Generator Row Sums and Probability Conservation

Why Generator Rows Must Sum to Zero

A generator records instantaneous rates of movement between states. Its entries are not probabilities: an off-diagonal entry is an intensity, measured in inverse time, while the diagonal entry balances all possible exits from that state. This balance is required for probabilities to remain conserved as a process evolves.

The key check is simple but essential: each row of a valid generator sums to zero. This is a row-by-row condition because a row describes the possible destinations from one particular current state. Checking it catches many sign, omission, and arithmetic errors before a generator is used in a model.

Definition: Consider a finite-state continuous-time Markov chain with states indexed by $1,\ldots,n$. At time $t$, for distinct states $i$ and $j$, let $q_{ij}(t)$ denote the instantaneous intensity of a transition from state $i$ to state $j$. The generator $Q(t)=(q_{ij}(t))$ has diagonal entries $$ q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t). $$ Thus $q_{ii}(t)\le 0$, all off-diagonal entries are nonnegative, and every row sums to zero. In a time-homogeneous model, the entries do not depend on $t$, and the generator is written $Q$.

The interpretation of the diagonal is particularly useful. The total intensity of leaving state $i$ is the sum of the intensities of all transitions from $i$ to a different state. Therefore, $$ \text{total exit intensity from state }i=\sum_{j\ne i}q_{ij}(t)=-q_{ii}(t). $$ The diagonal is not an additional transition intensity to another state. It is a negative balancing entry.

A finite-state generator must also have nonnegative off-diagonal entries: an intensity cannot be negative. A state with no possible exits has all off-diagonal entries in its row equal to zero, so its diagonal entry is zero too. Such a state is absorbing under the model.

Deriving the Row-Sum Condition

The row-sum condition follows from the fact that, starting in any state, the process must be in some state a short time later. The infinitesimal transition probabilities express this fact in terms of generator entries.

Infinitesimal transition probabilities: For a finite-state model with finite transition intensities at time $t$, over a small interval of length $h>0$, $$ P(X_{t+h}=j\mid X_t=i)=q_{ij}(t)h+o(h),\qquad j\ne i, $$ and $$ P(X_{t+h}=i\mid X_t=i)=1+q_{ii}(t)h+o(h). $$ Here $o(h)$ denotes a term that, divided by $h$, tends to zero as $h$ tends to zero. The formulas describe first-order behaviour, not exact probabilities for a finite interval.

Sum over all destinations.
Conditional on being in state $i$ at time $t$, the probabilities of being in the $n$ possible states at time $t+h$ must sum to one.

Insert the infinitesimal expressions.
The probability of remaining in $i$ is $1+q_{ii}(t)h+o(h)$. The probabilities of moving to other states sum to $h\sum_{j\ne i}q_{ij}(t)+o(h)$.

Use probability conservation.
The sum is $1+h\big(q_{ii}(t)+\sum_{j\ne i}q_{ij}(t)\big)+o(h)$, and it must equal one. Divide the difference from one by $h$ and let $h$ tend to zero. The coefficient of $h$ must be zero.

q_{ii}(t)+\sum_{j\ne i}q_{ij}(t)=0 \quad\Longrightarrow\quad q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t).

This argument also explains the probability interpretation. To first order, probability leaves state $i$ at total rate $\sum_{j\ne i}q_{ij}(t)$; the probability of remaining in $i$ decreases at that same rate. No probability is created or lost. For an interval of positive length, the infinitesimal expressions alone do not give exact transition probabilities, but the exact transition probabilities still sum to one.

Constructing and Checking a Generator

A reliable check is to work through each row separately. Add its off-diagonal entries to obtain the total exit intensity, put the negative of that amount on the diagonal, and verify that the row sums to zero. Then confirm that every off-diagonal entry is nonnegative and that all entries use the same time unit.

Worked Example: Building a Three-State Generator

Consider a hypothetical policyholder model with states Active, Disabled, and Dead. Assume a time-homogeneous continuous-time Markov chain, with all intensities measured per year. From Active, the transition intensity to Disabled is $0.04$ and to Dead is $0.01$. From Disabled, the intensity to Active is $0.20$ and to Dead is $0.05$. Dead is absorbing. Construct the generator and check its rows.

Method. For each state, add the intensities of all exits. The diagonal is the negative of that total.

For Active, the total exit intensity is $0.04+0.01=0.05$, so the diagonal entry is $-0.05$. Its row, in the stated state order, is $(-0.05,0.04,0.01)$. The row sum is $-0.05+0.04+0.01=0$.

For Disabled, the total exit intensity is $0.20+0.05=0.25$, so the diagonal entry is $-0.25$. Its row is $(0.20,-0.25,0.05)$. The row sum is $0.20-0.25+0.05=0$.

Dead has no exits. Its off-diagonal entries and its diagonal entry are all zero, giving row sum zero. Therefore, in the order Active, Disabled, Dead,

Q= \begin{pmatrix} -0.05&0.04&0.01\\ 0.20&-0.25&0.05\\ 0&0&0 \end{pmatrix}.

Check. All off-diagonal entries are nonnegative, each diagonal is the negative total exit intensity, and all three row sums are zero. The common mistake avoided here is assigning the absorbing state a negative diagonal despite there being no exits.

Worked Example: Finding and Correcting Two Row Errors

A hypothetical three-state model has the proposed generator, with entries in inverse years,

\widetilde Q= \begin{pmatrix} -0.12&0.08&0.05\\ 0.03&-0.10&0.06\\ 0&0.02&-0.02 \end{pmatrix}.

Determine whether it is valid and correct any errors, assuming the stated off-diagonal entries are the intended transition intensities.

Method. Compute each row sum and compare each diagonal with the negative sum of that row's off-diagonal entries.

Row 1 sums to $-0.12+0.08+0.05=0.01$, not zero. Its intended total exit intensity is $0.08+0.05=0.13$, so its diagonal should be $-0.13$.

Row 2 sums to $0.03-0.10+0.06=-0.01$, not zero. Its total exit intensity is $0.03+0.06=0.09$, so its diagonal should be $-0.09$.

Row 3 sums to $0+0.02-0.02=0$. Its only exit intensity is $0.02$, and its diagonal is correctly $-0.02$. Thus the corrected generator is

Q= \begin{pmatrix} -0.13&0.08&0.05\\ 0.03&-0.09&0.06\\ 0&0.02&-0.02 \end{pmatrix}.

The corrected row sums are $-0.13+0.08+0.05=0$, $0.03-0.09+0.06=0$, and $0+0.02-0.02=0$. All off-diagonal entries are nonnegative. The trap is to overlook that a diagonal can be negative and still be wrong; its magnitude must match the sum of the exits from that same state.

Time-Varying Intensities and Local Checks

When intensities vary with time, the same balance applies at each time $t$. The generator $Q(t)$ must have zero row sums at every time at which the rates are defined. A row that balances at one selected time is not enough to validate a proposed time-varying model.

Worked Example: Checking a Time-Varying Generator at a Specified Time

In a hypothetical model with states U, V, and W, suppose the intensities, in inverse years, are $q_{UV}(t)=0.02+0.01t$, $q_{UW}(t)=0.03$, $q_{VU}(t)=0.10$, and $q_{VW}(t)=0.04+0.01t$, where $t$ is measured in years. State W is absorbing. Find the generator at $t=2$ and check it.

Method. Evaluate each exit intensity at $t=2$, then set each diagonal to the negative sum of the exits from that state.

From U, $q_{UV}(2)=0.02+0.01(2)=0.04$, while $q_{UW}(2)=0.03$. The total exit intensity is $0.04+0.03=0.07$, so $q_{UU}(2)=-0.07$.

From V, $q_{VU}(2)=0.10$ and $q_{VW}(2)=0.04+0.01(2)=0.06$. The total is $0.10+0.06=0.16$, so $q_{VV}(2)=-0.16$. Since W is absorbing, its row is zero. Therefore,

Q(2)= \begin{pmatrix} -0.07&0.04&0.03\\ 0.10&-0.16&0.06\\ 0&0&0 \end{pmatrix}.

The row sums are $-0.07+0.04+0.03=0$, $0.10-0.16+0.06=0$, and $0$. All off-diagonal entries are nonnegative. The common mistake avoided is treating the time-varying rates as fixed at their value at time zero; here the diagonal entries are checked using rates evaluated at the same time $t=2$.

Probability Conservation in Use

The zero row sums are more than a convenient matrix check. They ensure that the local changes in probability balance across the possible destinations. If a process is currently in state $i$, the first-order probability decrease for staying in $i$ is $-q_{ii}(t)h$, while the first-order probability increases for moving to other states sum to $h\sum_{j\ne i}q_{ij}(t)$. Since these quantities are equal, their net contribution to total probability is zero.

This is the generator-level version of conservation: transition probabilities must account for all possible destinations, and their total remains one. It is useful to distinguish this principle from a numerical check of a transition probability matrix over a finite interval. The generator has rate entries and is checked by zero row sums; a transition probability matrix has probability entries and is checked by row sums of one.

Quantity	Off-diagonal entries	Diagonal entry	Row sum
Generator	Transition intensities, nonnegative	Negative total exit intensity	Zero
Transition probability matrix	Probabilities of ending in other states	Probability of ending in the same state	One

For a proposed generator, use this checklist:

Are all entries measured in a consistent time unit?
Are all off-diagonal entries nonnegative?
Does each diagonal equal the negative sum of the other entries in its row?
Does every row sum to zero, including rows for absorbing states?
If rates vary with time, do these checks hold at each relevant time?

A zero row sum alone does not validate a generator if an off-diagonal entry is negative. Conversely, nonnegative off-diagonal entries do not make a matrix valid if a diagonal fails to balance its exits. Both checks are required.

Key takeaway: In a valid finite-state continuous-time generator, each off-diagonal entry is a nonnegative transition intensity, and each diagonal entry is the negative sum of the other entries in its row. The resulting zero row sum expresses local probability conservation.

Check Your Understanding

For each problem, use row-wise exit totals and probability conservation to check your result.

A hypothetical model has states R, S, and D. The rates per year are R to S $0.07$, R to D $0.02$, S to R $0.11$, and S to D $0.03$, with D absorbing. Write the generator in the order R, S, D. (Answer: $\begin{pmatrix}-0.09&0.07&0.02\\0.11&-0.14&0.03\\0&0&0\end{pmatrix}$)
A proposed row of a generator is $(0.10,-0.19,0.08)$. Is it valid? If its off-diagonal entries are intended, give the corrected row. (Answer: No; its sum is $-0.01$, and the corrected row is $(0.10,-0.18,0.08)$.)
A state has three possible exits with intensities $0.025$, $0.015$, and $0.010$ per year. Find its diagonal entry and the first-order probability of remaining in the state over an interval of length $h$ years. (Answer: Diagonal $=-0.050$ per year; staying probability $=1-0.050h+o(h)$.)
At time $t=3$ years, a time-varying model has rates A to B $0.01+0.02t$, A to C $0.04$, and B to A $0.05$, B to C $0.03+0.01t$; C is absorbing. Give the generator at $t=3$. (Answer: $\begin{pmatrix}-0.11&0.07&0.04\\0.05&-0.11&0.06\\0&0&0\end{pmatrix}$)
A process has current-state probabilities $0.20, 0.50, 0.30$ in states A, B, C. From A, the total exit intensity is $0.12$; from B it is $0.04$; C is absorbing. What is the first-order total probability change over $h$ years, accounting for probability leaving and entering states? (Answer: Zero; the first-order outflow $0.20(0.12)h+0.50(0.04)h=0.044h$ is balanced by an equal inflow.)

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