Apply the Chapman-Kolmogorov equations to calculate discrete time transition probabilities in the Markov model.

A disability insurer needs the probability that a healthy life today is disabled three years from now. Chapman-Kolmogorov turns one-step rates into any horizon by summing over intermediate states.

Let $X_0, X_1, X_2, \ldots$ be a discrete-time Markov chain on a finite state space $\{0, 1, \ldots, r\}$. The one-step transition probability is

$$p_{ij} = P(X_{n+1} = j \mid X_n = i)$$

For a time-homogeneous chain, $p_{ij}$ does not depend on $n$. Collect entries in matrix $P$ with $p_{ij}$ in row $i$, column $j$. Each row sums to 1.

Let $p_{ij}^{(n)} = P(X_n = j \mid X_0 = i)$. Condition on the state $k$ at intermediate time $m$, apply the Markov property, and sum:

$$p_{ij}^{(m+n)} = \sum_{k=0}^{r} p_{ik}^{(m)} \, p_{kj}^{(n)}$$

In matrix form, $P^{(m+n)} = P^{(m)} \cdot P^{(n)}$. For a time-homogeneous chain this collapses to $P^{(n)} = P^n$.

KEY: CK is matrix multiplication. Row $i$ of $P^{(m)}$ dotted with column $j$ of $P^{(n)}$ yields the $ij$ entry of $P^{(m+n)}$.

Common mistakes

• Forgetting the path through Disabled. Writing ${}_{2}p_x^{01} = p_{00} \cdot p_{01}$ only. You must add $p_{01} \cdot p_{11}$ and any other reachable middle state.
• Squaring a time-inhomogeneous matrix. Using $P_{60}^2$ instead of $P_{60} \cdot P_{61}$ gives the wrong answer whenever rates differ by age.
• Wrong product order. $P^{(m,m+n)} = P_m P_{m+1} \cdots P_{m+n-1}$, left to right. Reversing flips the conditioning age.

Bottom line

• CK equation: ${}_{m+n}p_x^{ij} = \sum_k {}_{m}p_x^{ik} \cdot {}_{n}p_{x+m}^{kj}$, summing over every state $k$ the chain could occupy at the split time.
• Matrix form: time-homogeneous gives $P^{(n)} = P^n$; time-inhomogeneous gives the ordered product $P^{(m,m+n)} = P_m P_{m+1} \cdots P_{m+n-1}$, read left to right.
• Markov property is the engine: conditional on the current state, the future ignores the past, which is what lets you condition on $X_m = k$ and multiply.
• Row sums equal 1 in every transition matrix; absorbing states show 1 on the diagonal and zeros elsewhere, so their column mass only accumulates.

Exam shortcut

For one entry of $P^{(2)}$, write the explicit CK sum over middle state $k$ instead of forming the full product. Three multiplies beats nine when you only need one cell. DECISION: Rates depend on age, multiply distinct $P_x, P_{x+1}, \ldots$ in order. Rates constant across ages, raise the single $P$ to the power $n$. Check absorbing rows first.

The full lesson (about 1,544 words, 10 min read) adds 2 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 1f

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