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Transition Intensities and Probabilities · Tutorial 297 of 699

Deriving The Constant-Intensity Transition Matrix

See how the product-integral limit yields the matrix exponential, then derive and interpret its entries for a two-state system with constant transition intensities.

Advanced 10 min read

What You'll Learn

  • Connect the product-integral limit to repeated short-interval transition updates.
  • Derive the matrix exponential for constant transition-intensity matrices.
  • Obtain explicit transition probabilities for a two-state model.
  • Check the formula’s row sums, limiting behavior, and assumptions.
  • Recognize why time-varying intensity matrices generally require a product integral.

From Short-Interval Updates to a Matrix Exponential

A transition-intensity matrix describes instantaneous movement among states. A transition-probability matrix describes movement over a finite time interval. The connection between them is built from many short-interval updates: over a sufficiently short interval, the transition matrix is approximately the identity matrix plus the intensity matrix multiplied by the interval length. Taking the product of these updates over finer and finer partitions produces the product integral.

When the transition intensities are constant in time, every short-interval update uses the same matrix. The product-integral limit then reduces to a matrix exponential. This tutorial derives that result and works out its entries for a two-state system. The derivation also identifies the conditions under which this simplification is justified.

Definition: Let \(Q\) be a constant transition-intensity matrix for a finite-state process, with rows indexed by origin state and columns by destination state. Its off-diagonal entry \(q_{ij}\) is the transition intensity from state \(i\) to state \(j\), and each diagonal entry is \(q_{ii}=-\sum_{j\ne i}q_{ij}\). Write \(P(s,t)\) for the transition-probability matrix from time \(s\) to time \(t\), where \(P_{ij}(s,t)\) is the probability of being in state \(j\) at time \(t\), conditional on being in state \(i\) at time \(s\).

This tutorial uses the row-origin convention: rows of \(P(s,t)\) correspond to starting states, and each row sums to one. With this convention, a short transition update multiplies the current transition matrix on the right. In a time-homogeneous model, the length of the interval matters, but its starting time does not, so \(P(s,t)\) depends on \(t-s\).

The Product-Integral Limit for Constant Intensities

For a short interval of length \(h\), the definition of the transition intensities gives \(P(u,u+h)=I+Qh+o(h)\) as \(h\) approaches zero. The diagonal entries include the probability of remaining in the current state; the off-diagonal entries describe transitions to other states. The \(o(h)\) remainder represents terms that become negligible relative to \(h\) as the interval shrinks. In a finite-state continuous-time Markov model with bounded intensities, the probability of two or more jumps in an interval of length \(h\) is also of order \(o(h)\).

Partition an interval of length \(t\) into \(n\) equal pieces, each of length \(t/n\). Applying the short-interval update successively gives a product of \(n\) factors. For constant \(Q\), the first-order factors are identical, so the limiting product integral is

$$ P(0,t) =\lim_{n\to\infty}\left(I+Q\frac{t}{n}\right)^n =e^{Qt}. $$

The equality to the matrix exponential follows directly from the exponential series. Since every factor is a polynomial in the same matrix \(Q\), the binomial expansion applies:

$$ \left(I+Q\frac{t}{n}\right)^n =\sum_{k=0}^{n}\binom{n}{k}Q^k\left(\frac{t}{n}\right)^k. $$

For each fixed \(k\), the coefficient of \(Q^k\) tends to \(t^k/k!\), because \(\binom{n}{k}(t/n)^k\to t^k/k!\). Thus the limiting series is

$$ \sum_{k=0}^{\infty}\frac{(Qt)^k}{k!}=e^{Qt}. $$

For a finite matrix, this series converges. For example, in any submultiplicative matrix norm, the series is bounded term by term by the convergent scalar series \(\sum_{k=0}^{\infty}(\lVert Q\rVert t)^k/k!\). This justifies taking the limit termwise. The first-order short-interval factors give the same limit: over \(n\) intervals, the accumulated contribution from the \(o(t/n)\) remainders vanishes as the partition is refined under the stated boundedness and regularity conditions.

Theorem: For a finite-state, time-homogeneous continuous-time Markov model with constant transition-intensity matrix \(Q\), the transition-probability matrix over an interval of length \(t\ge 0\) is \(P(0,t)=e^{Qt}\). The matrix exponential is defined by \(e^{Qt}=\sum_{k=0}^{\infty}(Qt)^k/k!\). The result assumes that \(Q\) is a valid generator: its off-diagonal entries are nonnegative and its rows sum to zero.

Two States: An Explicit Formula

Consider two states, labelled 0 and 1. Let \(\alpha\) be the constant intensity from state 0 to state 1, and let \(\beta\) be the constant intensity from state 1 to state 0. Both rates are nonnegative and are measured in inverse-time units, such as per month. The intensity matrix is

$$ Q= \begin{pmatrix} -\alpha & \alpha\\ \beta & -\beta \end{pmatrix}. $$

Set \(\lambda=\alpha+\beta\). Direct multiplication gives \(Q^2=-\lambda Q\). Consequently, for every integer \(k\ge 1\), \(Q^k=(-\lambda)^{k-1}Q\). Substituting these powers into the exponential series yields

$$ e^{Qt} =I+\sum_{k=1}^{\infty}\frac{Q^k t^k}{k!} =I+Q\sum_{k=1}^{\infty}\frac{(-\lambda)^{k-1}t^k}{k!} =I+\frac{1-e^{-\lambda t}}{\lambda}Q, \qquad \lambda>0. $$

The scalar series in the middle is \((1-e^{-\lambda t})/\lambda\), as can be seen by factoring out \(1/\lambda\) and using the exponential series for \(e^{-\lambda t}\). Expanding the entries gives the transition probabilities:

$$ P(0,t)= \begin{pmatrix} \displaystyle \frac{\beta}{\lambda} +\frac{\alpha}{\lambda}e^{-\lambda t} & \displaystyle \frac{\alpha}{\lambda}(1-e^{-\lambda t}) \\[6pt] \displaystyle \frac{\beta}{\lambda}(1-e^{-\lambda t}) & \displaystyle \frac{\alpha}{\lambda} +\frac{\beta}{\lambda}e^{-\lambda t} \end{pmatrix}. $$

Each entry has a direct conditional interpretation. For example, \(P_{01}(0,t)\) is the probability of being in state 1 at time \(t\), given that the process started in state 0. It allows for any number of transitions during the interval, not only a single jump. The formula is valid for \(\lambda>0\); if \(\alpha=\beta=0\), no transitions occur and \(P(0,t)=I\).

Worked Example: Calculating All Four Probabilities

Suppose a hypothetical two-state process has \(\alpha=0.30\) per month and \(\beta=0.20\) per month. Calculate the probability matrix after 4 months.

The total rate is \(\lambda=\alpha+\beta=0.30+0.20=0.50\) per month, and \(\lambda t=(0.50\text{ per month})(4\text{ months})=2\). Thus \(e^{-\lambda t}=e^{-2}\approx0.1353\). Substituting into the formula gives

$$ P(0,4)= \begin{pmatrix} 0.20/0.50+(0.30/0.50)(0.1353) & (0.30/0.50)(1-0.1353) \\ (0.20/0.50)(1-0.1353) & 0.30/0.50+(0.20/0.50)(0.1353) \end{pmatrix} \approx \begin{pmatrix} 0.4812 & 0.5188\\ 0.3459 & 0.6541 \end{pmatrix}. $$

For a process starting in state 0, the probability of being in state 1 after 4 months is approximately 0.5188. For a process starting in state 1, the probability of being in state 0 after 4 months is approximately 0.3459. The rows sum to one, as required. The off-diagonal probabilities differ because the two transition intensities differ.

Checking the Formula and Its Assumptions

The formula has several useful checks. At \(t=0\), \(e^{-\lambda t}=1\), so the result is the identity matrix: the process has not had time to move. For \(t\ge0\), \(0<e^{-\lambda t}\le1\), so the entries are nonnegative. Adding either row gives one. For instance, the first row sum is

$$ \frac{\beta}{\lambda} +\frac{\alpha}{\lambda}e^{-\lambda t} +\frac{\alpha}{\lambda}(1-e^{-\lambda t}) =\frac{\alpha+\beta}{\lambda}=1. $$

If both rates are positive, the rows approach \((\beta/\lambda,\alpha/\lambda)\) as \(t\) increases. This is also a check on the long-run balance: the probability assigned to state 0 is \(\beta/\lambda\), and that assigned to state 1 is \(\alpha/\lambda\).

The assumptions matter as much as the algebra. The process must have constant intensities over the interval, and the Markov property means that, conditional on the current state, its future transition behavior does not depend on the earlier path. The rates must be valid intensities, so \(\alpha,\beta\ge0\). If these assumptions do not fit the process, the result may be an inaccurate model even when the matrix calculation is exact.

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

Suppose a hypothetical process can move from state 0 to state 1 at rate \(\alpha=0.40\) per week, and state 1 is absorbing, so \(\beta=0\). Calculate the transition matrix after 3 weeks.

Here \(\lambda=0.40\) per week and \(\lambda t=(0.40)(3)=1.20\). Since \(e^{-1.20}\approx0.3010\), the two-state formula gives

$$ P(0,3)= \begin{pmatrix} 0/0.40+(0.40/0.40)(0.3010) & (0.40/0.40)(1-0.3010) \\ (0/0.40)(1-0.3010) & 0.40/0.40+(0/0.40)(0.3010) \end{pmatrix} \approx \begin{pmatrix} 0.3010 & 0.6990\\ 0 & 1 \end{pmatrix}. $$

Starting in state 0, the probability of still being there after 3 weeks is approximately 0.3010; the probability of having entered state 1 is approximately 0.6990. Starting in state 1, the process remains there with probability one, as required by the absorbing-state assumption.

Approximations, Alternatives, and Failure Modes

The matrix exponential is the exact constant-intensity result, whereas a finite product of short-interval factors is an approximation to its limit. The distinction can be seen numerically.

Worked Example: Comparing a Finite Product with the Limit

For a hypothetical process, take \(\alpha=0.40\) and \(\beta=0.10\) per month over \(t=2\) months. Then \(\lambda=0.50\) per month and \(e^{-\lambda t}=e^{-1}\approx0.3679\). The exact matrix is

$$ e^{Qt}\approx \begin{pmatrix} 0.10/0.50+(0.40/0.50)(0.3679) & (0.40/0.50)(1-0.3679) \\ (0.10/0.50)(1-0.3679) & 0.40/0.50+(0.10/0.50)(0.3679) \end{pmatrix} = \begin{pmatrix} 0.4943 & 0.5057\\ 0.1264 & 0.8736 \end{pmatrix}. $$

Now divide the 2-month interval into four steps of length \(h=0.5\) month. One first-order update is

$$ I+hQ= \begin{pmatrix} 0.80 & 0.20\\ 0.05 & 0.95 \end{pmatrix}. $$

The four-step approximation is \((I+hQ)^4\). Its eigenvalues are 1 and \(1-h\lambda=0.75\), and it has the same long-run row \((0.20,0.80)\) as the exact matrix. Therefore its rows are \((0.20+0.80(0.75)^4,\;0.80[1-(0.75)^4])\) and \((0.20[1-(0.75)^4],\;0.80+0.20(0.75)^4)\), giving

$$ (I+hQ)^4= \begin{pmatrix} 0.4531 & 0.5469\\ 0.1367 & 0.8633 \end{pmatrix}. $$

The approximation differs from the exact matrix because four steps are not the limiting infinitely fine partition. For example, its \(P_{00}\) entry is about 0.0412 below the exact value. Refining the partition makes \((I+Qt/n)^n\) converge to \(e^{Qt}\).

A finite factor \(I+Qh\) is not automatically a valid probability matrix for an arbitrary step size: its diagonal entry for state \(i\) is \(1-h\sum_{j\ne i}q_{ij}\), which can be negative if \(h\) is too large. The exact matrix exponential remains a valid transition matrix under the generator assumptions. This is one reason not to treat a coarse first-order product as the exact answer.

For comparison, when intensities vary with time, each short-interval factor generally contains a different intensity matrix. Their order can then matter, and the product-integral formulation is the appropriate general expression. Replacing it by the exponential of the integrated intensity matrix is not generally valid unless additional conditions, such as commutation of the matrices at different times, hold. The constant-intensity case avoids this issue because all factors involve the same \(Q\).

Common warning signs include negative entries in a computed probability matrix, rows that fail to sum to one, or probabilities outside the interval from zero to one. These usually indicate an invalid intensity matrix, a coding or multiplication error, or an unsuitable finite-step approximation. A different error is to interpret \(e^{-\alpha t}\) as the full transition probability for a two-way system. It is the probability of no departure from state 0 over the interval only when starting in state 0 and considering its first departure; \(P_{00}(0,t)\) also includes paths that leave state 0 and return.

Key takeaway: Under constant, valid transition intensities, the product-integral limit is \(e^{Qt}\). For two states, the exponential has a closed form involving \(e^{-(\alpha+\beta)t}\); its entries account for paths with repeated transitions, and should be checked for nonnegativity and row sums of one.

Check Your Understanding

Answer each question, then compare your work with the solutions below.

  1. For the two-state model, state the intensity matrix when the rate from state 0 to state 1 is \(\alpha\) and the reverse rate is \(\beta\). What condition must its rows satisfy?
  2. Suppose \(\alpha=0.25\) per month and \(\beta=0.25\) per month. What is \(P_{01}(0,2)\)?
  3. If \(\alpha=0\) and \(\beta>0\), which state is absorbing? Explain using the transition matrix.
  4. Why does a product of identical first-order factors converge to a matrix exponential when the partition is refined?
  5. Why is the constant-intensity formula not generally obtained by simply exponentiating the integrated intensity matrix when intensities vary over time?

Solutions. (1) \(Q=\begin{pmatrix}-\alpha&\alpha\\\beta&-\beta\end{pmatrix}\); each row sums to zero, so the diagonal balances the rates of leaving that state. (2) Here \(\lambda=0.50\) per month and \(\lambda t=1\). Thus \(P_{01}(0,2)=(0.25/0.50)(1-e^{-1})\approx0.3161\), a probability with no units. (3) State 0 is absorbing because its departure rate is zero; the transition matrix has first row \((1,0)\). (4) With \(n\) equal steps the first-order product is \((I+Qt/n)^n\); its binomial-series coefficients converge to those of \(\sum_{k=0}^{\infty}(Qt)^k/k!=e^{Qt}\). (5) Different intensity matrices at different times need not commute, so their ordered product cannot generally be replaced by one ordinary exponential of the integrated matrix.

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