 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/9f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The normalizing constant satisfies

B^{- 1} = \sum_{n \in Z} e^{- n^{2} /2} .

k

E [e^{kY}] = B \sum_{n \in Z} e^{kn - n^{2} /2} = B e^{k^{2} /2} \sum_{n \in Z} e^{- (n - k)^{2} /2} .

k

merely permutes the summation indices, so the final sum is

B^{- 1}

. Consequently

E [e^{kY}] = e^{k^{2} /2} = E [e^{k X}] .

The last equality uses the moment-generating function of the standard normal variable from part (b).

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