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Exam FAM · Reserve Recursions and Premiums · Tutorial 493 of 988

Endowment Reserve Patterns at Maturity

Learn to set the maturity reserve correctly, apply the endowment recursion, and assess reserve patterns without assuming they must increase toward maturity.

Advanced 12 min read

What You'll Learn

  • Define an endowment reserve immediately before the premium or maturity benefit is paid
  • Apply the terminal reserve condition when a survival benefit is payable at maturity
  • Derive and use the one-year reserve recursion for death and survival outcomes
  • Identify when the final-year reserve rises or falls
  • Distinguish the reserve before the maturity payment from the value after it is paid
  • Check reserve patterns against premium timing and benefit timing

Why Maturity Changes the Reserve Pattern

A term insurance reserve near expiry is shaped by the remaining death benefit and the premiums still due. An endowment reserve has an additional boundary condition: if the insured survives to the end of the term, a maturity benefit is payable. Consequently, the reserve immediately before maturity is generally not zero. Getting that terminal value and its timing right is essential when working backward through the reserve recursion.

The pattern need not be steadily increasing. It depends on the death benefit, the maturity benefit, the premium, interest, and mortality. In particular, the reserve immediately before the final year can be above or below the reserve just before the maturity benefit. This tutorial develops a consistent timing convention and uses it to analyze those patterns. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Consider an \(n\)-year endowment issued at age \(x\). A death benefit \(B_{k+1}\) is paid at the end of policy year \(k+1\) if death occurs in that year, and a maturity benefit \(M\) is paid at time \(n\) if the insured survives. Assume annual premiums \(P\) are payable at times \(0,1,\ldots,n-1\) while the insured is alive. Define \({}_kV_x\) as the prospective reserve at time \(k\), conditional on survival to that time, immediately before any premium due at time \(k\) is paid. At time \(n\), the reserve is measured immediately before the maturity benefit is paid.

Assume an effective annual interest rate \(i\), with \(v=1/(1+i)\). At duration \(k\), write \({}_kq_x\) for the probability of death during the next year given survival to time \(k\), and \({}_kp_x=1-{}_kq_x\) for the corresponding one-year survival probability. The insurance model here is discrete: death benefits are paid at the end of the year of death. Premiums and benefits are certain contractual amounts conditional on the stated events; expenses are excluded.

The Recursion and the Maturity Condition

At time \(k<n\), the insurer receives premium \(P\) immediately. By the end of the year, interest has accumulated. A death claim \(B_{k+1}\) is paid if death occurs; otherwise, the insurer must have the reserve \({}_{k+1}V_x\) available at time \(k+1\). Conditional on being alive at time \(k\), the expected amount required at year-end is therefore \({}_kq_xB_{k+1}+{}_kp_x{}_{k+1}V_x\).

$$ ({}_kV_x+P)(1+i) = {}_kq_x B_{k+1}+{}_kp_x\,{}_{k+1}V_x, \qquad k=0,1,\ldots,n-1. $$

The maturity condition follows from the reserve definition: at time \(n\), survival to maturity has occurred, and the \(M\) benefit is due immediately. Thus the reserve just before that payment is \(M\). Once the benefit is paid, the contract has ended and its reserve is zero. These are two different instants, not competing terminal values.

$$ {}_nV_x=M. $$

The recursion can be applied backward from this terminal condition. It can also be rearranged to display the change from one reserve to the next. When \({}_kp_x>0\), solving for \({}_{k+1}V_x\) and subtracting \({}_kV_x\) gives the following result.

Theorem: Under the stated timing convention, the one-year reserve change is
$$ {}_{k+1}V_x-{}_kV_x = \frac{(i+{}_kq_x){}_kV_x+(1+i)P-{}_kq_xB_{k+1}}{{}_kp_x}. $$
If the numerator is positive, the reserve increases over that year; if it is negative, it decreases. The formula does not apply when \({}_kp_x=0\), because there is no surviving policy for which a next-year reserve is defined.

To justify the change formula, begin with the recursion and isolate the next reserve:

$$ {}_kp_x\,{}_{k+1}V_x = (1+i)({}_kV_x+P)-{}_kq_xB_{k+1}. $$

Subtract \({}_kp_x{}_kV_x\) from both sides. Since \(1+i-{}_kp_x=i+{}_kq_x\), the result is

$$ {}_kp_x({}_{k+1}V_x-{}_kV_x) = (i+{}_kq_x){}_kV_x+(1+i)P-{}_kq_xB_{k+1}. $$

Dividing by \({}_kp_x\) proves the formula. This form explains why a maturity benefit alone does not establish that reserves must rise: it affects the reserve levels through the terminal condition, while the year-to-year direction also depends on the premium, death benefit, interest, and current reserve.

Worked Reserve Patterns

Worked Example: A Net-Premium Endowment With Equal Benefits

A hypothetical two-year endowment pays \(100\) at the end of the year of death and \(100\) at maturity if the insured survives. The annual effective interest rate is \(10\%\), and the probability of death in each year, conditional on being alive at its start, is \(0.5\). Premiums are level and payable at times 0 and 1 while alive. Find the net premium that makes \({}_0V_x=0\), then calculate the reserves at times 1 and 2.

Method. Start with the terminal reserve, recurse backward, and choose \(P\) so that the reserve at issue is zero. This method avoids treating the maturity reserve as zero merely because the contract ends at time 2.

The terminal condition is \({}_2V_x=100\). At time 1, the death and survival outcomes both require \(100\) at the end of the year, so

$$ {}_1V_x=\frac{0.5(100)+0.5(100)}{1.10}-P =\frac{100}{1.10}-P. $$

At time 0, imposing \({}_0V_x=0\) gives

$$ 0 = \frac{0.5(100)+0.5\,{}_1V_x}{1.10}-P = \frac{50+0.5(100/1.10-P)}{1.10}-P. $$

The constant term on the right is \(10500/121\), and the coefficient of \(P\) is \(-16/11\). Hence \(P=(10500/121)(11/16)=2625/44\), or \(59.65909\ldots\). Substituting this exact premium,

$$ {}_1V_x=\frac{1000}{11}-\frac{2625}{44} =\frac{1375}{44}=31.25, \qquad {}_2V_x=100. $$

The reserves are therefore \({}_0V_x=0\), \({}_1V_x=31.25\), and \({}_2V_x=100\). As a check, at time 1 the accumulated amount is \((31.25+2625/44)(1.10)=100\), equal to the required amount under either outcome. At time 0, the accumulated premium is \( (2625/44)(1.10)=65.625\), and the expected year-end requirement is \(50+0.5(31.25)=65.625\). The calculations agree. The displayed reserve amounts are exact; the premium is shown to five decimal places for readability.

Worked Example: A Larger Maturity Benefit

A hypothetical two-year contract pays \(60\) at the end of the year of death and \(120\) on survival to maturity. The effective annual interest rate is \(0\%\), and each year's conditional death probability is \(0.25\). A level premium is due at times 0 and 1 while alive. Find the net premium and the reserves.

Method. Work backward using the survival benefit as the terminal reserve, then use the zero issue reserve condition to determine the premium. The common error avoided here is using the death benefit, rather than the survival benefit, for \({}_2V_x\).

At maturity, \({}_2V_x=120\). At time 1,

$$ {}_1V_x=0.25(60)+0.75(120)-P=105-P. $$

At time 0, require \({}_0V_x=0\):

$$ 0=0.25(60)+0.75(105-P)-P =93.75-1.75P. $$

Thus \(P=93.75/1.75=375/7\), or approximately \(53.5714\). The time-1 reserve is

$$ {}_1V_x=105-\frac{375}{7} =\frac{360}{7}\approx51.4286. $$

The reserve sequence is \(0\), \(360/7\), and \(120\). Check the two recursions: at time 1, \(0.25(60)+0.75(120)-375/7=360/7\); at time 0, \(15+0.75(360/7)-375/7=0\). The premium and reserves are calculated exactly before rounding; the approximations are to four decimal places.

Worked Example: A Reserve That Falls Toward Maturity

A hypothetical two-year endowment pays \(100\) at death and \(20\) on survival to maturity. The effective annual interest rate is \(0\%\), each year's conditional death probability is \(0.20\), and the annual premium is \(5\). The premium is specified by the contract and is not chosen to make the issue reserve zero. Calculate the reserves and determine the final-year direction.

Method. Apply the recursion backward from the maturity condition and compare adjacent reserves. This avoids assuming that a maturity benefit forces every reserve to rise.

The terminal reserve is \({}_2V_x=20\). At time 1, the death outcome requires \(100\), while survival leads to a reserve of \(20\):

$$ {}_1V_x=0.20(100)+0.80(20)-5=20+16-5=31. $$

At time 0,

$$ {}_0V_x=0.20(100)+0.80(31)-5 =20+24.8-5=39.8. $$

Thus the sequence is \(39.8\), \(31\), and \(20\): it falls as maturity approaches. In particular, the final-year change is \(20-31=-11\). The direct check is \(0.20(20-100)+5=-11\), the final-year change formula at zero interest. This is a valid reserve pattern for the specified premium, although the positive issue reserve shows that the premium is not an equivalence net premium under these assumptions.

Reading the Final-Year Boundary

The last recursion makes the role of the maturity benefit especially clear. Substituting \({}_{n}V_x=M\) into the recursion at time \(n-1\) gives

$$ {}_{n-1}V_x = \frac{{}_{n-1}q_xB_n+{}_{n-1}p_xM}{1+i}-P. $$

This is the reserve just before the premium at time \(n-1\). It accounts for the premium arriving at that time and for the two mutually exclusive year-end outcomes: death with benefit \(B_n\), or survival with benefit \(M\). The comparison with the terminal reserve is

$$ {}_nV_x-{}_{n-1}V_x = \frac{iM+{}_{n-1}q_x(M-B_n)}{1+i}+P. $$

A positive value means the reserve rises into maturity; a negative value means it falls. For example, if \(B_n\) is much larger than \(M\), the expected death claim can make the pre-maturity reserve substantial. Conversely, a sufficiently large maturity benefit can make the terminal reserve larger than the preceding reserve. Neither conclusion should be made without checking the amounts and timing.

Common Timing and Pattern Errors

  • Setting the pre-payment maturity reserve to zero. If \(M\) is payable at time \(n\), then \({}_nV_x=M\) immediately before payment. The reserve is zero immediately after the benefit is paid.
  • Putting the premium at the wrong time. Under the convention used here, \(P\) is due immediately at time \(k\). It is included with \({}_kV_x\) before interest accrues in that year.
  • Using the death probability for the maturity outcome. The death and survival probabilities are \({}_kq_x\) and \({}_kp_x\), respectively. The maturity benefit belongs to the survival term.
  • Assuming reserves always increase toward maturity. The final-year reserve change can be positive, zero, or negative. Use the recursion or the change formula rather than relying on a visual intuition about the approaching benefit.
  • Mixing reserve conventions. A reserve defined just after a premium will not use the same recursion as one defined just before that premium. State the measurement instant before calculating.
Key takeaway: For an endowment with a maturity benefit \(M\), work backward from \({}_nV_x=M\) when the reserve is measured immediately before that benefit. The reserve path is governed by both death and survival outcomes, and it need not be monotone.

Check Your Understanding

For each problem, use the convention that reserves are measured immediately before any premium due at that time. Premiums are payable while alive, and benefits are paid at the end of the policy year or at maturity as stated.

  1. A one-year endowment has death benefit \(80\), maturity benefit \(120\), premium \(4\), effective annual interest rate \(5\%\), and one-year death probability \(0.10\). Find \({}_1V_x\) and \({}_0V_x\). (Answer: \({}_1V_x=120\), \({}_0V_x\approx106.48\).)
  2. A two-year endowment has death benefit \(50\), maturity benefit \(100\), annual premium \(12\), zero interest, and conditional death probability \(0.20\) in each year. Find the reserves at times 0, 1, and 2. (Answer: \({}_0V_x=60.40\), \({}_1V_x=78\), \({}_2V_x=100\).)
  3. A two-year endowment has death benefit \(80\), maturity benefit \(40\), zero interest, and conditional death probability \(0.50\) in each year. Find the level net premium that makes \({}_0V_x=0\), and calculate \({}_1V_x\). (Answer: \(P=140/3\approx46.6667\), \({}_1V_x=40/3\approx13.3333\).)
  4. In the final year of a contract, the death benefit is \(200\), the maturity benefit is \(100\), the premium due at the start of the year is \(10\), the effective annual interest rate is \(4\%\), and the conditional death probability is \(0.10\). Find \({}_{n-1}V_x\) and state whether the reserve rises or falls into maturity. (Answer: \({}_{n-1}V_x\approx95.77\); it rises by approximately \(4.23\) to \(100\).)
  5. A two-year endowment has death benefit \(100\), maturity benefit \(20\), premium \(5\), zero interest, and conditional death probability \(0.20\) in both years. Find the reserve at time 1 and the change from time 1 to time 2. (Answer: \({}_1V_x=31\); the final-year change is \(-11\), so the reserve falls to \(20\).)

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