Lewis Option Pricing: Complex Fourier Transforms and Payoff Damping

# How do you derive this Carr-Madan-like equation?


# How do you derive this Carr-Madan-like equation?












How do you derive equation (3) below? The equation is tagged as equation (11) in this paper: http://janroman.dhis.org/finance/IR/Heston%E2%80%93Hull%E2%80%93White%20Model%20Part%20I.pdf

There are parts of this paper I don't understand. I suspect there are some tiny mistakes, which make things harder to unravel. This is how I am reading equation (3).

- $C(t)$ is the price of a call option at time $t$ with strike price $K$ and expiry time $T$.

- $\varphi(z)$ is the characteristic function of $x=\log(S(T))$, there $S$ is the price of the asset at time $T$. That is,

$$\varphi(z) = \mathbb{E}[e^{xt}|F(t)]\tag{1}$$

where $F$ is the filtration. The author does not state it, but I also believe that this expected value is computed under the risk neutral measure with respect to $F(t)$. If $q_T$ is the pdf of $x$ at time $T$ under the risk neutral measure, then

$$\varphi(z) = \int_{-\infty}^{\infty}e^{ixz}q_T(x)\mathrm{d}x\text{.}\tag{2}$$

- $f(x)$ is the payoff of the call option when the price of the asset at time $T$ is $e^{x}$. $e^{-cx}f(x)$ is a transformation of $f$ required to take the Fourier transform. $\hat{f}$ is the fourier transform of $e^{-cx}f(x)$.

Finally, our equation is the complex line integral

$$C(t) = \frac{e^{-r(T-t)}}{2\pi}\int_{c-\infty}^{c+\infty} \varphi(-z) \hat{f}(z)\mathrm{d}z\tag{3}\text{.}$$

This equation seems to be similar to equation (5) from the Carr & Madan paper here. That equation is just the inverse Fourier transform

$$C(t) = \frac{e^{-ck}}{2\pi}\int_{-\infty}^{\infty}e^{ivk}\psi_T(v)\mathrm{d}v\tag{4}$$

where $k=\log(K)$ and $\psi_T$ is the Fourier transform of $e^{ck}C(t)$. A big difference here with my equation (3) is the exchange of $\varphi(-z)$ for $e^{ivk}$.

How do I derive my equation (3)? How is it related to equation (4)? What is $f$ and $\hat{f}$?

## Answer by Kevin (score 5, accepted)

https://quant.stackexchange.com/a/64211

Equation (11) in Kammeyer and Kienitz' paper is a very well-known and popular option pricing formula. It goes back to the work from Lewis (2001), see Theorem 3.2 in Lewis' paper.

### Original Formula From Lewis (2001)

The formula in Lewis for the value of a European-style derivative is $$V(S_0) = \frac{e^{-rT}}{2\pi} \int_{\color{red}{i}\nu-\infty}^{\color{red}{i}\nu+\infty}\varphi_T(-z)\hat{w}(z)\text{d}z,$$ where

- $\nu$ is a real number. It defines the path along which we integrate in the complex plane: $\{z\in\mathbb{C}:\text{Im}(z)=\nu\}$. I give more information on this below.

- $\varphi_T$ is the (generalised) characteristic function of $\ln(S_T)$, the terminal log stock price on which our option payoff depends

- $w$ is the payoff function (as a function of $\ln(S_T)$). For a vanilla call option, $w(s)=\max\{e^s-K,0\}$. The function $\hat{w}$ is the (generalised) Fourier transform of the function $w$.

Note: both $\varphi_T$ and $\hat{w}$ are evaluated at points in the complex plane (not necessarily on the real line!)

### Your Formula

Kammeyer and Kienitz state that the time-$t$ value of a call option is $$C(t)=\frac{e^{-r(T-t)}}{2\pi} \int_{c-\infty}^{c+\infty} \varphi(-z)\hat{f}(z)\text{d}z.$$ First, two important points

- There is a tiny typo in the formula. The integral bounds should be $\color{red}{i}c-\infty$ and $\color{red}{i}c+\infty$.

- Your option pricing formula is for the time-$t$ option price. Thus, everything is conditional on $\mathcal{F}_t$, the filtration generated up to time $t$, see this answer. Lewis (2001) simply sets $t=0$.

The rest is identical to the original formula from Lewis. $\varphi$ is the characteristic function of $\ln(S_T)$, conditional on $\mathcal{F}_t$ and $f$ is the payoff function and $\hat{f}$ is its (generalized) Fourier transform.

### Fourier transform of the payoff function

Let $f(x)=\max\{e^x-e^k,0\}$ be the payoff of a vanilla call option with strike $K=e^k$. This function is not in $L^1$ and has no traditional Fourier transform! It has, however, a generalized Fourier transform. Normally, if $f:\mathbb{R}\to\mathbb{R}$, then we define the Fourier transform (in finance) to be $\hat{f}:\mathbb{R}\to\mathbb{C}, u\mapsto \int_\mathbb{R} e^{iux}f(x)\text{d}x$. For this integral to exist, $f$ needs to decay rapidly or be of compact support. The payoff function does not satisfy this.

The generalized Fourier transform of $f$ is $\hat{f}:\mathcal{S}_f\subset\mathbb{C}\to\mathbb{C}, u\mapsto \int_\mathbb{R} e^{iux}f(x)\text{d}x$. Thus, it is defined for a subset of the complex numbers! As it turns out this $\mathcal{S}_f$ is a horizontal strip in the complex plane. We can compute the transform for the payoff function as follows \begin{align} \hat{f}(u) &= \int_{-\infty}^\infty e^{iux} \left(e^x-e^k\right)^+ \text{d}x \\ &= \int_k^\infty \left(e^{x(iu+1)} - Ke^{iux}\right) \text{d}x \\ &= \left[ \frac{e^{x(iu+1)}}{iu+1} - K\frac{e^{iux}}{iu}\right]_{x=k}^{x=\infty} \\ &= −\frac{e^{ik(u−i)}}{u(u-i)}. \end{align} This last step is only valid if the term indeed vanishes as $x\to\infty$. This only happens if $\text{Im}(z)>1$. Thus, the strip for the call option payoff is $\mathcal{S}_f=\{z\in\mathbb{C}:\text{Im}(z)>1\}$. Similarly, for a put option, we have $\mathcal{S}_f=\{z\in\mathbb{C}:\text{Im}(z)<0\}$. These are the strips of integration which the payoff functions have valid Fourier transform. Note that both strips exclude the real line (i.e., there is no standard Fourier transform).

### Proof of Lewis' Option Pricing Formula

Starting with standard risk-neutral pricing, \begin{align} V &= e^{-rT}\mathbb{E}^\mathbb{Q}[w(\ln(S_T)] \\ &=e^{-rT}\mathbb{E}^\mathbb{Q}\left[\frac{1}{2\pi}\int_{i\nu-\infty}^{i\nu+\infty}e^{-iz\ln(S_T)}\hat{w}(z)\text{d}z\right] \\ &=\frac{e^{-rT}}{2\pi}\int_{i\nu-\infty}^{i\nu+\infty}\mathbb{E}^\mathbb{Q}\left[e^{i(-z)\ln(S_T)}\right]\hat{w}(z)\text{d}z \\ &=\frac{e^{-rT}}{2\pi}\int_{i\nu-\infty}^{i\nu+\infty}\varphi_T(-z)\hat{w}(z)\text{d}z \\ \end{align} Here, we are just using the definition of (inverse) generalized Fourier transforms and Fubini's theorem. A proof using Plancherel's theorem (or Parseval's theorem) is also possible. For Fubini to apply and the integrals to be well-defined, we need to integrate along a path in the complex where all terms are well-defined, hence the $\nu\in\mathcal{S}_V=\mathcal{S}_w\cap\mathcal{S}_f^*$ condition.

## Answer by ir7 (score 3)

https://quant.stackexchange.com/a/64220

Just a note to add to answer above. The damping parameter $c$, real number, becomes the imaginary part of a complex number due to this simple observation:

$$f_{c}(x) := {\rm e}^{-c x}f(x)$$

$$ \hat{f_c}(x) = \int {\rm e}^{ixy}f_c(y) dy = \int {\rm e}^{i(x+ic)y}f(y) dy = \hat{f}(x+ic) $$

(the hat sits on two different functions, $f$ and $f_c$). So (generically):

$$ E[f(X)] = \int {\rm e}^{cy} f_c(y) q_X(y) dy = \int {\rm e}^{cy} \left(1/2\pi \int {\rm e}^{-ixy} \hat{f_c}(x) dx \right) q_X(y) dy $$

$$ \stackrel{Fubini}{=} 1/2\pi\int \left( \int {\rm e}^{-i(x+ic)y} q_X(y) dy \right) \hat{f_c}(x) dx $$

$$ \stackrel{observation}{=} 1/2\pi \int \phi_X(-(x+ic))\hat{f}(x+ic) dx $$

$$ = 1/2\pi \int_{-\infty+ic}^{\infty +ic} \phi_X(-z)\hat{f}(z) dz $$

(With the calculation of $\hat{f}$ for call payoff in the answer above and the relationship (6) between $\phi_T$ and $\psi_T$ in Carr-Madan paper, we should get the reconciliation.)