Solving Whole-Life Recursion for Mortality

Why Solve the Recursion for Mortality?

A whole-life insurance value reflects both the interest rate and the chance of death at each age. When the interest rate and two consecutive insurance values are known, the one-year recursion can be rearranged to infer the mortality probability at the younger age. This is useful for checking whether a set of values is internally consistent with the assumed contract and interest basis.

The inferred probability is not automatically a valid mortality probability. The calculation must be checked against the range from zero to one, and the inputs must refer to the same benefit design, timing convention, and valuation basis. The method below is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: $q_x$ is the probability that a life aged $x$ dies within one year, conditional on being alive at age $x$. The corresponding one-year survival probability is $p_x=1-q_x$. $A_x$ denotes the present value of a unit whole-life insurance benefit payable at the end of the year of death, under the discrete annual model.

The Model and the Rearranged Formula

Assume an effective annual interest rate $i$, with discount factor $v=1/(1+i)$. The insured is alive at age $x$, the benefit is one unit, and it is paid at the end of the year of death. If the life survives to age $x+1$, the remaining contract has value $A_{x+1}$ at that age. The one-year recursion is

A_x=v\left(q_x+p_x A_{x+1}\right) =v\left(q_x+(1-q_x)A_{x+1}\right).

To infer $q_x$, first multiply by $1+i$. Then collect the terms containing $q_x$. This rearrangement requires $A_{x+1}\ne 1$; under the usual positive-interest unit-benefit setting, the values are ordinarily below one.

\begin{aligned} (1+i)A_x &=q_x+(1-q_x)A_{x+1}\\ &=A_{x+1}+q_x(1-A_{x+1}),\\[4pt] q_x &=\frac{(1+i)A_x-A_{x+1}}{1-A_{x+1}}. \end{aligned}

Theorem: Under the stated discrete unit whole-life assumptions, if $A_{x+1}\ne 1$, then $$ q_x=\frac{(1+i)A_x-A_{x+1}}{1-A_{x+1}}. $$ For $A_{x+1}<1$, the inferred value is a valid probability exactly when $$ A_{x+1}\le (1+i)A_x\le 1. $$

The bounds follow directly from the formula. When $A_{x+1}<1$, its denominator is positive. Thus $q_x\ge 0$ requires $(1+i)A_x-A_{x+1}\ge 0$, or $A_{x+1}\le(1+i)A_x$. Similarly, $q_x\le1$ requires $(1+i)A_x-A_{x+1}\le1-A_{x+1}$, which simplifies to $(1+i)A_x\le1$. These are feasibility checks, not optional rounding conventions.

A Short Derivation: Why the Recursion Holds

Condition on the outcome during the first year. With probability $q_x$, death occurs in that year and the unit benefit is paid at its end. Its present value at age $x$ is $v$. With probability $p_x$, the life survives; the contract then has value $A_{x+1}$ one year later, whose value at age $x$ is $vA_{x+1}$. Taking the probability-weighted present value gives

A_x=q_xv+p_xvA_{x+1} =v\left(q_x+(1-q_x)A_{x+1}\right).

The payment timing is essential to this argument: the death benefit is discounted one year because it is paid at the end of the year of death. A benefit payable immediately at death would have a different value and would not use this discrete recursion.

Worked Calculations

Worked Example: Infer Mortality from Two Consecutive Values

A hypothetical unit whole-life contract uses an effective annual interest rate of $5\%$. Its discrete insurance values are $A_x=0.60$ and $A_{x+1}=0.55$. Infer $q_x$, check that it is a valid probability, and verify the recursion. Treat the given values as exact.

Method. Use the rearranged one-year recursion. This avoids trying to infer mortality from either insurance value alone, which would not provide enough information.

First calculate $1+i=1.05$. The numerator is

(1+i)A_x-A_{x+1} =1.05(0.60)-0.55 =0.63-0.55 =0.08.

The denominator is $1-A_{x+1}=1-0.55=0.45$. Therefore

q_x=\frac{0.08}{0.45}=0.177777\ldots\approx 0.1778.

The result is between zero and one, so it is a valid probability. The implied survival probability is $p_x=1-q_x=0.822222\ldots$. Verify by substituting the unrounded probability into the recursion:

\begin{aligned} v\left(q_x+p_xA_{x+1}\right) &=\frac{0.177777\ldots+(0.822222\ldots)(0.55)}{1.05}\\ &=\frac{0.177777\ldots+0.452222\ldots}{1.05}\\ &=\frac{0.630000\ldots}{1.05} =0.60=A_x. \end{aligned}

Answer: $q_x\approx0.1778$, or about $17.78\%$, to four decimal places. The substitution recovers the stated $A_x$, as required.

Worked Example: Infer Mortality and Check Both Probability Bounds

For a hypothetical unit whole-life policy, the effective annual interest rate is $4\%$, $A_x=0.65$, and $A_{x+1}=0.60$. Determine the one-year death probability and show explicitly that it satisfies both probability bounds.

Method. Calculate the numerator and denominator separately, then check the result and substitute it back. The common mistake here is to use $iA_x$ in place of $(1+i)A_x$, omitting the original value $A_x$ when multiplying by $1+i$.

The numerator is

(1+i)A_x-A_{x+1} =1.04(0.65)-0.60 =0.676-0.60 =0.076.

The denominator is $1-A_{x+1}=0.40$. Hence

q_x=\frac{0.076}{0.40}=0.19.

This lies between zero and one, and $p_x=1-0.19=0.81$. The equivalent feasibility bounds also hold:

0.60\le 1.04(0.65)=0.676\le1.

For a direct check, evaluate the recursion using $v=1/1.04$:

\begin{aligned} v\left(q_x+p_xA_{x+1}\right) &=\frac{0.19+(0.81)(0.60)}{1.04}\\ &=\frac{0.19+0.486}{1.04}\\ &=\frac{0.676}{1.04} =0.65. \end{aligned}

Answer: $q_x=0.19$, or $19\%$. Both the probability bounds and the original recursion are satisfied.

Worked Example: Reject an Infeasible Pair of Values

A hypothetical valuation reports $i=4\%$, $A_x=0.80$, and $A_{x+1}=0.85$ for a unit whole-life insurance with end-of-year death benefit. Determine whether these values imply a valid $q_x$ under the stated model.

Method. Apply the rearrangement and interpret the sign of the answer. The common mistake is to report any algebraic result as a mortality probability without checking its range.

The numerator is

(1+i)A_x-A_{x+1} =1.04(0.80)-0.85 =0.832-0.85 =-0.018.

The denominator is $1-A_{x+1}=0.15$, so the algebraic result is

q_x=\frac{-0.018}{0.15}=-0.12.

A negative value is not a valid probability. The feasibility check shows the same problem: $0.85\le0.832$ is false, so the lower bound fails. Substitution would reproduce the reported $A_x$ only by using the impossible value $-0.12$; it cannot make the assumed model and data valid.

Answer: These values are inconsistent with the stated recursion. The inferred algebraic value is $-0.12$, not a valid mortality probability.

Interpretation, Assumptions, and Exam Traps

The formula isolates one annual mortality probability from two age-indexed insurance values. It does not estimate a full mortality table, nor does it remove the need to identify the contract assumptions. Before using it, confirm that both values are for the same unit whole-life benefit and the same interest and mortality basis, and that $A_x$ and $A_{x+1}$ use end-of-year death-benefit timing.

Forgetting the factor $1+i$. The rearrangement begins with $(1+i)A_x$, not $iA_x$. The factor arises from multiplying the recursion by $1+i$.
Reversing the age values. $A_x$ is the value at the younger age and $A_{x+1}$ is the continuation value after surviving one year. The numerator is $(1+i)A_x-A_{x+1}$.
Skipping the range check. A calculated number below zero or above one signals that the inputs cannot satisfy this model as exact values. It is not a valid estimate of $q_x$.
Mixing benefit timing or product types. The formula is for a discrete whole-life insurance with a unit death benefit payable at the end of the year of death. It should not be applied unchanged to a continuous benefit or a contract with a different benefit pattern.
Ignoring rounding. If the reported $A$-values have been rounded, a small apparent violation of a bound may reflect rounding. Do not silently alter the formula; state whether the values are treated as exact, or whether rounding could account for the discrepancy.

If $A_{x+1}=1$, the denominator is zero and the rearrangement cannot identify $q_x$. In fact, the recursion becomes $A_x=v$ regardless of $q_x$, so these two values do not determine mortality in that special case. Under ordinary positive-interest unit-benefit whole-life assumptions, this is not the usual situation, but the denominator condition should still be recognized.

Key takeaway: Compute $q_x$ as $\big((1+i)A_x-A_{x+1}\big)/(1-A_{x+1})$, provided $A_{x+1}\ne1$. Then check $0\le q_x\le1$, or equivalently, when $A_{x+1}<1$, check $A_{x+1}\le(1+i)A_x\le1$. Finally, substitute the result into the original recursion.

Check Your Understanding

For each problem, assume a discrete unit whole-life benefit payable at the end of the year of death. Treat all stated values as exact unless otherwise specified.

A hypothetical contract has $i=5\%$, $A_x=0.60$, and $A_{x+1}=0.50$. Infer $q_x$ and state whether it is valid. (Answer: $q_x=0.2600$, valid.)
For $i=3\%$, $A_x=0.50$, and $A_{x+1}=0.45$, calculate $q_x$ to four decimal places and give $p_x$ to four decimal places. (Answer: $q_x=0.1182$, $p_x=0.8818$.)
A valuation gives $i=6\%$, $A_x=0.70$, and $A_{x+1}=0.68$. Find $q_x$ to four decimal places and verify the feasibility bounds. (Answer: $q_x=0.1938$; $0.68\le0.742\le1$, so it is valid.)
With $i=4\%$, $A_x=0.70$, and $A_{x+1}=0.75$, determine whether the values are feasible. (Answer: No; $1.04(0.70)=0.728<0.75$, giving $q_x=-0.0880$.)
For $i=5\%$, $A_x=0.95$, and $A_{x+1}=0.80$, infer $q_x$, verify it is a probability, and state the implied survival probability. (Answer: $q_x=0.9875$, valid; $p_x=0.0125$.)

What You'll Learn