Why Invert an Insurance Recursion?
A one-year insurance recursion usually calculates a current-age value from an interest rate, a mortality probability, and a value at the next age. In an integrated calculation, however, the current value and some of those inputs may be known while one quantity is missing. Rearranging the recursion can recover that quantity, but only if the supplied values are consistent and the unknown is identifiable.
This tutorial concerns the recursion for a unit whole-life insurance benefit payable at the end of the year of death. The same discipline applies throughout: state the timing convention, isolate the unknown algebraically, and then check the result against the model's constraints.
The assumptions are important. The benefit is one unit, death benefits are paid at the end of the year of death, and if the insured survives the year, the remaining insurance has value \(A_{x+1}\) at time 1. Interest is earned at rate \(i\) over the year. The recursion is not the corresponding formula for a benefit payable immediately at death, nor does it apply unchanged to a different benefit amount or payment timing.
Rearranging the Recursion
Write \(B_x=q_x+(1-q_x)A_{x+1}\) for the bracketed one-year expected amount. Then \(A_x=vB_x\), or equivalently \(A_x(1+i)=B_x\). These two forms make different inversions convenient.
If \(i\), \(A_x\), and \(A_{x+1}\) are known and \(A_{x+1}\ne1\), solve \(q_x=\dfrac{A_x(1+i)-A_{x+1}}{1-A_{x+1}}\). Then check \(0\le q_x\le1\).
If \(q_x\), \(A_x\), and \(A_{x+1}\) are known and \(A_x>0\), solve \(i=\dfrac{q_x+(1-q_x)A_{x+1}}{A_x}-1\). Check that the result matches the assumed range of interest rates.
If \(i\), \(q_x\), and \(A_x\) are known and \(q_x<1\), solve \(A_{x+1}=\dfrac{A_x(1+i)-q_x}{1-q_x}\). Check that the result is compatible with a unit-benefit insurance value.
The rearrangements are useful only when their denominators are nonzero. A zero denominator is not a signal to divide by zero; it calls for checking the original recursion directly. In particular, the unknown may be unidentified, or the supplied quantities may be inconsistent.
Recovering the One-Year Mortality Probability
Worked Example: Infer Mortality from Two Insurance Values
A hypothetical whole-life contract has a unit benefit payable at the end of the year of death. The effective annual interest rate is \(4\%\), \(A_x=0.60\), and \(A_{x+1}=0.50\). Find \(q_x\). State the method and check the result.
Method. Use the mortality inversion, first expressing the recursion without the discount factor: $$ A_x(1+i)=q_x+(1-q_x)A_{x+1}. $$
Substitute the given values. The left side is \(0.60(1.04)=0.624\). The right side is \(q_x+(1-q_x)(0.50)=q_x+0.50-0.50q_x=0.50+0.50q_x\). Thus $$ 0.624=0.50+0.50q_x,\qquad q_x=\frac{0.124}{0.50}=0.248. $$
Therefore \(q_x=0.248\), which is within the probability range \([0,1]\). As a check, \(p_x=1-0.248=0.752\), so \(v(q_x+p_xA_{x+1})=(1/1.04)(0.248+0.752(0.50))=(1/1.04)(0.624)=0.60\), as required.
Common mistake avoided: Omitting the discount factor would equate \(A_x\) to the bracket directly and give a different mortality probability. The equation to use is \(A_x(1+i)=q_x+(1-q_x)A_{x+1}\), not \(A_x=q_x+(1-q_x)A_{x+1}\).
For this inversion, the denominator \(1-A_{x+1}\) reflects whether mortality changes the expected one-year amount. If \(A_{x+1}=1\), then the bracket \(q_x+(1-q_x)A_{x+1}\) equals 1 for every \(q_x\). The full recursion then requires \(A_x=v\). If \(A_x=v\), every \(q_x\in[0,1]\) is consistent and \(q_x\) is not identified. If \(A_x\ne v\), the supplied values are inconsistent with the recursion; there is no value of \(q_x\) that satisfies it.
Recovering the Interest Rate or Next-Age Value
Worked Example: Infer the Interest Rate
A hypothetical unit whole-life insurance has \(q_x=0.20\), \(A_{x+1}=0.50\), and \(A_x=0.50\). Assuming a nonnegative effective annual interest rate, find \(i\). Identify the inversion and verify the result in the original recursion.
Method. First calculate the undiscounted bracket: $$ q_x+(1-q_x)A_{x+1} =0.20+0.80(0.50) =0.20+0.40 =0.60. $$
Since \(A_x=vB_x\), \(v=A_x/B_x=0.50/0.60=5/6\). Therefore \(1+i=1/v=6/5\), and \(i=6/5-1=0.20\), or \(20\%\).
Check directly: \(v=1/1.20=5/6\), and \(v(0.60)=(5/6)(0.60)=0.50=A_x\). The inferred rate is nonnegative, as assumed.
Common mistake avoided: The bracket \(0.60\) is not itself the interest rate or the current insurance value. It must be divided by \(A_x\) to obtain \(1+i\), so \(i=B_x/A_x-1\).
Worked Example: Infer the Next-Age Insurance Value
For a hypothetical unit whole-life insurance, the effective annual interest rate is \(10\%\), \(q_x=0.20\), and \(A_x=0.55\). Find \(A_{x+1}\) and check the answer.
Method. Multiply the recursion by \(1+i\): \(A_x(1+i)=q_x+(1-q_x)A_{x+1}\). The left side is \(0.55(1.10)=0.605\), so $$ 0.605=0.20+0.80A_{x+1},\qquad A_{x+1}=\frac{0.605-0.20}{0.80} =\frac{0.405}{0.80} =0.50625. $$
Check the bracket using this value: \(0.20+0.80(0.50625)=0.20+0.405=0.605\). Discounting gives \(0.605/1.10=0.55\), which matches the supplied \(A_x\). The result is between 0 and 1, as expected for this unit-benefit insurance value under these assumptions.
Common mistake avoided: The survival probability is \(p_x=1-q_x=0.80\), not \(q_x\). The next-age value is weighted by survival, so dividing the remainder by \(0.80\) is essential.
Why the Recursion Holds
The recursion follows by conditioning on what happens during the first year. If death occurs in that year, probability \(q_x\), the unit benefit is paid at time 1 and has value \(v\) at time 0. If the life survives, probability \(p_x\), the contract continues and its value at time 1 is \(A_{x+1}\), worth \(vA_{x+1}\) at time 0. Taking the probability-weighted values gives $$ A_x=q_xv+p_x(vA_{x+1}) =v(q_x+p_xA_{x+1}). $$
This conditioning argument also explains why the inversion must preserve the timing. The death claim is discounted one year, and the surviving contract's next-age value is also discounted one year. If a benefit were paid at a different time, or if its amount differed from one, the equation and the resulting inverse calculations would need to reflect that change.
Feasibility, Identifiability, and Exam Traps
An algebraic answer is not automatically a valid actuarial answer. For a mortality inversion, the computed \(q_x\) must lie in \([0,1]\). For a next-age value under the specified unit whole-life insurance, the result should be compatible with the contract and mortality basis. If a calculation produces an impossible value, do not force it into the permitted range; first check whether the given quantities satisfy the recursion.
For the interest inversion, \(A_x\) must be nonzero to use \(i=B_x/A_x-1\). If \(A_x=0\), the original equation requires \(B_x=0\). When this equality holds, the recursion does not determine \(i\), because \(vB_x=0\) for every admissible \(v\); when it fails, the data are inconsistent. Under a nonnegative interest assumption, an inferred \(i\) must also satisfy \(i\ge0\).
For the next-age value inversion, \(1-q_x=p_x\) must be positive. If \(q_x=1\), the insured certainly dies during the year, so the recursion reduces to \(A_x=v\), independently of \(A_{x+1}\). If \(A_x=v\), the next-age value is not identified; if not, the supplied values are inconsistent.
| Unknown | Condition for the displayed inversion | What to check |
|---|---|---|
| \(q_x\) | \(A_{x+1}\ne1\) | Computed probability is in \([0,1]\); if \(A_{x+1}=1\), check whether \(A_x=v\) |
| \(i\) | \(A_x>0\) | Inferred rate is in the assumed range; if \(A_x=0\), test \(B_x=0\) |
| \(A_{x+1}\) | \(q_x<1\) | Computed value fits the model; if \(q_x=1\), check whether \(A_x=v\) |
A frequent exam trap is to rearrange correctly but forget to test the exceptional case. For example, substituting \(A_{x+1}=1\) into the mortality formula produces a zero denominator. The correct response is to return to the original recursion: it becomes \(A_x=v\). Thus the data either permit every \(q_x\in[0,1]\), if that equality holds, or permit no \(q_x\), if it does not. Calling \(q_x\) “unidentified” without first checking \(A_x=v\) overlooks the possibility of inconsistent inputs.
Check Your Understanding
For each problem, use the end-of-year-of-death whole-life recursion for a unit benefit. Unless stated otherwise, assume a nonnegative effective annual interest rate.
- A hypothetical policy has \(i=5\%\), \(A_x=0.57\), and \(A_{x+1}=0.40\). Find \(q_x\). (Answer: \(q_x=0.3308\), to four decimal places.)
- A contract has \(q_x=0.25\), \(A_{x+1}=0.60\), and \(A_x=0.65\). Find \(i\). (Answer: \(i=1/13\approx7.6923\%\).)
- Given \(i=4\%\), \(q_x=0.10\), and \(A_x=0.50\), find \(A_{x+1}\). (Answer: \(A_{x+1}=7/15\approx0.4667\).)
- Suppose \(A_{x+1}=1\), \(i=5\%\), and \(A_x=0.95\). Is there a value of \(q_x\) satisfying the recursion? (Answer: No; the recursion requires \(A_x=v=1/1.05\approx0.9524\), not \(0.95\).)
- A hypothetical contract has \(i=3\%\), \(A_x=0.80\), and \(A_{x+1}=0.70\). Find \(q_x\), then verify that it is a valid probability. (Answer: \(q_x=0.4133\), to four decimal places; it is in \([0,1]\).)