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I am trying to calculate numerically the price of a plain vanilla call through Fourier Transform, by applying the Gil-Pelaez formula. More precisely, we have that

\begin{equation} C(K) = S_0 \Pi_1 - K e^{-r T} \Pi_2, \end{equation}

where

\begin{eqnarray} \Pi_1 & = & \frac{1}{2} + \frac{1}{\pi} \int_0^\infty \mathfrak{Re} \left\{ \frac{\phi(u-i) e^{-\mathrm{i} u \ln(K)}}{\phi(-\mathrm{i}) \mathrm{i}u} \right\} \mathrm{d}u,\\ \Pi_2 & = & \frac{1}{2} + \frac{1}{\pi} \int_0^\infty \mathfrak{Re} \left\{ \frac{\phi(u) e^{-\mathrm{i} u \ln (K)}}{\mathrm{i} u} \right\} \mathrm{d}u \end{eqnarray}

and where

\begin{equation} \phi(u) = \exp \left\{ \mathrm{i} \left( \ln \left( S_0 \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right) u - \frac{1}{2} \sigma^2 u^2 T \right\}. \end{equation}

Although I do the algebra, it seems that that the integrands I get inside $\Pi_1$ and $\Pi_2$ are the same, something which is obviously false. Although I know it's a stupid question, could you please help me on which are the real parts of the integrands? You help is highly appreciated since I have tried calculating them several times.

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  • $\begingroup$ Could you please use latex? $\endgroup$
    – Richi Wa
    Oct 12, 2015 at 18:18

1 Answer 1

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Both integrand are different. One includes $\phi(u-i)$ and the other one simply $\phi(u)$. As one expects, in the Black-Scholes model, $\Pi_1$ and $\Pi_2$ collaps to $\Phi(d_1)$ and $\Phi(d_2)$.

Note firstly that if $X\sim N(\mu,\sigma^2)$, then \begin{align*} \phi_X(u) &= e^{iu\mu-\frac{1}{2}\sigma^2u^2}, \\ \phi_X(u-1) &=\phi_X(u) e^{\mu+\frac{1}{2}\sigma^2}e^{iu\sigma^2},\\ \phi_X(-i) &= e^{\mu+\frac{1}{2}\sigma^2}, \\ \frac{\phi_X(u-i)}{\phi_X(-i)} &= \phi_{\tilde{X}}(u), \end{align*} where $\tilde{X}\sim N(\mu+\sigma^2,\sigma^2)$. Thus, \begin{align*} \Pi_1 &= \frac{1}{2}+\frac{1}{\pi}\int_0^\infty \Re\left(\frac{e^{-i\ln(K)u}\varphi_{\tilde{\ln(S_T})}(u)}{iu}\right)\mathrm{d}u \\ &= 1-F_{\tilde{\ln(S_T)}}\big(\ln(K)\big) \\ &= 1-\Phi\left( \frac{\ln(K)-\left(\ln(S_0)+\left(r-q-\frac{1}{2}\sigma^2\right)T \right)- \sigma^2T}{\sigma\sqrt{T}}\right) \\ &= 1-\Phi\left( -\frac{\ln\left(\frac{S_0}{K}\right)+\left(r-q+\frac{1}{2}\sigma^2\right)T }{\sigma\sqrt{T}}\right) \\ &=1-\Phi(-d_1) \\ &=\Phi(d_1). \end{align*}

The second line applies the Gil-Pelaez formula which reads as follows $$ F_X(x) = \frac{1}{2}-\frac{1}{\pi} \int_0^\infty \Re\left(\frac{e^{-iux}\phi_X(u)}{iu}\right)\mathrm{d}u.$$

The case for $\Pi_2$ is the same and you can recover $\Phi(d_2)$.

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