As mentioned in Carr-Madan's paper, here, the European call option is: $$ C_T(k)=\frac{e^{\alpha k}}{\pi}\int_0^\infty\mathcal{Re}\left(e^{-iuk}\psi(u)\right)du $$ where $$ \psi(u)=e^{-rT}\frac{\phi_T(u-(\alpha+1)i)}{\alpha^2 + \alpha - u^2 + i(2\alpha+1)u} $$ and $\phi_T(u)$ is the characteristic function for a given process. Please refer to the paper all parameters.

So my question is, this derivation is based on the factor $S_0=1$ and define $k=\ln(K)$.

I am trying to derive the formula for any $S_0$ starting with $k=\ln(K/S_0)$ and $x=\ln(S_T/S_0)$, then end up with something like: $$ \psi(u)=e^{-rT}S_0\frac{\phi_T(u-(\alpha+1)i)}{\alpha^2 + \alpha - u^2 + i(2\alpha+1)u} $$ Intuitively, it looks OK to me. But is there any other sources with a general $S_0$? Any help will be appreciated! Thank you!

  • $\begingroup$ I may be wrong but I remember that formula to allow for a general $S_0$? If I remember correctly, Carr and Madan only assume $S_0=1$ for the second (time value) approach? I once wrote down a general formula for the second case. But the first ``standard'' approach from Carr and Madan incorporates a general $S_0$ via the characteristic function of $S_T$, i.e. $\phi_T(u)$? I may be wrong though and will look at the paper again. $\endgroup$
    – Kevin
    Apr 1, 2020 at 8:53
  • $\begingroup$ Thank you @KeSchn, I think you may be correct that Carr and Madan incorporate a general $S_0$ via the characteristic function of $S_T$. I'll take a look at details as well, but feel free to answer this question. $\endgroup$ Apr 1, 2020 at 12:40
  • $\begingroup$ Given that you can incorporate a general $S_0$ via $\phi_T(u)$, why would you still need the Fourier transform of the call option with respect to $\ln\left(\frac{K}{S_0}\right)$? Can't you simply take their formula and plug in your general $S_0$? $\endgroup$
    – Kevin
    Apr 1, 2020 at 12:51
  • $\begingroup$ I think I need to make sure the meaning of $k=\ln(K/S_0)$ is correct before I can plug into the formula? Here is the part I felt confused... $\endgroup$ Apr 1, 2020 at 13:07
  • $\begingroup$ Just take the formulae from the paper, use $\phi_T(u)$ according to your model which includes $S_0$ and set $k=\ln(K)$. If you follow the formulae from the paper, you’ll get correct option prices $\endgroup$
    – Kevin
    Apr 1, 2020 at 14:44


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