Reading Rows and Columns in a Transition Matrix

Introduction

A transition matrix organizes probabilities of moving between the states in a model over a specified time interval. Its entries are easy to misread if the row and column convention is not made explicit. Under the convention used here, the row identifies the state at the start of the interval, and the column identifies the state at the end.

This interpretation matters when using a state model for a hypothetical disability or long-term care contract. A matrix entry is a probability over a stated interval, conditional on the starting state. It is not an instantaneous transition intensity, and it does not by itself describe the sequence of events between the two endpoints.

Definition and Reading Convention

Definition: Let the possible states be labelled \(1,\ldots,n\). For calendar times \(t\) and \(t+s\), define \(p_{ij}(t,t+s)\) as the probability that the process is in state \(j\) at time \(t+s\), given that it is in state \(i\) at time \(t\). The transition probability matrix \(P(t,t+s)\) has entry \(p_{ij}(t,t+s)\) in row \(i\), column \(j\). Thus, row = starting state; column = ending state.

The time interval must be understood when interpreting an entry. For example, \(p_{ij}(t,t+1)\) refers to the probability of being in state \(j\) one year after time \(t\), conditional on being in state \(i\) at time \(t\). If transition behaviour depends on calendar time, the starting time \(t\) is relevant as well as the interval length.

The state model describes the process at its endpoints. An entry for starting in one state and ending in another specifies the probability of those endpoint states; it does not assert that the process made one direct jump or remained in just those two states during the interval.

A transition probability is different from a transition intensity. A probability applies to a stated interval and describes an endpoint event. An intensity describes an instantaneous rate of transition under the model. An intensity cannot be read directly as the probability in a matrix entry.

Reading a Matrix Entry and an Endpoint Event

Suppose a hypothetical long-term care model has three states: Home (H), Assisted living (A), and Nursing facility (N). The following entries are one-year transition probabilities, conditional on the state at the beginning of the year.

P(t,t+1)= \begin{pmatrix} 0.70 & 0.20 & 0.10\\ 0.15 & 0.65 & 0.20\\ 0.05 & 0.25 & 0.70 \end{pmatrix}, \qquad \text{state order }(H,A,N).

The entry in row A, column N is \(0.20\). It means that a person in Assisted living at time \(t\) has probability \(0.20\) of being in a Nursing facility at time \(t+1\). The entry in row N, column A is \(0.25\), which instead describes starting in a Nursing facility and ending in Assisted living. The two entries answer different questions.

To find the probability of ending in a set of states, add the probabilities in the starting-state row that correspond to those endpoint states. The events “end in Assisted living” and “end in a Nursing facility” are mutually exclusive, so their probabilities can be added when the question asks for either endpoint.

Worked Example: Starting in Assisted Living

Hypothetical scenario. Use the one-year matrix above. A person is known to be in Assisted living at the start of the year. Find (a) the probability of ending in a Nursing facility, (b) the probability of ending outside Assisted living, and (c) the probability of ending in a Nursing facility given that the person does not end in Assisted living.

Method. Fix the starting state first, so use row A. Its entries, in the stated column order, are \(0.15, 0.65, 0.20\).

(a) The probability of ending in a Nursing facility is the row A, column N entry:

P(\text{end in N}\mid\text{start in A})=0.20.

(b) Ending outside Assisted living means ending in either Home or a Nursing facility. These endpoint events are disjoint, so add their row A probabilities:

P(\text{end outside A}\mid\text{start in A}) =0.15+0.20=0.35.

(c) Given that the person does not end in Assisted living, the possible endpoints are Home and Nursing facility. The conditional probability is the probability of ending in N divided by the probability of ending outside A:

P(\text{end in N}\mid\text{start in A and end outside A}) =\frac{0.20}{0.35} =\frac{4}{7} \approx 0.5714.

The answer is \(0.20\), \(0.35\), and \(0.5714\), respectively. The conditional probability in part (c) is larger than the unconditional row entry because its denominator excludes the possibility of ending in Home. Common mistake avoided: the requested conditional probability in (c) is not just the row A, column N entry; the conditioning event changes the denominator.

Starting-State Distributions

Sometimes the state at the starting time is uncertain. Let \(\boldsymbol{\pi}(t)=(\pi_1,\ldots,\pi_n)\) be a row vector, where \(\pi_i\) is the probability of starting in state \(i\). To find the probability of an endpoint state, weight the relevant matrix entries by their starting-state probabilities and add.

Result: If the starting-state distribution is the row vector \(\boldsymbol{\pi}(t)\), then the probability of ending in state \(j\) is \(\sum_i \pi_i p_{ij}(t,t+s)\). In row-vector notation, the full endpoint distribution is \(\boldsymbol{\pi}(t)P(t,t+s)\).

This is the law of total probability. The possible starting states divide the population into mutually exclusive groups. For a fixed endpoint \(j\), the event “end in \(j\)” can occur together with any one of those starting states. The probability of the joint event “start in \(i\) and end in \(j\)” is \(\pi_i p_{ij}(t,t+s)\), by the definition of conditional probability. Adding those disjoint joint-event probabilities gives the result.

Worked Example: Combining an Uncertain Starting State

Hypothetical scenario. At the start of a year, a group of insured people has probabilities \(0.60\), \(0.30\), and \(0.10\) of being in Home, Assisted living, and a Nursing facility, respectively. For the coming year, use the one-year transition matrix shown earlier. Find the probability that a randomly selected person ends in each state.

Method. For each endpoint, multiply each starting-state probability by the transition probability from that starting state to the endpoint, then add the three contributions.

For Home, use column H:

\begin{aligned} P(\text{end in H}) &=(0.60)(0.70)+(0.30)(0.15)+(0.10)(0.05)\\ &=0.420+0.045+0.005\\ &=0.470. \end{aligned}

For Assisted living, use column A:

\begin{aligned} P(\text{end in A}) &=(0.60)(0.20)+(0.30)(0.65)+(0.10)(0.25)\\ &=0.120+0.195+0.025\\ &=0.340. \end{aligned}

For a Nursing facility, use column N:

\begin{aligned} P(\text{end in N}) &=(0.60)(0.10)+(0.30)(0.20)+(0.10)(0.70)\\ &=0.060+0.060+0.070\\ &=0.190. \end{aligned}

The resulting endpoint probabilities are \(0.470\) for Home, \(0.340\) for Assisted living, and \(0.190\) for a Nursing facility. As an arithmetic check, they add to \(0.470+0.340+0.190=1.000\). Common mistake avoided: using only the Home row would incorrectly assume everyone started in Home; each starting group must contribute according to its own probability.

Intervals, Endpoints, and Common Errors

A matrix must be tied to the interval for which its entries are defined. The same row and column labels can describe probabilities over different intervals, and those entries need not be the same. Read the time interval in the question before selecting an entry. If the matrix is specified for six months, an entry is not automatically a one-year probability.

Also distinguish “ends in state \(j\)” from “makes a direct transition to state \(j\).” A transition matrix entry is an endpoint probability. If the underlying process can change states more than once during the interval, the entry alone does not tell you the route taken. Questions about intermediate paths require information beyond simply identifying the row and column.

Key takeaway: First identify the starting state and select its row. Then identify the requested endpoint and select its column. For a set of endpoints, add the appropriate entries; for an uncertain start, weight entries by the starting-state probabilities before adding.

A Row-versus-Column Exam Trap

Reversing the convention can produce a plausible-looking answer that answers the opposite question. For example, in the displayed long-term care matrix, the probability of starting in Assisted living and ending in a Nursing facility is found in row A, column N. It is \(0.20\). The value \(0.25\), in row N, column A, describes starting in a Nursing facility and ending in Assisted living. Selecting that entry reverses both the origin and destination.

A reliable check is to write the requested transition in words before looking at the numbers: start in ___; end in ___. Locate the first blank among the rows and the second among the columns. This check is particularly useful when a problem presents a state order that differs from the order used in another part of the question.

Worked Example: A Six-Month Endpoint and a Conditional Probability

Hypothetical scenario. A separate model uses the states Home (H), Assisted living (A), and Nursing facility (N), in that order. Its entries describe six-month endpoint probabilities, conditional on the state at the start of the six months:

P(t,t+0.5)= \begin{pmatrix} 0.68 & 0.24 & 0.08\\ 0.18 & 0.62 & 0.20\\ 0.06 & 0.22 & 0.72 \end{pmatrix}.

A person starts in Home. Find (a) the probability of ending in Assisted living or a Nursing facility, and (b) the probability of ending in Assisted living conditional on not ending in a Nursing facility.

Method. Use row H, which is \((0.68,0.24,0.08)\). For (a), add the probabilities of the two disjoint requested endpoints:

P(\text{end in A or N}\mid\text{start in H}) =0.24+0.08=0.32.

For (b), the conditioning event “not ending in N” consists of the Home and Assisted living endpoints. Its probability is \(0.68+0.24=0.92\). The numerator, ending in Assisted living, has probability \(0.24\). Therefore,

P(\text{end in A}\mid\text{start in H and not end in N}) =\frac{0.24}{0.92} =\frac{6}{23} \approx 0.2609.

The answers are \(0.32\) and \(0.2609\). Both calculations use the six-month row for Home; neither is a one-year probability. Common mistake avoided: do not use \(0.24\) directly for part (b), because the question conditions on the event of not ending in N.

Check Your Understanding

For each question, use the stated row-start, column-end convention and pay attention to the interval and any conditioning event.

A hypothetical two-state model has state order Active, Disabled and one-year matrix \(\begin{pmatrix}0.90&0.10\\0.30&0.70\end{pmatrix}\). A person starts Disabled. What is the probability of being Active after one year? (Answer: \(0.30\))
A hypothetical care model has state order Home, Assisted, Facility. The row for a person starting Assisted is \((0.12,0.68,0.20)\). What is the probability of ending outside Assisted? (Answer: \(0.32\))
A hypothetical three-state model has state order A, B, C and the row for starting in A is \((0.55,0.30,0.15)\). Given that the endpoint is not C, find the probability that the endpoint is B. (Answer: \(0.30/(0.55+0.30)=0.3529\))
A hypothetical population starts in states A, B, C with probabilities \((0.50,0.30,0.20)\). For a specified interval, the probabilities of ending in C from A, B, and C are \(0.10, 0.25,\) and \(0.70\), respectively. Find the overall probability of ending in C. (Answer: \((0.50)(0.10)+(0.30)(0.25)+(0.20)(0.70)=0.265\))
A hypothetical model has state order Home, Assisted, Facility. Its six-month row for starting in Home is \((0.72,0.18,0.10)\). Find the probability of ending in Assisted or Facility, and state whether this is a probability of a direct jump to either state. (Answer: \(0.18+0.10=0.28\); it is an endpoint probability, not necessarily a direct-jump probability.)

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