Joint Probabilities of Starting and Ending States

From Conditional Transitions to Joint Probabilities

A transition probability answers a conditional question: if a process is in state $i$ at the start of an interval, what is the probability it will be in state $j$ at the end? If the starting state is uncertain, that conditional probability alone does not give the probability of the joint event “starts in $i$ and ends in $j$.” The starting-state probability must also be included.

This distinction is useful whenever a model begins with a population spread across several states. For example, a hypothetical portfolio might contain people in different health states at the valuation date. A transition matrix describes outcomes conditional on each starting state, while a joint table describes the probabilities of each starting-and-ending pair in the whole population.

Definition: Suppose a process can occupy states $1,\ldots,r$ at times $s$ and $t$, with $s<t$. Let $\pi_i=P(X_s=i)$ be the probability of starting in state $i$, and let $p_{ij}(s,t)=P(X_t=j\mid X_s=i)$ be the transition probability from $i$ to $j$ over the interval. The joint probability of starting in $i$ and ending in $j$ is $$ J_{ij}=P(X_s=i,\ X_t=j)=\pi_i p_{ij}(s,t). $$

Here $X_s$ denotes the state occupied at time $s$. We use row distributions: the initial distribution is the row vector $\boldsymbol{\pi}=(\pi_1,\ldots,\pi_r)$, and the transition matrix has entry $p_{ij}(s,t)$ in row $i$, column $j$. Thus row $i$ describes transitions conditional on starting in state $i$. Each $\pi_i$ is nonnegative and the initial probabilities sum to 1; each transition-matrix row is nonnegative and sums to 1.

Constructing the Joint Probability Table

The joint table has one row for each starting state and one column for each ending state. Its entry in row $i$, column $j$ is the initial probability $\pi_i$ multiplied by the corresponding conditional transition probability $p_{ij}(s,t)$. In matrix notation, this amounts to multiplying each row of the transition matrix by the initial probability for that row:

J=\operatorname{diag}(\boldsymbol{\pi})P(s,t), \qquad J_{ij}=\pi_i p_{ij}(s,t).

The initial and final marginal distributions can be recovered from the joint table. Summing a row gives the probability of its starting state, because the process must end in one of the states. Summing a column gives the probability of the corresponding ending state. In vector form, the final distribution is $\boldsymbol{\pi}P(s,t)$.

\sum_{j=1}^{r}J_{ij}=\pi_i, \qquad \sum_{i=1}^{r}J_{ij}=P(X_t=j), \qquad \boldsymbol{\pi}_t=\boldsymbol{\pi}P(s,t).

These identities provide important checks: the joint-table rows must reproduce the initial distribution, its columns must reproduce the final distribution, and all its entries must sum to 1. A valid joint table is not just a collection of transition probabilities; it is a probability distribution over pairs of states.

Worked Example: A Two-State Model Over One Interval

A hypothetical model has states 1 and 2. At time $s$, the state distribution is $\boldsymbol{\pi}=(0.60,0.40)$. Over the next year, the transition matrix is $$ P(s,t)= \begin{pmatrix} 0.80 & 0.20\\ 0.30 & 0.70 \end{pmatrix}. $$ Find the joint probabilities for each starting-and-ending pair and the final state distribution.

Method. Multiply each transition-matrix row by the probability of its starting state. The first row of the joint table is multiplied by $0.60$, and the second by $0.40$.

For a start in state 1, the joint probabilities are $0.60(0.80)=0.48$ for ending in state 1 and $0.60(0.20)=0.12$ for ending in state 2. For a start in state 2, they are $0.40(0.30)=0.12$ and $0.40(0.70)=0.28$, respectively. Thus $$ J= \begin{pmatrix} 0.48 & 0.12\\ 0.12 & 0.28 \end{pmatrix}. $$

The row sums are $0.48+0.12=0.60$ and $0.12+0.28=0.40$, which recover the initial distribution. The final probability of state 1 is the first column sum, $0.48+0.12=0.60$. The final probability of state 2 is $0.12+0.28=0.40$. Therefore the final distribution is $(0.60,0.40)$.

As a check, all four joint entries sum to $0.48+0.12+0.12+0.28=1.00$. The common mistake this method avoids is reporting $0.20$ as the probability of starting in state 1 and ending in state 2. That is only the conditional transition probability; the joint probability is $0.60(0.20)=0.12$.

Why the Product Gives the Joint Probability

The formula follows directly from the multiplication rule for probabilities. For any events $A$ and $B$ with $P(A)>0$, $P(A\cap B)=P(A)P(B\mid A)$. Set $A=\{X_s=i\}$ and $B=\{X_t=j\}$. Then $P(A)=\pi_i$ and $P(B\mid A)=p_{ij}(s,t)$, so

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

The argument does not require time-homogeneous transition probabilities. It requires only that the transition probabilities used correspond to the stated interval and conditioning event. If $\pi_i=0$, the joint probability is zero; the product formula remains valid even though conditional probabilities given a zero-probability starting event need to be specified by the model rather than inferred from observed probabilities.

Summing the joint probabilities across all possible ending states verifies the starting marginal: $$ \sum_{j=1}^{r}P(X_s=i,X_t=j) =\pi_i\sum_{j=1}^{r}p_{ij}(s,t) =\pi_i, $$ because each transition row sums to 1. Summing across starting states gives the final marginal: $$ P(X_t=j)=\sum_{i=1}^{r}P(X_s=i,X_t=j) =\sum_{i=1}^{r}\pi_i p_{ij}(s,t). $$ This is exactly the $j$th component of the row-vector product $\boldsymbol{\pi}P(s,t)$.

Worked Example: Three Starting and Ending States

In a hypothetical long-term care model, the states are Active (A), Receiving care (C), and Exited (E). At the start of a one-year interval, the probabilities are $0.50$, $0.30$, and $0.20$. The transition matrix for that interval, with rows and columns ordered A, C, E, is $$ P(s,t)= \begin{pmatrix} 0.70 & 0.20 & 0.10\\ 0.10 & 0.60 & 0.30\\ 0 & 0 & 1 \end{pmatrix}. $$ Calculate the joint table and the ending-state distribution.

Method. Scale each row by its starting-state probability. For a start in A, multiply the row by $0.50$, obtaining $(0.35,0.10,0.05)$. For a start in C, multiply the row by $0.30$, obtaining $(0.03,0.18,0.09)$. For a start in E, multiply the row by $0.20$, obtaining $(0,0,0.20)$. The joint table is therefore $$ J= \begin{pmatrix} 0.35 & 0.10 & 0.05\\ 0.03 & 0.18 & 0.09\\ 0 & 0 & 0.20 \end{pmatrix}. $$

The row sums are $0.35+0.10+0.05=0.50$, $0.03+0.18+0.09=0.30$, and $0+0+0.20=0.20$, as required. The ending probabilities are obtained from the column sums: $$ P(X_t=A)=0.35+0.03+0=0.38, \quad P(X_t=C)=0.10+0.18+0=0.28, \quad P(X_t=E)=0.05+0.09+0.20=0.34. $$

The final distribution is $(0.38,0.28,0.34)$, and its entries sum to 1. The joint probability of starting Active and ending in care is $0.10$, not $0.20$: the latter is the conditional probability of transition from A to C, before accounting for the $0.50$ probability of starting Active. The absorbing row for E means a process starting in E remains there over this interval; it does not change how the joint entries are calculated.

Combining Consecutive Intervals

If a problem supplies transition matrices for consecutive intervals rather than one matrix for the whole period, first obtain the transition matrix across the full period. Under the Markov property, the probability of moving from state $i$ to state $k$ over two intervals is found by summing over the intermediate state $j$: $p_{ik}(s,u)=\sum_j p_{ij}(s,t)p_{jk}(t,u)$. This is matrix multiplication in chronological order, $P(s,t)P(t,u)$. Once that full-interval matrix is available, the joint table is formed by scaling its rows by the initial distribution.

Worked Example: Joint Endpoints After Two Intervals

A hypothetical two-state process has initial distribution $(0.40,0.60)$. Its transition matrices over the first and second intervals are $$ P_1= \begin{pmatrix} 0.80 & 0.20\\ 0.25 & 0.75 \end{pmatrix}, \qquad P_2= \begin{pmatrix} 0.60 & 0.40\\ 0.10 & 0.90 \end{pmatrix}. $$ Find the joint distribution of the state at the initial time and the state at the end of the second interval.

Method. First multiply the interval matrices. Each entry is a sum over the two possible intermediate states: $$ P_1P_2= \begin{pmatrix} 0.80(0.60)+0.20(0.10) & 0.80(0.40)+0.20(0.90)\\ 0.25(0.60)+0.75(0.10) & 0.25(0.40)+0.75(0.90) \end{pmatrix} = \begin{pmatrix} 0.50 & 0.50\\ 0.225 & 0.775 \end{pmatrix}. $$

Now scale the first row by $0.40$ and the second row by $0.60$: $$ J= \begin{pmatrix} 0.40(0.50) & 0.40(0.50)\\ 0.60(0.225) & 0.60(0.775) \end{pmatrix} = \begin{pmatrix} 0.200 & 0.200\\ 0.135 & 0.465 \end{pmatrix}. $$ The row sums are $0.400$ and $0.600$, recovering the initial distribution. The ending distribution is the column sums $(0.200+0.135,\ 0.200+0.465)=(0.335,0.665)$.

As an independent check, propagate the initial distribution interval by interval. After the first interval it is $(0.40,0.60)P_1=(0.32+0.15,\ 0.08+0.45)=(0.47,0.53)$. After the second interval it is $(0.47,0.53)P_2=(0.282+0.053,\ 0.188+0.477)=(0.335,0.665)$, matching the joint-table column sums. Multiplying in the reverse order would not represent these chronological intervals and generally gives a different result.

Interpretation and Common Errors

A joint table describes endpoint pairs: it gives the probability of a particular state at time $s$ together with a particular state at time $t$. It does not by itself describe which states were occupied between those times. For instance, a start and end in the same state does not establish that the process remained there throughout the interval. Questions about whether a process ever entered a state require a different event calculation.

The following checks are useful when constructing or interpreting a joint table:

Check row scaling. Every entry in starting-state row $i$ is multiplied by the same initial probability $\pi_i$.
Check row totals. The sum of row $i$ must equal $\pi_i$, since all possible ending states are included.
Check column totals. The column sums must equal the final state distribution obtained by propagating the initial distribution.
Check the total. The entries of the full table must sum to 1.
Keep conditional and joint quantities distinct. $p_{ij}(s,t)$ is conditional on the starting state; $\pi_i p_{ij}(s,t)$ is the joint probability of that start and end.

Key takeaway: With a row-vector convention, form the joint table by multiplying each transition-matrix row by the probability of its starting state. Row sums recover the initial distribution; column sums recover the final distribution.

Check Your Understanding

For each question, construct or interpret the joint probabilities over the stated interval. Answers are given for checking.

A two-state model has initial distribution $(0.30,0.70)$ and transition matrix $\begin{pmatrix}0.90&0.10\\0.20&0.80\end{pmatrix}$. Find the joint table and final distribution. (Answer: $J=\begin{pmatrix}0.27&0.03\\0.14&0.56\end{pmatrix}$; final distribution $(0.41,0.59)$.)
A three-state process has initial distribution $(0.20,0.50,0.30)$ and transition matrix $\begin{pmatrix}0.50&0.30&0.20\\0.10&0.70&0.20\\0&0.40&0.60\end{pmatrix}$. Find the joint table and its column sums. (Answer: $J=\begin{pmatrix}0.10&0.06&0.04\\0.05&0.35&0.10\\0&0.12&0.18\end{pmatrix}$; column sums $(0.15,0.53,0.32)$.)
A process starts in state A with probability $0.25$ and in state B with probability $0.75$. The conditional probability of moving from A to B over the interval is $0.40$. What is the joint probability of starting in A and ending in B? (Answer: $0.25(0.40)=0.10$.)
The initial distribution is $(0.50,0.50)$, with consecutive transition matrices $\begin{pmatrix}0.70&0.30\\0.20&0.80\end{pmatrix}$ and $\begin{pmatrix}0.60&0.40\\0.10&0.90\end{pmatrix}$. Find the final distribution after both intervals. (Answer: the product matrix is $\begin{pmatrix}0.45&0.55\\0.20&0.80\end{pmatrix}$, so the final distribution is $(0.325,0.675)$.)
A joint table for two states is $J=\begin{pmatrix}0.24&0.16\\0.18&0.42\end{pmatrix}$. Recover the initial and final distributions, and state the conditional probability of ending in state 2 given a start in state 1. (Answer: initial distribution $(0.40,0.60)$; final distribution $(0.42,0.58)$; conditional probability $0.16/0.40=0.40$.)

What You'll Learn