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

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 211 / 5 / b / Solution by

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

z = x_{0} + i y

. Since

∣ M (z) ∣ \leq E e^{x_{0} ξ}

and

∣ z (z - 1) ∣^{- 1}

is integrable over

y \in R

, the proposed contour integral is absolutely convergent. The same bound justifies the Fubini's theorem when substituting

M (z) = E e^{z ξ}

:

\frac{1}{2 πi} \int_{x_{0} - i \infty}^{x_{0} + i \infty} \frac{M ( z )}{z ( z - 1 ) e ^{(z - 1) k}} d z = e^{k} E [\frac{1}{2 πi} \int_{x_{0} - i \infty}^{x_{0} + i \infty} \frac{e ^{z (ξ - k)}}{z ( z - 1 )} d z] = e^{k} E (e^{ξ - k} - 1)^{+} = C (k) .

The last equality uses the contour identity provided in the original PDF. Consequently

C (k) = \frac{1}{2 πi} \int_{x_{0} - i \infty}^{x_{0} + i \infty} \frac{M ( z )}{f ( z , k )} d z (x_{0} > 1) .

This is contour inversion for call prices. The supplied TeX corrupts the permitted identity into a reciprocal and drops the imaginary units from its endpoints; the PDF has

(e^{a} - 1)^{+}

and the vertical contour used above.

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