Checking Whether an Intensity Matrix Is Valid

Introduction

A proposed intensity matrix can look plausible while still describing no possible continuous-time state model. A transition rate may be negative, a diagonal entry may have the wrong sign, or the entries in a row may fail to balance. Checking these features before using the matrix prevents invalid inputs from being carried into later calculations.

This tutorial gives a practical test for a finite-state intensity matrix and shows how to diagnose each type of failure. It concerns whether the entries are structurally valid as intensities; it does not establish that the chosen rates are suitable for a particular hypothetical product or population.

Definition: For a finite-state continuous-time Markov model, an intensity matrix \(Q=(q_{ij})\) has one row and column for each state. For distinct states \(i\) and \(j\), \(q_{ij}\) is the instantaneous transition intensity from state \(i\) to state \(j\). The diagonal entry \(q_{ii}\) is not a transition to a different state. All entries in a matrix must use the same time unit.

The Three Checks

For a proposed finite-state intensity matrix, check every row and every entry. The conditions are:

Every off-diagonal entry is nonnegative: \(q_{ij}\geq 0\) when \(i\ne j\).
Every diagonal entry is nonpositive: \(q_{ii}\leq 0\).
Every row sums to zero: \(\sum_j q_{ij}=0\) for each state \(i\).

These are conditions on the instantaneous rates in the matrix, not on transition probabilities over a fixed interval. In particular, an off-diagonal intensity is not required to be at most 1. Its units are inverse time, such as per year or per month.

q_{ii}=-\sum_{j\ne i}q_{ij}

The displayed identity is the row-sum condition rearranged to isolate the diagonal. It is useful both for checking a matrix and for finding the diagonal entry when the off-diagonal intensities are specified. The sign check on the diagonal is still worth making explicitly, even though nonnegative off-diagonal entries and a zero row sum imply a nonpositive diagonal.

Check off-diagonal signs.
Scan entries outside the main diagonal. Any negative value cannot be a transition intensity.

Check diagonal signs.
Each diagonal entry must be zero or negative. A positive diagonal is invalid.

Calculate each row sum.
Add all entries in the row, including its diagonal. Each total must equal zero, subject only to an appropriate allowance for displayed rounding.

Why the Row Must Sum to Zero

The row-sum test follows from the requirement that probabilities sum to one. Consider a model at a fixed time, starting in state \(i\), and a short interval of length \(h\). For distinct states \(i\) and \(j\), the definition of intensity gives a transition probability \(q_{ij}h+o(h)\). The probability of being in state \(i\) at the end of the interval is \(1+q_{ii}h+o(h)\). Here \(o(h)\) denotes terms that become negligible relative to \(h\) as \(h\) tends to zero.

Adding the probabilities of ending in every state gives 1. The constant terms already sum to 1, so the coefficient of \(h\) must sum to zero:

\left(1+q_{ii}h+o(h)\right) +\sum_{j\ne i}\left(q_{ij}h+o(h)\right)=1.

Divide the difference from 1 by \(h\) and let \(h\) approach zero. The result is \(q_{ii}+\sum_{j\ne i}q_{ij}=0\), which is exactly the zero row-sum condition. This also explains why, once the off-diagonal intensities in a row are fixed, there is only one diagonal value that makes the row sum to zero.

Worked Example: Verifying a Three-State Matrix

A hypothetical model has three states, and its proposed intensities are measured per year. Determine whether the matrix is valid.

Q= \begin{pmatrix} -0.18&0.11&0.07\\ 0.05&-0.14&0.09\\ 0&0.06&-0.06 \end{pmatrix}.

Method. Check the off-diagonal entries first, then the diagonal entries, and finally the row sums. No mortality or other probability distribution assumption is needed: this is a structural check of the stated continuous-time model.

Off-diagonal entries. They are \(0.11, 0.07, 0.05, 0.09, 0,\) and \(0.06\). Each is nonnegative.

Diagonal entries. They are \(-0.18,-0.14,\) and \(-0.06\), all nonpositive.

Row sums. The first row sums to \(-0.18+0.11+0.07=0\). The second sums to \(0.05-0.14+0.09=0\). The third sums to \(0+0.06-0.06=0\).

All three conditions hold, so this is a valid intensity matrix. As a check on the first row, the diagonal is \(-(0.11+0.07)=-0.18\), as required. The common mistake this check avoids is to inspect only the diagonal signs and assume that a matrix with negative diagonals must be valid.

Diagnosing and Correcting Violations

A failed check identifies a problem, but it does not always identify a unique repair. If the off-diagonal rates are trusted and the diagonal is wrong, the row-sum condition determines the correction. If an off-diagonal rate is negative, the proposed rate itself must be reconsidered; changing the diagonal alone cannot make that transition intensity valid.

Worked Example: A Negative Intensity and an Incorrect Diagonal

A hypothetical three-state model gives the following proposed matrix, with all entries in rates per year. Identify every failure and give one valid repair. For the repair, suppose the stated positive off-diagonal rates are retained and the negative off-diagonal entry is determined to represent no allowed transition.

Q= \begin{pmatrix} -0.24&0.15&0.09\\ 0.08&-0.20&0.10\\ 0.04&-0.03&-0.01 \end{pmatrix}.

Row 1. Its off-diagonal entries are nonnegative, its diagonal is nonpositive, and its sum is \(-0.24+0.15+0.09=0\). This row passes.

Row 2. Its off-diagonal entries are nonnegative and its diagonal is nonpositive, but its sum is \(0.08-0.20+0.10=-0.02\). The row fails the zero-sum condition. Retaining the off-diagonal rates requires the diagonal to be \(-(0.08+0.10)=-0.18\), not \(-0.20\).

Row 3. Its entry \(-0.03\) is off the diagonal, so it is a negative transition intensity and is invalid. The row happens to sum to zero: \(0.04-0.03-0.01=0\). A zero sum does not excuse a negative off-diagonal entry. Under the stated repair assumption, replace \(-0.03\) with 0. Then the diagonal must be \(-(0.04+0)=-0.04\).

One repaired matrix is therefore:

Q_{\mathrm{repaired}}= \begin{pmatrix} -0.24&0.15&0.09\\ 0.08&-0.18&0.10\\ 0.04&0&-0.04 \end{pmatrix}.

The repaired row sums are \(0\), \(0.08-0.18+0.10=0\), and \(0.04+0-0.04=0\). All off-diagonal entries are nonnegative and all diagonal entries are nonpositive, so the repaired matrix passes. The repair depends on the stated assumption: without it, the negative transition rate must be investigated rather than silently replaced.

Worked Example: A Positive Diagonal Cannot Be Overlooked

A hypothetical model has proposed annual intensities:

Q= \begin{pmatrix} -0.12&0.08&0.04\\ 0.03&0.01&0.02\\ 0.05&0.02&-0.07 \end{pmatrix}.

Method. Apply all checks even when a row-sum calculation already points to a problem. This distinguishes the observed failures and shows how a row can be repaired if its off-diagonal rates are trusted.

Every off-diagonal entry is nonnegative. The first and third diagonal entries are nonpositive, but the second diagonal entry, \(0.01\), is positive, so the matrix fails the diagonal-sign condition. The second row also sums to \(0.03+0.01+0.02=0.06\), rather than zero.

Retain the second-row off-diagonal rates \(0.03\) and \(0.02\). Its diagonal must then be \(-(0.03+0.02)=-0.05\). The other rows already sum to zero: \(-0.12+0.08+0.04=0\) and \(0.05+0.02-0.07=0\). After replacing \(0.01\) with \(-0.05\), the second row sums to \(0.03-0.05+0.02=0\), and all diagonal signs are correct. The corrected matrix is valid.

The common mistake is to treat the positive diagonal as a harmless sign typo without checking the row. The row-sum calculation confirms the required correction when the off-diagonal entries are kept fixed.

Worked Example: A Rate Greater Than One Can Be Valid

A hypothetical model uses years as its time unit and proposes:

Q= \begin{pmatrix} -1.4&1.1&0.3\\ 0.2&-0.5&0.3\\ 0&0&0 \end{pmatrix}.

All off-diagonal entries are nonnegative, and the diagonal entries are nonpositive. The row sums are \(-1.4+1.1+0.3=0\), \(0.2-0.5+0.3=0\), and \(0+0+0=0\). The matrix is valid.

In particular, the entry \(1.1\) is an intensity of 1.1 per year, not a probability of 1.1. The matrix passes the structural tests even though that rate exceeds one. The short-interval probability approximation applies for sufficiently small intervals; an intensity is not itself a finite-interval probability. The common mistake this example avoids is imposing a probability bound on an intensity.

Rounding, Units, and Scope

A matrix reported to a limited number of decimal places may have a row sum that is close to, but not exactly, zero. For example, if unrounded entries sum to zero but are displayed after separate rounding, the printed row may not. When source values are available, check those unrounded values. If only rounded values are available, report the small discrepancy and its likely rounding explanation; do not silently alter a rate to force the row sum to zero.

Use one consistent time unit throughout a matrix. A row combining a rate per year with a rate per month cannot be checked meaningfully until the units are made consistent. A numerically valid matrix also does not verify that the states or allowed transitions are appropriate for the scenario; it verifies the basic mathematical conditions for an intensity matrix.

Key takeaway: A finite-state intensity matrix must have nonnegative off-diagonal entries, nonpositive diagonal entries, and zero row sums. Check all three conditions. If off-diagonal rates are fixed, the diagonal in row \(i\) must be \(-\sum_{j\ne i}q_{ij}\).

Check Your Understanding

For each problem, assume a finite-state continuous-time model and use the stated common time unit. Decide whether the proposed matrix is valid or identify the required correction.

For annual rates, check \(Q=\begin{pmatrix}-0.30&0.20&0.10\\0.05&-0.15&0.10\\0&0.04&-0.04\end{pmatrix}\). (Answer: Valid; all off-diagonal entries are nonnegative, all diagonals are nonpositive, and every row sums to zero.)
A row has off-diagonal entries \(0.07\) and \(0.03\), and diagonal entry \(-0.11\), all per year. Identify the failure and the corrected diagonal if the off-diagonal rates are retained. (Answer: The row sums to \(-0.01\); the corrected diagonal is \(-0.10\) per year.)
For annual rates, a row is \((0.30,-0.20,-0.10)\), with the middle entry on the diagonal. Is the zero row sum sufficient for validity? Give one valid repair if the \(0.30\) rate is retained and the negative off-diagonal entry represents no allowed transition. (Answer: No; the off-diagonal \(-0.10\) is invalid despite the zero sum. Replace it with \(0\) and set the diagonal to \(-0.30\).)
For monthly rates, consider \(Q=\begin{pmatrix}-0.06&0.04&0.02\\0.01&-0.03&0.02\\0&0&0\end{pmatrix}\). Is this a valid intensity matrix, and must the entry \(0.04\) be at most 1? (Answer: Valid; no, \(0.04\) is a monthly intensity, not a finite-interval probability.)
A proposed annual-rate matrix has rows \((-0.12,0.07,0.05)\), \((0.04,0.01,0.03)\), and \((0.02,0.03,-0.05)\). Keeping all off-diagonal entries fixed, identify the correction and verify the affected row. (Answer: Replace the second-row diagonal \(0.01\) with \(-0.07\); the corrected row sums to \(0.04-0.07+0.03=0\), and the matrix is valid.)

What You'll Learn