Commuting Intensity Matrices and Time-Varying Solutions

Why Commuting Generators Matter

For a time-homogeneous continuous-time Markov model, a constant generator \(Q\) gives a transition matrix through the matrix exponential. When intensities vary with time, the transition probabilities must account for the generator at each point in the interval. A useful special case occurs when the generators at different times commute: then the transition matrix can still be written as one matrix exponential, with the generator integrated over time.

The result is exact under its assumptions, not a numerical approximation. It can simplify calculations for models whose transition rates change over time but retain a compatible matrix structure. This tutorial is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: In a finite-state, time-inhomogeneous Markov model, \(Q(t)=(q_{ij}(t))\) is the generator at time \(t\), using the row convention: \(q_{ij}(t)\geq 0\) for \(i\ne j\), and each row sums to zero. Thus \(q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)\). Write \(P(s,t)\) for the transition matrix from time \(s\) to time \(t\). With this convention, the forward Kolmogorov equation is \(\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t)\), with \(P(s,s)=I\).

Two matrices \(A\) and \(B\) commute when \(AB=BA\). For a time-varying generator, the relevant condition is that generators at different times commute with one another. A simple, common sufficient pattern is \(Q(t)=g(t)Q_0\), where \(Q_0\) is fixed: every pair of such matrices commutes. But the condition also permits some families that are not scalar multiples of one fixed matrix.

The Integrated-Generator Formula

Suppose \(Q(t)\) is continuous on the interval from \(s\) to \(t\), is a valid generator at each time, and satisfies \(Q(u)Q(v)=Q(v)Q(u)\) for all \(u,v\) in the interval. Then

P(s,t)=\exp\!\left(\int_s^t Q(u)\,du\right).

The integral is taken entry by entry, and \(\exp(A)\) is the matrix exponential \(\sum_{k=0}^{\infty}A^k/k!\). The formula does not say that the matrices \(Q(u)\) are constant. It says that their accumulated effect can be represented by the exponential of their integral when they commute.

Regularity condition: If \(Q\) is continuous, the displayed formula is differentiable in the upper time \(t\) and satisfies the forward equation pointwise. If \(Q\) is only integrable, and is a valid generator almost everywhere with pairwise commutation holding almost everywhere, the same formula defines an absolutely continuous transition matrix that satisfies the forward equation almost everywhere. Integrability alone does not guarantee differentiability at every time.

Why the Formula Holds

Set \(A(t)=\int_s^t Q(u)\,du\). The pairwise commutation assumption implies that \(A(t)\) commutes with \(Q(t)\): each matrix in the integral commutes with \(Q(t)\), so their integral does as well. For continuous \(Q\), the fundamental theorem of calculus gives \(A'(t)=Q(t)\) at every time. When a differentiable matrix \(A(t)\) commutes with its derivative, differentiating its exponential gives

\frac{d}{dt}\exp(A(t))=\exp(A(t))A'(t).

One way to see why commutation matters is to differentiate the power series for the exponential. If \(A\) and \(A'\) commute, the derivative of \(A^k\) is \(kA^{k-1}A'\); summing the resulting series gives \(\exp(A)A'\). Consequently, for \(P(s,t)=\exp(A(t))\),

\frac{\partial}{\partial t}P(s,t) =\exp(A(t))Q(t) =P(s,t)Q(t).

Also \(A(s)=0\), so \(P(s,s)=\exp(0)=I\). Thus the proposed matrix satisfies both the forward equation and its initial condition; uniqueness of the solution to this finite-dimensional linear differential equation identifies it as the transition matrix. If \(Q\) is integrable rather than continuous, \(A\) is absolutely continuous and \(A'=Q\) almost everywhere. The same argument then gives an absolutely continuous solution satisfying the equation almost everywhere, not necessarily a pointwise derivative at every time.

Worked Example: Two States With a Quadratically Changing Rate

A hypothetical process has states 0 and 1. At time \(u\), the rate from 0 to 1 is \(0.2(1+u^2)\) per year, and the rate from 1 to 0 is \(0.1(1+u^2)\) per year. Find the transition matrix from time 0 to time 2. Assume a continuous-time Markov model with these continuous intensities.

Method and commutation check. The generator is \(Q(u)=(1+u^2)Q_0\), where

Q_0= \begin{pmatrix} -0.2&0.2\\ 0.1&-0.1 \end{pmatrix}.

Any two scalar multiples of \(Q_0\) commute, so the integrated-generator formula applies. The scalar integral over the two-year interval is

\int_0^2(1+u^2)\,du =\left[u+\frac{u^3}{3}\right]_0^2 =2+\frac{8}{3} =\frac{14}{3}.

Therefore \(\int_0^2 Q(u)\,du=\frac{14}{3}Q_0\). In the equivalent constant-rate two-state calculation, the accumulated rates are \(0.2(14/3)=14/15\) from 0 to 1 and \(0.1(14/3)=7/15\) from 1 to 0. Their sum is \(7/5=1.4\). The transition probability from 0 to 1 is the first accumulated rate divided by the total accumulated rate, multiplied by \(1-e^{-1.4}\):

P_{01}(0,2) =\frac{14/15}{7/5}(1-e^{-1.4}) =\frac{2}{3}(1-e^{-1.4}) \approx 0.5023.

Similarly, \(P_{10}(0,2)=\frac{1}{3}(1-e^{-1.4})\approx0.2511\). Completing each row to sum to one gives

P(0,2)\approx \begin{pmatrix} 0.4977&0.5023\\ 0.2511&0.7489 \end{pmatrix}.

The entries are rounded to four decimal places. The rows sum to one, and each entry is nonnegative. Common mistake avoided: replacing the varying rate by its value at the midpoint \(u=1\) would use a scalar multiplier of \(2\) for two years, or an accumulated multiplier of \(4\), instead of the correct integral \(14/3\). That midpoint approximation gives \(P_{01}=\frac{2}{3}(1-e^{-1.2})\approx0.4660\), which is not the exact transition probability.

Commutation Beyond Scalar Multiples

Commutation is a matrix condition, not a requirement that every intensity vary in the same proportion. The following three-state example uses two distinct generator matrices that commute.

Worked Example: Three States With Changing Directional Rates

Consider three states labelled 0, 1, and 2. At time \(u\in[0,1]\), transitions go clockwise at rate \(a(u)=0.1+0.1u\), and counterclockwise at rate \(b(u)=0.2-0.1u\). Assume these are the only transitions. Find \(P(0,1)\).

Method and commutation check. Define the clockwise and counterclockwise generators

C_+= \begin{pmatrix} -1&1&0\\ 0&-1&1\\ 1&0&-1 \end{pmatrix}, \qquad C_-= \begin{pmatrix} -1&0&1\\ 1&-1&0\\ 0&1&-1 \end{pmatrix}.

Then \(Q(u)=a(u)C_++b(u)C_-\). These matrices commute: they are circulant matrices, representing cyclic shifts and their reverse, and cyclic shifts commute. Hence \(Q(u)\) and \(Q(v)\) commute for every pair of times. Both rates remain positive on the interval: \(0.1\leq a(u)\leq0.2\) and \(0.1\leq b(u)\leq0.2\).

Integrating the two rate functions gives

\int_0^1a(u)\,du=0.1+\frac{0.1}{2}=0.15, \qquad \int_0^1b(u)\,du=0.2-\frac{0.1}{2}=0.15.

Thus the integrated generator is \(0.15(C_++C_-)\), where \(C_++C_-=J-3I\) and \(J\) is the \(3\times3\) matrix of ones. The matrix \(J/3\) projects onto the constant vector, while \(I-J/3\) projects onto the subspace whose coordinates sum to zero. Since \(J-3I\) has eigenvalue 0 on the first subspace and \(-3\) on the second,

P(0,1) =\exp\!\left(0.15(J-3I)\right) =\frac{J}{3}+e^{-0.45}\left(I-\frac{J}{3}\right).

Using \(e^{-0.45}\approx0.6376\), each diagonal entry is \(\frac13+\frac23(0.6376)\approx0.7584\), and each off-diagonal entry is \(\frac{1-0.6376}{3}\approx0.1208\). Therefore, to four decimal places,

P(0,1)\approx \begin{pmatrix} 0.7584&0.1208&0.1208\\ 0.1208&0.7584&0.1208\\ 0.1208&0.1208&0.7584 \end{pmatrix}.

Each row sums to \(0.7584+0.1208+0.1208=1.0000\), as required. The example shows that the instantaneous clockwise and counterclockwise rates change differently, while the generator matrices still commute.

Integrable Intensities and the Meaning of “Solution”

Continuity is a convenient condition for a pointwise differential equation, but it is not essential to the integrated formula. For example, intensities may change by a jump at a specified time. If the generator entries are integrable, the accumulated matrix \(\int_s^tQ(u)\,du\) is absolutely continuous in \(t\). Its derivative equals \(Q(t)\) almost everywhere, which is enough to define the usual absolutely continuous solution of the forward equation.

At a jump, the transition matrix remains continuous because it is the exponential of a continuous accumulated integral. Its derivative may nevertheless fail to exist at that instant. This distinction matters: an equation asserted almost everywhere is not the same as a pointwise derivative assertion at every time.

Worked Example: A Jump in the Intensity

For a hypothetical two-state process, let \(Q(u)=g(u)Q_0\) on \(0\leq u\leq2\), where

Q_0= \begin{pmatrix} -1&1\\ 1&-1 \end{pmatrix}, \qquad g(u)= \begin{cases} 0.1,&0\leq u<1,\\ 0.3,&1\leq u\leq2. \end{cases}

The scalar generators commute at all times, and the rate function is integrable. Its value at the single jump time does not affect the integral. The accumulated scalar multiplier is

\int_0^2g(u)\,du =(0.1)(1)+(0.3)(1)=0.4.

Therefore \(P(0,2)=\exp(0.4Q_0)\). The eigenvalues of \(Q_0\) are 0 and \(-2\), so the probability of changing state over the interval is \(\frac{1-e^{-0.8}}{2}\). Since \(e^{-0.8}\approx0.4493\), the probability of changing state is approximately \(0.2753\), and the probability of remaining in the same state is approximately \(0.7247\). Hence, to four decimal places,

P(0,2)\approx \begin{pmatrix} 0.7247&0.2753\\ 0.2753&0.7247 \end{pmatrix}.

The rows sum to one. The transition matrix is continuous at time 1, but its left and right derivatives there generally differ because \(g\) jumps. The forward equation therefore holds almost everywhere, not necessarily pointwise at the jump.

Common Exam Traps and Practical Checks

The commuting condition is what permits the ordinary exponential of an integral. Do not infer that condition merely because every \(Q(t)\) is a valid generator: valid generators need not commute with one another. Also keep the time interval and time units consistent; rates per year must be integrated over years. Before interpreting a computed exponential as a transition matrix, check that the model uses generator matrices and that the required commutation assumption holds.

Using a pointwise derivative under an integrability-only assumption. The formula is absolutely continuous and satisfies the equation almost everywhere; it need not be differentiable at a jump.
Replacing an integral by a midpoint rate without justification. This is generally a numerical approximation, not the exact commuting solution.
Forgetting the row convention. Here the forward equation is \(P'(s,t)=P(s,t)Q(t)\); mixing conventions can reverse matrix order in calculations.
Ignoring the interval. The exponent uses \(\int_s^tQ(u)\,du\), not an integral over an unrelated time origin.

The key practical sequence is to identify the generator, verify that it is valid, check pairwise commutation, integrate over the requested interval, and then evaluate the matrix exponential. As a final numerical check, confirm that each resulting row is a probability distribution.

Key takeaway: For a continuous, pairwise-commuting family of valid generators, \(P(s,t)=\exp(\int_s^tQ(u)\,du)\) is the pointwise solution of the forward equation. For merely integrable generators with the corresponding almost-everywhere commutation condition, the formula gives an absolutely continuous solution satisfying the equation almost everywhere.

Check Your Understanding

For each problem, apply the assumptions and time interval stated. Round transition probabilities to four decimal places where needed.

A two-state model has \(Q(u)=(2+u)Q_0\) for \(0\leq u\leq1\), where \(Q_0=\begin{pmatrix}-0.4&0.4\\0.2&-0.2\end{pmatrix}\). Find \(P(0,1)\). (Answer: \(\begin{pmatrix}0.4821&0.5179\\0.2589&0.7411\end{pmatrix}\))
A three-state model has \(Q(u)=(0.1+0.2u)(C_++C_-)\) for \(0\leq u\leq1\), with \(C_+\) and \(C_-\) as defined above. Find the diagonal and off-diagonal entries of \(P(0,1)\). (Answer: diagonal \(0.6992\); off-diagonal \(0.1504\))
For a two-state model, \(Q(u)=g(u)\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\), where \(g(u)=0.2\) for \(0\leq u<2\) and \(g(u)=0.1\) for \(2\leq u\leq5\). Find \(P(0,5)\) and state whether the forward equation holds pointwise at \(u=2\). (Answer: \(\begin{pmatrix}0.6233&0.3767\\0.3767&0.6233\end{pmatrix}\); it holds almost everywhere, but pointwise differentiability at the jump is not guaranteed.)
Let \(Q_1=\begin{pmatrix}-1&1\\0&0\end{pmatrix}\) for \(0\leq u<1\) and \(Q_2=\begin{pmatrix}0&0\\1&-1\end{pmatrix}\) for \(1\leq u\leq2\). Does the commuting integrated-generator formula apply? Compute the commutator \(Q_1Q_2-Q_2Q_1\). (Answer: No; \(Q_1Q_2-Q_2Q_1=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}\ne0\).)
A valid generator family is continuous and pairwise commuting on \([s,t]\). State the formula for \(P(s,t)\) and the differential equation it satisfies under the row convention. (Answer: \(P(s,t)=\exp(\int_s^tQ(u)\,du)\), and \(\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t)\) pointwise.)

What You'll Learn