.227
1.336
0.000
1.734
Vehicle Class
Convertible
Coupe.
Sedan
Truck
Minivan
Station wagon
Utility
6
1
1
0
1
1
1
1
1.200
1.300
0.000
1.406
1.875
2.000
2.500
Driver Age
Youth
Middle age
Old
2
1
0
1
2.000
0.000
1.800
Zone
1
2
3
4
5
Calculate the variance of a claim size for an observation from Zone 4,
with Vehicle Class Sedan and Driver Age Middle age
A. Less than 55
B. At least 55, but less than 60
C. At least 60, but less than 65
D. At least 65, but less than 70
E. At least 70
Page 34
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 35
32. D. For an observation from Zone 4, with Vehicle Class Sedan and Driver Age Middle age,
the mean is: exp[2.1] = 8.166.
Thus the variance is: φ V[µ] = (1) (µ2) = 8.1662 = 66.7.
Comment: We are given that the dispersion parameter φ is 1.
For the Gamma Distribution, V[µ] = µ2.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 36
33. You are given the following information for a GLM of customer retention:
Response variable
Retention
Response distribution
Binomial
Link
Logit
^
Parameter
df
β
Intercept
1
1.530
Number of Drivers
1
>1
€1
0
1
0.000
0.735
Last Rate Change
<0%
0%-10%
>10%
2
0
1
1
0.000
-0.031
·0.372
Calculate the probability of retention for a policy with 3 drivers and a prior rate change of 5%.
A. Less than 0.850
B. At least 0.850, but less than 0.870
C. At least 0.870, but less than 0.890
D. At least 0.890, but less than 0.910
E. At least 0.910
33. D.
€
exp[1.530 + 0.735 - 0.031]
= 90.33%.
1 + exp[1.530 + 0.735 - 0.031]
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 37
34. Determine which of the following statements are true.
I. The deviance is useful for testing the significance of explanatory variables in nested models.
II. The deviance for normal distributions is proportional to the residual sum of squares.
III. The deviance is defined as a measure of distance between the saturated and fitted model.
A. I only
B. II only
C. III only
D. All but III
E. All
34. E. All three statements are true. See pages 72-73 of de Jong and Heller.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 38
35. You are given the following information about three candidates for a Poisson frequency GLM on
a group of condominium policies:
Log
Model Variables in the Model
DF
Likelihood
AIC
BIC
1
Risk Class
5
-47,704
95,418
95,473.61182
2
Risk Class + Region
-47,495
3
Risk Class + Region + Claim Indicator 10
-47,365
94,750
€
€
€
€
• Insureds are from one of five Risk Class: A, B, C, D, E
• Condominium policies are located in several regions
• Claim Indicator is either Yes or No
• All models are built on the same data
Calculate the absolute difference between the AIC and the BIC for Model 2.
A. Less than 85
B. At least 85, but less than 95
C. At least 95, but less than 105
D. At least 105, but less than 115
E. At least 115
35. C. Since Model 2 has one fewer parameter than model 3, model 2 has 9 degrees of freedom.
AIC = (-2) (maximum loglikelihood) + (number of parameters)(2).
BIC = (-2) (maximum loglikelihood)) + (number of parameters) ln(number of data points).
Therefore, 95,473.61182 = (-2)(-47,704) + (5) ln(n). ⇒ n = 500,000.
For Model 2, AIC = -47,495 + (9)(2).
For Model 2, BIC = -47,495 + (9) ln(500,000).
The absolute difference between the AIC and
€ the BIC for Model 2 is:
| (9) ln(500,000) - 18 | = 100.1.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 39
36. You are given the following two graphs comparing the fitted values to the residuals of two
different linear models:
Determine which of the following statements are true.
I.
Graph 1 indicates the data is homoscedastic
II. .
Graph 1 indicates the data is heteroskedastic
III.
Graph 2 indicates the data is non-normal
A. I only
B. II only
C. III only
D. I and III
E. II and III
36. E. Graph one shows an increasing variance with fitted value.
Homoscedasticity would be constant variance, so statement 1 is false; statement 2 is true.
The residuals in Graph 2 are not symmetric around zero; there are more extreme positive values
than there are extreme negative values. This indicates that the residuals are not normally distributed.
Statement 3 is true.
Comment: In Graph 2 it is not clear the meaning of the horizontal lines.
Note that Graph 2 does not show normalized residuals. In that case one could compare to a
Standard Normal Distribution.
A Normal Q-Q Plot would have been much more useful than Graph 2.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
Page 40
37. Determine which of the following GLM selection considerations is true.
A. The model with the largest AIC is always the best model in model selection process.
B. The model with the largest BIC is always the best model in model selection process.
C. The model with the largest deviance is always the best model in model selection process.
D. Other things equal, when the number of observations > 1000, AIC penalizes more for
the number of parameters used in the model than BIC.
E. Other things equal, when number of observations > 1000, BIC penalizes more for
the number of parameters used in the model than AIC.
37. E. The model with the smallest AIC is usually the best model in model selection process,
all other things being equal. Statement A is not true.
The model with the smallest BIC is usually the best model in model selection process,
all other things being equal. Statement B is not true.
The model with the smallest deviance is usually the best model in model selection process,
all other things being equal. Statement C is not true.
AIC = (-2) (maximum loglikelihood) + (number of parameters)(2).
BIC = (-2) (maximum loglikelihood)) + (number of parameters) ln(number of data points).
The penalty for AIC is (number of parameters)(2).
The penalty for BIC is (number of parameters) ln(number of data points).
So the penalties are equal for: 2 = ln(number data Points). ⇒ number of data points = e2 = 7.4.
Thus, other things equal, when number of observations ≥ 8, BIC penalizes more for the number of
parameters used in the model than AIC. Thus statement E is true.
€ it is likely that one of them is true.
Comment: Since statements D and E are opposites,
Solutions Spring 2016 CAS Exam S,
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€
€
HCM 6/28/16,
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38. You are testing the addition of a new categorical variable into an existing GLM,
and are given the following information:
• A is the change in AIC and B is the change in BIC after adding the new variable.
• B > A + 25
• There are 1500 observations in the model.
Calculate the minimum possible number of levels in the new categorical variable.
A. Less than 3
B. 3
C.4
D. 5
E. More than 5
38. E. Change in AIC is: (2) (number of parameters added).
Change in BIC is: ln(1500) (number of parameters added).
We want: ln(1500) (number of parameters added) > (2) (number of parameters added) + 25.
⇒ Number of parameters added > 4.7. ⇒ Number of added parameters is at least 5.
⇒ Minimum possible number of levels in the new categorical variable is: 5 + 1 = 6.
€
€
€
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39. Determine which of the following statements about GLMs are true.
I. The saturated model has the highest possible deviance.
II. Deviance follows a χ2 distribution for all models in the exponential family.
III. Deviance is a useful measure of goodness of fit for all models in the exponential family.
A. I only
B. II only
C. III only
D. I, II and III
E. None of the above
39. E. The saturated model has the smallest possible deviance. Statement 1 is false.
Deviance approximately follows a χ2 distribution when the scale parameter does not have to be
estimated such as for the Poisson, but not for those cases such as the Gamma Distribution in which
the deviance parameter has to be estimated. Statement 2 is false.
Deviance is not a useful measure of goodness of fit for Bernoulli and Binomial models.
Statement 3 is false.
Comment: See pages 72 and 107 of de Jong and Heller.
Solutions Spring 2016 CAS Exam S,
HCM 6/28/16,
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40. You are given the following results from a fitted GLM on the frequency of accidents:
^
Parameter
df
β
se
Intercept
1
-11.2141
0.1826
0.0000
1.0874
0.3162
Location
Rural
City
€1
0
1
Calculate the Wald statistic for testing the null hypothesis of βcity = 0.
A. Less than 10.5
B. At least 10.5, but less than 11.5
C. At least 11.5, but less than 12.5
D. At least 12.5, but less than 13.5
E. At least 13.5
⎛ fitted parameter ⎞ 2
40. C. Wald Statistic = ⎜
⎟ = (1.0874/0.3162)2 = 11.826.
⎝ standard error ⎠
Comment: We would compare this statistic to a Ch-Square Distribution with 1 degree of freedom.
Using a computer, the p-value is 0.06%.
€
Solutions Spring 2016 CAS Exam S,
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41. A Poisson regression model with log link is used to estimate the number of diabetes deaths.
The parameter estimates for the model are:
Response variable
Number of Diabetes Deaths
Response distribution
Poisson
Link
Log
^
Parameter
df
β
p-value
Intercept
1
-15.000
<0.0001
-1.200
0.000
<0.0001
Gender: Female
Gender: Male
€1
1
Age
1
0.150
<0.0001
Age2
1
0.004
<0.0001
Age × Gender: Female
Age × Gender: Male
1
0
0.012
0.000
<0.0001
Calculate
the expected number of deaths for a population of 100,000 females age 25.
€
A.€Less than 3
B. At least 3, but less than 5
C. At least 5, but less than 7
D. At least 7, but less than 9
E. At least 9
41. C. 100,000 exp[-15 - 1.2 + (0.15)(25) + (0.004)(252 ) + (0.012)(25)]
= 100,000 e-9.65 = 6.44 deaths.
Solutions Spring 2016 CAS Exam S,
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42. You are building a model of claim counts, and in order to evaluate the risk of overdispersion,
you calculate the following:
• The log-likelihood of the Poisson regression is l p = 350.
€
€
• The log-likelihood of the Negative Binomial regression is l NB.
At a significance level of 0.5%, you reject the hypothesis that a Poisson response is more
€
appropriate than a Negative Binomial response.
€
Calculate the smallest possible value of l NB.
A. Less than 350
B. At least 350, but less than 352
€
C. At least 352, but less than 354
D. At least 354, but less than 356
E. At least 356
42. C. The test statistic is: (2) ( l NB - 350).
We compare to the (2)(0.5%) = 1% critical value
for the Chi-Square Distribution with one-degree of freedom: 6.63.
6.63 = (2) ( l NB - 350).€ ⇒ l NB = 353.315.
Comment: To compare a Poisson Regression and an otherwise similar Negative Binomial
Regression, the test statistic is twice the difference in loglikelihoods. For significance level α we
€
compare
to the€2α€tail of the Chi-Square Distribution with one-degree of freedom.
Will not be on the syllabus going forward.
Solutions Spring 2016 CAS Exam S,
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43. You are given the following information:
• The time series {Xt} is stationary in mean and variance.
• The first 8 observed values of x are as follow:
t
xt
€
1
2
3
4
5
6
7
8
€
13
10
12
9
8
8
7
5
Calculate the sample auto covariance function, ck, for lag k = 2.
A. Less than 1.80
B. At least 1.80, but less than 1.90
C. At least 1.90, but less than 2.00
D. At least 2.00, but less than 2.10
E. At least 2.10
43. B. Mean is: 72/8 = 9.
(13 - 9)(12 - 9) + (10 - 9)(9 - 9) + (12 - 9)(8 - 9) + (9 - 9)(8 - 9) + (8 - 9)(7 - 9) + (8 - 9)(5 - 9)
c2 =
8
= 1.875.
€
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44. You are given the following ARMA(p, q) model:
5
1
1
xt = xt-1 - 2xt-2 + xt-3 + wt - wt-1
2
2
2
The model is parameter redundant.
Which of the following statements is false?
€
€
A.€The model is non-stationary.
B. The model is ARMA(3,1).
C. The model is invertible.
D. The model can be restated as ARMA(2,0).
E. None of the above.
44. E. This is an ARMA(3,1) model. Statement B is true.
The characteristic equation is: 0 = 1 - 2.5B + 2B2 - 0.5B3 .
We can factor this: 0 = (1 - B/2) (1 - 2B + B2 ).
Thus the roots are 2, 1, and 1.
Since not all of these roots have absolute value greater than one, the series is not stationary.
Statement A is true.
φ(B) = 1 - B/2.
Setting φ(B) equal to zero: 0 = 1 - B/2. ⇒ B = 2.
Since this root has absolute value greater than one, the series is invertible. Statement C is true.
The series can be written as: θ(B) xt = φ(B) wt. ⇔ (1 - 2.5B + 2B2 - 0.5B3 ) xt = (1 - B/2) wt.
€
⇔ (1 - B/2) (1 - 2B + B2 ) xt = (1 - B/2) wt. ⇔ (1 - 2B + B2 ) xt = wt.
€
Thus we have restated the original model
€ as ARMA(2,0). Statement D is true.
Comment: A somewhat difficult and long question.
€
Solutions Spring 2016 CAS Exam S,
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45. You are given the following information:
• x and y are two time series that are stationary.
• The graphs below are generated using the ACF (auto correlation function) function of R.
• The dashed lines above and below zero indicate the range within which the ACF results
are considered not significantly different than zero.
• The symbol wt is white noise with zero mean.
€
Which of the following statements displays the model structure that best describes series x and
series y?
A. xt = 0.9xt-1 + wt and yt = 0.9yt-1 + wt + 0.6wt-1 - 0.3wt-2
B. xt = -0.9xt-1 + wt and yt = wt + 0.6wt-1 - 0.3wt-2
C. xt = 0.9xt-1 + wt and yt = wt + 0.6wt-1
D. xt = -0.9xt-1 + wt and yt = wt + 0.6wt-1 - 0.3wt-2 + 0.4wt-3
E. xt = 0.9Xxt-1 + wt and yt = wt + 0.6wt-1 - 0.3wt-2
Solutions Spring 2016 CAS Exam S,
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45. B. For an AR(1) model, with |α| < 1, the autocorrelations decline geometrically in absolute
value; for α < 0 the autocorrelations alternate sign. The first correlogram follows this pattern, indicating
an AR(1) model such as xt = -0.9xt-1 + wt is appropriate.
For an MA(2) model, the autocorrelations beyond lag two are zero. The second correlogram follows
this pattern, indicating an MA(2) model such as yt = wt + 0.6wt-1 - 0.3wt-2 is appropriate.
Comment: For an AR(1) model: ρk = αk.
Using the R function ARMAacf, one can compute the autocorrelations.
For the AR(1) series, xt = -0,9xt-1 + wt, the output from ARMAacf(ar=c(-0.9), lag.max = 10) is:
0
1.0000000
1
2
3
-0.9000000 0.8100000 -0.7290000
6
7
8
0.5314410
-0.4782969
0.4304672
For the MA(2) series, yt = wt + 0.6wt-1 - 0.3wt-2,
the output from ARMAacf(ma= c(.6,-.3), lag.max = 4) is:
0
1
2
1.0000000
0.2896552
-0.2068966
END OF THE EXAMINATION
4
5
0.6561000 -0.5904900
9
10
-0.3874205
0.3486784
3
0.0000000
4
0.0000000