 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2010/m/2/c/iii/solution

Within the class

{1, 2}

, the two-step matrix has eigenvalues one and

1/9

(1/2, 1/2)

. Diagonalizing this symmetric matrix, or solving the recursion

p_{n + 1} = 4/9 + p_{n} /9

from

p_{0} = 0

, gives

P (X_{2 n} = 2 ∣ X_{0} = 1) = \frac{1}{2} [1 - (\frac{1}{9})^{n}] .

Consequently the limit is $1/2$ . Restriction to even times is essential because the original chain has period two.

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