Closed-Form Values Under Constant Intensities

Why Constant Intensities Give Closed Forms

When transition intensities are constant, the probability of being disabled at a future time can be written as a combination of exponential terms. That makes continuous disability benefits amenable to direct integration rather than numerical valuation. The key is to distinguish time spent in the disabled state from the length of a single disability spell: recovery can return a person to active status, from which disability may occur again.

We use a hypothetical model with three states: active, disabled and dead. The model is a time-homogeneous continuous-time Markov chain. While active, a person becomes disabled with intensity $\lambda$ and dies with intensity $\mu$. While disabled, a person recovers with intensity $\rho$ and dies with intensity $\nu$. Death is absorbing. All intensities are constant and measured per year. Thus, conditional on the current state, transition rates do not depend on how long the person has already occupied that state.

Definition: Let $p_A(t)$ and $p_D(t)$ be the probabilities of being active and disabled at time $t$, respectively, for the specified initial state. The transition intensities are $\lambda$ (active to disabled), $\mu$ (active to death), $\rho$ (disabled to active) and $\nu$ (disabled to death). A continuous benefit of rate $b$ is paid per year while disabled; its expected present value over a payment interval is the integral of $b$ times the discounted disabled-state occupancy probability over that interval.

The total rate of leaving the active state is $\alpha=\lambda+\mu$, and the total rate of leaving the disabled state is $\beta=\rho+\nu$. These totals describe exits from each state; they do not by themselves describe the probability of occupying a state later, because a recovery can be followed by a new disability transition.

Deriving the Disabled-State Occupancy Probability

Over a short interval of length $h$, an active person has probability $\lambda h+o(h)$ of moving to disabled, and a disabled person has probability $\rho h+o(h)$ of moving to active. The forward equations for the two transient-state probabilities are therefore

\begin{aligned} p_A'(t)&=-\alpha p_A(t)+\rho p_D(t),\\ p_D'(t)&=\lambda p_A(t)-\beta p_D(t). \end{aligned}

The death state is not included as an unknown in this system because it is absorbing and there are no transitions back from death. Define the two positive decay rates

k_1=\frac{\alpha+\beta-\sqrt{(\alpha-\beta)^2+4\lambda\rho}}{2}, \qquad k_2=\frac{\alpha+\beta+\sqrt{(\alpha-\beta)^2+4\lambda\rho}}{2}, \qquad k_1<k_2.

For positive transition rates, these are distinct. The corresponding exponential terms solve the second-order equation obtained by differentiating the equation for $p_D$ and substituting for $p_A$. Specifically,

p_D''+(\alpha+\beta)p_D'+(\alpha\beta-\lambda\rho)p_D=0.

Its characteristic roots are $-k_1$ and $-k_2$. If the person starts active, then $p_D(0)=0$ and $p_D'(0)=\lambda$. If the person starts disabled, then $p_D(0)=1$ and $p_D'(0)=-\beta$. Applying these initial conditions gives both occupancy formulas.

Theorem: Write $\Delta=k_2-k_1$. For an initial active state, $$ p_D(t)=\frac{\lambda}{\Delta}\left(e^{-k_1t}-e^{-k_2t}\right). $$ For an initial disabled state, $$ p_D(t)=\frac{k_2-\beta}{\Delta}e^{-k_1t} +\frac{\beta-k_1}{\Delta}e^{-k_2t}. $$ These formulas apply to the stated constant-intensity model. They account for recovery and subsequent recurrence; they are not the probability that the initial disability spell is still in progress.

For example, in the active-start formula, the two initial conditions are verified directly: $p_D(0)=0$, and differentiating at zero gives $p_D'(0)=\lambda$. The decay rates also satisfy $k_1+k_2=\alpha+\beta$ and $k_1k_2=\alpha\beta-\lambda\rho$, as required by the second-order equation.

Continuous Benefit Values

Let the effective annual interest rate be $i$, and use the equivalent force of interest $\delta=\ln(1+i)$, so a payment at time $t$ is discounted by $e^{-\delta t}$. For a level benefit rate $b$ paid continuously while disabled from time $a$ to time $n$, the expected present value is $b\int_a^n e^{-\delta t}p_D(t)\,dt$. The interval endpoints satisfy $0\leq a\leq n$; taking $a=0$ gives immediate payment eligibility.

For concise expressions, define

J(k;a,n)=\int_a^n e^{-(\delta+k)t}\,dt =\frac{e^{-(\delta+k)a}-e^{-(\delta+k)n}}{\delta+k}.

Substituting the occupancy probabilities into the benefit integral gives closed forms for either initial state.

Result: For an initial active state, the value of a continuous benefit over $[a,n]$ is $$ V_A=b\frac{\lambda}{\Delta}\left[J(k_1;a,n)-J(k_2;a,n)\right]. $$ For an initial disabled state, it is $$ V_D=b\left[ \frac{k_2-\beta}{\Delta}J(k_1;a,n) +\frac{\beta-k_1}{\Delta}J(k_2;a,n) \right]. $$ For a benefit payable indefinitely, take $n\to\infty$; then $J(k;a,\infty)=e^{-(\delta+k)a}/(\delta+k)$. The formulas assume a constant benefit rate while disabled and no separate benefit cap or escalation.

The reason discounting combines with the decay rates is visible in the integrand: each occupancy term $e^{-kt}$, multiplied by the discount factor $e^{-\delta t}$, becomes $e^{-(\delta+k)t}$. In a model with recovery, however, $k_1$ and $k_2$ are the decay rates of the whole active-disabled system; simply replacing a disabled-state exit intensity by $\beta+\delta$ does not produce the full value.

Worked Example: Perpetual Benefit From an Initially Active State

A hypothetical policy pays a continuous benefit of $12,000 per year whenever its insured is disabled, with no maximum benefit period. Assume the insured starts active. The constant annual transition intensities are $\lambda=0.10$ for active-to-disabled, $\mu=0.10$ for active-to-death, $\rho=0.10$ for disabled-to-active and $\nu=0.10$ for disabled-to-death. Holding times are exponential as implied by the constant-intensity Markov model. The effective annual interest rate is $i=e^{0.05}-1$, equivalent to force $\delta=0.05$. Find the expected present value.

Method. Find the occupancy decay rates, then integrate the discounted active-start occupancy over the infinite payment interval.

First, $\alpha=\lambda+\mu=0.20$ and $\beta=\rho+\nu=0.20$. Therefore

k_1=\frac{0.40-\sqrt{0.04}}{2}=0.10, \qquad k_2=\frac{0.40+\sqrt{0.04}}{2}=0.30, \qquad \Delta=0.20.

At time $t=1$, for an occupancy check, the probability of being disabled is

p_D(1)=\frac{0.10}{0.20}(e^{-0.10}-e^{-0.30}) =0.5(0.904837-0.740818)=0.0820.

For the perpetual benefit, $J(k;0,\infty)=1/(\delta+k)$. Thus the discounted occupancy factor is

\frac{0.10}{0.20}\left(\frac{1}{0.05+0.10}-\frac{1}{0.05+0.30}\right) =0.5(6.666667-2.857143)=1.904762.

The expected present value is $12{,}000(1.904762)=\$22{,}857.14$, rounded to the nearest cent. The method includes benefits from later disability spells after recovery. The occupancy probability at one year is between zero and one, and the resulting value is positive, as required.

Worked Example: Finite Benefit From an Initially Disabled State

A hypothetical contract pays $1,000 per year continuously while disabled for at most four years. The insured is disabled at time zero. Assume $\lambda=\mu=\rho=\nu=0.10$ per year, with exponential holding times in each transient state, and effective annual interest $i=e^{0.05}-1$, so $\delta=0.05$. Find the expected present value over the four-year term.

Method. Use the disabled-start occupancy formula, since the person begins disabled and may recover, return to active status and become disabled again.

As in the preceding parameter calculation, $\alpha=\beta=0.20$, $k_1=0.10$, $k_2=0.30$, and $\Delta=0.20$. The coefficients for the disabled-start formula are

\frac{k_2-\beta}{\Delta}=\frac{0.30-0.20}{0.20}=0.5, \qquad \frac{\beta-k_1}{\Delta}=\frac{0.20-0.10}{0.20}=0.5.

For the first exponential term,

J(0.10;0,4)=\frac{1-e^{-(0.05+0.10)4}}{0.15} =\frac{1-e^{-0.60}}{0.15} =\frac{1-0.548812}{0.15}=3.007922.

For the second term,

J(0.30;0,4)=\frac{1-e^{-(0.05+0.30)4}}{0.35} =\frac{1-e^{-1.40}}{0.35} =\frac{1-0.246597}{0.35}=2.152580.

Hence the expected discounted disabled time is $0.5(3.007922)+0.5(2.152580)=2.580251$ years, and the expected present value is $1{,}000(2.580251)=\$2{,}580.25$, rounded to the nearest cent. This is less than the value of four years of certain continuous payment, which would be $1{,}000J(0;0,4)$; the difference reflects recovery, death and discounting.

Worked Example: Benefit Begins After a Waiting Period

A hypothetical policy pays $1,000 per year continuously while disabled, but pays nothing during the first year of a disability spell. For this valuation, assume the insured starts active and receives benefits only between policy times 1 and 4 if disabled. Use $\lambda=0.10$, $\mu=0.10$, $\rho=0.10$ and $\nu=0.10$ per year, exponential holding times, and effective annual interest $i=e^{0.05}-1$. The force is $\delta=0.05$. Find the expected present value over the stated payment interval.

Method. The occupancy model is unchanged; restrict the benefit integral to $[1,4]$. In particular, do not subtract a waiting period from the occupancy probabilities.

Here $k_1=0.10$, $k_2=0.30$, $\Delta=0.20$, and the insured starts active. The two discounted integrals are

\begin{aligned} J(0.10;1,4) &=\frac{e^{-0.15}-e^{-0.60}}{0.15} =\frac{0.860708-0.548812}{0.15}=2.079309,\\ J(0.30;1,4) &=\frac{e^{-0.35}-e^{-1.40}}{0.35} =\frac{0.704688-0.246597}{0.35}=1.308832. \end{aligned}

The expected discounted disabled time in the payment interval is $0.10/0.20\,(2.079309-1.308832)=0.385239$ years. Multiplying by the annual benefit rate gives $1{,}000(0.385239)=\$385.24$, rounded to the nearest cent. The value is positive and lower than the value for immediate eligibility over the same overall term, as expected.

Common Mistake: Confusing Spell Termination With State Occupancy

A common error is to treat the probability of being disabled at time $t$ as the probability that the original disability spell has not ended. For a person initially disabled, that incorrect calculation is $e^{-\beta t}$, where $\beta=\rho+\nu$. It ignores both recovery followed by re-entry into disability and the full system of transitions.

In the parameter setting above, at $t=1$ the incorrect calculation gives $e^{-0.20}=0.818731$. The correct disabled-state occupancy is

p_D(1)=0.5e^{-0.10}+0.5e^{-0.30} =0.5(0.904837+0.740818)=0.822828.

The correct expression includes all paths that end in the disabled state at time 1, not just the path with no transition out of the initial disability spell. Discounting is applied to the benefit at each payment time; it does not change the transition intensities or remove recovery and recurrence from the occupancy model.

Key takeaway: Under constant transition intensities, disabled-state occupancy is a sum or difference of exponential terms. Integrating each term after multiplying by $e^{-\delta t}$ gives a closed-form continuous benefit value. Use the correct initial-state formula and the actual interval during which benefits are payable.

Check Your Understanding

For each problem, assume the same three-state Markov structure defined above, with constant intensities per year and an absorbing death state.

Let $\lambda=0.08$, $\mu=0.02$, $\rho=0.10$ and $\nu=0.04$. Find $\alpha$, $\beta$ and the two decay rates $k_1,k_2$. (Answer: $\alpha=0.10,\ \beta=0.14,\ k_1=(0.24-\sqrt{0.0336})/2\approx0.0283,\ k_2\approx0.2117$.)
An initially active person has $\lambda=\mu=\rho=\nu=0.10$. Find the probability of being disabled at time 2. (Answer: $0.5(e^{-0.20}-e^{-0.60})\approx0.1350$.)
For the same intensities, an initially active person receives a continuous benefit of $500 per year while disabled for two years. The effective annual interest rate is $i=e^{0.05}-1$. Find the value. (Answer: $\$72.39$, using $250[J(0.10;0,2)-J(0.30;0,2)]$.)
An initially disabled person receives $750 per year continuously while disabled for life. Assume $\lambda=\mu=\rho=\nu=0.10$ and $i=e^{0.05}-1$. Find the value. (Answer: $\$3{,}571.43$, using $750[0.5/(0.15)+0.5/(0.35)]$.)
An initially active person has the same four intensities and receives $1,000 per year continuously while disabled, but only from time 1 to time 3. With $i=e^{0.05}-1$, find the value. (Answer: $\$236.81$, using $500[J(0.10;1,3)-J(0.30;1,3)]$.)

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