Derive One-Year Recursions for Whole-Life Insurance

Why a One-Year Recursion Is Useful

A whole-life insurance value includes possible payments at every future duration. Calculating each duration separately can be cumbersome. The one-year recursion reduces the valuation to two cases: death during the first policy year, or survival to the next policy year. It is useful only when the benefit amount and payment timing in those cases are made explicit.

This tutorial considers a discrete whole-life insurance issued at exact age $x$. The insured’s future lifetime is measured in years, and the benefit is paid at the end of the year of death. The effective annual interest rate is $i$, with discount factor $v=1/(1+i)$. Mortality probabilities may vary by age; they are conditional on the insured being alive at the start of the year.

Definition: $A_x$ is the expected present value, at age $x$, of a unit whole-life insurance benefit paid at the end of the year of death. Write ${}_1q_x$ for the probability of death during the year from age $x$ to age $x+1$, and ${}_1p_x=1-{}_1q_x$ for the probability of surviving that year. The recursion below assumes a level benefit, a fixed effective annual interest rate, and payment at the end of the year of death.

Deriving the One-Year Recursion

At the start of the policy year, there are two mutually exclusive possibilities. If the insured dies during that year, a unit benefit is paid at the end of that year. If the insured survives, the contract continues as a whole-life insurance at age $x+1$; its value at age $x+1$ is $A_{x+1}$. Both payments or values must be discounted from the end of the first year to the start of the year.

Separate the first-year outcomes.
Death during the first year has probability ${}_1q_x$; survival through the first year has probability ${}_1p_x$.

Assign the value conditional on each outcome.
If death occurs, the unit benefit is paid at time 1 and has present value $v$. If the insured survives, the remaining whole-life insurance has value $A_{x+1}$ at time 1 and present value $vA_{x+1}$ at time 0.

Take the probability-weighted average.
Multiply each conditional present value by the probability of its outcome, then add the two terms.

A_x={}_1q_xv+{}_1p_xvA_{x+1} =v\left({}_1q_x+{}_1p_xA_{x+1}\right).

This is a conditioning argument: the expected present value is the sum of the present values conditional on each first-year outcome, weighted by the probabilities of those outcomes. The death benefit is discounted by $v$ because it is paid at time 1. The continuing insurance value is also discounted by $v$, because $A_{x+1}$ is measured one year later. The probabilities sum to one, since ${}_1q_x+{}_1p_x=1$.

Theorem: For a discrete whole-life insurance with unit benefit paid at the end of the year of death, $$ A_x=v\left({}_1q_x+{}_1p_xA_{x+1}\right). $$ For a level benefit of $B$ dollars, the expected present value is $BA_x$. The recursion does not apply unchanged if the benefit is paid immediately on death, at a different time, or varies by duration.

When Mortality Is Constant After a Certain Age

A useful way to obtain a starting value for a sequence of recursions is to consider an age from which the one-year death probability is constant. Suppose the probability of death is $q$ at that age and at every future age, and let $p=1-q$. The value at that age is the same as the value one year later, because the interest and mortality assumptions are unchanged. Calling that common value $A$, the one-year recursion gives $A=v(q+pA)$. Solving this equation yields the constant-mortality whole-life value.

A=v(q+pA) \quad\Longrightarrow\quad A(1-vp)=vq \quad\Longrightarrow\quad A=\frac{vq}{1-vp}.

This expression is valid when the same $q$ and $p$ apply at every future integer age, $v$ is the same each year, and $1-vp\ne0$. With a positive interest rate and $0\le p\le1$, the denominator is positive. It is not a general formula for ages with changing mortality; in that case, use the one-year recursion with the appropriate age-specific probabilities.

Worked Example: A Different First-Year Mortality Rate

A hypothetical policy pays one unit at the end of the year of death. The effective annual interest rate is $25\%$. The one-year death probability is $0.10$ at age $x$, and it is $0.20$ at age $x+1$ and every integer age thereafter. Find $A_x$. State the method and identify a timing error to avoid.

Method and assumptions. Use the constant-mortality formula to value the insurance from age $x+1$, then apply the one-year recursion at age $x$. The discount factor is $v=1/1.25=0.8$. For ages $x+1$ onward, $q=0.20$ and $p=0.80$.

Value from age $x+1$.

A_{x+1} =\frac{vq}{1-vp} =\frac{0.8(0.20)}{1-0.8(0.80)} =\frac{0.16}{0.36} =\frac{4}{9}.

At age $x$, ${}_1q_x=0.10$ and ${}_1p_x=0.90$. Therefore,

A_x =0.8\left(0.10+0.90\left(\frac{4}{9}\right)\right) =0.8(0.10+0.40) =0.8(0.50) =0.40.

Check. The first-year death contribution is $0.8(0.10)=0.08$. The survival contribution is $0.8(0.90)(4/9)=0.32$. Their sum is $0.08+0.32=0.40$, as obtained by the recursion. The common timing error is to leave the death contribution undiscounted; the benefit is paid at time 1, not at time 0.

Applying the Recursion Across Several Ages

When mortality changes with age, calculate the value at a constant-mortality tail first, then work backward one age at a time. At each step, use that age’s own death and survival probabilities. The next-age value is an input to the recursion, not a substitute for the current age’s mortality probability.

Worked Example: Two Changing Years Before a Constant Tail

A hypothetical unit whole-life benefit is paid at the end of the year of death. The effective annual interest rate is $10\%$. The death probabilities are $0.10$ at age $x$, $0.20$ at age $x+1$, and $0.25$ at age $x+2$ and every integer age thereafter. Find $A_x$, rounding only the final answer to four decimal places.

Method and assumptions. Set $v=1/1.10=10/11$. First value the constant-mortality tail at age $x+2$, where $q=0.25$ and $p=0.75$. Then apply the one-year recursion at ages $x+1$ and $x$.

Tail value.

A_{x+2} =\frac{(10/11)(0.25)}{1-(10/11)(0.75)} =\frac{2.5/11}{3.5/11} =\frac{5}{7}.

Value at age $x+1$. Here ${}_1q_{x+1}=0.20$, so ${}_1p_{x+1}=0.80$.

A_{x+1} =\frac{10}{11}\left(0.20+0.80\left(\frac{5}{7}\right)\right) =\frac{10}{11}\left(\frac{1}{5}+\frac{4}{7}\right) =\frac{10}{11}\left(\frac{27}{35}\right) =\frac{54}{77}.

Value at age $x$. Here ${}_1q_x=0.10$ and ${}_1p_x=0.90$.

A_x =\frac{10}{11}\left(0.10+0.90\left(\frac{54}{77}\right)\right) =\frac{10}{11}\left(\frac{563}{770}\right) =\frac{563}{847} =0.6646989\ldots

Answer and check. Rounded to four decimal places, $A_x=0.6647$. At age $x$, the death contribution is $(10/11)(0.10)$, and the survival contribution is $(10/11)(0.90)(54/77)$; these are exactly the two terms in the displayed recursion. Using $0.25$ as the death probability at age $x$ would incorrectly apply the tail mortality to an earlier age.

Changing the Benefit Amount

If the benefit is a fixed amount $B$, the conditional death payment in the first year is $B$ rather than one unit. The continuing contract after survival still has expected present value $BA_{x+1}$, measured at age $x+1$. Thus the same conditioning argument gives $B A_x=v({}_1q_xB+{}_1p_xBA_{x+1})$. A fixed benefit scales the value; it does not change the probabilities or discounting.

Worked Example: A Dollar Benefit with Age-Dependent Mortality

A hypothetical policy pays $ \$50{,}000 $ at the end of the year of death. The effective annual interest rate is $4\%$. The death probabilities are $0.05$ at age $x$, $0.10$ at age $x+1$, and $0.20$ at age $x+2$ and every integer age thereafter. Find the expected present value at age $x$, rounded to the nearest cent.

Method and assumptions. Calculate the unit value backward from the constant-mortality tail, then multiply by the face amount. The discount factor is $v=1/1.04=25/26$.

Tail at age $x+2$. For $q=0.20$ and $p=0.80$,

A_{x+2} =\frac{(25/26)(0.20)}{1-(25/26)(0.80)} =\frac{5/26}{6/26} =\frac{5}{6}.

Value at age $x+1$. The survival probability is $0.90$.

A_{x+1} =\frac{25}{26}\left(0.10+0.90\left(\frac{5}{6}\right)\right) =\frac{25}{26}(0.85) =\frac{85}{104}.

Value at age $x$. The survival probability is $0.95$.

A_x =\frac{25}{26}\left(0.05+0.95\left(\frac{85}{104}\right)\right) =\frac{25}{26}\left(\frac{1719}{2080}\right) =\frac{42975}{54080} =0.7946560\ldots

Scale to the stated benefit.

\$50{,}000(0.7946560\ldots)=\$39{,}732.80.

The expected present value is $ \$39{,}732.80 $, rounded to the nearest cent. The unit value is less than one, as expected for a benefit paid in the future and discounted at a positive interest rate. The face amount is multiplied after calculating $A_x$; it is not added to the unit value.

Common Mistakes and Scope

The recursion is simple, but its terms represent different events and times. A calculation should identify the first-year death probability, the survival probability, the time of benefit payment, and the date at which the continuation value is measured.

Omitting discount on the death benefit. For example, with $i=25\%$, ${}_1q_x=0.10$, and $A_{x+1}=4/9$, the incorrect expression $0.10+0.90(0.8)(4/9)=0.42$ leaves the first-year death benefit undiscounted. The correct value is $0.8[0.10+0.90(4/9)]=0.40$.
Discounting the continuation value twice. $A_{x+1}$ is already the value at time 1. Discount it once to time 0, giving $vA_{x+1}$, and then multiply by the probability of survival.
Using the wrong age’s mortality. The first-year death probability is ${}_1q_x$; after survival, the next recursion uses ${}_1q_{x+1}$. A constant-mortality tail does not imply that mortality was constant at earlier ages.
Changing the payment timing without changing the formula. The factor $v$ on the death branch follows from payment at the end of the year of death. A benefit paid immediately on death has different timing and requires a different valuation.
Assuming the constant-tail formula works everywhere. $vq/(1-vp)$ requires constant mortality from the valuation age onward. For age-varying mortality, use the one-year recursion at each age.

The recursion is a basic calculation tool for discrete whole-life insurance, and a useful check on valuations assembled from year-by-year death probabilities. Correctly aligning conditional probabilities with payment dates is central to its use. The topic is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume a level benefit paid at the end of the year of death and an effective annual interest rate as stated. Use the one-year recursion and round only the final answer when requested.

A unit whole-life benefit has $i=20\%$, and the death probability is $0.20$ at age $x$ and every age thereafter. Find $A_x$. (Answer: $0.5000$)
A unit benefit has $i=25\%$, ${}_1q_x=0.10$, and death probability $0.20$ at age $x+1$ and every age thereafter. Find $A_x$. (Answer: $0.4000$)
A unit benefit has $i=10\%$, ${}_1q_x=0.30$, and death probability $0.10$ at age $x+1$ and every age thereafter. Find $A_x$, to four decimal places. (Answer: $0.5909$)
A unit benefit has $i=5\%$, ${}_1q_x=0.08$, ${}_1q_{x+1}=0.12$, and death probability $0.20$ at age $x+2$ and every age thereafter. Find $A_x$, to four decimal places. (Answer: $0.7638$)
A $ \$100{,}000 $ whole-life benefit has $i=4\%$, ${}_1q_x=0.15$, and $A_{x+1}=0.60$. Find its expected present value at age $x$, to the nearest cent. (Answer: $ \$63{,}461.54 $)

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