Using Gil-Pelaez Inversion to Price Black-Scholes Calls
Summary
The document explains how the two probability terms in a Fourier-based Black-Scholes call price are evaluated with the Gil-Pelaez inversion formula. It addresses why the integrands differ: the first uses the characteristic function shifted by an imaginary unit and normalized by its value at that point, while the second uses the unshifted characteristic function. For a normally distributed log price, the shift and normalization produce a distribution with the same variance and a mean increased by the variance.
Applying Gil-Pelaez to these distributions recovers the familiar normal probabilities associated with the Black-Scholes terms, Φ(d1) and Φ(d2). The derivation shows how the first term corresponds to a tilted log-price distribution and how the second follows similarly from the original distribution. The explanation is specific to the Black-Scholes setting and does not discuss numerical integration choices, convergence, or implementation error; the displayed setup also omits a dividend yield in its initial characteristic function, while the answer introduces one in its derivation.
Key ideas
- The call price is expressed using two probability terms obtained through Fourier inversion.
- The first probability term uses a shifted characteristic function, whereas the second uses the unshifted function.
- For a normal log price, the normalized shift changes the mean while preserving variance.
- The two inversion terms recover the standard Black-Scholes probabilities Φ(d1) and Φ(d2).
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# Gil-Palaez Inversion Formula in Black Scholes world
# Gil-Palaez Inversion Formula in Black Scholes world
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.
## Answer by Kevin (score 2)
https://quant.stackexchange.com/a/46651
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 $\Phi(d_2)$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.