Why Approximate Time-Varying Intensities?
When transition intensities change over time, a single constant-intensity transition matrix generally does not describe the whole interval. A practical approach is to divide the interval into short pieces, approximate the intensities within each piece, construct a local transition matrix, and combine the matrices. The method is useful when the rates are smooth but a closed-form transition matrix is unavailable or inconvenient.
This tutorial develops the short-step Euler approximation. It is a numerical method, not an exact transition model: its accuracy depends on the step size and the smoothness of the intensities. The examples use hypothetical state models.
Building and Combining the Short-Step Matrices
Suppose the period from \(s\) to \(t\) is split into \(n\) intervals, with endpoints \(t_0=s,t_1,\ldots,t_n=t\). Write \(h_k=t_k-t_{k-1}\), and choose a representative point \(t_k^*\) in each interval. The midpoint choice is \(t_k^*=(t_{k-1}+t_k)/2\). For each step, calculate
The approximation over the full period is the product in chronological order:
Here each matrix entry describes movement from the row state to the column state. The order follows the convention that a row vector of state probabilities is multiplied on the right: first by the matrix for the earliest interval, then by the matrix for the next interval. In general, the matrices do not commute, so changing their order can change the answer.
For an interval of length \(h\), the Euler construction approximates a direct transition from state \(i\) to state \(j\ne i\) by \(\lambda_{ij}(t^*)h\). The diagonal entry is one minus the sum of those outgoing probabilities. Thus, a necessary practical condition for this local approximation to give nonnegative entries is \(h\sum_{j\ne i}\lambda_{ij}(t^*)\leq 1\) for every state \(i\). If this fails, use smaller steps. Even when it holds, the method is approximate: a path with two or more transitions within one step is not represented explicitly in that local Euler matrix.
Why the Approximation Works
The short-step formula follows from the small-interval behavior of a continuous-time model. Over an interval of length \(h\), starting in state \(i\), the probability of one transition to a different state \(j\) is approximately \(\lambda_{ij}(t^*)h\). The probability of remaining in \(i\) is approximately one minus the total probability of leaving \(i\). For bounded intensities, the chance of two or more transitions during a sufficiently short interval is of order \(h^2\), so it is omitted from the first-order approximation.
These entries are exactly those of \(I+hQ(t^*)\). To combine steps, condition on the state at the end of the first step: the probability of starting in \(i\), being in an intermediate state \(k\), and ending in \(j\) contributes the product of the corresponding entries. Summing over intermediate states is matrix multiplication. Repeating this argument gives the chronological product of all the local matrices.
For smooth intensities, reducing the step size generally improves the approximation. The local Euler approximation omits terms of order \(h^2\); accumulated across the full interval, its error is typically of first order in the step size. This does not mean every individual probability must move monotonically toward its exact value as steps are refined, so a convergence check should compare successive refinements rather than assume improvement in a particular direction.
Worked Example: Two States With Intensities That Change Over Time
A hypothetical model has two states, \(A\) and \(B\), and runs from time 0 to time 1. The intensity from \(A\) to \(B\) is \(\lambda_{AB}(t)=0.20+0.20t\), and the intensity from \(B\) to \(A\) is \(\lambda_{BA}(t)=0.1625+0.05t\). Use two midpoint Euler steps of length \(0.5\) to approximate the transition matrix and the probability of starting in \(A\) and ending in \(B\).
Method. Evaluate both intensities at the two midpoints, construct the local matrices, and multiply them in chronological order. This method avoids treating rates that change over the year as if they were constant for the entire year.
The midpoints are \(t=0.25\) and \(t=0.75\). At \(t=0.25\), the intensities are \(0.20+0.20(0.25)=0.25\) from \(A\) to \(B\), and \(0.1625+0.05(0.25)=0.175\) from \(B\) to \(A\). Multiplying each rate by \(h=0.5\) gives transition probabilities \(0.125\) and \(0.0875\), respectively. Thus,
At \(t=0.75\), the intensities are \(0.20+0.20(0.75)=0.35\) from \(A\) to \(B\), and \(0.1625+0.05(0.75)=0.20\) from \(B\) to \(A\). The corresponding step probabilities are \(0.35(0.5)=0.175\) and \(0.20(0.5)=0.10\), giving
Multiply in time order. The \(A\)-to-\(B\) entry is \(0.875(0.175)+0.125(0.90)=0.153125+0.1125=0.265625\). The full product is
For example, the lower-left entry is \(0.0875(0.825)+0.9125(0.10)=0.0721875+0.09125=0.1634375\). Each row sums to 1. The requested approximate probability is therefore \(0.265625\), or \(0.2656\) to four decimal places.
Order check. If the matrices are mistakenly reversed, the \(A\)-to-\(B\) entry becomes \(0.825(0.125)+0.175(0.9125)=0.103125+0.1596875=0.2628125\), or \(0.2628\) to four decimal places. This is not the chronological calculation; reversing the matrices changes which transition probabilities are applied first.
Worked Example: A One-Way Transition to an Absorbing State
A hypothetical process starts in state \(H\) and can move only from \(H\) to an absorbing state \(D\). Its intensity is \(\lambda(t)=0.20+0.20t\) for \(0\leq t\leq1\). Use two midpoint Euler steps to estimate the probability of having entered \(D\) by time 1.
Method. Calculate the probability of staying in \(H\) during each step and multiply those stay probabilities. The probability of reaching \(D\) is one minus the probability of staying in \(H\) throughout.
The step length is \(0.5\). The midpoint intensities are \(\lambda(0.25)=0.25\) and \(\lambda(0.75)=0.35\). The first step has stay probability \(1-0.25(0.5)=0.875\), while the second has stay probability \(1-0.35(0.5)=0.825\). Therefore, the approximate probability of remaining in \(H\) for both steps is
The estimated probability of being in \(D\) at time 1 is \(1-0.721875=0.278125\), or \(0.2781\) to four decimal places. As a check, the transition matrix product is \(\begin{pmatrix}0.875&0.125\\0&1\end{pmatrix}\begin{pmatrix}0.825&0.175\\0&1\end{pmatrix}=\begin{pmatrix}0.721875&0.278125\\0&1\end{pmatrix}\).
Common mistake. The two per-step transition probabilities are \(0.125\) and \(0.175\), but they are not probabilities of disjoint events: a process that moves to \(D\) in the first step is already in \(D\) thereafter. Adding them gives the wrong result. The correct calculation is \(1-(0.875)(0.825)\).
For this one-way model the exact probability is also available from the integrated intensity: \(1-\exp(-\int_0^1(0.20+0.20t)\,dt)=1-\exp(-0.30)\), which is approximately \(0.2592\). The difference from \(0.2781\) illustrates that two coarse Euler steps provide an approximation, not an exact answer.
Worked Example: A Sequential Three-State Model
A hypothetical process has states \(0,1,2\). It can move from \(0\) to \(1\) with intensity \(\lambda_{01}(t)=0.10+0.10t\), and from \(1\) to absorbing state \(2\) with constant intensity \(\lambda_{12}(t)=0.20\). There are no other transitions. Use two midpoint Euler steps on \([0,1]\) to find the approximate probabilities of ending in each state, given that the process starts in state \(0\).
Method. Construct one local matrix at each midpoint and use their product. This accounts for the possibility that the process first moves from state \(0\) to state \(1\), then from state \(1\) to state \(2\) in the later step.
At \(t=0.25\), the \(0\)-to-\(1\) intensity is \(0.125\). With \(h=0.5\), its step probability is \(0.0625\), so the probability of staying in state \(0\) is \(0.9375\). The \(1\)-to-\(2\) step probability is \(0.20(0.5)=0.10\), leaving a probability of \(0.90\) in state \(1\). Thus,
At \(t=0.75\), the \(0\)-to-\(1\) intensity is \(0.175\), giving a step probability of \(0.0875\) and a stay probability of \(0.9125\). The \(1\)-to-\(2\) probabilities remain \(0.10\) and \(0.90\). Therefore,
The row for starting state \(0\) in the product is calculated as follows. Ending in state \(0\) requires staying in state \(0\) in both steps: \(0.9375(0.9125)=0.85546875\). Ending in state \(1\) can occur by staying in \(0\) in the first step and moving in the second, or by moving to \(1\) in the first step and staying in \(1\) in the second:
Ending in state \(2\) requires moving from \(0\) to \(1\) in the first step and then from \(1\) to \(2\) in the second, giving \(0.0625(0.10)=0.00625\). The approximate final distribution is therefore \((0.85546875,\,0.13828125,\,0.00625)\). The entries sum to \(0.85546875+0.13828125+0.00625=1\), as required. To four decimal places, the probabilities are \(0.8555\), \(0.1383\), and \(0.0063\).
Common mistake. The chance of ending in state \(2\) is not the product of the two transition probabilities from state \(0\). The second transition has intensity out of state \(1\), so the calculation must include the intermediate state and its probability at the start of the second step.
Practical Accuracy and Exam Traps
The method is most reliable when the intensities vary gradually within each step and the step length is small relative to the scale of the rates. If the rates change rapidly, a uniform grid may be inefficient; use shorter intervals where the rates change most or where the total outgoing intensity is large. A midpoint evaluation often represents a smoothly varying intensity better than evaluating only at the beginning of the interval, but it does not remove the Euler approximation error.
- Do not add overlapping transition probabilities. In a one-way model, use one minus the product of the stay probabilities, not the sum of the step transition probabilities.
- Keep the time units consistent. An intensity per year must be multiplied by a step length measured in years. If the step is a quarter-year, use \(h=0.25\), not \(h=3\) merely because the interval is three months.
- Use the midpoint belonging to each step. For a varying intensity, using the same rate in every step defeats the purpose of the time grid.
- Preserve chronological matrix order. With row-origin transition matrices, the earliest interval appears first in the product.
- Do not confuse an Euler matrix with an exact constant-rate matrix. The Euler matrix is \(I+hQ\). For a constant intensity matrix over a step, the exact local matrix is the matrix exponential \(\exp(hQ)\); these are generally different.
Check Your Understanding
For each problem, use the stated short-step approximation and keep the matrices in chronological order.
- A hypothetical process moves only from an active state to an absorbing state. Its intensity is \(\lambda(t)=0.10+0.20t\) on \([0,1]\). Using two midpoint Euler steps, calculate the probability of transition by time 1. (Answer: 0.1906)
- Two consecutive local transition matrices are \(P_1=\begin{pmatrix}0.90&0.10\\0.20&0.80\end{pmatrix}\) and \(P_2=\begin{pmatrix}0.85&0.15\\0.05&0.95\end{pmatrix}\). Starting in state \(B\), calculate the probability of ending in state \(A\). (Answer: 0.2100)
- A three-state process has consecutive local matrices \(P_1=\begin{pmatrix}0.90&0.10&0\\0&0.80&0.20\\0&0&1\end{pmatrix}\) and \(P_2=\begin{pmatrix}0.85&0.15&0\\0&0.90&0.10\\0&0&1\end{pmatrix}\). Starting in state \(0\), calculate the probabilities of ending in states \(0\), \(1\), and \(2\). (Answer: \(0.7650,\ 0.2250,\ 0.0100\))
- For a one-way transition to an absorbing state, \(\lambda(t)=0.10+0.20t\) on \([0,1]\). Use four equal midpoint Euler steps. The stay factors are \(0.96875,\ 0.95625,\ 0.94375,\ 0.93125\). Find the approximate transition probability. (Answer: 0.1858)
- For the two-state model in the first worked example, calculate the approximate probability of starting in \(A\) and ending in \(B\) if the two local Euler matrices are mistakenly multiplied in reverse order. (Answer: 0.2628)