Forward Equation for Time-Inhomogeneous Intensities

Why Calendar Time Matters

In a continuous-time multiple-state model, transition intensities may depend on more than the current state. They can also vary with calendar time: for example, a hypothetical disability model might use different transition rates during a waiting period and after it. When the rates change, the transition probabilities generally cannot be calculated by treating one generator as if it applied throughout the period.

The forward equation describes how probabilities change as the endpoint time moves forward. Its key feature in a time-inhomogeneous model is that the generator in the equation is the generator at the current endpoint time. This tutorial is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: For a finite-state continuous-time Markov model, the time-dependent generator \(Q(t)=(q_{ij}(t))\) has off-diagonal entries \(q_{ij}(t)\geq 0\), the transition intensity from state \(i\) to state \(j\) at time \(t\). Its diagonal entries satisfy \(q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)\), so every row sums to zero. We use row-vector distributions: \(p(t)\) is a row vector, and \(P(s,t)\) has entry \(P_{ij}(s,t)\), the probability of being in state \(j\) at time \(t\) given state \(i\) at time \(s\).

Assume the intensities are nonnegative, sufficiently regular for the differential equations below to hold (piecewise-continuous intensities are sufficient in the examples), and define a non-explosive process. The forward initial-value problem for transition probabilities is

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

Here \(s\) is fixed and \(t\) is the changing endpoint. If the initial distribution at time \(s\) is \(p(s)\), then \(p(t)=p(s)P(s,t)\), and therefore

\frac{d}{dt}p(t)=p(t)Q(t),\qquad p(s)\text{ specified}.

The product order matters under this convention: the probability row vector multiplies the generator on the right. The \(j\)th component of \(p(t)Q(t)\) is the instantaneous probability flow into state \(j\) from other states, minus the flow out of state \(j\).

Why the Forward Equation Uses \(Q(t)\)

The equation follows by considering a short interval at the endpoint. Over an interval of length \(h\), the probability of a transition from \(i\) to \(j\ne i\) is \(q_{ij}(t)h+o(h)\), and the probability of remaining in state \(i\) is \(1+q_{ii}(t)h+o(h)\). Thus

$$ P(t,t+h)=I+hQ(t)+o(h). $$

Starting at \(s\), first transition to time \(t\), then across this short interval. The Markov property gives \(P(s,t+h)=P(s,t)P(t,t+h)\). Subtract \(P(s,t)\), divide by \(h\), and let \(h\) decrease to zero:

\begin{aligned} \frac{P(s,t+h)-P(s,t)}{h} &=P(s,t)\frac{P(t,t+h)-I}{h}\\ &\longrightarrow P(s,t)Q(t). \end{aligned}

This establishes the forward equation and explains why it uses the generator at time \(t\), not a generator held fixed at the starting time \(s\). If \(Q(t)\) is constant, the equation reduces to the time-homogeneous forward equation. When it varies, that constant-generator solution does not generally apply over the whole interval.

Solving Across Changing Intensity Regimes

When the generator is constant within each of several consecutive intervals, solve within each interval using its own rates, then use the ending distribution as the next interval's initial distribution. In matrix form, if the intervals have generators \(Q_1,Q_2,\ldots,Q_m\) and lengths \(\Delta_1,\Delta_2,\ldots,\Delta_m\), the transition matrices are applied in chronological order from left to right:

p_{\mathrm{end}} =p_{\mathrm{start}}\exp(Q_1\Delta_1)\exp(Q_2\Delta_2)\cdots\exp(Q_m\Delta_m).

The expression is a product of interval transition matrices, not a single constant-generator calculation. In general, changing generators need not commute, so replacing this product with an exponential formed from an average generator is not justified.

Worked Example: A Reversible Process with Two Calendar-Time Regimes

A hypothetical process has states 0 and 1 and starts in state 0 at time 0. During year 1, the intensities \(0\to1\) and \(1\to0\) are \(0.2\) and \(0.1\) per year. During year 2, they are \(0.3\) and \(0.1\) per year. Find the probability of state 1 at time 2. The rates are constant within each year but change at time 1.

Method. For a two-state process with constant rates \(\alpha\) from 0 to 1 and \(\beta\) from 1 to 0, let \(r(t)\) be the probability of state 1. The forward equation gives \(r'=\alpha(1-r)-\beta r=\alpha-(\alpha+\beta)r\). Its equilibrium value is \(\alpha/(\alpha+\beta)\); over an interval of length \(u\), the solution is

r(u)=\frac{\alpha}{\alpha+\beta} +\left(r(0)-\frac{\alpha}{\alpha+\beta}\right)e^{-(\alpha+\beta)u}.

First year. Here \(\alpha=0.2\), \(\beta=0.1\), the interval length is 1, and \(r(0)=0\). Therefore,

r(1)=\frac{0.2}{0.3}+ \left(0-\frac{0.2}{0.3}\right)e^{-0.3} =\frac{2}{3}(1-e^{-0.3}) \approx 0.1728.

Second year. The new rates are \(\alpha=0.3\), \(\beta=0.1\). The new equilibrium is \(0.3/0.4=0.75\), and the year begins with \(r(1)\approx0.1727879\). Thus,

r(2)=0.75+(0.1727879-0.75)e^{-0.4} \approx 0.363083 \approx 0.3631.

The probability of state 0 at time 2 is \(1-0.363083\approx0.636917\); the two probabilities sum to 1. Common mistake avoided: using the first year's rates for both years ignores the change at time 1 and does not solve the stated model.

Continuously Varying Intensities

For a two-state model, the forward equation reduces to a scalar differential equation even when both rates vary continuously. Write \(\alpha(t)\) for the intensity from 0 to 1, \(\beta(t)\) for the intensity from 1 to 0, and \(r(t)\) for the probability of state 1. Then

r'(t)=\alpha(t)-\bigl(\alpha(t)+\beta(t)\bigr)r(t).

This is a linear differential equation. An integrating factor provides a direct solution: multiply by \(e^{\int_0^t(\alpha(u)+\beta(u))\,du}\), integrate, and then divide by that factor. For initial time 0, the result is

r(t)=e^{-\int_0^t(\alpha(u)+\beta(u))\,du}r(0) +\int_0^t e^{-\int_s^t(\alpha(u)+\beta(u))\,du}\alpha(s)\,ds.

The first term is the contribution from the initial probability of state 1 surviving the opportunity to leave that state. The integral accumulates probability flow from state 0 into state 1 at each time \(s\), weighted by the subsequent chance of not leaving state 1 before \(t\). This derivation depends on the two-state structure; in a general multistate model the forward system is a coupled system of equations.

Worked Example: Solving a Smoothly Changing Two-State Model

A hypothetical two-state process starts in state 0 at time 0. Over \(0\leq t\leq2\) years, the intensity from 0 to 1 is \(\alpha(t)=0.2+0.1t\) per year, and the intensity from 1 to 0 is \(\beta(t)=0.2-0.1t\) per year. Find the probability of state 1 at time 2. Both rates are nonnegative on this interval.

Method. Their sum is constant: \(\alpha(t)+\beta(t)=0.4\). The scalar forward equation is therefore

r'(t)+0.4r(t)=0.2+0.1t,\qquad r(0)=0.

To solve, try a particular solution \(r_p(t)=At+B\). Substitution gives \(A+0.4(At+B)=0.2+0.1t\). Matching the coefficients of \(t\) gives \(0.4A=0.1\), so \(A=0.25\). Matching constants gives \(A+0.4B=0.2\), so \(0.25+0.4B=0.2\), hence \(B=-0.125\). The general solution is

r(t)=0.25t-0.125+Ce^{-0.4t}.

The initial condition \(r(0)=0\) gives \(C=0.125\). At time 2,

r(2)=0.25(2)-0.125+0.125e^{-0.8} =0.375+0.125(0.449329) \approx0.431166 \approx0.4312.

The state-0 probability is \(1-r(2)\approx0.5688\), so the probabilities sum to 1. Common mistake avoided: the constant total intensity \(0.4\) does not make the generator constant; the individual intensities still vary with time, so the changing inflow term \(0.2+0.1t\) must be retained.

Use the Current Generator, Not a Frozen One

For a nonconstant generator, the distribution's instantaneous rate of change is found by evaluating \(p(t)Q(t)\) at that same time. The derivative is a local quantity; it does not by itself give the distribution at a later time. To obtain a later distribution, solve the initial-value problem across the interval, using the changing intensities throughout.

The following example focuses on correctly evaluating that local change in a three-state model.

Worked Example: Evaluating Probability Flow at the Current Time

A hypothetical model has states 0, 1, and 2. State 2 is absorbing. At time \(t=1\), the distribution is \(p(1)=(0.5,0.3,0.2)\). The current intensities are \(q_{01}(1)=0.1\), \(q_{02}(1)=0.2\), \(q_{10}(1)=0.3\), and \(q_{12}(1)=0.4\) per year. Find the three components of \(p'(1)\).

Method. Complete each generator row with its negative row sum. The current generator and forward-equation calculation are

Q(1)= \begin{pmatrix} -0.3&0.1&0.2\\ 0.3&-0.7&0.4\\ 0&0&0 \end{pmatrix}, \qquad p'(1)=p(1)Q(1).

The component for state 0 is \(0.5(-0.3)+0.3(0.3)+0.2(0)=-0.06\). The component for state 1 is \(0.5(0.1)+0.3(-0.7)+0.2(0)=-0.16\). The component for state 2 is \(0.5(0.2)+0.3(0.4)+0.2(0)=0.22\). Thus \(p'(1)=(-0.06,-0.16,0.22)\) per year. The components sum to zero, as required by conservation of total probability. Common mistake avoided: multiplying by a generator evaluated at some other time would give the wrong instantaneous flow.

Exam Checks and Common Traps

Before accepting a solution, check the time interval, the generator used on it, and the probability total. In particular, keep the time units consistent with the intensities: rates per year require interval lengths in years. At a change point, carry forward the full distribution—not just the probability of one selected state—unless a scalar equation has been justified for that model.

Freezing a time-varying generator: using \(Q(s)\) throughout \([s,t]\) is valid only if the generator is actually constant there.
Using an average rate without justification: an average intensity generally does not reproduce the transition probabilities generated by changing intensities.
Reversing multiplication order: under the row-vector convention, use \(p'(t)=p(t)Q(t)\), not \(Q(t)p(t)\).
Ignoring probability conservation: generator rows sum to zero, so the components of \(p'(t)\) sum to zero; a distribution's components must sum to 1.

Key takeaway: The time-inhomogeneous forward equation is \(P_t(s,t)=P(s,t)Q(t)\), with \(P(s,s)=I\). Evaluate the current generator at the current endpoint time, and propagate probabilities through each change in calendar-time rates.

Check Your Understanding

For each problem, use the forward equation with the stated row-vector convention. Probabilities are rounded to four decimal places where needed.

A process starts in state 0, and state 1 is absorbing. The intensity from 0 to 1 is \(\lambda(t)=0.05+0.05t\) per year for \(0\leq t\leq2\). Find the probability of state 1 at time 2. (Answer: 0.1813)
A reversible two-state process starts in state 0. In year 1, the intensities \(0\to1\) and \(1\to0\) are \(0.1\) and \(0.1\) per year; in year 2, they are \(0.2\) and \(0.1\) per year. Find the probability of state 1 at time 2. (Answer: 0.2399)
At time \(t=2\), a two-state process has state probabilities \(p_0=0.4\) and \(p_1=0.6\). The current intensities are \(q_{01}(2)=0.2\) and \(q_{10}(2)=0.1\) per year. Find the instantaneous derivative of the state-1 probability. (Answer: 0.02 per year)
A three-state model has current distribution \((0.5,0.3,0.2)\) and current generator \(\begin{pmatrix}-0.3&0.1&0.2\\0.3&-0.7&0.4\\0&0&0\end{pmatrix}\). Find the derivative of each state probability and verify its components sum to zero. (Answer: \((-0.06,-0.16,0.22)\) per year; sum \(0\))
A process has states 0 and 1, with state 1 absorbing. Starting in state 0 at time 0, its intensity from 0 to 1 is \(0.08+0.04t\) per year over \(0\leq t\leq3\). Find the probability of remaining in state 0 at time 3. (Answer: \(e^{-0.42}\approx0.6570\))

What You'll Learn