Transition Matrices With Time-Varying Intensities

Transition Probabilities Over Calendar-Time Intervals

In a continuous-time state model, transition intensities may change with calendar time. A hypothetical insurer might model the rate at which a policyholder moves from active to disabled as increasing over time, or a service provider might assume that transitions into different care states have different time patterns. When the rates vary, a transition probability must refer to a particular interval: knowing only the length of the interval may not identify the rate history within it.

Use \(s\) for the calendar-time start and \(t\) for the calendar-time end, with \(s\le t\). This notation keeps the interval endpoints visible. For example, \(P(1,3)\) represents transitions from calendar time 1 to calendar time 3; its duration is two time units, but the model uses the intensities at calendar times between 1 and 3.

Definition: For a finite-state continuous-time Markov model, \(p_{ij}(s,t)=P(X(t)=j\mid X(s)=i)\) is the probability of being in state \(j\) at calendar time \(t\), given state \(i\) at calendar time \(s\). The transition matrix \(P(s,t)\) has entries \(p_{ij}(s,t)\), with rows indexed by starting state and columns by ending state. Each row sums to 1.

The time-varying transition intensity from state \(i\) to state \(j\), for \(i\ne j\), is denoted here by \(\mu_{ij}(u)\), where \(u\) is calendar time. It describes the instantaneous transition rate at time \(u\), conditional on being in state \(i\) just before that time. Intensities are nonnegative and are assumed to be sufficiently well-behaved for the integrals below to exist. As in the earlier material on holding-time survival, the probability of avoiding a transition over an interval is determined by accumulated intensity.

One-Way Transitions: Integrate Before Exponentiating

Suppose a model has two states, active and exited. The only possible transition is active to exited, with intensity \(\lambda(u)\); exited is absorbing. Conditional on being active at time \(s\), the probability of remaining active until time \(t\) is the exponential of the negative accumulated intensity. The probability of exiting by \(t\) is its complement.

p_{\text{active,active}}(s,t) =\exp\!\left(-\int_s^t \lambda(u)\,du\right), \qquad p_{\text{active,exited}}(s,t) =1-\exp\!\left(-\int_s^t \lambda(u)\,du\right).

The derivation follows from survival over short intervals. Over a short interval of length \(du\) around time \(u\), the conditional probability of no exit is approximately \(1-\lambda(u)\,du\). Multiplying these no-exit probabilities across a fine partition of \([s,t]\) gives, in the limit, the exponential of the negative integral of \(\lambda\). Thus it is the accumulated intensity, not the intensity at just one point, that determines survival.

Worked Example: Active-to-Exited Probability With a Linear Intensity

A hypothetical model has active and exited states. The exit intensity is \(\lambda(u)=0.03+0.01u\) per year, where \(u\) is measured in years from a fixed origin. Find the transition matrix from calendar time 1 to calendar time 3. Use four decimal places.

Method. Integrate the calendar-time intensity over the specified interval, exponentiate the negative integral to obtain the probability of remaining active, and use the complement for exit. The exited state is absorbing.

The accumulated intensity is

\int_1^3 (0.03+0.01u)\,du =0.03(3-1)+0.005(3^2-1^2) =0.06+0.04 =0.10.

Therefore, the probability of remaining active is \(e^{-0.10}=0.904837\ldots\), and the probability of exiting is \(1-0.904837\ldots=0.095163\ldots\). With rows and columns ordered active, exited, the matrix is

P(1,3)= \begin{pmatrix} 0.9048 & 0.0952\\ 0 & 1 \end{pmatrix}.

Check: the active row sums to \(0.9048+0.0952=1.0000\), and the exited state remains exited. A common mistake is to use the rate at the start, \(\lambda(1)=0.04\), for the whole two-year interval, giving \(e^{-0.04(2)}=0.9231\). That calculation treats the rate as constant and ignores its increase. The integrated intensity is \(0.10\), so the correct survival probability is \(0.9048\).

More generally, if there are \(n\) states, the same row-and-column convention applies to \(P(s,t)\). A one-way transition model often produces a simple matrix because the destination is absorbing. For models with multiple possible destinations, the intensity out of the starting state is the sum of the intensities to each destination.

Competing Destinations From One State

Consider a state \(A\) with possible transitions to absorbing states \(B\) and \(C\). Let the corresponding intensities at calendar time \(u\) be \(\lambda_B(u)\) and \(\lambda_C(u)\). While in \(A\), the total intensity of leaving is \(\lambda_B(u)+\lambda_C(u)\). Hence the probability of staying in \(A\) throughout \([s,t]\) is the exponential of minus the integral of that total intensity.

In general, the probability of ending in \(B\) depends on the timing of the \(A\)-to-\(B\) intensity as well as the possibility of leaving for \(C\) first. A particularly useful exact case occurs when the two intensities maintain fixed proportions over the whole interval. If \(\lambda_B(u)=c_B g(u)\) and \(\lambda_C(u)=c_C g(u)\), for nonnegative constants \(c_B,c_C\) and a nonnegative function \(g\), then the share of exits going to \(B\) is \(c_B/(c_B+c_C)\). This statement relies on the fixed-proportion assumption; it should not be used when the relative intensities change over time.

Worked Example: Three-State Matrix With Proportional Intensities

A hypothetical model has an initial state \(A\) and absorbing states \(B\) and \(C\). From \(A\), the intensities are \(\lambda_B(u)=0.02+0.01u\) and \(\lambda_C(u)=0.04+0.02u\) per year. Find \(P(0,2)\), in the state order \(A,B,C\), to four decimal places.

Method. The second intensity is twice the first at every time, so the total exit intensity is three times \(\lambda_B(u)\), and the proportions of exits to \(B\) and \(C\) are \(1/3\) and \(2/3\).

First calculate the accumulated intensity to \(B\):

\int_0^2 (0.02+0.01u)\,du =0.02(2)+0.005(2^2) =0.04+0.02 =0.06.

The total accumulated exit intensity is \(3(0.06)=0.18\). Thus the probability of remaining in \(A\) is \(e^{-0.18}=0.835270\ldots\), and the probability of leaving \(A\) is \(1-0.835270\ldots=0.164730\ldots\). The probabilities of ending in \(B\) and \(C\) are

p_{AB}(0,2)=\frac{1}{3}(0.164730\ldots)=0.054910\ldots, \qquad p_{AC}(0,2)=\frac{2}{3}(0.164730\ldots)=0.109820\ldots.

Since \(B\) and \(C\) are absorbing, the full matrix is

P(0,2)= \begin{pmatrix} 0.8353 & 0.0549 & 0.1098\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}.

Check: the first row sums to \(0.8353+0.0549+0.1098=1.0000\), as required. The fixed one-third/two-thirds allocation is justified because the two intensities are proportional at every time. It would be a mistake to allocate exits in those proportions if the intensities were proportional only at the start or end of the interval.

Combining Consecutive Intervals

If \(s\le u\le t\), transition probabilities over \([s,t]\) can be calculated by conditioning on the state occupied at the intermediate calendar time \(u\). In matrix form, this gives

$$ P(s,t)=P(s,u)P(u,t). $$

The multiplication order matters: the first matrix carries probabilities from \(s\) to \(u\), and the second carries them from \(u\) to \(t\). With row vectors indexed by starting state, the intermediate-state probabilities are summed across the columns of the first matrix and rows of the second. This rule is useful for checking a direct calculation or assembling an interval from simpler intervals. The intensities used in each factor must correspond to that factor’s actual calendar-time interval.

Worked Example: Splitting an Interval With a Changing Exit Rate

A hypothetical two-state model has active-to-exited intensity \(\lambda(u)=0.02+0.01u\) per year, with exited absorbing. Calculate the transition matrix from time 0 to time 2 by splitting at time 1. Use four decimal places.

Method. Calculate the accumulated intensity separately on \([0,1]\) and \([1,2]\), construct each one-way transition matrix, and multiply them in chronological order.

For the first interval,

\int_0^1 (0.02+0.01u)\,du =0.02(1)+0.005(1^2) =0.025.

For the second interval,

\int_1^2 (0.02+0.01u)\,du =0.02(2-1)+0.005(2^2-1^2) =0.035.

Consequently, \(e^{-0.025}=0.9753099\ldots\) and \(e^{-0.035}=0.9656054\ldots\). The interval matrices are

P(0,1)= \begin{pmatrix} 0.9753099 & 0.0246901\\ 0 & 1 \end{pmatrix}, \qquad P(1,2)= \begin{pmatrix} 0.9656054 & 0.0343946\\ 0 & 1 \end{pmatrix}.

Multiply in chronological order:

P(0,2)=P(0,1)P(1,2) = \begin{pmatrix} (0.9753099)(0.9656054) & 0.0246901+(0.9753099)(0.0343946)\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0.9418 & 0.0582\\ 0 & 1 \end{pmatrix}.

As a direct check, the accumulated intensity over \([0,2]\) is \(0.025+0.035=0.060\), so the active-to-active probability is \(e^{-0.060}=0.9417645\ldots\), which rounds to \(0.9418\). The exited probability is its complement, \(0.0582355\ldots\), which rounds to \(0.0582\). The main trap is to use the first interval’s intensity for both years; the two intervals have different accumulated intensities because the rate changes with calendar time.

Model Checks and Common Traps

A transition matrix over a specified interval should satisfy basic probability checks. Every entry must be between 0 and 1, and each row must sum to 1. If a state is absorbing, its row has a 1 in its own-state column and 0 in every other column. These checks can reveal arithmetic or matrix-order errors, but they do not validate an inappropriate intensity assumption.

Using duration in place of calendar time. An intensity such as \(\lambda(u)=0.03+0.01u\) must be evaluated and integrated using the stated calendar-time variable \(u\), not reset to zero at the interval start unless the model explicitly defines it that way.
Using a constant-rate formula. The expression \(e^{-\lambda(t-s)}\) applies when the intensity is constant on the interval. For a time-varying intensity, use \(e^{-\int_s^t\lambda(u)\,du}\).
Confusing total exit with a particular destination. With competing destinations, the sum of the intensities determines the probability of remaining in the origin state. It does not, by itself, determine how exits are divided among destinations.
Reversing the interval matrices. Under the row-starting-state convention, use \(P(s,u)P(u,t)\), not the reverse product.

Key takeaway: Write both calendar-time endpoints, identify the intensity or intensities that apply over that interval, and integrate over those exact endpoints before constructing the transition probabilities. For consecutive intervals, multiply the corresponding matrices in chronological order and verify that each row is a probability distribution.

Check Your Understanding

For each problem, use the stated calendar-time variable and round probabilities to four decimal places.

A hypothetical one-way model has an active-to-exited intensity \(\lambda(u)=0.015+0.005u\) per year. Find the probability of remaining active from time 2 to time 4. (Answer: 0.9418)
From state \(A\), transitions to absorbing states \(B\) and \(C\) have intensities \(\lambda_B(u)=0.01+0.005u\) and \(\lambda_C(u)=2\lambda_B(u)\) per year. Find the first row of \(P(0,2)\). (Answer: \((0.9139,\ 0.0287,\ 0.0574)\))
A one-way active-to-exited model has \(\lambda(u)=0.01+0.01u\) per year. Find the transition matrix from time 1 to time 2, with state order active, exited. (Answer: \(\begin{pmatrix}0.9753&0.0247\\0&1\end{pmatrix}\))
From \(A\), the intensities to absorbing states \(B\) and \(C\) are \(\lambda_B(u)=0.02+0.01u\) and \(\lambda_C(u)=0.01+0.005u\) per year. Find the first row of \(P(1,3)\). (Answer: \((0.8869,\ 0.0754,\ 0.0377)\))
A one-way model has \(\lambda(u)=0.01+0.01u\) per year. Calculate \(P(0,2)\) by splitting at time 1, and state the product order. (Answer: \(P(0,1)P(1,2)=\begin{pmatrix}0.9608&0.0392\\0&1\end{pmatrix}\))

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