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  "body": "\n## Proof\n\nWe show by induction on $n$ that the $n$-step transition probabilities are the entries of the matrix power $P^n$. The equation then follows from $P^{m+n} = P^m P^n$.\n\n1. Let $P = (p_{ij})$ be the [[transition matrix|0F1KBVY161]] of the [[Markov chain|033TJBZDG0]], with $p_{ij} = P(X_{r+1} = j \\mid X_r = i)$ whenever $P(X_r = i) \u003e 0$. A row $i$ with $P(X_r = i) = 0$ for all $r$ is not determined by the chain; we fix it as any probability distribution on $S$.\n2. For matrices $A, B$ on $S$ with nonnegative entries, the [[product|076JGP6R3T]] $(AB)_{ij} = \\sum_{k \\in S} a_{ik} b_{kj}$ is a sum of nonnegative terms. Let $P^0$ be the identity matrix and $P^{n+1} = P^n P$.\n3. By induction, each $P^n$ has nonnegative entries and rows that sum to $1$: $\\sum_j (P^n P)_{ij} = \\sum_k (P^n)_{ik} \\sum_j p_{kj} = 1$. Exchanging the two sums is allowed for nonnegative terms, by [[Tonelli's theorem|067SYD2J4B]] for counting measure.\n4. For nonnegative matrices $A, B, C$ on $S$, $((AB)C)_{ij} = \\sum_l \\sum_k a_{ik} b_{kl} c_{lj} = (A(BC))_{ij}$, again by exchanging sums of nonnegative terms. By induction on $n$ with this rule, $P^{m+n} = P^m P^n$ for all $m, n \\ge 0$.\n5. Lemma: let $m \\le r$ and $i, k, j \\in S$. Then $P(X_m = i, X_r = k, X_{r+1} = j) = P(X_m = i, X_r = k) \\, p_{kj}$.\n6. Proof of the lemma: the event $\\set{X_m = i, X_r = k}$ is the countable disjoint union of the events $H = \\set{X_0 = i_0, \\dots, X_r = i_r}$ over the histories with $i_m = i$ and $i_r = k$.\n7. If $P(H) \u003e 0$, then $P(X_r = k) \u003e 0$. The [[Markov property|0HB8SXHPFA]] in the definition of the chain and time-homogeneity give $P(H \\cap \\set{X_{r+1} = j}) = P(H) \\, P(X_{r+1} = j \\mid X_r = k) = P(H) \\, p_{kj}$.\n8. If $P(H) = 0$, both sides of the equation in step 7 are $0$. Summing over all histories $H$, by countable additivity, proves the lemma.\n9. Claim: for $m, n \\ge 0$ and $i \\in S$ with $P(X_m = i) \u003e 0$, $P(X_{m+n} = j \\mid X_m = i) = (P^n)_{ij}$ for all $j$. We argue by induction on $n$.\n10. For $n = 0$, $P(X_m = j \\mid X_m = i)$ is $1$ if $j = i$ and $0$ otherwise. This is the entry $(P^0)_{ij}$ of the identity matrix.\n11. Assume the claim for $n$. By countable additivity over the value $k$ of $X_{m+n}$, $P(X_m = i, X_{m+n+1} = j) = \\sum_{k} P(X_m = i, X_{m+n} = k, X_{m+n+1} = j)$.\n12. By the lemma with $r = m + n$, this equals $\\sum_k P(X_m = i, X_{m+n} = k) \\, p_{kj}$.\n13. By the definition of [[conditional probability|0FJB06AD2P]] and the induction hypothesis, $P(X_m = i, X_{m+n} = k) = P(X_m = i) \\, (P^n)_{ik}$.\n14. So $P(X_m = i, X_{m+n+1} = j) = P(X_m = i) \\sum_k (P^n)_{ik} p_{kj} = P(X_m = i) \\, (P^{n+1})_{ij}$. Divide by $P(X_m = i)$ to get the claim for $n + 1$.\n15. By step 9, the $n$-step transition probability $p^{(n)}_{ij}$ does not depend on $m$, and $p^{(n)}_{ij} = (P^n)_{ij}$. Where no $m$ has $P(X_m = i) \u003e 0$, we set $p^{(n)}_{ij} = (P^n)_{ij}$, as in step 1. So $P^{(n)} = P^n$.\n16. By step 4, $p^{(m+n)}_{ij} = (P^{m+n})_{ij} = (P^m P^n)_{ij} = \\sum_{k \\in S} p^{(m)}_{ik} \\, p^{(n)}_{kj}$. In matrix form, $P^{(m+n)} = P^{(m)} P^{(n)}$.\n\n## Idea\n\nTo go from $i$ to $j$ in $m + n$ steps, the chain must pass through some state $k$ at time $m$. The Markov property says that the chain forgets how it reached $k$. So the probabilities of the two parts of the path multiply, and summing over $k$ is a matrix product.\n",
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