Why Transition Matrices Must Be Stochastic
A transition matrix records the probabilities of moving between states over a specified time interval. Its entries are not arbitrary model parameters: each must be a probability, and the possible destination states must account for the full probability of what happens. These two requirements give simple but powerful checks for a proposed matrix.
Consider a hypothetical benefit model with states such as active, disabled, and dead. Starting in any one state at time \(s\), the model describes the state occupied at a later time \(t\). The transition probabilities summarise those possible outcomes, whether or not a transition occurs during the interval.
The definition presumes that the listed states are mutually exclusive and collectively exhaustive at time \(t\): the process occupies exactly one of them. It also presumes that the probabilities are defined by a coherent model for each starting state. If the model is Markov, the transition probability depends on the current state and times, not on the earlier path. The stochastic-matrix checks themselves follow from probability rules; they do not require constant transition rates.
The Two Conditions, and Their Probability Basis
First, \(p_{ij}(s,t)\geq 0\), because it is a conditional probability. Second, fix a starting state \(i\). At time \(t\), the events \(\{X_t=1\},\ldots,\{X_t=n\}\) are disjoint and together cover every possible outcome, conditional on \(X_s=i\). Adding their conditional probabilities therefore gives one:
The second equality uses disjointness, and the last uses exhaustiveness. This is why a complete transition matrix is row-stochastic. It also explains when the row-sum rule does not apply as written: if the table omits a possible outcome, its listed entries need not sum to one. Such a table is incomplete as a transition matrix for the full model.
For a finite state model, these checks are necessary for every interval \(s\leq t\). They are not, by themselves, sufficient to show that a proposed matrix describes the intended product or that matrices for different intervals are mutually consistent. The checks answer a more basic question: can this table represent probabilities over the stated complete set of states?
Worked Example: Checking a Three-State Transition Matrix
A hypothetical contract model uses the ordered states active, disabled, and dead. For an interval from time \(s\) to \(t\), its proposed transition matrix is \(P(s,t)=\begin{pmatrix}0.70&0.20&0.10\\0.15&0.65&0.20\\0.05&0.25&0.70\end{pmatrix}\). Check whether it is stochastic, then interpret the row for a policy that is disabled at time \(s\).
Method. Check every entry for nonnegativity, then add the entries across each row. The stated states are assumed to be the complete set of states at time \(t\).
All nine entries are between zero and one, so none is negative. The row totals are \(0.70+0.20+0.10=1.00\), \(0.15+0.65+0.20=1.00\), and \(0.05+0.25+0.70=1.00\). Thus the matrix is stochastic.
The second row applies to a policy disabled at time \(s\). It assigns probability \(0.15\) to being active at \(t\), \(0.65\) to being disabled at \(t\), and \(0.20\) to being dead at \(t\). The probabilities total one, so the row accounts for every possible ending state under the model.
Common mistake avoided. The column totals are \(0.90\), \(1.10\), and \(1.00\). They are not all one, but that does not invalidate this matrix: starting states are arranged in rows, so it is the row totals that must equal one. Requiring column totals of one would impose an extra condition not required of a transition matrix.
Incomplete State Lists and Missing Probability
A row total below one can be evidence that the table has omitted a possible destination. It is not sound to assume automatically that the entries are wrong, or to rescale them so that they sum to one. First identify what the listed states mean and whether other outcomes are possible. For example, if surrender is possible but absent from a table listing only active, disabled, and dead, the missing probability may be the probability of surrender.
If a state is intentionally excluded from the recorded outcomes, the resulting probabilities can form a substochastic table: entries are nonnegative, but some row totals are less than one. Such a table does not describe transitions over the complete state space. The missing row probability has a specific interpretation only if the omitted outcomes are understood; it should not be silently treated as zero or redistributed.
Worked Example: Finding an Omitted Destination State
A hypothetical long-term care model has four possible states: active, disabled, dead, and surrendered. A table reports only the probabilities of ending active, disabled, or dead. Its active-starting row is \((0.82,0.10,0.03)\), and its disabled-starting row is \((0.12,0.70,0.10)\). The model states that these three listed destinations and surrender are exhaustive. Find the surrender probabilities and write the two complete rows in the stated state order.
Method. For each starting state, subtract the sum of the listed destination probabilities from one. This uses the assumption that the omitted outcome is precisely surrender, not an unspecified collection of unmodelled events.
For an active starting state, the listed probabilities total \(0.82+0.10+0.03=0.95\). The remaining probability is \(1.00-0.95=0.05\), so the complete row is \((0.82,0.10,0.03,0.05)\).
For a disabled starting state, the listed probabilities total \(0.12+0.70+0.10=0.92\). The remaining probability is \(1.00-0.92=0.08\), so the complete row is \((0.12,0.70,0.10,0.08)\).
Both complete rows are nonnegative and sum to one: \(0.95+0.05=1.00\) and \(0.92+0.08=1.00\). If surrender had not been specified as the omitted outcome, the same arithmetic would identify a missing total of \(0.05\) or \(0.08\), but would not establish which destination or event it represents.
Common mistake avoided. Dividing the three listed active-row entries by \(0.95\) would make them sum to one, but would change the question to probabilities conditional on ending in one of those three listed states. Those are not the original probabilities when surrender remains a possible outcome.
How a Stochastic Matrix Preserves Total Probability
The row condition has a useful consequence when a starting-state distribution is given. Use the row-vector convention: let \(\boldsymbol{\pi}\) list the probabilities of occupying each starting state, in the same order as the matrix rows. The distribution at time \(t\) is \(\boldsymbol{\pi}P(s,t)\). If the starting probabilities sum to one and the matrix is stochastic, the resulting probabilities also sum to one.
To see why, write \(q_j=\sum_i\pi_i p_{ij}(s,t)\) for the probability of ending in state \(j\). Summing over ending states and rearranging finite sums gives
The third equality uses the row sums. Nonnegative starting probabilities and nonnegative matrix entries also make every \(q_j\) nonnegative. Thus the resulting vector is again a probability distribution.
Worked Example: Updating a Starting-State Distribution
A hypothetical three-state portfolio has states active, disabled, and dead. At time \(s\), its state distribution is \(\boldsymbol{\pi}=(0.60,0.30,0.10)\). Over the interval to \(t\), use \(P(s,t)=\begin{pmatrix}0.70&0.20&0.10\\0.15&0.65&0.20\\0.05&0.25&0.70\end{pmatrix}\). Find the distribution at time \(t\), and check its total.
Method. Multiply the row-vector distribution by the transition matrix. Each ending-state probability is the sum over starting states of the starting probability multiplied by the corresponding transition probability.
The probability of being active at \(t\) is \(0.60(0.70)+0.30(0.15)+0.10(0.05)=0.42+0.045+0.005=0.47\).
The probability of being disabled at \(t\) is \(0.60(0.20)+0.30(0.65)+0.10(0.25)=0.12+0.195+0.025=0.34\).
The probability of being dead at \(t\) is \(0.60(0.10)+0.30(0.20)+0.10(0.70)=0.06+0.06+0.07=0.19\).
Therefore \(\boldsymbol{\pi}P(s,t)=(0.47,0.34,0.19)\), and \(0.47+0.34+0.19=1.00\). The result is a valid probability distribution, as required. This check can catch arithmetic or orientation errors, although a total of one alone does not prove that every component was calculated correctly.
Practical Checks and Exam Traps
When given a proposed transition matrix, use the following checks in order:
Identify what each row and column represents. Under the convention used here, rows are starting states and columns are ending states.
A negative entry is impossible as a probability, even if its row happens to sum to one. An entry greater than one is also impossible.
For a complete state list, add across each row. Each total must equal one, allowing for stated rounding in displayed numerical values.
Determine whether a destination is omitted, a value is misstated, or the table is intentionally conditional on a restricted set of outcomes. Do not normalize entries without a stated reason.
A row total above one is as problematic as a row total below one: probabilities for mutually exclusive destinations cannot collectively exceed one. Small differences may arise when entries have been rounded; in that case, report the displayed total and assess it against the precision given rather than treating rounding as evidence of a new state. But a materially negative entry or an unexplained row total is not repaired by rounding.
A second common trap is to test only the row sums. A matrix with row sums of one but a negative entry is still invalid. A third is to call a table complete because its named states seem plausible. Completeness is a model assumption that must be checked against the scenario: lapses, surrender, or other defined outcomes may require their own state.
Check Your Understanding
For each problem, state whether the proposed table is stochastic or identify the missing probability or calculation needed.
- A two-state hypothetical model has matrix \(\begin{pmatrix}0.91&0.09\\0.14&0.86\end{pmatrix}\). Are its entries consistent with a complete transition matrix? (Answer: Yes; both entries are nonnegative in each row, and the row sums are \(1.00\) and \(1.00\).)
- A three-state table reports the active-starting row as \((0.55,0.32,0.08)\). A fourth state, “withdrawn,” is known to be the only omitted destination. What is the probability of ending withdrawn? (Answer: \(1-(0.55+0.32+0.08)=0.05\).)
- For states \(A,B,C\), a proposed matrix is \(\begin{pmatrix}0.80&0.10&0.10\\0.20&0.60&0.20\\0&0.30&0.70\end{pmatrix}\). Starting probabilities are \((0.50,0.40,0.10)\). Find the ending distribution. (Answer: \((0.48,0.32,0.20)\).)
- A proposed matrix is \(\begin{pmatrix}0.74&0.28&-0.02\\0.10&0.75&0.15\\0&0.25&0.75\end{pmatrix}\). Its rows each sum to one. Is it stochastic, and why? (Answer: No; the entry \(-0.02\) is a negative probability, so the first row is invalid despite summing to one.)
- A hypothetical table covers only “active” and “disabled” destinations. One starting row is \((0.63,0.29)\), and surrender is possible but not listed. What can be concluded without renormalizing the row? (Answer: The listed probabilities total \(0.92\), leaving \(0.08\) for omitted outcomes; if surrender is the only omitted destination, its probability is \(0.08\), but the two-entry row is not a complete transition row.)