The <Mellin transform of call prices> can be inverted along a vertical contour with real part $x_0>1$: $C(k)=\frac1{2\pi i}\int_{x_0-i\infty}^{x_0+i\infty}\mathbb Ee^{z\xi}/(z(z-1)e^{(z-1)k})\,dz$. A finite <exponential moment> at $x_0$ and the quadratic denominator give absolute convergence.

Contour inversion for call prices

For a fixed realization $\xi=a$ and $\operatorname{Re}z>1$, direct integration gives
$$
\begin{aligned}
\int_{-\infty}^{\infty}(e^a-e^k)^+z(z-1)e^{(z-1)k}\,dk
&=z(z-1)\int_{-\infty}^a(e^a-e^k)e^{(z-1)k}\,dk\\
&=z(z-1)\left(\frac{e^{za}}{z-1}-\frac{e^{za}}z\right)=e^{za}.
\end{aligned}
$$
To interchange the integral and <expectation>, write $x=\operatorname{Re}z>1$ and estimate
$$
\mathbb E\int_{-\infty}^{\infty}\left|(e^\xi-e^k)^+z(z-1)e^{(z-1)k}\right|dk
=\frac{|z(z-1)|}{x(x-1)}\mathbb Ee^{x\xi}<\infty.
$$
The <Fubini's theorem> now proves the <Mellin transform of call prices> identity
$$
\boxed{M(z)=\int_{-\infty}^{\infty}C(k)z(z-1)e^{(z-1)k}\,dk\qquad(\operatorname{Re}z>1).}
$$
The required <exponential moment> is $\mathbb Ee^{x\xi}<\infty$ at the chosen real part; the assertion for every $x>1$ presupposes these <exponential moments> for every such $x$.

Solution

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

{c}

For $C(k)=\mathbb E(e^\xi-e^k)^+$ and $\operatorname{Re}z>1$ in the <exponential moment> domain, the <moment-generating function> satisfies $\mathbb Ee^{z\xi}=\int_{\mathbb R}C(k)z(z-1)e^{(z-1)k}dk$.

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