Reserves - Recursion
Dr. Handayani, S.Si, MM, MHP, HIA, FLMI, AFSI, AAK, AAIJ, AMRP, FSAI
Net Premium Reserve
Consider a fully discrete whole life insurance policy with face amount π issued to π₯ . Here is the diagram illustrating a single year of age from π₯ + π‘ to π₯ + π‘ + 1.
At the beginning of the year (time t), we have the reserves from the previous year, and we also collect premiums. At the end of the year (time t+1), we set up another reserve. What is the relationship between
π‘
π and
π‘+1π?
Suppose ππ₯ people purchase an identical policy at age x. After t years, the number of people who are alive at that time is ππ₯+π‘. At that time, each policy carries a reserve of π‘π, and each person pays a premium P. Thus, the reserves plus the premiums at time t is the total amount of money we have at time t for that group of policies. Accumulate this money with interest for one year:
ππ₯+π‘( π‘π + π)(1 + π)
At the end of the year (time t+1), this money will be used for two things:
1. For people who die, this money is used to pay death benefits. How many people die between age x+t and x+t+1? ππ₯+π‘. We will pay a death benefit of b to these people at time t+1.
2. For people who survive, this money is used to set up the next reserve. How many people survive to age x+t+1? ππ₯+π‘+1. We will set up a reserve of
π‘+1π for these people at time t+1.
The money we accumulate from time t to t+1 must be sufficient to pay for the two things described above:
ππ₯+π‘ π‘π + π 1 + π = ππ₯+π‘(π) + ππ₯+π‘+1 π‘+1π Dividing both sides of the equation by ππ₯+π‘, we have:
π‘π + π 1 + π = ππ₯+π‘(π) + ππ₯+π‘ π‘+1π
Expression above has the same interpretation as before, but it is on one policy rather than on a group of policies.
In words, we are saying:
At the beginning of the year, we have two things: (i) the reserve from the previous year, and (ii) the premium collected at the beginning of the year. We then accumulate both for one year.
At the end of the year, for the insured who dies, with probability ππ₯+π‘, we pay a benefit of b.
For the insured who survives, with probability ππ₯+π‘, we have to set up the next reserve.
We can also rewrite the equation above by expressing ππ₯+π‘ as 1 β ππ₯+π‘, and then rearranging the equation:
π‘π + π 1 + π = ππ₯+π‘ π β π‘+1π + π‘+1π
Notice the left hand side remains unchanged, but the right hand side has changed slightly. How should we interpret the right hand side?
β’ We set up a reserve at time t+1, regardless of whether or not the insured dies or survives during the year. This is represented by π‘+1π.
β’ For an insured who dies during the year, since we have the reserve of
π‘+1π to cover a portion of the death benefit, the remaining amount needed is π β π‘+1π, which is known as the net amount at risk (NAR).
REMARK
π‘π + π 1 + π = ππ₯+π‘ π β π‘+1π + π‘+1π is helpful when you want to solve for ππ₯+π‘.
When will reserve recursion be useful?
It is often useful to start the recursion at time 0, because π‘π = 0.
It is also useful to start the recursion at the end (when applicable). Note that:
β For n-year term insurance, ππ = 0.
β For n-year endowment insurance, ππ = endowment benefit.
REMARK
To start recursion off, you can start at time 0, since π‘π = 0. You can also start at the end, at which point the benefit reserve is 0. The following rules help:
1. For paid up insurance, the benefit reserve is the single benefit premium.
2. For term insurance, the benefit reserve at expiry is 0.
3. For endowment insurance, the benefit reserve right before maturity is the endowment benefit.
Example 1
A fully discrete 3-year endowment insurance with a face amount of 1,000 is issued on (π₯).
β π = 0.06
β The following has been taken from a mortality table
β The net premium is 332.51
Calculate the net premium reserves at time 1 and 2.
π ππ
π₯ 1,000
π₯ + 1 900
π₯ + 2 810
Solution 1
To find
1π, since we know
0π = 0, we can use the recursion involving
0π
and
1π:
0π + π 1 + π = ππ₯ 1,000 + ππ₯ 1π 0 + 332.51 1.06 = 1,000 β 900
1,000 1,000 + 900
1,000 1π
β΄
1π = 280.51Solution 1
To find
2π, we use the fact that
ππfor an n-year endowment insurance is equal to the endowment benefit, i.e.,
3π = 1,000. Thus, we can use the recursion involving
2πand
3π:
2π + π 1 + π = ππ₯+2 1,000 + ππ₯+2 3π
2π + 332.51 1.06 = 1,000
β΄
2π = 610.89Gross Premium Reserve
We now incorporate expenses and develop a recursion for gross premium reserves.
Let e denote the expense (per policy or per premium), and let E denote
the settlement expense. Below is a diagram illustrating a single year of
age from age x+t and x+t+1.
Note that at the time of death, we must take into account both the death benefit and the settlement expenses. Thus, the recursion becomes:
π‘
π
π + πΊ β π 1 + π = ππ₯+π‘ π + πΈ + ππ₯+π‘ π‘+1π
πIn words, we are saying:
At the beginning of the year, we have three things: (i) reserves from previous year, (ii) premiums collected, and (iii) expenses paid. Then, we accumulate the money for one year.
At the end of the year, for an insured who dies, with probability ππ₯+π‘, we pay a benefit of b and a settlement expense of E. For an insured who survives, with probability ππ₯+π‘, we have to set up the next reserve.
Example 2
For a fully discrete whole life insurance of 100,000 on (45) you are given
β The gross premium reserve at time 5 is 5,500 and at time 6 is 7,100.
β π50 = 0.009
β π = 0.05
β Renewal expenses at the start of each year are 50 plus 4% of the gross premium.
β Claim expenses are 200.
Calculate the gross premium
Solution 2
Since we are given
5π
πand
6π
π, we can use the recursion to connect them.
5
π
π + πΊ β π 1 + π = π50 π + πΈ + π50 6π
π5,500 + πΊ β 50 + 0.04πΊ 1.05 = 0.009 100,000 + 200 + 1 β 0.009 7,100 πΊ = 2,197.82
Recursion - Continuous
In the previous subsection, we discussed reserve recursion for policies with discrete cash flows.
In this subsection, we will derive a reserve recursion for policies with continuous cash flows (i.e., premiums/annuities are payable continuously and death benefits are payable at the moment of death).
The continuous time version of the reserve recursion is given by Thiele's
differential equation
Recursion - Continuous
π
ππ‘ π‘π = πΏπ‘ π‘π + πΊπ‘ β ππ‘ β ππ‘ + πΈπ‘ β π‘π ππ₯+π‘ where
β’ πΏπ‘ is the force of interest per year assumed earned at time t
β’ πΊπ‘ is the annual rate of premium payable at time t
β’ ππ‘ is the annual rate of premium-related expense payable at time t
β’ ππ‘ is the face amount payable at time t if the insured dies at exact time t
β’ πΈπ‘ is the expense of paying the face amount at time t (e.g., settlement/claim expense)
β’ ππ₯+π‘ is the force of mortality at age x+t.
What is the intuition behind Thiele's differential equation
The left hand side asks, "What is the rate of change in the reserve at time t?"
The right-hand side explains that the reserve is changing at a rate resulting from the following components:
β’ The reserve is earning interest a rate of πΏπ‘
β’ Premiums are received at a rate of πΊπ‘
β’ Expenses are paid at rate of ππ‘
β’ For insureds who die during the next instant (with "probabilityβ ππ₯+π‘), we have to pay the death benefit of ππ‘ and a settlement expense of πΈπ‘. However, we have reserves to help cover these costs. Thus, the amount needed is ππ‘ + πΈπ‘ β π‘π.
For a net premium reserve, drop all the expense-related terms, and substitute the net premium for πΊπ‘
Thiele's differential equation usually does not have a closed form solution, but it can be solved numerically using Euler's method. Using Euler's method, the rate of change in reserve can be approximated as the change in reserves adjusted by the period length:
π
ππ‘ π‘π β π‘+βπ β π‘π
β (a)
π
ππ‘ π‘π β π‘π β π‘ββπ
β (b)
(a) uses the derivative at time t to relate π‘π and π‘+βπ (i.e., we look forward). This approximation is called the Forward Euler Approximation.
(b) uses the derivative at time t to relate π‘π and π‘ββπ (i.e., we look backward). This approximation is called the Backward Euler Approximation.
Exercise 1
For a fully discrete 3-year endowment insurance is issued on ( π₯ ) with face amount 1,000. You are given:
β π = 0.06
β The following extract from the mortality table:
β 1,000π
π₯:3= 332.51
Calculate the increase in the benefit reserve in the second year,
2π β
1π
π ππ
π₯ 1,000
π₯ + 1 900
π₯ + 2 810
Exercise 2
For a fully discrete 3-year endowment insurance of 1,000 on ( π₯ ):
β π
π₯= π
π₯+1= 0.20
β π = 0.06
β π
π₯:3= 373.63
Calculate the increase in the benefit reserve in the second year,
2π β
1π
Exercise 3
For a fully discrete whole life insurance of 1,000 on (π₯ ):
β The benefit reserve at time 18 is 482
β The benefit premium is 18
β π = 0.06
β The benefit reserve at time 19 is 520
Determine π
π₯+18Exercise 4
For a fully discrete 10-pay whole life insurance of 1 on (40):
β π΄50 = 0.19
β π΄51 = 0.20
β π49 = π50
β π = 0.08
β The benefit premium is 0.02.
Calculate 9π