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Exam ALTAM · Transition Probabilities · Tutorial 59 of 999

Checking Numerical Transition Matrices

Learn to distinguish a valid transition matrix from an accurate numerical approximation, and to check both reliably.

Beginner 9 min read

What You'll Learn

  • Check nonnegativity and row sums under the row-to-column convention.
  • Use the identity matrix as the zero-duration boundary condition.
  • Recognize how absorbing states constrain the corresponding matrix rows.
  • Compare interval products with coarser and finer numerical approximations.
  • Avoid interpreting matrix multiplication probabilistically without the Markov condition.
  • Diagnose rounding discrepancies without silently changing numerical results.

Introduction

A numerical transition matrix can look plausible and still fail to represent probabilities. One entry may be negative, a row may not sum to one, or a calculation for a zero-length interval may not produce the identity matrix. These checks are quick, but they are meaningful only when the state space, time interval, and matrix convention are clear.

This tutorial uses the row-to-column convention: the entry in row \(i\), column \(j\) is the probability of being in state \(j\) at the end of an interval, given state \(i\) at its start. The checks below can identify structural problems in a numerical result. Passing them does not, by itself, prove that the approximation is accurate or that the model assumptions are appropriate.

Definition: For a finite state space containing all possible states, a transition matrix \(P(s,t)\) has entries \(P_{ij}(s,t)=P(X_t=j\mid X_s=i)\), for \(s\le t\). Each entry is a probability, and each row describes the possible ending states conditional on one starting state. In this tutorial, transition matrices are interpreted under a Markov model: conditional on the state at an intermediate time, the distribution of later states does not depend on the earlier history.

Basic Probability Checks

For a transition matrix on a complete finite state space, every entry must be between zero and one, inclusive, and every row must sum to one. The row condition follows because, conditional on starting in state \(i\), the events “the process ends in state \(j\)” over all states \(j\) are mutually exclusive and exhaustive. Their probabilities must therefore total one.

$$ 0\le P_{ij}(s,t)\le 1,\qquad \sum_j P_{ij}(s,t)=1\quad\text{for every starting state }i. $$

A row sum below one is not automatically acceptable merely because the entries are nonnegative. It may indicate an omitted state, a computation error, or a model that allows probability to leave the listed state space. If the intended model has a complete state space, the matrix is not a valid transition matrix as reported. A model that deliberately removes probability, such as one that stops tracking a process after a specified event, needs to be described accordingly; its matrix over the retained states may be substochastic rather than a transition matrix on a complete state space.

Similarly, do not assume that a displayed row sum of \(0.999\) must be wrong. If entries were rounded for presentation, a small discrepancy may be due to rounding. Check the precision and the size of the possible rounding error. If the calculation itself is under review, use unrounded values where possible. A row sum check is not permission to change an entry to force the total to one.

Worked Example: Auditing a Three-State Matrix

A hypothetical model has states Active, Disabled, and Dead. Dead is absorbing. A numerical calculation gives the one-year matrix below. Check its basic probability structure, including the absorbing-state condition.

$$ P(0,1)= \begin{pmatrix} 0.68&0.22&0.10\\ 0.12&0.73&0.15\\ 0&0&1 \end{pmatrix}. $$

Step 1: Check the entries. Each entry is between zero and one. In particular, there are no negative probabilities or entries greater than one.

Step 2: Add each row. The Active row sums to \(0.68+0.22+0.10=1.00\). The Disabled row sums to \(0.12+0.73+0.15=1.00\). The Dead row sums to \(0+0+1=1.00\). Thus all three rows satisfy the probability total.

Step 3: Check absorption. Since Dead is absorbing, a person starting Dead must remain Dead. The Dead row is \((0,0,1)\), as required. The matrix passes these structural checks. They do not establish that its probabilities are the correct values for the hypothetical model; that requires checking the underlying assumptions and numerical calculation.

Method and mistake avoided. Check each row conditional on its starting state, rather than adding columns. A column sum need not equal one. Also, do not infer that a valid-looking matrix is numerically accurate solely because its entries and row sums pass.

Boundary Behavior: Zero Time and Absorbing States

When the interval has zero duration, the process has no time to change state. Therefore, starting in state \(i\) implies that the ending state is \(i\) with probability one. The transition matrix at that boundary must be the identity matrix, whose diagonal entries are one and whose off-diagonal entries are zero.

$$ P(s,s)=I. $$

This is a boundary check, not a claim that the diagonal entries of a positive-duration transition matrix must equal one. During a positive interval, the process may leave its initial state and may later return; the same-state probability is then determined by the model and duration.

If a state is absorbing, its row must have a one in the column for that state and zeros elsewhere, for every interval under consideration. This condition applies only when the state is truly absorbing in the specified model. A state that is “usually terminal” but permits recovery or re-entry is not absorbing.

Worked Example: Checking a Zero-Duration Boundary

A two-state model has states Working and Retired. A program reports the following matrix for a zero-month interval:

$$ \widetilde P(s,s)= \begin{pmatrix} 0.998&0.002\\ 0.001&0.999 \end{pmatrix}. $$

Step 1: Check ordinary row conditions. The first row sums to \(0.998+0.002=1\), and the second sums to \(0.001+0.999=1\). All entries are nonnegative and no greater than one. The matrix passes those basic checks.

Step 2: Apply the zero-time condition. For zero elapsed time, the required matrix is \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\). The reported matrix has a \(0.002\) probability of moving from Working to Retired and a \(0.001\) probability of moving from Retired to Working, despite there being no elapsed time. It fails the boundary check.

Step 3: Diagnose rather than conceal. The discrepancies might result from a time-unit error, an interval that was not actually zero in the calculation, or a numerical implementation problem. The appropriate action is to inspect the input and computation. Renormalizing the rows would leave the off-diagonal probabilities in place, so it would not fix the zero-time violation.

Method and mistake avoided. Test the boundary condition separately from row sums. A stochastic matrix can still be the wrong matrix for the stated time interval.

Products Across Consecutive Intervals

Suppose \(P_1=P(s,u)\) describes the first interval and \(P_2=P(u,t)\) the next interval, with \(s\le u\le t\). Under the Markov assumption stated above, the transition matrix over the combined interval is \(P_1P_2\) in the row-to-column convention. The entry in row \(i\), column \(j\) is found by summing over each possible state \(k\) at the intermediate time.

$$ P_{ij}(s,t)=\sum_k P_{ik}(s,u)P_{kj}(u,t), \qquad P(s,t)=P(s,u)P(u,t). $$

The Markov condition matters. Conditioning on the intermediate state always gives a sum involving \(P(X_t=j\mid X_u=k,X_s=i)\). To replace that conditional probability with \(P_{kj}(u,t)\), it must not depend on the earlier starting state \(i\), given \(X_u=k\). Without this property, the product of the two matrices need not give the correct combined transition probabilities.

Justification: Under the Markov assumption, for each intermediate state \(k\), \(P(X_t=j\mid X_u=k,X_s=i)=P(X_t=j\mid X_u=k)=P_{kj}(u,t)\). Summing the joint probabilities over the mutually exclusive intermediate states gives \(P(X_t=j\mid X_s=i)=\sum_k P_{ik}(s,u)P_{kj}(u,t)\), which is exactly the \((i,j)\) entry of \(P(s,u)P(u,t)\).

There is also a useful numerical check on a product. If both factors are nonnegative and have row sums one, then their product has those properties too. For a row \(i\), \(\sum_j(P_1P_2)_{ij}=\sum_k P_{1,ik}\sum_j P_{2,kj}=\sum_k P_{1,ik}=1\). This algebraic fact checks the product’s probability structure; the Markov assumption is still needed to interpret that product as the model’s combined transition matrix.

Worked Example: Comparing a Coarse Result With a Finer Partition

For a hypothetical two-state model, a numerical method gives the following transition matrices for two consecutive half-year intervals. Assume the model is Markov and the row-to-column convention is used. Compare their product with a separately reported one-year coarse approximation.

$$ P_1= \begin{pmatrix} 0.90&0.10\\ 0.20&0.80 \end{pmatrix}, \qquad P_2= \begin{pmatrix} 0.80&0.20\\ 0.10&0.90 \end{pmatrix}, \qquad C= \begin{pmatrix} 0.74&0.26\\ 0.25&0.75 \end{pmatrix}. $$

Step 1: Check the interval matrices. The rows of \(P_1\) sum to \(0.90+0.10=1\) and \(0.20+0.80=1\). The rows of \(P_2\) sum to \(0.80+0.20=1\) and \(0.10+0.90=1\). Both matrices are nonnegative and have valid row sums. The coarse matrix \(C\) also has row sums \(0.74+0.26=1\) and \(0.25+0.75=1\).

Step 2: Multiply in interval order. The first interval is \(P_1\), followed by \(P_2\), so the combined matrix is \(P_1P_2\):

$$ P_1P_2= \begin{pmatrix} 0.90(0.80)+0.10(0.10)&0.90(0.20)+0.10(0.90)\\ 0.20(0.80)+0.80(0.10)&0.20(0.20)+0.80(0.90) \end{pmatrix} = \begin{pmatrix} 0.73&0.27\\ 0.24&0.76 \end{pmatrix}. $$

For example, the first entry is \(0.72+0.01=0.73\), while the first-row second entry is \(0.18+0.09=0.27\). The second row entries are \(0.16+0.08=0.24\) and \(0.04+0.72=0.76\). The product rows sum to \(0.73+0.27=1\) and \(0.24+0.76=1\).

Step 3: Compare entries. The absolute differences between the fine-partition product and \(C\) are \(0.01\) in each of the four entries. Thus the maximum absolute entrywise discrepancy is \(0.01\). Both results are structurally valid, but their difference is a sensitivity measure, not proof that either is exact.

Method and mistake avoided. Use the interval order \(P_1P_2\), and state the Markov assumption before interpreting the product. Reversing the factors generally changes the result. A finer partition can be useful for assessing numerical stability, but agreement alone does not verify the model or the method.

Interpreting Finer-Step Sensitivity

For a numerical method that approximates transitions over time-varying intervals, compare the result on a coarse partition with the result obtained after subdividing intervals more finely. The refined calculation uses the transition approximation for each smaller interval and combines them in chronological order. In a Markov model, this combination is a matrix product.

A smaller difference between successive refinements is evidence of numerical stability for the chosen calculation. It is not a universal guarantee of accuracy: the method may have a systematic error, the intensities or time units may be wrong, or the state space may omit a relevant state. Nor must every pair of successive estimates move monotonically toward the true value. Check several partitions when practical, use an appropriate tolerance, and verify the inputs and boundary conditions.

When assessing the difference, be explicit about the comparison. For two matrices \(A\) and \(B\) on the same states and interval, one simple measure is the largest absolute entrywise difference, \(\max_{i,j}|A_{ij}-B_{ij}|\). Compare corresponding entries, not unrelated row totals: both valid matrices have row totals of one, so row sums alone will not reveal whether their individual transition probabilities differ.

Common Exam Trap: A Product Without the Markov Condition

A common error is to multiply interval matrices automatically after conditioning on an intermediate state, without checking whether the later transition probabilities depend on the earlier history. Consider a process with states \(0,1,M\). Suppose \(X_0\) is equally likely to be \(0\) or \(1\), \(X_1=M\) in either case, and \(X_2=X_0\). The later outcome remembers the initial state even though the intermediate state is \(M\).

If one uses the marginal conditional probability given \(X_1=M\), then \(P(X_2=0\mid X_1=M)=0.5\) and \(P(X_2=1\mid X_1=M)=0.5\). Multiplying as if this were a Markov transition row would predict that a process starting in state \(0\) has probability \(0.5\) of ending in state \(0\). But the specified process has \(X_2=X_0\), so that probability is \(1\). Here \(P(X_2=0\mid X_1=M,X_0=0)=1\), not \(0.5\). The intermediate state does not contain all the information needed for the product formula.

Correction: Confirm the Markov assumption, or directly verify that the conditional distribution over the second interval, given the intermediate state, is independent of the earlier state and history. Only then use the standard matrix product as a transition calculation.

A Practical Audit Sequence

1
Fix the interpretation.
Identify the states, interval, and row-to-column convention. Confirm the state space is intended to include every possible ending state.
2
Check each entry and row.
Look for negative values, values above one, and row sums that differ from one. Account for displayed rounding without treating it as a reason to alter unrounded calculations.
3
Check the boundaries.
At zero duration, require the identity matrix. For any state specified as absorbing, check that its row puts probability one on that same state.
4
Check interval combinations and sensitivity.
For consecutive intervals, use chronological matrix order only under the Markov condition. Compare coarse and refined calculations as a diagnostic, not as an automatic proof of accuracy.

These checks are useful in applications where a numerical transition matrix feeds later actuarial calculations: an invalid probability structure can contaminate every subsequent result. They are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. The central distinction is between structural validity—the matrix obeys probability and boundary requirements—and numerical accuracy—its entries adequately approximate the intended model.

Check Your Understanding

For each problem, check the stated conditions and show the arithmetic before comparing with the answer.

  1. A two-state transition matrix is \(\begin{pmatrix}0.83&0.17\\0.06&0.94\end{pmatrix}\). Does it pass the nonnegativity and row-sum checks? (Answer: Yes; both rows are nonnegative and sum to \(1.00\).)
  2. In a three-state model, state 3 is absorbing. A reported matrix is \(\begin{pmatrix}0.40&0.50&0.10\\0.20&0.70&0.10\\0&0&1\end{pmatrix}\). Check the row totals and the absorbing-state row. (Answer: It passes; all three rows sum to one and the third row is \((0,0,1)\).)
  3. Under a Markov assumption, consecutive interval matrices are \(A=\begin{pmatrix}0.85&0.15\\0.30&0.70\end{pmatrix}\) and \(B=\begin{pmatrix}0.90&0.10\\0.20&0.80\end{pmatrix}\). Find the combined matrix in chronological order. (Answer: \(AB=\begin{pmatrix}0.795&0.205\\0.410&0.590\end{pmatrix}\).)
  4. A program reports \(P(s,s)=\begin{pmatrix}0.999&0.001\\0.002&0.998\end{pmatrix}\). Its entries are nonnegative and its rows sum to one. Is it valid for a zero-duration interval? (Answer: No; the required matrix is the identity, so both off-diagonal entries must be zero.)
  5. A coarse approximation is \(\begin{pmatrix}0.70&0.30\\0.40&0.60\end{pmatrix}\), and a refined approximation for the same interval is \(\begin{pmatrix}0.72&0.28\\0.38&0.62\end{pmatrix}\). Find the maximum absolute entrywise difference and state what it does and does not show. (Answer: \(0.02\); it measures the discrepancy between these approximations but does not, by itself, prove which one is accurate.)
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