Solving for a Missing Death Probability in a Term APV

Why a Missing Death Probability Requires Care

A discrete term insurance APV depends on both the chance of dying in each policy year and the chance of surviving long enough to reach that year. If one annual death probability is unknown, it therefore affects not only the benefit payable in that year but also the probabilities of receiving benefits in later years. The key is to build those dependencies into the equation before rearranging it.

Assume a hypothetical policy pays a benefit at the end of the policy year of death, provided death occurs within the term. Let $B_k$ be the benefit payable at the end of policy year $k$, and let $i$ be a fixed effective annual interest rate, with $v=1/(1+i)$. Assume mortality probabilities are conditional on being alive at the start of the relevant year, and that the model includes no expenses, premiums, or other benefits.

Definition: Write $q_{x+k-1}={}_1q_{x+k-1}$ for the probability that a life aged $x+k-1$ dies during the next year, conditional on being alive at that age. Write $p_{x+k-1}=1-q_{x+k-1}$. The probability of dying in policy year $k$, conditional on being alive at age $x$, is ${}_{k-1}p_x q_{x+k-1}$. For a term of $n$ years, the discrete insurance APV is $$ A=\sum_{k=1}^{n} v^k B_k\,{}_{k-1}p_x q_{x+k-1}. $$

In this expression, ${}_{k-1}p_x$ is the probability of surviving the first $k-1$ years from age $x$. The factor $v^k$ discounts a payment at the end of year $k$ to the start of the term. A benefit is paid only if death occurs in its corresponding policy year; survival through the whole term does not produce a payment under this model.

The One-Year Case

For a one-year term, the benefit is paid at the end of the year only if the insured dies during that year. There is no earlier survival factor, so the APV is simply the discounted benefit multiplied by the annual death probability. If $A$, $B_1$, and $i$ are known and $B_1>0$, the unknown probability follows directly.

A=vB_1q_x \qquad\Longrightarrow\qquad q_x=\frac{A}{vB_1}=\frac{A(1+i)}{B_1}.

Worked Example: Recovering a One-Year Death Probability

A hypothetical one-year policy pays $12,000 at the end of the year of death. The effective annual interest rate is 10%, and the insurance APV is $660. Find the annual death probability.

Method. For a one-year term, use $A=vB_1q_x$. The only unknown is $q_x$; there are no later policy years whose survival probabilities need to be adjusted.

First calculate the discount factor: $$ v=\frac{1}{1.10}=0.9090909\ldots $$ Then solve the APV equation: $$ 660=0.9090909\ldots(12{,}000)q_x, \qquad q_x=\frac{660}{0.9090909\ldots(12{,}000)} =\frac{660(1.10)}{12{,}000} =\frac{726}{12{,}000} =0.0605. $$

Thus the annual death probability is 0.0605, or 6.05%. Check: $0.9090909\ldots(12{,}000)(0.0605)=660$. The result lies between zero and one, as a probability must. A common mistake is to divide the APV by the benefit alone, which would omit the discount factor.

Short Terms: The Unknown Also Changes Later Survival

For a term longer than one year, the equation is still linear in a single unknown annual death probability, provided the other probabilities, benefits, and interest rate are fixed. However, if the unknown occurs before the end of the term, each later-year death probability includes survival through that unknown year. This creates an important correction: increasing the unknown probability raises the chance of an earlier claim but reduces the chance of a later claim.

Suppose $q_{x+j-1}=q$ is the one unknown probability, for policy year $j$. Define $$ C={}_{j-1}p_x $$ as the probability of surviving to the start of policy year $j$. Let $E$ be the contribution to the APV from policy years before $j$, and define $L$ as the sum of the discounted contributions from years after $j$, conditional on having survived to the start of year $j+1$, with the discounting still measured from the start of the term: $$ E=\sum_{k=1}^{j-1}v^kB_k\,{}_{k-1}p_xq_{x+k-1}, \qquad L=\sum_{k=j+1}^{n}v^kB_k \left(\prod_{r=j+1}^{k-1}p_{x+r-1}\right)q_{x+k-1}. $$ For $k=j+1$, the product in $L$ is empty and equals 1. The later-year contributions in the original APV must also include $C(1-q)$: survival to year $j$, followed by survival through year $j$.

Theorem: With exactly one unknown annual death probability $q=q_{x+j-1}$, and all other model inputs fixed, the term insurance APV is $$ A=E+C\left[v^jB_jq+(1-q)L\right] =E+CL+C\left(v^jB_j-L\right)q. $$ If $C(v^jB_j-L)\ne 0$, the unique algebraic solution is $$ q=\frac{A-E-CL}{C(v^jB_j-L)}. $$ A valid solution must also satisfy $0\le q\le 1$.

Why the Short-Term Equation Is Linear

The equation follows by separating benefits into three groups: benefits before year $j$, the benefit in year $j$, and benefits after year $j$. The earlier benefits cannot depend on $q$, because their policy years occur before the unknown probability applies. The year-$j$ benefit requires reaching the start of year $j$, which has probability $C$, and then dying in that year, which has probability $q$. A later benefit requires reaching year $j$ and then surviving it, with probability $C(1-q)$, before the later mortality probabilities apply.

Keep earlier years fixed.
Their total discounted contribution is $E$, because their death and survival probabilities precede the unknown year.

Write the year-$j$ contribution.
Reach year $j$ with probability $C$, then die with probability $q$, giving $Cv^jB_jq$.

Adjust later years for survival.
Later benefits require surviving year $j$, so their total is $C(1-q)L$, not $CL$ independently of $q$.

Collect the terms in $q$.
Adding the three groups gives $A=E+CL+C(v^jB_j-L)q$, which can be rearranged if its coefficient of $q$ is nonzero.

This also identifies when the unknown probability cannot be recovered uniquely. If $C(v^jB_j-L)=0$, the APV does not change with $q$. If the supplied APV equals the constant $E+CL$, every $q$ from zero to one fits; if it does not, no $q$ fits. Even when the coefficient is nonzero, the algebraic solution may fall outside the probability range, indicating inconsistent inputs under the stated model.

Worked Example: Solving for the First-Year Probability in a Two-Year Term

A hypothetical two-year term policy pays $10,000 at the end of policy year 1 if death occurs in that year, or $20,000 at the end of policy year 2 if death occurs in that year. The effective annual interest rate is 25%. The second-year death probability, conditional on survival to its start, is $q_{x+1}=0.04$. The APV is $1,260.80. Find $q_x$.

Method. Use the year-by-year APV. The probability of a year-1 claim is $q_x$. A year-2 claim requires survival through year 1, so its probability is $(1-q_x)q_{x+1}$.

The discount factor is $$ v=\frac{1}{1.25}=0.8, \qquad v^2=0.64. $$ The APV equation is $$ 1{,}260.80 =(0.8)(10{,}000)q_x +(0.64)(20{,}000)(1-q_x)(0.04). $$ Calculate each coefficient: $$ (0.8)(10{,}000)=8{,}000, \qquad (0.64)(20{,}000)(0.04)=512. $$ Therefore, $$ 1{,}260.80=8{,}000q_x+512(1-q_x) =512+7{,}488q_x, $$ so $$ q_x=\frac{1{,}260.80-512}{7{,}488} =\frac{748.80}{7{,}488} =0.1000. $$

Check the result in the original equation. The year-1 contribution is $8{,}000(0.10)=800$. The year-2 contribution is $512(0.90)=460.80$. Their sum is $1,260.80, as required. The solution $q_x=0.1000$ is a valid probability.

Worked Example with an Unknown Middle-Year Probability

Worked Example: Accounting for Both Earlier and Later Benefits

A hypothetical three-year term policy pays $5,000 at the end of year 1, $15,000 at the end of year 2, and $25,000 at the end of year 3 if death occurs in the corresponding year. The effective annual interest rate is 25%. The known probabilities are $q_x=0.10$ and $q_{x+2}=0.05$; the APV is $1,782.40. Find $q_{x+1}$.

Method. Write one term for each policy year. The year-3 claim requires surviving both years 1 and 2, so its probability includes the factor $1-q_{x+1}$.

Here $v=0.8$, $v^2=0.64$, and $v^3=0.512$. The year-1 contribution is $$ (0.8)(5{,}000)(0.10)=400. $$ The year-2 contribution is $$ (0.64)(15{,}000)(1-0.10)q_{x+1} =8{,}640q_{x+1}. $$ The year-3 contribution is $$ (0.512)(25{,}000)(1-0.10)(1-q_{x+1})(0.05) =576(1-q_{x+1}). $$ Thus, $$ 1{,}782.40 =400+8{,}640q_{x+1}+576(1-q_{x+1}) =976+8{,}064q_{x+1}. $$ Solving gives $$ q_{x+1}=\frac{1{,}782.40-976}{8{,}064} =\frac{806.40}{8{,}064} =0.1000. $$

Check: the three contributions are $400, $864, and $576(0.90)=\$518.40$. Their sum is $1,782.40. The inferred probability is valid. A common error is to calculate the year-3 contribution using the known $q_{x+2}$ but omit survival through year 2; doing so treats a year-3 claim as possible even for someone who died in year 2.

Common Exam Trap: Treating Later Claims as Independent

When the unknown probability occurs before another benefit year, the later death probability must be multiplied by survival through the unknown year. The two-year example makes the effect visible. If one incorrectly treats the year-2 claim probability as $q_{x+1}$ regardless of $q_x$, the mistaken equation becomes $$ 1{,}260.80=8{,}000q_x+512, \qquad q_x=\frac{748.80}{8{,}000}=0.0936. $$ This is not the model's answer. Substituting $q_x=0.0936$ into the correct APV equation gives $$ 8{,}000(0.0936)+512(1-0.0936) =748.80+464.0768 =1{,}212.8768, $$ not $1,260.80. The error is the omitted survival factor $1-q_x$ on the year-2 benefit. In a short-term APV, check the claim probability for every year by asking: what must happen before the insured can be alive to die in this year?

Finally, check the inferred probability before accepting it. A value below zero or above one is not a valid annual death probability. It can signal an incorrect equation, an omitted survival factor, or supplied APV and model inputs that cannot all be true together. Also check whether the coefficient of the unknown is nonzero; without that check, a rearrangement may divide by zero or falsely suggest a unique answer.

Check Your Understanding

For each hypothetical policy, use the stated effective annual interest rate and death-benefit timing. Check both the equation and whether the inferred probability is valid.

A one-year policy pays $8,000 at the end of the year of death. The effective annual interest rate is 25%, and the APV is $800. Find $q_x$. (Answer: $q_x=0.1250$.)
A two-year policy pays $5,000 at the end of year 1 and $10,000 at the end of year 2. The effective annual interest rate is 25%, $q_{x+1}=0.10$, and the APV is $1,312. Find $q_x$. (Answer: $q_x=0.2000$.)
A three-year policy pays $4,000, $10,000, and $20,000 at the ends of years 1, 2, and 3, respectively, if death occurs in those years. The effective annual interest rate is 25%, $q_x=0.10$, $q_{x+2}=0.05$, and the APV is $1,840.64. Find $q_{x+1}$. (Answer: $q_{x+1}=0.2000$.)
A two-year policy pays $5,000 at the end of year 1 and $10,000 at the end of year 2. The effective annual interest rate is 25%, $q_{x+1}=0.10$, and the stated APV is $4,336. Solve algebraically for $q_x$, then decide whether the inputs give a valid probability. (Answer: the equation gives $q_x=1.1000$, which is invalid; the stated APV is inconsistent with the model.)
A two-year policy pays $800 at the end of year 1 and $10,000 at the end of year 2. The effective annual interest rate is 25%, $q_{x+1}=0.10$, and the APV is $640. Determine whether the APV identifies a unique $q_x$. (Answer: no; the APV is $640 for every $q_x$ in $[0,1]$.)

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