# 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).

1. $$C(t)$$ is the price of a call option at time $$t$$ with strike price $$K$$ and expiry time $$T$$.
2. $$\varphi(z)$$ 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

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

1. $$f(x)$$ 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 $$\hat{f}$$?

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!)

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.

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.)

• Nice addition, +1! Wouldn't the inverse Fourier transform use $\frac{1}{2\pi}$ instead of $\frac{1}{\pi}$ though? May 25, 2021 at 20:32
• @Kevin I left the integrals without limits :) (I did say 'generically'). The root of such dilemma could be: if we have both $h$ and $\hat{h}$ integrable and $h$ is a real function, then $x \rightarrow e^{-ixy}\hat{h}(x)$ is even, so the limits can be $(-\infty$, $\infty )$ and $2\pi$ is used, or $(0$, $\infty )$ and $\pi$ is used. I'll edit it to $2\pi$.
– ir7
May 25, 2021 at 21:27
• Fully agreed, whenever we express a real-valued option price as $\frac{1}{2\pi}\int_\mathbb{R} e^{-iux}\phi(u)\text{d}u$, then symmetry typically kicks in and we get $\frac{1}{\pi}\int_0^\infty \text{Re}\left(e^{-iux}\phi(u)\right)\text{d}u$. I was only asking because the final line in the answer suggests we integral along the entire (shifted) real line, not just its positive part. Thanks very much for the clarification! :) May 25, 2021 at 23:14