Term Insurance Recursions at Expiry

Why Expiry Changes the Reserve Recursion

A term insurance contract pays a death benefit only while coverage is in force. If the insured survives to the end of the term, the coverage expires and, in the model considered here, there is no maturity benefit. That makes the reserve at expiry a known boundary value: zero. Starting from that value, the annual reserve equation can be applied backward through the covered years.

The boundary condition does not mean that the reserve in the final year is zero. During that year, the insurer may still have to pay a death benefit. The reserve one year before expiry therefore reflects both the chance of a final-year claim and any premium due at that time. Getting this distinction right is central to term insurance recursions.

Definition: Consider an $n$-year term insurance issued at age $x$, with level death benefit $B$, annual premium $P$, and effective annual interest rate $i$. Assume premiums are payable at the beginnings of policy years $0,\ldots,n-1$, including immediately at time $0$, and the death benefit is paid at the end of the policy year of death. There is no benefit on survival to expiry and no premium at time $n$. Let $V_k$ be the net premium reserve at time $k$, immediately before any premium due at that time. The terminal reserve is $V_n=0$.

For a life aged $x+k$, write $q_{x+k}={}_1q_{x+k}$ for the probability of death during the next year and $p_{x+k}={}_1p_{x+k}=1-q_{x+k}$ for the probability of survival through it. The recursion below uses annual, end-of-year death benefits and a reserve measured immediately before the beginning-of-year premium. If a problem instead defines the reserve immediately after that premium, its value differs by $P$.

The One-Year Cash-Flow Balance

At time $k$, the insurer holds the reserve $V_k$ and collects the premium $P$. That amount accumulates with interest to time $k+1$. With probability $q_{x+k}$, death occurs in the year and the insurer pays $B$ at the year’s end. With probability $p_{x+k}$, the insured survives and the insurer carries the reserve $V_{k+1}$ at time $k+1$, before the next premium. Equating the accumulated funds to the probability-weighted end-of-year liabilities gives the recursion.

(V_k+P)(1+i)=q_{x+k}B+p_{x+k}V_{k+1}, \qquad k=0,\ldots,n-1.

Solving for the reserve at time $k$ gives the backward form:

V_k=\frac{q_{x+k}B+p_{x+k}V_{k+1}}{1+i}-P =v\bigl(q_{x+k}B+p_{x+k}V_{k+1}\bigr)-P, \qquad v=\frac{1}{1+i}.

This is a net premium reserve recursion: the premium in the equation is the level premium for the contract, and expenses are not included. The formula depends on the stated timing. For example, it does not apply unchanged to a benefit payable immediately at death, a premium payable at year-end, or a contract that pays a benefit on survival to expiry.

Why the Terminal Reserve Is Zero

At time $n$, the insured has survived the full term. The policy has expired, there is no maturity payment, and no future premiums or death benefits remain. The prospective value of future benefits less future premiums is therefore zero. This establishes $V_n=0$; it does not set the reserve at time $n-1$ to zero.

Substituting the terminal value into the recursion for the final covered year yields an especially useful formula:

V_{n-1} =\frac{q_{x+n-1}B}{1+i}-P.

The discounted expected final-year death benefit contributes $q_{x+n-1}B/(1+i)$. The premium due at time $n-1$ is then subtracted because $V_{n-1}$ is measured immediately before that premium. If the reserve is requested immediately after the premium, it is $V_{n-1}+P$, or $q_{x+n-1}B/(1+i)$.

Key result: For a term policy with no maturity benefit, set $V_n=0$ and work backward using $V_k=v(q_{x+k}B+p_{x+k}V_{k+1})-P$. The final-year reserve before premium is generally not zero: it includes the discounted cost of the final-year death benefit, less the premium due then.

Backward Calculation in Practice

The recursion is efficient because each year’s reserve depends only on the next year’s reserve, the mortality probability for that year, the benefit, the premium, and interest. Begin with the expiry value, calculate the final covered year, and continue toward issue. This order also makes it easier to detect a mistaken terminal condition: a nonzero value at expiry should be questioned when the contract has neither a survival benefit nor continuing coverage.

Fix the valuation timing.
Confirm whether each reserve is measured before or after the premium due at that time.

Set the expiry value.
For term insurance with no maturity benefit, use $V_n=0$.

Recur backward.
For $k=n-1,n-2,\ldots,0$, use that year’s $q_{x+k}$ and $p_{x+k}$ in the reserve equation.

Check the cash-flow balance.
Verify that $(V_k+P)(1+i)=q_{x+k}B+p_{x+k}V_{k+1}$ for each calculated year.

Worked Example: A Two-Year Term Policy

A hypothetical two-year term policy has a death benefit of $B=\$10{,}000$, annual premium $P=\$200$, and effective annual interest rate $i=10\%$. The one-year death probabilities are $q_x=0.05$ in year 1 and $q_{x+1}=0.11$ in year 2. Calculate the reserves at times 1 and 0, immediately before any premium due at those times. Assume there is no benefit at expiry.

Method and timing. Use the backward recursion for reserves before premium, with $p_x=0.95$, $p_{x+1}=0.89$, and terminal reserve $V_2=0$.

Final covered year. In year 2, the expected end-of-year death benefit is $0.11(\$10{,}000)=\$1{,}100$. Thus

V_1=\frac{1{,}100+0.89(0)}{1.10}-200 =1{,}000-200 =\$800.00.

One year earlier. In year 1, the expected death benefit is $0.05(\$10{,}000)=\$500$, and the survival-weighted next reserve is $0.95(\$800)=\$760$. Therefore

V_0=\frac{500+760}{1.10}-200 =\frac{1{,}260}{1.10}-200 =\$945.45.

Check. At time 1, $(800+200)(1.10)=\$1{,}100$, which equals the final-year expected death benefit plus the survival-weighted terminal reserve. At time 0, $(945.4545\ldots+200)(1.10)=\$1{,}260$, matching $\$500+\$760$. The common mistake avoided is replacing the final-year calculation with $V_1=0$; zero applies at time 2, after coverage expires, not at time 1.

Worked Example: Following the Final-Year Benefit Backward

A hypothetical three-year term policy has a level death benefit of $\$50{,}000$, annual premium $P=\$1{,}000$, and effective annual interest rate $i=4\%$. The death probabilities in years 1, 2, and 3 are $0.01$, $0.02$, and $0.03$, respectively. Find the reserves at times 2, 1, and 0 before the premium due at each time. There is no payment on survival to expiry.

Method. Start with $V_3=0$. The survival probabilities for the three years are $0.99$, $0.98$, and $0.97$. Work backward; the death probability used at each step corresponds to that policy year.

Time 2. The final-year expected death benefit is $0.03(\$50{,}000)=\$1{,}500$. Hence

V_2=\frac{1{,}500+0.97(0)}{1.04}-1{,}000 =\$442.307692\ldots.

Time 1. The year-2 expected death benefit is $0.02(\$50{,}000)=\$1{,}000$. The survival-weighted reserve is $0.98(\$442.307692\ldots)=\$433.461538\ldots$. Therefore

V_1=\frac{1{,}000+433.461538\ldots}{1.04}-1{,}000 =\$378.328402\ldots.

Time 0. The year-1 expected death benefit is $0.01(\$50{,}000)=\$500$, and the survival-weighted next reserve is $0.99(\$378.328402\ldots)=\$374.545118\ldots$. Thus

V_0=\frac{500+374.545118\ldots}{1.04}-1{,}000 =-\$159.090271\ldots.

Check. At time 2, $(442.307692\ldots+1{,}000)(1.04)=\$1{,}500$. At time 1, $(378.328402\ldots+1{,}000)(1.04)=\$1{,}433.461538\ldots$, equal to $\$1{,}000+\$433.461538\ldots$. At time 0, $(-159.090271\ldots+1{,}000)(1.04)=\$874.545118\ldots$, equal to $\$500+\$374.545118\ldots$. The negative initial reserve is possible because this specified premium is high relative to the expected benefits; it is not evidence that the expiry reserve should be nonzero. The common mistake avoided is using the year-1 mortality rate in every backward step.

Timing and Arithmetic Traps

Two conventions can produce different-looking reserve values without changing the underlying cash flows. In this tutorial, $V_k$ is measured immediately before the premium at time $k$. If $W_k$ denotes the reserve immediately after that premium, then $W_k=V_k+P$. In particular, the final-year reserve immediately after premium is $q_{x+n-1}B/(1+i)$, while the reserve immediately before premium is that amount less $P$. Always match the reserve definition to the premium timing in the question.

A second trap is dropping the death benefit in the final covered year because the terminal reserve is zero. The correct final-year equation is $(V_{n-1}+P)(1+i)=q_{x+n-1}B$, not $(V_{n-1}+P)(1+i)=0$. The zero terminal reserve removes the survivor continuation liability; it does not remove the final-year claim.

Worked Example: Checking an Adjacent-Reserve Calculation

For a hypothetical term policy, the effective annual interest rate is $5\%$, the current-year death probability is $0.05$, the death benefit is $\$1{,}000$, and the annual premium is $\$100$. The reserve at time 1, immediately before the next premium, has already been determined as $V_1=10{,}700/147$. Find $V_0$, round the result to cents, and check the cash-flow equation.

Method. Apply the one-year backward recursion using $p_x=1-0.05=0.95$. The expected death benefit is $0.05(\$1{,}000)=\$50$.

\begin{aligned} V_0 &=\frac{50+0.95(10{,}700/147)}{1.05}-100\\ &=\frac{50+10{,}165/147}{1.05}-100\\ &=\frac{17{,}515/147}{1.05}-100\\ &=\frac{350{,}300}{3{,}087}-100\\ &=\frac{41{,}600}{3{,}087}\\ &=13.475866537\ldots. \end{aligned}

Check. Before rounding, $V_0+P=113.475866537\ldots$, so

1.05(V_0+P)=119.149659864\ldots =50+0.95(10{,}700/147).

The two sides agree. Rounding only at the final step gives $V_0=\$13.48$. A common arithmetic error is to round or transcribe $V_0$ incorrectly and then report a check that does not equal the right-hand side. The check should use the unrounded reserve, as above.

What to Carry into a Reserve Calculation

For term insurance, the expiry boundary is simple, but the years leading up to it are not interchangeable. Use the mortality probability for the specific year, include the death benefit while coverage remains in force, and subtract a premium only when its timing matches the reserve definition. The backward recursion is also a useful diagnostic: substituting each calculated reserve back into the one-year balance checks both the arithmetic and the timing.

These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In a multi-year calculation, write the terminal value first, label each year’s mortality assumptions, and keep unrounded values until the final requested answer.

Check Your Understanding

For each hypothetical contract, use the stated reserve timing and round dollar answers to the nearest cent when needed.

A one-year term policy has $i=5\%$, $q_x=0.02$, $B=\$20{,}000$, and $P=\$250$. Find the reserve at time 0 before premium. (Answer: $\$130.95$)
A two-year term policy has $i=4\%$, $B=\$10{,}000$, $P=\$100$, $q_x=0.03$, and $q_{x+1}=0.02$. Find $V_1$ and then $V_0$, both before premium. (Answer: $V_1=\$92.31,\;V_0=\$274.56$)
For a term policy’s final covered year, $i=5\%$, $q_{x+n-1}=0.04$, $B=\$25{,}000$, and $P=\$800$. Find the reserve immediately before the final premium. (Answer: $\$152.38$)
A three-year term policy has $i=10\%$, $B=\$5{,}000$, $P=\$100$, and yearly death probabilities $0.02$, $0.04$, and $0.10$. Starting from $V_3=0$, calculate backward and find $V_0$ before premium. (Answer: $\$339.47$)
In the final year of a term policy, $i=6\%$, $q_{x+n-1}=0.03$, $B=\$40{,}000$, and $P=\$900$. Find the reserve just before the premium and the reserve just after it. (Answer: Before premium, $\$232.08$; after premium, $\$1{,}132.08$)

What You'll Learn