The FFT paper for Option Pricing by Carr and Madan says that the reason for adding a dampener to the call price function is because it is not square-integrable: when the log of strike goes to -infinity, the call price goes to S0. I agree with this but I thought that the square integrable property meant that the integral of square of a function between -+ infinity is finite. So, I have 2 questions:

(1) How does the call price function break the square integrable property?

(2) If the call price is not square integrable why can't we take it's Fourier Transform?

Thank you


2 Answers 2


A real-valued function $f(x)$ is square integrable over $\Bbb{R}$, we write $f \in L^2(\Bbb{R})$, if and only if $$ \int_{-\infty}^{+\infty} f(x)^2 dx < \infty \tag{1} $$ A necessary condition for the above integral to be finite is $$\lim_{\vert x \vert\to\infty} f(x) = 0$$ To convince yourself, think of the interpretation of an integral as the area under a curve: what do you think happens to the integral $(1)$ when $f(x)$ tends to a non-zero asymptotic limit, knowing that the integrand $f(x)^2$ is positive everywhere?

Let $C(k,T)$ denote the price (as of today) of a European call option, expiring at $T$ and struck at $K=e^k$. Since $$\lim_{k\to-\infty} C(k,T) = \lim_{k\to-\infty} \Bbb{E}^\Bbb{Q}_t \left[ (S_T - e^k)^+ B_T^{-1} \right] = B(0,T) F(0,T) \ne 0$$ we have that $C(k,T) \notin L^2$ which answers your first question.

Your second question is more technical. Basically, you are facing a situation where the risk-neutral pdf associated to your diffusion framework is not analytically tractable, so that you cannot evaluate the expression $$ C(k,T) = \Bbb{E}^\Bbb{Q}_t \left[ (S_T - e^k)^+ B_T^{-1} \right] = \int_{-\infty}^{+\infty} (e^{s_T}-e^k)^+ B_T^{-1} \phi(T, s_T) ds_T $$

Still, your model being affine, you know that you can identify the characteristic function of $s_T=\ln(S_T)$ in closed-form, which happens to be the Fourier transform of the pdf $\phi(T,s_T)$: $$\mathcal{G}_g(u) = \int_{-\infty}^{+\infty} e^{ius} \phi(T,s) ds$$

From there you would like to appeal to a result known as the Parseval relation, which would allow you to write \begin{align} C(k,T) &= \int_{-\infty}^{+\infty} \underbrace{(e^{s_T}-e^k)^+ B_T^{-1}}_{f(s_T)} \underbrace{\phi(T, s_T)}_{g(s_T)} ds_T \\ & = \langle f(s_T), g(s_T) \rangle = \frac{1}{2\pi}\langle \mathcal{F}_f(u), \mathcal{G}_g(u) \rangle \tag{Parseval} \\ &= \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathcal{F}_f(u) \mathcal{G}_g(u) du \end{align} which would allow you to exploit your knowledge of the characteristic function $\mathcal{G}_g(u)$: you just need to find the Fourier transform of the discounted payoff. This is exactly what @MJ73550 did in his answer.

The thing is that the Parseval relationship only holds for functions of $f$ and $g$ in $L^2$. Using a similar argument as above it is easy to see that $f \notin L^2$

  • $\begingroup$ Thank you Quantuple for the detailed answer. My second question was slightly different actually. It was related to the fact that we add a dampener to the C(k,T) (exp(-alpha k) for atm options and 1/sinh(alpha k) for otm before fourier transform) because C(k,T) is not square integrable. But actually your first answer pointed to the necessary condition which I had missed. Thank you. $\endgroup$
    – sdeveloper
    Commented Aug 27, 2016 at 9:30

$$E\left[(e^{X_T}-K)^+\right]=E\left[e^{\lambda X_T}e^{-\lambda X_t}(e^{X_T}-K)^+\right]$$ for $\lambda>1$ it is ok to take the fourier transform of $e^{-\lambda x}(e^{x}-K)^+$. Let $\hat{f}_{\lambda,k}$ be the fourier transform of $e^{-\lambda x}(e^{x}-K)^+$ then you get:

$$E\left[(e^{X_T}-K)^+\right] = \int_{\mathbb{R}}E\left[e^{(\lambda +i2\pi\xi) X_T}\hat{f}_{\lambda,k}(\xi)\right]d\xi$$ and then you use Fourier transfrom pricing because you know well how to compute $\phi_{T}(u)=E[e^{u X_T}]$ thus, $$E\left[(e^{X_T}-K)^+\right] = \int_{\mathbb{R}}\phi_T\left(\lambda +i2\pi\xi\right)\hat{f}_{\lambda,k}(\xi)d\xi$$


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