Manual for SOA Exam MLC.

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1 Manual for SOA Eam MLC. Chapter 4. Life Insurance. Etract from: Arcones Manual for the SOA Eam MLC. Fall 2009 Edition. available at Manual for SOA Eam MLC. 1/9

2 Payments at the end of the m thly time interval It is unusual that claims are paid at the end of the year. A better model uses that claims are paid at the end of each month, or other period. In section, we consider the case when payments can be made at m different equally spaced times a year. Previous insurance quantities are defined as before. To indicate that payments are made m thly, a superinde is added to the actuarial notation of insurance variables. Manual for SOA Eam MLC. 2/9

In section, we consider the case when payments can be made at m different equally spaced times a year.

3 Suppose that a whole life insurance is paid at the end of the m thly ( time interval ] in which failure occurs. Let J = j if T j 1 m, j m, for some integer j 1. The present value of a whole life insurance paid the end of the m thly time interval in which failure occurs is Z = ν J /m Hence, A 2 A = E[Z ] = = E[(Z ) 2 ] = ν j/m P{J Var(Z ] = 2 A (A ) 2. ν 2j/m P{J Manual for SOA Eam MLC. 3/9

The present value of a whole life insurance paid the end of the m thly time interval in which

4 Suppose that an n year term life insurance guarantees a payment at the end of the m thly time interval in which failure occurs. The present value of this life insurance is Z 1 :1 = v J /m I (J nm) = In this case, we have that A1 nm :1 = E[Z 1 :1 ] = 2 A :n = E[(Z 1 :1 )2 ] = {v J /m if J nm, 0 if J ν j/m P{J nm Var(Z 1 :1 ) = 2 A1 :1 (A 1 ν 2j/m P{J :1 )2. > nm. Manual for SOA Eam MLC. 4/9

The present value of this life insurance is Z 1 :1 = v J /m I (J nm) = In this case, we have that

5 Suppose that an n year deferred life insurance guarantees a payment at the end of the m thly time interval in which failure occurs (if this death happens after n years). The present value of this life insurance is n Z = ν J /m I (J > nm) = {ν J /m if J > nm, 0 if J nm, In this case, we have that n A 2 n A = E[Z :n ] = j=nm+1 = E[(Z :n )2 ] = ν j/m P{J j=nm+1 Var( n Z ) = 2 n A ( n A ) 2. ν 2j/m P{J Manual for SOA Eam MLC. 5/9

The present value of this life insurance is n Z = ν J /m I (J > nm) = {ν J /m if J > nm, 0 if J nm, In this

6 Eample 1 Rose is 40 years old. She buys a whole life policy insurance which will pay $200,000 at the end of the month of her death. Suppose that the de Moivre model holds with terminal age 120. Find the mean and the standard deviation of the present value of this life insurance under the annual effective rate of interest of i = 10%. Manual for SOA Eam MLC. 6/9

her death. Suppose that the de Moivre model holds with terminal age 120.

7 Eample 1 Rose is 40 years old. She buys a whole life policy insurance which will pay $200,000 at the end of the month of her death. Suppose that the de Moivre model holds with terminal age 120. Find the mean and the standard deviation of the present value of this life insurance under the annual effective rate of interest of i = 10%. Solution: The actuarial present value of this life insurance is (200000)A (12) 40 = (200000) (12)(80) =(200000) (1.1) j/12 1 (12)(80) ν j/12 j 1 1 q = = (200000)a 960 (1.1) 1/ = Manual for SOA Eam MLC. 7/9

Find the mean and the standard deviation of the present value of this life insurance under the annual effective rate of interest of i = 10%.

8 Eample 1 Rose is 40 years old. She buys a whole life policy insurance which will pay $200,000 at the end of the month of her death. Suppose that the de Moivre model holds with terminal age 120. Find the mean and the standard deviation of the present value of this life insurance under the annual effective rate of interest of i = 10%. Solution: The second moment of the present value of this life insurance is (200000) 2 2A (12) 40 = (200000) 2 (12)(80) =(200000) 2 (1.1) j/6 1 (12)(80) ν 2j/12 j 1 1 q = = (200000) 2 1 a 960 (1.1) 1/ = Manual for SOA Eam MLC. 8/9

9 Eample 1 Rose is 40 years old. She buys a whole life policy insurance which will pay $200,000 at the end of the month of her death. Suppose that the de Moivre model holds with terminal age 120. Find the mean and the standard deviation of the present value of this life insurance under the annual effective rate of interest of i = 10%. Solution: The variance of the present value of this life insurance is Var(Z (12) 40 ) = ( )2 = The standard deviation of the present value of this life insurance is = Manual for SOA Eam MLC. 9/9

Manual for SOA Exam MLC.

Manual 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.