Building Rates From Competing Event Mechanisms

From Event Mechanisms to Transition Rates

A continuous-time Markov chain can be described by the rates at which it leaves a state and the probabilities of its possible destinations. In applications, these quantities often arise from several independent mechanisms: different failure modes, competing alarms, or distinct causes of a change. The key accounting rule is to add rates across mechanisms, but to allocate each mechanism’s rate only among the destinations it can produce.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $(X_t)_{t\ge0}$ be a continuous-time Markov chain on a finite state space $S$, with natural filtration $\mathcal{F}_t=\sigma(X_s:0\le s\le t)$. For a specified initial state $i$, write $\mathbb{P}_i$ for its law. A generator $Q=(q_{ij})$ has nonnegative off-diagonal entries and rows summing to zero; its exit rate at $i$ is $q_i=-q_{ii}=\sum_{j\ne i}q_{ij}$. Thus $q_{ij}$ is a rate, not a one-step transition probability.

Definition: At state $i$, suppose there are finitely many independent event mechanisms indexed by $r$. Mechanism $r$ has an exponential clock of rate $\lambda_{ir}\ge0$. When it rings, it moves the chain to $j\ne i$ with probability $a_{irj}$, where $a_{irj}\ge0$ and $\sum_{j\ne i}a_{irj}=1$ for each mechanism with $\lambda_{ir}>0$. Clocks are restarted on entering a state. The rate contribution of mechanism $r$ to destination $j$ is $\lambda_{ir}a_{irj}$.

A mechanism with rate zero can be omitted. The destination probabilities $a_{irj}$ may be deterministic—for example, a mechanism may always send the chain to one particular state—or randomized. If all active mechanisms at $i$ have zero rates, the chain has no event there and $i$ is absorbing.

Theorem (Rate aggregation): Under the definition above, the total rate of leaving $i$ is $\Lambda_i=\sum_r\lambda_{ir}$. If $\Lambda_i>0$, the probability that the next state is $j\ne i$ is $$ \frac{\sum_r\lambda_{ir}a_{irj}}{\Lambda_i}. $$ Consequently, the generator entries are $q_{ij}=\sum_r\lambda_{ir}a_{irj}$ for $j\ne i$, and $q_{ii}=-\Lambda_i$. If $\Lambda_i=0$, all off-diagonal entries in row $i$ are zero and $q_{ii}=0$.

Proof. Fix $i$, and first suppose $\Lambda_i>0$. Let $T_r$ be the clock time for mechanism $r$, and let $T=\min_r T_r$. Independence and the exponential survival probabilities give, for $t\ge0$, $\mathbb{P}_i(T>t)=\prod_r e^{-\lambda_{ir}t}=e^{-\Lambda_i t}$. Thus the first event time is exponential with rate $\Lambda_i$. For a mechanism with positive rate, the probability that it supplies the first event is $$ \mathbb{P}_i(T_r=T)=\int_0^\infty \lambda_{ir}e^{-\Lambda_i t}\,dt=\frac{\lambda_{ir}}{\Lambda_i}. $$ The integral follows by evaluating the antiderivative $-e^{-\Lambda_i t}/\Lambda_i$ at zero and infinity. Ties have probability zero because independent exponential variables have continuous distributions. Given that mechanism $r$ rings first, it chooses $j$ with probability $a_{irj}$. Summing over the mutually exclusive mechanisms, the probability of destination $j$ is $\sum_r(\lambda_{ir}/\Lambda_i)a_{irj}$.

To identify the generator rate, the probability that mechanism $r$ is first by time $h$ and chooses $j$ is $(\lambda_{ir}/\Lambda_i)(1-e^{-\Lambda_i h})a_{irj}=\lambda_{ir}a_{irj}h+o(h)$ as $h\downarrow0$. The chance of a second transition during this same interval is $O(h^2)$: after the first event the state is in the finite set $S$, whose exit rates have a finite maximum, so the probability of another event within the remaining time is bounded by a constant times $h$, while the probability of a first event by time $h$ is also bounded by a constant times $h$. Therefore $q_{ij}=\lim_{h\downarrow0}\mathbb{P}_i(X_h=j)/h=\sum_r\lambda_{ir}a_{irj}$. Summing these off-diagonal rates over $j\ne i$ gives $\sum_r\lambda_{ir}\sum_{j\ne i}a_{irj}=\Lambda_i$, so the row-sum-zero condition requires $q_{ii}=-\Lambda_i$. If $\Lambda_i=0$, every mechanism has rate zero, and the row is zero. $\square$

This result also gives a useful interpretation when multiple causes lead to the same observed destination. Their rates add in $q_{ij}$. The total rate determines how quickly an event occurs; the share of that rate assigned to a particular destination determines the destination probability. These are related quantities, but they are not interchangeable.

Worked Example: Several Mechanisms in a Three-State Chain

Let $S=\{A,B,C\}$. At $A$, two mechanisms move to $B$, at rates $\tfrac23$ and $\tfrac13$, while a third moves to $C$ at rate $\tfrac12$. At $B$, one mechanism moves to $A$ at rate $\tfrac12$, and two mechanisms move to $C$ at rates $\tfrac14$ each. At $C$, the mechanisms move to $A$ and $B$ at rates $\tfrac13$ and $\tfrac23$, respectively. Each mechanism has a deterministic destination.

At $A$, the rate to $B$ is $\tfrac23+\tfrac13=1$, and the rate to $C$ is $\tfrac12$. The total exit rate is $1+\tfrac12=\tfrac32$, so $q_{AA}=-\tfrac32$. At $B$, the rate to $A$ is $\tfrac12$, and the combined rate to $C$ is $\tfrac14+\tfrac14=\tfrac12$, giving exit rate $1$. At $C$, the total exit rate is $\tfrac13+\tfrac23=1$. The generator is therefore

Q= \begin{bmatrix} -\tfrac32&1&\tfrac12\\ \tfrac12&-1&\tfrac12\\ \tfrac13&\tfrac23&-1 \end{bmatrix}.

Check: The row sums are $-\tfrac32+1+\tfrac12=0$, $\tfrac12-1+\tfrac12=0$, and $\tfrac13+\tfrac23-1=0$. At $A$, the probability that the next state is $B$ is $1/(\tfrac32)=\tfrac23$, and the probability it is $C$ is $(\tfrac12)/(\tfrac32)=\tfrac13$; these probabilities sum to $1$.

Attributing a Destination to Its Cause

Sometimes the state change is observed but the mechanism that caused it is not. The rate decomposition still gives the conditional probability of each cause, provided the observed destination has positive total rate. For a destination $j\ne i$ with $q_{ij}>0$, that probability is $\lambda_{ir}a_{irj}/q_{ij}$. This is a conditional attribution among events that led to $j$, not the unconditional probability that mechanism $r$ wins the race.

Proposition (Cause attribution): If $\Lambda_i>0$ and $q_{ij}=\sum_r\lambda_{ir}a_{irj}>0$, then the probability that mechanism $r$ caused the next transition, conditional on its destination being $j$, is $$ \mathbb{P}_i(\text{mechanism }r\mid\text{next state }j) =\frac{\lambda_{ir}a_{irj}}{q_{ij}}. $$ The formula is not defined when $q_{ij}=0$, because that destination has probability zero.

Proof. The probability that mechanism $r$ is first and sends the chain to $j$ is $(\lambda_{ir}/\Lambda_i)a_{irj}$, by the competing-clock calculation. The probability that the next state is $j$ is $q_{ij}/\Lambda_i$. Since $q_{ij}>0$, division is valid, and the conditional probability is $$ \frac{(\lambda_{ir}/\Lambda_i)a_{irj}}{q_{ij}/\Lambda_i} =\frac{\lambda_{ir}a_{irj}}{q_{ij}}. $$ Summing this expression over mechanisms gives $\sum_r\lambda_{ir}a_{irj}/q_{ij}=1$, as required for the possible causes of destination $j$. $\square$

Worked Example: Two Causes of the Same Transition

Use the states $\{A,B,C\}$. At $A$, mechanisms $R_1$ and $R_2$ each lead to $B$, at rates $2$ and $1$, respectively; mechanism $R_3$ leads to $C$ at rate $3$. At $B$, the rates to $A$ and $C$ are $\tfrac12$ and $\tfrac32$. At $C$, the rates to $A$ and $B$ are $1$ and $\tfrac12$.

At $A$, the total rate to $B$ is $2+1=3$, and the rate to $C$ is $3$. The exit rate is $3+3=6$. The other two exit rates are $\tfrac12+\tfrac32=2$ at $B$ and $1+\tfrac12=\tfrac32$ at $C$. Thus

Q= \begin{bmatrix} -6&3&3\\ \tfrac12&-2&\tfrac32\\ 1&\tfrac12&-\tfrac32 \end{bmatrix}.

The first event from $A$ is caused by $R_1$ with probability $2/6=\tfrac13$, by $R_2$ with probability $1/6$, and by $R_3$ with probability $3/6=\tfrac12$. Hence the probability that the next state is $B$ is $\tfrac13+\tfrac16=\tfrac12$. Conditional on that destination, the cause probabilities are $2/3$ for $R_1$ and $1/3$ for $R_2$: equivalently, the rate contributions $2$ and $1$ are divided by the total rate $3$ to $B$.

Check: The generator row sums are $-6+3+3=0$, $\tfrac12-2+\tfrac32=0$, and $1+\tfrac12-\tfrac32=0$. The destination probabilities from $A$ are $3/6=\tfrac12$ for $B$ and $3/6=\tfrac12$ for $C$, summing to $1$.

Mechanisms with Multiple Possible Destinations

A single mechanism need not have a deterministic destination. Its rate is split according to its destination probabilities before contributions from different mechanisms are combined. For example, a mechanism with rate $2$ that chooses $B$ with probability $\tfrac14$ contributes rate $2\cdot\tfrac14=\tfrac12$ to $B$, not rate $2$. This distinction is important whenever an event has several possible outcomes.

Worked Example: Allocating a Randomized Mechanism

Let $S=\{A,B,C\}$. At $A$, mechanism $M$ has rate $2$ and sends the chain to $B$ with probability $\tfrac14$ or to $C$ with probability $\tfrac34$. A second mechanism $N$, of rate $1$, always sends the chain to $C$. At $B$, the rates to $A$ and $C$ are $\tfrac12$ each. At $C$, the rates to $A$ and $B$ are $\tfrac13$ and $\tfrac23$.

Mechanism $M$ contributes $2\cdot\tfrac14=\tfrac12$ to the rate from $A$ to $B$, and $2\cdot\tfrac34=\tfrac32$ to the rate from $A$ to $C$. Mechanism $N$ contributes another $1$ to the rate to $C$. Thus $q_{AB}=\tfrac12$, $q_{AC}=\tfrac32+1=\tfrac52$, and the exit rate at $A$ is $2+1=3$. The full generator is

Q= \begin{bmatrix} -3&\tfrac12&\tfrac52\\ \tfrac12&-1&\tfrac12\\ \tfrac13&\tfrac23&-1 \end{bmatrix}.

The next state from $A$ is $B$ with probability $(\tfrac12)/3=\tfrac16$, and $C$ with probability $(\tfrac52)/3=\tfrac56$. If the next state is $C$, the conditional probability that $M$ caused the transition is $(\tfrac32)/(\tfrac52)=\tfrac35$; the conditional probability that $N$ caused it is $1/(\tfrac52)=\tfrac25$.

Check: The row sums are $-3+\tfrac12+\tfrac52=0$, $\tfrac12-1+\tfrac12=0$, and $\tfrac13+\tfrac23-1=0$. The two destination probabilities from $A$ sum to $\tfrac16+\tfrac56=1$. Conditional cause probabilities given destination $C$ also sum to $\tfrac35+\tfrac25=1$.

A Common Rate-Accounting Error

A frequent mistake is to add all mechanism rates to obtain the exit rate and then assign that entire rate to every possible destination. In the second example, the total rate at $A$ is $6$, but the rates to $B$ and $C$ are each $3$. Writing $q_{AB}=6$ and $q_{AC}=6$ would give $q_{AA}=-12$, because the row-sum condition forces the diagonal to be $-12$. This contradicts the actual total event rate $2+1+3=6$, and it would incorrectly make each destination probability $6/12=\tfrac12$ for the wrong reason if the diagonal were also changed.

The correction is to group contributions by destination: $q_{AB}=2+1=3$ and $q_{AC}=3$, then set $q_{AA}=-(3+3)=-6$. Each row of a proposed generator should be checked against both facts: its off-diagonal entries are the sums of the relevant mechanism contributions, and its entries sum to zero.

The mechanism description can contain more detail than the state process reveals. If two causes lead to the same destination, their separate rates matter for causal attribution, but the state-process generator contains their sum. Conversely, knowing only the generator does not generally identify how its rates were divided among mechanisms.

Check Your Understanding

For each question, distinguish mechanism rates, destination rates, and destination probabilities.

At a state $i$, two mechanisms lead to $j$ at rates $\tfrac23$ and $\tfrac13$, and a third leads to $k$ at rate $1$. Find the exit rate and the probabilities of the two destinations.
A mechanism of rate $4$ chooses between two destinations with probabilities $\tfrac14$ and $\tfrac34$. What rate does it contribute to each destination?
If two mechanisms contribute rates $2$ and $3$ to the same destination, what is the conditional probability that the first mechanism caused the transition, given that this destination was reached?
Why is the cause-attribution formula undefined for a destination whose total transition rate is zero?
Explain why adding the total exit rate separately to every destination generally produces an incorrect generator.

Building Rates From Competing Event Mechanisms

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