Forward Equation for Time-Homogeneous Intensities

Introduction

In a continuous-time multiple-state model, a transition probability describes the chance of occupying one state after a specified duration, given a particular starting state. When the transition intensities are constant over time, the forward Kolmogorov equation gives a system of differential equations for how those probabilities change as the duration increases. The key practical step is expanding the matrix equation with the correct row and column indices.

This tutorial focuses on the forward equation for a time-homogeneous model. It establishes the boundary condition at duration zero, shows how to obtain and interpret individual transition-probability equations, and identifies a matrix-ordering trap. The material is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Model, Notation, and Boundary Condition

Suppose a process can occupy states \(1,\ldots,n\), and its transition intensities are constant over time. Use the row convention: \(p_{ij}(t)\) is the probability that the process is in state \(j\) after duration \(t\), conditional on starting in state \(i\). The transition-probability matrix is \(P(t)=[p_{ij}(t)]\).

Definition: The generator \(Q=[q_{ij}]\) has off-diagonal entries \(q_{ij}\geq 0\), representing the instantaneous intensity of transition from state \(i\) to a different state \(j\). Its diagonal entries are \(q_{ii}=-\sum_{j\ne i}q_{ij}\), so every row sums to zero. In a time-homogeneous model, these entries do not depend on time.

The diagonal is not a transition intensity to another state. It is the negative total intensity of leaving the state. Consequently, \(q_{ii}\leq 0\), while the off-diagonal entries are nonnegative. Intensities have units reciprocal to the time unit used in the model; for example, if time is measured in years, an intensity of \(0.3\) is per year.

Boundary condition: At duration zero, the process is still in its starting state with probability one. Therefore \(p_{ii}(0)=1\), \(p_{ij}(0)=0\) for \(i\ne j\), and \(P(0)=I\), the identity matrix.

This condition describes every row at once. It does not say that all transition probabilities are one at time zero: only the probability of remaining in the specified starting state is one.

The Forward Equation and Its Scalar Form

For a time-homogeneous generator, the forward equation is

\frac{\partial P(t)}{\partial t}=P(t)Q,\qquad P(0)=I.

Because \(Q\) is constant, the right side uses the generator for the destination-side transition during a short additional interval. To see why the multiplication order is \(P(t)Q\), consider an interval of length \(h\) added after duration \(t\). The transition probabilities over that short interval satisfy \(P(h)=I+hQ+o(h)\), where the remainder divided by \(h\) tends to zero as \(h\) tends to zero. Composition gives \(P(t+h)=P(t)P(h)\). Subtract \(P(t)\), divide by \(h\), and take the limit:

\frac{P(t+h)-P(t)}{h} =P(t)\frac{P(h)-I}{h} \longrightarrow P(t)Q.

Thus the derivative is \(P(t)Q\). In individual entries, the equation is

p_{ij}'(t)=\sum_{k=1}^{n}p_{ik}(t)q_{kj}.

For a fixed starting state \(i\), this formula says to take the current probabilities across row \(i\) of \(P(t)\), multiply them by column \(j\) of \(Q\), and add. The index \(k\) ranges over the state occupied just before the short additional interval. Terms with \(q_{kj}>0\) contribute to transitions into destination \(j\); the diagonal term \(p_{ij}(t)q_{jj}\) accounts for the negative total intensity of leaving \(j\).

At duration zero, \(P(0)=I\), so \(P'(0)=Q\). In particular, the initial slope of \(p_{ij}(t)\) is \(q_{ij}\). If there is no direct transition from \(i\) to \(j\), this initial slope is zero, even though the process might reach \(j\) later through intermediate states.

Worked Examples

Worked Example: Writing Equations for a Three-State Model

A hypothetical process has three states. From state 1 it moves to state 2 at intensity \(0.3\) per year. From state 2 it moves to state 1 at intensity \(0.1\) per year or to state 3 at intensity \(0.3\) per year. From state 3 it moves to state 2 at intensity \(0.2\) per year. Write the forward equations for a process starting in state 1, then calculate the initial slopes and the initial second derivative of \(p_{13}(t)\).

Method. First form each diagonal entry as the negative sum of the other intensities in its row. The generator is

Q= \begin{pmatrix} -0.3&0.3&0\\ 0.1&-0.4&0.3\\ 0&0.2&-0.2 \end{pmatrix}.

For a process starting in state 1, use row 1 of \(P(t)\). Applying \(p_{1j}'(t)=\sum_k p_{1k}(t)q_{kj}\) to each destination gives

\begin{aligned} p_{11}'(t)&=-0.3p_{11}(t)+0.1p_{12}(t),\\ p_{12}'(t)&=0.3p_{11}(t)-0.4p_{12}(t)+0.2p_{13}(t),\\ p_{13}'(t)&=0.3p_{12}(t)-0.2p_{13}(t). \end{aligned}

The boundary values are \(p_{11}(0)=1\), \(p_{12}(0)=0\), and \(p_{13}(0)=0\). Substitution into the equations gives \(p_{11}'(0)=-0.3\), \(p_{12}'(0)=0.3\), and \(p_{13}'(0)=0\), all per year. There is no direct transition from state 1 to state 3, which explains the zero initial slope for \(p_{13}\).

To calculate the initial second derivative, differentiate the equation for \(p_{13}'(t)\):

p_{13}''(0)=0.3p_{12}'(0)-0.2p_{13}'(0) =0.3(0.3)-0.2(0)=0.09\text{ per year}^2.

Thus \(p_{13}\) initially has zero slope but positive curvature: reaching state 3 requires first reaching state 2. Check and common mistake: the row of \(Q\) for the starting state has a zero in column 3, so \(p_{13}'(0)=0\), not \(0.3\). The intensity \(0.3\) is for the transition from state 2 to state 3 and affects the probability only after some probability has accumulated in state 2.

Worked Example: Evaluating Instantaneous Changes at a Model Time

Consider a hypothetical model with states \(A\) (active), \(D\) (disabled), and \(X\) (dead). The intensities are \(A\to D:0.20\), \(A\to X:0.10\), \(D\to A:0.05\), and \(D\to X:0.25\), all per year; state \(X\) is absorbing. At a particular duration, suppose the transition probabilities from starting state \(A\) are \(p_{AA}=0.60\), \(p_{AD}=0.25\), and \(p_{AX}=0.15\). Find the instantaneous derivatives of these three probabilities.

Method. The row sums of the generator must be zero. Therefore

Q= \begin{pmatrix} -0.30&0.20&0.10\\ 0.05&-0.30&0.25\\ 0&0&0 \end{pmatrix}.

The forward equation for the first row is

\begin{aligned} p_{AA}'&=-0.30p_{AA}+0.05p_{AD},\\ p_{AD}'&=0.20p_{AA}-0.30p_{AD},\\ p_{AX}'&=0.10p_{AA}+0.25p_{AD}. \end{aligned}

Insert the supplied probabilities. The active-state probability has derivative \(p_{AA}'=-0.30(0.60)+0.05(0.25)=-0.180+0.0125=-0.1675\) per year. The disabled-state probability has derivative \(p_{AD}'=0.20(0.60)-0.30(0.25)=0.120-0.075=0.0450\) per year. The dead-state probability has derivative \(p_{AX}'=0.10(0.60)+0.25(0.25)=0.060+0.0625=0.1225\) per year.

The three derivatives sum to \(-0.1675+0.0450+0.1225=0\), as required for probabilities in a row whose total remains one. The dead-state probability is increasing because transitions from both active and disabled states enter the absorbing state. Check and common mistake: the intensity \(0.25\) applies to probability currently in \(D\), so its contribution is \(0.25p_{AD}=0.0625\), not \(0.25p_{AA}\). An intensity must be weighted by the probability of occupying its departure state.

Worked Example: Applying the Boundary Condition to a Different Starting State

A hypothetical four-state process has transitions \(1\to2\) at intensity \(0.20\), \(1\to4\) at \(0.10\), \(2\to3\) at \(0.40\), \(2\to4\) at \(0.10\), \(3\to2\) at \(0.30\), and \(3\to4\) at \(0.20\), all per year. State 4 is absorbing. For a process starting in state 2, find the equations for \(p_{22}(t)\), \(p_{23}(t)\), and \(p_{24}(t)\), and calculate their initial derivatives.

Method. The generator, with rows and columns ordered \(1,2,3,4\), is

Q= \begin{pmatrix} -0.30&0.20&0&0.10\\ 0&-0.50&0.40&0.10\\ 0&0.30&-0.50&0.20\\ 0&0&0&0 \end{pmatrix}.

The boundary condition for a process starting in state 2 is row 2 of \(P(0)=I\): \(p_{22}(0)=1\), while \(p_{21}(0)=p_{23}(0)=p_{24}(0)=0\). Multiplying this transition-probability row by the relevant generator columns gives

\begin{aligned} p_{22}'(t)&=0.20p_{21}(t)-0.50p_{22}(t)+0.30p_{23}(t),\\ p_{23}'(t)&=0.40p_{22}(t)-0.50p_{23}(t),\\ p_{24}'(t)&=0.10p_{21}(t)+0.10p_{22}(t)+0.20p_{23}(t). \end{aligned}

At duration zero, these equations give \(p_{22}'(0)=-0.50\), \(p_{23}'(0)=0.40\), and \(p_{24}'(0)=0.10\), per year. The values sum to zero: \(-0.50+0.40+0.10=0\). The process begins in state 2, and its two possible exits have total intensity \(0.50\), divided between destinations 3 and 4 at intensities \(0.40\) and \(0.10\).

Check and common mistake: using the first row of \(Q\) would describe a process starting in state 1, not state 2. The boundary condition selects the starting row; it does not change which generator column corresponds to a destination.

Interpreting and Checking the Equations

Each equation is an instantaneous rate of change, not a probability by itself. For example, \(p_{ij}'(t)\) has units of probability per year when time is measured in years. A positive derivative means the transition probability is increasing at that duration; a negative derivative means it is decreasing. A derivative can be zero at an instant even when the probability is neither zero nor one.

The diagonal entry \(q_{jj}\) in the equation for destination \(j\) is especially important. It is negative and applies to the probability currently in state \(j\). Omitting it would ignore the chance that the process leaves the destination state during the short additional interval. Likewise, including a positive off-diagonal intensity without multiplying by the probability of its departure state would give the wrong rate.

Choose the starting state.
Use the corresponding row \(i\) of \(P(t)\); at duration zero, that row is row \(i\) of the identity matrix.

Choose a destination.
Use column \(j\) of \(Q\) to write \(p_{ij}'(t)=\sum_k p_{ik}(t)q_{kj}\).

Substitute and check.
Use the supplied transition probabilities, preserve the time units, and verify that the derivatives across a row sum to zero.

A frequent exam trap is reversing the matrix order and writing \(QP(t)\). With the row convention used here, the forward equation is \(P(t)Q\); its entry in row \(i\), column \(j\) is \(\sum_k p_{ik}(t)q_{kj}\). Reversing the order generally changes which probabilities and intensities are combined. The boundary condition \(P(0)=I\) can hide this error, since \(IQ=QI=Q\); checking only initial derivatives may not reveal it. Always expand the desired entry from the stated convention.

The equations can be used to find instantaneous changes without solving the entire system. Solving them to obtain transition probabilities at arbitrary durations is a separate task. For this tutorial, the essential skill is correctly setting up the differential equations, applying the identity boundary condition, and interpreting what each term says about the selected transition probability.

Key takeaway: For constant intensities and the row convention, \(\partial P(t)/\partial t=P(t)Q\), with \(P(0)=I\). In scalar form, \(p_{ij}'(t)=\sum_k p_{ik}(t)q_{kj}\); at duration zero, \(p_{ij}'(0)=q_{ij}\).

Check Your Understanding

For each problem, use the row convention and report derivatives in the stated time units.

A two-state model has intensities \(1\to2:0.12\) and \(2\to1:0.08\) per year. What is \(p_{12}'(0)\)? (Answer: \(0.12\) per year.)
A three-state model has transitions \(1\to2:0.40\), \(2\to3:0.20\), and \(3\to1:0.10\) per year. Write the forward equation for \(p_{13}'(t)\). (Answer: \(p_{13}'(t)=0.20p_{12}(t)-0.10p_{13}(t)\).)
In a model with \(A\to B:0.20\), \(A\to C:0.10\), \(B\to A:0.05\), and \(B\to C:0.25\) per year, with \(C\) absorbing, the transition probabilities from starting state \(A\) at a specified duration are \((0.70,0.20,0.10)\). Find \(p_{AC}'\) at that duration. (Answer: \(0.12\) per year.)
A model has \(1\to2:0.30\) and \(2\to3:0.40\) per year, with state 3 absorbing and no other transitions. Starting in state 1, find \(p_{13}'(0)\) and \(p_{13}''(0)\). (Answer: \(0\) per year and \(0.12\) per year squared.)
For a three-state model, \(Q=\begin{pmatrix}-0.50&0.30&0.20\\0.10&-0.40&0.30\\0&0&0\end{pmatrix}\). At a specified duration, the transition probabilities from starting state 2 are \((0.25,0.50,0.25)\). Find all three derivatives in that row. (Answer: \((-0.075,-0.125,0.200)\) per year.)

What You'll Learn