When State Probabilities Cannot Capture Payment Rules

Why Current State May Not Be Enough

A state model describes where a policy is at a particular time: for example, active, disabled, or deceased. A benefit that is payable whenever the policy is disabled at a payment date can be valued using the probability of being disabled at that date. But some hypothetical contracts impose a condition on how the policy reached its current state. A benefit might be paid only on the first disability onset, only after a return to active status, or only after two consecutive years of disability.

Two policies can be in the same current state but have different payment eligibility because their histories differ. The probability of being in that state does not, by itself, reveal how many policies satisfy the history condition. The solution is to track the relevant history—either by calculating probabilities of the required paths or by expanding the state description to include a memory of that history.

Definition: A payment rule is path-dependent when eligibility or amount depends on the sequence of states or transitions before the payment date, not only on the state occupied at that date. A payment rule is determined by the current state when the current state alone is sufficient to decide whether the payment is due and its amount.

Let $Y_t$ be the original state at time $t$, and let $p_j(t)=P(Y_t=j)$ be the probability of occupying state $j$ then. These are marginal state probabilities: they describe the distribution at one time, not the complete paths leading to that time. For a benefit $B$ payable at time $t$ only if event $C$ has occurred, its expected present value is $Bv^tP(C)$, where $v=1/(1+i)$ and $i$ is the effective annual interest rate. Substituting $p_j(t)$ for $P(C)$ is valid only if $C$ is exactly the event of occupying the eligible state at time $t$.

Path Probabilities and State Expansion

For a discrete-time Markov model, a transition probability gives the probability of the next state conditional on the current state. A path probability is found by multiplying the relevant transition probabilities, provided the model’s Markov assumption holds. If the payment depends on a particular path, its probability can be calculated from those path probabilities.

Another method is to add a history variable to the state. The augmented state records the original state and the minimum additional information needed to decide future eligibility. For example, instead of recording only “active,” distinguish “active, never disabled” from “active, previously disabled.” The augmented description is useful only if it retains all history that matters for the payment rule; irrelevant details need not be tracked.

State the eligibility rule.
Identify exactly which past events, transitions, or durations affect payment.

Check what the original state records.
If two different histories collapse into the same current state but lead to different eligibility, current-state probabilities alone do not suffice.

Retain the needed history.
Calculate the qualifying path probability directly, or define augmented states that distinguish histories relevant to future payments.

Value only qualifying payments.
Multiply each payment by the probability of its precise eligibility event and discount it to the valuation date.

Worked Examples

Worked Example: A Benefit on First Disability Onset at Year Two

A hypothetical contract starts with a policyholder active at time 0. At each year-end transition, the probabilities are $P(A\text{ to }A)=0.60$, $P(A\text{ to }D)=0.40$, $P(D\text{ to }A)=0.50$, and $P(D\text{ to }D)=0.50$, where $A$ is active and $D$ is disabled. The transitions follow a time-homogeneous Markov model. A benefit of $300 is paid at time 2 only if disability onset occurs at that transition, so the policyholder must be active at time 1 and disabled at time 2. Interest is 5% effective annually. Find the expected present value.

Method. Find the probability of the qualifying two-year path, then discount the payment at time 2. The payment event is the path $A$ at time 1 followed by $D$ at time 2.

The probability of being active at time 1 is $0.60$. Conditional on being active then, the probability of entering disability at time 2 is $0.40$. Thus

$$ P(A_1,D_2)=0.60(0.40)=0.24. $$

For comparison, the probability of being disabled at time 2 includes both paths that were already disabled at time 1 and paths entering disability at time 2:

P(D_2)=P(D_1)P(D_2\mid D_1)+P(A_1)P(D_2\mid A_1) =0.40(0.50)+0.60(0.40)=0.44.

The discount factor is $v=1/1.05$, so $v^2=1/1.1025$. The expected present value is therefore

300(0.24)v^2=\frac{72}{1.1025}=\$65.3061\ldots=\$65.31.

Check. The qualifying probability, $0.24$, is less than the probability of being disabled at time 2, $0.44$, because some policies disabled at time 2 were already disabled at time 1. The value is positive and less than the undiscounted expected payment, $300(0.24)=\$72$. The common mistake is to use $P(D_2)=0.44$ as the eligibility probability; that would give $132/1.1025=\$119.73$, incorrectly paying on paths that do not have a new onset at time 2.

Worked Example: Payment While Active After Prior Disability

A hypothetical contract begins in state $A$, active with no prior disability. Its annual transition probabilities are $P(A\text{ to }D)=0.30$, $P(A\text{ to }A)=0.70$, $P(D\text{ to }A)=0.40$, and $P(D\text{ to }D)=0.60$. At time 2, it pays $200 if the policyholder is active and has been disabled at least once by then. Interest is 4% effective annually. Find the expected present value and show how to encode the history.

Method. Use augmented states $A_0$ (active, never disabled), $D_1$ (disabled, so disability has occurred), and $A_1$ (active after at least one period in disability). The subscript distinguishes histories relevant to eligibility. The augmented transitions are $A_0$ to $A_0$ with probability $0.70$, $A_0$ to $D_1$ with probability $0.30$, $D_1$ to $A_1$ with probability $0.40$, $D_1$ to $D_1$ with probability $0.60$, $A_1$ to $A_1$ with probability $0.70$, and $A_1$ to $D_1$ with probability $0.30$.

With state order $(A_0,D_1,A_1)$, the transition matrix and initial distribution are

P= \begin{pmatrix} 0.70&0.30&0\\ 0&0.60&0.40\\ 0&0.30&0.70 \end{pmatrix}, \qquad \boldsymbol{\pi}_0=(1,0,0).

After one year, $\boldsymbol{\pi}_1=(0.70,0.30,0)$. Multiplying by the transition matrix gives

\boldsymbol{\pi}_2 =(0.70,0.30,0)P =(0.49,\;0.70(0.30)+0.30(0.60),\;0.30(0.40)) =(0.49,0.39,0.12).

Only $A_1$ satisfies the payment condition. The expected present value is

200(0.12)v^2=\frac{24}{1.04^2} =\frac{24}{1.0816} =\$22.1893\ldots=\$22.19.

Check. At time 2, the original active-state probability is $0.49+0.12=0.61$, but only $0.12$ of the probability is active after disability. The common mistake is to use the full active probability, which would incorrectly include $A_0$, the active policies that have never been disabled.

Worked Example: Benefit After Two Consecutive Years Disabled

A hypothetical contract starts active. Each year, the transition probabilities are $P(A\text{ to }D)=0.50$, $P(A\text{ to }A)=0.50$, $P(D\text{ to }D)=0.60$, and $P(D\text{ to }A)=0.40$. A $500 benefit is payable at time 2 only if the policyholder was disabled at both time 1 and time 2. Interest is 6% effective annually. Calculate the expected present value.

Method. Track the number of consecutive disabled years, capped at two, using states $A$, $D_1$ (one consecutive year disabled), and $D_{2+}$ (at least two consecutive years disabled). The transition from $D_1$ to $D_{2+}$ has probability $0.60$; a return to $A$ resets the consecutive-duration count.

After one year, the probabilities are $P(A_1)=0.50$, $P(D_{1,1})=0.50$, and $P(D_{2+,1})=0$. At time 2, the qualifying probability is the probability of the path $A,D,D$:

$$ P(D_1,D_2)=0.50(0.60)=0.30. $$

The current disabled-state probability at time 2 is larger. It includes both the qualifying path and paths that first enter disability at time 2:

$$ P(D_2)=0.50(0.60)+0.50(0.50)=0.30+0.25=0.55. $$

The benefit is paid at time 2, so the expected present value is

500(0.30)v^2=\frac{150}{1.06^2} =\frac{150}{1.1236} =\$133.4994\ldots=\$133.50.

Check. The expanded-state probability $P(D_{2+,2})=0.30$ is exactly the probability of two consecutive years disabled. Using the unexpanded probability $P(D_2)=0.55$ would value the benefit as $275/1.1236=\$244.75$, incorrectly including policies whose disability began only at time 2.

Why Augmented States Work

The construction is not merely a relabeling exercise. It works because the augmented state retains the information needed to update eligibility as the model moves forward. Suppose the original state process is Markov, and let $M_t$ record a relevant history feature, such as whether disability has ever occurred or how many consecutive periods have been spent disabled. Define $Z_t=(Y_t,M_t)$, updating $M_t$ deterministically from the previous memory and the next state. Then, when $M_t$ captures all history relevant to future payments, the next augmented state can be generated from the current augmented state without needing the full earlier path.

A payment condition expressed in terms of $Z_t$ can be valued using its state probabilities. If $b(z)$ is the amount payable at time $t$ in augmented state $z$, its expected present value at that date is

v^t\sum_z b(z)P(Z_t=z).

To see why, let $I_z$ equal 1 when $Z_t=z$ and 0 otherwise. The payment is the sum $\sum_z b(z)I_z$. Taking expectations gives $\sum_z b(z)E[I_z]=\sum_z b(z)P(Z_t=z)$, and discounting by $v^t$ gives the displayed value. This argument requires that the payment amount and eligibility really are determined by the augmented state. If not, the state must be expanded further, or the relevant path events must be valued directly.

Not every payment involving history requires a permanent state expansion. If a benefit is triggered by a specific transition, its expected value may be calculated from the probability of that transition, conditional on the state immediately before it. But an occupancy probability at the payment date alone does not generally identify how many policies made that transition or satisfy another historical condition.

Common Mistakes and Practical Checks

Replacing a path event with a state event. “Disabled at time 2” and “first becomes disabled at time 2” are different events. Write the eligibility event explicitly before using a probability.
Forgetting that histories can merge. An active policy that has never been disabled and one that has recovered may occupy the same original state but have different contractual eligibility.
Tracking too little or too much history. The augmented state must distinguish every history that can change a future payment, but need not record details that do not affect future eligibility or amount.
Discounting at the wrong date. In these examples the benefits are paid at the end of year 2, so their present values use $v^2$, not $v$.
Counting a payment more than once. For a once-only condition, the augmented memory must distinguish “not yet paid” from “already paid,” or the payment rule must otherwise ensure only the first qualifying event generates a payment.

Key takeaway: A marginal state probability answers “where is the policy now?” It does not necessarily answer “what happened before now?” When payment eligibility depends on history, identify the qualifying paths or expand the state to preserve exactly the history the rule requires.

Check Your Understanding

For each problem, use the stated payment date and interest convention. Where eligibility depends on history, distinguish that event from merely occupying the same current state.

A policy starts active. Each year $P(A\text{ to }A)=0.80$ and $P(A\text{ to }D)=0.20$. A $400 benefit is paid at time 2 only on a transition from active at time 1 to disabled at time 2. Interest is 5% effective annually. Find its expected present value. (Answer: $58.05)
A policy starts active, with $P(A\text{ to }D)=0.25$, $P(A\text{ to }A)=0.75$, $P(D\text{ to }A)=0.50$, and $P(D\text{ to }D)=0.50$. A $120 benefit is paid at time 2 if the policy is active then and has been disabled at least once. Interest is 5% effective annually. Find the expected present value. (Answer: $13.61)
A policy starts active. The annual probabilities are $P(A\text{ to }D)=0.40$, $P(A\text{ to }A)=0.60$, $P(D\text{ to }D)=0.50$, and $P(D\text{ to }A)=0.50$. A $1,000 benefit is paid at time 2 only if the policy was disabled at both time 1 and time 2. Interest is 5% effective annually. Find the expected present value. (Answer: $181.41)
A policy starts active, with $P(A\text{ to }D)=0.30$, $P(A\text{ to }A)=0.70$, $P(D\text{ to }A)=0.20$, and $P(D\text{ to }D)=0.80$. A $500 benefit is paid at time 2 if the policy is active then after having been disabled. Interest is 4% effective annually. Give the qualifying probability and expected present value. (Answer: Qualifying probability $0.30(0.20)=0.06$; expected present value $30/1.0816=\$27.74$)
A contract pays $250 at time 2 if the policy is active then and was disabled at time 1. Starting active, the probabilities are $P(A\text{ to }D)=0.30$, $P(A\text{ to }A)=0.70$, $P(D\text{ to }A)=0.40$, and $P(D\text{ to }D)=0.60$. Interest is 5% effective annually. What path qualifies, and what is the expected present value? (Answer: The path $A,D,A$ qualifies, with probability $0.30(0.40)=0.12$; expected present value $30/1.1025=\$27.21$)

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