A Multi-Step Identity and Recursion Calculation

Why Combine Identities and Recursions?

A life insurance value can often be found in more than one way. An identity separates a contract into simpler benefits, while a recursion relates the value at one age to the value one year later. A deferred-insurance factorization connects a value at the deferral date to the probability and discounting needed to reach that date. In a multi-step calculation, the main challenge is matching each formula to the benefit timing and survival condition in the contract.

This tutorial uses discrete annual insurance benefits. Unless a problem states otherwise, assume a level effective annual interest rate $i$, with $v=1/(1+i)$. Death benefits are paid at the end of the year of death. A maturity benefit is paid at the end of the stated term if the insured survives to then. Mortality probabilities are conditional on being alive at the start of the year.

Definition: For a unit-benefit $n$-year term insurance, $A^1_{x:\overline{n}|}$ is the expected present value of 1 paid at the end of the year of death if death occurs within $n$ years. The pure endowment value ${}_nE_x$ is the expected present value of 1 paid at time $n$ if the insured survives $n$ years. The $n$-year term-endowment value $A_{x:\overline{n}|}$ pays 1 at death within the term, or 1 at time $n$ if the insured survives to the end of the term.

The Term-Endowment Identity

The death and survival benefits of a term-endowment are mutually exclusive and cover all possible outcomes by time $n$: either death occurs during the term, or the insured survives to its end. The contract’s expected present value is therefore the sum of the values of those two benefits.

A_{x:\overline{n}|}=A^1_{x:\overline{n}|}+{}_nE_x

With annual benefit payments, the components can be written using the one-year death probabilities and survival probabilities:

A^1_{x:\overline{n}|} =\sum_{k=0}^{n-1}v^{k+1}\,{}_kp_x\,q_{x+k}, \qquad {}_nE_x=v^n\,{}_np_x

Here $q_{x+k}$ is the probability that a life aged $x+k$ dies during the following year, conditional on being alive at that age, and ${}_kp_x$ is the probability of surviving $k$ years from age $x$. The death benefit in year $k+1$ requires survival through $k$ years, followed by death in the next year; the discount factor is $v^{k+1}$, not $v^k$.

Worked Example: Splitting a Two-Year Term-Endowment

A hypothetical policy at age $x$ pays 1 at the end of the year of death if death occurs within two years; otherwise it pays 1 at the end of year 2. The effective annual interest rate is $25\%$, $q_x=0.20$, and $q_{x+1}=0.25$. Find the term-insurance, pure-endowment, and term-endowment values.

Method. Find the death-benefit value year by year, then add the survival benefit using the term-endowment identity. Here $v=1/1.25=0.8$, $p_x=1-q_x=0.8$, and $p_{x+1}=1-q_{x+1}=0.75$.

For death in year 1, the probability is $q_x=0.20$, and the payment is discounted one year: $v q_x=0.8(0.20)=0.16$.

For death in year 2, the probability is ${}_1p_xq_{x+1}=0.8(0.25)=0.20$, and the payment is discounted two years: $v^2{}_1p_xq_{x+1}=0.64(0.20)=0.128$. Thus $A^1_{x:\overline{2}|}=0.16+0.128=0.288$.

The probability of surviving two years is ${}_2p_x=p_xp_{x+1}=0.8(0.75)=0.60$. The pure-endowment value is ${}_2E_x=v^2{}_2p_x=0.64(0.60)=0.384$.

The identity gives $A_{x:\overline{2}|}=0.288+0.384=0.672$. As a check, the probabilities of the three possible outcomes are $0.20$ for death in year 1, $0.20$ for death in year 2, and $0.60$ for survival through year 2; they sum to 1. The death and survival benefit values use the correct payment times.

Common mistake avoided: Discounting the year-2 death benefit by $v$ or using $q_{x+1}$ without first multiplying by ${}_1p_x$. The year-2 payment requires both survival to age $x+1$ and death during the next year, and it is paid at time 2.

Whole-Life Recursion and Deferred Factorization

For a unit whole-life insurance paying at the end of the year of death, condition on what happens during the first year. A death in that year produces a discounted payment of 1. Survival through the year leads to a whole-life insurance at age $x+1$, whose value is discounted one year back to age $x$.

Recursion: For annual end-of-year death benefits and a level effective annual interest rate, $$ A_x=v\big(q_x+p_xA_{x+1}\big). $$

This recursion follows by conditioning on first-year death or survival. The first event has probability $q_x$, with a payment at time 1. The second has probability $p_x$; conditional on survival, the remaining contract has value $A_{x+1}$ at time 1. Multiplying the sum of these time-1 values by $v$ expresses the value at time 0. The formula assumes the same unit death benefit throughout and no benefit payable on survival at the end of the first year. If the death benefit varies by year or a maturity benefit is present, the first-year equation must reflect those contract terms.

A deferred whole-life insurance pays 1 at the end of the year of death only if death occurs after the first $n$ years; it pays nothing if death occurs during the deferral period. Conditional on survival to time $n$, its remaining value at that time is $A_{x+n}$. Reaching time $n$ requires both survival and discounting.

Factorization: For an $n$-year deferred whole-life insurance with unit benefit, $$ {}_{n|}A_x=v^n\,{}_np_x\,A_{x+n}={}_nE_x A_{x+n}. $$

This factorization does not include a payment for death during the waiting period. It also does not mean that the whole-life value at age $x+n$ is discounted by $v^n$ alone: the insured must survive to that age, so the factor ${}_np_x$ is essential.

Worked Example: Recovering a Missing Whole-Life Value

A hypothetical unit whole-life policy has an effective annual interest rate of $25\%$. At age $x$, $q_x=0.20$, and the whole-life value at age $x+1$ is $A_{x+1}=0.40$. Calculate $A_x$, then check the recursion by recovering $A_{x+1}$ from the equation.

Method. Apply the one-year whole-life recursion. The rate gives $v=0.8$, and $p_x=1-q_x=0.80$.

Substitute the values: $A_x=0.8\big(0.20+0.80(0.40)\big)$. The continuation term is $0.80(0.40)=0.32$, so the quantity in parentheses is $0.20+0.32=0.52$. Therefore $A_x=0.8(0.52)=0.416$.

For the check, rearrange the recursion: $A_{x+1}=(A_x/v-q_x)/p_x$. Using $A_x=0.416$, $v=0.8$, $q_x=0.20$, and $p_x=0.80$, this gives $(0.416/0.8-0.20)/0.80=(0.52-0.20)/0.80=0.40$, as stated.

Common mistake avoided: Adding $q_x$ directly to $A_{x+1}$ and discounting only one of the terms. Both first-year death and survival-continuation values are at time 1 before the entire sum is discounted.

Combining the Three Relationships

A useful multi-step strategy is to keep the benefit portions distinct. The term-endowment identity describes benefits during the first $n$ years and a payment on survival to time $n$. A deferred whole-life value describes death benefits after time $n$. Together, these are the death benefits of a whole-life policy over all possible years.

A_x = A^1_{x:\overline{n}|} + {}_{n|}A_x = A^1_{x:\overline{n}|} + {}_nE_x A_{x+n}

This is a decomposition of the whole-life death benefit, not an addition of the term-endowment value and the deferred-insurance value. The term-endowment already includes a survival payment at time $n$, while a whole-life policy has no such maturity payment. To use the term-endowment identity in this calculation, subtract the pure-endowment component first: $A^1_{x:\overline{n}|}=A_{x:\overline{n}|}-{}_nE_x$. Then add the deferred whole-life value.

Separate the term-endowment.
Use the death-benefit value $A^1_{x:\overline{n}|}=A_{x:\overline{n}|}-{}_nE_x$. Confirm that the pure endowment is discounted survival to time $n$.

Value the post-deferral insurance.
Use ${}_{n|}A_x={}_nE_xA_{x+n}$, provided the deferred benefit is paid only for death after the deferral period.

Combine and verify.
Add the death benefits before and after time $n$. If an age-indexed value is available, check the result against the one-year whole-life recursion.

Worked Example: Finding a Whole-Life Value from Three Components

A hypothetical policy has an effective annual interest rate of $25\%$, with $q_x=0.20$, $q_{x+1}=0.25$, and $A_{x+2}=0.416$. Its two-year unit term-endowment value is $0.672$. Find the unit whole-life value $A_x$ by combining the term-endowment identity, deferred-insurance factorization, and whole-life recursion.

Method. First obtain the two-year pure-endowment value and the term-insurance value. Then value the insurance deferred two years and add the death benefits before and after the deferral date. Finally check with two one-year recursions. The rate gives $v=0.8$, $p_x=0.8$, and $p_{x+1}=0.75$.

The two-year survival probability is ${}_2p_x=p_xp_{x+1}=0.8(0.75)=0.60$. Therefore the pure-endowment value is ${}_2E_x=v^2{}_2p_x=0.64(0.60)=0.384$.

The term-endowment identity gives $A^1_{x:\overline{2}|}=0.672-0.384=0.288$. The deferred whole-life value is ${}_{2|}A_x={}_2E_x A_{x+2}=0.384(0.416)=0.159744$. Thus the whole-life value is $A_x=0.288+0.159744=0.447744$.

Recursion check. First use the recursion at age $x+1$: $A_{x+1}=0.8\big(0.25+0.75(0.416)\big)$. The continuation term is $0.75(0.416)=0.312$, so $A_{x+1}=0.8(0.562)=0.4496$. Now apply the recursion at age $x$: $A_x=0.8\big(0.20+0.80(0.4496)\big)$. Since $0.80(0.4496)=0.35968$, this gives $A_x=0.8(0.55968)=0.447744$, matching the decomposition.

Common mistake avoided: Adding the full term-endowment value $0.672$ to the deferred value. That would count a maturity payment not present in whole-life insurance. The term-insurance part is $0.672-0.384=0.288$, and only that death-benefit value is combined with the deferred whole-life value.

Checks That Prevent Timing Errors

These calculations are most reliable when each term is tied to a specific event and payment time. Before substituting numbers, identify what must happen for each benefit to be paid. In particular, distinguish survival to the start of a year from survival through that year, and distinguish a death benefit paid at time $k+1$ from a survival benefit paid at time $n$.

Quantity	Event represented	Typical calculation
Death benefit in year $k+1$	Survive $k$ years, then die in the next year	$v^{k+1}{}_kp_xq_{x+k}$
Pure endowment at time $n$	Survive $n$ years and receive 1 then	$v^n{}_np_x$
Deferred whole-life benefit	Survive $n$ years, then receive the whole-life death benefit	$v^n{}_np_xA_{x+n}$

A practical check is to verify that the probabilities used for mutually exclusive outcomes cover the relevant possibilities. For an $n$-year term-endowment, death during the term and survival to its end exhaust the outcomes. For a whole-life decomposition at time $n$, death during the first $n$ years and death after time $n$ exhaust the possible death times. The maturity payment belongs only to the term-endowment, not to the whole-life policy.

Key takeaway: Use the term-endowment identity to isolate the death-benefit value, use the deferred factorization to value benefits after the deferral period, and use the whole-life recursion as an independent check. Keep benefit timing and survival conditions explicit at every step.

Check Your Understanding

For each problem, state the relevant event and payment timing before calculating.

A one-year term-endowment pays 1 at the end of the year on death or survival. If $i=10\%$, $q_x=0.10$, and $p_x=0.90$, find its term-insurance value, pure-endowment value, and total value. Give values to four decimal places. (Answer: $0.0909,\ 0.8182,\ 0.9091$)
For a unit whole-life policy, $i=25\%$, $q_x=0.30$, and $A_{x+1}=0.50$. Find $A_x$ using the one-year recursion. (Answer: $0.5200$)
At age $x$, $i=25\%$, $q_x=0.20$, and $q_{x+1}=0.25$. Find the value of a two-year unit term insurance with benefits paid at the end of the year of death. (Answer: $0.2880$)
A unit whole-life insurance is deferred for two years. If $i=25\%$, $p_x=0.80$, $p_{x+1}=0.75$, and $A_{x+2}=0.416$, find its value at age $x$. (Answer: $0.159744$)
A whole-life policy has $i=0\%$, $q_x=0.20$, $q_{x+1}=0.25$, and $A_{x+2}=0.40$. Use the two-year term-insurance value plus the two-year deferred whole-life value to find $A_x$. (Answer: $0.6400$)

Quantity	Event represented	Typical calculation
Death benefit in year \(k+1\)	Survive \(k\) years, then die in the next year	\(v^{k+1}{}_kp_xq_{x+k}\)
Pure endowment at time \(n\)	Survive \(n\) years and receive 1 then	\(v^n{}_np_x\)
Deferred whole-life benefit	Survive \(n\) years, then receive the whole-life death benefit	\(v^n{}_np_xA_{x+n}\)

What You'll Learn