Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 9F / c

Table of contents
- Solution c

Solution

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} .

Translation by the integer

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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