 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2/3f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

M_{Y} (t) = E [e^{t Y}]

for real

t

where this expectation is finite; a finite neighborhood of zero is usually required when speaking of an existing moment-generating function. For a standard normal distribution, completing the square gives

M_{Y} (t) = \frac{1}{2 π} \int_{R} e^{t y - y^{2} /2} d y = e^{t^{2} /2} \frac{1}{2 π} \int_{R} e^{- (y - t)^{2} /2} d y .

The translated Gaussian density integrates to one, so

M_{Y} (t) = e^{t^{2} /2} (t \in R) .

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