Recovering the Generator From the Semigroup

From Transition Probabilities to Instantaneous Rates

A transition probability \(p_{ij}(h)\) records the chance of being in state \(j\) after a time interval of length \(h\), given that the chain started in state \(i\). For small \(h\), a move from \(i\) to a different state \(j\) is usually unlikely, and its probability is approximately a rate multiplied by \(h\). The generator records these first-order rates. Recovering it from the transition semigroup makes precise how a family of finite-time transition probabilities determines instantaneous behavior.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_t)_{t\ge0}\) be a time-homogeneous continuous-time Markov chain on a finite state space \(S\). Write \(P(t)=(p_{ij}(t))\) for its transition matrix, where \(p_{ij}(t)=\mathbb{P}(X_{s+t}=j\mid X_s=i)\). Each \(P(t)\) is row-stochastic, and \(P(0)=I\). The semigroup property is \(P(s+t)=P(s)P(t)\).

The off-diagonal entry is a small-time transition rate: if \(q_{ij}\) exists, then \(p_{ij}(h)=q_{ij}h+o(h)\) as \(h\downarrow0\). Here \(o(h)\) denotes a quantity whose ratio to \(h\) tends to zero. The diagonal entry is different because \(p_{ii}(h)\) is close to \(1\), not to \(0\). It measures the first-order decrease in the probability of still being in \(i\).

What the Derivatives Imply

Proof. Fix \(i\). If \(j\ne i\), then \(p_{ij}(h)\ge0\) for every \(h>0\), so its limit after division by \(h\) is nonnegative: \(q_{ij}\ge0\). Because \(S\) is finite, we can differentiate the finite row-sum identity \(\sum_{j\in S}p_{ij}(h)=1\) at \(h=0\) from the right. The derivative of the constant on the right is zero, so \(\sum_{j\in S}q_{ij}=0\). Rearranging gives \(q_{ii}=-\sum_{j\ne i}q_{ij}\). Finally, the definition of each right derivative gives \(p_{ij}(h)=\mathbf{1}_{\{i=j\}}+h q_{ij}+o(h)\), including the diagonal case. \(\square\)

Finiteness matters in the row-sum argument: it permits differentiation of a finite sum. On a countable state space, coordinatewise derivatives alone do not justify interchanging a derivative and an infinite sum. A row-sum conclusion there needs conditions that justify that interchange. In the finite-state setting, the theorem shows that a valid generator has nonnegative off-diagonal entries and diagonal entries determined by the row-sum constraint.

The same derivatives have a direct interpretation through holding times and jump destinations. Suppose that, on leaving state \(i\), the chain waits an exponential time with rate \(q_i\ge0\), then jumps to a different state \(j\) with probability \(r_{ij}\), where \(\sum_{j\ne i}r_{ij}=1\) when \(q_i>0\). If \(q_i=0\), the state is absorbing. For a finite-state chain with these rates, the generator entries are \(q_{ij}=q_i r_{ij}\) for \(j\ne i\), and \(q_{ii}=-q_i\).

Proof. Let \(\Lambda=\max_{k\in S}q_k\), which is finite. Starting at \(i\), the probability of a first jump by time \(h\) is \(1-e^{-q_i h}\le q_i h\le\Lambda h\). Conditional on the time and destination of that first jump, the chance of a further jump before time \(h\) is at most \(\Lambda h\). Therefore the probability of at least two jumps by time \(h\) is at most \(\Lambda^2h^2\), which is \(o(h)\). For \(j\ne i\), the probability of exactly one jump, from \(i\) to \(j\), is \(\int_0^h q_i e^{-q_i s}r_{ij}e^{-q_j(h-s)}\,ds\). Dividing by \(h\) and letting \(h\downarrow0\) gives \(q_i r_{ij}\), since the exponential factors tend to \(1\) uniformly for \(0\le s\le h\). Events with two or more jumps contribute at most \(\Lambda^2h^2=o(h)\), so \(p_{ij}(h)/h\to q_i r_{ij}\). For the diagonal, the probability of no jump is \(e^{-q_i h}=1-q_i h+o(h)\). A path that leaves \(i\) and returns by time \(h\) must make at least two jumps, an event of probability \(o(h)\). Hence \((p_{ii}(h)-1)/h\to-q_i\). If \(q_i=0\), the chain stays at \(i\), and these same limits give zero for every entry in its row. \(\square\)

Thus, when \(q_i>0\), the total rate of leaving \(i\) is \(\sum_{j\ne i}q_{ij}=q_i\), and the destination probabilities are \(r_{ij}=q_{ij}/q_i\). The division is meaningful only when \(q_i>0\); for an absorbing state there is no next jump destination to describe.

A Common Mistake and Its Correction

A frequent error is to divide the diagonal transition probability itself by \(h\). In the two-state example, this would give \(p_{11}(h)/h=e^{-2h}/h\), which diverges as \(h\downarrow0\); it is not the diagonal generator entry. The chain is in state \(1\) with probability close to \(1\) at a very short time, so the relevant quantity is the change from \(1\): \((p_{11}(h)-1)/h=(e^{-2h}-1)/h\to-2\). For off-diagonal entries, by contrast, \(p_{ij}(0)=0\), so no subtraction is needed.

The derivative definition is useful because it separates immediate behavior from later transitions. An off-diagonal \(q_{ij}\) measures the first-order chance of moving from \(i\) to \(j\), while \(q_{ii}\) balances the total rate of leaving \(i\). These are local, time-zero quantities; a transition probability \(p_{ij}(t)\) for a nonzero \(t\) can include paths with several jumps and should not be read as a single jump rate.