Stochastic Continuity of Transition Semigroups

What Continuity at Time Zero Means

A continuous-time transition semigroup describes how transition probabilities depend on elapsed time. At elapsed time zero, a process that starts in state \(i\) is still in state \(i\). Stochastic continuity asks whether, over a sufficiently short positive interval, the probability of being somewhere else becomes small. This is a statement about probabilities, not a claim that the process makes no jumps.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(S\) be a finite or countable state space, and let \((X_t)_{t\ge0}\) be a time-homogeneous Markov process with filtration \((\mathcal{F}_t)_{t\ge0}\). Its transition probabilities are \(p_{ij}(t)=\mathbb{P}(X_{s+t}=j\mid X_s=i)\), for \(i,j\in S\); time homogeneity makes this conditional probability independent of \(s\). Write \(P(t)=(p_{ij}(t))\). Each row is a probability distribution, \(P(0)=I\), and the semigroup property is \(P(s+t)=P(s)P(t)\). The row convention means \(p_{ij}(s+t)=\sum_{k\in S}p_{ik}(s)p_{kj}(t)\).

For a time-homogeneous process, this condition also gives continuity in probability over a short interval starting at any fixed time \(t\). Indeed, conditional on \(X_t=i\), the probability of a change over \([t,t+h]\) is \(1-p_{ii}(h)\). If the initial state is random, the state at time \(t\) has a probability distribution over \(S\), and the same short-interval statement follows by averaging over that distribution.

Leaving the Starting State Controls the Limit

On a countable state space, the probability of being away from \(i\) after time \(h\) is \(1-p_{ii}(h)=\sum_{j\ne i}p_{ij}(h)\). Thus it controls every off-diagonal transition probability at once: for \(j\ne i\), \(0\le p_{ij}(h)\le1-p_{ii}(h)\). It also gives a total-variation description. If the total-variation distance between probability distributions \(\nu\) and \(\eta\) is \(\sup_{A\subseteq S}|\nu(A)-\eta(A)|\), then the distance between row \(i\) of \(P(h)\) and the point mass at \(i\) is exactly \(1-p_{ii}(h)\).

Proof. Fix \(i\), and let \(\nu_h\) be row \(i\) of \(P(h)\), while \(\delta_i\) is the probability distribution concentrated at \(i\). For any \(A\subseteq S\), if \(i\notin A\), then \(|\nu_h(A)-\delta_i(A)|=\nu_h(A)\le1-\nu_h(\{i\})\). If \(i\in A\), then \(|\nu_h(A)-\delta_i(A)|=1-\nu_h(A)=\nu_h(S\setminus A)\le1-\nu_h(\{i\})\). Taking \(A=S\setminus\{i\}\) attains this upper bound. Therefore the total-variation distance is \(1-p_{ii}(h)\). If this quantity tends to zero, then \(p_{ii}(h)\to1\), and for every \(j\ne i\), \(0\le p_{ij}(h)\le1-p_{ii}(h)\to0\). Conversely, the definition of stochastic continuity includes \(p_{ii}(h)\to1\), which makes the distance tend to zero. \(\square\)

This equivalence uses the probability-row property of the transition kernel. It does not require a finite state space: the countable sum of probabilities away from \(i\) is exactly \(1-p_{ii}(h)\).

Proof. Fix \(i\). If \(q_i>0\), the holding time before the first jump from \(i\) is exponential with rate \(q_i\), so the probability of no jump during an interval of length \(h\), when starting in \(i\), is \(e^{-q_i h}\). Ending in \(i\) is at least as likely as making no jump, so \(p_{ii}(h)\ge e^{-q_i h}\). Since \(p_{ii}(h)\le1\), we obtain \(0\le1-p_{ii}(h)\le1-e^{-q_i h}\to0\). If \(q_i=0\), the state is absorbing and \(p_{ii}(h)=1\), so the same conclusion holds. For any \(j\ne i\), \(0\le p_{ij}(h)\le1-p_{ii}(h)\), which gives \(p_{ij}(h)\to0\). These are exactly the required limits. \(\square\)

The proof needs only the first holding time: later jumps cannot increase the probability of having left the starting state by time \(h\) beyond the probability of at least one jump. The finiteness of \(q_i\) ensures that this first holding time is positive with probability one and that its probability of ending before \(h\) tends to zero.

Worked Examples

What the Semigroup Property Does Not Guarantee

Stochastic continuity is an additional regularity property; it does not follow from the semigroup identity alone. For example, on \(S=\{1,2\}\), define \(P(0)=I\), and for every \(t>0\) set

Every row sums to \(1\), and the matrix \(R\) displayed here satisfies \(R^2=R\), so \(P(s+t)=P(s)P(t)\): if both times are positive the product is \(R^2=R\), and if one time is zero the product is unchanged. Nevertheless, \(p_{11}(h)=\tfrac12\) for every \(h>0\), not a quantity tending to \(1\). This semigroup is not stochastically continuous at zero. It cannot describe a CTMC with finite holding rate at each state, which is why the finite-exit-rate condition in the theorem matters.

Common mistake. In the symmetric two-state example, one might write \(p_{12}(h)=1-e^{-2h}\), reasoning that this is the probability of at least one jump. That is the probability of leaving state \(1\) at least once, not the probability of ending in state \(2\): two jumps return the chain to state \(1\). The correct endpoint probability counts an odd number of jumps, giving \(p_{12}(h)=(1-e^{-4h})/2\). Both expressions tend to zero, but only the latter is the transition probability.

Stochastic continuity is useful because it formalizes the idea that short elapsed times cause only small changes in probability. For CTMCs with finite exit rates, it follows directly from the first holding time and gives convergence of each transition row to its time-zero point mass. It does not assert that a path is constant on an interval around zero, nor does it identify the rate of change of a transition probability; those are different questions.