Common Errors in Insurance Identities and Recursions

Why Recursion Errors Are Easy to Miss

Life insurance identities often look similar even when they describe different contracts. A one-year shift in duration, a missing maturity payment, or a discount factor placed on the wrong cash flow can produce a plausible-looking number that is nevertheless wrong. The safest approach is to identify the contract’s covered years, the payment time for each benefit, and the value at the end of the term before calculating.

The annual recursions in this tutorial use a discrete model: the insured is alive at the start of a policy year, death during that year has probability $q_{x+k}$, and survival through that year has probability $p_{x+k}=1-q_{x+k}$. A death benefit is paid at the end of the year of death. The effective annual interest rate is $i$, with $v=1/(1+i)$. Unless stated otherwise, benefits are paid only when the relevant event occurs.

Definition: Let $V_k$ be the value at time $k$, conditional on survival to that time, of the benefits still covered by the contract. For a death benefit $B_{k+1}$ paid at the end of policy year $k+1$, the one-year recursion is $$ V_k=v\left(q_{x+k}B_{k+1}+p_{x+k}V_{k+1}\right). $$ The terminal value $V_n$ is the value at time $n$ of any benefit payable then, conditional on survival to time $n$. For a term policy with no maturity benefit, $V_n=0$; for a pure endowment of amount $M$ payable at time $n$, $V_n=M$.

The recursion follows by conditioning on what happens during the next year. A death in that year produces the benefit $B_{k+1}$ at its end; survival leads to the remaining value $V_{k+1}$, also measured at the end of the year. Both outcomes are one year in the future, so their expected value at time $k$ is discounted by $v$. This is why the discount factor multiplies both terms in this particular end-of-year benefit convention.

Audit the Contract Before Calculating

Before applying a recursion or an identity, write down the contract’s duration and cash flows. For an $n$-year policy with death benefit $B$ at the end of the year of death and maturity benefit $M$ at time $n$, the possible payments are:

A death benefit $B$ if death occurs during any covered year $1,\ldots,n$.
A maturity benefit $M$ if the insured survives to time $n$.

The maturity payment is separate from the death benefit. In an endowment policy it is paid on survival to the end of the term; it is not a death benefit in the final year. Thus, when $B$ and $M$ differ, a final-year death pays $B$, while survival through the final year pays $M$.

For a term policy, the terminal condition is zero because no benefit remains payable after the coverage ends. Setting the terminal value equal to the death benefit adds a payment on survival to the end of the term, turning the calculation into one that includes a pure endowment. Conversely, setting the terminal value to zero for an endowment policy omits its maturity benefit.

Mark the policy years.
Determine exactly how many annual mortality intervals are covered and which $q_{x+k}$ applies in each year.

List benefits and payment times.
Distinguish each year’s death benefit from any benefit payable only on survival to the term’s end.

Set the terminal value.
Use zero for a term contract with no maturity payment, and use the specified maturity amount for a pure endowment or endowment contract.

Discount from each payment time.
A benefit payable at the end of year $j$ is discounted by $v^j$ to the issue date.

Check by another representation.
Compare the recursion with a direct present-value sum using unconditional probabilities of each benefit-paying event.

Worked Calculations and Error Checks

Worked Example: Keeping a Maturity Benefit in the Calculation

A hypothetical two-year endowment policy pays $10,000 at the end of the year of death if death occurs during the term, and pays $4,000 at time 2 if the insured survives the term. Assume $i=5\%$, $q_x=0.10$, and $q_{x+1}=0.20$. Calculate the issue value. Name the error avoided.

Method. Work backward from the maturity date, using the specified maturity amount as the terminal value. This avoids treating the endowment as a term policy and omitting its pure endowment.

The one-year survival probabilities are $p_x=1-0.10=0.90$ and $p_{x+1}=1-0.20=0.80$. The survival probability through two years is $ {}_2p_x=0.90(0.80)=0.72$. At time 2, the insured receives $4,000 if alive, so $V_2=4{,}000$.

At time 1, death during year 2 pays $10,000, and survival through year 2 leads to the $4,000 maturity benefit:

V_1=\frac{0.20(10{,}000)+0.80(4{,}000)}{1.05} =\frac{5{,}200}{1.05} =4{,}952.380952\ldots

At issue, death during year 1 pays $10,000; survival through year 1 leads to $V_1$:

V_0=\frac{0.10(10{,}000)+0.90(4{,}952.380952\ldots)}{1.05} =5{,}197.278911\ldots

Check by direct present value. The year-1 death probability is $0.10$; the year-2 death probability is $0.90(0.20)=0.18$; and the probability of surviving to maturity is $0.72$. Therefore,

V_0= \frac{0.10(10{,}000)}{1.05} +\frac{0.18(10{,}000)}{(1.05)^2} +\frac{0.72(4{,}000)}{(1.05)^2} =5{,}197.278911\ldots

The two methods agree. Rounded to the nearest cent, the issue value is $5,197.28. Setting $V_2=0$ would incorrectly omit the $4,000 survival benefit.

Worked Example: Discounting Each Death Benefit at the Correct Time

A hypothetical two-year term policy pays $6,000 at the end of year 1 if death occurs in year 1, and $9,000 at the end of year 2 if death occurs in year 2. Assume $i=4\%$, $q_x=0.10$, and $q_{x+1}=0.20$. Find the issue value and check the discounting. Name the error avoided.

Method. Use the term recursion with terminal value zero, and then check the result by discounting each year’s benefit from its actual payment time. This avoids discounting a year-2 payment by only one year.

The year-2 conditional value is

V_1=\frac{0.20(9{,}000)}{1.04} =1{,}730.769230\ldots

The policy ends after year 2 with no survival payment, so $V_2=0$. At issue,

V_0=\frac{0.10(6{,}000)+0.90V_1}{1.04} =\frac{600+0.90(1{,}730.769230\ldots)}{1.04} =2{,}074.704142\ldots

For the direct check, the unconditional probability of death in year 1 is $0.10$, and the unconditional probability of death in year 2 is $0.90(0.20)=0.18$. The second benefit is payable two years after issue:

\frac{0.10(6{,}000)}{1.04} +\frac{0.18(9{,}000)}{(1.04)^2} =576.923076\ldots+1{,}497.781065\ldots =2{,}074.704142\ldots

The check agrees with the recursion. Rounded to the nearest cent, the issue value is $2,074.70. Discounting the year-2 benefit by only $v$ would value it at time 1 rather than at issue.

Worked Example: Applying the Zero-Term Boundary

A hypothetical two-year term policy pays $8,000 at the end of the year of death and nothing on survival to the end of the term. Assume $i=5\%$, $q_x=0.10$, and $q_{x+1}=0.20$. Calculate its issue value, then quantify the error if an analyst sets the terminal value to $8,000. Name the error avoided.

Method. Apply the term recursion backward from $V_2=0$, then interpret the effect of replacing that boundary with a nonzero amount. This avoids adding a survival benefit that the contract does not provide.

At time 1, the death benefit is payable with probability $0.20$, and there is no value after survival to the end of the term:

V_1=\frac{0.20(8{,}000)+0.80(0)}{1.05} =1{,}523.809523\ldots

At issue,

V_0=\frac{0.10(8{,}000)+0.90(1{,}523.809523\ldots)}{1.05} =2{,}068.027210\ldots

The direct check uses death probabilities $0.10$ in year 1 and $0.90(0.20)=0.18$ in year 2:

V_0=\frac{800}{1.05}+\frac{1{,}440}{(1.05)^2} =2{,}068.027210\ldots

If the analyst incorrectly sets $V_2=8{,}000$, the recursion gives

V_1^{\mathrm{wrong}}= \frac{0.20(8{,}000)+0.80(8{,}000)}{1.05} =7{,}619.047619\ldots, \qquad V_0^{\mathrm{wrong}}= \frac{800+0.90(7{,}619.047619\ldots)}{1.05} =7{,}292.517006\ldots.

The excess is $7{,}292.517006\ldots-2{,}068.027210\ldots=5{,}224.489796\ldots$, exactly the value of an unintended $8,000 pure endowment:

\frac{0.90(0.80)(8{,}000)}{(1.05)^2} =5{,}224.489795\ldots

Thus the correct term value, rounded to the nearest cent, is $2,068.03. The wrong boundary has silently changed the contract by adding a benefit on survival.

Common Traps and Their Corrections

A frequent error is using the conditional probability $q_{x+1}$ as though it were the issue-date probability of death in year 2. The probability of death in year 2 from issue is $p_xq_{x+1}$: the insured must first survive year 1. More generally, the probability of death in year $j$ is survival through the first $j-1$ years multiplied by the conditional death probability in year $j$.

For the endowment example, using $0.20$ instead of $0.90(0.20)=0.18$ for year-2 death would overstate that death benefit’s present value. This is not repaired by applying a correct discount factor; the event probability itself is wrong.

Another trap is applying $v$ to only one term of the one-year recursion. Under the stated end-of-year payment convention, both death during the year and survival to the next value date produce cash flows or values one year ahead. The correct expression is $v(qB+pV_{\mathrm{next}})$. If a contract uses a different payment timing, its cash flows must be valued at those times rather than forcing them into this convention.

Finally, check that an identity compares contracts with the same duration and mortality assumptions. A term insurance value covers death benefits during the term; a pure endowment covers survival to the term’s end. Their combination can represent an endowment policy only when the amounts, term, and payment times match. A duration mismatch is not a harmless notation change: it changes which death years and survival event are included.

Key takeaway: Before trusting an insurance identity or recursion, reconcile its covered years, event probabilities, payment dates, and terminal value with the contract. Then verify the result by a direct present-value sum whenever the cash flows are manageable.

Check Your Understanding

For each problem, use the stated annual end-of-year death-benefit convention and round dollar answers to the nearest cent.

A one-year endowment pays $3,000 at the end of the year if death occurs and $2,000 at time 1 if the insured survives. Assume $i=4\%$ and $q_x=0.10$. Find its issue value. (Answer: $2,019.23)
A two-year term policy pays $6,000 at the end of the year of death. Assume $i=5\%$, $q_x=0.10$, and $q_{x+1}=0.20$. Find the issue value, using the unconditional year-2 death probability. (Answer: $1,551.02)
A two-year endowment pays $5,000 at the end of the year of death and $3,000 at time 2 on survival. Assume $i=5\%$, $q_x=0.20$, and $q_{x+1}=0.10$. Find its issue value. (Answer: $3,274.38)
A three-year term policy pays $4,000 at the end of the year of death. Assume $i=4\%$, $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$. Find its issue value by summing the three death-benefit present values. (Answer: $1,690.37)
A one-year term policy pays $10,000 at the end of the year of death and has issue value $840. If $i=5\%$, infer $q_x$. (Answer: $q_x=0.0882$)

What You'll Learn