Endowment Insurance as Term Insurance Plus Pure Endowment

Why an Endowment Has Two Components

An n-year endowment insurance pays a benefit if the insured dies during the contract term and also pays a benefit if the insured survives to the end of the term. Its value can therefore be understood as the sum of two familiar types of coverage: term insurance for death during the term, and a pure endowment for survival to the end. This separation is useful both for calculating values and for checking whether the benefits and payment times have been treated consistently.

The identity in this tutorial assumes the same benefit amount in both cases and annual discrete benefits: a death benefit is paid at the end of the year of death, and a survival benefit is paid at the end of year n. A different maturity benefit amount can be handled by scaling the pure-endowment component separately.

Definition: For a life aged x and a term of n years, let $A^1_{x:\overline{n}|}$ denote the actuarial present value of a unit benefit payable at the end of the year of death, if death occurs within n years. Let ${}_nE_x=v^n{}_np_x$ denote the actuarial present value of a unit benefit payable at time n if the insured survives n years. The n-year endowment value $A_{x:\overline{n}|}$ is the actuarial present value of a unit benefit payable at the end of the year of death if death occurs within n years, or at time n if the insured survives.

Here $v=1/(1+i)$, where i is the effective annual interest rate, and ${}_np_x$ is the probability that a life aged x survives n years. The notation ${}_kp_xq_{x+k}$ gives the probability of surviving k years and then dying in the following year. For k=0, ${}_0p_x=1$. These values are unit-benefit present values; a benefit of amount B has value B times the corresponding unit-benefit value.

The Endowment Identity

The insured either dies during the n-year term, or survives the full n years. These outcomes are mutually exclusive and together cover all possibilities under the model. In the first outcome, the endowment pays exactly the benefit described by n-year term insurance. In the second, it pays exactly the pure-endowment survival benefit. Because present value is calculated outcome by outcome, adding the two values gives the endowment value.

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

In expanded form, the two components are

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

Theorem: With the same unit benefit and the payment times defined above, the n-year endowment value is the sum of the n-year term-insurance value and the n-year pure-endowment value. Thus, if any two of the three values are known, the third follows by addition or subtraction.

Why the Identity Holds

The formula can also be justified directly from the benefit random variables. For each possible outcome, the present value of the endowment payment equals the present value of the term payment plus the present value of the pure-endowment payment. If death occurs during the term, the term payment is made and the pure-endowment payment is zero. If the insured survives to time n, the term payment is zero and the pure-endowment payment is made. Taking expectations preserves this sum, giving the identity.

Z_{\text{endowment}}=Z_{\text{term}}+Z_{\text{pure endowment}} \quad\Longrightarrow\quad E[Z_{\text{endowment}}] =E[Z_{\text{term}}]+E[Z_{\text{pure endowment}}].

This argument relies on matching the benefits and timing. The term component must pay only on death within the n-year term, while the pure endowment must pay only on survival to time n. If the death and maturity benefits differ, the unscaled identity for equal unit benefits does not apply directly to the whole contract. Instead, calculate the death-benefit amount times the term value and add the maturity-benefit amount times the pure-endowment value.

Worked Example: Calculate All Three Values from a Two-Year Model

A hypothetical policy pays $100,000 at the end of the year of death if death occurs within two years, or $100,000 at the end of year 2 if the insured survives. The effective annual interest rate is 5%. The annual mortality model has $q_x=0.10$ and $q_{x+1}=0.20$. Calculate the term-insurance, pure-endowment, and endowment values. Use the endowment identity as a check.

Method. First find the survival probabilities and discount factors. Then value the death and survival outcomes separately. This avoids treating the maturity benefit as though it were paid at the end of the year of death.

The one-year survival probabilities are $p_x=1-0.10=0.90$ and $p_{x+1}=1-0.20=0.80$. Thus, the probability of surviving two years is $0.90(0.80)=0.72$. The discount factor is $v=1/1.05=0.95238095$, so $v^2=0.90702948$.

The unit term-insurance value has two death-payment terms:

A^1_{x:\overline{2}|} =v(0.10)+v^2(0.90)(0.20) =0.09523810+0.16326531 =0.25850340.

The unit pure-endowment value is

{}_2E_x=v^2(0.72)=0.90702948(0.72)=0.65306122.

The unit endowment value is therefore $0.25850340+0.65306122=0.91156462$. Multiplying each unit value by the $100,000 benefit gives a term value of $25,850.34, a pure-endowment value of $65,306.12, and an endowment value of $91,156.46. Amounts are rounded to the nearest cent only at the final step.

Check. The component values add to $25,850.34 + $65,306.12 = $91,156.46, matching the endowment value. The method also confirms that the survival component uses the two-year survival probability and a two-year discount factor.

Recovering a Missing Component

The identity is equally useful when an endowment or one of its components is already known. Subtraction is appropriate only when the values refer to the same age, term, interest and mortality assumptions, and unit benefit. Confirm those conditions before using a reported value; otherwise, a numerical subtraction may combine incompatible quantities.

Worked Example: Recover the Term Value from Two Reported Values

For a hypothetical two-year contract, the effective annual interest rate is 25%, with $q_x=0.10$ and $q_{x+1}=0.20$. A valuation report gives a unit endowment value of 0.6560 and a unit pure-endowment value of 0.4608. Find the unit term-insurance value and check it using the mortality assumptions.

Method. Subtract the pure-endowment value from the endowment value. Then calculate the term value directly from the annual death probabilities as an independent check. The common mistake to avoid is subtracting a pure-endowment value based on a different benefit or duration.

By the identity,

A^1_{x:\overline{2}|} =A_{x:\overline{2}|}-{}_2E_x =0.6560-0.4608 =0.1952.

For the check, $v=1/1.25=0.8$, $p_x=0.9$, and $p_{x+1}=0.8$. The term value calculated from death in year 1 and death in year 2 is

A^1_{x:\overline{2}|} =(0.8)(0.10)+(0.8)^2(0.90)(0.20) =0.0800+0.1152 =0.1952.

The pure-endowment value also agrees with the stated report: $(0.8)^2(0.90)(0.80)=0.4608$. Adding the checked components gives $0.1952+0.4608=0.6560$, the reported endowment value.

Worked Example: Scale the Components to a Face Amount

A hypothetical three-year policy pays $50,000 at the end of the year of death if death occurs within three years, or $50,000 at the end of year 3 if the insured survives. The effective annual interest rate is 25%. The mortality model has $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$. Calculate the three present values.

Method. Calculate the unit-benefit values first, using each year's conditional death probability and the probability of reaching that year. Then multiply each component by $50,000. This keeps the probability calculations separate from benefit scaling.

The annual survival probabilities are $0.90$, $0.80$, and $0.75$. The probabilities of surviving to the start of years 1, 2, and 3 are $1$, $0.90$, and $0.90(0.80)=0.72$; the probability of surviving all three years is $0.90(0.80)(0.75)=0.54$. With $v=0.8$, the discount factors are $v=0.8$, $v^2=0.64$, and $v^3=0.512$.

The unit term-insurance value is

A^1_{x:\overline{3}|} =(0.8)(0.10)+(0.64)(0.90)(0.20)+(0.512)(0.72)(0.25) =0.08000+0.11520+0.09216 =0.28736.

The unit pure-endowment value is ${}_3E_x=(0.512)(0.54)=0.27648$. Therefore, the unit endowment value is $0.28736+0.27648=0.56384$. Multiplying by $50,000 gives a term value of $14,368, a pure-endowment value of $13,824, and an endowment value of $28,192.

Check. The component values sum to $14,368 + $13,824 = $28,192. The death probabilities used in the term calculation sum to $0.10+0.90(0.20)+0.72(0.25)=0.46$; adding the 0.54 probability of surviving three years gives 1. This confirms that the death and survival outcomes have been partitioned completely.

Common Mistakes and Their Corrections

The most frequent error is using a probability or discount factor for the wrong time. A pure endowment pays at time n only if the life survives n years, so its unit value is $v^n{}_np_x$. A term-insurance payment for death in year $k+1$ uses the probability ${}_kp_xq_{x+k}$ and discount factor $v^{k+1}$. Those factors cannot be interchanged.

For example, in the two-year model from the first worked example, a mistaken calculation might use $v^2p_x$ for the pure endowment, giving $0.90702948(0.90)=0.81632653$. That uses one-year survival rather than two-year survival. The correct probability is $p_xp_{x+1}=0.90(0.80)=0.72$, so the correct value is $0.90702948(0.72)=0.65306122$.

Another trap is to add values for different face amounts as if they were unit values. If the death benefit is $100,000 and the maturity benefit is $60,000, the present value is $100{,}000A^1_{x:\overline{n}|}+60{,}000{}_nE_x$, not $100{,}000A_{x:\overline{n}|}$. The simple identity applies directly when both benefits have the same amount.

Key takeaway: For equal unit benefits and the stated annual payment timing, $A_{x:\overline{n}|}=A^1_{x:\overline{n}|}+{}_nE_x$. Check the term, duration, benefit amount, mortality assumptions, and payment time before adding or subtracting values. This component view is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated annual mortality and interest assumptions. Round dollar amounts to the nearest cent where needed.

A hypothetical one-year policy pays $40,000 at the end of the year of death or at the end of year 1 if the insured survives. The effective annual interest rate is 4% and $q_x=0.15$. Find the term, pure-endowment, and endowment values. (Answer: $5,769.23, $32,692.31, and $38,461.54, respectively.)
For a two-year unit-benefit contract, $i=0$, $q_x=0.20$, and $q_{x+1}=0.10$. Calculate the term, pure-endowment, and endowment values. (Answer: 0.2800, 0.7200, and 1.0000, respectively.)
For a three-year contract, $i=0$, $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$. If the unit endowment value is 1.0000 and the unit pure-endowment value is 0.5400, recover the unit term-insurance value and verify it from the death probabilities. (Answer: 0.4600; $0.10+0.90(0.20)+0.90(0.80)(0.25)=0.4600$.)
A hypothetical two-year policy has equal $200,000 death and maturity benefits. Let $i=5\%$, $q_x=0.05$, and $q_{x+1}=0.10$. Calculate the term and pure-endowment values, then use their sum for the endowment value. (Answer: $26,757.37, $155,102.04, and $181,859.41, respectively.)
A one-year contract has a $100,000 death benefit and a $60,000 maturity benefit. The effective annual interest rate is 25% and $q_x=0.20$. Calculate its present value and explain why multiplying the equal-benefit endowment value by either benefit amount would be incorrect. (Answer: $54,400; use $100{,}000(0.8)(0.20)+60{,}000(0.8)(0.80)$, since the death and survival benefits differ.)

What You'll Learn