Fourier Transform of a European Put and Its Convergence Strip

# Fourier transform of a European put


# Fourier transform of a European put












In book The concepts and practice of mathematical finance, in the context of illustrating the stochastic volatility model, the Fourier transform $\hat{P}(\xi, V, T)$ of a European put $P(x, V, T)$ is listed as

$$ \hat{P}(\xi, V, T) = - \frac{K^{i \xi + 1}}{\xi^2 - i \xi} $$

where $x = \log S$ is the logarithm of real world underlying stock values $S$, $\xi$ the respective Fourier variable, $V$ is square vol, $T$ the time maturity, $i$ the imaginary unit. My development of the Fourier integral, using $K - e^x$ as payout for the put, leads to a different result:

$$ \hat{P}(\xi, V, T) = \int e^{i \xi x} ( K - e^x ) dx =\\ \frac{K}{i \xi} e^{i \xi x} - \frac{1}{1 + i \xi} e^{(1 + i \xi) x} = e^{i \xi x} \frac{K + i \xi (K - e^x)}{i \xi - \xi^2} $$

Where have I gone wrong? How to reach the book result?

## Answer by Kevin (score 4, accepted)

https://quant.stackexchange.com/a/66094

The generalised Fourier transform $\hat{P}(z)$ of the payoff of a put option with $P(x)=\max\{e^k-e^x,0\}$ is \begin{align*} \hat{P}(z) &= \int_{-\infty}^\infty e^{izx} \left( e^k - e^x \right)^+ \mathrm{d}x \\ &= \int_{-\infty}^k \left( e^ke^{izx}-e^{i(z-i)x} \right)\mathrm{d}x \\ &= \left[ e^k\frac{e^{izx}}{iz} -\frac{e^{i(z-i)x}}{i(z-i)} \right]_{-\infty}^k \\ &= \left( e^k\frac{e^{izk}}{iz} - \frac{e^{i(z-i)k}}{i(z-i)} \right)-0\\ &= \frac{e^{i(z-i)k}}{iz} - \frac{e^{i(z-i)k}}{i(z-i)} \\ &= -\frac{e^{ik(z-i)}}{z(z-i)}. \end{align*} The computation above is only valid if the summand for $x=-\infty$ indeed equals zero. In general, if $z\in\mathbb{R}$, $\lim\limits_{x\to-\infty}e^{ixz}$ does not make sense since $e^{izx}$ merely describes points on the unit circle around the origin. However, if $z=a+ib$ is complex, $e^{izx}=e^{-bx} e^{iax}$ which at least converges to zero as $x\to-\infty$ if $b=\text{Im}(z)<0$ since it contracts the unit circle into the origin. Equivalently, $\lim\limits_{x\to-\infty} |e^{izx}|=\lim\limits_{x\to-\infty} e^{-bx}=0$ if $b=\text{Im}(z)<0$.

Thus, we require $\text{Im}(z)<0$ for the first summand and $\text{Im}(z-i)<0$ for the latter. Both conditions together lead to $\text{Im}(z)<0$. Consequently, the generalised Fourier transform $\hat{P}(z)$ is only well-defined in the open strip $\mathcal{S}_P=\{z\in\mathbb{C}:\text{Im}(z)<0\}$.

The strip of regularity for a call option is $\mathcal{S}_C=\{z\in\mathbb{C}:\text{Im}(z)>1\}$.

Note: it is no coincidence to have $i$ and $0$ as poles of the payoff transform. You can use inversion theorems to see how they relate to $N(d_1)$ and $N(d_2)$ (or more general exercise probabilities).