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

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 9F / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

No. A standard counterexample comes from the log-normal distribution. Let

f (x) = \frac{1}{x 2 π} exp (- \frac{( l o g x ) ^{2}}{2}), x > 0,

and, for a fixed

0 < ∣ ε ∣ \leq 1

, let

f_{ε} (x) = f (x) [1 + ε sin (2 π lo g x)] .

This is a nonnegative density distinct from

f

. For every nonnegative integer

n

, substituting

z = lo g x

makes the difference of the

n

th moments proportional to

E [e^{n Z} sin (2 π Z)], Z \sim N (0, 1) .

It is the imaginary part of

E [e^{(n + 2 πi) Z}] = exp (\frac{( n + 2 πi ) ^{2}}{2}),

which vanishes because its phase is

2 πn

. The case

n = 0

also proves that

f_{ε}

is normalized. Thus the two distributions have every finite moment equal but are different. Unbounded random variables need not be determined by their moments.

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