Solutions to the
Fall 2015
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 2015 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 2015 by Howard C. Mahler.
Study Aid 2015-S
Howard Mahler
[email protected]
www.howardmahler.com/Teaching
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Solutions Fall 2015 CAS Exam S,
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Page 1
1. You are given the following information:
• Buses depart from the bus stop at a Poisson rate of 5 per hour.
• Today, Joe arrives at the bus stop just as a bus is leaving and will have to wait for the next bus
to depart.
• Yesterday, Joe arrived at the bus stop 5 minutes after the prior bus departed and had to wait
for the subsequent bus to depart.
Calculate how much longer Joe's expected wait time is today compared to yesterday.
A. Less than 1 minute
B. At least 1 minute, but less than 2 minutes
C. At least 2 minutes, but less than 3 minutes
D. At least 3 minutes, but less than 4 minutes
E. At least 4 minutes
1. A. The expected waiting time is independent of when the last bus left, since a Poisson Process
is memoryless. Joe's expected wait time today is equal to yesterday.
Comment: Similar to CAS3L, 5/11, Q.9.
Both expected waiting times are: 1/5 hour = 12 minutes.
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Page 2
2. You are given the following:
• A Poisson process with λ = 1.2 has its 4th event occur at time 10.
• This process began operating at time 0.
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Calculate the expected value of the time at which the 3rd event occurred.
A. Less than 2
B. At least 2, but less than 4
C. At least 4, but less than 6
D. At least 6, but less than 8
E. At least 8
2. D. Given the fourth event has occurred at time 10, the expected times of the first three events
are spread evenly over the interval (0, 10).
Conditional expected time of 1st event is: 2.5.
Conditional expected time of 2nd event is: 5.
Conditional expected time of 3rd event is: 7.5.
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3. You are given the following information:
• Lucy finds coins at a Poisson rate of 1 coin per 10 minutes.
• The denominations are randomly distributed as follows:
• 65% of the coins are worth 1 each;
• 20% of the coins are worth 5 each;
• 15% of the coins are worth 10 each.
€ Calculate the probability that in the first 30 minutes she finds at least 1 coin worth 10 each
€ and in the first hour finds at least 2 coins worth 10 each.
€ A. Less than 0.165
B. At least 0.165, but less than 0.175
C. At least 0.175, but less than 0.185
D. At least 0.185, but less than 0.195
E. At least 0.195
3. C. Dimes are Poisson with rate: (0.1)(0.15) = 0.015 per minute.
Number of dimes in 30 minutes is Poisson with mean: (30)(0.015) = 0.45.
Prob[at least 1 dime first 30 minutes and at least 2 dimes in first 60 minutes] =
Prob[1 dime in first 30 minutes] Prob[at least 1 dime in next 30 minutes] +
Prob[at least 2 dimes in first 30 minutes] =
(0.45 e-0.45) (1 - e-0.45) + 1 - e-0.45 - 0.45 e-0.45 = 0.1794.
Comment: What happens over the first 30 minutes is independent of what happens over the
second 30 minutes.
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4. You are given the following information on a workers' compensation policy:
• Claims occur according to a Poisson process with rate λ = 10 per week.
• Claims are classified independently into the following claim types:
Claim Type
Probability
1
0.2
2
0.3
3
0.5
• For each claim a random amount Xi, i = 1, 2, 3, is added to a fund.
• The fund starts with an amount of zero at time zero.
• The random variables Xi follow the Exponential distribution with the following means:
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€
Claim Type
1
2
3
Mean
500
300
200
Calculate the 95th percentile of the amount of the fund after 13 weeks,
using the Normal approximation.
A. Less than 43,000
B. At least 43,000, but less than 45,000
C. At least 45,000, but less than 47,000
D. At least 47,000, but less than 49,000
E. At least 49,000
4. C. The severity is a mixture of three Exponentials.
Mean of the mixture is the mixture of the means:
(0.2)(500) + (0.3)(300) + (0.5)(200) = 290.
The second moment of the mixture is the mixture of the second moments:
(0.2)(2)(5002 ) + (0.3)(2)(3002 ) + (0.5)(2)(2002 ) = 194,000.
Mean of the compound Poisson is: (13)(10)(290) = 37,700.
Variance of the compound Poisson is: (13)(10)(194,000) = 25,220,000.
95th percentile of the amount of the fund after 13 weeks: 37,700 + 1.645
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25,220,000 = 45,961.
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5. You are given the following information:
• The amount of damage involved in a home theft loss is an Exponential random variable
with mean 2,000.
• The insurance company only pays the amount exceeding the deductible amount of 500.
• The insurance company is considering changing the deductible to 1,000.
Calculate the absolute value of the change in the expected value of the amount the insurance
company pays per theft loss by changing the deductible from 500 to 1,000.
A. Less than 330
B. At least 330, but less than 350
C. At least 350, but less than 370
D. At least 370, but less than 390
E. At least 390
5. B. Due to the memoryless property of the Exponential, in each case the average payment per
(non-zero) payment is the same 2000.
With the 500 deducible, the probability of a payment is: e-500/2000 = 0.7788, and thus the average
payment per theft loss is: (0.7788)(2000) = 1557.6.
With the 1000 deducible, the probability of a payment is: e-1000/2000 = 0.6065, and thus the
average payment per theft loss is: (0.6065)(2000) = 1213.0.
|1557.6 - 1213.0| = 344.6.
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6. You are given the following information:
• An engine system consists of a series of 4 independent pumps, each of which is in operation
at time zero.
• The engine system will fail when anyone of the 4 pumps stops running.
• The amount of time a pump runs (in hours) is distributed uniformly over (0, 100).
Calculate the expected run time of the engine system in hours.
A. Less than 17.5
B. At least 17.5, but less than 18.5
C. At least 18.5, but less than 19.5
D. At least 19.5, but less than 20.5
E. At least 20.5
6. D. The expected value of the minimum of a sample of size 4 from a uniform distribution on
(0, 100) is: (100) / (4+1) = 20.
Comment: The expected value of the orders statistics are:
(100) (1/5, 2/5, 3/5, 4/5) = 20, 40, 60, 80.
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7. You are given the following information about a series system:
• There are three components: Component 1, Component 2, and Component 3.
• The lifetimes of Component 1 and Component 2 are each distributed U(0, 6).
• The lifetime of Component 3 is distributed U(0, 12).
• The component lifetimes are independent.
• U(a, b) is used to define a uniform distribution between a and b.
Calculate the expected system lifetime.
A. Less than 2.00
B. At least 2.00, but less than 2.25
C. At least 2.25, but less than 2.50
D. At least 2.50, but less than 2.75
E. At least 2.75
7. A. Since it is a series system, it only continues to function while all three components function.
For t ≤ 6, the probability that the system is still functioning is:
(1 - t/6)(1 - t/6)(1 - t/12) = 1 - 5t/12 + t2 /18 - t3 /432.
For t > 6, the system has failed.
The expected system lifetime is the integral of the survival function:
6
∫0 1 - 5t / 12 + t2 / 18 - t3 / 432 dt = 6 - (62)(5/24) + (63)(1/54) - (64)(1/1728) = 1.75.
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8. Ben and Allison each decide to wager 1 unit against the other person on flips of a fair coin, until
one of them runs out of money. At the start of the contest, Ben has 20 units and Allison has 55 units.
Calculate the variance of Ben's final wealth.
A. Less than 1
B. At least 1, but less than 10
C. At least 10, but less than 100
D. At least 100, but less than 1,000
E. At least 1,000
8. E. Ben either ends with nothing or all the money 75.
Since they each have an equal chance to win each coin flip,
Benʼs chance of ending with all the money is: 20/75 = 4/15.
Variance of 75 times a Bernoulli Distribution = (752 ) q (1-q) = 752 (4/15)(1 - 4/15) = 1100.
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9. You are given the following Markov chain transition matrix:
⎛ 0.7 0.2 0.1 0.0 0.0⎞
⎜ 0.4 0.4 0.2 0.0 0.0⎟
⎜
⎟
P = ⎜ 0.5 0.5 0.0 0.0 0.0 ⎟ .
⎜ 0.4 0.2 0.1 0.3 0.0 ⎟
⎜
⎟
⎝ 0.0 0.0 0.0 0.0 1.0 ⎠
Determine the number of transient states in this Markov chain.
A.€0
B. 1
C. 2
D. 3
E. 4
9. B. State 5 is an absorbing state and thus recurrent.
One can go from 1 to 2 to 3 to 1, so these are a class.
Once one is in this class, you remain in that class, and thus these states are recurrent.
Once one leave state 4, one can not return to state 4, so it is transitory.
There is only one transient state, 4.
Comment: A diagram of this Markov Chain; state 5 does not communicate with any other state:
4
3
1
2
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Page 10
10. An insurance company classifies its customers into three categories:
1. Good Risk
2. Acceptable Risk
3. Bad Risk
Customers independently transition between categories at the end of each year according to a
Markov process with the following transition matrix:
⎛ 0.5 0.5 0 ⎞
P = ⎜ 0.4 0.4 0.2⎟ .
⎜
⎟
⎝ 0.0 0.5 0.5⎠
Calculate the stationary probability of a customer being classified as a Good Risk.
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
10. B. We want πP = π. The equations for the stationary distribution are:
π 1 = 0.5 π1 + 0.4 π2. ⇒ π1 = 0.8 π2.
π 2 = 0.5 π1 + 0.4 π2 + 0.5 π3.
π 3 = 0.2 π2 + 0.5 π3. ⇒ π3 = 0.4 π2.
1 = π1 + π 2 €
+ π3. ⇒ π2 = 1/2.2 = 45.5%. ⇒ π1 = 0.8/2.2 = 36.4% ⇒ π2 = 0.4/2.2 = 18.2%.
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11. You are given the following information:
• For a group of automobile insurance policies, there are three premium levels: P, 1.1P, and 1.2P.
Each premium level corresponds to a State 0, 1 or 2 that is determined by claim filed by the
policyholder with movement to the next state taking place at the beginning of the policy term.
• The transition of policyholders between these levels is modeled by a Markov Chain as follows:
Current State
0
1
2
Current Annual
Premium
P
1.1P
1.2P
No Claims
0
0
1
Next State if
One Claim Two Claims
1
2
2
2
2
2
• For example, if a policyholder is in State 0, the premium the policyholder would pay is P.
If that policyholder has a claim, next year the policyholder would move to State 1
and pay 1.1 P for the premium.
• For each policyholder, the probability of number of claims is:
Number of Claims Probability
No Claim
0.5
One Claim
0.4
Two Claims
0.1
Calculate the long run mean value of the annual premium.
A. Less than 1.10P
B. At least 1.10P, but less than 1.12P
C. At least 1.12P, but less than 1.14P
D. At least 1.14P, but less than 1.16P
E. At least 1.16P
0
11. B. The transition matrix is: 1
2
⎛ 0.5 0.4 0.1⎞
⎜ 0.5 0 0.5⎟ .
⎜
⎟
⎝ 0 0.5 0.5⎠
We want πP = π. The equations for the stationary distribution are:
π 0 = 0.5 π0 + 0.5 π1. €
⇒ π0 = π1.
€
π 1 = 0.3 π0 + 0.5 π2. ⇒ π2 = 1.2π1.
π 2 = 0.1 π0 + 0.5 π1 + 0.4 π2.
1 = π0 + π 1 €
+ π2. ⇒ π1 = 1/3.2. ⇒ π0 = 1/3.2. ⇒ π2 = 1.2/3.2.
€
Thus the long run mean value of the annual premium is:
(1/3.2)P + (1/3.2)(1.1P) + (1.2/3.2)(1.2P) = 1.10625 P.
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Page 12
12. You are given the following information for a policyholder age 65:
• A 3-year term insurance policy on (65) provides for a death benefit of 1,000 payable
at the end of the year of death.
• This policy is purchased by a single premium, P, at time 0.
• If (65) lives to age 68, the single premium is returned without interest.
• Mortality rates are:
x
qx
65
0.15
66
0.20
67
0.25
• i = 0.10.
Calculate P using the equivalence principle.
A. Less than 640
B. At least 640, but less than 650
C. At least 650, but less than 660
D. At least 660, but less than 670
E. At least 670
12. C. Actuarial Present Value of benefits is:
(1000) {0.15/1.1 + (0.85)(0.2)/1.12 + (0.85)(0.8)(0.25)/1.13 } = 404.58.
APV of single premium including possible return: P {1 - (0.85)(0.8)(0.75)/1.13 } = 0.61683 P.
404.58 = 0.61683 P. ⇒ P = 655.90.
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13. You are given the following information:
• A life insurance company issues a special 3-year discrete insurance to a life (x).
• If the policyholder dies, at the end of the year of death, there is a random drawing.
With probability 0.2, the death benefit is 50,000. With probability 0.8, the death benefit is 0.
• At the beginning of each year the policy is in effect while (x) is alive, there is a random drawing.
With probability 0.8, the premium π is paid. With probability 0.2, no premium is paid.
• The random drawings are independent.
• The probability of surviving an additional k years is kp x = 0.9k, for k = 0, 1, 2, ...
• i = 0.06.
Calculate π using the equivalence principle.
A. Less than 1,100
B. At least 1,100, but less than 1,200
C. At least 1,200, but less than 1,300
D. At least 1,300, but less than 1,400
E. At least 1,400
13. B. Actuarial Present Value of benefits is:
(50,000)(0.2) {0.1/1.06 + (0.9 - 0.92 )/1.062 + (0.92 - 0.93 )/1.063 } = 2424.49.
APV of premiums:
π (0.8) {1 + 0.9 /1.06 + 0.92 /1.062 } = 2.05596 π.
2.05596 π = 2424.49. ⇒ π = 1179.
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14. You are given the following information:
• You draw a random sample of five observations from a distribution with density function:
1
f(y | α, θ) =
yα−1 exp[-y/θ], y > 0
Γ(α) θα
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• The sampled values are 0.5, 2.0, 10.0, 1.5, 7.0
• α = 2.
Calculate€
the maximum likelihood estimate of θ.
A. Less than 2.00
B. At least 2.00, but less than 2.25
C. At least 2.25, but less than 2.50
D. At least 2.50, but less than 2.75
E. At least 2.75
14. B. X = (0.5, 2.0, 10.0, 1.5, 7.0) / 5 = 4.2.
For a Gamma Distribution with alpha fixed, maximum Iikelihood is equal
to the method of moments.
€αθ = X . ⇒ θ = 4.2/2 = 2.1.
€ €
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Page 15
15. You are given a random sample of five observations from a Poisson distribution with an
unknown parameter, λ:
0, 0, 1, 4, 5
Calculate the maximum likelihood estimate of 1/λ.
A. Less than 0.4
B. At least 0.4, but less than 0.8
C. At least 0.8, but less than 1.2
D. At least 1.2, but less than 1.6
E. At least 1.6
15. B. For a Poisson Distribution, maximum Iikelihood is equal to the method of moments.
λ = (0 + 0 + 1+ 4 + 5)/5 = 2. ⇒ 1/λ = 0.5.
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16. A sample of independent and identically distributed random variables, (X1 , X2 , ..., Xn ), is drawn
from a distribution with the following probability density:
Pr(X = x) =
e - λ λx
, λ > 0, x ∈ {0, 1, 2, ... }
x!
Determine which of the following is a sufficient statistic for λ.
n
€
€
B. min(X1 , X2 , ..., Xn )
A. X1
C.
n
∑ Xi
D.
i=1
∑
n
Xi2
E.
i=1
Xi!
∏
i=1
16. C. X is a sufficient statistic for the Poisson Distribution, thus so is the sum of the sample.
€
€
€
n
€Alternately, the likelihood is:
n
f(Xi) = ∏
∏
i=1
i=1
e- λ
λ
Xi !
Xi
=
n
∑ Xi
{e-λn λ i=1 }
n
{
1
}.
∏
Xi!
i=1
We have factored the likelihood into 2 terms, the 2nd
€ of which is not a function of the parameter λ.
n
€
€
The first term is a function of λ and
Xi .
€
∑
i=1
n
Thus
∑ Xi is a sufficient €statistic for λ.
i=1
Comment: The Poisson is a linear exponential family; X is a sufficient statistic for any linear
€ exponential family.
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17. Let (X1 , X2 , ..., Xn ) be a sample of independent and identically distributed random variables
from an inverse exponential distribution with parameter θ.
Calculate the Rao-Cramer lower bound on the variance of the estimator of the mode.
A.
€
€
θ4
8n
B.
θ4
4n
C.
θ2
8n
17. D. f(x) = θ e-θ/x / x2 .
€
€
ln f(x) = ln[θ] - θ/x - 2 ln[x].
∂ lnf(x)
= 1/θ - 1/x.
∂θ
D.
€
E.
θ2
n
€
∂2 lnf(x)
= -1/θ2.
∂θ2
-1
Rao-Cramer lower bound for θ is:
€
θ2
4n
n E[
∂2 lnf(x)
∂θ2
= θ2/n.
]
As shown in the Tables attached to the exam, for the Inverse Exponential Distribution: Mode = θ/2.
^
2 = Var[ θ
⇒ Var[Mode] = Var[ θ^ ]/2€
]/4 =
€
θ2
.
4n
Comment: The Rao-Cramer lower bound is the minimum variance of any unbiased estimator of a
parameter. €
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Page 18
18. You are given the following information:
• The time to failure of a machine is uniformly distributed on [0, ω].
• The time to failure of each machine is independent.
• An inspection of 20 randomly selected machines revealed the following:
o The average time to failure of the sample was 10.5 years.
o The minimum time to failure of the sample was 2.3 years.
o The maximum time to failure of the sample was 15.6 years.
o The median time to failure of the sample was 12 years.
€
€ Calculate the maximum likelihood estimate for ω based on this sample.
€ A. Less than 14 years
€ B. At least 14 years, but less than 16 years
C. At least 16 years, but less than 18 years
D. At least 18 years, but less than 20 years
E. At least 20 years
18. B. The maximum likelihood estimate is the maximum of the sample = 15.6.
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Page 19
19. You are testing whether actuarial students average more than 6 hours of sleep per night.
You assume that they get at least 6 hours of sleep (H0 : µ ≥ 6) and take a random sample to test if
they actually get less than six hours (Ha: µ < 6).
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• The chosen significance level (α) is 0.10.
• The p-value of this test is 0.07 using the random sample.
• After the test, you are given that the distribution underlying the number of hours that actuarial
students sleep and find out the true underlying average is average 6.2 hours of sleep per night.
• In this case, you were asked to accept or reject the null hypothesis before the true underlying
distribution is known using the results of the random sample and the chosen significance level.
Determine which of the following statements is true once you know the true distribution:
A. You have correctly rejected the null hypothesis.
B. You have correctly failed to reject the null hypothesis.
C. You have correctly accepted the alternative hypothesis.
D. You have committed a Type I error.
E. You committed a Type II error.
19. D. Since the p-value of 7% is less than the significance level of 10%,
you have rejected the null hypothesis.
Since the true underlying average is 6.2 ≥ 6, the null hypothesis is true.
Thus you have committed a Type I error, rejecting the null hypothesis when it is true.
Comment: “accepted the alternative hypothesis” is not correct statistical language.
One should instead say “rejected the null hypothesis.”
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Page 20
20. X is a single observation from the probability density function:
f(x) = 2θx + 1 - θ, for 0 < x < 1.
You are testing the hypothesis:
H0 : θ = 0 versus H1 : θ = 1.
For the most powerful test of significance of size α, determine the critical region for which you would
reject the null hypothesis.
A. x < α
B. x > α
C. x < 1- α
D. x > 1 - α
E. x < α/2
20. D. For θ = 0, f(x) = 1, for 0 < x < 1. For θ = 0, we get a uniform distribution from 0 to 1.
For θ = 1, f(x) = 2x, for 0 < x < 1.
Using the Neyman-Pearson Lemma, the critical region is of the form:
(Likelihood for H1 ) / (Likelihood for H0 ) ≥ k. ⇔ 2x / 1 ≥ k. ⇔ x ≥ c.
α = significance level = Prob[x ≥ c | H0 ] = 1 - c. ⇒ c = 1 - α.
Critical region is: x > 1 - α.
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21. For Jones's political poll the question is what is the probability, p, that a voter would vote for
Jones:
• 15 voters were sampled to test:
H0 : p = 0.5 vs. H1 : p < 0.5.
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• The test statistic Y is the number of sampled voters favoring Jones.
• RR = {y < 3} is selected as the rejection region.
• Assume a binomial distribution for the probability that a voter will favor Jones.
Calculate the probability of a Type I error, α.
A. Less than 0.010
B. At least 0.010, but less than 0.015
C. At least 0.015, but less than 0.020
D. At least 0.020, but less than 0.025
E. At least 0.025
21. C. The significance level is F(3) for a Binomial Distribution with m = 15 and q = 1/2:
⎛ 15⎞
⎛ 15⎞
(1/2)15 {1 + 15 + ⎜ ⎟ + ⎜ ⎟ } = (1 + 15 + 105 + 455) / 32,768 = 0.0176.
⎝ 2 ⎠
⎝ 3 ⎠
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22. A company is evaluating its customer satisfaction through an anonymous survey of 100
randomly selected customers.
Given the following information:
• Surveyed customers score the company continuously between 1 (least satisfied) to
5 (most satisfied).
• The company is considered "best in class" if average customer satisfaction is 4.3 or higher.
• The null hypothesis is that the company is not "best in class".
• The standard deviation of the sample observations is 1.3.
• The selected significance level is 5%.
Use the normal approximation to determine the lowest sample mean, µ,
such that the null hypothesis is rejected.
A. Less than 4.4
B. At least 4.4, but less than 4.5
C. At least 4.5, but less than 4.6
D. At least 4.6, but less than 4.7
E. At least 4.7
22. C. We reject when X > c.
For a sample size of 100, the standard deviation of X is 1.3 /
For a 5% significance
level: c = 4.3 + (1.645) (1.3 /
€
100 .
100 ) = 4.514.
Comment: Usually the sample mean is denoted
by X rather than µ.
€
€
For a smaller sample size, we would use the t-distribution rather than the Normal Distribution.
€
H0 : µ < 4.3 vs. H1 : µ ≥ 4.3.
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23. We are testing the hypothesis that policyholders have no preference concerning the choice of
one of 3 big insurers. The table below shows a random sample of the number of policyholders for
the 3 insurers, for a total of n = 120 insureds:
Insurer 1
Insurer 2
Insurer 3
Observed Insureds
46
36
38
Calculate the χ2 test statistic.
A. Less than 1.00
B. At least 1.00, but less than 1.50
C. At least 1.50, but less than 2.00
D. At least 2.00, but less than 2.50
E. At least 2.50
23. B. The expected numbers of insureds are each: 120/3 = 40.
χ 2 test statistic is: (46 - 40)2 / 40 + (36 - 40)2 / 40 + (38 - 40)2 / 40 = 1.40.
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24. The following data represent a tutor's ranking of 10 psychology students as to their suitability for
their career and their knowledge of psychology:
Student
Career
Psychology
1
4
5
2
10
8
3
3
6
4
1
2
5
9
10
6
2
3
7
6
9
8
7
4
9
8
7
10
5
1
Calculate the Kendall's Tau correlation statistic for Career and Psychology.
A. Less than 0.35
B. At least 0.35, but less than 0.55
C. At least 0.55, but less than 0.75
D. At least 0.75, but less than 0.95
E. At least 0.95
24. B. Arrange the career column from smallest to largest.
Then for the psychology column, count up the numbers concordant and discordant.
For example, of the values below 6 in the psychology column, 4 are greater and 3 are less.
Career
Psychology
Concordant
Discordant
1
2
8
1
2
3
7
1
3
6
4
3
4
5
4
2
5
1
5
0
6
9
1
3
7
4
3
0
8
7
2
0
9
10
0
1
10
8
34
11
Kendall's Tau = (C - D) / (C + D) = (34 - 11) / (34 + 11) = 0.511.
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25. A statistician wishes to know whether longevity is related to heavy smoking.
The table below shows the number of cigarettes per day in the past 5 years paired with the number
of years lived for 9 subjects:
Subject
Cigarettes Years Lived
1
25
63
2
35
68
3
10
72
4
40
62
5
85
65
6
75
46
7
60
51
8
45
60
9
50
55
Calculate the Spearman rho correlation statistic for Cigarettes and Years Lived.
A. Less than -0.70
B. At least -0.70, but less than -0.65
C. At least -0.65, but less than -0.60
D. At least -0.60, but less than -0.55
E. At least -0.55
25. C. Rank the values in each of the columns.
Cigarettes Rank
Years Lived Rank
25
2
63
6
35
3
68
8
10
1
72
9
40
4
62
5
85
9
65
7
75
8
46
1
60
7
51
2
45
5
60
4
50
6
55
3
Then use the electronic calculator to compute the correlation between the 2 sets of ranks: -0.6167.
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26. Losses are separated into two types of claims: Type W and Type X.
• The values of 18 randomly selected claims are provided in the table below:
Type W
Type X
Loss Amount
Rank
Loss Amount
Rank
4,500
9
3,450
1
4,600
10
3,475
2
4,700
11
3,625
3
4,800
12
3,650
4
4,900
13
3,675
5
7,000
14
4,100
6
8,000
15
4,200
7
10,000
16
4,300
8
11,000
17
15,000
18
• H0 : MedianType W = MedianType X
• HA: MedianType W ≠ MedianType X
Calculate the minimum significance level, α, for which H0 can be rejected using
the Mann-Whitney Wilcoxon Test.
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
26. C. U = n1 n2 + n1 (n1 +1)/2 - (sum of the ranks of the first sample of size n1 ) =
(8)(10) + (8)(9)/2 - 100 = 16.
Consulting the table, the p-value for this two-sided test is: (2)(0.17) = 3.4%.
Alternately, totaling the number elements of the first sample less than that of the each of the
elements of the second sample
U = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 8 + 8 = 16. Proceed as before.
Comment: n1 = 8 and n2 = 10.
In order to use the tables attached to the exam, we must arrange things so that n1 ≤ n2 .
The null-hypothesis should be that the two distributions are the same, while the alternative
hypothesis for a two-sided test is that they have different locations.
Given the ranks, we make no use of the given loss amounts.
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27. You are asked to test the hypothesis that the median claim severity is different in year 1 than it is
in year 2.
Claim severities in years 1 and 2 for 5 randomly selected policyholders are in the table below:
Policyholder
Year 1 Severity
Year 2 Severity
1
250
150
2
350
300
3
1,000
100
4
100
85
5
35
90
Calculate the p-value for this hypothesis using the Sign Test.
A. Less than 0.20
B. At least 0.20, but less than 0.25
C. At least 0.25, but less than 0.30
D. At least 0.30, but less than 0.35
E. At least 0.35
27. E. The differences of Year 1 minus Year 2 are: 250 - 150 = 100, 50, 900, 15, -55.
4 out of 5 are positive.
We need to add up densities at 4 and 5 of the Binomial Distribution with m = 5 and q = 1/2,
and then double this sum for this two-sided test.
The p-value for this two-sided test is: (2) (5 + 1)/25 = 0.375.
Comment: One could take the differences in either order, as long as one works consistently.
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28. A manufacturer of gunpowder has developed a new powder, which was tested in 8 shells.
• Let y1 , ... , y8 denote the resulting muzzle velocities.
•
• Muzzle velocities are normally distributed with unknown mean µ and variance of 225.
• µ follows a normal distribution with mean of 2,800 and variance of 2,500.
€
€
€
€
∑ y = 23,672.
€
Calculate the weight assigned to the Maximum Likelihood Estimate, Y , in the Bayes estimator.
A. Less than 0.20
B. At least 0.20, but less than 0.40
€
C. At least 0.40, but less than 0.60
D. At least 0.60, but less than 0.80
E. At least 0.80
28. E. K = (Expected Value of the Process Variance) / (Variance of the Hypothetical Means) =
EPV / VHM = 225 / 2500 = 0.09.
The weight assigned to the observed mean is: Z = N / (N + K) = 8 / (8 + 0.09) = 98.9%.
Comment: We make no use of the given sum of the observations.
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29. You are given the following information for an insurance policy:
• Daily claim frequencies follow a Poisson process with parameter λ.
• The prior distribution of λ follows a gamma distribution with parameters α = 4 and β = 5.
• This policy had 3 claims per day in the first n days.
• The posterior mean of daily claims for this policy after n days is 10/3.
Calculate n.
A. Less than 5.5
B. At least 5.5, but less than 7.5
C. At least 7.5, but less than 9.5
D. At least 9.5, but less than 11.5
E. At least 11.5
Note: I have rewritten this past exam question.
29. D. For this Gamma-Poisson, the posterior distribution is also Gamma with parameters:
αʼ = α + # claims = 4 + 3n, and 1/θʼ = 1/θ + # days = 1/5 + n.
10/3 = posterior mean = αʼ θʼ = (4 + 3n) / (0.2 + n).
⇒ 2 +10 n = 12 + 9n. ⇒ n = 10.
Comment: The original question read:
”The posterior mean of daily claims for this policy after five days is 10/3.”
The parameter€of the Gamma that the statistics books call β is θ in the tables attached to the exam.
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30. You are given the following information about the random variable Y:
• Y is binomial with parameters m and q = θ.
• The prior distribution of θ is beta with parameters a = 9 and b = 5.
• A sample of size 40 is drawn.
• The number of successes in the sample was 15.
Calculate the mean of the posterior distribution of θ.
A. Less than 0.3
B. At least 0.3, but less than 0.4
C. At least 0.4, but less than 0.5
D. At least 0.5, but less than 0.6
E. At least 0.6
30. C. Posterior distribution is also Beta with:
aʼ = a + # successes = 9 + 15 = 26, and bʼ = b + # failures = 5 + 25 = 30.
Posterior mean is: aʼ (aʼ + bʼ) = 26 / (26 + 30) = 0.464.
Comment: They meant to say binomial with parameters m = 1, in other words we have a series of
Bernoulli trials with Y equal to one corresponding to a success.
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31. Given the following information:
• Y is a random variable in the exponential family
yθ - a(θ)
f(y) = c(y, φ) • exp[
]
φ
€
€
€
€
• a(θ) = - -2θ
• θ = -0.3
€
• φ = 1.6
€Calculate E(Y).
A. Less than -1
B. At least -1, but less than 0
C. At least 0, but less than 1
D. At least 1, but less than 2
E. At least 2
31. D. a(θ) = -(-2θ)1/2.
aʼ(θ) = -(-2)(1/2) /
-2θ = 1/ -2θ . E[Y] = aʼ(-0.3) = 1/ 0.6 = 1.291.
Comment: This is an Inverse Gaussian, with θ = -1/(2µ2).
-0.3 = -1/(2µ2). ⇒ µ = 1/ 0.6 = 1.291.
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32. A GLM is used to model claim size. You are given the following information about the model:
• Claim size follows a Gamma distribution.
• Log is the selected link function.
• Scale parameter is estimated to be 2.
• Model Output:
Variable
^
β
(Intercept)
2.32
Location - Urban
0.00
Location - Rural
€
-0.64
Gender - Female
0.00
Gender - Male
0.76
Calculate the variance of the predicted claim size for a rural male.
A. Less than 25
B. At least 25, but less than 100
C. At least 100, but less than 175
D. At least 175, but less than 250
E. At least 250
32. E. Estimated mean severity for a rural male is: exp[2.32 - 0.64 + 0.76] = 11.473.
For the Gamma Distribution, V[µ] = µ2.
Var[Y] = φ V[µ] = (2)(11.4732 ) = 263.3.
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33. You are given the following output from a GLM to estimate the probability of a claim:
• Distribution selected is Binomial.
• Link selected is Logit.
Parameter
β
Intercept
-1.485
Vehicle Body
Coupe
Roadster
Sedan
Station wagon
Truck
Utility
-0.881
-1.047
-1.175
-1.083
-1.118
-1.330
Driver's Gender
Male
-0.025
Area
B
C
D
0.094
0.037
-0.101
Calculate the estimated probability of a claim for:
• Driver Gender: Female
• Vehicle Body: Sedan
• Area: D
€ A. Less than 0.045
€ B. At least 0.045, but less than 0.050
€ C. At least 0.050, but less than 0.055
D. At least 0.055, but less than 0.060
E. At least 0.060
33. D. exp[βx] = exp[-1.485 + 0 - 1.175 - 0.101] = e-2.761 = 0.06323.
eβx
For the logit link function: µ = βx
= 0.06323 / (0.06323 + 1) = 5.95%.
e + 1
€
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34. You are given the following information for a model of vehicle claim counts by policy:
• The response distribution is Poisson and the model has a log link function.
• The model uses two categorical explanatory variables: Number of Youthful Drivers and
Number of Adult Drivers.
• The parameters of the model are given:
Parameter
Degrees of Freedom
Intercept
Number of Youthful Drivers
0
1
Number of Adult Drivers
1
2
1
^
β
-2.663
€
1
0.132
1
-0.031
Calculate the predicted claim count for a policy with one adult driver and one youthful driver.
A. Less than 0.072
B. At least 0.072, but less than 0.074
C. At least 0.074, but less than 0.076
D. At least 0.076, but less than 0.078
E. At least 0.078
34. E. µ = exp[-2.633 + 0.132 + 0] = 0.07957.
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35. You are given a GLM of liability claim size with the following potential explanatory variables
only:
• Vehicle price, which is a continuous variable modeled with a third order polynomial
• Average driver age, which is a continuous variable modeled with a first order polynomial
• Number of drivers, which is a categorical variable with four levels
• Gender, which is a categorical variable with two levels
• There is only one interaction in the model, which is between gender and average driver age.
Determine the maximum number of parameters in this model.
A. Less than 9
B. 9
C. 10
D. 11
E. At least 12
35. C.
Variable
Number of Parameters
Vehicle Price
4
Driver age
2-1=1
Number of drivers
4-1=3
Gender
2-1=1
Interaction Gender & Driver Age 1
Maximum number of parameter is: 4 + 1 + 3 + 1 +1 = 10.
Comment: A model with only Vehicle Price would involve: β0 + β1 (vp) + β 2 (vp)2 + β3 (vp)3 .
The interaction of gender and driver age only uses one parameter since each of gender and driver
age only use one parameter.
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36. You are given the following information for two potential logistic models used to predict the
occurrence of a claim:
• Model 1: (AlC = 262.68)
^
Parameter
€
(Intercept)
Vehicle Value ($000s)
Gender-Female
Gender-Male
• Model 2: (AlC = 263.39)
Parameter
€
€
β
€
-3.264
0.212
0.000
0.727
^
β
(Intercept)
-2.894
Gender-Female
0.000
Gender-Male
€
0.727
• AIC is used to select the most appropriate model.
Calculate the probability of a claim for a male policyholder with a vehicle valued $12,000 by using
the selected model.
A. Less than 0.15
B. At least 0.15, but less than 0.30
C. At least 0.30, but less than 0.45
D. At least 0.45, but less than 0.60
E. At least 0.60
36. D. Smaller AIC is better, so we prefer Model 1.
exp[βx] = exp[-3.264 + (12)(0.212) + 0.727] = e0.007 = 1.007.
eβx
For the logit link function: µ = βx
= 1.007 / (1.007 + 1) = 50.2%.
e + 1
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37. You are given the following table for model selection:
Number of
Parameters (p)
5
11
X
Y
Model
Deviance (= - 2 l_ )
Intercept + Age
A
Intercept + Vehicle Body
392
Intercept + Age + Vehicle Value
€392
Intercept + Age + Vehicle Body + Vehicle Value
B
Calculate Y.
A. Less than 36
B. 36
C. 37
D. 38
E. At least 39
AIC
435
414
446
501
37. C. AIC = (-2) (maximum loglikelihood for model) + 2p.
For example, 414 = 392 + (2)(11).
Thus 446 = 392 + 2X. ⇒ X = 27.
Since the model with Intercept + Vehicle Body has 11 parameters, adding vehicle body to the third
model adds 11 - 1 = 10 parameters.
Thus Y = 27 +€10 = 37.
Comment: Deviance = (2) (saturated max. loglikelihood - maximum loglikelihood for model).
However, in this exam question they have instead used:
deviance = (-2) (maximum loglikelihood for model).
Variable
Number of Parameters
Intercept
1
Age
4
Vehicle Body
10
Vehicle Value
22
Total
37
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38. You are testing the addition of a new categorical variable into an existing GLM.
You are given the following information:
• The change in model deviance after adding the new variable is -53.
• The change in AIC after adding the new variable is -47.
• The change in BIC after adding the new variable is -32.
• Prior to adding the new variable, the model had 15 parameters.
Calculate the number of observations in the model.
A. Less than 1,000
B. At least 1,000, but less than 1,100
C. At least 1,100, but less than 1,200
D. At least 1,200, but less than 1,300
E. At least 1,300
€
€
€
€
€
38. B. Let x be the number of additional parameters for the new model.
Let l1 be the loglikelihood for the original model, and l2 be the loglikelihood for the model including
the new variable.
Deviance = (2) (saturated max. loglikelihood - maximum likelihood for model).
Thus the change in model deviance is: -2( l2€- l1) = -53.
AIC = (-2) (maximum loglikelihood) + (number of parameters)(2).
Thus the change in AIC is: (-2)( l2 - l1) + 2x = -53 + 2x = -47. ⇒ x = 3.
BIC = (-2) (maximum loglikelihood)
€ +€(number of parameters) ln(number of data points).
Thus the change in BIC is: (-2)( l2 - l1) + x ln(n) = -53 + 3 ln(n) = -32. ⇒ n = e7 = 1097.
€
€ €
€
€
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39. You are given the following model output from five candidate models:
Model #
Included Parameters
R2
AIC
BIC
1
2
3
4
5
Gender
Age
Age + Gender
Age + Gender + Income
Age + Gender + Income + Uninsured
0.0162
0.0230
0.0390
0.0755
0.0762
7532
7524
7508
7469
7471
7899
7865
7885
7645
7659
Determine which statement is correct based on the selection criteria above.
A. Model 1 is the best according to R2 .
B. Model 1 is the best according to AIC.
C. Model 4 is the best according to R2 .
D. Model 4 is the best according to AIC.
E. Model 5 is the best according to BIC.
39. D. Smaller AIC is better, so model 4 has the best AIC.
Smaller BIC is better, so model 4 has the best BIC.
Comment: Model 5 has the biggest R2 .
However, one should not compare R2 between models with different numbers of parameters;
R2 always increases when one adds parameters, whether or not that increase is significant.
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40. You are given the following information related to the following linear model:
y ~ x1 + x2 + x3 .
€
• x1 ~ x2 + x3
Source
Regression
Error
Degrees of Freedom
2
503
Sum of Squares
3.73
3.04
€
• x2 ~ x1 + x3
Source
Regression
Error
Degrees of Freedom
2
503
Sum of Squares
24.74
226.34
€
• x3 ~ x1 + x2
Source
Regression
Error
Degrees of Freedom
2
503
Sum of Squares
214,258
85,883
Calculate the variable inflation factor for the variable which exhibits the greatest collinearity in
the original model.
A. Less than 1.0
B. At least 1.0, but less than 2.0
C. At least 2.0, but less than 3.0
D. At least 3.0, but less than 4.0
E. At least 4.0
40. D. If we regress the ith independent variable against all of the other independent variables,
then the Variance Inflation Factor is: VIFi = 1/(1 - Ri2 ) ≥ 1.
If one or more of the VIFs is large, that is an indication of multicollinearity.
R2 = RSS/TSS = RSS / (RSS + ESS).
For the first case: R1 2 = 3.73 / (3.73 + 3.04) = 0.5510. VIF1 = 1/(1 - 0.5510) = 2.23.
For the second case: R2 2 = 24.74 / (24.74 + 226.34) = 0.0985. VIF2 = 1/(1 - 0.0985) = 1.11.
For the third case: R3 2 = 214,258 / (214,258 + 85,883) = 0.7139. VIF3 = 1/(1 - 0.7139) = 3.49.
Comment: y ~ x1 + x2 + x3 is intended to mean a linear model with x1 , x2 and x3 as the
independent variables: Y = β0 + β1 X1 + β2 X2 + β3 X3 .
Highest R2 . ⇔ Most collinear with the other variables.
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41. You are building a model of claim counts.
To evaluate the risk of overdispersion, you find the following:
• The log-likelihood of the Poisson regression is lP = 350.
• The log-likelihood of the Negative Binomial regression is lNB .
• 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 lNB .
€
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
41. C. The test statistic is twice the difference in loglikelihoods: (2)( lNB - 350).
For this particular test comparing a Negative Binomial to its limit a Poisson, for significance level α we
compare to the 2α tail of the Chi-Square Distribution, or in this case (2)(0.5%) = 1%.
For one degree of freedom, the 1% critical value is 6.63. €
Thus we reject when (2)( lNB - 350) > 6.63. lNB > 353.3.
Comment: See page 91 of Generalized Linear Models for Insurance Data by de Jong and Heller.
Since the Poisson is a limit rather than a special case of the Negative Binomial, we need to modify
the usual likelihood
€ ratio test. The test
€ statistic is zero half of the time and follows a Chi-Square
Distribution with one degree of freedom the other half of the time.
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42. An AR(1) model was fit to the following time series data through time t = 7.
• The mean was subtracted from the data before the parameter was estimated.
• The estimated parameter for the model is -0.79.
time(t)
xt
€
€
1
2
3
4
5
6
7
mean
8.70
7.00
8.60
7.40
8.30
7.60
8.50
8.01
Calculate the forecast for the observation at t = 9, x^9 .
A. Less than 8.25
B. At least 8.25, but less than 8.35
C. At least 8.35, but less than 8.45
D. At least 8.45, but less than 8.55
E. At least 8.55
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42. B. Since the mean was subtracted from the data before the parameter was estimated,
for this AR(1) model: (xt - µ) = -0.79(xt-1 - µ) + wt.
Thus x^ 8 = 8.01 + (-0.79)(8.50 - 8.01) = 7.6229.
x^9 = 8.01 + (-0.79)(7.6229 - 8.01) = 8.316.
€
€
Comment: E[wt] = 0. The forecasts are for the expected future values; the observed future values
will vary about their mean due to their error terms.
The time series and forecasted values:
x
8.5
8.0
Forecasted
7.5
7.0
2
4
6
8
Time
HCMSA-2015-S
Solutions Fall 2015 CAS Exam S,
1/14/16,
Page 44
43. The three models AR(2), MA(1) and ARMA(2,1) are fitted to the following time series:
Year
Quarter 1
Quarter 2
Quarter 3
Quarter 4
2010
112
104
100
96
2011
101
101
105
94
2012
106
106
108
110
The results using R are as follows:
AR(2)
ar1
0.2618
Std Err
0.3289
σ2 = 25.02
Std Err
intercept
104.227
2.3972
log likelihood = -36.4
MA(1)
ARMA(2,1)
ar2
0.0868
0.3326
Std Err
ma1
0.1899
0.2574
σ2 = 25.66
log likelihood = -36.51
ar1
-0.5547
0.3329
σ2 = 21.31
intercept
103.798
1.7373
ar2
0.4437
0.3327
ma1
0.9779
0.235
intercept
104.4542
2.4584
log likelihood = -36.02
The models are ranked using the Akaike Information Criterion (AIC).
Determine the order from best to worst model.
A. AR(2), MA(1), ARMA(2,1)
B. AR(2), ARMA(2,1), MA(1)
C. MA(1), AR(2), ARMA(2,1)
D. MA(1), ARMA(2, 1), AR(2)
E. ARMA(2, 1), AR(2), MA(1)
43. C. AIC = (-2) (loglikelihood) + (2) (number of parameters).
AIC for the AR(2) model is: (-2)(-36.4) + (2)(4) = 80.8.
AIC for the MA(1) model is: (-2)(-36.51) + (2)(3) = 79.02.
AIC for the ARMA(2,1) model is: (-2)(-36.02) + (2)(5) = 82.04
Smaller AIC is better. Thus MA(1) is best and ARMA(2,1) is worst.
Comment: The AR(2) model is: xt = 0.2618xt-1 + 0.0868xt-2 + wt.
The MA(1) model is: xt = wt + 0.1899wt-1.
The ARMA(2,1) model is: xt = -0.5547xt-1 + 0.4437xt-2 + wt + 0.9779wt-1.
In each case we have fit σ2 and the intercept. However, when using AIC for purposes of comparing
models one could ignore these two fitted parameters since they are fit in every model.
HCMSA-2015-S
Solutions Fall 2015 CAS Exam S,
1/14/16,
Page 45
44. The two time series {xt} and {yt} are stationary in the mean and the variance.
The following 8 sample values are available:
t
xt
yt
1
10
8
2
12
9
3
11
11
4
16
12
5
19
15
6
22
16
7
25
20
8
29
21
Calculate the sample cross covariance function for lag 4.
A. Less than -9
B. At least -9, but less than-8
C. At least -8, but less than -7
D. At least -7, but less than -6
E. At least -6
44. B. X = 18. Y = 14.
c4 (x, y) = {(x5 - X )(y1 - Y ) + (x6 - X )(y2 - Y ) + (x7 - X )(y3 - Y ) + (x8 - X )(y4 - Y )} / 8 =
{(19 - 18)(8 - 14) + (22 - 18)(9 - 14) + (25 - 18)(11 - 14) + (29 - 18)(12 - 14)} / 8 = -8.625.
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Comment: This exam question should have specified whether they wanted c4 (x, y), where x is the
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€
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input and y is the response, or instead c4 (y, x).
c4 (y, x) = {(y5 - Y )(x1 - X ) + (y6 - Y )(x2 - X ) + (y7 - Y )(x3 - X ) + (y8 - Y )(x4 - X )} / 8 =
{(1)(-8) + (2)(-6) + (6)(-7) + (7)(-2)} / 8 = -9.5.
1
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Equation
with€
R: ck(x,€y) =
n
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n-k
∑ (x€t + k - €x) (yt - y) .
t=1
HCMSA-2015-S
Solutions Fall 2015 CAS Exam S,
1/14/16,
Page 46
45. You are given the following AR processes:
1
Model I: xt = 2 xt-1 + wt
Model II: xt = xt-1 + 2 xt-2 + wt
1
1
Model
€ III: xt = 2 xt-1 + 2 xt-2 + wt
Determine which processes are stationary.
A. Model I only
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B. Model II only
C. Model III only
D. Model I and II only
E. Model I and III only
45. A. All roots of the characteristic equation must exceed one in absolute value, in order for an AR
process to be stationary.
For Model I, the characteristic equation is: 1 - B/2 = 0. ⇒ B = 2. |B| > 1. Yes.
For Model II, the characteristic equation is: 1 - B - 2B2 = 0. ⇒
B=
1 ±
1 + 8
= 1/2 or -1. No.
-4
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€
For Model III, the characteristic equation is: 1 - B/2 - B2 /2 = 0. ⇒
€
€
B=
1/ 2 ±
1/ 4 + 2
= 1 or -2. No.
-1
END OF EXAMINATION
€