 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2011/m/2/c/solution

The finite irreducible aperiodic Markov chain converges to its stationary distribution, regardless of its initial law. Therefore

E [X_{j}] \to π_{+} - π_{-}

n \to \infty lim \frac{1}{n} E [S_{n}] = \frac{1 - p - q}{1 - p + q} .

X_{j}

to be independent.

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