Why the Deferment Boundary Matters
A deferred insurance contract has a period in which the insured is covered by the policy but a death benefit is not payable. That distinction changes the reserve recursion: during deferment, the death outcome has no benefit payment, but survival carries the policy forward. Once deferment ends, the death benefit enters the recursion. The reserve at that boundary must reflect the coverage and premiums still due after that point.
The key question is therefore not simply whether a contract is “deferred.” At each policy duration, identify whether a death benefit is payable, whether a premium is due, and what future contract remains if the insured survives. The examples below use annual, discrete reserves and premiums payable at the beginning of each year.
Unless stated otherwise, assume annual effective interest \(i\), \(v=1/(1+i)\), no expenses, and a level annual premium \(P\) payable in advance while the insured is alive and the contract remains in force. Death terminates the policy. A benefit for death during a covered year is paid at the end of that year. These assumptions fix the timing in each recursion.
The Recursion During Deferment
For a year wholly inside the no-benefit period, the premium is received at the start of the year. It accumulates with interest to the end of the year. If the insured dies, the death benefit is zero; if the insured survives, the accumulated amount must support the reserve at the next policy duration.
Thus the forward and backward forms during deferment are
The forward form requires \({}_1p_{x+k}>0\). The absence of a death benefit does not mean the reserve is zero during deferment. Premiums may accumulate, and the future coverage may have substantial value. Nor does it imply that the reserve at the end of deferment is zero: that value depends on the contract that starts paying benefits at the boundary.
What Happens at the Boundary?
At time \(n\), the no-benefit period is over. The reserve \({}_nV_x\) is the value, at that time, of the contract that remains, using its actual future benefit and premium schedule. For deferred whole-life insurance with premiums continuing after time \(n\), the boundary value is the ordinary whole-life reserve at attained age \(x+n\) under that premium schedule. For deferred term insurance, it is the reserve for the remaining term. It is not automatically zero merely because deferment has ended.
If a death benefit \(B\) is payable for death in the year from time \(k\) to time \(k+1\), the one-year equation is
The benefit term appears only in years for which the contract pays on death. For a deferred whole-life contract with benefit \(B\) after a deferment of \(n\) years, the no-benefit recursion applies for \(k<n\), and the benefit-bearing recursion applies for \(k\ge n\). For a deferred term contract, it applies only during the post-deferment coverage term; at the term’s expiry, the reserve is zero if there is no maturity benefit.
Why the One-Year Equation Holds
The recursion is a conditional cash-flow balance from the viewpoint of a policy in force at time \(k\). The reserve and premium available at the start of the year accumulate to \(({}_kV_x+P)(1+i)\) by the end of the year. On the death path, the insurer pays the benefit for that year. On the survival path, the insurer holds the next reserve. Taking the probability-weighted value of those two possible end-of-year obligations gives the equation.
The reserve plus premium is \({}_kV_x+P\), so its end-of-year value is \(({}_kV_x+P)(1+i)\).
Death in the year has probability \({}_1q_{x+k}\) and produces the year’s benefit, if any. Survival has probability \({}_1p_{x+k}\) and leaves reserve \({}_{k+1}V_x\).
The accumulated funds equal the expected benefit plus the expected next reserve. During deferment the benefit is zero; after deferment it is included when payable.
Worked Examples
Worked Example: A Deferred Whole-Life Reserve Across the Boundary
A hypothetical insurer issues a two-year deferred whole-life policy with benefit \(B=1331\). The annual effective interest rate is \(i=0.10\), and the annual mortality assumption is \({}_1q_{x+k}=0.10\) at every duration, so \({}_1p_{x+k}=0.90\). A level net premium \(P=81\) is payable annually in advance while the insured is alive, including after deferment. Given \({}_0V_x=0\), find \({}_1V_x\), \({}_2V_x\), and \({}_3V_x\).
Method. Apply the no-benefit recursion in the first two years. At time 2, the contract has become whole-life insurance, so the death-benefit recursion applies in the third year.
First year of deferment. There is no death benefit. Therefore,
Second year of deferment.
First year after deferment. Death in this year pays \(1331\). Using the recursion,
The left side is \(301(1.10)=331.10\), and the death-benefit term is \(133.10\). Hence
Check. Substituting \({}_3V_x=220\) gives \(133.10+0.90(220)=331.10\), matching the accumulated amount. The reserve at the boundary is not zero: it supports the whole-life coverage that begins there. A common mistake is to use a zero death benefit in the year after deferment just because the policy began as deferred insurance.
Worked Example: Working Backward from the Post-Deferment Reserve
A hypothetical deferred whole-life policy has a two-year deferment, benefit \(B=1000\), annual premium \(P=40\), annual effective interest \(i=0.25\), and constant annual death probability \(q=0.25\). Premiums continue for life while the insured is alive. Find the reserves at times 2, 1, and 0.
Method. First find the ordinary whole-life reserve at the boundary, where the remaining contract is whole life. Then use the backward no-benefit recursion for the two deferment years.
At the boundary, the whole-life reserve is constant from year to year under these constant assumptions. Denote that reserve by \(V\). The death-benefit recursion gives
Expanding and collecting terms gives \(1.25V+50=250+0.75V\), so \(0.50V=200\) and \(V=400\). Therefore, \({}_2V_x=400\).
In the year from time 1 to time 2, death still produces no benefit. Use the backward form:
For the first year, also within deferment,
Check. In the first deferment year, \((80+40)(1.25)=150=0.75(200)\). In the second, \((200+40)(1.25)=300=0.75(400)\). Both balances have no death-benefit term. The common mistake avoided here is starting from zero at time 2 rather than valuing the whole-life contract that remains.
Worked Example: A Deferred Term Policy with a Finite Coverage Boundary
A hypothetical policy pays \(B=1000\) for death during the two years after a two-year deferment. It pays nothing for death during deferment, and there is no maturity benefit at the end of the two-year coverage term. The annual effective interest rate is \(i=0\), the annual death probability is \(q=0.50\) at each duration, and the level premium \(P=100\) is payable at the start of each of the four policy years while the insured is alive. Calculate the reserves at times 0 through 4.
Method. Start at the end of coverage, where the reserve is zero because there is no benefit after expiry. Work backward through the covered years, then through deferment.
At time 4, the term has expired and no maturity benefit is payable, so \({}_4V_x=0\). In the final covered year, from time 3 to time 4,
In the first covered year, from time 2 to time 3,
In the second year of deferment, death still produces no benefit:
In the first year of deferment,
Check. The initial reserve is zero, consistent with the premium being level and the policy’s expected benefits and premiums balancing at issue. Specifically, the expected benefit is \(1000(0.25)(0.50)+1000(0.125)(0.50)=125+62.50=187.50\). The expected number of premiums is \(1+0.50+0.25+0.125=1.875\), giving expected premiums \(100(1.875)=187.50\). The method avoids treating the end of deferment as the end of the policy: coverage starts at time 2, and the reserve there is \(600\), not zero.
Common Timing Errors
A reliable calculation begins by marking the policy durations at which the contract changes. For each one-year interval, write the death benefit and premium that actually apply in that interval before substituting probabilities. This is particularly important when the boundary falls exactly at a policy anniversary.
- Putting the benefit in too early. For a deferment of \(n\) years, death during years \(0\) through \(n-1\) has no benefit under the contract described here. The benefit-bearing recursion begins with the year from time \(n\) to \(n+1\).
- Setting the boundary reserve to zero automatically. At time \(n\), whole-life or term coverage may still remain. Use the value of that remaining contract. A zero boundary reserve is appropriate only when the contract has ended with no further benefit or other obligation.
- Using the wrong branch probability. In the deferment recursion, the next reserve is multiplied by the probability of survival, not by the probability of death. The death branch contributes zero only when the policy truly pays no benefit on death in that year.
- Forgetting the premium timing. These recursions assume the premium is paid at the start of the year. If a contract does not collect a premium in a particular year, omit \(P\) for that year rather than applying the same equation unchanged.
Check Your Understanding
For each problem, identify which years are in deferment before applying the recursion.
- A deferred whole-life policy has one year of deferment, \({}_0V_x=0\), \(P=50\), \(i=0.10\), and \({}_1p_x=0.90\). Find \({}_1V_x\). (Answer: \(61.11\))
- After deferment, a whole-life policy has \(B=1000\), \(P=100\), \(i=0.10\), and constant \(q=0.20\). Find its constant post-deferment reserve. (Answer: \(300\))
- A policy has one year of deferment followed by two covered years, \(B=800\), \(P=50\), \(i=0\), and constant \(q=0.25\). There is no benefit at expiry. Given \({}_3V_x=0\), find \({}_2V_x\). (Answer: \(150\))
- For the policy in question 3, find \({}_1V_x\), the reserve at the start of the first covered year. (Answer: \(262.50\))
- A deferred whole-life policy has two years of deferment, \(B=1000\), constant \(q=0.20\), \(i=0\), and premiums payable while alive throughout the policy. Find the level net premium that gives \({}_0V_x=0\). (Answer: \(128\))