 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-201/3/a/solution

ξ_{k} = S_{k} - S_{k - 1}

, then

μ = E ξ_{k} = p - (1 - p) = 2 p - 1.

The centered increments

ξ_{k} - μ

are independent of the past and have mean zero, so

M_{n} = S_{n} - μ n

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