 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009/m/2/a/solution

For

n = 0

, the identity matrix gives the assertion. Suppose it holds at

n

. Conditioning on the state at time

n

and using the Markov property gives

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

The nonnegative countable sum is justified by the law of total probability. Induction proves the $n$ -step transition matrix is $P^{n}$ .

 New to topics? Read the docs here!