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Reserves - Recursion

Dr. Handayani, S.Si, MM, MHP, HIA, FLMI, AFSI, AAK, AAIJ, AMRP, FSAI

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Net Premium Reserve

Consider a fully discrete whole life insurance policy with face amount 𝑏 issued to π‘₯ . Here is the diagram illustrating a single year of age from π‘₯ + 𝑑 to π‘₯ + 𝑑 + 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

𝑑

𝑉 and

𝑑+1

𝑉?

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Suppose 𝑙π‘₯ people purchase an identical policy at age x. After t years, the number of people who are alive at that time is 𝑙π‘₯+𝑑. At that time, each policy carries a reserve of 𝑑𝑉, 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:

𝑙π‘₯+𝑑( 𝑑𝑉 + 𝑃)(1 + 𝑖)

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? 𝑑π‘₯+𝑑. 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? 𝑙π‘₯+𝑑+1. We will set up a reserve of

𝑑+1𝑉 for these people at time t+1.

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The money we accumulate from time t to t+1 must be sufficient to pay for the two things described above:

𝑙π‘₯+𝑑 𝑑𝑉 + 𝑃 1 + 𝑖 = 𝑑π‘₯+𝑑(𝑏) + 𝑙π‘₯+𝑑+1 𝑑+1𝑉 Dividing both sides of the equation by 𝑙π‘₯+𝑑, we have:

𝑑𝑉 + 𝑃 1 + 𝑖 = π‘žπ‘₯+𝑑(𝑏) + 𝑝π‘₯+𝑑 𝑑+1𝑉

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 π‘žπ‘₯+𝑑, we pay a benefit of b.

For the insured who survives, with probability 𝑝π‘₯+𝑑, we have to set up the next reserve.

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We can also rewrite the equation above by expressing 𝑝π‘₯+𝑑 as 1 βˆ’ π‘žπ‘₯+𝑑, and then rearranging the equation:

𝑑𝑉 + 𝑃 1 + 𝑖 = π‘žπ‘₯+𝑑 𝑏 βˆ’ 𝑑+1𝑉 + 𝑑+1𝑉

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 𝑑+1𝑉.

β€’ For an insured who dies during the year, since we have the reserve of

𝑑+1𝑉 to cover a portion of the death benefit, the remaining amount needed is 𝑏 βˆ’ 𝑑+1𝑉, which is known as the net amount at risk (NAR).

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REMARK

𝑑𝑉 + 𝑃 1 + 𝑖 = π‘žπ‘₯+𝑑 𝑏 βˆ’ 𝑑+1𝑉 + 𝑑+1𝑉 is helpful when you want to solve for π‘žπ‘₯+𝑑.

When will reserve recursion be useful?

It is often useful to start the recursion at time 0, because 𝑑𝑉 = 0.

It is also useful to start the recursion at the end (when applicable). Note that:

βœ“ For n-year term insurance, 𝑛𝑉 = 0.

βœ“ For n-year endowment insurance, 𝑛𝑉 = endowment benefit.

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REMARK

To start recursion off, you can start at time 0, since 𝑑𝑉 = 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.

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Example 1

A fully discrete 3-year endowment insurance with a face amount of 1,000 is issued on (π‘₯).

βœ“ 𝑖 = 0.06

βœ“ The following has been taken from a mortality table

βœ“ The net premium is 332.51

Calculate the net premium reserves at time 1 and 2.

π’š π’π’š

π‘₯ 1,000

π‘₯ + 1 900

π‘₯ + 2 810

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Solution 1

To find

1𝑉

, since we know

0𝑉 = 0

, we can use the recursion involving

0𝑉

and

1𝑉

:

0𝑉 + 𝑃 1 + 𝑖 = π‘žπ‘₯ 1,000 + 𝑝π‘₯ 1𝑉 0 + 332.51 1.06 = 1,000 βˆ’ 900

1,000 1,000 + 900

1,000 1𝑉

∴

1𝑉 = 280.51

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Solution 1

To find

2𝑉

, we use the fact that

𝑛𝑉

for an n-year endowment insurance is equal to the endowment benefit, i.e.,

3𝑉 = 1,000

. Thus, we can use the recursion involving

2𝑉

and

3𝑉

:

2𝑉 + 𝑃 1 + 𝑖 = π‘žπ‘₯+2 1,000 + 𝑝π‘₯+2 3𝑉

2𝑉 + 332.51 1.06 = 1,000

∴

2𝑉 = 610.89

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Gross 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.

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Note that at the time of death, we must take into account both the death benefit and the settlement expenses. Thus, the recursion becomes:

𝑑

𝑉

𝑔 + 𝐺 βˆ’ 𝑒 1 + 𝑖 = π‘žπ‘₯+𝑑 𝑏 + 𝐸 + 𝑝π‘₯+𝑑 𝑑+1

𝑉

𝑔

In 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 π‘žπ‘₯+𝑑, we pay a benefit of b and a settlement expense of E. For an insured who survives, with probability 𝑝π‘₯+𝑑, we have to set up the next reserve.

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Example 2

For a fully discrete whole life insurance of 100,000 on (45) you are given

βœ“ The gross premium reserve at time 5 is 5,500 and at time 6 is 7,100.

βœ“ π‘ž50 = 0.009

βœ“ 𝑖 = 0.05

βœ“ Renewal expenses at the start of each year are 50 plus 4% of the gross premium.

βœ“ Claim expenses are 200.

Calculate the gross premium

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Solution 2

Since we are given

5

𝑉

𝑔

and

6

𝑉

𝑔

, we can use the recursion to connect them.

5

𝑉

𝑔 + 𝐺 βˆ’ 𝑒 1 + 𝑖 = π‘ž50 𝑏 + 𝐸 + 𝑝50 6

𝑉

𝑔

5,500 + 𝐺 βˆ’ 50 + 0.04𝐺 1.05 = 0.009 100,000 + 200 + 1 βˆ’ 0.009 7,100 𝐺 = 2,197.82

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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

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Recursion - Continuous

𝑑

𝑑𝑑 𝑑𝑉 = 𝛿𝑑 𝑑𝑉 + 𝐺𝑑 βˆ’ 𝑒𝑑 βˆ’ 𝑏𝑑 + 𝐸𝑑 βˆ’ 𝑑𝑉 πœ‡π‘₯+𝑑 where

β€’ 𝛿𝑑 is the force of interest per year assumed earned at time t

β€’ 𝐺𝑑 is the annual rate of premium payable at time t

β€’ 𝑒𝑑 is the annual rate of premium-related expense payable at time t

β€’ 𝑏𝑑 is the face amount payable at time t if the insured dies at exact time t

β€’ 𝐸𝑑 is the expense of paying the face amount at time t (e.g., settlement/claim expense)

β€’ πœ‡π‘₯+𝑑 is the force of mortality at age x+t.

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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 𝛿𝑑

β€’ Premiums are received at a rate of 𝐺𝑑

β€’ Expenses are paid at rate of 𝑒𝑑

β€’ For insureds who die during the next instant (with "probability” πœ‡π‘₯+𝑑), we have to pay the death benefit of 𝑏𝑑 and a settlement expense of 𝐸𝑑. However, we have reserves to help cover these costs. Thus, the amount needed is 𝑏𝑑 + 𝐸𝑑 βˆ’ 𝑑𝑉.

For a net premium reserve, drop all the expense-related terms, and substitute the net premium for 𝐺𝑑

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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:

𝑑

𝑑𝑑 𝑑𝑉 β‰ˆ 𝑑+β„Žπ‘‰ βˆ’ 𝑑𝑉

β„Ž (a)

𝑑

𝑑𝑑 𝑑𝑉 β‰ˆ 𝑑𝑉 βˆ’ π‘‘βˆ’β„Žπ‘‰

β„Ž (b)

(a) uses the derivative at time t to relate 𝑑𝑉 and 𝑑+β„Žπ‘‰ (i.e., we look forward). This approximation is called the Forward Euler Approximation.

(b) uses the derivative at time t to relate 𝑑𝑉 and π‘‘βˆ’β„Žπ‘‰ (i.e., we look backward). This approximation is called the Backward Euler Approximation.

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Exercise 1

For a fully discrete 3-year endowment insurance is issued on ( π‘₯ ) with face amount 1,000. You are given:

βœ“ 𝑖 = 0.06

βœ“ The following extract from the mortality table:

βœ“ 1,000𝑃

π‘₯:3

= 332.51

Calculate the increase in the benefit reserve in the second year,

2

𝑉 βˆ’

1

𝑉

π’š π’π’š

π‘₯ 1,000

π‘₯ + 1 900

π‘₯ + 2 810

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Exercise 2

For a fully discrete 3-year endowment insurance of 1,000 on ( π‘₯ ):

βœ“ π‘ž

π‘₯

= π‘ž

π‘₯+1

= 0.20

βœ“ 𝑖 = 0.06

βœ“ 𝑃

π‘₯:3

= 373.63

Calculate the increase in the benefit reserve in the second year,

2

𝑉 βˆ’

1

𝑉

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Exercise 3

For a fully discrete whole life insurance of 1,000 on (π‘₯ ):

βœ“ The benefit reserve at time 18 is 482

βœ“ The benefit premium is 18

βœ“ 𝑖 = 0.06

βœ“ The benefit reserve at time 19 is 520

Determine π‘ž

π‘₯+18

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Exercise 4

For a fully discrete 10-pay whole life insurance of 1 on (40):

βœ“ 𝐴50 = 0.19

βœ“ 𝐴51 = 0.20

βœ“ π‘ž49 = π‘ž50

βœ“ 𝑖 = 0.08

βœ“ The benefit premium is 0.02.

Calculate 9𝑉

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