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Transition Intensities and Probabilities · Tutorial 299 of 699

Initial-State Distributions In State-Probability Estimation

Learn to propagate a single-state or mixed initial distribution through an estimated transition matrix and interpret the resulting state probabilities.

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What You'll Learn

  • Distinguish a transition-matrix row from a population’s initial-state distribution.
  • Calculate state probabilities by multiplying an initial row vector by an estimated transition matrix.
  • Interpret the result as a mixture weighted by the starting population’s state proportions.
  • Check dimensions, state ordering, probability constraints, and the matrix’s time interval.
  • Recognize why changing the initial mix can change overall state probabilities even when the transition matrix stays fixed.

From Transition Matrix to Population State Probabilities

A transition matrix describes how state membership changes over a specified interval. A population, however, may begin that interval spread across several states. To describe its state distribution at the end, the transition information must be combined with the population’s starting composition. This distinction matters in applications: two populations can have the same transition matrix but different state probabilities at follow-up because they began in different proportions of the states.

The calculation is a matrix-vector multiplication. Its interpretation depends on using a consistent convention for whether origins index rows or columns, and on aligning the initial proportions with the matrix’s state ordering. The result is a vector of estimated probabilities, one for each destination state.

Definition: Let the model have \(K\) states, indexed in a fixed order. Write \(\hat{P}(s,t)\) for an estimated transition matrix from time \(s\) to time \(t\), with origin states in rows and destination states in columns. Its \((j,k)\) entry, \(\hat{P}_{jk}(s,t)\), represents the estimated transition probability from state \(j\) at \(s\) to state \(k\) at \(t\). Let \(\hat{\boldsymbol{\pi}}(s)\) be a row vector of initial-state probabilities in that same state order. The estimated state-probability vector at \(t\) is
$$ \hat{\boldsymbol{\pi}}(t) = \hat{\boldsymbol{\pi}}(s)\hat{P}(s,t). $$
Its \(k\)th entry is the estimated probability of occupying state \(k\) at \(t\) for a population with the specified initial distribution.

The initial vector may represent everyone starting in one state, or a mixture. For example, \((1,0,0)\) represents a population entirely in state 1, while \((0.50,0.30,0.20)\) represents a population with 50% in state 1, 30% in state 2, and 20% in state 3. The entries must be nonnegative and sum to one. A matrix estimated from transition data is not itself the population’s state distribution: it describes how each origin state contributes to the destination states.

Why the Multiplication Gives the Mixture

For a fixed destination state \(k\), a person can be in that state at \(t\) after starting in any of the \(K\) states. The law of total probability adds those mutually exclusive possibilities. If the initial probability of state \(j\) is \(\pi_j(s)\), then the probability of starting in \(j\) and ending in \(k\) is \(\pi_j(s)P_{jk}(s,t)\). Summing over origins gives the destination probability:

$$ \pi_k(t) = \sum_{j=1}^{K}\pi_j(s)P_{jk}(s,t). $$

This is exactly the \(k\)th component of the row-vector product \(\boldsymbol{\pi}(s)P(s,t)\). Replacing the population quantities and transition probabilities by their estimates gives the plug-in estimate in the definition. No new transition model is created by this step: the initial distribution supplies the weights, and the matrix supplies the origin-specific transition information.

The calculation assumes that the states are mutually exclusive and collectively exhaustive for the population at each time, that the initial vector and matrix use the same state labels and ordering, and that the matrix refers to the interval and target population of interest. Each row of a valid transition-probability matrix is nonnegative and sums to one. The initial vector must also be a probability vector. These conditions make the resulting vector a probability distribution. The interpretation of the estimated matrix itself relies on the assumptions of the method used to estimate it; multiplication does not remove those assumptions.

Worked Example: Everyone Starts in One State

Suppose a hypothetical three-state model uses the order “well,” “ill,” and “dead.” For a specified interval, an estimated transition matrix is

$$ \hat{P}(s,t)= \begin{pmatrix} 0.70 & 0.20 & 0.10\\ 0.10 & 0.60 & 0.30\\ 0.20 & 0.20 & 0.60 \end{pmatrix}. $$

Each row is nonnegative and sums to one: \(0.70+0.20+0.10=1\), \(0.10+0.60+0.30=1\), and \(0.20+0.20+0.60=1\). Suppose the hypothetical population all begins well, so \(\hat{\boldsymbol{\pi}}(s)=(1,0,0)\). Then

$$ \hat{\boldsymbol{\pi}}(t) = (1,0,0) \begin{pmatrix} 0.70 & 0.20 & 0.10\\ 0.10 & 0.60 & 0.30\\ 0.20 & 0.20 & 0.60 \end{pmatrix} = (0.70,0.20,0.10). $$

For instance, the estimated probability of being ill at \(t\) is \(1(0.20)+0(0.60)+0(0.20)=0.20\). The result selects the first row because the initial distribution places all its weight on the first origin state. Its entries sum to one and retain the state order: 70% well, 20% ill, and 10% dead at \(t\), according to this estimated matrix and specified starting population.

Worked Example: A Mixed Starting Population

Now suppose a hypothetical population begins with 50% well, 30% ill, and 20% dead. Use the same estimated matrix and interval as above. The initial vector is \(\hat{\boldsymbol{\pi}}(s)=(0.50,0.30,0.20)\); its entries are nonnegative and sum to one. The destination-state probabilities are obtained by weighting each column by the corresponding initial proportions:

$$ \begin{aligned} \hat{\boldsymbol{\pi}}(t) &=(0.50,0.30,0.20) \begin{pmatrix} 0.70 & 0.20 & 0.10\\ 0.10 & 0.60 & 0.30\\ 0.20 & 0.20 & 0.60 \end{pmatrix}\\ &=\bigl( 0.50(0.70)+0.30(0.10)+0.20(0.20),\\ &\qquad 0.50(0.20)+0.30(0.60)+0.20(0.20),\\ &\qquad 0.50(0.10)+0.30(0.30)+0.20(0.60) \bigr)\\ &=(0.42,0.32,0.26). \end{aligned} $$

For the first component, \(0.35+0.03+0.04=0.42\). For the second, \(0.10+0.18+0.04=0.32\). For the third, \(0.05+0.09+0.12=0.26\). The estimated probabilities sum to \(0.42+0.32+0.26=1\). Under this model and starting composition, the estimated probability of being dead at \(t\) is 0.26. It is a probability for the mixed population, not the probability for people who began in any one particular state.

Changing the Initial Mix Changes the Result

The matrix entries describe how each origin state contributes to destination states; the initial distribution determines how much each row contributes. Consequently, the same estimated matrix can yield different population state probabilities when applied to populations with different starting compositions. This is not a contradiction: the populations are being assigned different weights across the origin states.

Worked Example: Comparing Two Starting Compositions

Consider a hypothetical two-state setting with states “event-free” and “event.” For the same specified interval and target model, let the estimated matrix be

$$ \hat{P}(s,t)= \begin{pmatrix} 0.80 & 0.20\\ 0.25 & 0.75 \end{pmatrix}. $$

Both rows sum to one. Population A begins with 80% event-free and 20% with the event, so its initial vector is \((0.80,0.20)\). Its estimated follow-up distribution is

$$ (0.80,0.20) \begin{pmatrix} 0.80 & 0.20\\ 0.25 & 0.75 \end{pmatrix} = \bigl(0.80(0.80)+0.20(0.25),\ 0.80(0.20)+0.20(0.75)\bigr) =(0.69,0.31). $$

Thus the estimated probability of being in the event state is \(0.16+0.15=0.31\). Population B instead begins with 40% event-free and 60% with the event, giving

$$ (0.40,0.60) \begin{pmatrix} 0.80 & 0.20\\ 0.25 & 0.75 \end{pmatrix} = \bigl(0.40(0.80)+0.60(0.25),\ 0.40(0.20)+0.60(0.75)\bigr) =(0.47,0.53). $$

The second population’s estimated event-state probability is 0.53, rather than 0.31, even though the matrix is unchanged. The difference follows from the initial mix: more of Population B starts in the origin state whose row assigns a larger probability to the event state at \(t\). These are model-based comparisons of marginal state probabilities; by themselves they do not establish a causal effect of population membership or starting composition.

Checks, Alternatives, and Common Errors

Before interpreting a product, verify that it represents the intended calculation. The initial vector has \(K\) entries, the matrix is \(K\times K\), and both use exactly the same state ordering. The interval in the matrix must match the question, and the starting proportions must describe the population to which the result will apply. Check that the vector and matrix contain no negative probabilities, that the vector sums to one, and that each matrix row sums to one. Small discrepancies may arise from displayed rounding; large discrepancies suggest a coding, transcription, or estimation problem. A resulting vector that fails to sum to one is a warning to inspect these inputs and the multiplication.

If the target population starts entirely in one state, the calculation reduces to selecting that origin row. For a mixed population, multiplying by the initial vector is the weighted average of all rows. An unweighted average of rows is appropriate only when the starting proportions are equal. Likewise, weighting the columns instead of the rows reverses the intended roles under the row-origin convention. When available, directly estimating state proportions at follow-up is another way to obtain a marginal distribution; it answers the same kind of population question only if it refers to the same target population and time. It does not replace the transition matrix when the purpose is to apply transition information to a specified, possibly different, initial mix.

A further limitation is that a correct multiplication cannot repair an unsuitable matrix. If transition behavior differs across groups in ways not represented by the matrix, applying one common matrix to a mixed population may not describe that target adequately. Also, the calculation produces a point estimate; it does not, by itself, quantify sampling uncertainty. Uncertainty assessment requires methods beyond this multiplication.

Common pitfall: “The population starts 50% in state 1 and 50% in state 2, so I should average the entries in each column without checking the state order.” The correction is to represent the initial mix as a row vector and compute \(\hat{\boldsymbol{\pi}}(s)\hat{P}(s,t)\). Each origin row receives the weight of the corresponding starting-state proportion. An unweighted row average is valid only for equal initial proportions, and the labels must match the matrix order.
1
Fix the state order.
Write the same ordered list of states for the initial vector and for the matrix rows and columns.
2
Check the inputs.
Confirm the initial proportions are nonnegative and sum to one, and check that each transition-matrix row is a probability distribution.
3
Multiply and interpret.
Compute the row-vector product, then report each component as an estimated probability for its named destination state at the matrix’s end time.

Check Your Understanding

Answer each question using the row-origin, column-destination convention. Complete solutions follow the questions.

  1. With \(\hat{P}=\begin{pmatrix}0.9&0.1\\0.2&0.8\end{pmatrix}\) and \(\hat{\boldsymbol{\pi}}=(1,0)\), what is the estimated destination distribution?
  2. Using the same matrix, calculate the destination distribution for \(\hat{\boldsymbol{\pi}}=(0.25,0.75)\).
  3. For a valid three-state matrix, what must be true of each row, and why does a valid initial vector and matrix produce a vector that sums to one?
  4. Explain why two populations can have different estimated state distributions at \(t\) when the same transition matrix is used.

Solutions. (1) The initial vector selects the first row: \((1,0)\hat{P}=(0.9,0.1)\). (2) Multiply each column by the corresponding initial weights: the first component is \(0.25(0.9)+0.75(0.2)=0.375\), and the second is \(0.25(0.1)+0.75(0.8)=0.625\). Thus the result is \((0.375,0.625)\), which sums to one. (3) Every matrix entry must be nonnegative and each row must sum to one. If the initial entries \(\pi_j\) are nonnegative and sum to one, then the resulting entries are nonnegative, and their sum is \(\sum_k\sum_j\pi_jP_{jk}=\sum_j\pi_j\sum_kP_{jk}=\sum_j\pi_j=1\). (4) The initial vectors supply different weights to the matrix’s origin rows. Different weights can produce different mixtures of the rows, even though the transition matrix is unchanged.

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