北美精算师MLC考试题答案2013-05

k =0 19 k +1
20
s k +1
k
q40 = 100, 000 20 E40 + π v
k =0 k
19
k +1
(1 + i ) k +1 − 1 q k 40 d
1 q40 − A40:20 )
= 100, 000 20 E40 + = 100, 000
20
π π
k q40 − v d k =0
May 2013 MLC Solutions
1. Key: D
Let k be the policy year, so that the mortality rate during that year is q30+ k −1 . The objective is to determine the smallest value of k such that v k −1 ( k −1 p30 ) (1000 P30 ) < v k ( k −1 p30 ) q30+ k −1 (1000)
9.
4.5
Key: E
V = v 0.5
0.5
0.5
p x + 4.5 5V + v 0.5 =
0.5
q x + 4.5 b, where b = 10, 000 is the death benefit during year 5
0.5
q x + 4.5 =
q x+4
1 − 0.5 q x + 4
P30 < vq30+ k −1 0.10248 q29+ k < 15.8561 1.06 q29+ k > 0.00685 29 + k > 51 k > 22 Therefore, the smallest value that meets the condition is 23.
2.
Key: A
19
k +1
q40 = 100, 000
20
E40 +
π
( d
20
E40 +
( d
20
40:20 + v 20 20 p40 q40 − 1 + da 1 − v 20 d 20 6%π . − 20 p40 a
)
40:20 − π 20 p40 = 100, 000 20 E40 + π a = 100, 000
, K < 20 π a = K +1 K ≥ 20 0, − 0, K < 20 π a = K +1 . 20 − π a 20 , K ≥ 20 π a
The first term is the random variable that corresponds to a 20-year temporary annuity. The 20 second term is the random variable that corresponds to a payment with a present value of π a
where T50:55 = min(T50 , T55 ) is the time until the first death.
L = 100e
−0.05T 50:55
− 60 > 0 e−0.05T50:55 > 0.6 T50:55 < −20 ln(0.6)
Pr( L > 0) = Pr T50:55 < −20 ln(0.60) = 1 − Pr [T50:55 ≥ −20 ln(0.60)] = 1 − Pr [T50 ≥ −20 ln(0.60)] Pr [T55 ≥ −20 ln(0.60) ] −20 ln(0.60) −0.04[ −20ln(0.60)] = 1 − 1 − e 50 = 0.4712
Expected Present Value of Benefits: 5, 000(10 / 1, 000) / 1.06 + 7,500(15 / 1, 000) / 1.062 + 10, 000(18 / 1, 000) / 1.063 = 47.17 + 100.12 + 151.13 = 298.42. Expected Present Value of Premiums: [1 + (870 / 1, 000) / 1.06 + (701 / 1, 000) / 1.06 2 ]P = 2.4446 P. The annual premium is P = 298.42/2.4446 = 122
0.5(0.04561) = 0.02334 1 − 0.5(0.04561)
0.5
p x + 4.5 = 0.97666
5
V= V=
( 4V + P) (1.03) − q x + 4b p x+4
(1, 405.08 + 647.46) (1.03) − 0.04561(10, 000) = 1, 737.25 0.95439 − 0.5 (0.97666) (1, 737.25) + (1.03) −0.5 (0.02334) (10, 000) 4.5V = (1.03)
1 1 40 1 fT (t )dt = dt 0 (1 + 0.2t ) 40 0 1 + 0.2t 1 1 1 40 = ln (1 + 0.2t ) 0 = ln(9) = 0.27465 40 0.2 8 2 2 −2 2 E ( Z ) = E{(1 + 0.2T ) [(1 + 0.2T ) ] } = E[(1 + 0.2T ) −2 ] =
5.
10
Key: A
02 p 65:60 = 10 0 10 t 00 22 p 65:60 μ 02 10 − t p 65 + t :60 + t dt
= e −0.01t (0.005)e −0.008(10−t ) dt
0
= 0.005e −0.08 e −0.002t dt
0
10
40:20 − 20 p40 a 20 π . contingent on surviving 20 years. The expected present value is then π a
The third approach takes the most steps.
40:20 = 100, 000 πa E40 + π v
5
= 1, 671.81 + 229.98 = 1,902
4.5
V can also be calculated retrospectively:
0.5
px + 4 = 0.5(0.04561) = 0.02281
A60:3 − q60 v − (1 − q60 )v 3
0.86545 −
0.01 0.99(0.017) 0.99(0.983) + + 1.045 1.045 2 1.0453 = 0.87777 =
8.
Key: A
Var ( Z ) = E ( Z 2 ) − E ( Z ) 2
−2 −1 E (Z ) = E (1 + 0.2T )(1 + 0.2T ) = E (1 + 0.2T )
20
40:20 E40 + π a
4.
Key: C
There are four career paths Joe could follow. Each is of the form: p35 ( p36 )( p37 )(transition probability 1)(transition probability 2) where the survival probabilities depend on each year’s employment type. For example, the first entry below corresponds to Joe being an actuary for all three years. The probability that Joe is alive on January 1, 2016 is: (0.9)(0.85)(0.8)(0.4)(0.4)
+ (0.9)(0.85)(0.65)(0.4)(0.6) + (0.9)(0.7)(0.8)(0.6)(0.8) + (0.9)(0.7)(0.65)(0.6)(0.2) = 0.50832 The expected present value of the endowment is (100, 000)0.50832 / (1.08) 3 = 40, 352
40
1 1 −1 1 = fT (t ) = 0 (1 + 0.2t ) 2 40 0.2 (1 + 0.2t ) 0 1 1 1 = 1 − = = 0.11111 8 9 9
40
40
Var ( Z ) = 0.11111 − (0.27465) 2 = 0.03568
= 0.005e −0.08
1 − e −0.02 = 0.0457 0.002
6.
Key: B
t
t
t p 50 = 1 − , 0 ≤ t ≤ 50 50 p 55 = e −0.04t , t ≥ 0
t
p 50:55 = Pr(T50:55
t −0.04t , 0 ≤ t ≤ 50 1 − e > t ) = 50 0, t > 50
3.
Key: C
There are three ways to approach this problem. In all cases, let π denote the benefit premium., The first approach is an intuitive result based on the solution to Sample Question 309. The key is that in addition to the pure endowment, there is a benefit equal in value to a temporary annuity due with annual payment π. However, if the insured survives the 20 years, the value of the annuity is not received. 40:20 = 100, 000 20 E40 + π a 40:20 − 20 p40 a 20 6%π . πa Based on this equation, 100, 000 20 E40 100, 000v 20 100, 000 100, 000 π= = = = = 2,565 20 38.993 a s20 20 p40 a20 The second approach uses random variables to derive the expected present value of the return of premium benefit. Let K be the curtate future lifetime of (40). The present value random variable is then π s v K +1 , K < 20 Y = K +1 K ≥ 20 0,
7.
Key: D
A60:3 = q60 v + (1 − q60 ) q60 +1v 2 + (1 − q60 )ห้องสมุดไป่ตู้1 − q60 +1 )v 3 = 0.86545
0.01 0.99 − 1.05 1.05 3 = 0.017 when v = 1/1.05. = q 60+1 = 0.99 0.99 (1 − q60 )v 2 − (1 − q60 )v 3 − 2 1.05 1.05 3 The primes indicate calculations at 4.5% interest. ′ = q60 v′ + (1 − q60 ) q60+1v′ 2 + (1 − q60 ) (1 − q60 +1 )v′ 3 A60:3