Generator Matrices with Competing Exits and State Changes

Why the Generator Matrix Matters

An integrated benefit model may allow a person to move between benefit states and also to leave the model through competing causes such as death, retirement, or withdrawal. A generator matrix is a compact accounting of the instantaneous rates for all these possible moves. Constructing it correctly is essential: a missing transition or a diagonal with the wrong sign changes the model’s state dynamics.

The focus here is the matrix construction itself. We assume a finite-state continuous-time Markov model: conditional on the current state, the rates governing the next infinitesimal period do not depend on the earlier path. Intensities are nonnegative, expressed in a common time unit, and are finite. They may be constant or may vary with time or duration. The matrices below use the row-vector convention: rows identify the current state and columns identify the destination state.

Definition: For states numbered \(1,\ldots,n\), the generator \(Q(t)=(q_{ij}(t))\) has, for \(i\ne j\), \(q_{ij}(t)\) equal to the instantaneous transition intensity from state \(i\) to state \(j\). Its diagonal entry is \(q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)\). Thus every row of the full generator sums to zero. An absorbing state has no rates out, so its entire row is zero.

The phrase full generator matters. If only the non-absorbing benefit states are displayed, transitions out of those states are omitted from the displayed submatrix; its row sums are generally negative, not zero. To obtain a full generator, include each destination state that the model distinguishes, including absorbing states for death, retirement, and withdrawal.

How to Assemble a Generator

First list the states and fix their order. Then record every permitted move and its intensity. A move that is not permitted has rate zero. For each state, add all rates to different destination states; the negative of that total is the diagonal entry. Exit causes should remain separate destinations when the model needs to distinguish them. For example, death and withdrawal are both exits from a benefit state, but they are not the same transition if cause-specific outcomes matter.

Fix the state order.
Use the same order for rows and columns, and label it clearly. Include the absorbing destinations needed to represent the exits.

Enter off-diagonal rates.
For each permitted move from state \(i\) to a different state \(j\), place its intensity in \(q_{ij}\). Put zero in \(q_{ij}\) if the move is not allowed.

Complete the diagonal.
For each row, add its off-diagonal entries and put the negative of that sum on the diagonal.

Check the model and arithmetic.
Confirm that each off-diagonal rate is nonnegative, each row sums to zero, and absorbing-state rows contain only zeros.

Why the Diagonal Is the Negative Total Rate

The diagonal entry is not an additional transition intensity. It accounts for the probability of remaining in the current state over a very short interval. If the process is currently in state \(i\), then over an interval of length \(h\), the probability of a single transition from \(i\) to \(j\ne i\) is \(q_{ij}(t)h+o(h)\). The probability of leaving \(i\) for any different state is therefore the sum of these terms.

For a finite-state continuous-time Markov model, the probability of two or more transitions in an interval of length \(h\) is of smaller order than \(h\). Consequently, the probability of still being in \(i\) is one minus the sum of the first-order probabilities of leaving \(i\). The infinitesimal change in the probability of remaining in \(i\), divided by \(h\), is therefore the negative of the total rate out.

P(\text{remain in }i\text{ over }h) =1-\left(\sum_{j\ne i}q_{ij}(t)\right)h+o(h), \qquad q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t).

This argument also explains the row-sum check. The diagonal cancels the total of the off-diagonal rates in that row. It does not mean that the process has a negative transition intensity: only the off-diagonal entries represent rates of moving to another state.

Worked Example: Benefit-State Changes with Three Competing Exits

Consider a hypothetical continuous-time model with six states, in the order Active (\(A\)), Disabled (\(D\)), Extended Care (\(C\)), Dead (\(X\)), Retired (\(R\)), and Withdrawn (\(W\)). Assume all rates are constant per year. From \(A\), the rates to \(D,R,X,W\) are \(0.08,0.12,0.01,0.04\), respectively. From \(D\), the rates to \(A,C,X,W\) are \(0.30,0.10,0.02,0.03\). From \(C\), the rates to \(D,X,R,W\) are \(0.05,0.04,0.02,0.01\). The last three states are absorbing. Construct the generator and verify its row sums.

Method. Use the stated state order for both rows and columns. Enter the listed rates in the corresponding off-diagonal positions; all unlisted moves have rate zero.

For \(A\), the total rate out is \(0.08+0.12+0.01+0.04=0.25\), so \(q_{AA}=-0.25\). For \(D\), the total is \(0.30+0.10+0.02+0.03=0.45\), so \(q_{DD}=-0.45\). For \(C\), the total is \(0.05+0.04+0.02+0.01=0.12\), so \(q_{CC}=-0.12\). Each absorbing state has a zero row.

Q= \begin{pmatrix} -0.25&0.08&0&0.01&0.12&0.04\\ 0.30&-0.45&0.10&0.02&0&0.03\\ 0&0.05&-0.12&0.04&0.02&0.01\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{pmatrix}.

Check. The first three row sums are \(-0.25+0.08+0+0.01+0.12+0.04=0\), \(0.30-0.45+0.10+0.02+0+0.03=0\), and \(0+0.05-0.12+0.04+0.02+0.01=0\). The final three row sums are zero because those states are absorbing. Thus every row sums to zero.

Common mistake avoided: Treating all exits as a single “leave” state would remove the distinctions among death, retirement, and withdrawal. Here each has its own column, while the diagonal still uses the total rate of all destinations.

Rates Must Use a Common Time Unit

Every entry in a generator must be expressed per the same unit of time. A rate given per month cannot be added directly to a rate given per year. For example, a constant intensity of \(0.005\) per month is \(12(0.005)=0.06\) per year. This is a conversion of an intensity, not a conversion of a one-month transition probability.

Worked Example: Converting Monthly Intensities Before Adding Competing Rates

A hypothetical model has states Employed (\(E\)), Disabled (\(I\)), Dead (\(X\)), Retired (\(R\)), and Withdrawn (\(W\)), in that order. All states except \(E\) and \(I\) are absorbing. In \(E\), the monthly intensities to \(I,X,R,W\) are \(0.005,0.0005,0.003,0.002\). In \(I\), the monthly intensities to \(E,X,W\) are \(0.020,0.001,0.0015\). Assume these are constant forces per month. Construct the annual-rate generator.

Step 1: Convert each nonzero intensity. Multiplication by 12 gives the annual rates from \(E\): to \(I\), \(12(0.005)=0.060\); to \(X\), \(12(0.0005)=0.006\); to \(R\), \(12(0.003)=0.036\); and to \(W\), \(12(0.002)=0.024\). From \(I\), the annual rates are \(12(0.020)=0.240\) to \(E\), \(12(0.001)=0.012\) to \(X\), and \(12(0.0015)=0.018\) to \(W\).

Step 2: Calculate the diagonals. The total annual rate out of \(E\) is \(0.060+0.006+0.036+0.024=0.126\), so \(q_{EE}=-0.126\). The total out of \(I\) is \(0.240+0.012+0.018=0.270\), so \(q_{II}=-0.270\).

Q= \begin{pmatrix} -0.126&0.060&0.006&0.036&0.024\\ 0.240&-0.270&0.012&0&0.018\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{pmatrix} \quad\text{(rates per year).}

Step 3: Verify. The row for \(E\) sums to \(-0.126+0.060+0.006+0.036+0.024=0\). The row for \(I\) sums to \(0.240-0.270+0.012+0+0.018=0\). The absorbing rows also sum to zero.

Common mistake avoided: Adding the monthly rates to the annual-rate matrix without conversion would mix units. The matrix would not represent a coherent set of rates, even if its rows were forced to sum to zero.

Time-Dependent Intensities and Matrix Checks

The definition of a generator applies at a specified time even when intensities vary with time. In that case, write \(Q(t)\); calculate the diagonal using the off-diagonal rates at the same time \(t\). The row-sum condition is checked at each time at which the intensities are defined. Do not use a rate from one duration for one transition and a rate from a different duration for another unless the model explicitly specifies that combination.

Worked Example: Completing a Duration-Dependent Generator

A hypothetical contract has states Active (\(A\)), Disabled (\(D\)), Dead (\(X\)), and Retired (\(R\)), in that order. Duration \(t\) is measured in years. The rates per year are \(q_{AD}(t)=0.04+0.005t\), \(q_{AR}(t)=0.02\), \(q_{AX}(t)=0.01\), \(q_{DA}(t)=0.15\), \(q_{DX}(t)=0.02+0.002t\), and \(q_{DR}(t)=0.01\). Death and retirement are absorbing. Find \(Q(3)\) and check it.

Step 1: Evaluate the rates at \(t=3\). The rates from \(A\) are \(q_{AD}(3)=0.04+0.005(3)=0.055\), \(q_{AR}(3)=0.020\), and \(q_{AX}(3)=0.010\). Their sum is \(0.055+0.020+0.010=0.085\), so \(q_{AA}(3)=-0.085\).

From \(D\), the rates are \(q_{DA}(3)=0.150\), \(q_{DX}(3)=0.02+0.002(3)=0.026\), and \(q_{DR}(3)=0.010\). Their sum is \(0.150+0.026+0.010=0.186\), so \(q_{DD}(3)=-0.186\).

Q(3)= \begin{pmatrix} -0.085&0.055&0.010&0.020\\ 0.150&-0.186&0.026&0.010\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix} \quad\text{(rates per year).}

Step 2: Check. The first row sums to \(-0.085+0.055+0.010+0.020=0\), and the second sums to \(0.150-0.186+0.026+0.010=0\). Both absorbing rows are zero. The matrix is therefore a valid generator at duration three, subject to the stated model assumptions.

Common mistake avoided: Leaving the diagonal entries at their values for another duration would violate the row-sum condition at \(t=3\). The diagonal must be recalculated whenever the relevant off-diagonal rates change.

Exam Checks and Frequent Errors

A row-sum check is a useful arithmetic test, but it is not a complete validation of a model. A matrix can have zero row sums and still put a rate in the wrong column, omit a permitted move, or assign a transition that the state map disallows. Check the state order and allowed moves as well as the arithmetic.

Wrong diagonal sign: The diagonal is the negative total rate out, not the positive total.
Missing competing exit: Each cause-specific exit that must remain distinct needs its own destination column.
Mixing units: Convert every intensity to the same time basis before summing a row.
Forcing the displayed submatrix to sum to zero: If absorbing destinations have been omitted, the transient-state submatrix is not the full generator and its rows need not sum to zero.
Nonzero absorbing-state row: A state declared absorbing has no outgoing transition intensities in this model.

Key takeaway: In a full generator, enter each permitted transition intensity off the diagonal, set each diagonal to the negative sum of the other entries in its row, and verify the state map, units, and zero row sums.

Check Your Understanding

For each problem, use the stated row-vector convention and include absorbing destinations in the full generator.

A hypothetical model has states Active (\(A\)), Disabled (\(D\)), Dead (\(X\)), Retired (\(R\)), in that order. Per year, \(A\) has rates \(0.03\) to \(D\), \(0.01\) to \(X\), and \(0.04\) to \(R\); \(D\) has rates \(0.20\) to \(A\) and \(0.02\) to \(X\). Find the generator. (Answer: \(Q=\begin{pmatrix}-0.08&0.03&0.01&0.04\\0.20&-0.22&0.02&0\\0&0&0&0\\0&0&0&0\end{pmatrix}\).)
In a hypothetical model, state \(P\) has monthly intensities \(0.004\) to \(B\), \(0.001\) to death, and \(0.002\) to withdrawal. Convert them to annual rates and give the diagonal for row \(P\). (Answer: Annual rates \(0.048,0.012,0.024\) per year; diagonal \(q_{PP}=-0.084\) per year.)
At a specified duration, a full generator row has off-diagonal entries \(0.07,0,0.03,0.02\). What must its diagonal be, and what is the row sum after completion? (Answer: Diagonal \(-0.12\); row sum \(0\).)
A benefit state \(C\) can move to \(L\) at rate \(0.06\), to death at rate \(0.015\), and to withdrawal at rate \(0.025\), all per year. In the order \(C,L,X,W\), give row \(C\), including its diagonal. (Answer: \((-0.10,\ 0.06,\ 0.015,\ 0.025)\); its row sum is \(0\).)
A proposed full generator uses state order \((A,D,X,W)\). Its \(A\)-row is \((-0.09,0.04,0.02,0.04)\), with all rates per year. Identify the error and correct the diagonal if the listed off-diagonal rates are intended. (Answer: The row sums to \(0.01\), not \(0\); the off-diagonal total is \(0.10\), so the corrected diagonal is \(-0.10\).)

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