Why Intensity-Matrix Rows Sum to Zero

Introduction

An intensity matrix records the instantaneous rates of movement between states. Its off-diagonal entries describe transitions from one state to another; its diagonal entry has a different role. It balances the rates of leaving that state so the entries in each row add to zero.

This identity follows from a basic probability fact: conditional on starting in a given state, the process must be in some state a short time later. The diagonal entry accounts for remaining in the starting state, while the off-diagonal entries account for moving elsewhere. The identity is useful both for understanding the matrix and for checking calculations.

Definition: For a finite-state continuous-time Markov model, let $q_{ij}$ be the intensity of a transition from state $i$ to a different state $j$, where $i\ne j$. With the row-origin convention, the intensity matrix is $Q=(q_{ij})$. Its diagonal entry $q_{ii}$ is defined by the instantaneous change in the probability of remaining in state $i$. For every state $i$, $$ q_{ii}=-\sum_{j\ne i}q_{ij}, \qquad \sum_j q_{ij}=0. $$

Why Each Row Must Sum to Zero

Fix a state $i$, and suppose the process is in state $i$ at time $t$. Over a short interval of length $h$, the conditional probabilities of being in each possible state at time $t+h$ must sum to one. For $j\ne i$, the definition of transition intensity gives $P(X(t+h)=j\mid X(t)=i)=q_{ij}h+o(h)$. Here $o(h)$ denotes a term that, divided by $h$, tends to zero as $h$ tends to zero.

In a finite-state model with the usual continuous-time transition assumptions, the probability of two or more jumps during a sufficiently short interval is of smaller order than $h$. Thus, to first order, the probability of still being in state $i$ is one minus the probabilities of moving to each other state. This gives the diagonal entry.

\begin{aligned} P(X(t+h)=i\mid X(t)=i) &=1-\sum_{j\ne i}P(X(t+h)=j\mid X(t)=i)\\ &=1-\left(\sum_{j\ne i}q_{ij}\right)h+o(h),\\[4pt] q_{ii} &=\lim_{h\to0}\frac{P(X(t+h)=i\mid X(t)=i)-1}{h}\\ &=-\sum_{j\ne i}q_{ij}. \end{aligned}

Adding the diagonal entry to the off-diagonal entries now gives $q_{ii}+\sum_{j\ne i}q_{ij}=0$. The argument applies to each starting state, so it establishes the row-sum identity for every row. It depends on the entries in a row sharing the same starting state: row $i$ describes the possible destinations conditional on starting in state $i$.

Interpretation: The diagonal entry is not an intensity of a jump from a state to itself. It is a negative balancing entry. The outgoing transition intensities are the off-diagonal entries; their sum determines the diagonal entry.

Worked Examples

Worked Example: Checking a Three-State Model

Consider a hypothetical continuous-time Markov model with states $A$ (active), $D$ (disabled), and $X$ (dead). The model allows transitions $A\to D$ at $0.07$ per year, $A\to X$ at $0.02$ per year, $D\to A$ at $0.04$ per year, and $D\to X$ at $0.06$ per year. State $X$ is absorbing. Construct the intensity matrix and check its row sums.

Assumptions and method. This is a continuous-time Markov model with the stated constant intensities, all measured per year. Death is represented by entry into $X$; no separate mortality distribution is assumed. Interest discounting is not involved. For each row, use the negative sum of the outgoing intensities as the diagonal entry.

For state $A$, the outgoing intensities sum to $0.07+0.02=0.09$, so $q_{AA}=-0.09$. For state $D$, they sum to $0.04+0.06=0.10$, so $q_{DD}=-0.10$. Since $X$ has no outgoing transitions, $q_{XX}=0$. The matrix is therefore

Q= \begin{pmatrix} -0.09&0.07&0.02\\ 0.04&-0.10&0.06\\ 0&0&0 \end{pmatrix}.

Check the rows separately: row $A$ sums to $-0.09+0.07+0.02=0$; row $D$ sums to $0.04-0.10+0.06=0$; and row $X$ sums to $0+0+0=0$. Each row satisfies the identity.

Method and mistake avoided. Balancing each row by its own outgoing intensities gives the correct diagonal. A common mistake is to use the outgoing sum from another state: for example, using $-0.10$ as $q_{AA}$ would make row $A$ sum to $-0.01$, not zero.

Worked Example: Finding a Missing Diagonal Entry

A hypothetical long-term care model has states $W$ (working), $C$ (receiving care), and $D$ (dead). From $C$, the intensity of recovery to $W$ is $0.03$ per year and the intensity of transition to $D$ is $0.05$ per year. Find $q_{CC}$, and check the result by adding the entries in row $C$.

Assumptions and method. The specified values are the only outgoing transition intensities from $C$ in this continuous-time Markov model, and both are per year. Death is a state, not a separately specified lifetime distribution. No interest rate or discounting assumption is needed. Apply the row-sum identity.

The sum of the off-diagonal entries in row $C$ is $q_{CW}+q_{CD}=0.03+0.05=0.08$. Therefore

q_{CC}=-(q_{CW}+q_{CD})=-(0.03+0.05)=-0.08.

The row sum check is $q_{CW}+q_{CC}+q_{CD}=0.03-0.08+0.05=0$. Thus the required diagonal entry is $-0.08$ per year.

Method and mistake avoided. The diagonal is the negative of the sum of all off-diagonal entries in the row, not the negative of one selected transition intensity. Setting $q_{CC}=-0.05$, for example, would give $0.03-0.05+0.05=0.03$, which fails the row-sum check.

Worked Example: Connecting the Row Sum to Short-Interval Probabilities

Consider a hypothetical model with states $R$ (recovering), $L$ (long-term care), and $D$ (dead). From $L$, the intensity of recovery is $0.03$ per year and the intensity of death is $0.07$ per year. Find the first-order short-interval probabilities of remaining in $L$, moving to $R$, and moving to $D$ over $h=0.02$ years. Check that they sum to one to first order.

Assumptions and method. Use the continuous-time Markov small-interval approximation, retaining first-order terms in $h$. The rates are constant for this calculation and measured per year; $h$ is in years. Death is represented by the transition to $D$, with no separate mortality distribution. No interest rate is involved.

First, the sum of outgoing intensities from $L$ is $0.03+0.07=0.10$ per year, so $q_{LL}=-0.10$ per year. The first-order probability of recovery is $0.03(0.02)=0.0006$, and the first-order probability of death is $0.07(0.02)=0.0014$. The probability of remaining in $L$, to first order, is $1-0.10(0.02)=0.998$.

$$ 0.998+0.0006+0.0014=1.0000. $$

The first-order probabilities balance as required. More precisely, the short-interval approximations omit terms of order $o(h)$; they are not asserted to be exact finite-interval probabilities.

Method and mistake avoided. Multiply each off-diagonal intensity by the interval length, then obtain the stay probability by subtracting the sum of the outgoing first-order probabilities from one. Do not treat the negative diagonal entry itself, $-0.10$, as a probability or multiply it by $h$ and report a negative probability of staying.

Using the Identity Carefully

The row-sum identity is a consequence of probability conservation, not an additional transition that the model must allow. It is also specific to the row-origin convention: the entries in row $i$ all describe outcomes conditional on starting in state $i$. With a different matrix convention, the corresponding balance would be expressed differently.

When checking a calculation, keep rates and probabilities distinct. An intensity has units of inverse time, such as per year. A first-order probability over an interval of $h$ years is approximately the intensity multiplied by $h$, and is dimensionless. The zero row sum is an identity among intensities; it does not say that the entries of an intensity-matrix row are themselves probabilities.

A useful check is to add the entries across each row, including the diagonal. If the result is not zero, either a diagonal entry has not balanced its outgoing rates, an off-diagonal entry has been put in the wrong row or column, or an arithmetic error has occurred. The check does not by itself replace careful specification of which state changes the model allows.

Key takeaway: For each starting state, the probabilities of remaining there or moving to any other state exhaust the possibilities. Consequently, the diagonal intensity entry is the negative sum of the off-diagonal entries in its row, and every row sums to zero.

Check Your Understanding

For each problem, use the stated row-origin convention and report rates in the stated units.

A hypothetical model has states $A,B,C$. From $A$, transitions to $B$ and $C$ have intensities $0.018$ and $0.027$ per year. Find $q_{AA}$ and verify the row sum. (Answer: $q_{AA}=-0.045$ per year; row sum $=0$.)
In a hypothetical three-state model, a state $M$ has outgoing intensities $0.11$ and $0.04$ per year to two different states. Find its diagonal entry and the first-order probability of remaining in $M$ over $0.5$ years. (Answer: $q_{MM}=-0.15$ per year; stay probability $\approx 0.925$.)
In a hypothetical model, state $B$ can move to $A,C,D$ at intensities $0.025,0.015,0.010$ per year, respectively. Determine $q_{BB}$, then check the sum of row $B$. (Answer: $q_{BB}=-0.050$ per year; row sum $=0$.)
A row of a hypothetical intensity matrix is $(-0.12, 0.03, x, 0.05)$, where the first entry is the diagonal and the rates are per year. Find $x$ and state the row-sum check. (Answer: $x=0.04$ per year; $-0.12+0.03+0.04+0.05=0$.)
State $Z$ in a hypothetical model is absorbing. Write its row in an intensity matrix and give the first-order probability of remaining in $Z$ over an interval of $0.1$ years. (Answer: all entries in row $Z$ are $0$; the first-order probability of remaining in $Z$ is $1$.)

What You'll Learn