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I am working through Madan-Carr's issue Option valuation using the fast Fourier transform (a copy of said paper can be found online here). On page 6 (equation 11) the authors introduce: \begin{align} \tag{11} z_T(k)=e^{-rT}\int_{-\infty}^{+\infty} \left[ (e^k-e^s){1}_{\{s<k,k<0 \}}+(e^s-e^k){1}_{\{ s>k,k>0 \}} \right] q_T(s)ds, \end{align} where $q_T$ is the risk-neutral density of the log price $s:= \ln(S_T)$.

The authors then proceed by using the Fourier transform of said function $z_T$ above, which brings me to my question.

Question: Is the function $z_T$ integrable? i.e. is it in $L^1( \mathbb{R},dx)$? Or in what sense do the authors permit the usage of Fourier transform in their paper.


My approach: My approach was to focus on \begin{align} C_T(k)&=\int_{k}^{+\infty}1_{k >0} e^{-rT}(e^s-e^k)q_T(s)ds \end{align} and check for is integrability. My crude bounds (using Tonelli's theorem): \begin{align} \label{eq:12} \int_0^\infty |C_T(k)|dk & \leq \int_0^\infty \left( \int_k^{+ \infty} |e^s-e^k| q_T(s)ds \right)dk \nonumber \\ & \leq 2 \int_0^\infty \left( \int_k^{+ \infty} e^s q_T(s)ds \right) dk = 2 \int_0^\infty \left( \int_0^s e^s dk \right) q_T(s)ds \nonumber \\ & \leq 2 \int_0^\infty e^{2s} q_T(s)ds \leq \int_{-\infty}^\infty e^{u} q_T(u/2)du. \end{align} According to the authors (page 3 on the pdf), we have \begin{align} e^{-rT}\int_{- \infty}^\infty e^s q_T(s)ds = S_0 \end{align} which looks promising, since I could use that in my estimate above, however, there is no relationship between $q_T(u/2)$ and $q_T(u)$ I am aware of in order to conclude. What am I missing?

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Assuming that $S_T = e^s$ is lognormal, the density is of the form

$$q_T(s) =\frac{1}{\sqrt{2 \pi}\sigma e^s}\exp\left[ {-\frac{(s - \mu)^2}{2\sigma^2}}\right] = \mathcal{O} \left(\frac{e^{-as^2}}{e^s}\right),$$

and $e^{2s} q_T(s)$ is integrable. Thus,

$$\int_0^\infty |C_T(k)| \, dk \leqslant 2\int_0^\infty e^{2s} q_T(s)\,ds < \infty$$

The authors of the paper are proposing a Fourier-based numerical approach to option pricing that they assert is applicable to a variety of stochastic models other than GBM. A few are mentioned, e.g., jump-diffusion and variance-gamma.

You could check these case by case to ensure the density is sufficiently well-behaved. On the other hand, the equation on page three ensures a finite expectated value $E(S_T)$, and, generally, an assumption of finite variance where $E(S_T^2)= E(e^{2s}) < \infty$ is typical.

Also you could have arrived at a tighter bound

$$\int_0^\infty |C_T(k)| \, dk \leqslant 2\int_0^\infty se^{s} q_T(s)\,ds, $$

which would only fail in the unlikely circumstance that $e^sq_T(s)$ is integrable, but $se^sq_T(s)$ is not.

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  • $\begingroup$ Thanks a lot for your input, I see that under the assumption of lognormal law that would complete the argument. Unfortunately I don't have a good background in mathematical finance and the authors don't seem to explicitly make such an assumption. Intuitively, as the spot price at maturity $T$ of some underlying asset I could imagine that a geometric Brownian motion might be a good model for $S_T$. $\endgroup$
    – Spaced
    Jun 25, 2018 at 15:11
  • $\begingroup$ @Spaced: I added to the answer to address your concern. Don't forget that the motivation behind this paper is to introduce a numerical technique that may be applicable to a broader class of models. $\endgroup$
    – RRL
    Jun 25, 2018 at 15:58
  • $\begingroup$ Thanks for your clarification, that's been really helpful. $\endgroup$
    – Spaced
    Jun 25, 2018 at 16:28

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