1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could you please use latex? $\endgroup$ – Ric Oct 12 '15 at 18:18
1
$\begingroup$

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 $\Phi(d_2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.