solutions to the fall 2016 cas exam showardmahler.com/teaching/s_files/solsf16casexams.pdfusing the...

Solutions to the Fall 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 Sprepared by

Howard C. Mahler, FCASCopyright 2016 by Howard C. Mahler.

Howard [email protected]/Teaching

1. You are given the following information:

€

• An insurance company pays claims according to a Poisson process at a rate of 5 per day.

€

• Claims are sub-divided into three categories: Minor, Major, and Severe, with claim amounts provided below: Category Claim AmountMinor 1 Major 4 Severe 10

€

• It is known that the proportion of claims in the Severe category is 0.15.

€

• The total expected claim payment amount in one day is 17. Calculate the proportion of Major claims. A. Less than 0.25 B. At least 0.25, but less than 0.40 C. At least 0.40, but less than 0.55 D. At least 0.55, but less than 0.70 E. At least 0.70

1. B. Let the proportion of Major claims be p.17 = (5) {(1)(0.85 - p) + 4p + (10)(0.15)}.

€

⇒ p = 0.35.

Solutions Fall 2016 CAS Exam S, HCM 11/30/16, Page 1

2. For an auto policy, claims occur according to a Poisson process with rate λ = 10 per day. Once reported, claims are independently classified as Property Damage (PD) only, Bodily Injury (BI) only, or both PD and BI. You are given that 30 percent of claims are PD only, 20 percent are BI only, and the remaining 50 percent are both PD and BI. Calculate the probability that in a given day there will be exactly 4 PD only claims, 2 BI only claims, and 6 both PD and BI claims. A. Less than 0.63 percent B. At least 0.63 percent, but less than 0.68 percent C. At least 0.68 percent, but less than 0.73 percent D. At least 0.73 percent, but less than 0.78 percent E. At least 0.78 percent

2. B. Thinning, we get three independent Poisson Processes:PD with λ = 3, BI with λ = 2, and Both with λ = 5.Prob[4 PD] = e-3 34 / 4! = 0.1680. Prob[2 BI] = e-2 22 / 2! = 0.2707.Prob[6 Both] = e-5 56 / 6! = 0.1462.Thus, the desired probability is: (0.1680)(0.2707)(0.1462) = 0.665%.


3. The time X to wait in line is an exponentially distributed random variable with mean 5 minutes. Calculate the probability that the total waiting time will be longer than 30 minutes from the time that individual arrived in line, given that the wait has already been 20 minutes. A. Less than 0.1 B. At least 0.1, but less than 0.2 C. At least 0.2, but less than 0.3 D. At least 0.3, but less than 0.4 E. At least 0.4

3. B. Since the process is memoryless, we want the probability of waiting at least 10 more minutes: e-10/5 = 0.135.


4. Steve catches fish at a Poisson rate of 3 per hour. The price Steve gets at the market for each fish is randomly distributed as follows:

Price Probability $10 20% $20 60% $30 20%

Using the normal approximation without a continuity correction, calculate the probability that Steve will receive at least $300 for fish caught in a four hour-period. A. Less than 0.19 B. At least 0.19, but less than 0.20 C. At least 0.20, but less than 0.21 D. At least 0.21, but less than 0.22 E. At least 0.22

4. C. Mean severity is: (20%)(10) + (60%)(20) + (20%)(30) = 20.Second moment of severity is: (20%)(102) + (60%)(202) + (20%)(302) = 440.For 4 hours, mean aggregate is: (3)(4)(20) = 240.Variance of aggregate is: (3)(4)(440) = 5280.Without using the continuity correction:

Prob[at least 300] = 1 - Φ[

€

300 - 2405280

] = 1 - Φ[0.83] = 1 - 0.7967 = 0.2033.


5. A call center currently has 2 representatives and 2 interns who can handle customer calls. If all representatives including interns are currently on a call, an incoming call will be placed on hold until a representative or intern is available. You are given the following information:

€

• For each representative, the time taken to handle each call is given by an exponential distribution with a mean value equal to 1.

€

• For each intern, the time taken to handle each call is given by an exponential distribution with mean 2.

€

• Handle times are independent A customer calls the call center and is placed on hold, and is the first person in line. Calculate the expected time to complete the call (including both hold time and service). A. Less than 1.2 B. At least 1.2, but less than 1.4 C. At least 1.4, but less than 1.6 D. At least 1.6, but less than 1.8 E. At least 1.8

5. D. The time until he is served, is the minimum of 4 independent Exponentials with hazard rates: 1, 1, 1/2, 1/2. This is an Exponential with hazard rate: 1 + 1 + 1/2 + 1/2 = 3.

€

⇒ The mean time waiting is 1/3.

The probability that the first person available is a representative is:

€

(2)(1)(2)(1) + (2)(1/ 2) = 2/3.

€

⇒ The mean service time is: (2/3)(1) + (1 - 2/3)(2) = 4/3.

€

⇒ Expected total time is: 1/3 + 4/3 = 5/3.



€

• At the time of installation a given machine has the following survival function:

S(x) =

€

1 - x100

⎛

⎝ ⎜

⎞

⎠ ⎟ 1/ 2

, for 0 ≤ x ≤ 100; x in months

€

•After 16 months, that machine is still functioning. Calculate the probability that the machine will stop working between months 36 and 51. A. Less than 0.095 B. At least 0.095, but less than 0.105 C. At least 0.105, but less than 0.115 D. At least 0.115, but less than 0.125 E. At least 0.125

6. C. S(16) =

€

1 - 16 /100 = 0.9165.

S(36) =

€

1 - 36 /100 = 0.8. S(51) =

€

1 - 51/100 = 0.7

Prob[fails between 36 and 51 | survives to 16] =

€

S(36) - S(51)S(16) =

€

0.8 - 0.70.9165 = 0.1091.


7. You are given the following information about a watch with 6 different parts:

€

• There are 3 red wires with expected lifetimes of 50, 75, and 100.

€

• There are 3 yellow wires with expected lifetimes of 25, 50, and 75.

€

• The lifetimes of all wires are independent and exponentially distributed. Calculate the probability that a red wire will break down before a yellow wire. 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

7. E. Minimum of the failure times for red wires is Exponential with hazard rate: 1/50 + 1/75 + 1/100 = 0.0433.Minimum of the failure times for yellow wires is Exponential with hazard rate: 1/25 + 1/50 + 1/75 = 0.0733.

Prob[red before yellow] =

€

0.04330.0433 + 0.0733 = 37.1%.


8. For a parallel system with two independent machines, you are given the following information:

€

• The hazard rate for each machine is:

µx =

€

1100 - x , for 0 ≤ x < 100; x in months

€

• One machine has worked for 40 months and the other machine has worked for 60 months. Calculate the probability that the system will function for 20 more months. 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

8. D. H(x) =

€

1100 - t0

x∫ dt = ln[100] - ln[100 - x].

S(x) = Exp[-H(x)] = (100-x)/100 = 1 - x/100, for 0 ≤ x < 100.

Prob[Machine 1 fails in next 20 months] =

€

S(40) - S(60)S(40) =

€

0.6 - 0.40.6 = 1/3.

Prob[Machine 2 fails in next 20 months] =

€

S(60) - S(80)S(60) =

€

0.4 - 0.20.4 = 1/2.

Prob[systems functions for next 20 months] = 1 - (1/3)(1/2) = 5/6.

Comment: For a uniform distribution from 0 to ω, µx =

€

1ω - x .



€

•A system has two minimal path sets: {1, 2, 4} and {1, 3, 4}.

€

• Reliability for components 1 and 2 is uniformly distributed from 0 to 1.

€

• Reliability for components 3 and 4 is uniformly distributed from 0 to 2.

€

• All components in the system are independent.

€

• You are starting at time 0. Calculate the probability that the lifetime of the system will be less than 0.25. A. Less than 0.30 B. At least 0.30, but less than 0.35 C. At least 0.35, but less than 0.40 D. At least 0.40, but less than 0.45 E. At least 0.45

9. C. The system function if either: 1, 2, and 4 function, or 1, 3, and 4 function.Thus the system fails if either: 1 fails, 4 fails, or 2 and 3 both fail.Prob[1 fails by 0.25] = 0.25.Prob[4 fails by time 0.25] = 0.125.Prob[2 and 3 both fail by time 0.25] = (0.25)(0.125) = 0.03125.

€

⇒ The probability that system functions until 0.25 is: (1 - 0.25) (1 - 0.125) (1 - 0.03125) = 0.6357.

€

⇒ The probability that the lifetime of the system will be less than 0.25 is: 1 - 0.6357 = 0.3643.Alternately, Prob[system functions until time 0.25] = Prob[1, 2, and 4 function until time 0.25] + Prob[1, 3, and 4 function until time 0.25]

- Prob[1, 2, 3, and 4 function until time 0.25] =(0.75)(0.75)(0.875) + (0.75)(0.875)(0.875) - (0.75)(0.75)(0.875)(0.875) = 0.6357.

€

⇒ The probability that the lifetime of the system will be less than 0.25 is: 1 - 0.6357 = 0.3643.Comment: S(t) = (1- t)(1 - t/2)(1 - t t/2) = 1 - 1.5t + 0.75t3 - 0.25t4, for t ≤ 1. S(1) = 0.Thus the mean lifetime of the system is:

€

(1 - 1.5t + 0.75t3 - 0.25t4) dt0

1∫ = 1 - 1.5/2 + 0.75/4 - 0.25/5 = 0.3875.


10. You are given the following information about a system:

€

• This is a 2-out-of-3 system.

€

• All components are independent.

€

• The probability of each component functioning is p = 0.90. Calculate the reliability of the system. A. Less than 0.970 B. At least 0.970, but less than 0.975 C. At least 0.975, but less than 0.980 D. At least 0.980, but less than 0.985 E. At least 0.985

10. B. System works if either exactly 2 or exactly 3 components work.Thus the probability that the system works is: (3)(0.92)(0.1) + 0.93 = 0.972.


11. A firm classifies its workers into one of four categories based on their employment status: 1. Full-Time 2. Part-Time 3. On-Leave 4. Retired

Workers independently transition between categories at the end of each year according to a Markov chain with the following transition probability matrix:

P =

€

0.6 0.1 0.1 0.20.4 0.4 0.1 0.10.1 0.1 0.7 0.10.0 0.0 0.0 1.0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

Calculate the probability a current full-time employee will ever go on leave. 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


11. C. Let x be the chance that if the chain is in state 1 it ever enters state 3 in the future.Let y be the chance that if the chain is in state 2 it ever enters state 3 in the future.(Since state 4 is an absorbing state, if the chain is in state 4 it can not enter state 3 in the future.)If the chain is in state 1, the probabilities of the next state are: 0.6, 0.1, 0.1, and 0.2.Therefore: x = 0.6x + 0.1y + 0.1.

€

⇒ 4x = y + 1If the chain is in state 2, the probabilities of the next state are: 0.4, 0.4, 0.1, and 0.1.Therefore: y = 0.4x + 0.4y + 0.1.

€

⇒ 6y = 4x + 1.

€

⇒ 4x = (4x+1)/6 + 1.

€

⇒ x = 7/20 = 0.35.

€

⇒ y = (4)(0.35) - 1 = 0.4. Comment: sij = the expected number of time periods the chain is in state j, if it starts in state i, for i and j both transient states.Let PT be the matrix of transition probabilities for transient states only. Then S = (I-PT)-1.fij = probability that the chain visits state j at some time in the future, if it starts in state i, for both i and j transient states. fij = (sij - δij)/sjj, where δij = 1 if i = j and 0 if i ≠ j.

PT =

€

0.6 0.1 0.10.4 0.4 0.10.1 0.1 0.7

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ . I-PT =

€

0.4 -0.1 -0.1-0.4 0.6 -0.1-0.1 -0.1 0.3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

Using a computer, (I-PT)-1 =

€

3.7778 0.8889 1.55562.8889 2.4444 1.77782.2222 1.1111 4.4444

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

Then f13 = (s13 - 0)/s33 = 1.5556/4.4444 = 0.35. f23 = (s23 - 0)/s33 = 1.7778/4.4444 = 0.40.



€

• Stock XYZ pays a quarterly dividend based on the current state of the economy, as follows: State Quarterly Dividend per Share 1: Economic Peak 0.30 2: Recession 0.01 3: Economic Trough 0.00 4: Economic Expansion 0.20

€

• The economy changes among these four states each quarter according to the transition probability matrix:

P =

€

0.20 0.80 0.00 0.000.00 0.70 0.30 0.000.00 0.00 0.40 0.600.40 0.00 0.00 0.60

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

€

• Annual interest rate i = 0%.

€

• Transition happens after the dividend is paid. Calculate the long-term quarterly dividend per share for Stock XYZ. A. Less than 0.08 B. At least 0.08, but less than 0.11 C. At least 0.11, but less than 0.14 D. At least 0.14, but less than 0.17 E. At least 0.17

12. B. Going down the columns of the transition matrix, the balance equations are:0.2 π1 + 0.4 π4 = π1.

€

⇒ π4 = 2π1.0.8 π1 + 0.7 π2 = π2.

€

⇒ π2 = 8π1/3.0.3 π2 + 0.4 π3 = π3.

€

⇒ π3 = π2/2 = 4π1/3.0.6 π3 + 0.6 π4 = π4.Also, π1 + π2 + π3 + π4 = 1.

€

⇒ π1 (1 + 8/3 + 4/3 + 2) = 1.

€

⇒ π1 = 1/7.

€

⇒ π2 = 8/21.

€

⇒ π3 = 4/21.

€

⇒ π4 = 2/7.

€

⇒ The long-term quarterly dividend per share is: (1/7)(0.3) + (8/21)(0.01) + (4/21)(0) + (6/21)(0.2) = 0.1038.


13. You are given the following information for ABC Insurance Company, which sells auto policies:

€

• Customers transition each policy period between three states. The average service cost per customer per policy period varies by state as follows: State Average Service Cost per Customer1: Gold 2 2: Silver 1 3: Copper 15

€

• The transition probability matrix is:

P =

€

0.80 0.20 0.000.20 0.60 0.200.00 0.40 0.60

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Calculate the long-term average service cost per customer per policy period. A. Less than 3.0 B. At least 3.0, but less than 5.0 C. At least 5.0, but less than 7.0 D. At least 7.0, but less than 9.0 E. At least 9.0

13. B. Going down the columns of the transition matrix, the balance equations are:0.8 π1 + 0.2 π2 = π1.

€

⇒ π1 = π2.0.2 π1 + 0.6 π2 + 0.4 π3 = π2. 0.2 π2 + 0.6 π3 = π3.

€

⇒ π3 = π2/2.Also, π1 + π2 + π3 = 1.

€

⇒ π2 (1 + 1 + 1/2) = 1.

€

⇒ π2 = 0.4.

€

⇒ π1 = 0.4.

€

⇒ π3 = 0.2.

€

⇒ The long-term average service cost per customer per policy period is: (0.4)(2) + (0.4)(1) + (0.2)(15) = 4.2.


14. Two models are used to estimate the probability of at least one claim for an auto policy in a given year. The models are defined as follows:

€

• Model I: Independent events with probability of occurrence of at least one claim in a given year equal to 0.1.

€

• Model II: Markov Chain with conditional probability of at least one claim given no claims in prior year equal to 0.08, and conditional probability of at least one claim given at least one claim in prior year equal to 0.12.

There are no claims in Year 1. The probability of at least one claim in Year 2 and no claims in Year 3 and at least one claim in Year 4 are denoted by PI and PII, for models I and II respectively. Calculate I PI - PII I.A. Less than 0.3 percent B. At least 0.3 percent, but less than 0.4 percent C. At least 0.4 percent, but less than 0.5 percent D. At least 0.5 percent, but less than 0.6 percent E. At least 0.6 percent

14. B. For Model I, the probability is: (0.1)(0.9)(0.1) = 0.009.For Model II, the probability is: (0.08)(0.88)(0.08) = 0.005632.The absolute difference is: | 0.009 - 0.005632 | = 0.3368%.Comment: Does not really make use of Markov Chain techniques.

For Model II, the transition matrix is:

€

NoYes

€

0.92 0.080.88 0.12⎛

⎝ ⎜

⎞

⎠ ⎟ .



€

• A new medical breakthrough is expected to reduce the probability of death between ages 70 and 75 by one-half.

€

• Prior to the breakthrough, mortality follows the Illustrative Life Table. Calculate the probability that a life (50) lives to age 75 after the breakthrough. A. Less than 0.64 B. At least 0.64, but less than 0.66 C. At least 0.66, but less than 0.68 D. At least 0.68, but less than 0.70 E. At least 0.70

15. C. For the Illustrative Life Table: 20p50 = S(70)/S(50) = 6,616155/8,950,901 = 0.7392.For the Illustrative Life Table: 5p70 = S(75)/S(70) = 5,396,081/6,616155 = 0.8156.However, the probability of death between ages 70 and 75 is only one-half of what it was or:(1 - 0.8156)/2 = 0.0922.Thus the probability that a life (50) lives to age 75 after the breakthrough is: (0.7392)(1 - 0.0922) = 0.6710.Comment: Multiplying the mortality rate by a constant, takes the survival function to the power of that constant. Thus if instead the mortality rates between ages 70 and 75 were reduced by a half,then the probability that a life (50) lives to age 75 would be: (0.7392)(0.81562) = 0.6676.



€

• There are three independent lives (50), (60), and (70).

€

• Mortality follows the Illustrative Life Table. Calculate the probability that (70) will be the only person alive after 20 years. A. Less than 0.0200 B. At least 0.0200, but less than 0.0220 C. At least 0.0220, but less than 0.0240 D. At least 0.0240, but less than 0.0260 E. At least 0.0260

16. B. 20p50 = S(70)/S(50) = 6,616155/8,950,901 = 0.7392.20p60 = S(80)/S(60) = 3,914,365/8,188,074 = 0.4781.20p70 = S(90)/S(70) = 1,058,491/6,616155 = 0.1560.The probability that (70) will be the only person alive after 20 years is:(0.1560) (1 - 0.7392) (1 - 0.4781) = 0.0212.


17. You are given an independent random sample of size n, x1 x2, ... , xn, from an exponential distribution with mean θ.

Calculate the Rao-Cramer lower bound for

€

θ , an unbiased estimator of θ.

A. θ B.

€

X C. 1/θ2 D. θ2/n E. 1/

€

θ 2

17. D. f(x) = (1/θ) e-x/θ. ln f(x) = - ln(θ) - x/θ.

€

∂ln[f(x)]∂θ

= -1/θ + (x/θ2).

€

∂2ln[f(x)]∂θ2

= 1/θ2 - (2x/θ3).

€

⇒ E[

€

∂2ln[f(x)]∂θ2

] = 1/θ2 - 2E[X]/θ3 = 1/θ2 - 2θ/θ3 = -1/θ2.

Rao-Cramer lower bound for

€

θ is:

€

-1

n E[∂2 ln[f(x)]∂θ2

] = θ2 / n.

Alternately, the Exponential Distribution is a linear exponential family, and thus maximum likelihood is equal to the method of moments and the variance of the estimated parameter is equal to the Rao-Cramer lower bound.

€

θ =

€

X .

€

⇒ Var[

€

θ ] = Var[

€

X ] = Var[X]/n = θ2 / n. Comment: The Rao-Cramer lower bound contains 1/n, eliminating all of the choices but D.


18. You are given an independent random sample of size 50, y1, y2, ... , y50, from a population

with mean µ and variance σ2. Two estimators for µ are:

€

•

€

µ1 = the sample mean

€

•

€

µ2 = the weighted average sample mean, where the first 10 observations receive twice as much weight as the other observations.

Calculate the efficiency of

€

µ1 relative to

€

µ2. A. Less than 1.025 B. At least 1.025, but less than 1.050 C. At least 1.050, but less than 1.075 D. At least 1.075, but less than 1.100 E. At least 1.100

18. E. Var[

€

µ1] = σ2/50 = 0.02σ2.

€

µ2 =

€

2 Xi + Xi 11

50∑

i=1

10∑

(2)(10) + 40 =

€

2 Xi + Xi 11

50∑

i=1

10∑

60 .

Var[

€

µ2] =

€

22 Var[ Xi] + Var[ Xi] 11

50∑

i=1

10∑

602 =

€

(4)(10σ2) + (40σ2)602 = σ2 80/3600 = 0.02222σ2.

The efficiency of

€

µ1 relative to

€

µ2 is: Var[

€

µ2]/Var[

€

µ1] = 0.02222 / 0.2 = 1.111. Comment: For two unbiased estimators of the same quantity, define the relative efficiency of estimator B compared to estimator A as Variance[A]/Variance[B].Estimator 1 has a smaller variance than estimator 2, thus estimator 1 is more efficient..


19. You are given: I. If

€

θ n is an unbiased estimator of θ and the variance of

€

θ n = 1/n, it must be a consistent estimator

of θ.

II. If

€

θ n is a consistent estimator of θ, it must be an unbiased estimator of θ.

III. If

€

θ n is a biased estimator of θ, it is not a consistent estimator of θ.

Determine which of the following is correct. A. I is true; II is false; III is true. B. I is false; II is false; III is true. C. I is true; II is false; III is false. D. I is true; II is true; III is false. E. I is false; II is true; III is true,

19. C. An asymptotically unbiased estimator, whose variance goes to zero as the number of data points goes to infinity, is consistent. Unbiased

€

⇒ Asymptotically unbiased. So Statement I is true.Statement II is not true.Statement III is not true.Comment: An estimator could have any of the four possible combinations of biased/unbiased and consistent/inconsistent.Statement 2: Consistent.

€

⇒ Unbiased.Statement 3: Not Unbiased.

€

⇒ Not Consistent.Statement 3 is the contrapositive of Statement 2; thus Statements 2 and Statements 3 are either both true or both false.


20. You are given an independent random sample from a normal distribution X with unknown mean, µ. You are given the following information:

€

• σ2(X) = 400

€

• H0: µ = 0

€

• H1: µ = 10

€

• You will reject the null hypothesis if

€

X > α for some value of α. Calculate the minimum sample size needed so that the probability of a Type I error and the probability of a Type II error are each no more than 10%. A. Less than 10 B. At least 10, but less than 15 C. At least 15, but less than 20 D. At least 20, but less than 25 E. At least 25

20. E. Variance of the average is 400/n.

10% = significance = Prob[

€

X > α | µ = 0] = 1 - Φ[

€

α

400 / n].

€

⇒

€

α

400 / n = 1.282.

10% = prob. of a Type II error = Prob[

€

X < α | µ = 10] = 1 - Φ[

€

α - 10400 / n

].

€

⇒

€

α - 10400 / n

= -1.282.

Dividing the two equations: α/(α-10) = -1.

€

⇒ α = 5.

€

⇒ n = 400 (1.282/5)2 = 26.3.Since the sample size is integer, the minimum sample size needed is 27.


21. You are given:

€

• A coin has two sides, Heads and Tails

€

• This coin is tossed 400 times

€

• H0: Probability of Heads = 0.50

€

• H1: Probability of Heads ≠ 0.50

€

• H0 is not rejected if the number of heads in the 400 tosses is between 184 and 216 inclusive, and rejected otherwise

€

• The normal approximation to the binomial distribution is used Calculate the probability of a Type I error. A. Less than 0.01 B. At least 0.01, but less than 0.03 C. At least 0.03, but less than 0.05 D. At least 0.05, but less than 0.07 E. At least 0.07

21. E. For q = 0.5, the mean is 200 and the variance is: (0.5)(1 - 0.5)(400) = 100.

Prob[reject | H0] = Φ[

€

216.5 - 200100

] - Φ[

€

183.5 - 200100

] = Φ[1.65] - Φ[-1.65]

= 0.9505 - 0.0495 = 0.9010.Probability of a Type I error is: 1 - 0.9010 = 0.0990.Without using the continuity correction:

Prob[reject | H0] = Φ[

€

216 - 200100

] - Φ[

€

184 - 200100

] = Φ[1.60] - Φ[-1.60] = 0.8904.

Probability of a Type I error is: 1 - 0.9010 = 0.1096.


22. You are given:

€

• A coin has two sides, Heads and Tails

€

• This coin is tossed 150 times

€


€


€

• H0 is not rejected if the number of heads in the 150 tosses is less than 79, and rejected otherwise

€

• The normal approximation to the binomial distribution is usedCalculate the probability of a Type II error. A. Less than 0.01 B. At least 0.01, but less than 0.03 C. At least 0.03, but less than 0.05 D. At least 0.05, but less than 0.06 E. At least 0.06

22. B or C. For q = 0.6, the mean is: (0.6)(150) = 90, and variance is: (0.6)(0.4)(150) = 36.

Probability of a Type II error = Prob[fail to reject | H1] = Φ[

€

78.5 - 9036

] = Φ[-1.92] = 2.74%.

Without using the continuity correction one could get either of two answers:

Φ[

€

79 - 9036

] = Φ[-1.83] = 3.36%, or Φ[

€

78 - 9036

] = Φ[-2.00] = 2.28%.


23. You are given: • H0: θ = θ0

€

• H1: θ = θ1

€

• An independent random sample is drawn from a distribution with probability mass function f(x|θ),for x = 1, 2, ... , 7. The values of the likelihood function at θ0 and θ1 are given in the table below.

x 1 2 3 4 5 6 7 f(x|θ0) 0.01 0.01 0.01 0.01 0.01 0.01 0.94 f(x|θ1) 0.06 0.05 0.04 0.03 0.02 0.01 0.79 The Neyman-Pearson lemma is used to find the most powerful test for H0 versus H1 with significance level α = 0.04. Calculate the probability of Type II error for this test. A. Less than 0.02 B. At least 0.02, but less than 0.32 C. At least 0.32, but less than 0.62 D. At least 0.62, but less than 0.92 E. At least 0.92

23. D. The critical region is where f(x|θ1)/f(x|θ0) is big.x 1 2 3 4 5 6 7 f(x|θ0) 0.01 0.01 0.01 0.01 0.01 0.01 0.94 f(x|θ1) 0.06 0.05 0.04 0.03 0.02 0.01 0.79 f(x|θ1)/f(x|θ0) 6 5 4 3 2 1 0.8404

We want the critical region to cover 4% probability when H0 is true.Thus we reject when x is 1, 2, 3, or 4.Thus the probability of Type II error = Prob[fail to reject | H1] = Prob[x ≥ 5 | H1]= 2% + 1% + 79% = 82%.Comment: We need to assume that a single observation is being used to test H0 vs. H1.If we are taking a sample of size greater than 1, things quickly get complicated.The power of the test is: 0.06 + 0.05 + 0.04 + 0.03 = 0.18 = 1 - 0.82.


24. You are given an independent random sample of 10 draws from an exponential distribution with mean θ. The lowest value is excluded, and Y is the expected average value of the remaining 9 draws. Calculate Y / θ. A. Less than 1.02 B. At least 1.02, but less than 1.07 C. At least 1.07, but less than 1.12 D. At least 1.12, but less than 1.17 E. At least 1.17

24. C. The expected value of the whole sample is θ.The expected value of the minimum is θ/10.The expected value of the remaining 9 claims is Y.Therefore, θ = (1/10) (θ/10) + (9/10) Y.

€

⇒ Y = 1.1θ.

€

⇒ Y/θ = 1.1. Alternately, E[X(2)] = θ(1/10 + 1/9). E[X(3)] = θ(1/10 + 1/9 + 1/8). E[X(10)] = θ(1/10 + 1/9 + 1/8 + ... + 1/1).Y = {E[X(2)] + E[X(3)] + ... + E[X(10)]} / 9 = θ{9/10 + 9/9 + 8/8 + 7/7 + ... + 1}/9 = θ9.9/9 = 1.1θ.

€

⇒ Y/θ = 1.1.


25. You are given an independent random sample of size 40, x1, x2, ... , x40, from an exponential distribution with mean 150. Calculate the probability that the second smallest observation (the observation with one observation smaller and 38 observations larger in the dataset) is greater than 20. A. Less than 0.02 B. At least 0.02, but less than 0.04 C. At least 0.04, but less than 0.06 D. At least 0.06, but less than 0.08 E. At least 0.08

25. B. Prob[X(2) > 20] = Prob[all 40 are greater than 20] + Prob[exactly 39 are greater than 20] =

S(20)40 + 40 F(20) S(20)39 = (e-20/150)40 + (40)(1 - e-20/150)(e-20/150)39 = 3.24%.

Alternately, gk(yk) =

€

n!(k -1)! (n- k)! [F(y)]k-1 [1-F(y)]n-k f(y).

Thus the density of the second order statistic (k = 2) is: n (n-1) F(y) f(y) S(y)n-2.In this case, this density is: (40)(39) (1 - e-y/150) (e-y/150/150) (e-y/150)38 = 10.4 (e-y39/150 - e-y40/150).

Prob[X(2) > 20] = 10.4

€

e- y39/ 150 - e- y40 / 150 dy20

∞

∫ =

(10.4) {exp[-(20)(39)/150] 150/39 - exp[-(20)(40)/150] 150/40} = (10.4) (0.0212176 - 0.0181048) = 3.24%.Comment: Similar to CAS3, 11/06, Q.9.Let W1 and W2 be two independent Exponential Distributions each with mean 150.Then X(2) is: W1/40 + W2/39. I do not see how this would help you answer this question.

€

y(e- y / θ / θ) dy0

∞

∫ = θ.

€

⇒

€

y(e- y / θ ) dy0

∞

∫ = θ2.

E[X(2)] = 10.4

€

y(e- y39/ 150 - e- y40 / 150) dy0

∞

∫ = (10.4) {(150/39)2 - (150/40)2} = 7.5962

= (150) (1/40 + 1/39).


26. You are testing the hypothesis that the median values from two populations, m1 and m2 are the same.

€

• H0: m1 = m2

€

• H1: m1 ≠ m2

€

• Sample size n1 = 6

€

• Sample size n2 = 8

€

• Wilcoxon rank sum of sample 1: W1 = 34 Calculate the p-value of this test. 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

26. D. U = n1n2 + n1(n1+1)/2 - (sum of ranks of first sample) = (6)(8) + (6)(7)/2 - 34 = 35.U has support on the integers from 0 to n1 n2 = (6)(8) = 48.In order to use the table attached to the exam, we need to subtract U from 48: 48 - 35 = 13.The two sided p-value is: 2 Prob[U ≤ 13] = (2)(9.0%) = 18.0%.


27. You are given the results of a matched pairs experiment by policy. Policy Sample Loss Modeled Loss S 0 5,000 T 0 10,000 U 0 15,000 V 0 20,000 W 0 25,000 X 0 30,000 Y 80,000 35,000 Z 0 50,000

€

• H0: MedianSample = MedianModeled

€

• H1: MedianSample > MedianModeled Calculate the smallest value for α for which H0 can be rejected using the Sign-Rank Wilcoxon test for a matched pairs experiment. 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

27. E. This is a one-sided test.The median of the sample is 0, while that of the model is 22,500.Thus there is no evidence that MedianSample > MedianModeled. Thus the p-value is much greater than 5%.Comment: I assume the question should have read H1: MedianSample < MedianModeled.In which case, one would take the differences; I chose to take model minus sample:5K, 10K, 15K, 20K, 25K, 30K, -45K, 50K.So the sum of the negative ranks is 7. Looking in the table attached to the exam, for n = 8: 7 > 5.

€

⇒ The one sided p-value is greater than 5%.


28. You are given the following random sample, x1, x2, ... , x7, of size 7: 4 2 0 0 1 1 a

from the following distributions:

f(x|λ) =

€

e- λ λx

x! , x

€

∈ {0, 1, 2, ...}

f(λ) =

€

λ2

2 e−λ, λ > 0, θ > 0

Calculate the value of a such that the Bayesian point estimate for λ = 3. A. Less than 2 B. At least 2, but less than 6 C. At least 6, but less than 10 D. At least 10, but less than 14 E. At least 14

28. D. This is a Gamma-Poisson. The prior Gamma has α = 3 and θ = 1.The Posterior Gamma has αʼ = 3 + 4 + 2 + 0 + 0 + 1 + 1 + a = 11 + a, and 1/θʼ = 1/1 + 7. θʼ = 1/8.Thus, 3 = αʼ θʼ = (11 + a)/8.

€

⇒ a = 13.Alternately, using Buhlmann Credibility, K = 1/θ = 1/1 = 1. Z = N/(N+K) = 7/(7 + 1) = 7/8.Observed mean = (8 + a)/7. Prior Mean = (3)(1) = 3.Thus 3 = (7/8)(8 + a)/7 + (1/8)(3).

€

⇒ a = 13.Comment: The density f(λ) did not need “θ > 0”.



€

• Policies are divided into two classes, Safe and Unsafe

€

• Claims per year per policy follow a Poisson distribution Classes Percent of Population Mean Claims Per Year Per PolicySafe 40% 0.4 Unsafe 60% 0.8

€

• A randomly selected policy had 1 claim in the first year. Calculate the probability that this randomly selected policy is an Unsafe policy. A. Less than 0.66 B. At least 0.66, but less than 0.68 C. At least 0.68, but less than 0.70 D. At least 0.70, but less than 0.72 E. At least 0.72

29. B. Chance of observation if Safe: 0.4e-0.4 = 0.2681.Chance of observation if Unsafe: 0.8e-0.8 = 0.3595.We apply of Bayes Theorem; posterior probabilities are proportional to: (a priori chance of risk type) (chance of observation given the risk type).Probability weight for Safe: (40%)(0.2681) = 0.1072.Probability weight for UnSafe: (60%)(0.3595) = 0.2157.Posterior probability of Unsafe is: 0.2157 / (0.1072 + 0.2157) = 66.8%.


30. For a large portfolio of insurance contracts,

€

ε represents the proportion of policies for which there were claims made within one year. The random variable

€

ε has a beta prior distribution with parameters a = 10, b = 5, and θ = 1. In a given year, for 100 randomly selected policies from the portfolio, there were 22 claims. Calculate the posterior mean of

€

ε . A. Less than 0.15 B. At least 0.15, but less than 0.25 C. At least 0.25, but less than 0.35 D. At least 0.35, but less than 0.45 E. At least 0.45

30. C. We have to assume no more than one claim per policy. Then this is a Beta-Bernoulli.aʼ = 10 + 22 = 32. bʼ = 5 + 100 - 22 = 83.Posterior mean is: aʼ/(aʼ + bʼ) = 32/(32 + 83) = 27.8%.Comment: As written, one can have more than one claim per policy.The question should be rewritten in either of two ways. Either one can have at most one claim per policy, or there are 22 out of of the 100 policies which had at least one claim.


31. Within the context of Generalized Linear Models, suppose that y has an exponential distribution with probability density function expressed as:

f(y) =

€

1µ

exp(-

€

yµ

); for y > 0

Determine the variance of y in terms of µ. A. 1/µ B.

€

µ C. µ D. µ2 E. Cannot be determined from the given information

31. D. An Exponential Distribution with mean µ has variance µ2.Comment: In the tables attached to the exam, the Exponential Distribution has mean θ, second moment 2θ2, and thus variance θ2.Does not depend on being within the context of Generalized Linear Models.For a Gamma Distribution: Var[Y] = αθ2 = (αθ)2 / α = E[Y]2 / α. Here α = 1.


32. You are given the following GLM output: Response variable Pure PremiumResponse distribution GammaLink log

Parameter df

€

β^

Intercept 1 4.78

Risk Group 2 Group 1 0 0.00 Group 2 1 -0.20 Group 3 1 -0.35

Vehicle Symbol 1 Symbol 1 0 0.00 Symbol 2 1 0.42

Calculate the predicted pure premium for an insured in Risk Group 2 with Vehicle Symbol 2. A. Less than 135 B. At least 135, but less than 140 C. At least 140, but less than 145 D. At least 145, but less than 150 E. At least 150

32. D. Exp[4.78 - 0.20 + 0.42] = e5 = 148.41.


33. You are given the following two probability density functions: (i) f(y ; θ) = θy-θ-1 ; for y > θ

(ii) f(y ; θ) = θe-yθ ; for y > 0 and θ > 0 Determine which of the following statements is true. A. (i) doesn't belong to the exponential family B. (i) and (ii) both belong to the exponential family C. Only (i) is in canonical form D. (i) and (ii) both are in canonical form E. None of above are true

33. A. The first “density” does not integrate to one over its support:

€

θ / yθ + 1 dyθ

∞

∫ =

€

-θyθ⎤

⎦ ⎥

y = θ

y = ∞

= 1/θθ. Thus it is not an exponential family.

The first distribution has the parameter as part of its support, and thus is not an exponential family. ln[f(y)] = ln[θ] + ln[y] (-θ-1). However, this does have the form a(y) b(θ) + c(θ) + d(y), with a(y) = ln[y], b(θ) = -θ - 1, c(θ) = ln[θ], and d(y) = 0.Since a(y) ≠ y, this is not in canonical form.The second distribution is an Exponential with θ the hazard rate; it is an exponential family in canonical form. ln[f(y)] = -yθ + ln[θ]. This has the form a(y) b(θ) + c(θ) + d(y), with a(y) = y, b(θ) = -θ, c(θ) = ln[θ], and d(y) = 0.Since a(y) = y, this is in canonical form.Comment: The CAS intended the answer to be B, that both densities are exponential families.Definition 7.5.1 in Hogg, McKean, and Craig states that the support can not depend on the parameter θ. However, Section 3.2 of Dobson and Barnett does not include that portion of the commonly used definition of exponential families. This exam question is taken from Exercise 3.3 in Dobson and Barnett.


34. You are given the following information for a fitted GLM: Response variable Status (1 or 0)Response distribution BernoulliLink Logit AIC 28.971

Parameter df

€

β^

Intercept 1 -12.7921x1 1 1.9104x2 1 0.1558

€

• The deviance residual value for the first observation is 0.4902.

€

• The corresponding hat matrix diagonal value is 0.0505. Calculate the standardized deviance residual for the first observation. A. Less than 0 B. At least 0, but less than 0.40 C. At least 0.40, but less than 0.80 D. At least 0.80, but less than 1.20 E. At least 1.20

34. C. The standardized deviance residual is the deviance residual divided by

€

1 - hii :

0.4902 /

€

1 - 0.0505 = 0.5031.Comment: The (nonstandardized) deviance residual of 0.4902 is also in answer range C.

For the Binomial, the deviance is: 2

€

{yi ln[ yiy i]

i=1

n∑ + (mi - yi) ln[ mi - yi

mi - y i] } .

Here we have a Bernoulli, so m = 1.If for example the first observation is 1 and

€

y 1 = 0.8868, then the first residual is: 1 - 0.8868 > 0.The contribution of the first observation to the deviance is: 2 {(1) ln[1/0.8868] + (1 -1) ln[0/(1-0.8868)]} = -2 ln[0.8868] = 0.2403. Since the ordinary residual is positive, the first deviance residual is:

€

0.2403 = 0.4902,matching the value given in the question.


35. You are given, y1 y2, ... , yn, independent and Poisson distributed random variables with respective means µi for i = 1, 2, ... , n. A Poisson GLM was fitted to the data with a log-link function expressed as:

E(yi) = exp[β0 + β1xi]where xi refers to the predictor variable. Analysis of a set of data provided the following output:

xi yi

€

y i yi log(yi/

€

y i) 0 7 6.0 1.0791 0 9 6.0 3.6492 0 2 6.0 -2.1972 1 3 6.6 -2.3654 1 10 6.6 4.1552 1 8 6.6 1.5390 1 5 6.6 -1.3882 1 7 6.6 0.4119

Calculate the observed deviance for testing the adequacy of the model. A. Less than 4.0 B. At least 4.0, but less than 6.0 C. At least 6.0, but less than 8.0 D. At least 8.0, but less than 10.0 E. At least 10.0


35. D. For the Poisson, the deviance is: 2

€

{yi ln[yi / λi ]i=1

n

∑ - (yi - λ i ) }

€

yi ln[yi / λ i ]i=1

n∑ = 1.0791 + 3.6492 + ... = 4.8836.

€

yii=1

n∑ = 7 + 9 + ... = 51.

€

λii=1

n∑ = 6 + 6 + ... = 51.

2

€

{yi ln[yi / λi ]i=1

n

∑ - (yi - λ i ) } = (2) (4.8836 - 51 + 51) = 9.7672.

Comment: See Table 5.1 in Dobson and Barnett.

For the Poisson, provided there is an intercept in the model:

€

yii=1

n∑ =

€

λii=1

n∑ .

Thus the deviance is equal to: 2

€

yi ln[yi / λ i ]i=1

n∑ .

For example, the first contribution to the deviance is: (2) {1.0791 - (7 - 6)} = 0.1582.Since 7 - 6 = 1 > 0, the first deviance residual is positive:

€

0.1582 = 0.398.The deviance residuals are: 0.398, 1.139, -1.899, -1.571, 1.229, 0.527, -0.651, 0.154.


36. You are given the following information for a fitted model: Response variable Rating Response distribution NormalLink IdentityResidual Std. Error 7.139

Parameter df

€

β^

Intercept 1 11.011Complaints 1 0.692Privileges 1 -0.104 Learning 1 0.249Raises 1 -0.033Critical 1 0.015

• The first record in the data has the following values:

ID Rating Complaints Privileges Learning Raises Critical 1 43 51 30 39 61 92

• The corresponding hat matrix diagonal value is 0.3234. Calculate the DFITS for the observation above. A. Less than -3 B. At least -3, but less than -2 C. At least -2, but less than -1 D. At least -1, but less than 0 E. At least 0

36. C. Observed value of the rating is 43. Using the Identity link, the fitted value of the rating is: 11.011 + (51)(0.692) + (30)(-0.104) + (39)(0.249) + (61)(-0.033) + (92)(0.015) = 52.261.Residual is: 43 - 52.261 = -9.261.

Standardized residual is:

€

ε is 1 - hii

=

€

- 9.2617.139 1 - 0.3234

= -1.577.

DFFITSi = (ith standardized residual)

€

hii1- hii

= (-1.577)

€

0.32341 - 0.3234

= -1.090.


37. You are given the outputs from two GLMs fitted to the same data from a trial of a new drug. Model 1 Model 2

Response variable Number Response variable Number Response distribution Poisson Response distribution Negative binomial Link Log Link Log AIC 273.877 AIC 164.880

Parameter

€

β^ s. e. (

€

β^ ) Parameter

€

β^ s. e. (

€

β^ )

Intercept 4.529 0.147 Intercept 4.526 0.595

Treatdrug Treatdrug Placebo 0.000 0.000 Placebo 0.000 0.000 Drug -1.359 0.118 Drug -1.368 0.369

Age -0.039 0.006 Age -0.039 0.021

Determine which of the following statements is false using the Wald test. A. Under Model 1, the Treatdrug coefficient has a p-value less than 0.01. B. Under Model 1, the Age coefficient has a p-value less than 0.01. C. Under Model 2, the Treatdrug coefficient has a p-value less than 0.01. D. Under Model 2, the Age coefficient has a p-value less than 0.01. E. Under both models, the intercept coefficient has a p-value of less than 0.01.

37. D. The Wald statistic for Treatdrug coefficient for Model 1 is: -1.359/0.118 = -11.516. p-value = 2 Φ[-11.516] < 0.01.The Wald statistic for Age coefficient for Model 1 is: -0.039/0.006 = -6.5. p-value = 2 Φ[-6.5] < 0.01.The Wald statistic for Treatdrug coefficient for Model 2 is: -1.368/0.369 = -3.707. p-value = 2 Φ[-3.707] < 0.01.The Wald statistic for Age coefficient for Model 2 is: -0.039/0.021 = -1.857. p-value = 2 Φ[-1.86] = (2)(0.0314) = 6.28%.The Wald statistic for the intercept coefficient for Model 1 is: 4.529/0.147 = 30.81. p-value = 2 (1 - Φ[30.81]) < 0.01.The Wald statistic for the intercept coefficient for Model 2 is: 4.526/0.595 = 7.6. p-value = 2 (1 - Φ[7.6]) < 0.01.Comment: One has to assume that in each case the null hypothesis is that the parameter is zero.One could square my Wald Statistics and instead compare to the Chi-Square Distribution with one degree of freedom, and get the same p-values.For example, (-0.039/0.021)2 = 3.45. 3.45 < 3.84.

€

⇒ p-value is greater than 5%.It would have been better if “false” had been underlined in the question.


38. You are given the following probability density function for a single random variable, X:

f(xIθ) =

€

θ2πx3⎛

⎝ ⎜

⎞

⎠ ⎟ 1/ 2

€

exp[- θ(x -1)2

2x ]

Consider the following statements: 1. f(x) is a member of the exponential family of distributions II. The score function, U(θ), is:

U(θ) =

€

12θ -

€

(x -1)2

2x

III. The Fisher Information, I(θ), is: I(θ) = 2θ2

Determine which of the above statements are true. A. I only B. II only C. III only D. I, II and III E. The correct answer isn't given by A, B, C or D

38. E. lnf(x) = ln[θ]/2 - ln[2πx3]/2 -

€

θ(x -1)2

2x .

We want: lnf(x) = a(x) b(θ) + c(θ) + d(x). This density is an exponential family with: a(x) = -(x-1)2/(2x), b(θ) = θ, c(θ) = ln[θ]/2, and d(x) = -ln[2πx3]/2. Statement I is true.

U(θ) =

€

∂lnf(x) ∂θ

=

€

12θ -

€

(x -1)2

2x . Statement II is true.

Alternately, U(θ) = a(x) bʼ(θ) + cʼ(θ) = {-

€

(x -1)2

2x }(1) +

€

12θ . Statement II is true.

€

∂2lnf(x) ∂θ2 = 1/(2θ2). I(θ) = -E[

€

∂2lnf(x) ∂θ2 ] = 1/(2θ2). Statement III is not true.

Alternately, I(θ) =

€

b' '(θ) c'(θ)b'(θ) - c”(θ) = (0)(

€

12θ ) / 1 - {-1/(2θ2)} = 1/(2θ2). Statement III is not true.

Comment: An Inverse Gaussian Distribution with µ = 1, for x > 0. It is an exponential family, but not a linear exponential family since a(x) ≠ x.


39. You are performing a two-factor analysis of variance with interactions, and are given the following ANOVA table:

Source of variation Degrees of freedom Sum of squares Levels of Factor 1 4 1514.94 Levels of Factor 2 1 240.25 Interactions 4 190.30 Residuals 90 722.30

Calculate the F-statistic for testing the significance of the interaction terms. A. Less than 2.0 B. At least 2.0, but less than 4.0 C. At least 4.0, but less than 6.0 D. At least 6.0, but less than 8.0 E. At least 8.0

39. C. The mean sum of squares for interactions is: 190.30/4 = 47.575.The mean sum of squares for residuals is: 722.30/90 = 8.026.F-Statistic is: 47.575/8.026 = 5.928, with 4 and 90 degrees of freedom.Comment: Similar to Table 6.12 in Dobson and Barnett.Using a computer, the p-value is 0.03%.The mean sum of squares for Factor 1 is: 1514.94/4 = 378.74.In order to test whether the level of Factor 1 are significantly different, the F-Statistic is:378.74/8.026 = 47.19, with 4 and 90 degrees of freedom. The p-value is less than 10-12.The mean sum of squares for Factor 2 is: 240.35/1 = 240.35.In order to test whether the level of Factor 2 are significantly different, the F-Statistic is:240.35/8.026 = 29.95, with 1 and 90 degrees of freedom. The p-value is 4 x 10-7.


40. You are given the following information to find the evidence of interaction between two variables, α and β. Multiple models are fitted to a dataset and their results are shown in the following table:

Model Scaled deviance Degrees of Freedom µ 26 9 µ + αj + βk 6.07 6 µ + αj + βk + (αβ)jk 5 4

Calculate the F-statistic for testing the significance of the interaction terms. A. Less than 0.5 B. At least 0.5, but less than 2.5 C. At least 2.5, but less than 3.0 D. At least 3.0, but less than 3.5 E. At least 3.5

40. A. For a regression, the scaled deviance is another name for the Error Sum of Squares.To test whether there is an interaction of effects, compare the model with interactions to the model without interactions. Compute the F-Statistic:

€

(ESSR - ESSUR)/ (difference in number of parameters for the two models)ESSUR / (total number of data points - number of fitted parameters unrestricted model) =

=

€

(6.07 - 5) / (6 - 4)5 / 4 = 0.428, with 2 and 4 degrees of freedom.

Comment: Using a computer, the p-value is 67.9%.


41. A modeler is considering revising a GLM for claim counts with age as an explanatory variable. It is currently being included in the model as a continuous variable with no interactions. Determine which of the following statements is false. A. Including age as a categorical variable with more than two levels would increase the model

degrees of freedom. B. Including a polynomial term age2 may decrease model deviance. C. A plot of age against the residuals may be used to assess if a transformation of age is

necessary. D. One way of assessing multicollinearity is by performing a regression of age against

all other explanatory variables. E. Including several interaction terms involving age may make the model more parsimonious.

41. E. The degrees of freedom of the model are the number of fitted betas.The continuous variable age adds one degree of freedom to the model. A discrete age would instead add levels - 1 to the degrees of freedom to the model. Thus Statement A is true.Adding an additional term to a model usually improves the fit and thus decreases the deviance. Thus Statement B is true.If there is a pattern of the residuals with age, then we have not explained all of the variation of the response with age. In that case, a transformation of age may be useful.Statement C is true. See page 35 and Figures 2.3 and 2.4 of Dobson and Barnett.The variance inflation factor, one way of assessing multicollinearity, is calculated by performing a regression of age against all other explanatory variables.VIFj = 1 / {1 - R(j)2}, where R(j)2 is the coefficient of determination from this regression. If one or more of the VIFs is large, that is an indication of multicollinearity.Thus statement D is true.A parsimonious model uses the fewest number of parameters that get the job done.Adding terms to a model makes it less parsimonious. Thus Statement E is false.Comment: It would have been better if “false” had been underlined in the question.Parsimonious is used here to mean stingy or frugal. See page 36 of Dobson and Barnett.Occamʼs razor or the law of parsimony states that no more causes should be assumed than are needed to account for an effect; in modeling we should not use more parameters than are needed.


42. You are given the following ordered sample of size 6 from a time series: 1 1.5 1.6 1.4 1.5 1.7

Calculate the sample lag 2 autocorrelation. A. Less than -0.6 B. At least -0.6, but less than -0.3 C. At least -0.3, but less than 0.0 D. At least 0.0, but less than 0.3 E. At least 0.3

42. C. Mean is 1.45. c0 =

€

(1-1.45)2 + (1.5 -1.45)2 + (1.6 -1.45)2 + (1.4 -1.45)2 + (1.5 - 1.45)2 + (1.7 - 1.45)26 = 0.0492.

c2 =

€

(1-1.45)(1.6 -1.45) + (1.5 -1.45)(1.4 -1.45) + (1.6 -1.45)(1.5 -1.45) + (1.4 -1.45)(1.7 -1.45)6

= -0.125.r2 = c2/ c0 = -0.0125 / 0.0492 = -0.254.

Comment: One can use the stat function on the calculator to determine σx2, the biased estimator of the variance, which is c0.


43. An ARMA(3,0) model is fit to the following quarterly time series: Year Quarter 1 Quarter 2 Quarter 3 Quarter 4 2013 3.53 1.33 1.85 0.61 2014 0.98 3.61 3.44 3.38 2015 2.91 2.12 4.62 2.93

The estimated coefficients are: ar1 ar2 ar3 Intercept 0.252 0.061 -0.202 2.637

Forecast the value for Quarter 1 of 2016. A. Less than 3.00 B. At least 3.00, but less than 3.25 C. At least 3.25, but less than 3.50 D. At least 3.50, but less than 3.75 E. At least 3.75

43. A. (See comment) This AR(3) model, where each value has the intercept subtracted:xt - 2.637 = (0.252)(xt-1 - 2.637) + (0.061)(xt-2 - 2.637) + (-0.202)(xt-3 - 2.637).Forecasted value for Quarter 1 of 2016 is:2.637 + (0.252)(2.93 - 2.637) + (0.061)(4.62 - 2.637) + (-0.202)(2.12 - 2.637) = 2.936.Comment: “The formula used to describe the results of an autoregressive time series model could be interpreted as a simple OLS model with lagged variables, leading to answer B. The intent of the problem was to test candidatesʼ understanding of how a mean reverting time series model is constructed using an autoregressive model works, leading to the original posted answer A. However, given the ambiguity in the problem stem, the CAS decided to accept both A and B as correct answers.” Personally, I do not see any ambiguity in this exam question.Answer B was gotten as follows: 2.637 + (0.252)(2.93) + (0.061)(4.62) + (-0.202)(2.12) = 3.229.The output of fitting in R an AR(3) model to the given time series: ar1 ar2 ar3 intercept 0.2521 0.0615 -0.2024 2.6374s.e. 0.2837 0.3679 0.3368 0.3751sigma^2 estimated as 1.229: log likelihood = -18.36, aic = 46.72Other than the intercept, none of the coefficients is significantly different from zero.The output of fitting in R an AR(1) model to the given time series: ar1 intercept 0.2478 2.6415s.e. 0.2748 0.4252sigma^2 estimated as 1.28: log likelihood = -18.54, aic = 43.08The AR(1) model has a better (smaller) AIC than the AR(3) model.


44. You are given the AR(3) model below for zt, a company's revenue for year t: zt = 5 + 0.85zt-1 - 0.02zt-3 + wt where wt is white noise, with:

€

• E(wt) = 0

€

• Var(wt) = σ2 The revenues for the last 4 years are as follows:

Year Revenue 2012 20 2013 15 2014 22 2015 19

Forecast the expected revenue for 2017. A. Less than 19 B. At least 19, but less than 20 C. At least 20, but less than 21 D. At least 21, but less than 22 E. At least 22

44. E. Forecasted value for 2016: 5 + (0.85)(19) + (-0.02)(15) = 20.85.Then the forecasted value for 2017: 5 + (0.85)(20.85) + (-0.02)(19) = 22.28.Comment: One could rewrite the given model as:zt - 29.412 = 0.85(zt-1 - 29.412) - 0.02(zt-3 - 29.412) + wt,where the intercept of 29.412 = 5 / (1 - 0.85 + 0.02).



€

• xt is modeled as an ARIMA(1, 0, 0) given by: xt - 300.53 = 0.95(xt-1 - 300.53) + wt where wt is white noise, with:

€

• E(wt) = 0

€

• Var(wt) = σ2

€

• x2015 = 310.12 Calculate the forecast for the time period 2027, x2027. A. Less than 306.0 B. At least 306.0, but less than 306.4 C. At least 306.4, but less than 306.8 D. At least 306.8, but less than 307.2 E. At least 307.2

45. A.

€

x2016 - 300.53 = (0.95) (310.12 - 300.53).

€

x2017 - 300.53 = (0.95) (

€

x2016 - 300.53) = (0.952) (310.12 - 300.53).

Similarly,

€

x2027 - 300.53 = (0.9512) (310.12 - 300.53).

€

⇒

€

x2027 = 300.53 + 5.18 = 305.71.Comment: 300.53 is the intercept of the given model.


solutions to the fall 2016 cas exam showardmahler.com/teaching/s_files/solsf16casexams.pdfusing the...

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