November 2012 MLC Solutions

November 2012 MLC Solutions Question # 1 Answer: E Pr(last survivor dies in the third year)  2 p 80:90  3 p 80:90   2 p 80  2 p 90  2 p 80:90    3 p 80  3 p 90  3 p 80:90   [0.9(0.8)  0.6(0.5)  0.9(0.8)(0.6)(0.5)]  [0.9(0.8)(0.7)  0.6(0.5)(0.4)  0.9(0.8)(0.7)(0.6)(0.5)(0.4)]  (0.72  0.30  0.216)  (0.504  0.12  0.06048)  0.804  0.56352  0.24048

Question # 2 Answer: B Under constant force over each year of age, l x  k  (lx )1 k (l x 1 ) k for x an integer and 0  k  1. l [60]  2.75  l [60]  5.75 q  2 3 [60]  0.75 l [60]  0.75 l [60]  0.75   80, 0000.25  79, 0000.75   79, 249 l [60]  2.75   77, 0000.25  74, 0000.75   74, 739 l [60]  5.75   67, 0000.25  65, 0000.75   65, 494 23

q [60]  0.75 

74, 739  65, 494 79, 249

 0.1167 1000 2 3 q [60]  0.75  1000(0.11679)  116.8

Question # 3 Answer: B 1

1  t t 4  . Since S0 (t )  1  F0 (t )   1   , we have ln S0 (t )  ln  4   d 1 1 1 1 1     110. and 65  Then t   ln S0 (t )  dt 4  t 180 4   65 3

e106   t p106 , since 4 p106  0 t 1

 106  t  1 1/4  S0 (106  t )  110   4t      1/ 4 S0 (106)  4   106  1    110  1/4

t

p106

1

 4t 4    t 1  4  1  0.25  3 0.25  2 0.25  10.25  4  2.4786 3

e106

Question # 4 Answer: D 10

V  50, 000  A50  10 E50 A60   1116  a50  10 E 50 a60   50, 000[0.24905  0.51081(0.36913)]  1116[13.2668  0.51081(11.1454)]  13, 428

Question # 5 Answer: C

V 0

0

V  2000

2

Year 1:  0V  P  1  i   qx  2000  1V   px 1V

P(1.1)  0.15(2000  1V )  0.85( 1V ) 1.1P  300  1V Year 2:  1V  P  1  i   qx 1 (2000  2V )  px 1 (2000) (1.1P  300  P )(1.1)  0.165(2000  2000)  0.835(2000) 2.31P  330  2330 2330  330 P  1152 2.31

Question # 6 Answer: B d  tV   G  E  S   9.6V (    ) where G, E, S and  are evaluated at t = 9.6 and dt t  9.6 where S includes claims-related expenses. Then, d  tV   450  0.02(450)  (106, 000  200)(0.01)  9.6V (0.01  0.05) dt t 9.6

 621  0.06 9.6V V  9.8V  0.2

9.6

d  tV   126.68  0.2(621  0.06 9.6V ) dt t  9.6

 250.88  0.012 9.6V V

9.6

250.88  247.91 1.012


Account Value at end of year 5  (30  20  9)1.06  43.46 Surrender Value at end of year 5  43.46  20  23.46 Asset Share  [(20  (20  2))1.08  0.001(1000)  0.05(23.46)] / (1  0.001  0.05)  38.867 / 0.949  40.96

Question # 8 Answer: A

Let CSVk and SCk denote the cash surrender value and surrender charge at time k. In this solution, COIk denotes the cost of insurance for month k, to be deducted from the account value at time k – 1, that is, the beginning of month k.

i (12)  0.054  i per month  0.0045 CSV13  AV13  SC13 1802.94  AV13  125  AV13  1927.94

AV12   AV11  300(1  W )  10  COI12 1.0045

COI12  50, 000(0.002) / 1.0045  99.55 AV11  CSV11  SC11  1200  500  1700

AV12  1700  300(1  W )  10  99.551.0045

AV12  1890.45  300W 1.0045  1898,96  301.35W COI13  50, 000(0.003) / 1.0045  149.33

AV13  [ AV12  300(1  0.15)  10  149.33](1.0045)  (1898.96  301.35W  95.67)(1.0045)  2003.61  302.71W  1927.94 W  (2003.61  1927.94) / 302.71  0.25

Question # 9 Answer: A Year Premium

Expense Charge

1

5000

100

2

5000

100

COI

Interest

EOY AV

200(5.40/1.06) = 1019 200(6.00/1.06) = 1132

0.06(5000–100 –1019) = 233 0.06(4114+5000 –100–1132) = 473

5000–100–1019+233 = 4114 4114+5000–100 –1132+473 = 8355

PV expected surrender cost in year 2  0.06(1  0.0034)(1  0.0038)(1  0.06)(8355)0.93 / 1.07 2  380

Question # 10 Answer: C p x  k : y  k  px  k  p y  k  px  k : y  k  0.84366  0.86936  0.77105  0.94197 for k  0,1, 2 . qx  k : y  k  1  px  k : y  k  1  0.94197  0.05803 for k  0,1, 2 .

pxy  pxy px 1: y 1  px  k 1: y  k 1   pxy   0.77105k for k  0,1, 2 . k

k

p xy

qx  k : y  k

Discount factor

Product

1 0.77105 0.59452

0.05803 0.05803 0.05803

1/1.03=0.97087 1/1.082 = 0.85734 1/1.13 = 0.75131

0.05634 0.03836 0.02592

k 0 1 2

k

EPV = (100,000)(0.05634 + 0.03836 + 0.02592) = 12,062


x

qx

t

70 71 72

0.03318 0.03626

0 1 2

t

p70

1 0.96682 0.93176

t

q70

0.03318 0.03506

Spot rate 0.016 0.026

v(t 1)

EPV

0.984252 0.94996

0.03266 0.03331

Sum =

0.06597

The EPV of a two-year deferred insurance is 0.93176(0.94996) A72  0.88513(0.54560)  0.48293 , where A72 is from the ILT at 6% (6% is correct since all forward rates are 6% after two years). Then the answer is 1000(0.06597  0.48293)  548.90

Question # 12 Answer: B

Because it is impossible to return to state 0, 1 p000 and 1 p000 are the same. Then,

t

p  tp e 00 0

00 0

  



 0  j1  0  s ds  t

2

0j

t01  t02  0.01  0.02(2t )   0.5 0.01  0.02(2t )   0.015  0.03(2t ) which is Makeham’s law with A  0.015, B  0.03, c  2 1

1

p000  1 p000  e 0.015(1) e

0.03( 2 1) ln(2)

 0.943

It is not necessary to recognize that this is Makeham’s law. The value can be calculated directly as t 00 00  1  1 p0  1 p0  exp   0 0.015  0.03(2 ) dt    1   0.03(2t )    exp    0.015t    ln 2  0      exp[(0.015  0.08656  0  0.04328)]  exp(0.05828)  0.943

Question # 13 Answer: D 00  p 50

l51( ) 90,365   0.90365 l50( ) 100, 000 03 p 50

00 q51(3)  q50(3)  1  p50(3)  1  p 50

d

30 51

( ) 51

l

 90,365

p

03 51

( ) 51

l

ln p51(3) 00 ln p 51

0 p 50

00  1  p 50

0 p 51  l51( )

 d ( 3)   50   ( )  l   50 

00   1 p 50   

 1  0.90365

1100 100,000 1 0.90365

 0.0115

ln(1  0.0115)  l52( )  1    l52( )   l51( )  ln  ( )   l51 

ln(1  0.0115) (1  80, 000 / 90,365)  984 ln(80, 000 / 90,365)

d 51(1)  l51( )  l52( )  d 51(2)  d 51(3)  90,365  80, 000  8200  984  1181 01 p 51

00 q51(1)  1  p51(1)  1  p 51

0 p 51

00  1  p 51

 d (1)   51   ( )  l   51 

00   1 p 51   

 1  (80, 000 / 90,365)

1181 90,365 80,000 1 90,365

 0.0138

10, 000q51(1)  10, 000(0.0138)  138

Note: This solution uses multi-state notation for dependent probabilities. There is alternative notation for these when the context is strictly multiple decrement, as it is here. Question # 14 Answer: A

2v K 1 , K  20 Z   K 1 v , K  20 E[ Z ]  2 A40  20 E40 A60  2(0.36987)  0.51276(0.62567)  0.41892 Var ( Z )  E[ Z 2 ]  ( E[ Z ])2  0.24954  0.418922  0.07405 SD( Z )  0.07405  0.27212 1 An alternative way to obtain the mean is E[ Z ]  2 A40:20  20| A40 . Had the problem asked for the evaluation of the second moment, a formula is 1 E[ Z 2 ]  (22 )  2 A40:20   (v2 )20  20 p40   2 A60  .

Question # 15 Answer: D

Benefit 300,000 330,000 360,000 390,000

Half-year 1 2 3 4

Under CF assumption, 0.5

p x 1 

0.5

0.5

PV of Benefit 275,229 277,754 277,986 276,286 px 

0.5

PV  277, 000 if and only if (x) dies in the 2nd or 3rd half years.

p x  0.5  (0.84) 0.5  0.9165 and

p x 1.5  (0.77) 0.5  0.8775 . Then the probability of dying in the 2nd or 3rd half-

years is  0.5 p x 1  0.5 p x0.5    p x 1  0.5 p x1   (0.9165)(0.0835)  (0.84)(0.1225)  0.1794


For t  0 and h  0.5, 0.5

p10  0 p10  0.5  0 p10   01   02   0 p11 10  0 p12  20   0  0.5  0  1 10  0   0.5 10  0.03.

Similarly

0.5

p12  0.5 12  0.05 .

Then, 0.5 p11  1  0.03  0.05  0.92 . For t  0.5 and h  0.5, 1

p10 

0.5

p10  0.5



0.5

p10   01   02   0.5 p11 10  0.5 p12  20

 0.03  0.5[0.03(0.02)  0.92(0.06)  0]  0.0573.



Question # 17 Answer: E

i (4)  0.08 means an interest rate of j  0.02 per quarter. This problem can be done with two quarterly recursions or as a single calculation. Using two recursions: 800  706 1000  10.5V  60(1  0.1)1.02  800  V 10.75 706 800  V  541.02  117.5 753.72  10.5 0.8825 10.5V  713.31 V

898  800 1000 898 800 898  10.25V 1.02  109.13

 10.25V 1.02 

10.5

713.31 

0.8909

10.25V  730.02.

Using a single step: 10.25V is the EPV of cash flows through time 10.75 plus flows thereafter (that is,

0.5

E80.25 times the EPV of cash

V ).

10.75

 898  800 800  706  800 706 V  1000    60(1  0.1)  753.72 2 898(1.02) 898(1.02) 2  898(1.02) 898(1.02)   730.

10.25

Question # 18 Answer: B

EPV of benefits at issue  1000 A40  4 11 E40 (1000 A51 )

 161.32  (4)(0.50330)(259.61)  683.97 EPV of expenses at issue  100  10( a40  1)  100  10(13.8166)  238.17   (683.97  238.17) / a40  922.14 / 14.8166  62.24 G  1.02  63.48

EPV of benefits at time 1  1000 A41  4 10 E 41 (1000 A51 )

 168.69  (4)(0.53499)(259.61)  724.25 EPV of expenses at time 1  10(a41 )  10(14.6864)  146.86 Gross Prem Reserve  724.25  146.86  Ga41  871.11  63.48(14.6864)  61.18

Question # 19 Answer: E

Let Xi be the present value of a life annuity of 1/12 per month on life i for i  1, 2, , 200 . 200

Let S   180 X i be the present value of all the annuity payments. i 1

(12) E[ X i ]  a62 

2

Var  X i  

1  A62(12) 1  0.4075   10.19267 d (12) 0.05813



12  12  A62  A62

d    12

2



2



0.2105  0.47052  13.15255 0.058132

E[ S ]   200 180 10.19267   366,936.12

Var  S   200 180  13.15255   85, 228,524 2

With the normal approximation, for Pr S  M   0.90 , M  E[ S ]  1.282 Var  S   366,936.12  1.282 85, 228,524  378, 771.45 . So  

378, 771.45  1893.86 200


Let C be the annual contribution, then C 

20

E 45 a65 a45:20

Let K65 be the future curtate lifetime of (65). The required probability is  E a a45:20   Ca45:20  Pr   aK 1   Pr  20 45 65  aK 1  65 65  E   a45:20 20 E45   20 45   

   Pr  9.8969  a   Pr a65  aK

65 1

K65 1

Thus, since a14  9.8527 and a15  10.2950 we have



Pr aK

65 1



 9.8969  Pr  K 65  1  14   1  14 p 65

 1

l 79 l 65

 1

4, 225,163  0.439 7,533,964

Note that it is not necessary to know the mortality or interest rates before age 65 because the values of 20 E45 and a45:20 cancel out in determining the actuarial accumulated value of the contributions. Question # 21 Answer: A

Let   benefit premium, then  a xy  100 A xy  3 1 1 3 1 1  a xy   e 0.05t  e 0.01t  e 0.03t  dt    13.54167 0 4 4 0.06 4 0.08 4  A xy  1   a xy  0.3229167

0.01 1  0.06 6 0.02 2 Ay   0.07 7 A xy  A x  A y  A xy  0.1294643 Ax 



100(0.1294643)  0.956044 13.54167


The equation of value is given in words by: Premiums for at most two years = 1000 for dying within two years + return of premium without interest (pay P if die in first year, 2P if die in second year). In symbols, the equation is: 1 1 Pa80:2  1000 A80:2  P ( IA)80:2 Solving for P we obtain P 

1 1000 A80:2 1 a80:2  ( IA) 80:2

.

Then, a80:2  1  vp 80  1 

0.91970  1.90388 1.0175

 0.08030 0.91970(0.08764)  1 1000 A80:2  1000  vq 80  v 2 p 80 q 81   1000     156.77 1.0175 2  1.0175 0.08030 0.91970(0.08764) 1  vq 80  2v 2 p 80 q 81  2  0.23463 ( IA) 80:2 1.0175 1.0175 2

Then, P 

156.77  93.92 1.90388  0.23463

Question # 23 Answer: A

Net amount at risk  1000  3V  987.82 Expected deaths  (10, 000  30) q47  9970(0.00466)  46.46 Actual deaths = 18 Mortality Gain/Loss = (Expected deaths – Actual deaths)(Net amount at risk)  (46.46  18)(987.82)  28,113


possible transitions H Z H L H Z D H LD H H Z H H L

probability 0.05 0.04 0.05(0.7)  0.035 0.04(0.6)  0.024 0.90(0.05)  0.045 0.90(0.04)  0.036

discounted benefits 250 v 250 v 1000 v 2 1000 v 2 250 v 2 250 v 2

APV 11.904762 9.523810 31.746032 21.768707 10.204082 8.163265

Sum of these gives total APV

=93.31066

Question # 25 Answer: E Ga x:10  100, 000 A1x:10  G ( IA)1x:10  0.45G  0.05Gax :10  200ax :10 G

100, 000(0.17094)  200(6.8865) 18, 471.3   3604.23 (1  0.05)(6.8865)  0.96728  0.45 5.124895