Chapter 8
• Basic Concepts
• Commutation Functions
• Annuities Payable m thly
• Varying Life Annuities
• Annual Premiums and Premium Reserves

• We know how to compute present value of contingent payments
• Life tables are sources of probabilities of surviving
• We can use data from life tables to compute present values of payments which are contingent on either survival or death

Example (pure endowment), p. 155
• Yuanlin is 38 years old. If he reaches age 65, he will receive a single payment of 50,000. If i = .12, find an expression for the value of this payment to Yuanlin today. Use the following entries in the life table: l
38
= 8327, l
65
= 5411

• Pure endowment: 1 is paid t years from now to an individual currently aged x if the individual survives
• Probability of surviving is t p x
• Therefore the present value of this payment is the net single premium for the pure endowment, which is: t
E x
= ( t p x
) ( 1 + t )
– t
= v t t p x

• Aretha is 27 years old. Beginning one year from today, she will receive 10,000 annually for as long as she is alive. Find an expression for the present value of this series of payments assuming i = .09
• Find numerical value of this expression if p x
= .95 for each x

Life annuity present value
( net single premium ) of annuity a x
Series of payments of 1 unit as long as individual is alive
1 1 age probability x x + 1 x + 2 p x 2 p x
…..
1 x + n n p x
…..
a x
 vp x
 v
2
2 p x
 v
3
3 p x
   v n n p x
   t



1 v t t p x

Temporary life annuity present value
Series of n payments of 1 unit
(contingent on survival) a x : n| last payment
1 1 1 age probability x x + 1 x + 2 p x 2 p x
…..
x + n n p x a x : n |
 vp x
 v
2
2 p x
 v
3
3 p x
   v n n p x
 t n 

1 v t t p x

n - years deferred life annuity
Series of payments of 1 unit as long as individual is alive in which the first payment is at x + n + 1 present value n
|a x first payment
1 1 age x x + 1 x + 2
… x + n x + n +1 n+1 p x x + n + 2 n+2 p x
… probability n
| a x
 v n

1 n

1 p x
 v n

2 n

2 p x
 t



1 v n
 t n
 t
 v n

3 n

3 p x p x
 s




1 n v s s
   v n
 t n
 t p x p x
  
Note: n
| a x
 a x
 a x : n |

ä x
1 1 1
Life annuities-due
1
…..
x + n n p x
…
 a  x

1
 a x

1
 t



1 v t t p x
ä x : n| x x + 1 x + 2 p x 2 p x
1 1 1
…..
1 x + n-1 n-1 p x x + n
 a  x : n |

1
 a x : n

1 |

1
 n t

1 

1 v t t p x n
|ä x x x + 1 x + 2 p x 2 p x
1 1 x x + 1 x + 2
… x + n n p x x + n +1 n+1 p x
1 x + n + 2 n+2 p x
… n
|  a  x
 n

1
| a x

a  x : n |

1
 a x : n

1 | but
  x : n |

( 1
 i ) a x : n | a x

1 : n |
 vp x

1
 a  x : n |

Recall: present value of a pure endowment of 1 to be paid n years hence to a life currently aged x n
E x
 v n n p x
 v n

 l x
 l x n


 v x
 n l x
 n v x l x
Denote D x
Then n
E x
= v x l x
= D x+n
/ D x

Life annuity and commutation functions a x
 t



1 v t t p x
 t



1 t
E x
Since n
E we have x
= D x+n
/ D x a x
 t



1
D x
 t
D x

1
D x

D x

1

D x

2

D x

3
 

Define commutation function N x as follows:
N x
 t



0
D x
 t
 t



0 v x
 t l x
 t
Then: a x

N x

1
D x

Identities for other types of life annuities a x : n | temporary life annuity
 t n 

1
D
D x
 t x

N x

1

N x
 n

1
D x n-years delayed l. a.
n
| a x

N x
 n

1
D x temporary l. a.-due
  x : n |

N x

N x
 n
D x

Accumulated values of life annuities temporary life annuity a x : n |
 n
E x
 s x : n | since we have a x : n |

N x

1

N x
 n

1
D x s x : n |
 and
N x

1

N x
 n

1
D x
 n n
E x

D x
 n
D x similarly for temporary life annuity-due:
 a  x : n |
 n
E x
  s  x : n | and  s  x : n |

N x

N x
 n
D x
 n

Examples (p. 162 – p. 164)
• ( life annuities and commutation functions ) Marvin, aged
38, purchases a life annuity of 1000 per year. From tables, we learn that N
38
= 5600 and N
39
= 5350. Find the net single premium Marvin should pay for this annuity
– if the first 1000 payment occurs in one year
– if the first 1000 payment occurs now
• Stay verbally the meaning of ( N
35
– N
55
) / D
20
• ( unknown rate of interest ) Given N x
N x+2
= 4810 and q x
= .005, find i
= 5000, N x+1
=4900,

• Select group of population is a group with the probability of survival different from the probability given in the standard life tables
• Such groups can have higher than average probability of survival (e.g. due to excellent health) or, conversely, higher mortality rate
(e.g. due to dangerous working conditions)

• Suppose that a person aged x is in the first year of being in the select group
• Then p
[x] and q
[x] denotes the probability of survival for
= 1 – p
[x]
1 year denotes the probability of dying during
1 year for such a person
• If the person stays within this group for subsequent years, the corresponding probabilities of survival for 1 more year are denoted by p
[x]+1
, p
[x]+2
, and so on
• Similar notations are used for life annuities: a
[x] denotes the net single premium for a life annuity of 1
(with the first payment in one year) to a person aged x in his first year as a member of the select group
• A life table which involves a select group is called a select-and-ultimate table

Examples (p. 165 – p. 166)
• ( select group ) Margaret, aged 65, purchases a life annuity which will provide annual payments of 1000 commencing at age 66. For the next year only, Margaret’s probability of survival is higher than that predicted by the life tables and, in fact, is equal to p
65
+ .05, where p
65 is taken from the standard life table. Based on that standard life table, we have the values D
65
= 300, D
66
= 260 and N
67
= 1450. If i = .09, find the net single premium for this annuity
• ( select-and-ultimate table ) A select-and-ultimate table has a select period of two years. Select probabilities are related to ultimate probabilities by the relationships p
[x]
= (11/10) p x p
[x]+1
D
61
= (21/20) p x+1
= 1500, and ä
. An ultimate table shows D
60:20|
= 1900,
= 11, when i = .08. Find the select temporary life annuity
ä
[60]:20|
60 and

• The following values are based on a unisex life table:
N
38
= 5600, N
39
N
42
= 4625.
= 5350, N
40
= 5105, N
41
= 4865,
It is assumed that this table needs to be set forward one year for males and set back two years for females. If Michael and Brenda are both age 40, find the net single premium that each should pay for a life annuity of 1000 per year, if the first payment occurs immediately.

thly
• Payments every m th part of the year
• Problem: commutation functions reflect annual probabilities of survival
• First, we obtain an approximate formula for present value
• Assume for a moment that the values D y are also given for non-integer values of y

a x
Usual life annuity
1 1 age x x + 1 x + 2
…..
1 x + n
…..
Annuity payable every 1/m part of the year a (m) x
1/m 1/m age x x +
1/m x +
2/m
1/m 1/m
…..
x +
(m-1)/m x + 1
…..

a (m) x
Annuity payable every 1/m part of the year
1/m 1/m 1/m 1/m age x x +
1/m x +
2/m
…..
x +
(m-1)/m x + 1
…..
 a x
( m ) 
1 m
 v
1 / m
1 / m
1 m

 v
1 / m l x

1 / m l x p x
 v
2 / m
2 / m
 v
2 / m l x

2 / m l x p x
   v i
 j / m i
 j / m
   v i
 j / m l x
 i
 j / m l x p x
 


 




1 m

 v x

1 / m l v x l x x

1 / m
 v x

2 / m l x

2 / m v x l x
   v x
 i
 j / m l x
 i
 j / m v x l x
 

 

1 m


D x

1 / m
D x

D x

2 / m
D x
  
D x
 i
 j / m
D x
 



Annuity payable every 1/m part of the year a (m) x
1/m 1/m 1/m 1/m age x x +
1/m x +
2/m
…..
x +
(m-1)/m x + 1
…..

 a x
( m )
1
 mD x
1 m



 




D x
D x
D x

1 /
D x

1 / m

1

1 / m m



D x
D x

2
D x

D x
/
2 m

1

2
/ m

/ m







D x

D x

(
D x
 i

D x m

1 )

1

(
/ j / m m
 m

1 )

/




D x

1 m




D x

2

  




1 mD x


 j m 

1
D x

1
 j / m
 j m 

1
D x

1
 j / m
 


 
1 mD x i
 m 

0 j

1
D x
 i
 j / m

Using linear interpolation for D x+i+j/m
D x
 i
 j / m

D x
 i
 j m

D x
 i

1

D x
 i
 j m 

1
D x
 i
 j / m
 j m 

1
D x
 i
 j m

D x
 i

1

D x
 i


 mD x
 i


D x
 i

1

D x
 i
 j m 

1 j m
 mD x
 i


D x
 i

1

D x
 i
 m ( m
2 m

1 )
 mD x
 i


D x
 i

1

D x
 i
 ( m
2

1 ) i
 m 

0 j

1
D x
 i
 j / m
 mD x
 m



D x

1

D x
 ( m
2

1 )
D x

1

D x

2
 


 mD x

1


D x

2 mD x

( m

1 )  
D x

1
2

D x

1
 ( m
2

1 )

D x
D x

2
 mN x

1
 m
2

1
D x
 mD x

2

D x

1



D x

3

D x

2
 ( m
2

1 )
 
D x

3

D x

2

 


Using linear interpolation for D x+i+j/m i
 m 

0 j

1
D x
 i
 j / m
 mN x

1
 m

1
D x
2 a x
( m ) 
1 mD x

 mN x

1
 m
2

1
D x



N x

1
D x
 m

1

2 m a x
 m

1
2 m a x
( m )  a x
 m

1
2 m

Continuous life annuity a x
 lim m
  a x
( m )  m lim
   a x
 m

1
2 m
 a x

1
2 a x
 lim m
  a x
( m ) 
 lim m
 
1 m
 v
1 / m
1 / m p x
 v
2 / m
2 / m p x
   v i
 j / m i
 j / m p x
 

 
0
 v t t p x dt a x
 a x

1
2 a x
 
0
 v t t p x dt

n|a (m) x age
Annuity payable m-thly, deferred a (m) x+n x
… x + n
1/m 1/m 1/m 1/m x + n+1/m x + n+2/m
…..
x +n+
(m-1)/m x + n+1
…..
n a x
( m )  v n n p x a
( x m

) n

D x
 n a
( x m

) n
D x


D x
D
 n x a x
 n
 m

1
2 m
 n a x

D x
D x
 n
2 m
1 n a x
( m )  n a x

D x
 n
D x

 2 m
1



Annuity payable m-thly, temporary a (m) x:n|
1/m 1/m 1/m 1/m age x x +
1/m x +
2/m
…..
x +n+
(m-1)/m x + n
( m ) a x : n |
 a x
( m )  n a x
( m )  a x
 m

1
2 m


 n a x

D x
 n
D x
2 m

 a x
 n a x




1

D x
 n
D x

 m

1
2 m
 a x : n |


 1

D x
 n
D x

 m

1
2 m
1



( m a x : n |
)  a x : n |



1

D x
 n
D x

 m

1
2 m

• Page 168, 8.10

• Arithmetic increasing annuities
• It is sufficient to look at the sequence 1,2,3,….
• Temporary decreasing annuities

• Ernest, aged 50, purchases a life annuity, which pays 5,000 for 5 years,
3,000 for 5 subsequent years, and
8,000 each year after. If the first payment occurs in exactly 1 year, find the price in terms of commutation functions.

Arithmetic increasing annuity
(Ia) x
1 2 n age x x + 1 x + 2
…..
x + n
…..
probability p x 2 p x
  x
 vp x
 t



1 v t t p x

2 v
2
2
 t



2 v t t p x p x

3 v
3
3 p x
 t



3 v t t p x
   n p x
   nv n n p x
   a x

1
| a x

2
| a x
 t



0 t
| a x
 t



0
N x
 t

1
D x
 
S x
 t



0
N x
 t
  x

S x

1
D x

Arithmetic increasing annuity, temporary
(Ia) x:n|
1 2 n age probability x x + 1 x + 2 p x 2 p x
…..
x + n n p x x + n+1 x : n |

S x

1

S x
 n

1
 nN x
 n

1
D x

Arithmetic decreasing annuity, temporary
(Da) x:n| n n-1 1 x x + 1 x + 2
…..
x + n x + n+1

Arithmetic decreasing annuity, temporary
(Da) x:n| n n-1 1
(Ia) x:n| x x + 1 x + 2
1 2
…..
x + n n x + n+1 x x + 1 x + 2
…..
x + n x + n+1

Arithmetic decreasing annuity, temporary
(Da) x:n| n n-1 1
(Ia) x:n| x x + 1 x + 2
1 2
…..
x + n n x + n+1
(n+1)a x : n| x x + 1 x + 2 n+1 n+1
…..
x + n n+1 x x + 1 x + 2
…..
x + n
  x : n |

  x : n |

( n

1 ) a x : n | x + n+1

Arithmetic decreasing annuity, temporary
(Da) x:n| age probability n n-1 1 x x + 1 x + 2 p x 2 p x
…..
x + n n p x
  x : n |

  x : n |

( n

1 ) a x : n | x + n+1 x : n |

S x

1

S x
 n

1
 nN x
 n

1
D x a x : n |

N x

1

N x
 n

1
D x
  x : n |
 nN x

1

( S x

2

S x
 n

2
)
D x

• Georgina, aged 50, purchases a life annuity which will pay her 5000 in one year, 5500 in two years, continuing to increase by 500 per year thereafter. Find the price if S
51
5000, N
51
= 450, and D
50
= 60
=
• Redo the previous example if the payments reach a maximum level of 8000, and then remain constant for life.
Assume S
58
= 2100
• Two annuities are of equal value to Jim, aged 25. The first is guaranteed and pays him 4000 per year for 10 years, with the first payment in 6 years. The second is a life annuity with the first payment of X in one year. Subsequent payments are annual, increasing by .0187 each year.
If i = .09, and from the 7% -interest table, N
26
=930 and D
25
=
30, find X.

• Paying for deferred life annuity with a series of payments instead of a single payment
• Premium reserve is an analog of outstanding principal
• Premiums often include additional expenses and administrative costs
• In such cases, the total payment is called gross premium
• Loading = gross premium – net premium
• General approach : actuarial present values of two sequences of payments must be the same (equation of value)

x
P P
Annual premiums P = t
P( n
|ä x
)
P
1 1 1 x + 1
… x + t-1 x +t
… x + n x + n +1 x + n + 2 age
• t is the number of premium payments
• Present value of premiums is P ä x:t|
• Present value of benefits is n
|ä x
• Therefore P ä x:t|
= n
|ä x
…
P
 n
  x
  x : t |

N x
 n
N x
D x

N x
 t
D x

N x
N
 x
 n
N x
 t t
P ( n
  x
)

N x
N
 x

N n x
 t

• Arabella, aged 25, purchases a deferred life annuity of 500 per month, with the first benefit coming in exactly 20 years.
She intends to pay for this annuity with a series of annual payments at the beginning of each year for the next 20 years. Find her net annual premium if D
5000, ä
25
= 15 and ä
45
= 11.5
25
= 9000, D
45
=

Reserves
P P P P 1 1 1 age x x + 1
… x + t -1
… x +n-1 x + n x + n +1 x + n + 2
… n t
V ( n
|ä x
)
• Analog of outstanding principal immediately after premium t has been paid
• Assume that the number of premium payments is n
• Reserve n t
V ( n
|ä x
) = PV of all future benefits – PV of all future premiums
 t n
V ( n
 a  x
)


 n
 t
|   x
 t

P  a  x
 t : n
  x
 t |
 t


N
N
D x x
 x
 t n
 t

, t
P ( N
 x

D x
 t t n

N x
 n
)
, t
 n

• Arabella, aged 25, purchases a deferred life annuity of 500 per month, with the first benefit coming in exactly 20 years.
She intends to pay for this annuity with a series of annual payments at the beginning of each year for the next 20 years. Assume that 50% of her first premium is required for initial underwriting expenses, and 10% of all subsequent premiums are needed for administration costs. In addition,
100 must be paid for issue expenses. Find Arabella’s annual gross premium, if D and ä
45
= 11.5
25
= 9000, D
45
= 5000, ä
25
= 15,

Chapter 9
• Basic Concepts
• Commutation Functions and Basic Identities
• Insurance Payable at The Moment of Death
• Varying Insurance
• Annual Premiums and Premium Reserves

• Benefits are paid upon the death of the insured
• Types of insurance
– Whole life policy
– Term insurance
– Deferred insurance
– Endowment insurance

• Benefit ( the face value ) is paid to the beneficiary at the end of the year of death of inured person
• If the face value is 1 and insurance is sold to a person aged x , the premium is denoted by A x
A x age probability x x + 1 x + 2 p x 2 p x
1
…..
x + t t q x x + t+1

A x
 t



0 t p x
 q x
 t
 v t

1
A x age probability x x + 1 x + 2 p x 2 p x
1
…..
x + t x + t+1 q x+t

• Benefit ( the face value ) is paid to the beneficiary at the end of the year of death of inured person, only if the death occurs within n years
• If the face value is 1 and insurance is sold to a person aged x , the premium is denoted by A 1 x
A
1 x : n |
 t n

1 

0 t p x
 q x
 t
 v t

1

• Does not come into force until age x+n
• If the face value is 1 and insurance is sold to a person aged x , the premium is denoted by A 1 x:n|
A x

A
1 x : n |
 n
| A x n
| A x

A x

A
1 x : n |

• Benefit (the face value) is paid to the beneficiary at the end of the year of death of inured person, if the death occurs within n years
• If the insured is still alive at the age x+n , the face value is paid at that time
• If the face value is 1 and insurance is sold to a person aged x , the premium is denoted by A x:n|
A x : n |

A
1 x : n |
 n
E x
Exercise:
A
1 x : n |

A x

A x : n |

• Rose is 38 years old. She wishes to purchase a life insurance policy which will pay her estate 50,000 at the end of the year of her death. If i=.12, find an expression for the actuarial present value of this benefit and compute it, assuming p x
= .94 for all x.
• Michael is 50 years old and purchases a whole life policy with face value 100,000. If lx= 1000(1-x/105) and i=.08, find the price of this policy.
• Calculate the price of Rose’s and Michael’s policies if both policies are in force for a term of only 30 years.
• Calculate the price of Rose’s and Michael’s policies if both policies are to be 30 years endowment insurance.

• Recall:
D x
 v x l x n
E x

D x
 n
D x
N x
 t



0
D x
 t

• Recall:
A x
 t



0 t p x
 q x
 t
 v t

1
• So we need: t p x
 q x
 t
 v t

1  l x
 t l x
 d x
 t l x
 t
 v t

1  d x
 t l x
 v t

1  d x
 t v x
 t

1 l x v x
C x
 d x v x

1 t p x
 q x
 t
 v t

1 
C x
 t
D x

A x
 t



0 t p x
 q x
 t
 v t

1  t



0
C x
 t
D x

1
D x t



0
C x
 t
M x
 t



0
C x
 t
A
 x
M x
D x

A
1 x : n |
 t n

1 

0 t p x
 q x
 t
 v t

1
A
1 x : n |
 t n

1 

0
C x
 t
D x
 t



0
C x
D x
 t
 t


 n
C x
 t
D x

M x
D x

M x
 n
D x
A
1 x : n |

M x

M x
 n
D x

A x : n |

A
1 x : n |
 n
E x
A x : n |

M x

M x
 n
D x

D x
 n
D x

M x

M x
 n

D x
 n
D x

• We can represent insurance premiums in terms of actuarial present values of annuities, e.g. A x
= 1 – d ä x
• Hence they also can be found using
“old” commutation functions

• Juan, aged 40, purchases an insurance policy paying
50,000 if death occurs within the next 20 years, 100,000 if death occurs between ages 60 and 70, and 30,000 if death occurs after that. Find the net single premium for this policy in terms of commutation functions.
• Phyllis, aged 40, purchases a whole life policy of 50,000. If
N
40
= 5000, N
41
= 4500, and i = .08, find the price.

• We consider scenario when the benefit is paid at the end of the year of death
• Alternatively, the benefit can be paid at the moment of death

Divide each year in m parts
A x
( m )  v
1 / m
1 / m q x v
1

/ m
 l v i x
 l x j / m
 l x v
2 / m
 l x

1 / m
 i

(
 l x
1 / m v
2 p
/ m j

1 ) / m x l

1 / m x

1 / l x q x

1 / m m

 l x

1 / m

 l x

1 /
 v l x

2 m
 l x
 i

( j

1 ) / m l x
 i

(
 l x j

1 ) / m
 i
 i

/ m j / m j / m i

( j

1 ) / m

   p x

1 / m q x

( j

1 ) / m
  

v
1 /
 m l x v i
 j / m
 l x

1 / m l l x x
 i

(
 v
2 / m j

1 ) / m l x l x

1 / m

 l x
 i
 j / m l x l x

2 / m
  
1 ml x
 




 v


1 / v m i
 l j x
/ m

1 / l x

1 / m l x
 i

( m
 v j

1 ) / m
1 /
2 / m
 m l x

1 / m l x
 i

1 j / m
/

 l m x

2 /
 m


1 ml x

1 ml x
 v
1




/ v m i

 j l
/ x
' m

 s ( 0 , 1 ) l
' x
 s
/ m
( i ,
 v j ) / m
2

/ m 
 l
' x
 s ( 0 , 2 ) / m i
 m 

0 j

1 v i
 j / m  l
' x
 s ( i , j ) / m
 





 






Taking the limit as m →∞ we get:
A x
 m lim
 
A x
( m )


0
 v t  l
' x
 t l x
 m lim
 

1 ml x i
 m 

0 j

1 v i
 j / m  l
' x
 s ( i , j ) / m dt
 

0 v t  l x
 t l x



 l
' x
 t l x
 t


 dt
 

0 v t  t p x
  x
 t dt

Premium for insurance payable at the moment of death
• Whole life policy:
A x
 

0 v t
 t p x
  x
 t dt
• Term policy:
A : n |
  n
0 v t
 t p x
  x
 t dt

• Find the net single premium for a 100,000 life insurance policy, payable at the moment of death, purchased by a person aged 30 if i = .06 and t p
30
= (.98) t for all t
• Solve the previous example if it is 20 years endowment l insurance, force of interest is .06 and x
= 105 – x, 0 ≤ x ≤ 105.

• Using integration by parts, we can get
Ā x
= 1 – δ ā x
• Approximate formula:
Ā x
≈ (i/δ) A x
• To obtain it, use linear interpolation in the following expression:
A

1 l x
 v
1 / m
( l x
 l x

1 / m
)
 v
2 / m
( l x

1 / m
 l x

2 / m
)


 v i
 j / m
( l x
 i

( j

1 ) / m
 l x
 i
 j / m
)


