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 $\hat{f}$?


2 Answers 2


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.


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.)

  • $\begingroup$ Nice addition, +1! Wouldn't the inverse Fourier transform use $\frac{1}{2\pi}$ instead of $\frac{1}{\pi}$ though? $\endgroup$
    – Kevin
    Commented May 25, 2021 at 20:32
  • $\begingroup$ @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 $2\pi$. $\endgroup$
    – ir7
    Commented May 25, 2021 at 21:27
  • $\begingroup$ 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! :) $\endgroup$
    – Kevin
    Commented May 25, 2021 at 23:14

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