Competing Exponential Clocks and Destination Probabilities

One Waiting Time, Several Possible Events

At a state of a continuous-time Markov chain, several different events may be able to occur next. For example, a system may fail, be repaired, or receive a new arrival. If each mechanism has its own independent exponential clock, the next change is determined by whichever clock rings first. Two questions then matter: how long does the system wait, and which event occurs?

We work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). An exponential clock with rate \(\lambda>0\) is a random time \(E\) with survival function \(\mathbb{P}(E>t)=e^{-\lambda t}\) for \(t\ge0\), and density \(\lambda e^{-\lambda t}\) for \(t>0\). The rate has units of inverse time; it is not itself a probability.

For a finite family, the survival calculation follows directly from independence. With countably many clocks, one must take care: the infimum of values all greater than \(t\) could equal \(t\). The finite-total-rate assumption rules out that boundary issue almost surely, as the proof shows.

Proof. First suppose \(J\) is countable, and list its elements as \(j_1,j_2,\ldots\) (the finite case is included by stopping the list). For fixed \(t\ge0\), let \(A_t=\{E_j>t\text{ for every }j\in J\}\). The events requiring the first \(m\) clocks to exceed \(t\) decrease to \(A_t\). Independence and continuity of probability for decreasing events give

It remains to relate \(A_t\) to \(\{T>t\}\). The event \(\{T>t\}\) implies \(A_t\), while the only possible discrepancy is \(A_t\cap\{T=t\}\). Define \(N_{t+1}=\sum_{j\in J}\mathbf{1}_{\{E_j\le t+1\}}\), the number of clocks no later than \(t+1\). By monotone convergence and \(1-e^{-x}\le x\) for \(x\ge0\),

A nonnegative integer-valued random variable with finite expectation is finite almost surely, so \(N_{t+1}<\infty\) almost surely. If \(A_t\) holds and \(T=t\), then for every positive integer \(m\) there must be a clock with value in \((t,t+1/m)\). Otherwise all clock values would be at least \(t+1/m\), contradicting \(T=t\). These clocks must include infinitely many distinct indices: any finite collection of values strictly greater than \(t\) has a minimum strictly greater than \(t\). Thus \(N_{t+1}=\infty\), an event of probability zero. Therefore \(\mathbb{P}(T>t)=\mathbb{P}(A_t)=e^{-\Lambda t}\) for each \(t\ge0\), which is the survival function of an exponential variable with rate \(\Lambda\).

For a fixed \(j\), the probability that clock \(j\) is strictly earlier than every other clock is obtained by integrating over its ringing time \(s\). Independence and the same survival calculation applied to the other clocks give

For a one-clock family the survival factor for the empty set of other clocks is \(1\), and this same calculation gives probability \(1\). For any finite or countable family, the events that different clocks are strictly first are disjoint, and their probabilities sum to \(\sum_j\lambda_j/\Lambda=1\). Hence one of them occurs almost surely, proving existence and uniqueness of the first clock. Summing these disjoint probabilities over \(j\in C\) proves the subset formula. \(\square\)

Using the Formulas

The total rate describes the waiting time, while the relative rates describe the event type. For a finite family, this also has a useful density interpretation: the probability density that clock \(j\) rings first at time \(t>0\) is \(\lambda_j e^{-\Lambda t}\). Indeed, clock \(j\) contributes density \(\lambda_j e^{-\lambda_j t}\), and every other clock must survive to time \(t\), contributing the product \(e^{-(\Lambda-\lambda_j)t}\). Integrating that density over all \(t>0\) gives \(\lambda_j/\Lambda\).

Destination Probabilities at a CTMC State

Suppose a continuous-time Markov chain \((X_t)_{t\ge0}\) has countable state space \(S\), and consider its event mechanisms while it is at a state \(i\). If a mechanism moves the chain from \(i\) to \(j\ne i\) at rate \(q_{ij}>0\), represent that mechanism by an independent exponential clock of rate \(q_{ij}\). The total rate of these active clocks is \(q_i=\sum_{j\ne i}q_{ij}\), provided \(0<q_i<\infty\). The diagonal generator entry is \(q_{ii}=-q_i\), so the generator row sums to zero. The competing-clock result says that the time to the first event is exponential with rate \(q_i\), and the probability that its destination is \(j\) is \(q_{ij}/q_i\).

If \(q_i=0\), there is no active outgoing clock in this description: the state has no event of these types, and a destination probability is not defined. If the sum of outgoing rates is infinite, the finite-total-rate theorem does not apply. These conditions matter; a ratio of rates is a probability only when the total rate is finite and positive.

Why the Rate Ratios Matter

When event mechanisms compete, doubling one rate while leaving the others fixed increases that event’s share of the first-event probability, but it also increases the total rate and therefore shortens the mean waiting time. These are distinct effects: the rate sum controls the clock for “something happens,” and the normalized individual rates control what happens.

Common mistake. Rates are not probabilities. For rates \(2\) and \(3\), an incorrect calculation might assign first-event probabilities \(2\) and \(3\). These values exceed \(1\) and do not sum to \(1\). The correction is to divide each rate by the total rate \(2+3=5\), giving probabilities \(2/5\) and \(3/5\). More generally, the rate for event \(j\) must be normalized by the sum of all active rates.

Independence is also essential to the product survival calculation. Without it, the marginal exponential distributions alone do not determine the distribution of the minimum or the probability of which clock is first. For instance, if two rate-\(1\) clocks are equal to the same exponential random time, their minimum is exponential with rate \(1\), not \(2\); neither clock is uniquely first. The theorem applies to independent clocks, not merely clocks with exponential marginal distributions.