Why Intensity-Matrix Checks Matter
An intensity matrix is a compact description of the instantaneous transition rates in a continuous-time state model. A single sign error or misplaced entry can change the model being represented, even when the numbers look plausible. Some errors are caught by checking row sums; others, such as swapping two off-diagonal rates in the same row, can leave every row sum correct. A reliable audit therefore checks the matrix against both its mathematical requirements and the transition structure specified by the hypothetical contract or population.
Throughout this tutorial, use the row-vector convention: rows identify the current state, columns identify the destination state, and a row distribution is multiplied on the right by the intensity matrix. For a time-homogeneous model, write the matrix as \(Q\); for a time-inhomogeneous model, write \(Q(t)\). Each off-diagonal entry is a transition intensity, with units such as per year. The diagonal entry records the negative total intensity of leaving that state.
A Reliable Audit: Structure Before Arithmetic
Before calculating, write down the state order and list the allowed exits from each state. Then enter each allowed rate in the row of its origin and the column of its destination. Only after those off-diagonal entries are in place should you calculate the diagonal. This order reduces the risk of using a diagonal value to conceal a missing or incorrectly entered exit rate.
Record the state labels in a fixed order. Under the row-vector convention, row \(i\), column \(j\) describes movement from \(i\) to \(j\).
Use the model description to identify which off-diagonal entries may be nonzero. An omitted route is zero; an impossible route must stay zero.
Put each nonnegative rate in its origin-destination cell. Confirm its time unit and whether it is an intensity rather than a probability.
For each row, negate the sum of its off-diagonal entries. Check that the row sums to zero and compare every entry with the allowed-transition list.
Why the Diagonal Is the Negative Exit Total
The diagonal rule is not an arbitrary matrix convention. It follows from the fact that, over a short time interval, the probabilities of all possible outcomes from a state must sum to one. For distinct states \(i\) and \(j\), the defining local behavior of an intensity is that the chance of a transition from \(i\) to \(j\) in an interval of length \(h\), conditional on being in \(i\) at its start, is \(q_{ij}(t)h+o(h)\). Here \(o(h)\) denotes a term that becomes negligible relative to \(h\) as \(h\) approaches zero.
Adding over the possible destinations other than \(i\), the total chance of leaving \(i\) during this short interval is \(\sum_{j\ne i}q_{ij}(t)h+o(h)\). Consequently, the chance of remaining in \(i\) is \(1-\sum_{j\ne i}q_{ij}(t)h+o(h)\). The coefficient of \(h\) for that diagonal position is therefore the negative sum of the exit intensities:
This reasoning also explains two useful checks. First, an intensity matrix cannot have a positive diagonal entry when its off-diagonal intensities are nonnegative. Second, a zero row sum is necessary, but it does not establish that the off-diagonal entries describe the right transitions or use the right rates.
Worked Example: Constructing and Checking a Three-State Matrix
Consider a hypothetical time-homogeneous disability model with states Active (A), Disabled (D), and Dead (X), in that order. The specified constant annual intensities are \(0.08\) from A to D, \(0.02\) from A to X, \(0.10\) from D to A, and \(0.04\) from D to X. Death is absorbing. Construct and audit \(Q\). Assume all rates are per year.
Method and transition list. Use the row-vector convention. The possible exits from A are to D and X; those from D are to A and X; X has no exits. Thus the off-diagonal entries are \(q_{AD}=0.08\), \(q_{AX}=0.02\), \(q_{DA}=0.10\), \(q_{DX}=0.04\), and all other off-diagonal entries are zero.
Calculate the diagonal entries. From A, the total exit intensity is \(0.08+0.02=0.10\), so \(q_{AA}=-0.10\). From D, it is \(0.10+0.04=0.14\), so \(q_{DD}=-0.14\). Since X has no exit, \(q_{XX}=0\).
Check. The row totals are \(-0.10+0.08+0.02=0\), \(0.10-0.14+0.04=0\), and \(0+0+0=0\). Every off-diagonal entry is nonnegative, the diagonal entries are nonpositive, and the absorbing row is all zero. The matrix matches the stated transition list.
Common mistake avoided. Putting \(0.08\) in the D-to-A cell because it is the first rate mentioned would misplace a valid rate. A row-sum check might still pass if the diagonal were adjusted. The transition list and state order must therefore be checked as well as the arithmetic.
Errors That a Zero Row Sum Cannot Catch
A row-sum check tests a relationship among the entries in one row; it does not verify their meanings. In particular, swapping two off-diagonal entries in a row preserves their sum. Similarly, a matrix may have zero row sums while including a transition that the model rules out. In an audit, keep two questions separate: is this row arithmetically balanced? and does each entry represent the specified route?
Worked Example: Repairing a Plausible but Incorrect Matrix
A hypothetical long-term-care model has states Independent (I), Receiving Care (C), and Dead (X), in that order. Its constant annual intensities are I-to-C \(0.12\), I-to-X \(0.03\), C-to-I \(0.05\), and C-to-X \(0.08\). Death is absorbing, and there are no other transitions. A proposed matrix is shown below. Identify its defects and give the corrected matrix.
Step 1: Check row sums. The first row sums to \(-0.15+0.03+0.12=0\). The second sums to \(0.05-0.08+0.03=0\). The third sums to \(0.01+0-0.01=0\). Thus, all rows pass the arithmetic check.
Step 2: Compare each row with the transition list. In row I, the I-to-C and I-to-X rates have been swapped: the required entries are \(0.12\) and \(0.03\), respectively. The diagonal \(-0.15\) is nevertheless the correct negative total exit intensity for I. In row C, the C-to-X rate \(0.08\) has been entered as \(0.03\); as a result, the diagonal \(-0.08\) is wrong as well. The required C exit total is \(0.05+0.08=0.13\). In row X, the proposed matrix permits X-to-I at rate \(0.01\), contrary to the absorbing-state assumption.
Step 3: Rebuild from the specified rates. The I row has diagonal \(-(0.12+0.03)=-0.15\). The C row has diagonal \(-(0.05+0.08)=-0.13\). The absorbing row is all zero. Therefore:
Check. The row sums are \(-0.15+0.12+0.03=0\), \(0.05-0.13+0.08=0\), and \(0\). The corrected entries agree with every stated route. This example shows why passing a row-sum test is not enough: the proposed matrix passed that test but represented incorrect rates and an impossible transition.
Common mistake avoided. Do not “repair” a matrix by changing only the diagonal until each row sums to zero. The diagonal must be determined from the correct off-diagonal rates, and those rates must first be checked against the model.
Rates Are Not Probabilities
An intensity is a rate per unit time, not the probability of a transition over a whole year. In a one-exit model with a constant intensity \(\lambda\), the probability of no exit during a full year is \(e^{-\lambda}\), so the probability of at least one exit is \(1-e^{-\lambda}\). This relation depends on the stated constant-intensity, single-exit assumptions; a probability over a year in a more complex model cannot generally be inserted into an intensity matrix as though it were a rate.
Worked Example: Converting a One-Year Exit Probability Under Explicit Assumptions
A hypothetical two-state model has Healthy (H) and Dead (X), with death absorbing. Suppose the probability that a Healthy life remains Healthy for one year is \(0.80\). Assume the only possible exit from H is to X and that the transition intensity is constant over the year. Find the annual intensity and construct the intensity matrix. Report the rate to four significant figures.
Method. Under the stated single-exit, constant-intensity assumptions, the probability of no exit in one year is \(e^{-\lambda}\). Set this equal to \(0.80\):
To four significant figures, \(\lambda=0.2231\) per year. There is only one exit from H, so the H diagonal is \(-0.2231\). Since X is absorbing, its row is zero:
Check. The first row sums to \(-0.2231+0.2231=0\); the second row sums to zero. The unrounded intensity gives \(e^{-0.22314355\ldots}=0.80\), as required. The displayed rate is rounded only at the final step.
Common mistake avoided. Entering \(0.20\) as the intensity would confuse the one-year exit probability \(1-0.80\) with a rate. Under these assumptions, the intensity is \(-\ln(0.80)\), not \(0.20\). Without the constant-intensity and single-exit assumptions, the one-year probability alone does not justify this conversion.
A Practical Checklist and Common Traps
Before using an intensity matrix in further calculations, perform a row-by-row review. In particular, do not rely on visual plausibility: a negative number may be appropriate on the diagonal, but a negative off-diagonal intensity is not. Confirm that all entries use the same time unit and that a rate stated per month has not been treated as a rate per year. If a model is time-inhomogeneous, the checks apply at each time \(t\); every off-diagonal intensity must be nonnegative and every row must sum to zero at that time.
| Check | What should be true? | What it does not establish |
|---|---|---|
| Off-diagonal signs | Each allowed transition intensity is nonnegative. | That the route or numerical value is correct. |
| Diagonal | It equals the negative sum of the row's off-diagonal entries. | That no exit route has been omitted or added. |
| Row total | Every row sums to zero. | That rates have not been swapped within a row. |
| Transition structure | Only transitions allowed by the model have nonzero entries. | That the rates use the correct time unit. |
| Units and interpretation | Entries are intensities with a consistent time basis. | That a probability over a finite period can be used directly as an intensity. |
A further convention trap is mixing row-vector and column-vector formulations. This tutorial uses rows for origins and row sums of zero. A different convention may place origins in columns and use column sums; do not combine one convention's entry placement with the other's multiplication rule. The state labels and convention should be written beside the matrix whenever there is any risk of ambiguity.
These checks are useful because an intensity matrix is the foundation for later state-model calculations: an error can propagate even if subsequent arithmetic is flawless. The construction and diagnostic skills here are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use the row-vector convention and state any assumptions needed.
- A hypothetical two-state model has Working (W) and Dead (X), with X absorbing and a constant W-to-X intensity of \(0.04\) per year. Write its intensity matrix in the order W, X. (Answer: \(\begin{pmatrix}-0.04&0.04\\0&0\end{pmatrix}\).)
- A care model has states Independent (I), Care (C), and Dead (X). Its only nonzero transition intensities are I-to-C \(0.07\), I-to-X \(0.02\), C-to-I \(0.03\), and C-to-X \(0.06\), all per year; X is absorbing. Give the matrix in the order I, C, X. (Answer: \(\begin{pmatrix}-0.09&0.07&0.02\\0.03&-0.09&0.06\\0&0&0\end{pmatrix}\).)
- A proposed row for a transient state has entries \((-0.11,0.07,0.04)\). Its stated allowed exits are to the second state at \(0.05\) per year and to the third state at \(0.06\) per year. Is the row valid, and what is the corrected row? (Answer: It sums to zero but has the two exit rates misplaced; the corrected row is \((-0.11,0.05,0.06)\).)
- A hypothetical single-exit state has a one-year probability of no exit equal to \(0.90\). Assuming a constant intensity over the year, find the intensity to four significant figures and give the row entries for that state and its absorbing destination. (Answer: \(\lambda=-\ln(0.90)=0.1054\) per year; row \((-0.1054,0.1054)\).)
- A proposed absorbing-state row in a three-state model is \((0,0.02,-0.02)\), and the state order is Active, Disabled, Dead. The contract rules out any transition out of Dead. Explain the defect and state the correct row. (Answer: The row incorrectly allows Dead-to-Disabled at intensity \(0.02\); the correct row is \((0,0,0)\).)