Numerical Valuation with Piecewise-Constant Transition Rates

Why Piecewise-Constant Rates Need Careful Valuation

In a multi-state model, transition intensities can change with duration, calendar time, or an assumed change in the insured's circumstances. A single transition matrix is then not enough to describe the whole valuation period. Divide the period into intervals over which the intensities are constant, calculate a transition matrix for each interval, and combine those matrices in time order. Discount cash flows at the time they occur, rather than treating all payments as if they were made at an interval boundary.

The calculations below use a continuous-time Markov model. The state set is fixed, transition intensities are constant within each interval, and the transition process follows the Markov property: conditional on the current state, future transitions do not depend on the earlier path. Interest is represented by a constant force of interest $\delta$. The examples are hypothetical.

Definition: For states $1,\ldots,n$, the generator $Q$ has off-diagonal entries $q_{ij}$, the instantaneous intensity of transition from state $i$ to state $j$. Its diagonal entries are $q_{ii}=-\sum_{j\ne i}q_{ij}$. For constant $Q$ over an interval of length $h$, the transition matrix is $P(h)=\exp(Qh)$. With row-vector state probabilities, the probability row vector at the end is the starting row vector multiplied on the right by $P(h)$.

The off-diagonal entries of a transition matrix are probabilities, not intensities. They account for all possible paths during the interval, including multiple transitions. In particular, multiplying an intensity by the interval length gives only an approximation to a transition probability, not the exact probability except as a short-interval approximation.

Calculating and Composing Transition Matrices

For two states, with intensity $\alpha$ from state A to state D and intensity $\beta$ from D to A, the generator and the transition matrix over $h$ years are

Q= \begin{pmatrix} -\alpha & \alpha\\ \beta & -\beta \end{pmatrix}, \qquad P(h)= \begin{pmatrix} \frac{\beta}{\alpha+\beta}+\frac{\alpha}{\alpha+\beta}e^{-(\alpha+\beta)h} & \frac{\alpha}{\alpha+\beta}\left(1-e^{-(\alpha+\beta)h}\right) \\[4pt] \frac{\beta}{\alpha+\beta}\left(1-e^{-(\alpha+\beta)h}\right) & \frac{\alpha}{\alpha+\beta}+\frac{\beta}{\alpha+\beta}e^{-(\alpha+\beta)h} \end{pmatrix}.

If the rates change after an interval, calculate a new matrix using the new generator. For intervals of lengths $h_1,\ldots,h_m$, with generators $Q_1,\ldots,Q_m$, the overall transition matrix is $P_1(h_1)P_2(h_2)\cdots P_m(h_m)$. The order matters: the matrix for the first interval is on the left.

Worked Example: Composing Two Different Rate Intervals

A hypothetical two-state model starts in A. For the first year, the intensities are $\alpha_1=0.2$ and $\beta_1=0.3$ per year. For the next half-year, they are $\alpha_2=0.4$ and $\beta_2=0.1$ per year. Find the probability of being in D after 1.5 years.

Method and setup. Use the two-state matrix formula in each interval, then multiply the matrices in chronological order. In the first interval, $\alpha_1+\beta_1=0.5$, so $e^{-0.5(1)}=0.6065307$. Thus

P_1= \begin{pmatrix} 0.8426123 & 0.1573877\\ 0.2360816 & 0.7639184 \end{pmatrix}.

For the second interval, $\alpha_2+\beta_2=0.5$ and $h_2=0.5$, so $e^{-0.5(0.5)}=0.7788008$. Therefore

P_2= \begin{pmatrix} 0.8230406 & 0.1769594\\ 0.0442398 & 0.9557602 \end{pmatrix}.

The initial distribution is $(1,0)$. The probability of ending in D is the D entry of $(1,0)P_1P_2$, or equivalently the sum over the possible states at the first interval boundary:

P(D\text{ at }1.5) =0.8426123(0.1769594)+0.1573877(0.9557602) =0.2995.

The check is that each displayed matrix has nonnegative entries and each row sums to 1, up to rounding. The common mistake is to use only the second interval's transition probability from A to D, ignoring the probability of already being in D at the first boundary.

Valuing State Payments and Transition Payments

Let $c_i$ be the continuous payment rate per year while in state $i$, and let $b_{ij}$ be a lump sum paid immediately on a transition from $i$ to $j$. Within a constant-rate interval, a useful expected payment rate while in state $i$ is

g_i=c_i+\sum_{j\ne i}q_{ij}b_{ij}.

The term $q_{ij}b_{ij}$ is an expected payment rate: conditional on currently being in state $i$, transitions to $j$ occur with intensity $q_{ij}$, and each such transition pays $b_{ij}$. The expected present value over an interval of length $h$, starting with row-vector distribution $\pi$, is

\pi\int_0^h e^{-\delta s}\exp(Qs)g\,ds.

Here $g$ is a column vector. The matrix exponential gives the state probabilities at time $s$; multiplying by $g$ gives the expected payment rate then; and $e^{-\delta s}$ discounts that rate to the start of the interval. If a separate payment is made at the interval endpoint, value it separately using the endpoint state distribution and the endpoint discount factor.

Why the integral formula holds: Over a short time $ds$, conditional on being in state $i$, a state payment contributes approximately $c_i\,ds$. A transition $i\to j$ has probability $q_{ij}\,ds+o(ds)$ and contributes $b_{ij}$, so its expected payment is $q_{ij}b_{ij}\,ds+o(ds)$. Thus $g_i$ is the instantaneous expected payment rate in state $i$. At time $s$, the distribution is $\pi\exp(Qs)$; multiplying by the payment rate and discount factor, then integrating over the interval, gives the stated expected present value.

Worked Example: State Benefits, a Transition Lump Sum, and an Endpoint Payment

A hypothetical contract starts in A and runs for one year. The constant transition intensities are $\alpha=0.2$ from A to D and $\beta=0.3$ from D to A per year. The force of interest is $\delta=0.05$. It pays a continuous benefit of $1,000 per year in D, $500 immediately on each A-to-D transition, and an additional $200 at the end of the year if the insured is then in D. Find the expected present value.

Method and setup. First value payments made during the year, then value the endpoint payment. Starting in A, the probability of being in D at time $s$ is $0.4(1-e^{-0.5s})$. The discounted expected time in D is therefore

\int_0^1 e^{-0.05s}0.4(1-e^{-0.5s})\,ds =0.4\left(\frac{1-e^{-0.05}}{0.05} -\frac{1-e^{-0.55}}{0.55}\right) =0.0824917.

The continuous benefit has present value $1{,}000(0.0824917)=\$82.49$. For the transition lump sum, the probability of being in A at time $s$, given the initial state A, is $0.6+0.4e^{-0.5s}$. The discounted expected number of A-to-D transitions, with intensity $0.2$, is

0.2\int_0^1 e^{-0.05s}(0.6+0.4e^{-0.5s})\,ds =0.2\left(0.6\frac{1-e^{-0.05}}{0.05} +0.4\frac{1-e^{-0.55}}{0.55}\right) =0.1785839.

Multiplying by $500 gives a transition-payment present value of $500(0.1785839)=\$89.29$. At the endpoint, the probability of D is $0.4(1-e^{-0.5})=0.1573877$. The discounted endpoint payment is $200e^{-0.05}(0.1573877)=\$29.94$. Adding the three components gives

82.49+89.29+29.94=\$201.72.

The total is checked by keeping the endpoint benefit separate from the continuous-time integral: it is paid only at time 1, and only if the state at that time is D. The common mistake is to count the endpoint payment as if it were payable throughout the year.

Accumulating Values Across Rate Changes

For several intervals, first propagate the state distribution to each boundary. Then calculate the expected present value of payments during each interval using the distribution at its start and that interval's generator and payment assumptions. If an interval value is calculated at its own start, discount it back to the original valuation time by the accumulated discount factor. With a constant force of interest, an interval beginning at time $t$ has discount factor $e^{-\delta t}$.

Set the state and payment conventions.
List the states, transition directions, continuous payment rates, transition lump sums, and any endpoint payments.

Build each interval's generator.
Use that interval's intensities, set each diagonal entry to the negative sum of the other entries in its row, and calculate its transition matrix.

Propagate probabilities and value interval payments.
Carry the row-vector distribution forward by multiplying by each interval matrix. Value within-interval payments at their actual times, and discount interval-start values back to time 0.

Check the result.
Check that each transition matrix has rows summing to 1, that state probabilities sum to 1 at each boundary, and that each payment is included exactly once.

Worked Example: Valuation with a Rate Change at One Year

A hypothetical model has two states, Active (A) and Disabled (D). D is absorbing. The insured starts in A. The A-to-D intensity is $0.2$ per year during year 1 and $0.4$ per year during year 2. The force of interest is $\delta=0.1$. A continuous benefit of $1,200 per year is paid in D, and $2,000 is paid immediately on transition from A to D. Find the expected present value over two years.

Method and setup. In an interval with intensity $\lambda$, the probability of still being in A after $s$ years, given A at the interval start, is $e^{-\lambda s}$. Define $A_x=(1-e^{-x})/x$. The discounted time in D during an interval of length 1, given that it starts in A, is $A_{0.1}-A_{0.1+\lambda}$. The discounted expected number of transitions from A to D is $\lambda A_{0.1+\lambda}$. If the interval starts in D, its continuous-benefit value is $1{,}200A_{0.1}$.

For year 1, $\lambda=0.2$, $A_{0.1}=0.9516258$, and $A_{0.3}=0.8639393$. Starting in A, the continuous-benefit value is $1{,}200(0.9516258-0.8639393)=\$105.22$. The transition-payment value is $2{,}000(0.2)(0.8639393)=\$345.58$. Thus the year-1 value is $\$450.80$.

At time 1, the probabilities are $P(A)=e^{-0.2}=0.8187308$ and $P(D)=1-e^{-0.2}=0.1812692$. For year 2, $\lambda=0.4$ and $A_{0.5}=0.7869387$. Starting in A at time 1, the continuous-benefit value is $1{,}200(0.9516258-0.7869387)=\$197.62$, and the transition-payment value is $2{,}000(0.4)(0.7869387)=\$629.55$. The total conditional value is $\$827.18$. Starting in D, the value is $1{,}200(0.9516258)=\$1{,}141.95$.

Weight these two conditional values by the state probabilities at time 1. The year-2 value at time 1 is $0.8187308(827.1755)+0.1812692(1{,}141.9510)=\$884.23$. Discounting that amount one year gives $e^{-0.1}(884.23)=\$800.09$ at time 0. Therefore the total expected present value is $450.80+800.09=\$1{,}250.89$.

The check is that the second-year value includes both people already in D at time 1 and those who become disabled during year 2. The common mistake is to discount the second-year value using a two-year factor: it is already valued at the start of year 2, time 1, so the factor back to time 0 is $e^{-0.1}$, not $e^{-0.2}$.

Common Numerical Traps

Mixing up intensities and probabilities. An intensity is a rate per unit time; calculate $\exp(Qh)$ to obtain interval probabilities.
Reversing the matrix order. With row-vector probabilities, multiply by the first interval matrix, then the second: $\pi P_1P_2$.
Using the wrong time unit. If intensities are annual, express interval lengths in years, including fractional years.
Discounting at the wrong date. Payments within an interval are discounted from their payment times; an interval value measured at its start must then be discounted from that start to time 0.
Omitting or duplicating cash flows. Continuous state payments, transition lump sums, and endpoint payments are different payment types and should be identified separately.

These calculations are useful whenever transition assumptions or product payments change by duration or period. Keeping the probability propagation separate from the cash-flow valuation makes the model easier to audit and helps identify whether an error comes from the intensities, the timing, or the payment amounts. This method is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, calculate the requested probability or present value. Rates are per year.

A person starts in a healthy state, which transitions to a care state at intensity $0.1$ for two years and then at intensity $0.3$ for one year. The care state is absorbing. What is the probability of being in care after three years? (Answer: $0.3935$)
In a two-state model, the intensities are $0.1$ from A to D and $0.2$ from D to A for one year. Starting in A, what is the probability of being in D after one year? (Answer: $0.0864$)
A person starts active and can transition only from active to disabled, at constant intensity $0.2$ for one year. The force of interest is $0.05$. The contract pays $1,000 continuously per year while disabled and $500 immediately on disability. Find the expected present value. (Answer: $\$179.09$)
A person starts in A. The only transition is A to absorbing D, with intensity $0.25$ for two years. The force of interest is $0.08$, and a $4,000 lump sum is paid immediately on transition. Find its expected present value. (Answer: $\$1,464.09$)

What You'll Learn