 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-2/12f/iii/solution

Using part ii and symmetry of the simple random walk,

E X_{n}^{2} - E M_{n}^{2} = \sum_{k \geq 1} (2 k - 1) P (X_{n} = k) > 0

for

n \geq 1

. Thus $E M_{n}^{2} < E X_{n}^{2} = n$ .

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