Solution. Let us write s for the policy year. Then the mortality rate during year s is q 30+s 1. q 30+s 1

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Solutions to the May 2013 Course MLC Examination

by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net

Dr. Ostaszewski’s manual for Course MLC is available from Actex Publications

(http://www.actexmadriver.com) and the Actuarial Bookstore

(http://www.actuarialbookstore.com)

Exam MLC seminar at Illinois State University:

http://math.illinoisstate.edu/actuary/exams/prep_courses.shtml

May 2013 Course MLC Examination, Problem No. 1

For a fully discrete whole life insurance of 1000 on (30), you are given: (i) Mortality follows the Illustrative Life Table.

(ii) i = 0.06.

(iii) The premium is the benefit premium.

Calculate the first year for which the expected present value at issue of that year’s premium is less than the expected present value at issue of that year’s benefit. A. 11 B. 15 C. 19 D. 23 E. 27

Solution.

Let us write s for the policy year. Then the mortality rate during year s is q_30+s−1. We are looking for the smallest value of s such that the expected present value at issue of that year’s premium is less than the expected present value at issue of that year’s benefit, or

1.06− s−1( )⋅s−1p30⋅1000P30 expected present value at issue of year s premium

< 1.06−s_⋅

s−1p30⋅q30+s−1⋅1000 expected present value at issue of year s benefit

.

Since the interest rate is 6%, and mortality follows the Illustrative Life Table, we can get the value of _1000P₃₀ from values given in the table

1000P₃₀=1000 A30 a₃₀ =

102.48 15.8561. Hence we have this inequality

1.06− s−1( )⋅s−1p30⋅ 102.48 15.8561< 1.06 −s_⋅ s−1p30⋅q30+s−1⋅1000, or 1000q_30+s−1> 1.06 ⋅ 102.48 15.8561≈ 6.85091542.

The first age in the table where 1000 times mortality rate exceeds 6.85091542 is age 52. We set 30 + s – 1 = 52, and conclude that s = 23.

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P&C Insurance Company is pricing a special fully discrete 3-year term insurance policy on (70). The policy will pay a benefit if and only if the insured dies as a result of an automobile accident. You are given:

(i) x _l x τ ( ) _d x 1 ( ) _d x 2 ( ) _d x 3 ( ) _Benefit 70 1000 80 10 40 5.000 71 870 94 15 60 7,500 72 701 108 18 82 10,000 where dx 1

( )_{represents deaths from cancer,}_d

x

2

( )_{represents deaths from automobile} accidents, and dx

3

( )_{represents deaths from all other causes.} (ii) i = 0.06.

(iii) Level premiums are determined using the equivalence principle. Calculate the annual premium.

A. 122 B. 133 C. 144 D. 155 E. 166 Solution.

The expected present value of benefits is 5000 1.06 ⋅ 10 1000+ 7500 1.062⋅ 15 1000+ 10000 1.063 ⋅ 18 1000 ≈ 298.42.

Let P be the annual premium sought. The expected present value of premiums is P+ P 1.06⋅ 870 1000+ P 1.062⋅ 701 1000≈ 2.4446P.

Since the premium is set by the equivalence principle, those two quantities must be equal, and therefore

P≈298.42

2.4446≈ 122. Answer A.

For a special fully discrete 20-year endowment insurance on (40), you are given: (i) The only death benefit is the return of annual benefit premiums accumulated with interest at 6% to the end of the year of death.

(ii) The endowment benefit is 100,000.

(iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06.

Calculate the annual benefit premium.

A. 2365 B. 2465 C. 2565 D. 2665 E. 2765 Solution.

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Let us write P for the annual benefit premium sought. Note that in case of death, the death benefit is the accumulated value of all premiums paid up to that point, with interest, at the end of the year of death, so all dying policyholders simply get their own premiums with interest back, and any remaining premiums are those for surviving policyholders. On the other hand, reserve is only calculated for surviving policyholders. This is the central idea of the most efficient solution. At policy duration 20, the benefit reserve equals, prospectively, the actuarial present value of future benefits minus the actuarial present value of future premiums, and since there are no future premiums, it simply equals

20V = 100,000. On the other hand, retrospectively, the reserve equals the actuarial accumulated value of past premiums minus the accumulated actuarial value of past benefits. But the past benefits were paid to those policyholders who died by a return of their own premiums with interest, which means that any past premiums left are the ones paid by policyholders surviving till policy duration 20, and since the reserve is calculated only for surviving policyholders, the reserve equals 20V= Ps20 6%. Equating the two formulas for the same reserve at policy duration 20, we obtain

100,000= Ps_{20 6%}. Based on this,

P=100,000

s_{20 6%} ≈ 2,564.58. Answer C.

Employment for Joe is modeled according to a two-state homogeneous Markov model with states: Actuary (Ac) and Professional Hockey Player (H). You are given:

(i) Transitions occur December 31 of each year. The one-year transition probabilities are: Ac H Ac H 0.4 0.6 0.8 0.2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥

(ii) Mortality for Joe depends on his employment: q_35+kAc = 0.10 + 0.05k for k = 0, 1, 2, q_35+kH = 0.25 + 0.05k for k = 0, 1, 2.

(iii) i = 0.08.

On January 1, 2013, Joe turned 35 years old and was employed as an actuary. On that date, he purchased a 3-year pure endowment of 100,000. Calculate the expected present value at issue of the pure endowment.

A. 32,510 B. 36,430 C. 40,350 D. 44,470 E. 48,580 Solution.

The specific mortalities given by the formulas in (ii) are: q₃₅Ac= 0.10, q₃₆Ac = 0.15, q₃₇Ac= 0.2, q 35 H = 0.25, q 36 H = 0.3, q 37

H = 0.35. Over the next three years, Joe has the following paths to collection of the endowment benefit at the end of that period (with

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probabilities for each step indicated below):

1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15 Actuary Alive Actuary Alive Actuary Alive

0.9 0.4 0.85 0.4 0.8

1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15 Actuary Alive Actuary Alive Hockey Alive

0.9 0.4 0.85 0.6 0.65

1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15 Actuary Alive Hockey Alive Hockey Alive

0.9 0.6 0.7 0.2 0.65

1/1/13  12/31/13  1/1/14  12/31/14  1/1/15  12/31/15 Actuary Alive Hockey Alive Actuary Alive

0.9 0.6 0.7 0.8 0.8 Therefore, the probability of collecting the 100,000 endowment benefit is

0.9⋅0.4 ⋅0.85 ⋅0.4 ⋅0.8 + 0.9 ⋅0.4 ⋅0.85 ⋅0.6 ⋅0.65 +

+ 0.9 ⋅0.6 ⋅0.7 ⋅0.2 ⋅0.65 + 0.9 ⋅0.6 ⋅0.7 ⋅0.8 ⋅0.8 = 0.50832. The expected present value at issue of the pure endowment is

0.50832⋅100,000

1.083 ≈ 40,352.08. Answer C.

The joint lifetime of Kevin, age 65, and Kira, age 60, is modeled as:

You are given the following constant transition intensities: (i) µ01 = 0.004.

(ii) µ02= 0.005.

Exam MLC: Spring 2013 – 9 – GO ON TO NEXT PAGE

5.

The joint lifetime of Kevin, age 65, and Kira, age 60, is modeled as: State 0 Kevin alive Kira alive 01 µ State 1 Kevin alive Kira dead µ02 µ13 State 2 Kevin dead Kira alive 23

µ _{Kevin dead}State 3

Kira dead You are given the following constant transition intensities:

(i) _µ01₌_0.004 (ii) _µ02 ₌_0.005 (iii) _µ03₌_0.001 (iv) _µ13₌_0.010 (v) _µ23₌_0.008 Calculate 02 10 p65:60. (A) 0.046 (B) 0.048 (C) 0.050 (D) 0.052 (E) 0.054 03 µ

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(iii) µ03= 0.001. (iv) µ13= 0.010. (v) µ23 = 0.008. Calculate ₁₀p_65:6002 _. A. 0.046 B. 0.048 C. 0.050 D. 0.052 E. 0.054 Solution.

The probability ₁₀p_65:6002 _{is the probability of transition from state 0 to state 2 in such a} way that the joint life status (65:60) is in state 0 at time 0 and is in state 2 at time 10. The subscript 65:60 indicates that the probability refers to the joint life status (65:60). The probability sought equals (remember that under constant forces of transition, the

probability of remaining in state j over period of time t equals the expression of the form exp(–t times the sum of forces of transition out of state j) )

10p65:60 02 ₌ tp65:60 00 Remain in state 0 for t years  ⋅ µ02dt Transition from state 0 to state 2 in period t ,t+dt( )  ⋅10−tp65+t:60+t22 Remain in state 2 til the end of 10 years

   0 10

∫

= = e−(µ01+µ02+µ03)t ⋅µ02_⋅e−µ23(10−t) dt 0 10

∫

= e−0.01t_{⋅0.005 ⋅e}−0.008 10−t( )_dt 0 10

∫

= = 0.005 ⋅ e−0.01t_⋅e−0.08_⋅e0.008t dt 0 10

∫

=0.005 e0.08 ⋅ e −0.01t_⋅e0.008t dt 0 10

∫

= = 0.005 e0.08 ⋅ e −0.002t_dt 0 10

∫

=0.005 e0.08 ⋅a10 0.2% = 0.005 e0.08 ⋅ 1− e−0.02 0.002 ≈ 0.04569732. Answer A.

For a wife and husband ages 50 and 55, with independent future lifetimes, you are given: (i) The force of mortality on (50) is µ_50+t = 1

50− t, for 0≤ t < 50. (ii) The force of mortality on (55) is µ_55+t = 0.04, for t> 0.

(iii) For a single premium of 60, an insurer issues a policy that pays 100 at the moment of the first death of (50) and (55).

(iv) δ = 0.05.

Calculate the probability that the insurer sustains a positive loss on the policy. A. 0.45 B. 0.47 C. 0.49 D. 0.51 E. 0.53

Solution.

The force of mortality on (50) represents De Moivre’s Law, so that tp50 = 50− t

50 = 1− t 50

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for 0≤ t ≤ 50. The force of mortality on (55) is constant, so that _tp₅₅ = e−0.04t for t≥ 0. Since the future lifetimes are independent, _tp_50:55= 1− t

50 ⎛

⎝⎜ ⎞⎠⎟e−0.04t for 0≤ t ≤ 50, but for t > 50, tp50:55 = 0 because (50) will be dead with certainty for t> 50. Let us write (using two possible notations for the future lifespan)

T_50:55= T 50 :55

(

)

= min T

(

₅₀,T₅₅

)

= min T 50

(

( )

,T 55

( )

)

for the future lifespan of the joint life status (50:55). The loss of the insurer on this policy is given by the formula

L= 100e−0.05T50:55 − 60.

We want to know the probability that L > 0. We have

Pr L

(

> 0

)

= Pr 100e

(

−0.05T50:55 − 60 > 0

)

= Pr 100e

(

−0.05T50:55 > 60

)

= = Pr e

(

−0.05T50:55 > 0.6

)

= Pr lne

(

−0.05T50:55 > ln0.6

)

= = Pr −0.05T

(

50:55> ln0.6

)

= Pr T50:55 < ln 0.6 −0.05 ⎛ ⎝⎜ ⎞⎠⎟ = = 1− Pr T50:55 ≥ ln 0.6 −0.05 ⎛ ⎝⎜ ⎞⎠⎟ =1− 1− ln 0.6 −0.05 50 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ e−0.04⋅ ln 0.6 −0.05 ⎛ ⎝⎜ ⎞⎠⎟ ₌ = 1− 1+ln 0.6 2.5 ⎛ ⎝⎜ ⎞⎠⎟

( )

eln 0.6 0.04 0.05 = 1− 1+ln 0.6 2.5 ⎛ ⎝⎜ ⎞⎠⎟0.60.8 ≈ 0.47124578. Answer B.

You are given: (i) q60 = 0.01. (ii) Using i = 0.05, A_60:3 = 0.86545. Using i = 0.045 calculate A_60:3. A. 0.866 B. 0.870 C. 0.874 D. 0.878 E. 0.882 Solution. Using i = 0.05, 0.86545= A_60:3 = q60 1.05+ 1− q₆₀

(

)

q₆₁ 1.052 + 1− q₆₀

(

)

(

1− q₆₁

)

1.053 . But we are given that q60 = 0.01, and using that, we obtain

0.86545= 0.01 1.05 + 0.99q₆₁ 1.052 + 0.99 1

(

− q61

)

1.053 , so that 0.86545⋅1.053 = 0.01⋅1.052+ 0.99q 61⋅1.05 + 0.99 1− q

(

61

)

,

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q61=

0.86545⋅1.053_{− 0.01⋅1.05}2_{− 0.99}

0.99⋅1.05 − 0.99 ≈ 0.01700114. Now we calculate using i = 0.045

A_60:3 = q60 1.045+ 1− q60

(

)

q61 1.0452 + 1− q60

(

)

(

1− q61

)

1.0453 ≈ ≈ 0.01 1.045+ 0.99⋅0.01700114 1.0452 + 0.99⋅0.98299886 1.0453 ≈ 0.87776672. Answer D.

For a special increasing whole life insurance on (40), payable at the moment of death, you are given:

(i) The death benefit at time t is bt = 1+ 0.2t, t≥ 0.

(ii) The interest discount factor at time t is v t

( )

= 1+ 0.2t

(

)

−2, t≥ 0. (iii) _tp₄₀⋅µ_40+t = 0.025, 0≤ t < 40 0, otherwise. ⎧ ⎨ ⎪ ⎩⎪

(iv) Z is the present value random variable for this insurance. Calculate Var(Z). A. 0.036 B. 0.038 C. 0.040 D. 0.042 E. 0.044 Solution. We have E Z

( )

= E b

(

_T⋅v T

( )

)

= E 1+ 0.2T

(

)

(

1+ 0.2T

)

−2

)

= E 1+ 0.2T

(

)

−1

)

= = fT

( )

t 1+ 0.2tdt 0 40

∫

= 1 40 1+ 0.2tdt 0 40

∫

= 1 40⋅0.2ln 1

(

+ 0.2t

)

t=0 t=40 = = 0.125 ln9 − ln1

(

)

≈ 0.27465307. Also, E Z

( )

2 _{= E b} T⋅v T

( )

(

)

2

(

)

= E 1+ 0.2T

(

)

(

1+ 0.2T

)

−2

)

2

)

= E 1+ 0.2T

(

)

−2

)

= = fT

( )

t 1+ 0.2t

(

)

2dt 0 40

∫

= 1 40

(

1+ 0.2t

)

−2 dt 0 40

∫

= 1 40 1+ 0.2t

(

)

−1 −1⋅0.2 ⎛ ⎝⎜ ⎞ ⎠⎟_t₌₀ t=40 = = 0.125 9−1 −1− 1−1 −1 ⎛ ⎝⎜ ⎞ ⎠⎟ = 0.125 − 1 9+1 ⎛ ⎝⎜ ⎞⎠⎟ ≈0.11111111. Therefore, Var Z

( )

= E Z

( )

2 − E Z

(

( )

2

)

≈ 0.11111111− 0.274653072≈ 0.0356768. Answer A.

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For a fully discrete whole life insurance of 10,000 on (x), you are given: (i) Deaths are uniformly distributed over each year of age.

(ii) The benefit premium is 647.46.

(iii) The benefit reserve at the end of year 4 is 1405.08. (iv) qx+4= 0.04561.

(v) i= 0.03.

Calculate the benefit reserve at the end of 4.5 years.

A. 1570 B. 1680 C. 1750 D. 1830 E. 1900 Solution.

Let us write _tV for the benefit reserve at policy duration t, and P for the bene. We use a slightly modified standard recursive reserve formula

4V+ P

(

)

⋅ 1+ i

( )

0.5 Reserve from policy duration 4 with premium accumulated with interest pays for the items on the right-hand side

 = 0.5px+4⋅4.5V Reserve at policy duration 4.5, of course only held for surviving policyholders

   +0.5qx+4⋅10,000 ⋅ 1+ i

( )

−0.5

Death benefits for those who die by policy duration 4.5, note that we have to discount for half a year, as the death benefit is paid at the end of the year .

We substitute known values and obtain 1405.08+ 647.46

(

)

⋅1.030.5 =

0.5px+4⋅4.5V+0.5qx+4⋅10,000 ⋅1.03

−0.5_.

We are not given 0.5qx+4 but we know qx+4= 0.04561 and we know that the UDD assumption applies. Hence

0.5qx+4_UDD= 0.5qx+4= 0.022805 and

0.5px+4= 1−0.5qx+4= 0.977195.

Substituting this, we obtain 1405.08+ 647.46

(

)

⋅1.030.5 _{= 0.977195 ⋅} 4.5V+ 0.022805 ⋅10,000 ⋅1.03 −0.5_, so that 4.5V = 1405.08+ 647.46

(

)

⋅1.030.5_{− 0.022805 ⋅10,000 ⋅1.03}−0.5 0.977195 ≈ 1,901.77. Answer E.

A multi-state model is being used to value sickness benefit insurance:

For a policy on (x) you are given:

(i) Premiums are payable continuously at the rate of P per year while the policyholder is healthy.

Exam MLC: Spring 2013 – 19 – GO ON TO NEXT PAGE

10.

A multi-state model is being used to value sickness benefit insurance: healthy (h) sick (s)

dead (d) For a policy on (x) you are given:

(i) Premiums are payable continuously at the rate of P per year while the policyholder is healthy.

(ii) Sickness benefits are payable continuously at the rate of B per year while the policyholder is sick.

(iii) There are no death benefits. (iv) ij

x t

µ ₊ denotes the intensity rate for transition from i to j, where , , or .

i j s h= d

(v) δ is the force of interest. (vi) ( )i

tV is the reserve at time t for an insured in state i where i s h= , or .d

Which of the following gives Thiele’s differential equation for the reserve that the insurance company needs to hold while the policyholder is sick?

(A) ( )s ( )s sh

(

( )h ( )s

)

t t x t t t d V V B V V dt =δ + −µ + − (B) ( )s ( )s sh

(

( )h ( )s

)

t t x t t t d V V B V V dt =δ − −µ + − (C) ( )s ( )s sh

(

( )h ( )s

)

sd ( )s t t x t t t x t t d V V B V V V dt =δ + −µ + − −µ + (D) ( )s ( )s sh

(

( )h ( )s

)

sd ( )s t t x t t t x t t d V V B V V V dt =δ − −µ + − −µ + (E) ( )s ( )s sh

(

( )h ( )s

)

sd ( )s t t x t t t x t t d V V B V V V dt =δ − −µ + − +µ +

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(ii) Sickness benefits are payable continuously at the rate of B per year while the policyholder is sick.

(iii) There are no death benefits.

(iv) µ_xij_+t_{denotes the intensity rate for transition from i to j, where i, j = s, h or d.}

(v) δ is the force of interest. (vi) tV

i

( )_{is the reserve at time t for an insured in state i where i = s, h or d.}

Which of the following gives Thiele’s differential equation for the reserve that the insurance company needs to hold while the policyholder is sick?

A. d dt tV s ( )₌_δ tV s ( )_{+ B −}_µ x+t sh tV h ( )₋ tV s ( )

(

)

B. d dt tV s ( )₌_δ tV s ( )_{− B −}_µ x+t sh tV h ( )₋ tV s ( )

(

)

C. d dt tV s ( ) ₌_δ tV s ( )_{+ B −}_µ x+t sh tV h ( )₋ tV s ( )

(

)

−µx+t sd tV s ( ) D. d dt tV s ( )₌_δ tV s ( )_{− B −}_µ x+t sh tV h ( )₋ tV s ( )

(

)

−µx+t sd tV s ( ) E. d dt tV s ( )₌_δ tV s ( )_{− B −}_µ x+t sh tV h ( )₋ tV s ( )

(

)

+µx+t sd tV s ( ) Solution.

The rate of change of reserve held while the policyholder is sick, d dt tV s ( )_,_{has the} following components:

• Instantaneous interest on the current reserve: δ _tV( )s ,

• Rate of premiums received while in state s: 0 (premium is paid only while the policyholder is healthy),

• Rate of benefits paid (outgo, so a negative component) while in state s: –B,

• Transition intensity for transition to state h, times the change in reserve upon transition (as a result of such transition, the insurer must hold reserve for h and release reserve for s): −µxsh+t⋅ tV h ( )₋ tV s ( )

(

)

,

• Transition intensity for transition to state d, times the change in reserve upon transition (as a result of such transition, the insurer releases reserve for s, and there is no reserve for state d, because the policy ends): −µx+t

sd _{⋅ 0 −} tV

s

( )

(

)

.

By adding all of those terms, we obtain d dt tV s ( )₌_δ tV s ( )_{− B −}_µ x+t sh tV h ( )₋ tV s ( )

(

)

+µx+t sd tV s ( )_. Answer E.

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model, you are given:

(i) The death benefit for cause (1) is 1000 and for cause (2) is F. (ii) Death benefits are payable at the end of the year of death. (iii) q45 1 ( ) _{= 0.04}_and_q 45 2 ( )_{= 0.20.} (iv) i = 0.06

(v) Z is the present value random variable for this insurance. Calculate the value of F that minimizes Var(Z).

A. 0 B. 50 C. 167 D. 200 E. 500

Solution.

Z can have the following values:

1000⋅1.06−1 with probability 0.04, F⋅1.06−1 with probability 0.20,

0 with probability 1 – 0.04 – 0.20 = 0.76. Therefore, the mean of Z is

0.04⋅1000 ⋅1.06−1+ 0.20 ⋅ F ⋅1.06−1, while the second moment of Z is

0.04⋅ 1000 ⋅1.06

(

−1

)

2+ 0.20 ⋅ F ⋅1.06

(

−1

)

2. The variance of Z is Var Z

( )

= = 0.04 ⋅ 1000 ⋅1.06

(

−1

)

2+ 0.20 ⋅ F ⋅1.06

(

−1

)

2− 0.04 ⋅1000 ⋅1.06

(

−1+ 0.20 ⋅ F ⋅1.06−1

)

2 = =0.04⋅ 1000

(

)

2 + 0.20 ⋅ F2− 0.042⋅10002− 0.202⋅ F2− 2 ⋅0.04 ⋅0.20 ⋅1000F 1.062 = =0.16F 2_{−16F + 38400} 1.062 .

This numerator of this expression is a quadratic function of F, which is minimized for F= − −16

2⋅0.16 = 50. Answer B.

Russell entered a defined benefit pension plan on January 1, 2000, with a starting salary of 50,000. You are given:

(i) The annual retirement benefit is 1.7% of the final three-year average salary for each year of service.

(ii) His normal retirement date is December 31, 2029.

(iii) The reduction in the benefit for early retirement is 5% for each year prior to his normal retirement date.

(iv) Every January 1, each employee receives a 4% increase in salary. (v) Russell retires on December 31, 2026.

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Calculate Russell’s annual retirement benefit.

A. 49,000 B. 52,000 C. 55,000 D. 58,000 E. 61,000 Solution.

The final three-year average salary is 50,000⋅1.04

26_+1.0425_+1.0424

3 ≈ 133,360.20. Based on this, the annual retirement benefit is approximately

0.017 1.7% for each year of service  ⋅ 27 Number of years of service  ⋅133,360.20

Final average salary   ⋅ 0.95 3 5% reduction for each of three years of early retirement

 ≈ 52,030.

Answer B.

An automobile insurance company classifies its insured drivers into three risk categories. The risk categories and expected annual claim costs are as follows:

Risk Category Expected Annual Claim Cost

Low 100

Medium 300

High 600

The pricing model assumes:

• At the end of each year, 75% of insured drivers in each risk category will renew their insurance.

• i = 0.06.

• All claim costs are incurred mid-year.

For those renewing, 70% of Low Risk drivers remain Low Risk, and 30% become Medium Risk. 40% of Medium Risk drivers remain Medium Risk, 20% become Low Risk, and 40% become High Risk. All High Risk drivers remain High Risk. Today the Company requires that all new insured drivers be Low Risk. The present value of expected claim costs for the first three years for a Low Risk driver is 317. Next year the company will allow 10% of new insured drivers to be Medium Risk. Calculate the percentage increase in the present value of expected claim costs for the first three years per new insured driver due to the change.

A. 14% B. 16% C. 19% D. 21% E. 23%

Solution.

There are three states: Low Risk, Medium Risk and High Risk. The transition matrix is Q= 0.7 0.3 0 0.2 0.4 0.4 0 0 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥

(12)

Q2 ₌ 0.55 0.33 0.12 0.22 0.22 0.56 0 0 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ .

We know that the present value of expected claim costs for the first three years for a Low Risk driver is 317. The present value of expected claim costs for the first three years for a Medium Risk driver is

300⋅1.06−0.5+ 100 ⋅0.2 + 300 ⋅0.4 + 600 ⋅0.4

(

)

⋅ 0.75 Probability of renewing  ⋅1.06−1.5+ + 100 ⋅0.22 + 300 ⋅0.22 + 600 ⋅0.56

(

)

⋅ 0.752 Probability of renewing twice  ⋅1.06−2.5≈ 758.7025.

Therefore, the present value of costs for the new portfolio is 0.9⋅ 317 + 0.1⋅ 758.7025 ≈ 361.170254.

The rate of increase is approximately 361.170254

317 −1 ≈ 13.933834. Answer A.

For a universal life insurance policy with a death benefit of 150,000, you are given: (i)

Policy Monthly Percent of Monthly Cost Monthly Year Premium Premium of Insurance Expense Charge Rate per 1000 Charge

1 2000 3.5% 1.00 50

(ii) i( )12 = 0.06.

(iii) The account value at the end of month 11 is 25,000. Calculate the account value at the end of month 12.

A. 26,830 B. 26,850 C. 26,870 D. 26,890 E. 26,910 Solution.

We have the standard recursive formula (we have only one interest rate here, no difference between credited rate and rate used for discounting)

AV_End = AV

(

_Start+ P 1− f

(

)

− e − COI

)

( )

1+ i , where

COI = DBEnd− AVEnd

1+ i ⋅ COI rate

(

)

.

Let us write AVt for the account value at the end of month t. Then

AV12 = AV

(

11+ P 1− f

(

)

− e − COI

)

( )

1+ i =

(13)

Also, COI =150,000− 27014.40 −1.005COI

(

)

1.005 ⋅0.001 ≈ 122.373731− 0.001COI, so that COI ≈122.373731 1.001 ≈ 122.25148. We conclude that AV12 = 27014.40 −1.005COI ≈ 27014.40 −1.005 ⋅122.25148 ≈ 26891.5373. Answer D.

For fully discrete whole life insurance policies of 10,000 issued on 600 lives with independent future lifetimes, each age 62, you are given:

(i) Mortality follows the Illustrative Life Table. (ii) i = 0.06.

(iii) Expenses of 5% of the first year gross premium are incurred at issue. (iv) Expenses of 5 per policy are incurred at the beginning of each policy year. (v) The gross premium is 102% of the benefit premium.

(vi) ₀L is the aggregate present value of future loss at issue random variable. Calculate Pr( 0L < 60,000), using the normal approximation.

A. 0.74 B. 0.78 C. 0.82 D. 0.86 E. 0.90 Solution.

The aggregate present value of future loss random variable is the sum of 600 individual policies present value of future loss random variables, which are independent and

identically distributed, thus by the Central Limit Theorem the aggregate present value of future loss random variable can be approximated by a normal random variable with the same mean and variance. Let us begin by finding the mean and variance of 0L. We will need to know the gross premium for this calculation. The benefit premium is

10,000A₆₂ a₆₂ ILT=

10⋅ 396.70

10.6584 ≈ 372.1947. Therefore, the gross premium is

G≈ 1.02 ⋅ 372.1947 ≈ 379.638595. Let us denote by ₀LSingle

the present value of future loss at issue random variable for a single policy. Then

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0L

Single = 10000 ⋅1.06− K+1( )−

G− 5

(

)

Gross premium after recurrent expense of 5

 ⋅ aK+1 + 0.05G

One time expense of 5% of gross premium at issue, it contributes to the loss

 ≈ ≈ 10000 ⋅1.06− K+1( )_{− 374.638595 ⋅}1−1.06− K+1( ) 0.06 1.06 a_K₊₁=1−vK+1 d    + 0.05 ⋅ 379.638595 ≈ ≈ 10000 +374.638595 0.06 1.06 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⋅1.06− K+1( )₋ 374.638595 0.06 1.06 − 0.05 ⋅ 379.638595 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ≈ ≈ 16618.6152 ⋅1.06− K+1( )_{− 6599.63325.}

The expected value of ₀LSingle is E 0L Single

(

)

≈ 16618.6152 ⋅ A62− 6599.63325 = ILT = ILT16.6186152⋅ 396.7 − 6599.63325 ≈ −7.0286073. The variance of ₀LSingle_is

Var 0L Single

(

)

≈ 16618.61522⋅ 2 A62− A62 2

(

)

= ILT = ILT16618.6152 2_{⋅ 0.19941− 0.3967}

(

2

)

_{≈ 11,610,292.90,}

so that the standard deviation is approximately 11,610,292.90 ≈ 3,407.3880. The aggregate loss present value of future loss at issue random variable 0L is the sum of six hundred independent identically distributed random variables with the mean and variance we have just calculated. Therefore,

E

( )

₀L ≈ 600 ⋅ −7.0286073

(

)

≈ −4217.1644, and

Var

( )

0L ≈ 600 ⋅ 3,407.3880 ≈ 83,463.6194.

Therefore, if we write Z for a standard normal random variable, and Φ for its cumulative distribution function, we obtain

Pr

(

0L< 60,000

)

≈ Pr 0L− −4217.1644

(

)

83, 463.6194 < 60,000− −4217.1644

(

)

83, 463.6194 ⎛ ⎝⎜ ⎞ ⎠⎟ ≈ ≈ Pr Z <60,000− −4217.1644

(

)

83, 463.6194 ⎛ ⎝⎜ ⎞ ⎠⎟ ≈ Pr Z < 0.76940306

(

)

≈ ≈ Φ 0.76940306

(

)

≈ 0.7792. Answer B.

For a fully discrete whole life insurance policy of 2000 on (45), you are given: (i) The gross premium is calculated using the equivalence principle.

(15)

(ii) Expenses, payable at the beginning of the year, are:

% of Premium Per 1000 Per Policy

First year 25% 1.5 30

Renewal years 5% 0.5 10

(iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06.

Calculate the expense reserve at the end of policy year 10.

A. −2 B. −10 C. −14 D. −19 E. −27 Solution.

While this is not stated clearly, it is implied by the structure of expenses that all premiums are level annual premiums. The expense reserve equals the actuarial present value of future expenses minus the actuarial present value of future expense premiums. Or, equivalently, it equals the actuarial accumulated value of past expense premiums minus the actuarial accumulated value of past expenses. In order to know the expense premiums, we must know the gross and net (i.e., benefit) premium. The benefit premium is simpler to calculate: 2000A₄₅ a45 = ILT 2⋅201.20 14.1121 ≈ 28.5145.

Since the gross premium is calculated using the equivalence principle, it must satisfy the equation Ga45 = 2000A45+ 1⋅ 2000 1000+ 20 ⎛ ⎝⎜ ⎞⎠⎟

First year extra expenses, per thousand and per policy

 + 0.5 ⋅2000 1000+10 ⎛ ⎝⎜ ⎞⎠⎟ Recurring expenses, per thousand and per policy 

⋅ a45+

+ 0.20G

First year percentage of premium expenses in excess of such expenses in all renewal years

 + 0.05Ga45 Percentage of premium expenses that are the same in all years

   .

From this, we can solve for G and obtain

Ga45 = 2000A45+ 22 +11a45+ 0.20G + 0.05Ga45, and

G= 2000A45+ 22 +11a45 0.95a₄₅− 0.20 ILT=

2⋅201.20 + 22 +11⋅14.1121

0.95⋅14.1121− 0.20 ≈ 43.8900.

Based on this, the expense premium is approximately 43.8900 – 28.5145 = 15.3755. At policy duration 10, the expense reserve equals

0.5⋅2000 1000+10 ⎛ ⎝⎜ ⎞

⎠⎟⋅ a55+ 0.05Ga55−15.3755a55≈

≈ 11+ 0.05⋅43.8900 −15.3755

(

)

⋅ a₅₅ = ILT =

ILT

(

11+ 0.05⋅43.8900 −15.3755

)

⋅12.2758 ≈ −26.77.

Of course, you should know that there is nothing strange about having a negative expense reserve.

(16)

You are profit testing a fully discrete whole life insurance of 10,000 on (70). You are given:

(i) Reserves are benefit reserves based on the Illustrative Life Table and 6% interest. (ii) The gross premium is 800.

(iii) The only expenses are commissions, which are a percentage of gross premiums. (iv) There are no withdrawal benefits.

(v)

Policy Year k q_70+k−1(death) _q_70+k−1(withdrawal) Commission Rate Interest Rate

1 0.02 0.20 0.80 0.07

2 0.03 0.04 0.10 0.07

Calculate the expected profit in policy year 2 for a policy in force at the start of year 2. A. 180 B. 190 C. 200 D. 210 E. 220

Solution.

The benefit premium (used for benefit reserves calculations) is 10000P₇₀ =10,000A70 a70 = ILT 10⋅514.95 8.5693 ≈ 600.92423. The benefit reserve at policy duration 1 is

1V70 = 10000 A

(

71− P71a71

)

_ILT=10⋅530.26 − 600.92423⋅8.2988 ≈ 315.65. The benefit reserve at policy duration 2 is

2V70 = 10000A72− Pa71 =

ILT10⋅545.60 − 600.92423⋅8.0278 ≈ 631.90.

The general recursive formula, which shows emergence of the expected profit in policy year t is:

(

P+t−1V− Et

)

( )

1+ i = Sqx+t−1+tVpx+t−1+ Prt.

The version of it that applies to policy year 2 in this problem is:

(

P+1V − E1

)

( )

1+ i = Sq71 death ( )_{+ 0⋅q} 70+k−1 withdrawal ( )₊ 2Vp71+ Pr2. Therefore, Pr₂ = P +

(

₁V − E₁

)

( )

1+ i − Sq₇₁(death) −₂V 1− q₇₁(death) − q_70+k−1(withdrawal)

(

)

= ≈ 800 + 315.65− 0.1⋅800

(

)

⋅1.07 −10,000⋅0.03− 631.90⋅ 1− 0.03− 0.04

(

)

≈ ≈ 220.48. Answer E.

An insurance company sells special fully discrete two-year endowment insurance policies to smokers (S) and non-smokers (NS) age x. You are given:

(i) The death benefit is 100,000. The maturity benefit is 30,000.

(ii) The level annual premium for non-smoker policies is determined by the equivalence principle.

(17)

(iii) The annual premium for smoker policies is twice the non-smoker annual premium. (iv) µx+t NS _{= 0.1, t > 0.} (v) qx+k S _{= 1.5q} x+ j NS , for k = 0,1. (vi) i = 0.08.

Calculate the expected present value of the loss at issue random variable on a smoker policy.

A. −30,000 B. −29,000 C. −28,000 D. −27,000 E. −26,000 Solution.

Given that the force of mortality for non-smokers is constant, qx NS= q x+1 NS = 1− e−0.1≈ 0.09516258. And, of course, qx+k S _{= 1.5q} x+ j NS _{≈ 1.5 ⋅0.09516258 ≈ 0.14274387.}

Since we are not told that the premium for smokers is determined by the equivalence principle, while it is the case for non-smokers, we must conclude that the premium for smokers is not derived from the equivalence principle, so the only way to find it is by finding the premium for smokers and doubling it. Let us find the premium for non-smokers, PNS

, which is determined by the equivalence principle: PNS_a x:2 = P NS 1+1.08−1⋅ px NS

(

)

= = 100,000 ⋅1.08−1⋅qx NS_{+100,000 ⋅1.08}−2_{⋅ p} x NS_⋅q x+1 NS _{+ 30,000 ⋅1.08}−2_{⋅ p} x NS_{⋅ p} x+1 NS , so that PNS₌100,000⋅1.08−1⋅qx NS_{+100,000⋅1.08}−2_{⋅ p} x NS_⋅q xNS+1+ 30,000⋅1.08−2⋅ px NS_⋅q x+1 NS 1+1.08−1⋅ p_xNS .

We calculate the numerator of the above as 100,000 1.08 ⋅0.09516258 + 100,000 1.082 ⋅ 1− 0.09516258

(

)

⋅0.09516258 + +40,000 1.082 ⋅ 1− 0.09516258

(

)

2 ≈ 37.251.4986.

We calculate the denominator as 1+1− 0.09516258 1.08 ≈ 1.83781242. Therefore, P NS_≈37.251.4986 1.83781242 ≈ 20,269.48, and PS= 2PNS≈ 2⋅20,269.48 ≈ 40,538.96.

The expected present value of the loss at issue random variable on a smoker policy is therefore, approximately,

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100,000⋅1.08−1_⋅q x S_{+100,000⋅1.08}−2_{⋅ p} x NS_⋅q x+1 NS _{+ 30,000⋅1.08}−2_{⋅ p} x NS_⋅q x+1 NS ₋ − PS_{− P}S_⋅1.08−1_{⋅ p} x NS_≈100,000 1.08 ⋅0.14274387 + +100,000 1.082 ⋅0.14274387 ⋅ 1− 0.14274387

(

)

+ 30,000 1.082 ⋅ 1− 0.14274387

(

)

2 − − 40,538.96 −40,538.96 1.08 ⋅ 1− 0.14274387

(

)

≈ −30,107.43. Answer A.

You are given:

(i) The following extract from a mortality table with a one-year select period:

x l_{[ ]}_x d_{[ ]}_x l_x₊₁ x + 1

65 1000 40 -- 66

66 955 45 -- 67

(ii) Deaths are uniformly distributed over each year of age. (iii) eo _{[ ]}₆₅ = 15.0. Calculate e_{[ ]}₆₆ o . A. 14.1 B. 14.3 C. 14.5 D. 14.7 E. 14.9 Solution.

We note immediately that l₆₆ = l_{[ ]}₆₅₊₁= 1000 − 40 = 960, and l₆₇ = l_{[ ]}₆₆₊₁= 955 − 45 = 910, so that d66 = l66− l67 = 960 − 910 = 50. We also have

15= e_{[ ]}₆₅ o = tp[ ]65 dt 0 1

∫

+ p_{[ ]}₆₅ ⋅ tp66dt 0 1

∫

+ p_{[ ]}₆₅ ⋅ p66⋅e67 o = UDD = UDD

(

1− tq[ ]65

)

dt 0 1

∫

+ p_{[ ]}₆₅ ⋅ 1− tq

(

66

)

dt 0 1

∫

+ p_{[ ]}₆₅ ⋅ p66⋅e67 o = = 1−1 2q[ ]65 ⎛ ⎝⎜ ⎞⎠⎟ +p[ ]65 ⋅ 1− 1 2q66 ⎛ ⎝⎜ ⎞⎠⎟ +p[ ]65 ⋅ p66⋅e67 o = = 1−1 2⋅ 40 1000 ⎛ ⎝⎜ ⎞⎠⎟ + 960 1000⋅ 1− 1 2⋅ 50 960 ⎛ ⎝⎜ ⎞⎠⎟ + 960 1000⋅ 910 960⋅e67 o . Therefore, e67 o = 15− 1−1 2⋅ 40 1000 ⎛ ⎝⎜ ⎞⎠⎟ −1000960 ⋅ 1− 1 2⋅ 50 960 ⎛ ⎝⎜ ⎞⎠⎟ 960 1000⋅ 910 960 ≈ 14.3791209. We also have

(19)

e_{[ ]}₆₆ o = tp[ ]66 dt 0 1

∫

+ p_{[ ]}₆₆ ⋅e67 o = UDD

(

1− tq[ ]66

)

dt 0 1

∫

+ p_{[ ]}₆₆ ⋅e67 o = = 1−1 2q[ ]66 ⎛ ⎝⎜ ⎞⎠⎟ +p[ ]66 ⋅e67 o = 1−1 2⋅ 45 955 ⎛ ⎝⎜ ⎞⎠⎟ +910955⋅14.3791209 ≈ 14.6780105. Answer D.

Scientists are searching for a vaccine for a disease. You are given: (i) 100,000 lives age x are exposed to the disease.

(ii) Future lifetimes are independent, except that the vaccine, if available, will be given to all at the end of year 1.

(iii) The probability that the vaccine will be available is 0.2. (iv) For each life during year 1, qx= 0.02.

(v) For each life during year 2, qx+1= 0.01, if the vaccine has been given, and qx+1= 0.02,

if it has not been given.

Calculate the standard deviation of the number of survivors at the end of year 2. A. 100 B. 200 C. 300 D. 400 E. 500

Solution.

The probability distribution of the random number of survivors is mixed, with probability 0.2 of having lower mortality due to the vaccine, and with probability 0.8 of having the same mortality as the first year. Let A be the event that the vaccine becomes available. Let N be the random number of survivors at the end of year 2. If event A happens, for each of the original 100,000 lives, that person is still alive at the end of year 2 with probability 0.98⋅0.99. This means that for each person, the number of survivors for that person is 1 with probability 0.98⋅0.99, and 0 with probability 1− 0.98 ⋅0.99, so it is a Bernoulli Trial with p = 0.98⋅0.99. The number of survivors for all 100,000 lives insured is binomial with n = 100,000 and p = 0.98⋅0.99. This means that its first moment is 100,000⋅0.98 ⋅0.99 = 97020, and the second moment is

100,000⋅0.98 ⋅0.99 ⋅ 1− 0.98 ⋅0.99

(

)

Variance

+ 97020 2 Square of the first monent 

= 9412883291.196. If event A does not happen, the number of survivors for all 100,000 lives insured is binomial with n = 100,000 and p = 0.98⋅0.98, with mean 100,000⋅0.98 ⋅0.98 = 96040, and the second moment

100,000⋅0.98 ⋅0.98 ⋅ 1− 0.98 ⋅0.98

(

)

Variance

+_{Square of the first monent}96040 2 = 9223685403.184. This implies that

E N

( )

= 0.2 ⋅97020 + 0.8 ⋅96040 = 96236,

E N

( )

2 _{= 0.2 ⋅9412883291.196 + 0.8 ⋅9223685403.184 = 9261524980.7864,} and therefore

(20)

Var N

( )

= E N

( )

2 _{− E N}

(

( )

)

2 _{= 9261524980.7864 − 96236}2 , and the standard deviation of N is

Var N

( )

≈ 9261524980.7864 − 962362 _{≈ 396.5914603.} Answer D.

You are given: (i) δt = 0.06, t ≥ 0.

(ii) µx

( )

t = 0.01, t ≥ 0.

(iii) Y is the present value random variable for a continuous annuity of 1 per year, payable for the lifetime of (x) with 10 years certain.

Calculate Pr Y

(

> E Y

( )

)

. A. 0.705 B. 0.710 C. 0.715 D. 0.720 E. 0.725 Solution. We have E Y

( )

= a₁₀ + e−0.06⋅10⋅e−0.01⋅10⋅ax+10= 1− e−0.06⋅10 0.06 + e −0.7_⋅ 1 0.06+ 0.01≈ 14.61388183. Therefore, Pr Y

(

> E Y

( )

)

≈ Pr Y > 14.61388183

(

)

= Pr 1− e −0.06T 0.06 > 14.61388183 ⎛ ⎝⎜ ⎞ ⎠⎟= = Pr T > ln 1− 0.06 ⋅14.61388183

(

)

−0.06 ⎛ ⎝⎜ ⎞ ⎠⎟ ≈ Pr T > 34.9035565

(

)

= = e−0.01⋅34.9035565≈ 0.705368043. Answer A.

For a whole life insurance of 10,000 on (x), you are given: (i) Death benefits are payable at the end of the year of death. (ii) A premium of 30 is payable at the start of each month. (iii) Commissions are 5% of each premium.

(iv) Expenses of 100 are payable at the start of each year. (v) i = 0.05.

(vi) 1000Ax+10 = 400.

(vii) ₁₀V is the gross premium reserve at the end of year 10 for this insurance. Calculate 10V using the two-term Woolhouse formula for annuities.

(21)

Solution.

The most common form of the Woolhouse formula is

ax m ( )_{≈ a} x− m−1 2m − m2₋₁ 12m2

(

δ+µx

)

.

But this problem asks us to use the two-term Woolhouse formula, i.e., ax m ( )_{≈ a} x− m−1 2m . In this problem, we have

ax+10 = 1− A_x₊₁₀ d = 1− 0.4 0.05 1.05 = 12.6, and ax+10 12 ( ) _{≈ a} x+10− 12−1 2⋅12 = 12.6 − 11 24≈ 12.14166667. The gross premium reserve sought is

10V = 10,000Ax+10

Actuarial present value of future benefits

   + 100ax+10

Actuarial present value of future expenses

  − 12 ⋅ 30

(

)

⋅ 1− 0.05

(

)

⋅ ax+10

12 ( ) Actuarial present value of future premiums after commissions

 ≈ ≈ 10,000 ⋅0.4 +100 ⋅12.6 − 360 ⋅0.95 ⋅12.14166667 ≈ 1107.55. Answer D.

For an increasing two-year term insurance on (x), you are given: (i) The death benefit during year k is 2000k, k = 1, 2.

(ii) Death benefits are payable at the end of the year of death. (iii) qx+k−1= 0.02k, k= 1,2.

(iv) The following information about zero coupon bonds of 100 at t = 0: Maturity (in years) Price

1 97.00

2 92.00

(v) Z is the present value random variable for this insurance. Calculate Var(Z).

A. 569,600 B. 570,600 C. 571,600 D. 572,600 E. 573,600 Solution.

The possible values of the random variable Z are: • In case of death in the first year,

Z = 2000 ⋅ 97

100 with probability 0.02, • In case of death in the second year,

Z = 4000 ⋅ 92

(22)

and otherwise Z = 0. Therefore, E Z

( )

= 2000 ⋅ 97 100⋅0.02 + 4000 ⋅ 92 100⋅0.98 ⋅0.04 = 183.056, E Z

( )

2 = 2000 ⋅ 97 100 ⎛ ⎝⎜ ⎞⎠⎟ 2 ⋅0.02 + 4000 ⋅ 92 100 ⎛ ⎝⎜ ⎞⎠⎟ 2 ⋅0.98 ⋅0.04 = 606134.08, Var Z

( )

= E Z

( )

2 − E Z

(

( )

)

2 = 606134.08 −183.0562 = 572624.580864. Answer D.

For a fully discrete whole life insurance, you are given:

(i) First year expenses are 10% of the gross premium and 5 per policy. (ii) Renewal expenses are 3% of the gross premium and 2 per policy. (iii) Expenses are incurred at the start of each policy year.

(iv) There are no deaths or withdrawals in the first two policy years. (v) i = 0.05.

(vi) The asset share at time 0 is 0. The asset share at the end of the second policy year is 64.11.

Calculate the gross premium.

A. 32.7 B. 34.2 C. 35.7 D. 37.2 E. 38.7 Solution.

The basic asset shares recursive formula is

h−1AS+ G 1− c

(

h−1

)

− eh−1

(

)

⋅ 1+ i

( )

= q_x( )1_+h−1_{+ q} x+h−1 2 ( ) _⋅ hCV+ hAS⋅ px+h−1 τ ( ) _.

First we use it for year 1, with h = 1, and obtain 0AS+ G 1− 0.10

(

)

− 5

(

)

⋅1.05 = 0 + 0⋅1CV+1AS⋅ 1− 0 − 0

(

)

, resulting in

1AS= 0.9G − 5

(

)

⋅1.05 = 0.945G − 5.25.

Then we write the recursive formula for the second policy year and obtain 1AS+ G 1− 0.03

(

)

− 2

(

)

⋅1.05 = 0 + 0⋅₂CV+₂AS⋅ 1− 0 − 0

(

)

, or 1.915G− 7.25

(

)

⋅1.05 = 2AS= 64.11. From this G= 64.11+ 7.623 2.01075 ≈ 35.67474823. Answer C.

For a fully discrete whole life insurance on (60), you are given: (i) Mortality follows the Illustrative Life Table

(23)

(ii) i = 0.06.

(iii) The expected company expenses, payable at the beginning of the year, are: 50 in the first year, 10 in years 2 through 10, 5 in years 11 through 20, 0 after year 20.

Calculate the level annual amount that is actuarially equivalent to the expected company expenses.

A. 8.5 B. 11.5 C. 12.0 D. 13.5 E. 15.0 Solution.

The expected company expenses are:

50+1E60⋅10a61:9 +10E60⋅5a70:9 = 40 + 5a60:10 + 5a60:20 = = 40 + 5 a

(

60−10E60⋅ a70

)

+ 5 a

(

60−20E60⋅ a80

)

=

ILT = ILT40+ 5 11.1454 − 0.4512 ⋅8.5693

(

)

≈7.27893184 + 5 11.1454 − 0.14906 ⋅5.905

(

)

≈10.2652007 ≈ ≈ 127.720663.

If this were paid as a level amount instead, the annual payment would be approximately 127.720663

a60

=127.720663

11.1454 ≈ 11.4594956. Answer B.