Simultaneous Reserve Equations for Unknown Premium and Benefit

Why Use Multiple Reserve Equations?

A one-year reserve recursion links a reserve at one duration to the reserve at the next duration, the premium collected, and the benefit payable if death occurs. When both the level premium and the death benefit are unknown, one such equation is generally not enough to determine both. Reserve information at multiple durations can supply additional equations, provided the mortality probabilities differ enough to distinguish the two unknowns.

The method is useful only when the reserve timing and contract assumptions are consistent across the equations. In particular, a premium collected at the start of the year accumulates interest during that year. Omitting that accumulation changes the equation and can produce an incorrect premium even when the benefit estimate happens to look plausible.

Definition: Consider a hypothetical whole-life policy with a level annual premium $P$ payable at the start of each year while the insured is alive and a level death benefit $B$ payable at the end of the year of death. Let ${}_kV_x$ be the reserve at duration $k$, immediately before the premium due at that time. Assume an effective annual interest rate $i$, no expenses, and one-year mortality probabilities ${}_1q_{x+k}$ and ${}_1p_{x+k}=1-{}_1q_{x+k}$. The policy continues throughout the year. These assumptions give the one-year reserve recursion below.

\bigl({}_kV_x+P\bigr)(1+i) = {}_1q_{x+k}B + {}_1p_{x+k}{}_{k+1}V_x

The equation is conditional on the insured being alive at duration $k$. The reserve and premium are available at the start of the year; they accumulate at interest until the end of the year. With probability ${}_1q_{x+k}$, death occurs and the benefit is paid. With probability ${}_1p_{x+k}$, the insured survives and the reserve at the next duration is held. The recursion does not apply unchanged if the premium is paid at a different time, the benefit has different timing or amount, expenses are included, or the reserve is defined at a different point relative to the premium.

Turning Reserve Data into a Linear System

For compactness, write $V_k={}_kV_x$ and $q_k={}_1q_{x+k}$, so that $p_k=1-q_k$. Rearranging the recursion gives a linear equation in the two unknowns $B$ and $P$. The quantity on the right, calculated from the known reserve data, is denoted $c_k$.

q_k B-(1+i)P = (1+i)V_k-p_kV_{k+1} = c_k

This rearrangement explains why multiple durations are informative. The coefficient of $B$ is the mortality probability $q_k$, which may vary by duration. The coefficient of $P$ is the same at every duration when the effective interest rate and premium timing are fixed. Subtracting two duration equations eliminates the premium.

\begin{aligned} q_1B-(1+i)P&=c_1,\\ q_2B-(1+i)P&=c_2,\\ (q_2-q_1)B&=c_2-c_1. \end{aligned}

Theorem: Under the stated assumptions, two reserve equations at durations with different one-year mortality probabilities determine a unique death benefit and level premium. If $q_1\ne q_2$, then $$ B=\frac{c_2-c_1}{q_2-q_1}, \qquad P=\frac{q_1B-c_1}{1+i}. $$ The equations do not identify a unique pair when $q_1=q_2$: they are either inconsistent or redundant. Additional equations can check consistency, but cannot resolve this lack of identification if their mortality probabilities are also identical.

The derivation is direct: subtracting the equations cancels the common premium term, leaving $(q_2-q_1)B=c_2-c_1$. If the mortality probabilities differ, division gives $B$. Substituting that result into either original equation gives $P$. If the probabilities are equal, subtraction either yields $0=c_2-c_1$, which adds no information, or a contradiction.

A Reliable Solution Process

Fix the timing and assumptions.
Identify when the premium is paid, when the benefit is paid, how the reserve is timed, and which interest and mortality assumptions apply to each duration.

Calculate the reserve-data quantity.
For each observed transition, compute $c_k=(1+i)V_k-p_kV_{k+1}$. Keep the reserve and mortality data paired at the same duration.

Solve the equations.
Subtract two equations with different mortality probabilities to find $B$, then substitute to find $P$.

Check all available durations.
Substitute the results into every original equation. An extra equation is a consistency check, not a reason to ignore a mismatch.

Worked Applications

Worked Example: Solve from Two Consecutive Durations

A hypothetical whole-life policy has an unknown level annual premium $P$ and level end-of-year death benefit $B$. The premium is paid at the start of each year. The reserve is measured immediately before that premium, there are no expenses, and the effective annual interest rate is $i=0.10$. At duration 0, ${}_1q_x=0.20$, $V_0=\$38$, and $V_1=\$41$. At duration 1, ${}_1q_{x+1}=0.50$, and $V_2=\$12.20$. Find $P$ and $B$.

Method. Compute $c_k$ for each transition, form the two linear equations, and subtract them. This uses the same premium timing in both equations.

At duration 0, ${}_1p_x=1-0.20=0.80$, so

$$ c_0=1.10(38)-0.80(41)=41.80-32.80=9.00. $$

At duration 1, ${}_1p_{x+1}=1-0.50=0.50$, so

$$ c_1=1.10(41)-0.50(12.20)=45.10-6.10=39.00. $$

The two equations are $0.20B-1.10P=9.00$ and $0.50B-1.10P=39.00$. Subtracting the first from the second gives

0.30B=30.00, \qquad B=\frac{30.00}{0.30}=\$100.00.

Substitute $B=100$ into the first equation:

0.20(100)-1.10P=9, \qquad 20-9=1.10P, \qquad P=\frac{11}{1.10}=\$10.00.

Check. At duration 0, $0.20(100)-1.10(10)=20-11=9$. At duration 1, $0.50(100)-1.10(10)=50-11=39$. Both match the calculated reserve-data quantities. The answer is $B=\$100.00$ and $P=\$10.00$.

Common mistake avoided. The premium is collected at the start of the year, so its accumulated amount in the recursion is $1.10P$, not $P$. Replacing $1.10P$ by $P$ would give $P=\$11.00$ from the first equation, which fails the original reserve recursion.

Worked Example: Use a Third Duration as a Consistency Check

For a second hypothetical policy, suppose the level premium and end-of-year death benefit are unknown. Premiums are paid at the start of each year; reserves are measured immediately before premiums; expenses are excluded; and $i=0.10$. The known reserve sequence is $V_0=\$10.00$, $V_1=\$12.50$, $V_2=\$13.50$, and $V_3=\$9.25$. The successive one-year death probabilities are $0.20$, $0.50$, and $0.80$. Find $P$ and $B$, and check the third transition.

Method. Calculate all three values of $c_k$, solve using the first two equations, and substitute into the third. This tests whether the additional reserve observation is consistent with the same contract parameters.

For the first transition, $p_0=0.80$:

$$ c_0=1.10(10.00)-0.80(12.50)=11.00-10.00=1.00. $$

For the second transition, $p_1=0.50$:

$$ c_1=1.10(12.50)-0.50(13.50)=13.75-6.75=7.00. $$

The first two equations are $0.20B-1.10P=1.00$ and $0.50B-1.10P=7.00$. Their difference gives $0.30B=6.00$, hence $B=\$20.00$. Substitution into the first equation gives

0.20(20)-1.10P=1, \qquad P=\frac{4-1}{1.10} =\frac{30}{11} \approx \$2.73.

For the third transition, $p_2=0.20$. Its reserve-data quantity is

$$ c_2=1.10(13.50)-0.20(9.25)=14.85-1.85=13.00. $$

The fitted parameters give $0.80(20)-1.10(30/11)=16-3=13$, as required. The third equation is consistent. The premium is rounded to the nearest cent only at the final step; the exact value is $30/11$ dollars. The solution method avoids the mistake of treating a third observation as automatically valid: it must satisfy the same recursion.

Worked Example: Solve from Two Nonconsecutive Durations

A hypothetical policy has the same level-premium, level-benefit structure and reserve timing as above, with no expenses. The effective annual interest rate is $i=0.04$. At one observed duration, the one-year death probability is $0.10$, the reserve is $\$40$, and the next reserve is $\$44$. At another duration, the one-year death probability is $0.30$, the reserve is $\$15$, and the next reserve is $\$8$. The observations need not be consecutive. Find the common premium and benefit.

Method. Apply the recursion separately to each observed year. The durations may be nonconsecutive because each equation relates a reserve to its own next-duration reserve.

For the first observation, $p=0.90$, and for the second, $p=0.70$. Thus

$$ c_1=1.04(40)-0.90(44)=41.60-39.60=2.00, $$

$$ c_2=1.04(15)-0.70(8)=15.60-5.60=10.00. $$

The equations are $0.10B-1.04P=2.00$ and $0.30B-1.04P=10.00$. Subtracting yields $0.20B=8.00$, so $B=\$40.00$. Substituting into the first equation gives

0.10(40)-1.04P=2, \qquad P=\frac{4-2}{1.04} =\frac{25}{13} \approx \$1.92.

Check. The first equation gives $4-1.04(25/13)=4-2=2$. The second gives $12-2=10$. Therefore both observations support a $\$40.00$ benefit and a premium of $\$1.92$ when rounded to the nearest cent. The exact premium used in the check is $25/13$ dollars.

Identification, Checks, and Exam Traps

The difference $q_2-q_1$ controls whether two equations can separate the unknown benefit from the unknown premium. When it is zero, both equations have the same coefficients. When it is nonzero but small, the algebra gives a unique answer, but small errors in the reserve or mortality inputs can have a comparatively large effect on the estimated benefit. Always report whether the equations agree and whether the resulting values are plausible under the stated hypothetical contract.

A negative calculated premium or benefit is not automatically an algebra error, but it is a signal to recheck the assumptions, reserve timing, mortality probabilities, and arithmetic. Under a conventional policy with a positive level premium and positive death benefit, such a result may indicate that the supplied observations are not compatible with the assumed model.

Premium accumulation: with a start-of-year premium and an end-of-year reserve equation, use $(1+i)P$, not $P$ or $iP$.
Survival probability: the next-duration reserve is weighted by $p_k=1-q_k$, not by $q_k$.
Sign when subtracting: keep the equation in the form $q_kB-(1+i)P=c_k$; subtracting two such equations gives $(q_2-q_1)B=c_2-c_1$.
Consistent timing: do not combine a reserve measured before a premium with an equation derived for a reserve measured after that premium.
Extra observations: check every available equation. A mismatch means the assumed model and data do not all fit, even if two equations produce a numerical answer.

Key takeaway: Compute $c_k=(1+i)V_k-p_kV_{k+1}$ for each observed transition, solve $q_kB-(1+i)P=c_k$, and verify the results against all available reserve equations. A unique solution from two durations requires different one-year mortality probabilities.

Check Your Understanding

For each problem, use the reserve timing and assumptions stated in the question. Keep exact values through the calculations and round money only where requested.

A hypothetical policy has a level start-of-year premium, a level end-of-year death benefit, no expenses, and $i=0.05$. At one duration, $q=0.20$, $V_k=\$20$, and $V_{k+1}=\$20$. At another, $q=0.50$, $V_j=\$30$, and $V_{j+1}=\$20$. Find $P$ and $B$. (Answer: $B=\$55.00$, $P\approx \$5.71$.)
For a similar hypothetical policy, $i=0.04$. One observed transition has $q=0.10$, $V_k=\$10$, and $V_{k+1}=\$10$. Another has $q=0.30$, $V_j=\$20$, and $V_{j+1}=\$10$. Find the common premium and benefit. (Answer: $B=\$62.00$, $P\approx \$4.62$.)
A hypothetical contract has no expenses, start-of-year premiums, end-of-year benefits, and $i=0.08$. Three observed transitions are $(q,V_k,V_{k+1})=(0.20,20,19.90)$, $(0.50,30,23.44)$, and $(0.80,40,37.60)$. Solve using the first two transitions and check the third. (Answer: $B=\$50.00$, $P=\$4.00$; the third transition is consistent.)
Two observed transitions for a hypothetical policy have the same mortality probability $q=0.20$ and the same calculated value $c_k=3.00$. The effective annual interest rate is $0.06$, and both equations have the stated reserve timing and premium structure. Can these equations determine a unique $P$ and $B$? (Answer: No. Both reduce to $0.20B-1.06P=3.00$, so they provide only one independent equation.)

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