 Solution

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

Let

p_{n} = P (X_{n} = 1 ∣ X_{0} = 1)

. Conditioning on the previous state gives

p_{n + 1} = (1 - a) p_{n} + b (1 - p_{n}) = b + (1 - a - b) p_{n}

, with

p_{0} = 1

b / (a + b)

; subtracting it leaves a geometric recursion. Therefore

p_{n} = \frac{b}{a + b} + \frac{a}{a + b} (1 - a - b)^{n} .

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