Reserve Changes at Premium-Paying and Paid-Up Boundaries

Why the Final Premium Is a Boundary

A limited-pay policy can continue to provide insurance long after its scheduled premiums end. At the date of the final premium, the reserve changes because the insurer receives one last payment; in later years, the reserve develops without any premium income. Keeping these two effects separate is essential when using annual reserve recursions.

The key is to say precisely whether a reserve is measured just before or just after a premium is received. In this tutorial, premiums are level and paid annually in advance while due. Death benefits are paid at the end of the year of death. Unless stated otherwise, reserves are prospective net reserves: expected present values of future benefits minus expected present values of future premiums, conditional on the insured being alive at the valuation time.

Definition: Write ${}_kV_x$ for the reserve at duration $k$, immediately after any premium due at time $k$ has been received. Write ${}_kV_x^{-}$ for the reserve immediately before that premium. If the premium due at time $k$ is $P_k$, then $$ {}_kV_x^{-}={}_kV_x-P_k. $$ If no premium is due at time $k$, the before- and after-payment reserves are the same. At the time of the final premium, the after-payment reserve is the reserve for the policy on a paid-up basis: no further premiums are included in its future cash flows.

The minus sign in the boundary relation reflects the premium’s status as future income to the insurer just before it is received. Before payment, the insurer’s net prospective liability includes that income; after payment, it does not. Thus the reserve immediately before payment is lower by the amount of the premium.

The Two One-Year Recursions

Let $i$ be the effective annual interest rate. For a life aged $x+k$, let $q_{x+k}$ be the probability of death during the next year and $p_{x+k}=1-q_{x+k}$ the probability of survival. Let $B_{k+1}$ be the death benefit paid at the end of that year. The following formulas assume annual premiums, annual valuation times, and benefits paid at the end of the year of death.

While premiums are still payable, suppose a premium $P_{k+1}$ is due at time $k+1$, if the insured survives to that date. The value at time $k+1$ for a surviving policy, just before that premium, is ${}_{k+1}V_x-P_{k+1}$. The recursion is therefore

{}_kV_x(1+i) =q_{x+k}B_{k+1} +p_{x+k}\bigl({}_{k+1}V_x-P_{k+1}\bigr).

After the final premium, there is no premium at time $k+1$ or later. The recursion becomes

{}_kV_x(1+i) =q_{x+k}B_{k+1} +p_{x+k}\,{}_{k+1}V_x.

In both equations, the current reserve accumulates with interest over the year. At year-end, the death branch produces the benefit, while the survival branch produces the reserve for a policy that remains in force. The only difference is whether a premium is due in that survival branch.

Theorem: If the final premium $P_m$ is due at time $m$, then ${}_mV_x^{-}={}_mV_x-P_m$. For the year ending at time $m$, if a premium is due at time $m$ on survival, use the premium-paying recursion with ${}_mV_x-P_m$ in the survival term. For each subsequent year, use the paid-up recursion with no premium term.

Why the Recursions Hold

The recursions follow from a one-year conditional expected-value calculation. Consider a policy just after the premium at time $k$, and accumulate its reserve for one year. Its accumulated amount is ${}_kV_x(1+i)$. If death occurs during the year, the insurer pays $B_{k+1}$ at year-end. If the insured survives, the insurer begins the next year with the reserve at time $k+1$. When a premium is due then, the value just before receiving it is the post-premium reserve minus that premium.

Equating the accumulated reserve to the probability-weighted year-end amounts gives

{}_kV_x(1+i) =q_{x+k}B_{k+1} +p_{x+k}\,{}_{k+1}V_x^{-}.

Substituting ${}_{k+1}V_x^{-}={}_{k+1}V_x-P_{k+1}$ gives the premium-paying formula. If no premium is due, ${}_{k+1}V_x^{-}={}_{k+1}V_x$, giving the paid-up formula. This reasoning also shows why the premium term belongs only in the survival branch: a policyholder who dies during the year does not reach the next premium date.

Locate the valuation date.
Decide whether the reserve is before or after the premium due at that date.

Check whether another premium is due.
If the insured survives to the next valuation date and a premium is payable, subtract it from the next post-premium reserve.

Apply the appropriate recursion.
Include the premium in the survival branch only while premiums remain payable; omit it after the final premium.

Check the boundary.
At the final premium date, verify that the before-premium reserve equals the after-premium reserve minus the premium due.

Worked Boundary and Recursion Calculations

Worked Example: A Three-Payment Whole-Life Policy

A hypothetical whole-life policy pays a death benefit of $900 at the end of the year of death. The annual effective interest rate is $i=0.25$, and the probability of death in each year is $q=0.20$, independently of duration; thus $p=0.80$. Level premiums of $100 are due at times 0, 1, and 2, with the payment at time 2 the final premium. Calculate the reserve just before and just after that final premium, and then calculate the reserves at times 1 and 0 just after their premiums. Use the method of working backward from the paid-up boundary.

Step 1: Find the paid-up reserve at time 2. With no future premiums, the constant reserve $V$ for this whole-life policy satisfies the paid-up recursion

$$ 1.25V=0.20(900)+0.80V. $$

Therefore, $0.45V=180$, so ${}_2V_x=V=\$400$.

Step 2: Find the reserve just before the final premium. The premium due at time 2 is $100, so

{}_2V_x^{-}=400-100=\$300.

Step 3: Work back to time 1. A premium is due at time 2 if the policyholder survives the year. The premium-paying recursion gives

{}_1V_x(1.25) =0.20(900)+0.80(400-100) =180+240 =420, \qquad {}_1V_x=\frac{420}{1.25}=\$336.

Step 4: Work back to time 0. The next premium, due at time 1, is also $100:

{}_0V_x(1.25) =0.20(900)+0.80(336-100) =180+188.80 =368.80, \qquad {}_0V_x=\frac{368.80}{1.25}=\$295.04.

Check. The final boundary satisfies $300=400-100$, and each backward step uses the death benefit in the death branch and the next reserve less the premium due in the survival branch. The common mistake avoided here is adding $100 to the paid-up reserve to obtain the before-premium reserve; the correct before-premium value is $100 lower.

Worked Example: A Paid-Up Two-Year Endowment

A hypothetical two-year endowment policy pays $1,000 at the end of the year of death, or $1,000 at time 2 if the insured survives to maturity. It has one premium of $700 due at time 0 and no later premiums. The annual effective interest rate is $i=0.25$, and the annual mortality probabilities are $q=0.20$ and $p=0.80$ in both years. Find the reserve immediately after the premium at time 0, the reserve at time 1, and the reserve immediately before the premium at time 0.

Step 1: Value the policy at time 1. If the insured is alive at time 1, a payment of $1,000 occurs at time 2 whether death occurs in year 2 or the insured survives to maturity. Thus the time-1 reserve is

{}_1V_x=\frac{1,000}{1.25}=\$800.

Step 2: Use the paid-up recursion for the first year. There is no premium due at time 1, so

{}_0V_x(1.25) =0.20(1,000)+0.80(800) =200+640 =840, \qquad {}_0V_x=\frac{840}{1.25}=\$672.

Step 3: Find the before-premium reserve. The $700 premium is due at time 0, so

{}_0V_x^{-}=672-700=-\$28.

Check. Directly from time 0, the expected present value of the benefits is $0.20(1,000)(0.80)+0.80(1,000)(0.80^2)=160+512=\$672$, agreeing with the recursion. The negative before-premium reserve is consistent with a premium due at that instant; it does not change the post-premium reserve. The common mistake avoided is inserting a premium into the survival branch at time 1 when no premium is payable.

Worked Example: A Term Policy Crossing Its Final-Premium Date

A hypothetical three-year term policy pays $1,000 at the end of the year of death and has premiums of $100 at times 0 and 1 only. The annual effective interest rate is $i=0.25$. The probabilities of death in years 1, 2, and 3 are respectively $q_0=0.10$, $q_1=0.20$, and $q_2=0.25$; write $p_j=1-q_j$. Find the reserve just before and after the final premium at time 1, then find the post-premium reserve at time 0. Use the paid-up recursion first, then cross the premium boundary.

Step 1: Find the reserve at time 2. At time 2, the only remaining benefit is the year-3 death benefit. Hence

{}_2V_x(1.25)=0.25(1,000)+0.75(0), \qquad {}_2V_x=\frac{250}{1.25}=\$200.

Step 2: Find the reserve after the final premium at time 1. No premium is due at time 2, so

{}_1V_x(1.25) =0.20(1,000)+0.80(200) =200+160 =360, \qquad {}_1V_x=\frac{360}{1.25}=\$288.

Step 3: Move across the final-premium boundary. The reserve just before the $100 premium at time 1 is

{}_1V_x^{-}=288-100=\$188.

Step 4: Find the reserve at time 0. A premium is due at time 1 if the insured survives year 1. Therefore,

{}_0V_x(1.25) =0.10(1,000)+0.90(288-100) =100+169.20 =269.20, \qquad {}_0V_x=\frac{269.20}{1.25}=\$215.36.

Check. The boundary difference is $288-188=\$100$, exactly the final premium. In year 2, the recursion used no premium term; in year 1, the survival branch included the premium due at time 1. The common mistake avoided is applying the paid-up recursion one year too early, which would omit the final premium from the time-0 survival branch.

Common Timing Traps

At a premium-paying boundary, the premium is due only if the insured survives to the payment date. It is therefore subtracted from the next post-premium reserve in the survival branch, not from the death benefit. Once premiums stop, that subtraction disappears. The same distinction applies even if the policy continues for many years after becoming paid-up.

For example, in the first worked example, the incorrect boundary calculation ${}_2V_x^{-}={}_2V_x+100=500$ reverses the cash-flow effect. The correct calculation is ${}_2V_x^{-}=400-100=300$. Also check whether a stated reserve is before or after a premium: a reserve reported immediately after the final premium already reflects receipt of that premium, so it must not be reduced again when using a later paid-up recursion.

The transition is useful whenever a reserve calculation crosses the end of a premium-payment period. A reliable solution identifies the exact premium dates, labels the reserve convention, and changes the recursion only after the final payment date. These boundary checks help prevent sign and timing errors in limited-pay, term, endowment, and whole-life calculations. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify whether the next premium is payable before selecting a recursion. Unless stated otherwise, benefits are paid at the end of the year of death.

A final premium of $80 is due at time 4. The reserve immediately after it is $460. Find the reserve immediately before it. (Answer: $380)
After premiums have ceased, $i=0.05$, $q=0.10$, and the death benefit is $1,000. If the next reserve is $300, find the current reserve. (Answer: $352.38)
After premiums have ceased, $i=0$, $q=0.25$, the death benefit is $800, and the next reserve is $240. Find the current reserve. (Answer: $380)
A two-year endowment pays $500 at death during either year or at time 2 on survival. There are no premiums after time 0; $i=0.25$ and $q=0.20$ in both years. Find the reserve at time 1. (Answer: $400)
A premium of $40 is due at time 1 if the insured survives. At time 1, the reserve just after that premium is $300. For the preceding year, $i=0.10$, $q=0.20$, and the death benefit is $500. Find the reserve at time 0 just after its premium. (Answer: $280.00)

What You'll Learn