Interpreting Diagonal and Off-Diagonal Rates

What the Entries of a Generator Mean

A transition matrix describes probabilities over a fixed time step. A generator instead describes how those probabilities begin to change over a very short time interval. Its off-diagonal entries distinguish possible destinations from a given state; its diagonal records the total first-order loss of probability from that state. This interpretation makes it possible to read rates directly from a table and to check whether the table is internally consistent.

Throughout, let \((X_t)_{t\ge0}\) be a time-homogeneous continuous-time Markov chain on a finite state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Its transition probabilities are \(p_{ij}(t)=\mathbb{P}(X_t=j\mid X_0=i)\), and its transition matrix is \(P(t)=(p_{ij}(t))\). The generator is \(Q=(q_{ij})\), with \(q_{ij}=\lim_{h\downarrow0}(p_{ij}(h)-\mathbf{1}_{\{i=j\}})/h\). These limits exist for the finite-state chains considered here. In particular, for distinct states \(i\) and \(j\), \(q_{ij}=\lim_{h\downarrow0}p_{ij}(h)/h\): it is the instantaneous rate of moving from \(i\) to \(j\).

Definition: A finite-state generator, or \(Q\)-matrix, has \(q_{ij}\ge0\) for \(i\ne j\) and \(\sum_{j\in S}q_{ij}=0\) for every \(i\). The total exit rate from \(i\) is \(q_i=-q_{ii}=\sum_{j\ne i}q_{ij}\). Thus \(q_i\) is nonnegative, while \(q_{ii}\) is nonpositive.

The entries \(q_{ij}\) for \(i\ne j\) give separate rates for separate destinations. The total exit rate \(q_i\) combines them: it measures the first-order rate of leaving \(i\) for any different state. The diagonal is determined by those outgoing rates, rather than being an additional rate of a transition from \(i\) to itself. All rates have units of inverse time. If time is measured in a different unit, the numerical values of the rates change accordingly.

Why the Diagonal Is the Negative Exit Rate

The row-sum constraint is more than a convenient algebraic check. It expresses conservation of probability at first order. For a short interval of length \(h\), the chance of being in any state must still sum to one. Consequently, the first-order changes in a row of \(P(h)\) must sum to zero: probability gained in other states balances probability lost from the initial state.

Proposition: For a finite-state CTMC with generator \(Q\), started at state \(i\), as \(h\downarrow0\), \(p_{ij}(h)=q_{ij}h+o(h)\) when \(j\ne i\), and \(p_{ii}(h)=1-q_i h+o(h)\). Therefore \(\mathbb{P}_i(X_h\ne i)=q_i h+o(h)\), where \(q_i=\sum_{j\ne i}q_{ij}=-q_{ii}\).

Proof. By the definition of the generator, for \(j\ne i\), \(p_{ij}(h)/h\to q_{ij}\), which gives \(p_{ij}(h)=q_{ij}h+o(h)\). For \(j=i\), the definition gives \((p_{ii}(h)-1)/h\to q_{ii}\), so \(p_{ii}(h)=1+q_{ii}h+o(h)=1-q_i h+o(h)\). Since \(S\) is finite, we can sum the off-diagonal expansions over all \(j\ne i\): \(\mathbb{P}_i(X_h\ne i)=\sum_{j\ne i}p_{ij}(h)=h\sum_{j\ne i}q_{ij}+o(h)=q_i h+o(h)\). The generator row sum gives \(\sum_{j\ne i}q_{ij}=-q_{ii}=q_i\), as claimed. \(\square\)

The proposition concerns probabilities over a small time interval, not a full description of the time spent in a state. In particular, \(q_{ii}\) is not a negative probability and is not the rate of a self-transition. It is the first-order change in the probability of being in the starting state. The separate treatment of holding times comes later.

Worked Rate Tables

Worked Example: Outgoing Rates in a Three-State Chain

Suppose a finite-state CTMC on \(S=\{1,2,3\}\) has generator

Q= \begin{bmatrix} -5&2&3\\ 1&-4&3\\ 0&6&-6 \end{bmatrix}.

Check the rate constraints and interpret the outgoing rates from each state.

Every off-diagonal entry is nonnegative. The row sums are \(-5+2+3=0\), \(1-4+3=0\), and \(0+6-6=0\), so this is a valid rate table. From state \(1\), the rate to state \(2\) is \(2\), and the rate to state \(3\) is \(3\). Its total exit rate is \(q_1=2+3=5=-q_{11}\). From state \(2\), the rates to states \(1\) and \(3\) are \(1\) and \(3\), so \(q_2=1+3=4=-q_{22}\). From state \(3\), the rate to state \(1\) is \(0\) and the rate to state \(2\) is \(6\), so \(q_3=0+6=6=-q_{33}\).

Check: The diagonal entries are respectively \(-5,-4,-6\), the negatives of the outgoing-rate sums \(5,4,6\). Starting from state \(1\), for example, \(p_{12}(h)=2h+o(h)\), \(p_{13}(h)=3h+o(h)\), and \(\mathbb{P}_1(X_h\ne1)=5h+o(h)\). These are small-time approximations, not exact probabilities for arbitrary \(h\).

The same entries also determine instantaneous probability flow when the initial state is random. If the initial distribution is the row vector \(\mu\), then the distribution at time \(h\) is \(\mu P(h)\). In a finite-state chain, \(P(h)=I+hQ+o(h)\) entrywise, so the first-order change is \(\mu Q\). The next result spells out what that product means for an individual state.

Proposition: Let \((X_t)_{t\ge0}\) be a finite-state CTMC with generator \(Q\) and initial distribution \(\mu\), a row vector. For each state \(j\), the instantaneous rate of change in its probability is \(\left.\frac{d}{dt}\mathbb{P}(X_t=j)\right|_{t=0}=\sum_{i\ne j}\mu_iq_{ij}-\mu_jq_j\). The first term is incoming probability flow to \(j\); the second is outgoing probability flow from \(j\).

Proof. The distribution at time \(h\) is \(\mu P(h)\). Since \(S\) is finite and \(P(h)=I+hQ+o(h)\) entrywise, its \(j\)-th coordinate is \(\mu_j+h\sum_i\mu_iq_{ij}+o(h)\). The derivative at zero is therefore \(\sum_i\mu_iq_{ij}\). Separate the term \(i=j\) from the rest to obtain \(\sum_{i\ne j}\mu_iq_{ij}+\mu_jq_{jj}\). By the definition \(q_{jj}=-q_j\), this equals \(\sum_{i\ne j}\mu_iq_{ij}-\mu_jq_j\). Each term has the stated interpretation: \(\mu_i\) weights the rate from \(i\) into \(j\), while \(\mu_jq_j\) is the mass currently in \(j\) times its total exit rate. \(\square\)

Worked Example: Incoming and Outgoing Probability Flow

Use the generator from the previous example, and suppose the initial distribution is \(\mu=(\tfrac12,\tfrac13,\tfrac16)\). Find the instantaneous changes in the probabilities of the three states.

The generator’s off-diagonal entries are nonnegative and its row sums are \(-5+2+3=0\), \(1-4+3=0\), and \(0+6-6=0\), as checked above. The initial probabilities sum to \(\tfrac12+\tfrac13+\tfrac16=1\). By the proposition, the derivative vector is \(\mu Q\). Its first coordinate is \(\tfrac12(-5)+\tfrac13(1)+\tfrac16(0)=-\tfrac52+\tfrac13=-\tfrac{15}{6}+\tfrac{2}{6}=-\tfrac{13}{6}\). Its second coordinate is \(\tfrac12(2)+\tfrac13(-4)+\tfrac16(6)=1-\tfrac43+1=2-\tfrac43=\tfrac23\). Its third coordinate is \(\tfrac12(3)+\tfrac13(3)+\tfrac16(-6)=\tfrac32+1-1=\tfrac32\).

Thus the probability derivatives at time zero are \((-\tfrac{13}{6},\tfrac23,\tfrac32)\). The negative derivative for state \(1\) means its probability initially decreases; the positive derivatives for states \(2\) and \(3\) mean their probabilities initially increase. This does not say those probabilities must keep changing in the same direction later.

Check: The derivatives sum to \(-\tfrac{13}{6}+\tfrac{4}{6}+\tfrac{9}{6}=0\), as required because total probability stays equal to one. For state \(3\), incoming flow is \(\tfrac12(3)+\tfrac13(3)=\tfrac32+1=\tfrac52\); outgoing flow is \(\tfrac16(6)=1\); their difference is \(\tfrac52-1=\tfrac32\), agreeing with the matrix product.

Worked Example: A State with No Outgoing Rates

Consider a CTMC on \(S=\{1,2,3\}\) with generator

Q= \begin{bmatrix} -3&1&2\\ 4&-4&0\\ 0&0&0 \end{bmatrix}.

Interpret state \(3\) and calculate the instantaneous change in its probability when the initial distribution is \((\tfrac12,\tfrac14,\tfrac14)\).

All off-diagonal entries are nonnegative. The row sums are \(-3+1+2=0\), \(4-4+0=0\), and \(0+0+0=0\). The total exit rates are \(q_1=1+2=3\), \(q_2=4+0=4\), and \(q_3=0+0=0\). Thus the table assigns no outgoing rate from state \(3\). In particular, \(q_{31}=q_{32}=0\), and its diagonal must be \(q_{33}=0\).

For the specified initial distribution, the derivative vector is \((\tfrac12,\tfrac14,\tfrac14)Q\). The first coordinate is \(\tfrac12(-3)+\tfrac14(4)+\tfrac14(0)=-\tfrac32+1=-\tfrac12\). The second is \(\tfrac12(1)+\tfrac14(-4)+\tfrac14(0)=\tfrac12-1=-\tfrac12\). The third is \(\tfrac12(2)+\tfrac14(0)+\tfrac14(0)=1\). Therefore, probability initially accumulates in state \(3\) at rate \(1\), entirely from the rate \(2\) out of state \(1\) weighted by its initial probability \(\tfrac12\).

Check: The derivatives sum to \(-\tfrac12-\tfrac12+1=0\). The incoming flow into state \(3\) is \(\tfrac12(2)+\tfrac14(0)=1\), and its outgoing flow is \(\tfrac14(0)=0\), so the net rate is \(1-0=1\), as calculated.

Checking a Rate Interpretation

A proposed \(Q\)-matrix must pass two checks before its entries can be interpreted as rates: every off-diagonal entry must be nonnegative, and every row must sum to zero. Once these conditions hold, the diagonal is not a free parameter. For each row, add the off-diagonal entries and negate the result. This gives the diagonal entry and the total exit rate in one step.

Worked Example: Correcting an Inconsistent Diagonal

Suppose a model specifies a rate of \(3\) from state \(1\) to state \(2\), and a rate of \(1\) from state \(1\) to state \(3\). A proposed row for state \(1\) is \((-3,3,1)\). Is it consistent, and what must the diagonal be?

The off-diagonal entries are nonnegative, but the proposed row sum is \(-3+3+1=1\), not zero. The row therefore cannot be a generator row. The total exit rate specified by the two outgoing rates is \(3+1=4\), so the diagonal must be \(-4\). The corrected row is \((-4,3,1)\), whose sum is \(-4+3+1=0\). For a complete example, the matrix \(\begin{bmatrix}-4&3&1\\2&-5&3\\1&0&-1\end{bmatrix}\) has nonnegative off-diagonal entries and row sums \(-4+3+1=0\), \(2-5+3=0\), and \(1+0-1=0\). Its total exit rates are \(4,5,1\).

Check: In the corrected first row, the total exit rate is \(3+1=4=-(-4)\). The original diagonal \(-3\) would incorrectly report an exit rate of \(3\), even though the listed outgoing rates add to \(4\).

Common mistake. It is tempting to read \(q_{ii}\) as a rate of staying in state \(i\), or to treat it as a separate rate alongside the off-diagonal entries. That interpretation fails: in the inconsistent row above, adding the purported “stay rate” to the outgoing rates gives no probability-conserving rate table. The diagonal is instead fixed by conservation, \(q_{ii}=-\sum_{j\ne i}q_{ij}\). A second related error is to call \(-q_{ii}\) the rate to one particular destination. It is the total rate to all different states; the individual destination rates are the separate entries \(q_{ij}\).

These interpretations are useful whenever a model is specified by a table of event rates. They let you detect inconsistent rows, identify which transitions contribute to an exit rate, and determine whether probability in a state is initially increasing or decreasing under a given distribution. They also keep distinct two quantities that are easy to confuse: a transition-specific rate \(q_{ij}\), and the total exit rate \(q_i\).

Check Your Understanding

For each question, use the off-diagonal entries as destination-specific rates and check row sums before interpreting a proposed generator.

A row has rates \(2\) to one state and \(5\) to another. What diagonal entry and total exit rate are required?
Why is \(q_{ii}\) not a rate of a self-transition?
For the matrix in the worked example with a state having no outgoing rates, what is the net instantaneous probability flow into state \(3\) if the initial distribution is \((\tfrac12,\tfrac14,\tfrac14)\)?
For a finite-state generator and initial distribution \(\mu\), explain the incoming and outgoing terms in the derivative of the probability of state \(j\).
A proposed row is \((-6,2,3)\). Does it satisfy the generator row-sum constraint? If not, what diagonal is required by the two off-diagonal rates?