Connecting Reserves Across Adjacent Durations

Why Adjacent-Duration Reserves Are Connected

A prospective reserve at a given duration is the expected present value of future insurer outgo, conditional on the policy being in a specified state at that duration. Over the next period, the policy may remain in that state, move to another state, or terminate. Each possible path can have its own cash flow and its own reserve at the end of the period. The reserve today must therefore reflect all those possibilities, weighted by their transition probabilities and discounted to today.

This connection is useful both for calculating reserves recursively and for checking them against a prospective valuation. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. The key is to keep the current state, the transition during the period, and the state at the next valuation time distinct.

Definition: Let $V_t(i)$ be the prospective reserve at time $t$, conditional on the policy being in state $i$ immediately before the period from $t$ to $t+1$. Let $p_{ij}(t)$ be the probability that a policy in state $i$ at time $t$ is in state $j$ at time $t+1$. Let $C_{ij}(t+1)$ be the net insurer outgo at time $t+1$ for that transition, with benefits treated as positive outgo and premiums received as negative outgo. If $v=1/(1+i)$ is the one-period discount factor, then, under the assumptions below, $$ V_t(i)=v\sum_j p_{ij}(t)\bigl(C_{ij}(t+1)+V_{t+1}(j)\bigr). $$

Assumptions and Interpretation

This one-period relation assumes a discrete valuation grid and a constant effective interest rate $i$ over the period. The transition probabilities are conditional on the state at time $t$, and the listed possibilities for the next state are mutually exclusive and exhaustive, so $\sum_j p_{ij}(t)=1$. The cash flow $C_{ij}(t+1)$ is paid at the end of the period on the transition from $i$ to $j$. The reserve $V_{t+1}(j)$ is measured at that same time, after that end-of-period cash flow.

A cash flow that depends on more than the destination state must still be indexed by the transition. For example, a payment on disability entry is associated with the transition from active to disabled, not with every policy that is disabled at the next valuation time. If a contract has a cash flow at time $t$, before the period begins, that cash flow must be handled separately according to the valuation-time convention; it should not be silently included as an end-of-period payment.

The formula also makes clear why reserves are state-specific. $V_t(i)$ conditions on being in state $i$, and the relevant transition probabilities are the row of probabilities starting from $i$. A population-average reserve may be calculated from state-specific reserves and a state distribution, but it is not a substitute for the conditional reserve.

Why the Recursion Holds

Condition on the state reached at time $t+1$. If the policy moves from state $i$ to state $j$, the insurer pays net outgo $C_{ij}(t+1)$ at that time and then has a liability $V_{t+1}(j)$. Their sum is the value, at time $t+1$, of the period-end cash flow and the remaining contract. Its value at time $t$ is multiplied by $v$. Taking the conditional expectation over the possible destination states gives the recursion.

List possible transitions.
For the current state $i$, identify every possible state $j$ at the next valuation time and its probability $p_{ij}(t)$.

Value each path at time $t+1$.
For each transition, add its net outgo $C_{ij}(t+1)$ to the next-state reserve $V_{t+1}(j)$.

Average and discount.
Weight each path value by its transition probability, sum, and multiply by $v$.

Equivalently, multiplying through by $1+i$ gives a useful accumulation form:

(1+i)V_t(i)=\sum_j p_{ij}(t)\bigl(C_{ij}(t+1)+V_{t+1}(j)\bigr).

This form says that the reserve accumulated for one period equals the expected period-end outgo plus the expected next reserve. It is a direct check on the timing: a cash flow at $t+1$ is not discounted relative to $V_{t+1}$, since both are valued at $t+1$.

Worked Example: Calculating Reserves in an Active–Disabled–Death Model

A hypothetical policy is valued annually at an effective annual interest rate of $5\%$. At duration $t$, an active policy has probabilities $0.80$ of remaining active, $0.15$ of becoming disabled, and $0.05$ of death. The end-of-year benefits on these transitions are respectively $0$, $\$10$, and $\$100$. At duration $t+1$, the reserves are $\$40$ if active, $\$70$ if disabled, and $\$0$ after death. Calculate the active reserve at time $t$.

Method. Apply the transition-weighted one-period recursion. The discount factor is $v=1/1.05$. The path values at time $t+1$ are $\$0+\$40=\$40$ for active-to-active, $\$10+\$70=\$80$ for active-to-disabled, and $\$100+\$0=\$100$ for active-to-death.

\begin{aligned} V_t(\mathrm{active}) &=\frac{1}{1.05}\left[0.80(40)+0.15(10+70)+0.05(100)\right]\\ &=\frac{1}{1.05}\left[32+12+5\right]\\ &=\frac{49}{1.05}\\ &=\$46.67. \end{aligned}

Check. Accumulating the reserve for one year gives $1.05(46.6667)=49$, equal to the probability-weighted end-of-year value $32+12+5=49$, up to rounding. The common mistake avoided here is to use only the next-state reserves and omit the $\$10$ disability-entry benefit and $\$100$ death benefit.

Transition Cash Flows and Premiums

The transition cash flow is net outgo, so a premium received at the end of the period is negative. It should be included only on the transitions on which it is actually payable. The following example illustrates why a premium received only when the policy remains active cannot be subtracted from every possible outcome.

Worked Example: A Premium Payable Only on Continued Active Status

A hypothetical policy has a one-year effective interest rate of $4\%$. From the active state, the next state is active with probability $0.90$, disabled with probability $0.08$, or dead with probability $0.02$. At the end of the year, a $\$12$ premium is received only if the policy remains active. A $\$30$ benefit is paid on transition to disability and a $\$150$ benefit on death. The next-duration reserves are $\$20$ in the active state, $\$55$ in the disabled state, and $\$0$ after death. Find the active reserve.

Method. Write the net outgo separately for each transition. Active-to-active has net outgo $-\$12$; active-to-disabled has net outgo $\$30$; and active-to-death has net outgo $\$150$. The corresponding path values are $-12+20=8$, $30+55=85$, and $150+0=150$.

\begin{aligned} V_t(\mathrm{active}) &=\frac{1}{1.04}\left[0.90(8)+0.08(85)+0.02(150)\right]\\ &=\frac{1}{1.04}\left[7.2+6.8+3.0\right]\\ &=\frac{17}{1.04}\\ &=\$16.35. \end{aligned}

Check and trap. Accumulating the unrounded reserve gives $1.04(16.3461538)=17$, the expected net period-end value. A common wrong calculation subtracts the $\$12$ premium on every outcome, as though it were received even after disability or death. That would give $(27.8-12)/1.04=\$15.19$, which is incorrect: the premium applies only to the active-to-active transition.

Using the Relation in Reverse

The same equation can be rearranged to solve for an unknown next-duration reserve, a missing transition payment, or a reserve check. In particular, when the transition probability to a state is positive, the corresponding next-state reserve can be isolated algebraically. This is useful when a reserve table is given at one duration and a single entry at the next duration is unknown.

Worked Example: Recovering an Unknown Disabled-State Reserve

A hypothetical contract has an effective annual interest rate of $5\%$. At time $t$, an active policy remains active with probability $0.80$, becomes disabled with probability $0.15$, and dies with probability $0.05$. The period-end net outgo is $-\$5$ on active-to-active, $\$25$ on active-to-disabled, and $\$100$ on death. The next-duration active reserve is $\$30$, the death reserve is $\$0$, and the current active reserve is $\$40$. Find the next-duration disabled reserve $x$.

Method. Use the accumulated form of the recursion. The path values are $-5+30=25$ on active-to-active, $25+x$ on active-to-disabled, and $100+0=100$ on death.

\begin{aligned} 1.05(40)&=0.80(25)+0.15(25+x)+0.05(100)\\ 42&=20+3.75+0.15x+5\\ 42&=28.75+0.15x\\ 13.25&=0.15x\\ x&=\frac{13.25}{0.15}=\$88.33. \end{aligned}

Check. Substituting the unrounded value $x=88.3333$ gives $20+0.15(113.3333)+5=20+17+5=42$, matching the accumulated current reserve. The result is positive and is consistent with the larger benefit and liability represented by the disabled-state path. The common mistake avoided is dividing by the transition probability before accounting for the other destination-state contributions.

State-Specific Checks and Common Traps

A reserve recursion is more than a computational shortcut. It provides a consistency check between the reserve at time $t$ and the cash flows and reserves at time $t+1$. For each current state, verify that its own probability row is used, its own transition-dependent payments are included, and the complete expected value is discounted exactly one period.

Mixing conditional and unconditional values. Start with the probabilities conditional on the current state. Do not use an overall population probability unless the question explicitly asks for a population-level quantity.
Applying a transition benefit to the wrong path. A benefit on entry to disability is attached to transitions into disability, not automatically to every disabled-state reserve.
Discounting the next reserve twice. The cash flow and next reserve are both valued at $t+1$, so add them first and then apply one factor $v$.
Ignoring the timing of premiums. A premium is negative net outgo only on transitions where it is payable, and only at the specified payment time.
Using an incomplete transition row. Include termination or death as a destination when it is possible. Its next reserve may be zero, but its terminal benefit need not be zero.

If there is a terminal duration $n$, the boundary reserve $V_n(j)$ must be specified according to the contract’s maturity or termination terms. The one-period relation then links earlier reserves back from that boundary. It does not determine the boundary condition by itself.

Key takeaway: For each current state, add the transition-specific net outgo to the reserve for the destination state, weight by the corresponding transition probability, and discount the expectation one period. The resulting conditional value is the reserve at the current duration.

Check Your Understanding

For each problem, apply the one-period relation and keep the stated cash-flow timing and current state explicit.

A single-state policy remains in force with probability $0.80$ and terminates with probability $0.20$. At the end of the year, the continuing-state reserve is $\$30$, the termination benefit is $\$50$, and the termination reserve is zero. If $i=5\%$, find the current reserve. (Answer: $\$32.38$)
From state A, a policy moves to A, D, or death with probabilities $0.70$, $0.25$, and $0.05$. The end-of-year benefits are respectively $\$0$, $\$20$, and $\$100$; next-state reserves are $\$40$, $\$60$, and $\$0$. If $i=4\%$, find $V_t(\mathrm{A})$. (Answer: $\$50.96$)
At $i=5\%$, an active policy has probabilities $0.80$, $0.15$, and $0.05$ of remaining active, becoming disabled, or dying. Net end-of-year outgo is $-\$5$, $\$25$, and $\$100$, respectively. Given $V_t(\mathrm{active})=\$40$, $V_{t+1}(\mathrm{active})=\$30$, and $V_{t+1}(\mathrm{death})=0$, find $V_{t+1}(\mathrm{disabled})$. (Answer: $\$88.33$)
A policy in state A moves to A with probability $0.60$ or D with probability $0.40$. At the end of the period, the transition outgo is $\$0$ on A-to-A and $\$10$ on A-to-D. The next reserves are $\$20$ in A and $\$50$ in D. If $i=0\%$, find $V_t(\mathrm{A})$. (Answer: $\$36.00$)
A population has $60\%$ of policies in state A and $40\%$ in state D. Their conditional reserves are $\$40$ and $\$70$, respectively. Find the population-average reserve. (Answer: $\$52.00$)

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