 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2010/m/m5/b/solution

For

n = 0

, the transition probabilities are the entries of the identity

P^{0}

. Suppose the formula holds for

n

. Conditioning on

X_{n} = k

and using the Markov property gives

P (X_{n + 1} = j ∣ X_{0} = i) = \sum_{k \in S} P (X_{n} = k ∣ X_{0} = i) P_{kj} = \sum_{k} (P^{n})_{ik} P_{kj} = (P^{n + 1})_{ij} .

The countable sums have nonnegative terms, so they are justified by the law of total probability. Induction proves

P (X_{n} = j ∣ X_{0} = i) = (P^{n})_{ij} .

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