Mixtures of Initial States and Transition Probabilities

Why Initial-State Mixtures Matter

A population may begin at time $s$ in more than one possible state. For example, a hypothetical group of policyholders might include people who are healthy, disabled, or already in a long-term care state. If the starting state is not the same for everyone, the probability of being in a given state at a later time is not simply one row of a transition matrix. It is a weighted combination of the probabilities from each possible starting state.

The weights are the probabilities of the initial states. The quantities being combined are the transition probabilities conditional on those initial states. Keeping those two ingredients distinct is the central calculation skill in this topic.

Definition: Let the possible states be $1,\ldots,m$. An initial-state distribution at time $s$ is a row vector $\boldsymbol{\pi}(s)=(\pi_1(s),\ldots,\pi_m(s))$, where $\pi_i(s)=P(X_s=i)$, each $\pi_i(s)\geq 0$, and $\sum_{i=1}^m\pi_i(s)=1$. Let $p_{ij}(s,t)=P(X_t=j\mid X_s=i)$ be the transition probability from state $i$ at time $s$ to state $j$ at time $t$. The unconditional probability of state $j$ at time $t$ is $\sum_{i=1}^m\pi_i(s)p_{ij}(s,t)$.

Write $\mathbf{P}(s,t)$ for the transition matrix whose $(i,j)$ entry is $p_{ij}(s,t)$. Under the row-vector convention, the endpoint distribution is

\boldsymbol{\pi}(t)=\boldsymbol{\pi}(s)\mathbf{P}(s,t), \qquad \pi_j(t)=\sum_{i=1}^m \pi_i(s)p_{ij}(s,t).

This calculation assumes that the given initial probabilities describe the population at time $s$, and that the transition probabilities apply to individuals conditional on their state at $s$. The transition probabilities must refer to the same interval and the same state definitions. If the model or transition probabilities differ across groups for reasons beyond their starting states, use the appropriate group-specific transition probabilities rather than treating one common matrix as applicable to everyone.

Why the Weighted Sum Is the Right Formula

The events $X_s=i$, for $i=1,\ldots,m$, are mutually exclusive and cover all possible starting states. To finish in state $j$, an individual must start in one of those states and then make the transition to $j$. Therefore, the law of total probability gives

P(X_t=j) =\sum_{i=1}^m P(X_t=j\mid X_s=i)P(X_s=i) =\sum_{i=1}^m p_{ij}(s,t)\pi_i(s).

The matrix product expresses all these sums at once: its $j$th entry multiplies the initial probability of each state $i$ by the transition probability from $i$ to $j$, then adds the contributions. This also explains why the initial distribution is written as a row vector when the transition matrix has starting states in rows and destination states in columns.

Check the initial distribution.
Confirm that its entries are nonnegative and add to 1.

Identify each transition probability.
For endpoint state $j$, use column $j$: the entry in row $i$ is the probability of reaching $j$ from initial state $i$.

Weight and add.
Multiply each entry in that column by the corresponding initial-state probability, and sum the products.

Check the endpoint distribution.
Its entries should be nonnegative and add to 1, subject to rounding.

One-Interval Mixtures

The following example uses a hypothetical three-state model. The given matrix summarizes the transition probabilities over one year; no further assumptions about the transition process are needed to perform the mixture calculation.

Worked Example: A Mixed Starting Population and an Endpoint Payment

A hypothetical portfolio has states Healthy (H), Disabled (D), and Dead (X). At the start of a one-year period, the state distribution is $(0.60,0.30,0.10)$, in that order. The one-year transition matrix is

\mathbf{P}= \begin{pmatrix} 0.70&0.20&0.10\\ 0.10&0.70&0.20\\ 0&0&1 \end{pmatrix}.

Calculate the unconditional probabilities of each endpoint state. Also calculate the expected hypothetical payment of \$12,000 if payable only when the person is Disabled at the end of the year.

Method. Apply the law of total probability to each destination state by weighting its transition probabilities with the initial-state distribution. The row sums are each 1, and the initial probabilities sum to $0.60+0.30+0.10=1.00$.

Healthy at year-end. The transition probabilities to H are $0.70$, $0.10$, and $0$ from H, D, and X, respectively. Thus

P(X_1=\mathrm{H})=(0.60)(0.70)+(0.30)(0.10)+(0.10)(0) =0.42+0.03+0=0.45.

Disabled at year-end. The corresponding probabilities are $0.20$, $0.70$, and $0$, so

P(X_1=\mathrm{D})=(0.60)(0.20)+(0.30)(0.70)+(0.10)(0) =0.12+0.21+0=0.33.

Dead at year-end. The probabilities are $0.10$, $0.20$, and $1$, giving

P(X_1=\mathrm{X})=(0.60)(0.10)+(0.30)(0.20)+(0.10)(1) =0.06+0.06+0.10=0.22.

The endpoint probabilities check: $0.45+0.33+0.22=1.00$. The expected payment is the payment amount multiplied by the probability that it is payable:

E[\text{payment}]=\$12{,}000(0.33)=\$3{,}960.

The result is \$3,960. It is an expected payment per person in the stated starting population, not a payment of \$3,960 to every person. The common mistake this method avoids is averaging the three rows equally despite the unequal initial proportions.

Mixtures Across Consecutive Intervals

If the endpoint is later than one interval, first obtain the transition probabilities over the full time span. For consecutive intervals, the transition matrix for the combined interval is the product of the interval matrices in chronological order under the row-vector convention. The initial mixture is then applied to that combined matrix. Equivalently, one can apply the initial distribution to the first matrix and then to the next.

Worked Example: Two Intervals and a Mixed Initial Distribution

A hypothetical model has two states, Active (A) and Inactive (I). At time 0, $40\%$ of the population is Active and $60\%$ is Inactive. The transition matrices for the first and second intervals are

\mathbf{P}_{01}= \begin{pmatrix} 0.80&0.20\\ 0.10&0.90 \end{pmatrix}, \qquad \mathbf{P}_{12}= \begin{pmatrix} 0.70&0.30\\ 0.20&0.80 \end{pmatrix}.

Find the unconditional state distribution at time 2.

Method. Compose the interval matrices, then weight the resulting rows by the initial distribution $(0.40,0.60)$. The combined transition probabilities from each initial state are

\mathbf{P}_{02} =\mathbf{P}_{01}\mathbf{P}_{12} = \begin{pmatrix} (0.80)(0.70)+(0.20)(0.20)&(0.80)(0.30)+(0.20)(0.80)\\ (0.10)(0.70)+(0.90)(0.20)&(0.10)(0.30)+(0.90)(0.80) \end{pmatrix} = \begin{pmatrix} 0.60&0.40\\ 0.25&0.75 \end{pmatrix}.

For example, the probability of being Active at time 2 given Active at time 0 is $0.60$: the two possible states at time 1 contribute $0.80(0.70)=0.56$ and $0.20(0.20)=0.04$. The combined row sums are $0.60+0.40=1$ and $0.25+0.75=1$.

Now apply the initial mixture. The unconditional probability of Active at time 2 is

$$ (0.40)(0.60)+(0.60)(0.25)=0.24+0.15=0.39. $$

The unconditional probability of Inactive is

$$ (0.40)(0.40)+(0.60)(0.75)=0.16+0.45=0.61. $$

The probabilities sum to $0.39+0.61=1.00$, as required. As a check, propagating interval by interval gives the time-1 Active probability $0.40(0.80)+0.60(0.10)=0.38$, and the time-1 Inactive probability $0.40(0.20)+0.60(0.90)=0.62$. The time-2 Active probability is then $0.38(0.70)+0.62(0.20)=0.266+0.124=0.39$, agreeing with the combined-matrix calculation. This check avoids reversing the order of the interval matrices.

Recovering an Initial Mixture from an Endpoint Probability

Sometimes the initial distribution is described by an unknown mixture proportion, while an unconditional endpoint probability is given. The mixture formula can then be rearranged to find the starting proportion. This is still a weighted-average calculation: it does not require treating an endpoint state as though it revealed an individual's starting state.

Worked Example: Solving for an Unknown Starting Proportion

In a hypothetical two-group population, each person starts in either state A or state B. The one-period probabilities of reaching endpoint state T are $0.20$ from A and $0.50$ from B. The overall probability of being in T at the endpoint is $0.38$. Find the proportion that started in A, assuming everyone starts in one of these two states and the stated transition probabilities apply to both groups.

Method. Let $w$ be the initial proportion in A, so $1-w$ is the initial proportion in B. The unconditional endpoint probability is

$$ 0.20w+0.50(1-w)=0.38. $$

Expanding and solving,

0.20w+0.50-0.50w=0.38, \qquad -0.30w=-0.12, \qquad w=0.40.

Thus $40\%$ started in A and $60\%$ started in B. Substitution checks the answer: $0.20(0.40)+0.50(0.60)=0.08+0.30=0.38$. The result is feasible because $0\leq w\leq1$. A common error is to subtract the two endpoint probabilities in the wrong order and obtain a negative weight; checking the answer in the original weighted-average equation catches that sign error.

Common Mistakes and Interpretation

An unconditional endpoint probability describes the population as a whole, with its actual initial mix. It is not the transition probability from any one starting state unless the population starts entirely in that state. In particular, an endpoint distribution can change if the initial mix changes even when the transition matrix remains the same.

Taking an unweighted average of rows. Rows receive weights from the initial distribution; they are equally weighted only when the initial probabilities are equal.
Using the wrong column. To calculate the probability of one destination state, select that destination's column and weight entries by the probabilities of the corresponding starting states.
Mixing up the interval. Use transition probabilities for the full requested interval, or compose the matrices for consecutive intervals before applying the initial mix.
Ignoring the distribution check. A valid endpoint distribution has nonnegative entries summing to 1. A failure often signals a misplaced weight, an incorrect matrix order, or arithmetic error.

This calculation matters whenever results are required for a heterogeneous starting population, including hypothetical disability, care, or pension-state models. Separating the starting proportions from the conditional transitions makes the calculation transparent and makes it easier to test how a different initial mix would change the overall result. The topic is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the initial-state weights and the stated transition probabilities to find the requested unconditional endpoint result.

A two-state population starts with probabilities $0.70$ in state U and $0.30$ in state V. The one-period transition matrix, with rows and columns ordered U, V, is $\begin{pmatrix}0.80&0.20\\0.10&0.90\end{pmatrix}$. Find the endpoint probabilities. (Answer: $P(U)=0.5900,\ P(V)=0.4100$)
A three-state hypothetical model has initial distribution $(0.20,0.50,0.30)$ and transition matrix $\begin{pmatrix}0.60&0.30&0.10\\0.20&0.50&0.30\\0&0&1\end{pmatrix}$. Find the probability of each endpoint state. (Answer: $(0.2200,0.3100,0.4700)$)
A two-state population has initial distribution $(0.25,0.75)$, followed by consecutive transition matrices $\begin{pmatrix}0.90&0.10\\0.40&0.60\end{pmatrix}$ and $\begin{pmatrix}0.80&0.20\\0.30&0.70\end{pmatrix}$. Find the endpoint distribution after both intervals. (Answer: $(0.5625,0.4375)$)
Initially, $65\%$ of a hypothetical population is in state A and $35\%$ in state B. The probabilities of reaching endpoint state C are $0.40$ from A and $0.70$ from B. Find the unconditional probability of C and the expected payment per person if a payment of \$2,000 is made only in C. (Answer: $P(C)=0.5050$; expected payment $=\$1{,}010.00$)
A two-state starting population is a mixture of A and B. The endpoint probabilities of reaching T are $0.25$ from A and $0.65$ from B. If the unconditional probability of T is $0.45$, find the initial proportion in A. (Answer: $0.5000$, or $50\%$)

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