Manual for SOA Exam MLC.


 Damon Thompson
 2 years ago
 Views:
Transcription
1 Chapter 6. Benefit premiums. Extract from: Arcones Fall 2010 Edition, available at 1/77
2 Fully discrete benefit premiums In this section, we will consider the funding of insurance products paid at the end of the year of death with annual benefits premiums made at the beginning of the year. The funding is made as far as the individual is alive and the term of the insurance has not expired. 2/77
3 Fully discrete whole life insurance. Definition 1 The loss random random variable for a unit whole life insurance paid at the end of the year of death funded with an annual benefit premium at the beginning of the year while the individual is alive is denoted by L x. The insurance contract described in the previous definition is called a fully discrete whole life insurance. 3/77
4 Theorem 1 For a fully discrete whole insurance, (i) The loss random variable is L x = v Kx Pä Kx = Z x PŸx ( = Z x 1 + P d =Z x 1 Z x d ) P d. (ii) The actuarial present value of the loss L x is ( E[L x ] = A x Pä x = A x 1 + P ) P d d. (iii) The variance of the loss L x is Var(L x ) = ( 1 + P ) 2 ( Var(Z x ) = 1 + P ) 2 ( 2 2 A x A ) x. d d 4/77
5 The loss L x is a random variable. The loss is positive for some insurees and negative for another ones. The loss random variable L x is a discrete random variable taking the values v k Pä k i, k = 1,..., ω x. Since L x = v ( Kx 1 + P ) d P d, the loss is decreasing with K x. The biggest loss is attained when K x = 1, and it is v 1 Pä 1 = v P. The smallest loss is attained when K x = ω x, and it is v ω x Pä ω x. Hence, v ω x Pä ω x L x v P. To make a profit, an insurer would like that insurees will die as late as possible. To find the probability that the loss is positive, we manipulate P{L x > 0} = P{v Kx ( 1 + P d ) P d > 0} to get P{K x k 0 } = k0 q x. 5/77
6 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = /77
7 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = (i) John entered this contract and died 10 years, 5 months and 5 days after the issue of this contract. Find the insurer s loss at the time of the issue of the policy. 7/77
8 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = (i) John entered this contract and died 10 years, 5 months and 5 days after the issue of this contract. Find the insurer s loss at the time of the issue of the policy. Solution: (i) John made annual benefit premiums of $46 at times 0, 1,..., 10 years. John s state received a death benefit of $10000 at time 11 years. The insurer s loss at issue time is (10000)e (11)(0.075) (46)ä 11 e = /77
9 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = (ii) Peter entered this contract and died 42 years, 2 months and 20 days after the issue of this contract. Find the insurer s loss at the time of the issue of the policy. 9/77
10 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = (ii) Peter entered this contract and died 42 years, 2 months and 20 days after the issue of this contract. Find the insurer s loss at the time of the issue of the policy. Solution: (ii) Peter made annual benefit premiums of $46 at times 0, 1,..., 42 years. Peter s state received a death benefit of $10000 at time 43 years. The insurer s loss at issue time is (10000)e (43)(0.075) (46)ä 43 e = /77
11 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = (iii) Calculate the probability that the loss at issue is positive. 11/77
12 Example 1 Consider a fully discrete whole life insurance with face value The annual benefit premium paid at the beginning of each year which the insuree is alive is $46. Suppose that the force of mortality is δ = (iii) Calculate the probability that the loss at issue is positive. Solution: (iii) The loss at issue random variable is 10000L x = (10000)v Kx (46)ä Kx = (10000)v Kx (46) 1 v Kx =v Kx ( e ) 46 1 e =( )e (0.075)Kx d We need to find P{10000L x > 0}. 12/77
13 10000L x = ( )e (0.075)Kx > 0 is equivalent to e (0.075)Kx > = and to ln ( ) K x < = The probability that the loss is positive is P{K x 37} = P{T x 37} = 1 e (37)(0.005) = Notice that if K x = 37, the loss is (10000)e (37)(0.075) (46)ä 37 e = If K x = 38, the loss is (10000)e (38)(0.075) (46)ä 38 e = /77
14 Definition 2 The benefit premium of a fully discrete whole insurance funded under the equivalence principle is denoted by P x. If a whole insurance is funded under the equivalence principle, then E[L x ] = 0. P x is the annual benefit premium at which the insurer expects to break even. Theorem 2 If a fully discrete whole insurance is funded using the equivalence principle, then P x = A x ä x = da x 1 A x = 1 ä x d and Var(L x ) = 2 A x A x 2 2 (1 A x ) 2 = A x A x (dä x ) /77
15 Theorem 3 If a fully discrete whole insurance is funded using the equivalence principle, then P x = A x ä x = da x 1 A x = 1 ä x d and Var(L x ) = 2 A x A x 2 2 (1 A x ) 2 = A x A x (dä x ) /77
16 Theorem 3 If a fully discrete whole insurance is funded using the equivalence principle, then P x = A x ä x = da x 1 A x = 1 ä x d and Var(L x ) = 2 A x A x 2 2 (1 A x ) 2 = A x A x (dä x ) 2. 2 Proof: Since 0 = E[L x ] = A x Pä x, we have that P x = Ax ä x. Using that ä x = 1 Ax d, we get that P x = da x 1 A x = 1 ä x d. 16/77
17 Theorem 3 If a fully discrete whole insurance is funded using the equivalence principle, then P x = A x ä x = da x 1 A x = 1 ä x d and Var(L x ) = 2 A x A x 2 2 (1 A x ) 2 = A x A x (dä x ) 2. 2 Proof: We have that 1 A x 1 + P dax x d = 1 + d = 1 = 1. 1 A x dä x Hence, Var(L x ) = ( 1 + P ) 2 x ( 2 2 A x A ) 2 A x A 2 2 x x = d (1 A x ) 2 = A x A 2 x (dä x ) 2. 17/77
18 Example 2 Consider the life table x l x An 80 year old individual signs a whole life policy insurance which will pay $50000 at the end of the year of his death. The insuree will make level benefit premiums at the beginning of the year while he is alive. Suppose that i = 6.5%. 18/77
19 (i) Find the net single premium for this policy. Solution: (i) A 80 = (1.065) + (1.065) (1.065) + (1.065) (1.065) 250 = (1.065) 250 The net single premium for this policy is 250 (50000)A 80 = (50000)( ) = /77
20 (ii) Find the benefit annual premium for this policy using the equivalence principle. Solution: (ii) We have that ä 80 = /1.065 = , P 80 = A 80 ä 80 = = The benefit annual premium for this policy using the equivalence principle is (50000)P 80 = (50000)( ) = /77
21 (iii) Suppose that year old individuals enter this insurance contract and they die according with the deterministic group. Suppose that the insurer makes a fund to paid for death benefits. Benefit annual premiums are deposited in this fund. Let BO k be the balance in this fund at time k before making depostis/withdrwals of death benefits/benefit premiums. Let l x+k P 80 be the total amount of annual benefit premiums received at time k. Let d x+k 1 B be the total amount of death benefits paid at time k. Let B k be the balance in this account at time k after making the deposits/withdrawals for death benefits/benefit premiums. Make a table with BO k, l x+k P 80, d x+k 1 B and B k, for k = 0, 1, 2,..., 6. 21/77
22 Solution: (iii) We have that BO 0 = 0 and BO k = (1 + i)b k 1, for k = 1, 2,..., 6. We have that B k = BO k + l x+k P 80 d x+k 1 B. We have that k BO k l x+k P d x+k 1 B B k /77
23 (iv) Find the probability mass function of the random loss when the benefit premium follows the equivalence principle. Solution: (iv) The loss is We have that L 80 = (50000)Z 80 ( )Ÿ 80 =(50000)v K 1 v K ( ) 0.065/1.065 =( )v K P{L 80 = ( )v k } = l 80+k 1 l 80+k l 80, k = 1, 2,..., 6. 23/77
24 In particular, P{L 80 = } = = , P{L 80 = } = = , P{L 80 = } = = , P{L 80 = } = = , P{L 80 = } = = , P{L 80 = } = = /77
25 (v) Find the probability that the loss is positive. Solution: (v) The probability that the loss is positive is Alternatively, we have that = P{( )v K > 0} =P{K 80 < ln( / ) } ln(1.065) =P{K 80 < } = P{K 80 3} = = /77
26 (vi) Find the variance of the loss. Solution: (vi) We got that A 80 = We have that A 80 = (1.065) + (1.065) (1.065) + (1.065) (1.065) + (1.065) = Hence, Var(L 80 ) = (50000) 2 2 A 80 (A 80 ) 2 (1 A 80 ) ( )2 =(50000) (1 ( )) 2 = /77
27 Theorem 4 Under constant force of mortality µ for life insurance funded through the equivalence principle, P x = vq x. Proof: Since A x = qx q x +i and ä x = 1+i q x +i, P x = qx qx +i 1+i qx +i = vq x. 27/77
28 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = /77
29 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = (i) Find the net single premium for this policy. 29/77
30 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = (i) Find the net single premium for this policy. Solution: (i) A 30 = q x q x + i = 1 e e = and the net single premium is (70000)( ) = /77
31 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = (ii) Find the benefit annual premium for this policy. 31/77
32 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = (ii) Find the benefit annual premium for this policy. Solution: (ii) We have that P x = vq x = (1.06) 1 (1 e 0.02 ) = The benefit annual premium for this policy is (70000)( ) = /77
33 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = (iii) Find the variance of the present value of the loss for this insurance contract. 33/77
34 Example 3 Maria is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Maria is alive. Assume that i = 6% and death is modeled using the constant force of mortality µ = (iii) Find the variance of the present value of the loss for this insurance contract. Solution: (iii) We have that 2 A 30 = 1 e e (1.06) 2 1 = , Var(L x ) = (70000) 2 2 A x A x 2 (1 A x ) ( )2 =(70000) ( ) 2 = /77
35 Besides the equivalence principle, there are other ways to determine annual benefit premiums. Unless said otherwise, we will assume that the equivalence principle is used. An insurer expects to break even if it uses the equivalence principle. However, an insurer should be compensated for facing losses. Often, the annual benefit premium in an insurance contract is bigger than the annual benefit premium obtained using the equivalence principle. The risk charge (or security loading) is the excess of the benefit annual premium over the benefit annual premium found using the equivalence principle. 35/77
36 The 100α th percentile annual premium for a fully discrete whole life insurance to (x) is the largest P α such that { } α P{Z x P α Ÿ x > 0} = P{v Kx P α ä Kx > 0} = P 1 > P α. s Kx P α is a (1 α) th quantile of 1 s Kx. P α = 1 s kα, where k α satisfies that P{K x < k α } α < P{K x < k α + 1}. Then, P α = 1 s kα satisfies that { P{Z x P α Ÿ x > 0} = P and for P > P α, P{Z x PŸ x > 0} = P =P{K x k α } > α. 1 s Kx { 1 s Kx } > 1 = P{K x < k α } α s kα > P } P { 1 s Kx } 1 s kα 36/77
37 Example 4 Michael is 50 year old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of each year while Michael is alive. Assume that i = 6% and death is modeled using De Moivre s model with terminal age 100. (i) Calculate the benefit annual premium if the probability of a loss is at most (ii) Calculate the benefit annual premium using the equivalence principle. (iii) Calculate the probability of a loss is positive for the benefit annual premium that uses the equivalence principle. 37/77
38 Solution: (i) If 0.25 = P{T 50 t} = t 50 then t = Hence, P{K 50 < 13} = P{K 50 12} < 0.25 < P{K 50 13} = P{K 50 < 14}. We take k 0.25 = 13 and P 0.25 = 1 s = The benefit annual premium is (100000)( ) = /77
39 Solution: (ii) A 50 = a = , P 50 = da 50 = (0.06/1.06)( ) = A The benefit annual premium is (100000)( ) = /77
40 Solution: (iii) The probability that the loss random variable is positive is { ( P v K P ) 50 P } { 50 d d > 0 = P v K 50 > P } 50 d + P 50 { } =P v K > (0.06/1.06) { } { =P v K 50 > = P K 50 < ln( ) } ln(1.06) =P{K 50 < } = P{K 50 19} = = /77
41 Next, we discuss the percentile annual premium for an aggregate of policies. Suppose that an insurer offers a whole life insurance of B to n lives aged x paid at the end of the year of death. Each of these live insurances is funded by a level annual benefit premium of P paid at the beginning of the year. The loss for a policy is Hence, BL x = BZ x PŸx = Bv Kx Pä Kx = BZ x P 1 Z x ( d =Z x B + P ) P d d. E[BL x ] = BA x Pä x and Var(BL x ) = ( B + P d ) 2 Var(Z x ). Let BL x,1,..., BL x,n be the losses for these n insurees. The aggregate loss is L Agg = n j=1 BL x,1. Hence, ( E[L Agg ] = n(ba x Pä x ) and Var(L Agg ) = n B + P ) 2 Var(Z x ). d 41/77
42 Suppose that P is chosen so that the aggregate loss is positive with probability α, where α is small. Then, zero is a (1 α) th quantile of L Agg and 0 = E[L Agg ] + z 1 α Var(L Agg ) =n(ba x Pä x ) + z 1 α nvar(zx ) ( B + P ). d From this linear equation in P, we can find the aggregate percentile annual premium P. 42/77
43 Example 5 An insurance company company offers a whole life insurance to lives aged 20 paying at the end of the year of death. Each insuree will make an annual premium of P at the beginning of year while he is alive to fund this insurance. Suppose that 1000 policyholders enter this insurance product. The insurance company sets a fund with the proceeds earning a rate of 6% to pay future benefit claims. Use i = 6% and the life table for the USA population in 2004 to calculate P so that the probability that the aggregate loss is positive is less than or equal to /77
44 We have that A x = and 2 A x = Hence, ä x = /1.06 = , Var(Z x ) = ( ) 2 = , Each loss is (75000)L x = (75000)Z x PŸx = (75000)Z x P 1 Z x ( d =Z x P ) 1.06P We have that E[(75000)L x ] = (75000)A x Pä x = P, ( Var((75000)L x ) = P ) 2 ( ) /77
45 For the aggregate model, 1000 E (75000)L x = P, j= ( Var (75000)L x = (1000) P ) 2 ( ) 0.06 j=1 =( ) We have that ( P ) = P + ( ) ( P ) 0.06 = P Hence, P = = /77
46 Example 6 An insurer offers a fully discrete whole life insurances of on independent lives age 30, you are given: (i) i = 0.06 (ii) Mortality follows the life table for USA population in (iii) The annual contract premium for each policy is 1.25P x. Using the normal approximation, calculate the minimum number of policies the insurer must issue so that the probability that the aggregate loss for the issued policies is less than /77
47 Solution: Let π be the annual contract premium. We have that π = (1.25)(10000) A x = (1.25)(10000) ä x = , E[(10000)L x ] = (10000)A x πä x =(10000)( ) ( )(16.213) = , E[L Aggreg ] = n, ( Var((10000)L x ) = π ) 2 ( 2 A x A 2 d x) ( = ) 2 ( ( ) 2 ) = , 6/106 Var(L Aggreg ) = n( ) We have that 0 = n + (1.6449) n( ). So, n = (1.6449)2 ( ) ( ) 2 = Since n is a positive integer, n = /77
48 Suppose that the funding scheme is limited to the first t years. The present value of the loss is L = v Kx Pä min(kx,t) = Z x PŸ x:t. The present value of the loss is A x Pä x:t. Definition 3 The benefit premium for a fully discrete whole life insurance funded for the first t years that satisfies the equivalence principle is denoted by tp x = A x ä x:t. 48/77
49 Example 7 Ethan is 30 years old and purchases a whole life insurance policy with face value of payable at the end of the year of death. This policy will be paid by a level benefit annual premium at the beginning of the next 30 years while Ethan is alive. Assume that δ = 0.05 and death is modeled using the constant force of mortality µ = Find the benefit annual premium for this policy. Solution: We have that A 30 = q x q x + i = 1 e e e = , ä 30:30 = 1 e (30)(0.08) 1 e 0.08 = ( ) The benefit annual premium for this policy is 30P 30 = (50000)( ) = /77
50 n year term insurance. An n year term insurance paid at the end of the year of death funded at the beginning of the year while the insuree is alive is a called a fully discrete n year term insurance. Definition 4 The loss random variable for a fully discrete n year term insurance is denoted by L 1 x:n. Theorem 5 L 1 x:n = v Kx I (K x n) Pä min(kx,n) = Z x:n 1 PŸ x:n =Zx:n 1 P 1 Z x:n, d E[L 1 x:n ] = A1 x:n Pä x:n. 50/77
51 We got that L 1 x:n = Z 1 x:n PŸx:n. Definition 5 The benefit premium for a fully discrete n year term insurance obtained using the equivalence principle is denoted by Px:n 1 and by P(A 1 x:n ). We have that P 1 x:n = P(A1 x:n ) = A1 x:n ä x:n. 51/77
52 Theorem 6 We have that Var(L 1 x:n ) = ( 1 + P d ) 2 2A 1 x:n + P2 d 2 2A 1 x:n (E[L 1 x:n ] + P d ) 2. 52/77
53 Proof: Using that Z x:n = Zx:n 1 + Z x:n 1, we get that L 1 x:n = Z x:n 1 P 1 Z x:n = P ( d d P ) Zx:n 1 d + P d Z x:n 1. Since Zx:n 1 Z x:n 1 = v Kx I (K x n)v n I (n < K x ) = 0, [ ( E E[L 1 x:n ] + P ) ] [ 2 (( = E 1 + P d d ( = 1 + P ) 2 2A 1 x:n d + P2 ( Var(L 1 x:n ) = Var x:n, d 2 2A 1 ) ) Z 1 x:n + P d Z 1 x:n L 1 x:n ] + P d [ ( =E E[L 1 x:n ] + P ) ] 2 ( E[L 1 x:n d ] + P ) 2 d ( = 1 + P ) 2 2A 1 x:n d + P2 d 2 2A x:n (E[L 1 1 x:n ] + P ) 2. d ) 2 ] 53/77
54 Example 8 An insurer offers a four year life insurance of 25 years old. Mortality is given by the table: x q x This life insurance is funded by benefit payments at the beginning of the year. The benefit payment is i = 5%. Calculate the annual benefit premium using the equivalence principle. Solution: We have to find (10000)P 1 25:4 = (10000) A1 25:4 ä 25:4. 54/77
55 Solution: We have that A 1 25:4 = (0.01)(1.05) 1 + (0.99)(0.02)(1.05) 2 + (0.99)(0.98)(0.03)(1.05) 3 + (0.99)(0.98)(0.97)(0.04)(1.05) 4 = , A 1 25:4 = (0.99)(0.98)(0.97)(0.96)(1.05) 4 = , A 25:4 = = , ä 25:4 = /1.05 = , (10000)P 1 25:4 = (10000) = /77
56 The present value of the loss for a t year funded n year term insurance is v Kx I (K x n) Pä min(kx,t) = Z 1 x:n PŸ x:t. The actuarial present value of the loss for a t year funded n year term insurance is A 1 x:n Pä x:t. The benefit premium which satisfies the equivalence principle is tp 1 x:n = P( ta 1 x:n ) = A1 x:n ä x:t. 56/77
57 Example 9 A 20 year term life insurance policy to (x) with face value of payable at the end of the year of death is funded by a level benefit annual premium at the beginning of the next 10 years while (x) is alive. Assume that δ = 0.05 and the constant force of mortality is µ = Find the benefit annual premium for this policy. Solution: We have that q x A 1 x:20 = (1 e (20)(δ+µ) ) q x + i =(1 e (20)( ) 1 e 0.02 ) 1 e e = , ä x:10 = (1 e (10)( ) 1 ) = , 1 e ( ) A 1 P = (10000) 10 P 1 x:20 = (10000) x:20 = (10000) ä x: = /77
58 n year pure endowment. The loss for a fully discrete n year pure endowment is Lx:n 1 = v n I (K x > n) Pä min(kx,n) = Z x:n 1 PŸ x:n = Zx:n 1 P 1 Z x:n. d The actuarial present value of the loss for a fully discrete n year term insurance is E[L 1 x:n ] = A 1 x:n Pä x:n. The benefit premium which satisfies the equivalence principle is Px:n 1 = P(A x:n 1 ) = A x:n 1. ä x:n 58/77
59 Theorem 7 We have that Var(L 1 x:n ) = P2 d 2 2A 1 x:n + ( 1 + P ) 2 2A [ ] x:n (E 1 d Lx:n 1 + P ) 2. d 59/77
60 Proof: We have that Chapter 6. Benefit premiums. L 1 x:n = Z 1 x:n PŸx:n = Z 1 x:n P 1 Z x:n d Hence, E [ ( Lx:n 1 + P ) ] 2 = E d = P2 d 2 2A ( 1 x:n P d [ (P d Z 1 x:n ) 2 2A 1 x:n ) 2 ] = P ( d +P d Z x:n P ) Zx:n 1 d + E [ ( 1 + P d ) ] 2 ( ) 2 Zx:n 1 and Var(L 1 x:n ) = Var ( L 1 = P2 d 2 2A 1 x:n + ( 1 + P d ) x:n + P d ) 2 2A [ ] x:n (E 1 Lx:n 1 + P ) 2. d 60/77
61 Example 10 Austin is 40 years old and purchases a 20 year pure endowment policy with face value of This policy will be paid by a level benefit annual premium at the beginning of the next 20 years while Austin is alive. Assume that v = 0.94 and death is modeled using the De Moivre model with terminal age 110. (i) Find the net single premium for this policy. (ii) Find the benefit annual premium for this policy using the equivalence principle. (iii) Calculate the standard deviation of the loss for this policy. 61/77
62 Solution: (i) We have that A 1 40:20 = v p 40 = (0.94) = The net single premium for this policy is (50000)( ) = /77
63 Solution: (ii) We have that A 40:20 = a 20 6/94 70 ä 40:20 = 1 A 40:20 d (0.94) = , 70 = = The benefit annual premium for this policy using the equivalence principle is P 1 20:40 = = /77
64 Solution: (iii) We have that 2 A 1 40:20 = a 20 6/94(2+6/94) = , 70 2 A = (0.94)40 = , 40:20 70 P 1 40:20 = = , d 0.06 Var(L 1 40:20 ) = ( )2 ( ) + ( ) 2 ( ) ( ) 2 = The standard deviation of the loss for this policy is (50000) = /77
65 If a n year pure endowment is funded only t years where 1 t n, then the present value of the loss for a n year pure endowment is v n I (K x > n) Pä min(kx,t) = Z 1 x:n PŸ x:t. The actuarial present value of the loss for a t year funded n year term insurance is A 1 x:n Pä x:t. The benefit premium which satisfies the equivalence principle is tpx:n 1 = P( tax:n 1 ) = A x:n 1. ä x:t 65/77
66 n year endowment. The loss for a fully discrete n year endowment is L x:n = v min(n,kx ) Pä min(kx,n) = Z x:n PŸx:n = Z x:n P 1 Z x:n ( d = 1 + P ) Z d x:n P d. The actuarial present value of the loss for a fully discrete n year term insurance is We have that Var(L x:n ) = E[L x:n ] = A x:n Pä x:n. ( 1 + P d ) 2 Var(Z x:n ) = ( 1 + P d ) 2 ( 2 A x:n (A x:n ) 2). The benefit premium which satisfies the equivalence principle is P x:n = P(A x:n ) = A x:n ä x:n. 66/77
67 Example 11 A 20 year endowment insurance policy to (x) with face value of payable at the end of the year of death is funded by a level benefit annual premium at the beginning of the next 20 years while (x) is alive. Assume that δ = 0.05 and death is modeled using the constant force of mortality µ = (i) Find the benefit annual premium for this policy. (ii) Calculate the variance for the loss at issue random variable. 67/77
68 Solution: (i) We have that A x:20 = (1 e (20)( ) 1 e 0.02 ) 1 e e e (20)( ) = , ä x:20 = (1 e (20)(δ+µ) 1 ) 1 vq x =(1 e (20)( ) 1 ) 1 e ( ) = , P = (10000) P x:20 = (10000) A x:20 = (10000) ä x: = /77
69 Solution: (ii) 2 A x:20 =(1 e (20)((2) ) 1 e 0.02 ) 1 e e (2) e (20)( ) = , ( Var(L x:20 ) = 1 + P ) 2 (2 A d (A x:20 x:20 )2) ( = ) 2 ( ( ) 2 ) 1 e 0.05 = /77
70 The present value of the loss for a t year funded n year endowment is v min(n,kx ) Pä min(kx,t) = Z x:n PŸ x:t. The actuarial present value of the loss for a t year funded n year term insurance is A x:n Pä x:t. The benefit premium which satisfies the equivalence principle is tp x:n = P( t A x:n ) = A x:n ä x:t. 70/77
71 Theorem 8 P x:n = P 1 x:n + P 1 x:n. Proof: We have that Px:n 1 + P x:n 1 = A1 x:n + A x:n 1 = A x:n = P ä x:n ä x:n ä x:n. x:n 71/77
72 Theorem 9 np x = P 1 x:n + P 1 x:n A x+n. Proof: We have that Px:n 1 + P x:n 1 A x+n = A1 x:n + A x:n 1 A x+n = A1 x:n + n A x ä x:n ä x:n ä x:n = A x ä x:n = n P x. 72/77
73 n year deferred insurance. The present value of the loss for a n year deferred insurance is n L x = v Kx I (K x > n) Pä Kx = n Z x PŸ x. The actuarial present value of the loss for a n year term insurance is n A x Pä x. The benefit premium which satisfies the equivalence principle is P( n A x ) = n A x ä x. 73/77
74 Theorem 10 We have that Var( n L x ) = P2 d 2 2A 1 x:n + ( 1 + P ) 2 ( 2 n A x E [ n L x ] + P ) 2. d d 74/77
75 Proof: Using that Z x = Z 1 x:n + n Z x, n L x = n Z x PŸ x = n Z x P 1 Z x d = P ( d +P d Z x:n P ) n Z x. d Since Z 1 x:n n Z x = v Kx I (n < K x )v Kx I (K x n) = 0, Hence, E [ ( n L x + P ) ] 2 = P2 d d 2 2A ( 1 x:n P ) 2 2 n A x d ( Var( n L x ) = Var = P2 d 2 2A ( 1 x:n P d n L x + P d ) ) 2 2 n A x ( E [ n L x ] + P d ) 2. 75/77
76 The present value of the loss for a t year funded n year deferred insurance is v Kx I (K x > n) Pä min(kx,t) = n Z x PŸx:t. The actuarial present value of the loss for a t year funded n year term insurance is n A x Pä x:t. The benefit premium which satisfies the equivalence principle is tp( n A x ) = n A x ä x:t. 76/77
77 Plan Whole life insurance t year funded whole life insurance n year term insurance t year funded n year term insurance n year pure endowment insurance P 1 t year funded n year pure endowment insurance tp 1 n year endowment t year funded n year endowment insurance n year deferred insurance t year funded n year deferred insurance Premium P x = Ax ä x tp x = Ax ä x:t P 1 x:n = A1 x:n ä x:n tp 1 x:n = A1 x:n ä x:t x:n = A x:n 1 ä x:n x:n = A x:n 1 ä x:n Table: Annual benefit premiums in the fully discrete case P x:n = A x:n ä x:n tp x:n = A x:n ä x:t P( n A x ) = n Ax ä x tp( n A x ) = n Ax ä x:t 77/77
Manual for SOA Exam MLC.
Chapter 6. Benefit premiums Extract from: Arcones Fall 2010 Edition, available at http://www.actexmadriver.com/ 1/90 (#4, Exam M, Spring 2005) For a fully discrete whole life insurance of 100,000 on (35)
More informationManual for SOA Exam MLC.
Chapter 6. Benefit premiums. Extract from: Arcones Fall 2010 Edition, available at http://www.actexmadriver.com/ 1/24 Nonlevel premiums and/or benefits. Let b k be the benefit paid by an insurance company
More informationManual for SOA Exam MLC.
Chapter 5. Life annuities. Extract from: Arcones Manual for the SOA Exam MLC. Spring 2010 Edition. available at http://www.actexmadriver.com/ 1/114 Whole life annuity A whole life annuity is a series of
More informationManual for SOA Exam MLC.
Chapter 4. Life insurance. Extract from: Arcones Fall 2009 Edition, available at http://www.actexmadriver.com/ (#1, Exam M, Fall 2005) For a special whole life insurance on (x), you are given: (i) Z is
More informationManual for SOA Exam MLC.
Chapter 4. Life Insurance. c 29. Miguel A. Arcones. All rights reserved. Extract from: Arcones Manual for the SOA Exam MLC. Fall 29 Edition. available at http://www.actexmadriver.com/ c 29. Miguel A. Arcones.
More informationPremium Calculation. Lecture: Weeks 1214. Lecture: Weeks 1214 (STT 455) Premium Calculation Fall 2014  Valdez 1 / 31
Premium Calculation Lecture: Weeks 1214 Lecture: Weeks 1214 (STT 455) Premium Calculation Fall 2014  Valdez 1 / 31 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is
More informationManual for SOA Exam MLC.
Chapter 4. Life Insurance. Extract from: Arcones Manual for the SOA Exam MLC. Fall 2009 Edition. available at http://www.actexmadriver.com/ 1/44 Properties of the APV for continuous insurance The following
More informationManual for SOA Exam MLC.
Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at http://wwwactexmadrivercom/ 1/70 Due n year deferred annuity Definition 1 A due n year deferred
More informationPremium Calculation. Lecture: Weeks 1214. Lecture: Weeks 1214 (Math 3630) Annuities Fall 2015  Valdez 1 / 32
Premium Calculation Lecture: Weeks 1214 Lecture: Weeks 1214 (Math 3630) Annuities Fall 2015  Valdez 1 / 32 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is funded
More informationManual for SOA Exam MLC.
Chapter 4. Life Insurance. Extract from: Arcones Manual for the SOA Exam MLC. Fall 2009 Edition. available at http://www.actexmadriver.com/ 1/14 Level benefit insurance in the continuous case In this chapter,
More informationManual for SOA Exam MLC.
Chapter 5. Life annuities Extract from: Arcones Fall 2009 Edition, available at http://www.actexmadriver.com/ 1/60 (#24, Exam M, Fall 2005) For a special increasing whole life annuitydue on (40), you
More informationManual for SOA Exam MLC.
Chapter 4 Life Insurance Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at http://wwwactexmadrivercom/ 1/13 Nonlevel payments paid at the end of the year Suppose that a
More informationManual for SOA Exam MLC.
Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at http://wwwactexmadrivercom/ 1/94 Due n year temporary annuity Definition 1 A due n year term annuity
More informationMath 630 Problem Set 2
Math 63 Problem Set 2 1. AN nyear term insurance payable at the moment of death has an actuarial present value (i.e. EPV) of.572. Given µ x+t =.7 and δ =.5, find n. (Answer: 11) 2. Given: Ā 1 x:n =.4275,
More informationPremium Calculation  continued
Premium Calculation  continued Lecture: Weeks 12 Lecture: Weeks 12 (STT 456) Premium Calculation Spring 2015  Valdez 1 / 16 Recall some preliminaries Recall some preliminaries An insurance policy (life
More informationInsurance Benefits. Lecture: Weeks 68. Lecture: Weeks 68 (STT 455) Insurance Benefits Fall 2014  Valdez 1 / 36
Insurance Benefits Lecture: Weeks 68 Lecture: Weeks 68 (STT 455) Insurance Benefits Fall 2014  Valdez 1 / 36 An introduction An introduction Central theme: to quantify the value today of a (random)
More informationPlease write your name and student number at the spaces provided:
MATH 3630 Actuarial Mathematics I Final Examination  sec 001 Monday, 10 December 2012 Time Allowed: 2 hours (6:008:00 pm) Room: MSB 411 Total Marks: 120 points Please write your name and student number
More information1. Revision 2. Revision pv 3.  note that there are other equivalent formulae! 1 pv 16.5 4. A x A 1 x:n A 1
Tutorial 1 1. Revision 2. Revision pv 3.  note that there are other equivalent formulae! 1 pv 16.5 4. A x A 1 x:n A 1 x:n a x a x:n n a x 5. K x = int[t x ]  or, as an approximation: T x K x + 1 2 6.
More informationSolution. Let us write s for the policy year. Then the mortality rate during year s is q 30+s 1. q 30+s 1
Solutions to the May 213 Course MLC Examination by Krzysztof Ostaszewski, http://wwwkrzysionet, krzysio@krzysionet Copyright 213 by Krzysztof Ostaszewski All rights reserved No reproduction in any form
More informationTABLE OF CONTENTS. 4. Daniel Markov 1 173
TABLE OF CONTENTS 1. Survival A. Time of Death for a Person Aged x 1 B. Force of Mortality 7 C. Life Tables and the Deterministic Survivorship Group 19 D. Life Table Characteristics: Expectation of Life
More informationManual for SOA Exam MLC.
Chapter 5. Life annuities. Section 5.7. Computing present values from a life table Extract from: Arcones Manual for the SOA Exam MLC. Spring 2010 Edition. available at http://www.actexmadriver.com/ 1/30
More information4. Life Insurance. 4.1 Survival Distribution And Life Tables. Introduction. X, Ageatdeath. T (x), timeuntildeath
4. Life Insurance 4.1 Survival Distribution And Life Tables Introduction X, Ageatdeath T (x), timeuntildeath Life Table Engineers use life tables to study the reliability of complex mechanical and
More informationSome Observations on Variance and Risk
Some Observations on Variance and Risk 1 Introduction By K.K.Dharni Pradip Kumar 1.1 In most actuarial contexts some or all of the cash flows in a contract are uncertain and depend on the death or survival
More informationAnnuities. Lecture: Weeks 911. Lecture: Weeks 911 (STT 455) Annuities Fall 2014  Valdez 1 / 43
Annuities Lecture: Weeks 911 Lecture: Weeks 911 (STT 455) Annuities Fall 2014  Valdez 1 / 43 What are annuities? What are annuities? An annuity is a series of payments that could vary according to:
More informationJANUARY 2016 EXAMINATIONS. Life Insurance I
PAPER CODE NO. MATH 273 EXAMINER: Dr. C. BoadoPenas TEL.NO. 44026 DEPARTMENT: Mathematical Sciences JANUARY 2016 EXAMINATIONS Life Insurance I Time allowed: Two and a half hours INSTRUCTIONS TO CANDIDATES:
More informationPremium calculation. summer semester 2013/2014. Technical University of Ostrava Faculty of Economics department of Finance
Technical University of Ostrava Faculty of Economics department of Finance summer semester 2013/2014 Content 1 Fundamentals Insurer s expenses 2 Equivalence principles Calculation principles 3 Equivalence
More informationSOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS The following questions or solutions have been modified since this document was prepared to use with the syllabus effective
More informationMATH 3630 Actuarial Mathematics I Class Test 2 Wednesday, 17 November 2010 Time Allowed: 1 hour Total Marks: 100 points
MATH 3630 Actuarial Mathematics I Class Test 2 Wednesday, 17 November 2010 Time Allowed: 1 hour Total Marks: 100 points Please write your name and student number at the spaces provided: Name: Student ID:
More informationNovember 2012 Course MLC Examination, Problem No. 1 For two lives, (80) and (90), with independent future lifetimes, you are given: k p 80+k
Solutions to the November 202 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net Copyright 202 by Krzysztof Ostaszewski All rights reserved. No reproduction in
More information2 Policy Values and Reserves
2 Policy Values and Reserves [Handout to accompany this section: reprint of Life Insurance, from The Encyclopaedia of Actuarial Science, John Wiley, Chichester, 2004.] 2.1 The Life Office s Balance Sheet
More informationMathematics of Life Contingencies MATH 3281
Mathematics of Life Contingencies MATH 3281 Life annuities contracts Edward Furman Department of Mathematics and Statistics York University February 13, 2012 Edward Furman Mathematics of Life Contingencies
More informationContents. 3 Survival Distributions: Force of Mortality 37 Exercises Solutions... 51
Contents 1 Probability Review 1 1.1 Functions and moments...................................... 1 1.2 Probability distributions...................................... 2 1.2.1 Bernoulli distribution...................................
More informationManual for SOA Exam MLC.
Manual for SOA Eam MLC. Chapter 4. Life Insurance. Etract from: Arcones Manual for the SOA Eam MLC. Fall 2009 Edition. available at http://www.actemadriver.com/ Manual for SOA Eam MLC. 1/9 Payments at
More informationMay 2012 Course MLC Examination, Problem No. 1 For a 2year select and ultimate mortality model, you are given:
Solutions to the May 2012 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net Copyright 2012 by Krzysztof Ostaszewski All rights reserved. No reproduction in any
More informationChapter 2. 1. You are given: 1 t. Calculate: f. Pr[ T0
Chapter 2 1. You are given: 1 5 t F0 ( t) 1 1,0 t 125 125 Calculate: a. S () t 0 b. Pr[ T0 t] c. Pr[ T0 t] d. S () t e. Probability that a newborn will live to age 25. f. Probability that a person age
More informationSOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE WRITTENANSWER QUESTIONS AND SOLUTIONS
SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE WRITTENANSWER QUESTIONS AND SOLUTIONS Questions February 12, 2015 In Questions 12, 13, and 19, the wording was changed slightly
More informationHeriotWatt University. BSc in Actuarial Mathematics and Statistics. Life Insurance Mathematics I. Extra Problems: Multiple Choice
HeriotWatt University BSc in Actuarial Mathematics and Statistics Life Insurance Mathematics I Extra Problems: Multiple Choice These problems have been taken from Faculty and Institute of Actuaries exams.
More informationFurther Topics in Actuarial Mathematics: Premium Reserves. Matthew Mikola
Further Topics in Actuarial Mathematics: Premium Reserves Matthew Mikola April 26, 2007 Contents 1 Introduction 1 1.1 Expected Loss...................................... 2 1.2 An Overview of the Project...............................
More informationACTS 4301 FORMULA SUMMARY Lesson 1: Probability Review. Name f(x) F (x) E[X] Var(X) Name f(x) E[X] Var(X) p x (1 p) m x mp mp(1 p)
ACTS 431 FORMULA SUMMARY Lesson 1: Probability Review 1. VarX)= E[X 2 ] E[X] 2 2. V arax + by ) = a 2 V arx) + 2abCovX, Y ) + b 2 V ary ) 3. V ar X) = V arx) n 4. E X [X] = E Y [E X [X Y ]] Double expectation
More informationO MIA009 (F2F) : GENERAL INSURANCE, LIFE AND
No. of Printed Pages : 11 MIA009 (F2F) kr) ki) M.Sc. ACTUARIAL SCIENCE (MSCAS) N December, 2012 0 O MIA009 (F2F) : GENERAL INSURANCE, LIFE AND HEALTH CONTINGENCIES Time : 3 hours Maximum Marks : 100
More informationSolution. Because you are not doing business in the state of New York, you only need to do the calculations at the end of each policy year.
Exercises in life and annuity valuation. You are the valuation actuary for the Hard Knocks Life Insurance Company offering life annuities and life insurance to guinea pigs. The valuation interest rate
More informationACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS
ACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS DAVID C. M. DICKSON University of Melbourne MARY R. HARDY University of Waterloo, Ontario V HOWARD R. WATERS HeriotWatt University, Edinburgh CAMBRIDGE
More information1. Datsenka Dog Insurance Company has developed the following mortality table for dogs:
1 Datsenka Dog Insurance Company has developed the following mortality table for dogs: Age l Age l 0 2000 5 1200 1 1950 6 1000 2 1850 7 700 3 1600 8 300 4 1400 9 0 Datsenka sells an whole life annuity
More informationActuarial mathematics 2
Actuarial mathematics 2 Life insurance contracts Edward Furman Department of Mathematics and Statistics York University January 3, 212 Edward Furman Actuarial mathematics MATH 328 1 / 45 Definition.1 (Life
More informationSOA EXAM MLC & CAS EXAM 3L STUDY SUPPLEMENT
SOA EXAM MLC & CAS EXAM 3L STUDY SUPPLEMENT by Paul H. Johnson, Jr., PhD. Last Modified: October 2012 A document prepared by the author as study materials for the Midwestern Actuarial Forum s Exam Preparation
More information**BEGINNING OF EXAMINATION**
November 00 Course 3 Society of Actuaries **BEGINNING OF EXAMINATION**. You are given: R = S T µ x 0. 04, 0 < x < 40 0. 05, x > 40 Calculate e o 5: 5. (A) 4.0 (B) 4.4 (C) 4.8 (D) 5. (E) 5.6 Course 3: November
More information1 Cashflows, discounting, interest rate models
Assignment 1 BS4a Actuarial Science Oxford MT 2014 1 1 Cashflows, discounting, interest rate models Please hand in your answers to questions 3, 4, 5 and 8 for marking. The rest are for further practice.
More informationREGS 2013: Variable Annuity Guaranteed Minimum Benefits
Department of Mathematics University of Illinois, UrbanaChampaign REGS 2013: Variable Annuity Guaranteed Minimum Benefits By: Vanessa Rivera Quiñones Professor Ruhuan Feng September 30, 2013 The author
More informationInstitute of Actuaries of India
Institute of Actuaries of India Subject CT5 General Insurance, Life and Health Contingencies May 2008 Examination INDICATIVE SOLUTION Introduction The indicative solution has been written by the Examiners
More informationPractice Exam 1. x l x d x 50 1000 20 51 52 35 53 37
Practice Eam. You are given: (i) The following life table. (ii) 2q 52.758. l d 5 2 5 52 35 53 37 Determine d 5. (A) 2 (B) 2 (C) 22 (D) 24 (E) 26 2. For a Continuing Care Retirement Community, you are given
More informationEXAMINATION. 6 April 2005 (pm) Subject CT5 Contingencies Core Technical. Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE
Faculty of Actuaries Institute of Actuaries EXAMINATION 6 April 2005 (pm) Subject CT5 Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and
More informationActuarial Science with
Actuarial Science with 1. life insurance & actuarial notations Arthur Charpentier joint work with Christophe Dutang & Vincent Goulet and Giorgio Alfredo Spedicato s lifecontingencies package Meielisalp
More informationManual for SOA Exam FM/CAS Exam 2.
Manual for SOA Exam FM/CAS Exam 2. Chapter 6. Variable interest rates and portfolio insurance. c 2009. Miguel A. Arcones. All rights reserved. Extract from: Arcones Manual for the SOA Exam FM/CAS Exam
More informationSOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS Copyright 5 by the Society of Actuaries Some of the questions in this study note are taken from past SOA examinations. M95 PRINTED
More informationb g is the future lifetime random variable.
**BEGINNING OF EXAMINATION** 1. Given: (i) e o 0 = 5 (ii) l = ω, 0 ω (iii) is the future lifetime random variable. T Calculate Var Tb10g. (A) 65 (B) 93 (C) 133 (D) 178 (E) 333 COURSE/EXAM 3: MAY 0001
More informationSafety margins for unsystematic biometric risk in life and health insurance
Safety margins for unsystematic biometric risk in life and health insurance Marcus C. Christiansen June 1, 2012 7th Conference in Actuarial Science & Finance on Samos Seite 2 Safety margins for unsystematic
More informationChapter 2 Premiums and Reserves in Multiple Decrement Model
Chapter 2 Premiums and Reserves in Multiple Decrement Model 2.1 Introduction A guiding principle in the determination of premiums for a variety of life insurance products is: Expected present value of
More informationINSTRUCTIONS TO CANDIDATES
Society of Actuaries Canadian Institute of Actuaries Exam MLC Models for Life Contingencies Friday, October 31, 2014 8:30 a.m. 12:45 p.m. MLC General Instructions 1. Write your candidate number here. Your
More informationOn Simulation Method of Small Life Insurance Portfolios By Shamita Dutta Gupta Department of Mathematics Pace University New York, NY 10038
On Simulation Method of Small Life Insurance Portfolios By Shamita Dutta Gupta Department of Mathematics Pace University New York, NY 10038 Abstract A new simulation method is developed for actuarial applications
More informationSOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS The following questions or solutions have been modified since this document was prepared to use with the syllabus effective
More informationSOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS Copyright 005 by the Society of Actuaries Some of the questions in this study note are taken from past SOA examinations. M0905 PRINTED
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More informationMathematics of Life Contingencies. Math 3281 3.00 W Instructor: Edward Furman Homework 1
Mathematics of Life Contingencies. Math 3281 3.00 W Instructor: Edward Furman Homework 1 Unless otherwise indicated, all lives in the following questions are subject to the same law of mortality and their
More informationLecture Notes on Actuarial Mathematics
Lecture Notes on Actuarial Mathematics Jerry Alan Veeh May 1, 2006 Copyright 2006 Jerry Alan Veeh. All rights reserved. 0. Introduction The objective of these notes is to present the basic aspects of the
More informationHomework #2 Solutions
MAT Spring Problems Section.:, 8,, 4, 8 Section.5:,,, 4,, 6 Extra Problem # Homework # Solutions... Sketch likely solution curves through the given slope field for dy dx = x + y...8. Sketch likely solution
More informationIntroduction to polynomials
Worksheet 4.5 Polynomials Section 1 Introduction to polynomials A polynomial is an expression of the form p(x) = p 0 + p 1 x + p 2 x 2 + + p n x n, (n N) where p 0, p 1,..., p n are constants and x os
More informationINSTITUTE OF ACTUARIES OF INDIA
INSTITUTE OF ACTUARIES OF INDIA EXAMINATIONS 17 th November 2011 Subject CT5 General Insurance, Life and Health Contingencies Time allowed: Three Hours (10.00 13.00 Hrs) Total Marks: 100 INSTRUCTIONS TO
More informationINSTITUTE AND FACULTY OF ACTUARIES EXAMINATION
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 8 October 2015 (pm) Subject CT5 Contingencies Core Technical
More information1 Interest rates, and riskfree investments
Interest rates, and riskfree investments Copyright c 2005 by Karl Sigman. Interest and compounded interest Suppose that you place x 0 ($) in an account that offers a fixed (never to change over time)
More informationActuarial Mathematics and LifeTable Statistics. Eric V. Slud Mathematics Department University of Maryland, College Park
Actuarial Mathematics and LifeTable Statistics Eric V. Slud Mathematics Department University of Maryland, College Park c 2006 Chapter 4 Expected Present Values of Insurance Contracts We are now ready
More informationIntroduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 4.4 Homework
Introduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 4.4 Homework 4.65 You buy a hot stock for $1000. The stock either gains 30% or loses 25% each day, each with probability.
More informationUnobserved heterogeneity; process and parameter effects in life insurance
Unobserved heterogeneity; process and parameter effects in life insurance Jaap Spreeuw & Henk Wolthuis University of Amsterdam ABSTRACT In this paper life insurance contracts based on an urnofurns model,
More informationEXAM 3, FALL 003 Please note: On a onetime basis, the CAS is releasing annotated solutions to Fall 003 Examination 3 as a study aid to candidates. It is anticipated that for future sittings, only the
More informationValuation of the Minimum Guaranteed Return Embedded in Life Insurance Products
Financial Institutions Center Valuation of the Minimum Guaranteed Return Embedded in Life Insurance Products by Knut K. Aase SveinArne Persson 9620 THE WHARTON FINANCIAL INSTITUTIONS CENTER The Wharton
More information4. Continuous Random Variables, the Pareto and Normal Distributions
4. Continuous Random Variables, the Pareto and Normal Distributions A continuous random variable X can take any value in a given range (e.g. height, weight, age). The distribution of a continuous random
More information3. The Economics of Insurance
3. The Economics of Insurance Insurance is designed to protect against serious financial reversals that result from random evens intruding on the plan of individuals. Limitations on Insurance Protection
More informationPolynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if
1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c
More information6 Insurances on Joint Lives
6 Insurances on Joint Lives 6.1 Introduction Itiscommonforlifeinsurancepoliciesandannuitiestodependonthedeathorsurvivalof more than one life. For example: (i) Apolicywhichpaysamonthlybenefittoawifeorotherdependentsafterthedeathof
More informationMath 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 2 Solutions
Math 70/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 2 Solutions About this problem set: These are problems from Course /P actuarial exams that I have collected over the years,
More informationAssurancevie 2. Manuel Morales Department of Mathematics and Statistics University of Montreal. February 2006
Assurancevie 2 Manuel Morales Department of Mathematics and Statistics University of Montreal February 2006 Chapter 9 Multiple Life Functions 1. Joint Distributions 2. Jointlife status 3. Lastsurvivor
More informationMINIMUM SURRENDER VALUES AND PAIDUP VALUES
MARCH 2002 Actuarial Standard 4.02 MINIMUM SURRENDER VALUES AND PAIDUP VALUES Life Insurance Actuarial Standards Board TABLE OF CONTENTS INTRODUCTION PAGE The Standard 2 Application of the Surrender Value
More informationUniversidad Carlos III de Madrid. Licenciatura en Ciencias Actuariales y Financieras Survival Models and Basic Life Contingencies
Universidad Carlos III de Madrid Licenciatura en Ciencias Actuariales y Financieras Survival Models and Basic Life Contingencies PART II. Lecture 4: Net Premium Calculations This is a summary of formulae
More informationHeriotWatt University. M.Sc. in Actuarial Science. Life Insurance Mathematics I. Tutorial 5
1 HeriotWatt University M.Sc. in Actuarial Science Life Insurance Mathematics I Tutorial 5 1. Consider the illnessdeath model in Figure 1. A life age takes out a policy with a term of n years that pays
More informationLecture Notes on the Mathematics of Finance
Lecture Notes on the Mathematics of Finance Jerry Alan Veeh February 20, 2006 Copyright 2006 Jerry Alan Veeh. All rights reserved. 0. Introduction The objective of these notes is to present the basic aspects
More informationST 371 (IV): Discrete Random Variables
ST 371 (IV): Discrete Random Variables 1 Random Variables A random variable (rv) is a function that is defined on the sample space of the experiment and that assigns a numerical variable to each possible
More information**BEGINNING OF EXAMINATION**
Fall 2002 Society of Actuaries **BEGINNING OF EXAMINATION** 1. Given: The survival function s x sbxg = 1, 0 x < 1 x d i { }, where s x = 1 e / 100, 1 x < 45. = s x 0, 4.5 x Calculate µ b4g. (A) 0.45 (B)
More informationLIFE INSURANCE AND PENSIONS by Peter Tryfos York University
LIFE INSURANCE AND PENSIONS by Peter Tryfos York University Introduction Life insurance is the business of insuring human life: in return for a premium payable in one sum or installments, an insurance
More informationWorksheet 4.7. Polynomials. Section 1. Introduction to Polynomials. A polynomial is an expression of the form
Worksheet 4.7 Polynomials Section 1 Introduction to Polynomials A polynomial is an expression of the form p(x) = p 0 + p 1 x + p 2 x 2 + + p n x n (n N) where p 0, p 1,..., p n are constants and x is a
More information4. Life Insurance Payments
4. Life Insurance Payments A life insurance premium must take into account the following factors 1. The amount to be paid upon the death of the insured person and its present value. 2. The distribution
More informationAnnuities, Insurance and Life
Annuities, Insurance and Life by Pat Goeters Department of Mathematics Auburn University Auburn, AL 368495310 1 2 These notes cover a standard oneterm course in Actuarial Mathematics with a primer from
More informationModeling Insurance Markets
Modeling Insurance Markets Nathaniel Hendren Harvard April, 2015 Nathaniel Hendren (Harvard) Insurance April, 2015 1 / 29 Modeling Competition Insurance Markets is Tough There is no wellagreed upon model
More informationRandom variables P(X = 3) = P(X = 3) = 1 8, P(X = 1) = P(X = 1) = 3 8.
Random variables Remark on Notations 1. When X is a number chosen uniformly from a data set, What I call P(X = k) is called Freq[k, X] in the courseware. 2. When X is a random variable, what I call F ()
More informationMath 151. Rumbos Spring 2014 1. Solutions to Assignment #22
Math 151. Rumbos Spring 2014 1 Solutions to Assignment #22 1. An experiment consists of rolling a die 81 times and computing the average of the numbers on the top face of the die. Estimate the probability
More informationProbability and Statistical Methods. Chapter 4 Mathematical Expectation
Math 3 Chapter 4 Mathematical Epectation Mean of a Random Variable Definition. Let be a random variable with probability distribution f( ). The mean or epected value of is, f( ) µ = µ = E =, if is a discrete
More information3. Continuous Random Variables
3. Continuous Random Variables A continuous random variable is one which can take any value in an interval (or union of intervals) The values that can be taken by such a variable cannot be listed. Such
More informationCHAPTER 4. Test Bank Exercises in. Exercise Set 4.1
Test Bank Exercises in CHAPTER 4 Exercise Set 4.1 1. Graph the quadratic function f(x) = x 2 2x 3. Indicate the vertex, axis of symmetry, minimum 2. Graph the quadratic function f(x) = x 2 2x. Indicate
More informationTRANSACTIONS JUNE, 1969 AN UPPER BOUND ON THE STOPLOSS NET PREMIUMACTUARIAL NOTE NEWTON L. BOWERS, JR. ABSTRACT
TRANSACTIONS OF SOCIETY OF ACTUARIES 1969 VOL. 21 PT. 1 NO. 60 VOL. XXI, PART I M~TINO No. 60 TRANSACTIONS JUNE, 1969 AN UPPER BOUND ON THE STOPLOSS NET PREMIUMACTUARIAL NOTE NEWTON L. BOWERS, JR. ABSTRACT
More informationLiving to 100: Survival to Advanced Ages: Insurance Industry Implication on Retirement Planning and the Secondary Market in Insurance
Living to 100: Survival to Advanced Ages: Insurance Industry Implication on Retirement Planning and the Secondary Market in Insurance Jay Vadiveloo, * Peng Zhou, Charles Vinsonhaler, and Sudath Ranasinghe
More informationMath 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 5 Solutions
Math 370/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 5 Solutions About this problem set: These are problems from Course 1/P actuarial exams that I have collected over the
More informationProfit Measures in Life Insurance
Profit Measures in Life Insurance Shelly Matushevski Honors Project Spring 2011 The University of Akron 2 Table of Contents I. Introduction... 3 II. Loss Function... 5 III. Equivalence Principle... 7 IV.
More information