Why a Term-Insurance Value Can Reveal Mortality
A term-insurance value reflects two possible events during the next year: death, which triggers the current year’s benefit, and survival, which leaves a shorter term contract in force. If the interest rate and the value of that surviving contract are known, the current value contains information about the one-year probability of death. Recovering that probability requires keeping the death and survival branches separate.
The calculation applies to a discrete, level-benefit term insurance: the benefit is one unit, payable at the end of the year of death, and nothing is paid if the insured survives to the end of the term. The values used are actuarial present values per unit of benefit, with no premiums included. The result depends on this timing and benefit specification; it is not a formula for a continuous benefit or a contract with a different death benefit.
The Term-Insurance Recursion and Its Solution
For a term of at least two years, let \(A=A^{1}_{x:\overline{n}|}\). At the end of the first year, death produces a unit benefit, while survival produces a contract worth \(C\) at age \(x+1\). Discounting either possible amount back one year gives the recursion below.
The value \(C\) is conditional on survival: it is not itself multiplied by a probability until it is placed in the first-year recursion. Rearranging the recursion isolates \(q_x\).
This rearrangement also explains the denominator. A higher \(q_x\) replaces some of the survival-continuation outcome \(C\) with the death benefit of one unit. Each increase in the death probability changes the undiscounted first-year expected value by \(1-C\), not by one. Dividing by \(1-C\) accounts for that difference.
The lower bound follows because \(1-C\) is positive: \(q_x\ge0\) exactly when \((1+i)A-C\ge0\). For the upper bound, \(q_x\le1\) is equivalent to \((1+i)A-C\le1-C\), or \((1+i)A\le1\). If \(C=1\), the first-year expected amount is \(q_x+(1-q_x)=1\) regardless of mortality. The recursion then gives \(A=v\) but cannot identify \(q_x\). If \(C>1\), the continuation amount exceeds the unit death benefit, and the usual bounds above do not apply in that form; for the stated unit term-insurance model, such a continuation value would itself require investigation.
Worked Calculations
Worked Example: Inferring Mortality from a Term Value
A hypothetical six-year unit term insurance has a current actuarial present value \(A=5/13\). At the end of the first year, if the insured survives, the remaining five-year contract has value \(C=1/4\). The effective annual interest rate is \(i=4\%\). Find the implied one-year mortality probability and check the recursion.
Method. Use the one-year term recursion and solve for \(q_x\). The value \(C\) is already conditional on survival, so it is weighted by \(1-q_x\) in the recursion.
First calculate the undiscounted current value:
Now substitute the current value and continuation value into the solved formula:
Thus the implied one-year mortality probability is \(0.2000\), or \(20\%\). To check, the survival probability is \(1-0.20=0.80\). The recursion gives
The values satisfy the recursion. A common mistake this example avoids is subtracting \(C\) from \((1+i)A\) and stopping: that difference is \(0.15\), not \(q_x\). It must be divided by \(1-C=0.75\).
Worked Example: Using Decimal Values and Checking the Survival Branch
A hypothetical four-year unit term insurance has current value \(A=37/105\), while the value at age \(x+1\) of the surviving three-year contract is \(C=0.30\). The effective annual interest rate is \(5\%\). Infer the first-year mortality probability and verify the result.
Method. Multiply the current value by \(1+i\), subtract the conditional continuation value, and divide by \(1-C\).
Therefore,
The inferred mortality probability is \(0.1000\), and the survival probability is \(0.9000\). Substitution checks the result:
The calculation reproduces the supplied current value. In particular, the \(0.30\) continuation value is multiplied by the survival probability \(0.90\); using it as an unconditional amount would change the model.
Worked Example: Detecting Values That Imply an Impossible Probability
A hypothetical unit term insurance is reported to have current value \(A=0.99\), continuation value \(C=0.80\), and effective annual interest rate \(i=4\%\). Determine whether these figures can describe the stated contract.
Method. Calculate the implied probability, then test it against the probability bounds. Here \(C<1\), so the solved formula applies.
Thus the algebraic value of \(q_x\) would be
This is not a valid probability because it exceeds one. The bounds confirm the problem: they require \(0.80\le1.0296\le1\), but the upper bound fails. Equivalently, even if death were certain, the undiscounted first-year amount could be at most the unit benefit, so the current value could be at most \(v=1/1.04\), approximately \(0.9615\). The reported value \(0.99\) is larger. The figures are inconsistent with the stated model; rounding \(1.148\) to a probability would not repair that inconsistency.
Common Errors and Checks
The algebra is short, but the inputs represent different quantities and have different roles. Before calculating, identify the current value, the value conditional on survival, the annual interest rate, and the timing and amount of the death benefit. Then test the result rather than assuming any supplied values must be mutually consistent.
Use \(A\) for the current term-insurance value and \(C\) for the surviving contract’s value one year later, with one fewer year of coverage.
Calculate \((1+i)A\). Do not use \(A\) in place of this undiscounted first-year expected amount.
For \(C<1\), calculate \(q_x=((1+i)A-C)/(1-C)\). The denominator reflects the difference between the unit death benefit and the continuation value.
Verify \(0\le q_x\le1\), or equivalently \(C\le(1+i)A\le1\), and substitute the result back into the recursion.
A frequent exam trap is to treat the continuation value as though it were paid with certainty. For the first worked example, that shortcut gives
and incorrectly labels \(0.15\) as \(q_x\). But the actual undiscounted expected amount is \(q_x+(1-q_x)C\). Rearranging this expression gives \((1+i)A-C=q_x(1-C)\), so the correct calculation is \(0.15/0.75=0.20\). The shortcut omits the factor \(1-C\).
Another issue is the case \(C=1\): the denominator in the formula is zero, and the recursion supplies no information about mortality. A zero denominator is not a cue to approximate; it signals that mortality is not identifiable from these two values and the interest rate alone. Finally, a result outside \([0,1]\) is evidence that at least one input, timing convention, or model assumption is inconsistent with the others.
Check Your Understanding
For each problem, use the stated discrete unit-benefit model. Check the inferred probability against its bounds where relevant.
- A hypothetical term contract has \(i=5\%\), current value \(A=8/21\), and one-year continuation value \(C=0.20\). Find \(q_x\) and verify it by substitution. (Answer: \(q_x=0.2500\).)
- A hypothetical policy has \(i=6\%\), \(A=23/53\), and \(C=0.40\). Infer the one-year mortality probability and check the current value using the recursion. (Answer: \(q_x=0.1000\).)
- For a hypothetical term contract, \(i=3\%\), \(A=0.72\), and \(C=0.65\). Calculate \(q_x\) to four decimal places and state whether the values pass the probability-bound check. (Answer: \(q_x=0.2617\); the values pass the bounds.)
- A hypothetical policy has \(i=4\%\), current value \(A=0.60\), and continuation value \(C=0.50\). Find \(q_x\) and confirm whether substitution returns \(A=0.60\). (Answer: \(q_x=0.2480\); substitution returns \(A=0.6000\).)
- A hypothetical unit term contract is reported to have \(i=2\%\), \(A=0.88\), and \(C=0.90\). Calculate the algebraic value of \(q_x\) and decide whether it is a possible mortality probability. (Answer: \(q_x=-0.0240\); the values are inconsistent with a probability.)