Changing the Clock, Not the Route
A continuous-time Markov chain has two distinct features: the sequence of states it visits and the time it spends between successive jumps. Multiplying every transition rate by the same positive constant changes the speed of the clock. It does not favor one possible destination over another, because all rates out of a given state are multiplied by that same constant. This distinction is useful when rates are expressed in different time units, or when comparing a process that runs faster or slower than another.
Let \((X_t)_{t\ge0}\) be a non-explosive, time-homogeneous continuous-time Markov chain on a finite or countable state space \(S\), defined on a probability space \((\Omega,\mathcal F,\mathbb P)\). Write its generator as \(Q=(q_{ij})\), with finite exit rates \(q_i=-q_{ii}=\sum_{j\ne i}q_{ij}\). Thus \(q_{ij}\ge0\) for \(i\ne j\) and each row of \(Q\) sums to zero. When \(q_i>0\), the holding time in state \(i\) is exponential with rate \(q_i\), and the probability that the next state is \(j\ne i\) is \(q_{ij}/q_i\). When \(q_i=0\), the state is absorbing.
To make the conditional statements precise, use the natural filtration \(\mathcal F_t^X=\sigma(X_u:0\le u\le t)\) of \(X\). The natural filtration of \(Y\) is \(\mathcal G_t=\mathcal F_{ct}^X\). In particular, conditioning on the observed history of \(Y\) up to time \(t\) means conditioning on the history of \(X\) up to time \(ct\), not on an arbitrary larger filtration.
The Time-Change Result
Proof. Let \(p_{ij}(t)=\mathbb P(X_t=j\mid X_0=i)\), and let \(p^c_{ij}(t)\) denote the transition probabilities of \(Y\). For any \(s,t\ge0\), the Markov property of \(X\) relative to its natural filtration gives, almost surely,
The conditional probability depends on the past only through \(Y_t\), so \(Y\) is time-homogeneous Markov and \(p^c_{ij}(s)=p_{ij}(cs)\). This proves \(P_c(s)=P(cs)\). By the definition of the generator, for each \(i,j\),
For the holding-time and jump assertions, consider a visit to state \(i\) with \(q_i>0\). If the original holding time is \(H\), then the time spent in \(i\) by \(Y\) is \(H/c\): the rescaled path reaches the same next state when its clock has advanced by the original holding time \(H\). For \(u\ge0\), \(\mathbb P(H/c>u)=\mathbb P(H>cu)=e^{-cq_i u}\), so the rescaled holding time is exponential with rate \(cq_i\). The next state has not changed under the time change, and its probability is \(q_{ij}/q_i\). Equivalently, the scaled rates give \((cq_{ij})/(cq_i)=q_{ij}/q_i\). If \(q_i=0\), the original path stays in \(i\) forever, so does the rescaled path. Non-explosion ensures the processes are defined as paths for all \(t\ge0\). \(\square\)
The semigroup identity also follows directly from the pathwise definition: observing \(Y\) after duration \(t\) is observing \(X\) after duration \(ct\). The generator result says that the local rate of change is scaled by the same factor. The jump chain, in contrast, records only successive destinations and therefore does not use the holding times.
Worked Example: A Two-State Chain Runs Three Times Faster
On \(S=\{0,1\}\), let \(Q=\begin{bmatrix}-2&2\\1&-1\end{bmatrix}\). The row sums are \(-2+2=0\) and \(1-1=0\), so this is a valid generator. Find the holding rates and jump probabilities before and after taking \(c=3\). Also compare the probability of being in state \(1\) at time \(t\), starting from state \(0\).
In the original chain, \(q_0=2\) and \(q_1=1\). From state \(0\), the next jump must be to \(1\), with probability \(2/2=1\); from state \(1\), the next jump must be to \(0\), with probability \(1/1=1\). The holding times have distributions \(\operatorname{Exp}(2)\) and \(\operatorname{Exp}(1)\), with means \(1/2\) and \(1\).
The rescaled generator is \(3Q=\begin{bmatrix}-6&6\\3&-3\end{bmatrix}\). Its row sums are \(-6+6=0\) and \(3-3=0\). The new exit rates are \(6\) and \(3\), so the holding-time means are \(1/6\) and \(1/3\). The jump probabilities are \(6/6=1\) and \(3/3=1\), unchanged.
For completeness, in the original two-state chain the probability \(f(t)=p_{01}(t)\) satisfies \(f'(t)=2(1-f(t))-f(t)=2-3f(t)\), with \(f(0)=0\): the rate into state \(1\) is \(2\) when in state \(0\), and the rate out is \(1\) when in state \(1\). Solving gives \(f(t)=\tfrac23(1-e^{-3t})\). The rescaled chain therefore has probability \(p^c_{01}(t)=p_{01}(3t)=\tfrac23(1-e^{-9t})\).
Check: At \(t=0\), both formulas give \(0\). Their derivatives at zero are \(2\) and \(6\), respectively, agreeing with the original and rescaled rates from \(0\) to \(1\). The destination probabilities are still \(1\) in both chains.
Worked Example: A State-Dependent Chain Keeps Its Jump Choices
Consider \(S=\{0,1,2\}\) and \(Q=\begin{bmatrix}-2&2&0\\1&-4&3\\0&4&-4\end{bmatrix}\). Each row sums to zero: \(-2+2+0=0\), \(1-4+3=0\), and \(0+4-4=0\). Take \(c=2\). Find the jump probabilities from each state and the mean times for the first two jumps when the chain starts in state \(0\) and follows \(0\to1\to2\).
The exit rates are \(q_0=2\), \(q_1=4\), and \(q_2=4\). From state \(0\), the next state is \(1\) with probability \(2/2=1\). From state \(1\), the probabilities of moving to \(0\) and \(2\) are \(1/4\) and \(3/4\). From state \(2\), the next state is \(1\) with probability \(4/4=1\). Thus the probability that the first two jumps follow \(0\to1\to2\) is \(1\cdot\tfrac34=\tfrac34\).
The scaled generator is \(2Q=\begin{bmatrix}-4&4&0\\2&-8&6\\0&8&-8\end{bmatrix}\). Its row sums are \(-4+4+0=0\), \(2-8+6=0\), and \(0+8-8=0\). From state \(1\), the scaled probabilities are \(2/8=1/4\) for state \(0\) and \(6/8=3/4\) for state \(2\); from states \(0\) and \(2\) the only destinations remain \(1\). The same two-jump route still has probability \(1\cdot\tfrac34=\tfrac34\).
In the original chain, the mean holding time at \(0\) is \(1/2\), and at \(1\) it is \(1/4\). The mean time to make the first two jumps is therefore \(1/2+1/4=3/4\), conditional on this route (the holding-time means depend on the current state, not on which destination is chosen). In the rescaled chain those means are \(1/4\) and \(1/8\), giving \(1/4+1/8=3/8\).
Check: The rate of each row doubled, while each row's normalized off-diagonal entries stayed the same: at state \(1\), \(2/8=1/4\) and \(6/8=3/4\). The calculated mean time \(3/8\) is exactly half of \(3/4\), as required for \(c=2\).
Invariant Distributions and a Further Consequence
Uniform time rescaling changes the rate at which a chain moves, but not its invariant distributions. Here an invariant distribution means a probability row vector \(\pi\) satisfying \(\pi P(t)=\pi\) for every \(t\ge0\). If \(\pi\) is invariant for \(X\), then \(\pi P_c(t)=\pi P(ct)=\pi\), so it is invariant for \(Y\) as well. The converse holds too: for any \(u\ge0\), set \(t=u/c\); then \(\pi P(u)=\pi P_c(u/c)\). This uses \(c>0\).
Worked Example: A Three-State Cycle Has the Same Invariant Distribution
Let \(S=\{0,1,2\}\) and \(Q=\begin{bmatrix}-1&1&0\\0&-2&2\\3&0&-3\end{bmatrix}\), a cycle with exit rates \(1,2,3\). The row sums are \(-1+1+0=0\), \(0-2+2=0\), and \(3+0-3=0\). Find a probability vector \(\pi\) satisfying \(\pi Q=0\), and check that the same vector satisfies \(\pi(5Q)=0\).
Write \(\pi=(\pi_0,\pi_1,\pi_2)\). The three coordinates of \(\pi Q=0\) require \(-\pi_0+3\pi_2=0\), \(\pi_0-2\pi_1=0\), and \(2\pi_1-3\pi_2=0\). These give \(\pi_0=3\pi_2\) and \(\pi_0=2\pi_1\), so the proportions are \(\pi_0:\pi_1:\pi_2=6:3:2\). Their sum is \(11\), hence \(\pi=(6/11,3/11,2/11)\), whose entries sum to \(1\).
Substitution verifies the equations: \(-6/11+3(2/11)=0\), \(6/11-2(3/11)=0\), and \(2(3/11)-3(2/11)=0\). Therefore \(\pi Q=0\). For the scaled generator \(5Q\), each coordinate of \(\pi(5Q)\) is five times the corresponding zero above, so \(\pi(5Q)=0\) as well. In this finite-state example, \(\pi Q=0\) implies invariance for the semigroup: using the matrix exponential \(e^{tQ}=\sum_{n\ge0}t^nQ^n/n!\), \(\pi Q=0\) gives \(\pi Q^n=0\) for every \(n\ge1\), and thus \(\pi e^{tQ}=\pi\). The same calculation applies to \(5Q\).
Check: The vector is normalized since \(6/11+3/11+2/11=1\), and each coordinate of \(\pi Q\) is zero by the displayed substitutions. Rate scaling changes the speed of circulation but leaves this invariant distribution unchanged.
Common Pitfall: Scaling Only Part of the Generator
A generator's diagonal entries are not independent of its off-diagonal entries: each diagonal is the negative of the total exit rate, so every row must sum to zero. For the first example, a wrong calculation would multiply the off-diagonal rate \(q_{01}=2\) by \(3\) but leave \(q_{00}=-2\). That would produce the row \((-2,6)\), whose sum is \(-2+6=4\), not zero; it is not a valid generator. The correct scaled row is \((-6,6)\). The jump probability from \(0\) is still \(6/6=1\), but its holding rate is \(6\), not \(2\).
Keep the two effects separate: multiplying rates by \(c\) divides each positive-mean holding time by \(c\), while leaving each embedded jump probability unchanged. The assertion concerns uniform scaling by one constant \(c>0\). If rates are changed by different factors in different states, the destination probabilities at a given state may still be unchanged when all rates out of that state share a common factor, but there is no single global deterministic time change in general.
Check Your Understanding
Use the definitions and verify each rate calculation by checking row sums and jump probabilities.
- If a state has exit rate \(5\), what are its holding-time distribution and mean after scaling all rates by \(c=4\)?
- From a state, suppose the rates to two destinations are \(2\) and \(6\). Compute the two jump probabilities before and after scaling by \(c=3\).
- For a generator \(Q\), explain why the transition semigroup of \(Y_t=X_{ct}\) is \(P(ct)\).
- Explain why a probability vector invariant for \(P(t)\) is also invariant for \(P_c(t)\).
- What is wrong with multiplying only the off-diagonal entries of a generator by \(c\) while leaving its diagonal unchanged?