Recovering Missing Values from Multiple Recursions

Why Link More Than One Recursion?

A single insurance recursion relates a value at one age to the next age. In a problem with several related contract values, the most useful information may come from linking recursions: a whole-life value can reveal a mortality probability, which can then determine a term value; a term value and an endowment value can in turn determine a pure-endowment value. The calculation is not complete until the recovered quantities also agree with an independent identity.

This tutorial assumes annual discrete benefits, with death benefits paid at the end of the year of death and any maturity benefit paid at the end of the term. These timing assumptions matter: changing the payment time changes the present values and their recursions. The techniques here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $i$ be the effective annual interest rate, $v=1/(1+i)$, ${}_1q_x$ the probability a life aged $x$ dies within one year, and ${}_1p_x=1-{}_1q_x$ the probability of surviving one year. Let $A_x$ be the expected present value of a unit whole-life insurance benefit. Let $T_{x,n}$ be the expected present value of a unit $n$-year term insurance, $E_{x,n}$ the expected present value of a unit pure endowment payable at time $n$ if the life survives, and $M_{x,n}$ the expected present value of a unit $n$-year endowment insurance. In this tutorial, the term insurance pays on death within $n$ years, while the endowment insurance pays on death within $n$ years or on survival to time $n$, but not both.

The Linked Recursions and Identities

The following formulas apply to the annual discrete benefit timing just defined. The term recursion ends with $T_{x,0}=0$: after the term has expired, there is no remaining term insurance benefit. The pure-endowment boundary is $E_{x,0}=1$, since a payment due immediately is not discounted and requires no future survival. Similarly, $M_{x,0}=1$.

\begin{aligned} A_x &= v\left({}_1q_x+{}_1p_x A_{x+1}\right),\\ T_{x,n} &= v\,{}_1q_x+v\,{}_1p_x T_{x+1,n-1},\\ E_{x,n} &= v\,{}_1p_x E_{x+1,n-1},\\ M_{x,n} &= v\,{}_1q_x+v\,{}_1p_x M_{x+1,n-1}. \end{aligned}

In the whole-life recursion, death during the first year produces a unit payment at year-end, while survival leaves a whole-life contract valued at age $x+1$. The other recursions use the same first-year split but have different benefits after survival. A term policy has $n-1$ years remaining; a pure endowment requires survival through the year; and an endowment policy retains both its death and maturity benefits.

Theorem: With unit benefits and the payment timing above, the $n$-year endowment value can be calculated in either of two ways: $$ M_{x,n}=T_{x,n}+E_{x,n}. $$ If the whole-life insurance continues after the $n$-year period, its value also satisfies $$ A_x=T_{x,n}+E_{x,n}A_{x+n}. $$

The first identity follows because the endowment benefit occurs on exactly one of two disjoint outcomes: death within $n$ years or survival to time $n$. The second identity partitions a whole-life contract into death within the first $n$ years and survival to time $n$, followed by whole-life coverage from age $x+n$. The pure endowment factor $E_{x,n}$ accounts for both survival and discounting to time $n$.

A Reliable Recovery Process

When several values or probabilities are missing, first identify which equation contains the unknown and use only the values that are given. Then substitute the recovered quantity into the remaining equations. Avoid rounding intermediate values: a small rounding error can make two correct identities appear inconsistent.

State the model.
Record the annual effective interest rate, the discount factor, the one-year mortality probabilities, and the benefit timing.

Choose an equation containing the unknown.
For example, isolate ${}_1q_x$ from the whole-life recursion or isolate a remaining term value from the term recursion.

Recover the linked values.
Substitute the result into the term, pure-endowment, or endowment equation as needed.

Check independently.
Verify the endowment sum identity or the whole-life decomposition, not just the equation used to find the unknown.

Worked Calculations

Worked Example: Recovering Two Mortality Probabilities and an Endowment Value

A hypothetical insurer values unit benefits at an effective annual interest rate of $25\%$, so $v=0.8$. The reported whole-life values are $A_x=0.416$ and $A_{x+1}=0.400$. A two-year term value is $T_{x,2}=0.288$. Recover ${}_1q_x$, ${}_1q_{x+1}$, $E_{x,2}$, and $M_{x,2}$, and verify the endowment value by recursion.

Method and first recovery. Use the whole-life recursion at age $x$. Write $q={}_1q_x$, so ${}_1p_x=1-q$:

0.416=0.8\left(q+(1-q)(0.400)\right) =0.8(0.400+0.600q) =0.320+0.480q.

Thus $0.096=0.480q$, giving ${}_1q_x=0.200$ and ${}_1p_x=0.800$.

Recover the next mortality probability. Put $r={}_1q_{x+1}$. The two-year term recursion gives:

0.288=0.8(0.200)+0.8(0.800)\left[0.8r\right] =0.160+0.512r.

Therefore $0.128=0.512r$, so ${}_1q_{x+1}=0.250$ and ${}_1p_{x+1}=0.750$.

Calculate the pure endowment. Survival through both years is $0.800(0.750)=0.600$, and the two-year discount factor is $v^2=0.640$. Hence:

E_{x,2}=v^2\,{}_1p_x\,{}_1p_{x+1} =0.640(0.800)(0.750)=0.384.

Calculate and check the endowment. The sum identity gives $M_{x,2}=0.288+0.384=0.672$. For an independent recursion check, the one-year endowment value at age $x+1$ is $M_{x+1,1}=0.8[0.250+0.750(1)]=0.800$. Then:

M_{x,2}=0.8\left(0.200+0.800(0.800)\right) =0.8(0.840)=0.672.

Both calculations agree. The method avoids treating $T_{x,2}$ as if it were the endowment value: the pure-endowment component, $0.384$, must also be included.

Worked Example: Recovering a Pure Endowment and Testing a Whole-Life Identity

At an effective annual interest rate of $25\%$, a hypothetical two-year contract has a term insurance value of $T_{x,2}=0.1952$ and an endowment value of $M_{x,2}=0.6560$. The one-year death probabilities are ${}_1q_x=0.10$ and ${}_1q_{x+1}=0.20$. Recover $E_{x,2}$, check it directly from survival, and verify the values against a whole-life continuation with $A_{x+2}=0.4000$.

Recover the missing value from the linked identity. Since an endowment is term insurance plus a pure endowment:

E_{x,2}=M_{x,2}-T_{x,2}=0.6560-0.1952=0.4608.

Check the term value from mortality. Here $v=1/1.25=0.8$, ${}_1p_x=0.90$, and ${}_1p_{x+1}=0.80$. The two-year term benefit can occur in year 1 or year 2:

T_{x,2}=0.8(0.10)+0.8(0.90)\left[0.8(0.20)\right] =0.0800+0.1152=0.1952.

Check the pure endowment from survival. Survival through both years has probability $0.90(0.80)=0.72$, and $v^2=0.64$. Therefore:

E_{x,2}=0.64(0.72)=0.4608.

The independently calculated components give $T_{x,2}+E_{x,2}=0.1952+0.4608=0.6560$, matching the reported $M_{x,2}$.

Verify the whole-life continuation. With $A_{x+2}=0.4000$, the whole-life value at age $x$ implied by the decomposition is:

A_x=T_{x,2}+E_{x,2}A_{x+2} =0.1952+0.4608(0.4000)=0.37952.

The result is consistent with the two-year split of the whole-life contract. The important check is that the continuation value is multiplied by the survival-and-discount factor $E_{x,2}$, not by survival probability alone.

Worked Example: Checking Linked Values Across Three Ages

A hypothetical contract uses an effective annual interest rate of $20\%$, so $v=5/6$, with ${}_1q_x=0.10$ and ${}_1q_{x+1}=0.20$. A continuation whole-life value is $A_{x+2}=0.4000$. Calculate $T_{x,2}$, $E_{x,2}$, and $M_{x,2}$, then find $A_x$ by two routes and check that the results agree.

Calculate the two-year term value. The year-1 contribution is $v{}_1q_x=(5/6)(0.10)=1/12$. The year-2 contribution is $v{}_1p_x\cdot v{}_1q_{x+1}=(5/6)(0.90)(5/6)(0.20)=1/8$. Thus:

T_{x,2}=\frac{1}{12}+\frac{1}{8} =\frac{5}{24}=0.208333\ldots.

Calculate the pure endowment and endowment. The two-year survival probability is $0.90(0.80)=0.72$, and $v^2=25/36$. Therefore $E_{x,2}=(25/36)(0.72)=0.5000$, and:

M_{x,2}=T_{x,2}+E_{x,2} =\frac{5}{24}+\frac{1}{2} =\frac{17}{24}=0.708333\ldots.

First route to $A_x$. Apply the whole-life recursion at age $x+1$, then at age $x$:

A_{x+1}=\frac{5}{6}\left(0.20+0.80(0.4000)\right) =\frac{13}{30}=0.433333\ldots,

$$ A_x=\frac{5}{6}\left(0.10+0.90\frac{13}{30}\right) =\frac{49}{120}=0.408333\ldots. $$

Second route and check. Split the whole-life contract after two years:

A_x=T_{x,2}+E_{x,2}A_{x+2} =\frac{5}{24}+\frac{1}{2}(0.4000) =\frac{49}{120}=0.408333\ldots.

The routes agree. This check tests the timing and survival discounting across the full two-year period, not merely the arithmetic in one recursion.

Common Exam Traps

A frequent error is to compute the second-year death benefit using only the probability of dying in that year. The life must first survive year 1, and the year-2 payment is discounted for two years. In the second worked example, the incorrect calculation $0.8(0.20)=0.1600$ omits both year-1 survival and the additional discount. The correct year-2 term contribution is $0.8(0.90)\cdot0.8(0.20)=0.1152$.

Another trap is to mix up the endowment and pure-endowment components. Under the stated benefit definitions, the term and pure-endowment events are mutually exclusive, so their present values add to the endowment value. Also, a whole-life continuation value after time $n$ is multiplied by $E_{x,n}$, which includes both discounting and survival. Every verification must use the same interest rate, mortality probabilities, and payment timing as the original calculation.

Key takeaway: Recover missing quantities by isolating them in a recursion, then verify the result with a different identity. In particular, check $M_{x,n}=T_{x,n}+E_{x,n}$ and, when a continuation whole-life value is available, $A_x=T_{x,n}+E_{x,n}A_{x+n}$. Agreement across identities is a strong check on both the arithmetic and the contract timing.

Check Your Understanding

Use annual discrete benefits and retain full precision until the final answer.

At $i=25\%$, suppose ${}_1q_x=0.10$ and ${}_1q_{x+1}=0.20$. Calculate $T_{x,2}$, $E_{x,2}$, and $M_{x,2}$. (Answer: $T_{x,2}=0.1952,\ E_{x,2}=0.4608,\ M_{x,2}=0.6560$)
At $i=20\%$, suppose ${}_1q_x=0.20$ and ${}_1q_{x+1}=0.10$. Calculate the two-year term, pure-endowment, and endowment values. (Answer: $T_{x,2}=2/9=0.2222\ldots,\ E_{x,2}=0.5000,\ M_{x,2}=13/18=0.7222\ldots$)
At $i=25\%$, $A_x=0.440$ and $A_{x+1}=0.500$. Also $T_{x,2}=0.224$. Recover ${}_1q_x$, ${}_1q_{x+1}$, $E_{x,2}$, and $M_{x,2}$. (Answer: ${}_1q_x=0.10,\ {}_1q_{x+1}=0.25,\ E_{x,2}=0.432,\ M_{x,2}=0.656$)
At $i=10\%$, ${}_1q_x=0.20$ and ${}_1q_{x+1}=0.30$. Calculate $T_{x,2}$, $E_{x,2}$, and $M_{x,2}$, giving the exact fractions. (Answer: $T_{x,2}=46/121,\ E_{x,2}=56/121,\ M_{x,2}=102/121$)
At $i=25\%$, with ${}_1q_x=0.10$ and ${}_1q_{x+1}=0.20$, a report lists $T_{x,2}=0.1952$, $E_{x,2}=0.4608$, and $M_{x,2}=0.6688$. Identify the inconsistency and give the corrected endowment value. (Answer: The components sum to $0.1952+0.4608=0.6560$, so $M_{x,2}=0.6688$ is inconsistent; the corrected value is $0.6560$.)

What You'll Learn