Section 18 Forward and backward equations

• Markov jump process in infinitesimal time periods
• Transition semigroup, Chapman–Kolmogorov equations, and forward and backward equations
• Matrix exponential as a solution to the forward and backward equations

18.1 Transitions in infinitesimal time periods

In the last section, we defined the continuous time Markov jump process with generator matrix $\mathsf Q$ in terms of the holding times and the jump chain: from state $i$, we wait for a holding time exponentially distributed with rate $q_i = -q_{ii}$, then jump to state $j$ with probability $r_{ij} = q_{ij}/q_i$.

So what happens starting from state $i$ in a very small amount of time $\tau$? The remaining time until a move is still $T \sim \operatorname{Exp}(q_i)$, by the memoryless property. So the probability we don’t move in the small amount of time $\tau$ is $$\mathbb P(T > \tau) = \mathrm{e}^{-q_i \tau} = 1 - q_i\tau + o(\tau) .$$ Here we have used the tail probability of the exponential distribution and the first two terms of the Taylor series. $$\mathrm{e}^x = 1 + x + o(x) \qquad \text{as $x \to 0$.}$$ The probability that we do move in the small amount of time $\tau$, and that the move is to state $j$ is $$\mathbb P(T \leq \tau)r_{ij} = \big(q_i\tau + o(\tau)\big) \frac{q_{ij}}{q_i} = q_{ij}\tau + o(\tau) .$$ The probability we make two or more jumps is a lower order term $o(\tau)$. Thus we have $$\mathbb P\big(X(t+\tau) = j \mid X(t) = i \big) = \begin{cases} 1 - q_i\tau + o(\tau) & \text{for } i = j \\ q_{ij}\tau + o(\tau) & \text{for } i \neq j. \end{cases}$$

This is an equivalent definition of the Markov jump process.

18.2 Transition semigroup and the forward and backward equations

Let us write $p_{ij}(t) = \mathbb P(X(t) = j \mid X(0) = i)$ for the transition probability over a time $t$. This is the continuous-time equivalent to the $n$-step transition probability $p_{ij}(n)$ we had before in discrete time, but now $t$ can be any positive real value. It can be convenient to use the matrix form $\mathsf P(t) = (p_{ij}(t) : i,j \in \mathcal S)$.

In discrete time, we were given a transition matrix $\mathsf P$, and it was easy to find $\mathsf P(n) = \mathsf P^n$ as a matrix power. In continuous time, we are given the generator matrix $\mathsf Q$, so how can we find $\mathsf P(t)$ from $\mathsf Q$?

First, let’s consider $p_{ij}(s+t)$. To get from $i$ to $j$ in time $s + t$, we could first go from $i$ to some $k$ in time $s$, then from that $k$ to $j$ in time $t$, and the intermediate state $k$ can be any state. So we have $$p_{ij}(s+t) = \sum_{k\in \mathcal S} p_{ik}(s)p_{kj}(t) .$$ You should recognise this as the Chapman–Kolmogorov equations again. In matrix form, we can write this as $$\mathsf P(s+t) = \mathsf P(s) \mathsf P(t) .$$ Pure mathematicians sometimes call this equation the “semigroup property”, so we sometimes call the matrices $(\mathsf P(t) : t \geq 0)$ the transition semigroup.

Second, as in previous examples, we can try to get a differential equation for $p_{ij}(t)$ by looking at an infinitesimal increment $p_{ij}(t+\tau)$.

We can start with the Chapman–Kolmogorov equations. We have $$\begin{align*} p_{ij}(t+\tau) &= \sum_k p_{ik}(t)p_{kj}(\tau) \\ &=p_{ij}(t)(1 - q_j\tau) + \sum_{k \neq j} p_{ik}(t)q_{kj}\tau + o(\tau) \\ &= p_{ij}(t) + \sum_k p_{ik}(t)q_{kj}\tau + o(\tau) , \end{align*}$$ where we have treated the $k = j$ term of the sum separately, and taken advantage of the fact that $q_{jj} = - q_j$.

As we have done many times before, we take a $p_{ij}(t)$ to the left hand side and divide through by $\tau$ to get $$\frac{p_{ij}(t + \tau) - p_{ij}(t)}{\tau} = \sum_k p_{ik}(t)q_{kj} + \frac{o(\tau)}{\tau} .$$ Sending $\tau$ to 0 gives us the (Kolmogorov) forward equation $$p'_{ij} (t) = \sum_k p_{ik}(t)q_{kj} .$$ The initial condition is, of course, $p_{ii}(0) = 1$ and $p_{ij}(0) = 0$ for $j \neq i$.

Writing $\mathsf P(t) = (p_{ij}(t))$, and recognising the right hand side as a matrix multiplication, we get the convenient matrix form $$\mathsf P'(t) = \mathsf{P}(t) \mathsf{Q} \qquad \mathsf P(0) = \mathsf I ,$$ where $\mathsf I$ is the identity matrix.

Alternatively, we could have started with the Chapman–Kolmogorov equations the other way round as $$p_{ij}(t+\tau) = \sum_k p_{ik}(\tau)p_{kj}(t) ,$$ with the $\tau$ in the first term rather than the second. Following through the same argument would have given the Kolmogorov backward equation $$\mathsf P'(t) = \mathsf{Q} \mathsf{P}(t) \qquad \mathsf P(0) = \mathsf I ,$$ where the $\mathsf Q$ and $\mathsf P$ are also the other way round.

The forward and backward equations both define the transition semigroup $(\mathsf P(t))$ in terms of the generator matrix $\mathsf Q$.

18.3 Matrix exponential

In discrete time, when we are given the transition matrix $\mathsf P$, we could find the $n$-step transition probabilities $\mathsf P(n)$ using the matrix power $\mathsf P^n$. (At least when the state space was finite – one has to be careful with what the matrix power means if the state space is infinite.) In continuous time (and finite state space), when we are given the generator matrix $\mathsf Q$, it will turn out that a similar role is played by the matrix exponential $\mathsf P(t) = \mathrm{e}^{t\mathsf Q}$, which solves the forward and backward equations. In this way, the generator matrix $\mathsf Q$ is a bit like “the logarithm of the transition matrix $\mathsf P$” – although this is only a metaphor, rather than a careful mathematical statement.

Let’s be more formal about this. You may remember that the usual exponential function is defined by the Taylor series $$\mathrm{e}^x = \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots.$$ Similarly, we can define the matrix exponential for any square matrix by the same series $$\mathrm{e}^{\mathsf A} = \exp({\mathsf A}) = \sum_{n=0}^\infty \frac{{\mathsf A}^n}{n!} = {\mathsf I} + {\mathsf A} + \tfrac12 \mathsf A^2+ \tfrac16 \mathsf A^3 + \cdots,$$ where we interpret $\mathsf A^0 = \mathsf I$ to be the identity matrix.

In the same way that a computer can easily calculate a matrix power $\mathsf P^n$ very quickly, a computer can also calculate matrix powers very quickly and very accurately – see Problem Sheet 9 Question 5 for more details about how.

Since a matrix commutes with itself, we have $\mathsf {A}^n \mathsf A = \mathsf{AA}^n$, so a matrix commutes with its own exponential, meaning that $\mathsf A \mathrm{e}^{\mathsf A} = \mathrm{e}^{\mathsf A} \mathsf A$. This will be useful later.

Some properties of the standard exponential include $$(\mathrm{e}^x\big)^n = \mathrm{e}^{nx} \qquad \mathrm{e}^{x+y} = \mathrm{e}^x \mathrm{e}^y \qquad \frac{\mathrm d}{\mathrm d x} \mathrm{e}^{ax} = a \mathrm{e}^{ax} .$$ Similarly, we have for the matrix exponential $$(\mathrm{e}^{\mathsf A}\big)^t = \mathrm{e}^{t{\mathsf A}} \qquad \mathrm{e}^{\mathsf A + \mathsf B} = \mathrm{e}^{\mathsf A} \mathrm{e}^{\mathsf B} \qquad \frac{\mathrm d}{\mathrm d t} \mathrm{e}^{t{\mathsf A}} = {\mathsf A} \mathrm{e}^{t{\mathsf A}} = \mathrm{e}^{t{\mathsf A}} {\mathsf A} .$$

From this last expression, we get that $$\frac{\mathrm d}{\mathrm d t} \mathrm{e}^{t{\mathsf Q}} = \mathrm{e}^{t{\mathsf Q}} {\mathsf Q} = {\mathsf Q} \mathrm{e}^{t{\mathsf Q}} .$$ Comparing this to the forward equation $\mathsf P'(t) = \mathsf{P}(t) \mathsf Q$ and backward equation $\mathsf{P}'(t) = \mathsf Q \mathsf{P}(t)$, we see that we have a solution to the forward and backward equations of $\mathsf P(t) = \mathrm{e}^{t\mathsf Q}$, which also satisfies the common initial condition $\mathsf P(0) = \mathrm{e}^{0\mathsf Q} = \mathsf I$.

We also see that we have the semigroup property $$\mathsf P(s+t) = \mathrm{e}^{(s+t)\mathsf Q} = \mathrm{e}^{s \mathsf Q + t \mathsf Q} = \mathrm{e}^{s \mathsf Q}\mathrm{e}^{t \mathsf Q} = \mathsf P(s) \mathsf P(t) .$$

As a formal summary, we have the following.

Theorem 18.1 Let $(X(t))$ be a time homogeneous continuous time Markov jump process with generator matrix $\mathsf Q$. Let $(\mathsf P(t))$ be the transition semigroup, where $\mathsf P(t) = (p_{ij}(t))$ is defined by $p_{ij}(t) = \mathbb P(X(t) = j \mid X(0) = i)$.

Then $(\mathsf P(t))$ is the minimal nonnegative solution to the forward equation $$\mathsf P'(t) = \mathsf P(t) \mathsf Q \qquad \mathsf P(0) = \mathsf I ,$$ and is also the minimal nonnegative solution to the backward equation $$\mathsf P'(t) = \mathsf Q \mathsf P(t) \qquad \mathsf P(0) = \mathsf I .$$

When the state space $\mathcal S$ is finite, the forward and backward equations both have a unique solution given by the matrix exponential $\mathsf P(t) = \mathrm{e}^{t \mathsf Q}$.

In the next section, we develop the theory we already know in discrete time: communicating classes, hitting times, recurrence and transience.