Backward Equation with Time-Dependent Intensities

Why the Starting Time Matters

In a time-inhomogeneous Markov model, transition intensities can change with calendar time. The probability of moving between states during the next short interval therefore depends on when that interval begins. To calculate a transition probability from time $s$ to a fixed terminal time $t$, the backward equation uses the generator at the starting time $s$. This is different from the forward equation, which changes the ending time.

Throughout, use a finite state space and the row-origin convention: $P(s,t)$ has entry $P_{ij}(s,t)$, the probability of being in state $j$ at time $t$ given state $i$ at time $s$. The generator $Q(s)$ has off-diagonal entry $q_{ij}(s)$ equal to the instantaneous intensity of a transition from state $i$ to state $j$, and diagonal entries chosen so each row sums to zero. Assume the intensities are deterministic and continuous, or piecewise continuous with the equations applied on each interval of continuity.

Definition: For $s\leq t$, the backward Kolmogorov equation in the row-origin convention is $$ \frac{\partial}{\partial s}P(s,t)=-Q(s)P(s,t), \qquad P(t,t)=I. $$ The derivative is with respect to the starting time $s$; the terminal time $t$ is held fixed. In particular, the generator multiplying $P(s,t)$ is $Q(s)$, not $Q(t)$.

The boundary condition $P(t,t)=I$ says that over a zero-length interval the state is unchanged with probability one. The backward equation evolves the transition probabilities from this terminal condition toward earlier starting times. By contrast, the forward equation is $\partial P(s,t)/\partial t=P(s,t)Q(t)$: it changes the terminal time while holding the starting time fixed.

Why the Backward Equation Holds

Fix the terminal time $t$, and consider a short initial interval from $s$ to $s+h$. Conditional on the state at $s+h$, the future transition probabilities are the entries of $P(s+h,t)$. Over the short first interval, the transition matrix is $I+hQ(s)+o(h)$, where $o(h)/h$ tends to zero as $h$ tends to zero. Applying the Markov property across this first interval gives the needed local relation.

P(s,t)=\bigl(I+hQ(s)+o(h)\bigr)P(s+h,t).

Subtract $P(s+h,t)$ from both sides, divide by $h$, and let $h$ tend to zero. The resulting derivative is with respect to the earlier starting time, which accounts for the minus sign.

\frac{P(s+h,t)-P(s,t)}{h} =-Q(s)P(s+h,t)+o(1) \quad\Longrightarrow\quad \frac{\partial}{\partial s}P(s,t)=-Q(s)P(s,t).

This argument also explains why substituting $Q(t)$ is generally incorrect: the short interval being expanded is $[s,s+h]$, so its intensity is determined by the generator at $s$. If $Q$ is constant, its time argument does not matter, but that special case can conceal the distinction.

Using a Terminal Probability Vector

It is often simpler to work with one column of the transition matrix, or with a terminal event. Let $g$ be a column vector whose entry $g_j$ is the specified terminal value in state $j$. Define $u(s)=P(s,t)g$. Its entry $u_i(s)$ is the expected terminal value conditional on being in state $i$ at time $s$. In particular, for a terminal-state probability, set $g_j=1$ for the desired terminal state and $g_j=0$ for the others.

Result: For fixed terminal time $t$, the vector $u(s)=P(s,t)g$ satisfies $$ u'(s)=-Q(s)u(s), \qquad u(t)=g. $$ For a probability calculation, $g$ is an indicator vector, and the solution is worked backward from $t$ to the desired starting time.

In coordinates, if the current state is $i$, then $$ u_i'(s)=\sum_j q_{ij}(s)u_j(s) =\sum_{j\ne i}q_{ij}(s)\bigl(u_j(s)-u_i(s)\bigr). $$ The second equality uses $q_{ii}(s)=-\sum_{j\ne i}q_{ij}(s)$. It shows that the derivative depends on the differences between the terminal values associated with the possible next states and the current state. These equations must be paired with the terminal values at $s=t$, not with initial values at an earlier time.

Worked Equations from Simple Models

Worked Example: One Transition to an Absorbing State

A hypothetical two-state process has state 0 and absorbing state 1. The only transition is from 0 to 1, with intensity $\lambda(s)=1+s$ per year, where $s$ is measured in years. Find the probabilities of being in states 0 and 1 at time 2, given state 0 at time 1.

Method. Use the backward equation for the terminal event “be in state 0 at time 2.” The generator and terminal vector are $$ Q(s)= \begin{pmatrix} -(1+s)&1+s\\ 0&0 \end{pmatrix}, \qquad g= \begin{pmatrix} 1\\0 \end{pmatrix}. $$

Since state 1 is absorbing, $u_1(s)=0$. The state-0 equation is therefore $$ u_0'(s)=-(1+s)u_0(s), \qquad u_0(2)=1. $$ Separating variables and integrating from $s$ to 2 gives $$ \ln u_0(2)-\ln u_0(s)=-\int_s^2(1+r)\,dr, \qquad u_0(s)=\exp\left(-\int_s^2(1+r)\,dr\right). $$ At $s=1$, the cumulative intensity is $$ \int_1^2(1+r)\,dr =\left[r+\frac{r^2}{2}\right]_1^2 =4-1.5=2.5. $$ Thus $P_{00}(1,2)=e^{-2.5}=0.0821$, to four decimal places. Since the two states are exhaustive, $P_{01}(1,2)=1-0.0821=0.9179$. The probabilities sum to $1.0000$, as required.

Check. The chance of remaining in state 0 is the exponential of minus the integrated exit intensity over the interval, and the complementary probability is the chance of reaching the absorbing state. The common mistake of integrating to get 2 rather than 2.5 would give $e^{-2}=0.1353$; it is wrong because $\int_1^2(1+r)\,dr=2.5$, not 2.

Worked Example: Symmetric Switching with Time-Varying Intensities

A hypothetical process switches between states 0 and 1. In each direction the intensity at time $s$, measured in years, is $s$ per year. Find the probabilities of being in each state at time 1, given state 0 at time 0.

Method. Calculate the probability of terminal state 0 with $g=(1,0)^\mathsf{T}$. The generator is $$ Q(s)= \begin{pmatrix} -s&s\\ s&-s \end{pmatrix}. $$ The backward equations are $u_0'=s(u_0-u_1)$ and $u_1'=s(u_1-u_0)$, with $u_0(1)=1$ and $u_1(1)=0$.

Set $C(s)=u_0(s)+u_1(s)$ and $D(s)=u_0(s)-u_1(s)$. Adding the equations gives $C'(s)=0$, so $C(s)=C(1)=1$. Subtracting them gives $$ D'(s)=2sD(s), \qquad D(1)=1. $$ Integrating this scalar equation backward, $$ D(s)=\exp\left(\int_1^s 2r\,dr\right)=e^{s^2-1}. $$ Consequently $D(0)=e^{-1}=0.3679$, to four decimal places. Since $u_0=(C+D)/2$ and $u_1=(C-D)/2$, $$ u_0(0)=\frac{1+e^{-1}}{2}=0.6839,\qquad u_1(0)=\frac{1-e^{-1}}{2}=0.3161. $$

Check. The probabilities are nonnegative and sum to $1.0000$. The symmetric switching process can be in state 0 at time 1 after an even number of switches, so the probability is not simply the probability of no switch. The common mistake avoided here is treating one-way survival as the terminal-state probability when transitions can occur in both directions.

Working Across a Change in the Generator

If intensities are piecewise continuous, solve the backward system separately on each interval and pass the terminal vector at the boundary to the preceding interval. The vector at the boundary is the value of the continuation problem from that time onward. It is not generally the original terminal vector.

Worked Example: A Three-State Process with Two Time Periods

Consider a hypothetical process with states 0, 1, and 2, where state 2 is absorbing. During years $0\leq s<1$, the only transition is $0\to1$, at intensity $1+s$ per year. During years $1\leq s\leq2$, the only transition is $1\to2$, at intensity $1+s$ per year. Find the probability of being in state 2 at time 2, given state 0 at time 0.

Method. Start with the terminal indicator $g=(0,0,1)^\mathsf{T}$ at time 2. On the second interval, state 1 reaches state 2 with probability one minus the probability of no transition: $$ u_1(1)=1-\exp\left(-\int_1^2(1+r)\,dr\right) =1-e^{-2.5}=0.9179150. $$ State 0 cannot move during this second interval, so $u_0(1)=0$; state 2 remains absorbing, so $u_2(1)=1$.

Now solve backward on $0\leq s<1$. In that interval state 1 stays in state 1, so $u_1(s)=u_1(1)$. The state-0 equation is $u_0'(s)=(1+s)(u_0(s)-u_1(1))$, with $u_0(1)=0$. Its solution is $$ u_0(s)=u_1(1)\left(1-\exp\left(-\int_s^1(1+r)\,dr\right)\right). $$ At $s=0$, the first-period cumulative intensity is $\int_0^1(1+r)\,dr=1.5$. Therefore $$ u_0(0)=(1-e^{-1.5})(1-e^{-2.5}) =(0.7768698)(0.9179150) =0.7131, $$ to four decimal places.

Check. The result is the product of the chance of moving from 0 to 1 by time 1 and, conditional on reaching 1, the chance of moving from 1 to 2 by time 2. Numerically, the factors are $0.7768698$ and $0.9179150$, so the probability is $0.7131$, within the possible range from 0 to 1. The common mistake avoided is using the terminal value $g$ unchanged at time 1; the continuation probability $u_1(1)=0.9179150$ must first be calculated.

Checks and Common Exam Traps

A backward-equation calculation is easiest to audit when the time argument, direction of integration, and boundary values are stated explicitly. The derivative is taken with respect to $s$, but the solution is often evaluated by integrating from the known terminal condition at $t$ toward a smaller $s$. A negative derivative sign in the matrix equation does not mean that probabilities must decrease as the starting time moves earlier; the sign and direction of the change must be interpreted together.

Using the wrong generator. The backward equation has $-Q(s)P(s,t)$. The forward equation has $P(s,t)Q(t)$. Substituting the ending-time generator into the backward equation is generally wrong when intensities vary.
Reversing the matrix order. With row-origin probabilities, the backward equation multiplies on the left by $Q(s)$. Do not switch to a different convention midway through a calculation.
Using an initial condition instead of a terminal condition. For fixed $t$, specify $P(t,t)=I$, or $u(t)=g$, and solve toward earlier starting times.
Miscomputing cumulative intensity. Integrate the time-varying rate over exactly the interval in question. For example, on $[1,2]$, the rate $1+s$ has cumulative intensity $2.5$, not 2.
Forgetting probability checks. For each starting state, transition probabilities must be nonnegative and sum to one. For a terminal indicator, each component of $u(s)$ must lie between zero and one.

Key takeaway: Fix the terminal time, identify the terminal vector, and solve $u'(s)=-Q(s)u(s)$ backward in the starting time. The local generator is evaluated at the time $s$ being moved, and each time interval must use its own intensities.

Check Your Understanding

For each problem, use the backward equation and the stated terminal time. Probabilities in the answers are rounded to four decimal places.

A two-state process has state 1 absorbing and intensity $\lambda(s)=3s$ from state 0 to state 1. Find $P_{00}(0,1)$ and $P_{01}(0,1)$. (Answer: $0.2231$ and $0.7769$)
In a hypothetical two-state absorbing model, the intensity from state 0 to state 1 is $2+s$ per year. Starting in state 0 at time 1, find the probabilities of states 0 and 1 at time 2. (Answer: $P_{00}(1,2)=0.0302,\ P_{01}(1,2)=0.9698$)
A two-state process switches in both directions, each at intensity $1+s$ per year. Starting in state 0 at time 0, find the probabilities of the two states at time 1. (Answer: $P_{00}(0,1)=0.5249,\ P_{01}(0,1)=0.4751$)
A three-state process has state 2 absorbing. On $[0,1)$, only $0\to1$ is possible, at constant intensity 2 per year. On $[1,2]$, only $1\to2$ is possible, at intensity $1+s$ per year. Find the probability of state 2 at time 2 given state 0 at time 0. (Answer: $(1-e^{-2})(1-e^{-2.5})=0.7937$)
A two-state process has state 1 absorbing and intensity $\lambda(s)=2s$. Find the probability of state 0 at time 1 given state 0 at time $0.5$, and state which time interval is integrated. (Answer: $e^{-0.75}=0.4724$, integrating over $[0.5,1]$)

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