 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-211/5/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

For a fixed realization

ξ = a

and

Re z > 1

, direct integration gives

\int_{- \infty}^{\infty} (e^{a} - e^{k})^{+} z (z - 1) e^{(z - 1) k} d k = z (z - 1) \int_{- \infty}^{a} (e^{a} - e^{k}) e^{(z - 1) k} d k = z (z - 1) (\frac{e ^{z a}}{z - 1} - \frac{e ^{z a}}{z}) = e^{z a} .

To interchange the integral and expectation, write

x = Re z > 1

and estimate

E \int_{- \infty}^{\infty} (e^{ξ} - e^{k})^{+} z (z - 1) e^{(z - 1) k} d k = \frac{∣ z ( z - 1 ) ∣}{x ( x - 1 )} E e^{x ξ} < \infty.

M (z) = \int_{- \infty}^{\infty} C (k) z (z - 1) e^{(z - 1) k} d k (Re z > 1) .

E e^{x ξ} < \infty

at the chosen real part; the assertion for every

x > 1

presupposes these exponential moments for every such

x

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