The Matrix Exponential for Constant Intensities

From Intensities to Finite-Time Probabilities

A transition intensity describes an instantaneous rate of movement between states. A transition probability describes where a process may be after a finite duration. For a model with constant intensities, the matrix exponential provides the link between these quantities. It combines the possible sequences of transitions during the interval; it is not obtained by exponentiating each intensity separately.

Use the row convention: the entry in row \(i\), column \(j\) of a transition matrix is the probability of being in state \(j\) after the interval, given that the process began in state \(i\). In particular, each row describes one possible starting state. This convention determines the order of the matrices in the equations below.

Definition: For a finite-state, time-homogeneous continuous-time Markov model with constant intensity matrix \(Q\), the transition-probability matrix over duration \(t\) is \(P(t)=\exp(Qt)\). The matrix exponential is defined by the power series \(\exp(Qt)=I+Qt+(Qt)^2/2!+(Qt)^3/3!+\cdots\), where \(I\) is the identity matrix. Its entry \(P_{ij}(t)\) is the probability of being in state \(j\) at time \(t\), conditional on starting in state \(i\) at time zero.

Here \(Q\) is a generator: its off-diagonal entries are nonnegative transition intensities, and each diagonal entry is the negative sum of the other entries in its row. Thus every row of \(Q\) sums to zero. The model assumes a fixed set of states and rates that do not vary with time. The duration \(t\) must use the same time unit as the rates; for example, rates per year require \(t\) in years. If intensities vary with time, the constant-matrix formula \(P(t)=\exp(Qt)\) does not in general apply.

Why the Matrix Exponential Gives the Transition Matrix

The reason for the formula is that the transition matrix evolves according to the forward differential equation \(P'(t)=P(t)Q\), with \(P(0)=I\). The initial condition says that at zero duration each state is unchanged with probability one. The differential equation describes how probabilities change as the interval is extended.

For a constant \(Q\), differentiate the defining power series term by term. Since every term is a power of the same matrix \(Q\), its derivative is \(\frac{d}{dt}(Q^n t^n/n!)=Q^n t^{n-1}/(n-1)!\) for \(n\geq 1\). Therefore,

\frac{d}{dt}\exp(Qt) =Q+Q^2t+\frac{Q^3t^2}{2!}+\cdots =\exp(Qt)Q.

At \(t=0\), all positive powers of \(t\) vanish, so \(\exp(Q\cdot 0)=I\). Thus the matrix exponential satisfies both the forward differential equation and its initial condition. The solution to this constant-coefficient linear differential equation is unique, which justifies identifying the solution with \(P(t)\).

The power series also gives useful interpretation. The identity term accounts for no change over a zero-length interval; the \(Qt\) term reflects first-order movement over a short duration; higher powers account for the effects of further transitions. These contributions are matrix products, so they incorporate intermediate states and paths through the model.

Worked Calculations

Worked Example: A One-Way Transition to an Absorbing State

Consider a hypothetical two-state model with states Active and Dead. A person moves from Active to Dead at a constant intensity of \(0.4\) per year, and Dead is absorbing. Find the transition matrix for a duration of \(1.5\) years. Use the row convention.

Method. Form the generator by placing the transition intensity off the diagonal and choosing diagonal entries so each row sums to zero. Since there is no transition out of Dead,

Q= \begin{pmatrix} -0.4&0.4\\ 0&0 \end{pmatrix}.

The duration is \(t=1.5\) years, so the product of the intensity and duration is \(0.4(1.5)=0.6\). For this one-way model, the probability of remaining Active is \(e^{-0.6}\), and the probability of reaching Dead is \(1-e^{-0.6}\). The Dead row remains unchanged because that state is absorbing. Therefore,

P(1.5)= \begin{pmatrix} e^{-0.6}&1-e^{-0.6}\\ 0&1 \end{pmatrix} \approx \begin{pmatrix} 0.5488&0.4512\\ 0&1.0000 \end{pmatrix}.

The entries are rounded to four decimal places only in the final matrix. As a check, the Active row sums to \(e^{-0.6}+1-e^{-0.6}=1\), the Dead row sums to one, and all entries lie between zero and one. The common mistake avoided here is treating \(0.4\) as a probability; it is a rate per year, and the finite-duration probability depends on the product of rate and time.

Worked Example: Transitions in Both Directions

A hypothetical two-state model has states In Service and On Leave. The intensity from In Service to On Leave is \(0.25\) per year, and the intensity from On Leave to In Service is also \(0.25\) per year. Find the transition matrix over two years.

Method. The generator and duration are

Q= \begin{pmatrix} -0.25&0.25\\ 0.25&-0.25 \end{pmatrix}, \qquad t=2.

The two states are symmetric: at any time, either state has the same intensity of leaving for the other. The difference between the two state probabilities decays at rate \(0.25+0.25=0.5\) per year. Equivalently, the generator has one zero eigenvalue, corresponding to equal long-run probabilities, and one eigenvalue \(-0.5\). The zero-eigenvalue component gives probability \(1/2\) in each state; the other component is multiplied by \(e^{-0.5t}\). Hence the transition matrix has diagonal entries \((1+e^{-0.5t})/2\) and off-diagonal entries \((1-e^{-0.5t})/2\).

Substitute \(t=2\), so \(0.5t=1\):

P(2)= \begin{pmatrix} (1+e^{-1})/2&(1-e^{-1})/2\\ (1-e^{-1})/2&(1+e^{-1})/2 \end{pmatrix} \approx \begin{pmatrix} 0.6839&0.3161\\ 0.3161&0.6839 \end{pmatrix}.

The entries are rounded to four decimal places at the end. Each row sums to one, and the symmetry of the model is reflected in the symmetry of the matrix. Notice that the probability of being in the starting state is greater than \(1/2\), but less than one: reversals are possible, while the process has not completely lost its dependence on the starting state after two years. The common mistake avoided is to use only the intensity out of the initial state and ignore that transitions back can occur during the interval.

Worked Example: A Three-State Sequential Model

Consider a hypothetical model with states Working, Disabled and Dead. A transition can occur from Working to Disabled at a constant intensity of \(0.5\) per year, and from Disabled to Dead at the same intensity. Dead is absorbing, and there are no other transitions. Find the transition matrix over two years.

Method. The generator is

Q= \begin{pmatrix} -0.5&0.5&0\\ 0&-0.5&0.5\\ 0&0&0 \end{pmatrix}.

For a starting state of Working, remaining Working for two years has probability \(e^{-0.5(2)}=e^{-1}\). To be Disabled at the end, the process must make its first transition during the interval and must not make the second one before the endpoint. With equal transition intensities, the resulting probability is \(0.5(2)e^{-1}=e^{-1}\). The remaining probability is in Dead. Starting in Disabled, the one-way transition formula gives probabilities \(e^{-1}\) of still being Disabled and \(1-e^{-1}\) of being Dead. Starting in Dead gives probability one of remaining there. Thus,

P(2)= \begin{pmatrix} e^{-1}&e^{-1}&1-2e^{-1}\\ 0&e^{-1}&1-e^{-1}\\ 0&0&1 \end{pmatrix} \approx \begin{pmatrix} 0.3679&0.3679&0.2642\\ 0&0.3679&0.6321\\ 0&0&1.0000 \end{pmatrix}.

The entries are rounded to four decimal places only in the final matrix. The first row sums to \(e^{-1}+e^{-1}+1-2e^{-1}=1\); the second and third rows also sum to one. The zero entries are required because there is no path from Disabled back to Working or from Dead to another state. This example checks both path structure and row interpretation: the first row is conditional on starting Working, not on starting in any other state.

Checks and Common Exam Traps

The matrix exponential produces a transition-probability matrix, so its entries must be nonnegative and each row must sum to one. These checks can reveal a calculation or transcription error, although passing them does not by itself prove that every entry is correct. The zero-duration check is also immediate: \(\exp(Q\cdot 0)=I\).

A frequent error is to exponentiate each entry of \(Qt\) separately. In the reversible example, this would give an array with entries \(e^{-0.5}\) on the diagonal and \(e^{0.5}\) off the diagonal. Its rows do not sum to one, and some entries exceed one, so it cannot be a transition matrix. The correct expression is the matrix power series, which includes matrix multiplication and the effects of possible paths.

Another common error is a time-unit mismatch. If \(Q\) is measured per year but the requested duration is six months, use \(t=0.5\) years, not \(t=6\). Before computing, verify that every intensity and the duration use compatible units. Also keep the row convention consistent: \(P_{ij}(t)\) conditions on starting in row state \(i\) and gives the probability of ending in column state \(j\).

Key takeaway: For a finite-state model with constant intensities, build the generator \(Q\) using consistent time units and calculate \(P(t)=\exp(Qt)\). Check that \(P(0)=I\), that every row sums to one, and that all entries are valid probabilities.

Check Your Understanding

For each problem, use the row convention and round probabilities to four decimal places where needed.

A hypothetical two-state model has state A moving to absorbing state B at \(0.3\) per year. Find the transition matrix after two years and check its row sums. (Answer: \(\begin{pmatrix}0.5488&0.4512\\0&1.0000\end{pmatrix}\))
A two-state model has intensity \(0.4\) per year from X to Y and \(0.1\) per year from Y to X. Find the full transition matrix after two years. (Answer: \(\begin{pmatrix}0.4943&0.5057\\0.1264&0.8736\end{pmatrix}\))
A sequential model has states 1, 2 and 3, with transitions only from 1 to 2 at \(0.2\) per month and from 2 to absorbing state 3 at \(0.4\) per month. Find the transition matrix after three months. (Answer: \(\begin{pmatrix}0.5488&0.2476&0.2036\\0&0.3012&0.6988\\0&0&1.0000\end{pmatrix}\))
A two-state model has rates \(0.2\) per year from A to B and \(0.3\) per year from B to A. The initial row distribution is \((0.7,0.3)\). Find the probability of being in A after two years. (Answer: \(0.6368\))

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