Off-Diagonal Entries as Transition Intensities

Introduction

An intensity matrix organizes the rates at which a model can move directly from one state to another. Its off-diagonal entries have a simple but important job: each one records the intensity for a transition from the state named by its row to the state named by its column. Correctly locating these entries depends on tracking both the state ordering and the direction of each transition.

This tutorial focuses on constructing and interpreting those off-diagonal entries. It does not calculate the diagonal entries. The key habit is to read every entry as an ordered pair: from the row state to the column state.

Definition: For distinct states \(i\) and \(j\), the transition intensity \(\mu_{ij}(t)\) is the instantaneous rate of moving directly from state \(i\) to state \(j\) at time \(t\), conditional on being in state \(i\) at time \(t\). In the intensity matrix, \(\mu_{ij}(t)\) is placed in the row for state \(i\) and the column for state \(j\). These are the off-diagonal entries, because the source and destination states differ.

Reading a Cell as a Directed Transition

A matrix entry is not just a rate associated with two states; its position specifies the direction. If state 1 is active and state 2 is disabled, then \(\mu_{12}(t)\) represents active to disabled. The entry \(\mu_{21}(t)\), if that direct return transition is allowed, represents disabled to active. The rates need not be equal, and one direction being permitted does not imply that the reverse direction is permitted.

The convention can be summarized as follows:

Read the row.
The row identifies the state occupied immediately before the direct transition.

Read the column.
The column identifies the destination state reached by that transition.

Place the stated intensity.
Put the rate for that directed transition in the cell at the intersection of its source row and destination column.

This is a bookkeeping rule, but it has actuarial consequences. Reversing the order can change the model's interpretation: a recovery intensity is not an incidence intensity, and an intensity of entering a care state is not an intensity of leaving it.

Why the Entry Has This Meaning

The placement follows from the definition of a transition intensity. Let \(h\) be a small positive time interval. Conditional on being in state \(i\) at time \(t\), the probability of a direct transition to a different state \(j\) during that interval is

P(\text{state }j\text{ at }t+h\mid\text{state }i\text{ at }t) =\mu_{ij}(t)h+o(h), \qquad i\ne j.

Here \(o(h)\) denotes a term that becomes negligible relative to \(h\) as \(h\) approaches zero. Dividing the probability by \(h\) and taking the limit gives the intensity for the ordered movement \(i\) to \(j\). The source condition is state \(i\), and the destination event is state \(j\). Therefore the corresponding matrix cell must use \(i\) as its row and \(j\) as its column.

The formula is a small-interval approximation, not an exact probability for a general finite interval. Its units are also rates: if time is measured in years, \(\mu_{ij}(t)\) is expressed per year. The earlier material on transition intensities and their units establishes these distinctions; here they help interpret the entries consistently.

Worked Example: Disability and Death Transitions

Consider a hypothetical model with states \(H\) (healthy), \(D\) (disabled), and \(X\) (dead), ordered \(H,D,X\). Assume the model is Markov and time-homogeneous, with rates measured per year. The permitted direct transitions have intensities \(H\to D: 0.020\), \(H\to X: 0.004\), \(D\to H: 0.100\), and \(D\to X: 0.080\). Construct the off-diagonal entries.

Method. For each arrow, use its starting state as the row and its ending state as the column. The state order is \(H,D,X\), so row 1 is \(H\), row 2 is \(D\), and row 3 is \(X\); the columns follow the same order.

The transition \(H\to D\) has source \(H\) and destination \(D\), so it goes in row \(H\), column \(D\), with value \(0.020\) per year. The transition \(H\to X\) goes in row \(H\), column \(X\), with value \(0.004\) per year.

The transition \(D\to H\) goes in row \(D\), column \(H\), with value \(0.100\) per year. The transition \(D\to X\) goes in row \(D\), column \(X\), with value \(0.080\) per year. These are four distinct source-destination pairs, so they occupy four different off-diagonal cells.

\begin{array}{c|ccc} \text{Source}\backslash\text{Destination} & H & D & X\\ \hline H & & 0.020 & 0.004\\ D & 0.100 & & 0.080\\ X & & & \end{array}

Check. Reading the entries back by row gives \(H\to D\) and \(H\to X\) in the \(H\) row, and \(D\to H\) and \(D\to X\) in the \(D\) row. Each matches the stated directed transition and its rate. The blank diagonal cells are not being specified in this example.

Common mistake avoided: placing \(0.100\) in row \(H\), column \(D\) because it is the recovery rate. Recovery is \(D\to H\), so the correct location is row \(D\), column \(H\).

Constructing Entries from a State Order

Before filling cells, write down the precise state order. A graph may show the same states without indicating that they must appear in any particular matrix order. The order assigned to rows must also be used for columns. If the order changes, the meaning of a cell changes even though the numerical rate does not.

A reliable construction is to list each permitted arrow with its source, destination, and rate, then translate that list into cells. The example below uses long-term care states and puts the states in an order that does not follow the usual progression from independent living to death.

Worked Example: Long-Term Care With a Nonstandard State Order

A hypothetical long-term care model has states \(I\) (independent), \(C\) (receiving care at home), \(N\) (in a nursing facility), and \(X\) (dead). Use the order \(N,X,I,C\). Assume constant transition intensities per year and the following permitted direct transitions: \(I\to C: 0.030\), \(I\to X: 0.005\), \(C\to I: 0.120\), \(C\to N: 0.180\), \(C\to X: 0.040\), \(N\to C: 0.060\), and \(N\to X: 0.150\). Construct the off-diagonal entries.

Method. First translate the chosen order into row and column numbers: \(N\) is first, \(X\) second, \(I\) third, and \(C\) fourth. Then locate each arrow from its source row to its destination column.

For \(I\to C\), the source \(I\) is third and the destination \(C\) is fourth, so place \(0.030\) in row 3, column 4. For \(I\to X\), place \(0.005\) in row 3, column 2.

For \(C\to I\), place \(0.120\) in row 4, column 3. For \(C\to N\), place \(0.180\) in row 4, column 1. For \(C\to X\), place \(0.040\) in row 4, column 2.

For \(N\to C\), place \(0.060\) in row 1, column 4. For \(N\to X\), place \(0.150\) in row 1, column 2. Thus the populated off-diagonal cells are:

\begin{array}{c|cccc} \text{Source}\backslash\text{Destination} & N & X & I & C\\ \hline N & & 0.150 & & 0.060\\ X & & & &\\ I & & 0.005 & & 0.030\\ C & 0.180 & 0.040 & 0.120 & \end{array}

Check. There are seven stated direct transitions and exactly seven populated off-diagonal cells. In particular, the care-at-home row is the fourth row, not the second row, and the nursing-facility-to-care-at-home transition is in row \(N\), column \(C\). The empty row for \(X\) reflects that no outgoing transition from \(X\) was listed; this example makes no claim about any diagonal entry.

Common mistake avoided: using a familiar state order instead of the order specified in the problem. The labels, rather than an assumed ordering, determine each row and column.

Rates Must Match the Model's Time Unit

An off-diagonal cell contains an intensity, so its numerical value must be expressed in the time unit used by the model. A rate reported per month cannot be copied unchanged into a model whose time variable is in years. Convert the rate before placing it in the matrix; the row and column still come from the transition's source and destination.

Worked Example: Converting Monthly Rates Before Placement

A hypothetical rehabilitation model has states \(A\) (active) and \(R\) (in rehabilitation), ordered \(A,R\). It also has an absorbing state \(X\) (dead), with full order \(A,R,X\). The stated intensities are \(A\to R: 0.015\) per month, \(A\to X: 0.0005\) per month, \(R\to A: 0.020\) per month, and \(R\to X: 0.004\) per month. Construct the off-diagonal entries for a model with time measured in years. Assume constant rates. No interest or separate mortality table is needed for this transition-matrix task.

Method. There are 12 months in a year, so a constant intensity per month is multiplied by 12 to express it per year. Convert each rate first:

0.015(12)=0.180,\qquad 0.0005(12)=0.006,\qquad 0.020(12)=0.240,\qquad 0.004(12)=0.048.

The first rate is for \(A\to R\), so put \(0.180\) in row \(A\), column \(R\). The \(A\to X\) rate \(0.006\) goes in row \(A\), column \(X\). The \(R\to A\) rate \(0.240\) goes in row \(R\), column \(A\), and the \(R\to X\) rate \(0.048\) goes in row \(R\), column \(X\).

\begin{array}{c|ccc} \text{Source}\backslash\text{Destination} & A & R & X\\ \hline A & & 0.180 & 0.006\\ R & 0.240 & & 0.048\\ X & & & \end{array}

Check. Each entry is now in per-year units, and each rate is in the cell matching its arrow. For example, \(0.240\) per year in row \(R\), column \(A\) represents rehabilitation-to-active movement, not active-to-rehabilitation movement.

Common mistake avoided: copying \(0.015\) into the annual-rate matrix without conversion. That would retain a monthly numerical rate while labeling it per year, understating the annual intensity by a factor of 12.

Common Interpretation Traps

The following checks help catch mistakes before using a matrix in later calculations:

Reversing the arrow. A rate for \(i\to j\) belongs in row \(i\), column \(j\), not row \(j\), column \(i\).
Ignoring the state order. Identify the labels for both the row and column positions before entering numbers.
Assuming a reverse transition. A stated rate for one direction supplies no rate for the opposite direction.
Mixing time units. Convert a monthly or daily rate when the model uses years, or state the units clearly if the model uses a different time scale.
Calling an intensity a finite-period probability. An intensity is an instantaneous rate. For a short interval of length \(h\), the relevant transition probability is approximately \(\mu_{ij}(t)h\), not simply \(\mu_{ij}(t)\).

As a final audit, read every populated cell aloud as “from row state to column state,” and compare that statement with the original transition description. This check is especially useful when a problem supplies a graph in one order but requests a matrix in another.

Key takeaway: For every direct transition \(i\to j\) between distinct states, place \(\mu_{ij}(t)\) in the row for \(i\) and the column for \(j\). The transition direction, state ordering, and time unit must all agree with the model.

Check Your Understanding

For each problem, identify the requested off-diagonal entries using the stated ordering and units.

A model has states \(C\) (covered), \(D\) (disabled), and \(X\) (dead), ordered \(C,X,D\). The permitted rates per year are \(C\to D:0.025\), \(C\to X:0.006\), \(D\to C:0.080\), and \(D\to X:0.120\). Give the four cell locations and values. (Answer: \(C\to D\): row 1, column 3, \(0.025\); \(C\to X\): row 1, column 2, \(0.006\); \(D\to C\): row 3, column 1, \(0.080\); \(D\to X\): row 3, column 2, \(0.120\), all per year.)
States \(A\) (active), \(L\) (in care), and \(X\) (dead) are ordered \(L,A,X\). The constant rates are \(A\to L:0.012\) per month, \(L\to A:0.150\) per year, and \(L\to X:0.060\) per year. For an annual-time model, give the three cells and values. (Answer: \(A\to L\): row 2, column 1, \(0.144\) per year; \(L\to A\): row 1, column 2, \(0.150\) per year; \(L\to X\): row 1, column 3, \(0.060\) per year.)
A retirement model uses state order \(R,D,A\): retired, dead, and active. Its permitted rates per year are \(A\to R:0.040\), \(R\to D:0.090\), \(R\to A:0.200\), and \(A\to D:0.010\). State the location and value for each transition, and identify which rate describes a return to active status. (Answer: \(A\to R\): row 3, column 1, \(0.040\); \(R\to D\): row 1, column 2, \(0.090\); \(R\to A\): row 1, column 3, \(0.200\); \(A\to D\): row 3, column 2, \(0.010\), all per year; \(R\to A\) is the return-to-active rate.)
A hypothetical model has state order \(C,X,A\): receiving care, dead, active. The stated intensities are \(A\to C:24\) per 1,000 active person-years, \(A\to X:5\) per 1,000 active person-years, \(C\to A:0.160\) per year, and \(C\to X:0.090\) per year. Express every entry per year and place it. (Answer: \(A\to C\): row 3, column 1, \(0.024\); \(A\to X\): row 3, column 2, \(0.005\); \(C\to A\): row 1, column 3, \(0.160\); \(C\to X\): row 1, column 2, \(0.090\), all per year.)

What You'll Learn