Fourier Transform Methods for Black-Scholes¶

The heat equation approach solved the Black-Scholes PDE by transforming variables until the equation became the classical diffusion equation. Fourier transform methods take a more direct route: rather than reducing to a known PDE, they diagonalize the differential operator itself. In frequency space, derivatives become multiplication, the PDE collapses to an algebraic equation, and the solution is recovered by inverse transformation. Crucially, the characteristic function \(\phi_T(\omega) = \mathbb{E}^{\mathbb{Q}}[e^{i\omega \ln S_T}]\) that appears in the Fourier solution is the Fourier transform of the transition density --- the same Gaussian kernel derived in the heat equation section and the same density under which the Feynman-Kac expectation is computed. The three core methods are united here: the spectral viewpoint decomposes the same pricing operator into its frequency components.

This section is organized in two parts. Part A develops the analytic solution: transform of the PDE, characteristic functions, and the damping trick. Part B addresses the computational problem of evaluating the inverse Fourier integral efficiently via Carr-Madan, FFT, and Gil-Pelaez formulas.

Part A --- Analytic Solution¶

Fundamental Theory¶

Fourier Transform Definition¶

For a function \(f: \mathbb{R} \to \mathbb{C}\):

\[ \boxed{\hat{f}(\omega) = \mathcal{F}[f](\omega) = \int_{-\infty}^{\infty} f(x)e^{-i\omega x}dx} \]

Inverse transform:

\[ \boxed{f(x) = \mathcal{F}^{-1}[\hat{f}](x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega x}d\omega} \]

Key Properties¶

Linearity:

\[ \mathcal{F}[\alpha f + \beta g] = \alpha\hat{f} + \beta\hat{g} \]

Derivative:

\[ \boxed{\mathcal{F}\left[\frac{df}{dx}\right](\omega) = i\omega\hat{f}(\omega)} \]

\[ \boxed{\mathcal{F}\left[\frac{d^2f}{dx^2}\right](\omega) = -\omega^2\hat{f}(\omega)} \]

Shift:

\[ \mathcal{F}[f(x-a)](\omega) = e^{-i\omega a}\hat{f}(\omega) \]

Convolution theorem:

\[ \boxed{\mathcal{F}[f * g] = \hat{f} \cdot \hat{g}} \]

where \((f * g)(x) = \int_{-\infty}^{\infty}f(x-y)g(y)dy\).

Parseval's Identity¶

\[ \boxed{\int_{-\infty}^{\infty}|f(x)|^2 dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}|\hat{f}(\omega)|^2 d\omega} \]

Energy is preserved under the transform.

Black-Scholes PDE in Fourier Space¶

Log-Price Formulation¶

As in the heat equation section, define \(x = \ln(S/K)\) and \(\tau = T - t\). The Black-Scholes PDE becomes:

\[ \boxed{\frac{\partial V}{\partial \tau} = \frac{\sigma^2}{2}\frac{\partial^2 V}{\partial x^2} + \left(r - \frac{\sigma^2}{2}\right)\frac{\partial V}{\partial x} - rV} \]

with terminal condition \(V(x,0) = \Phi(Ke^x)\).

Apply Fourier Transform¶

Transform in the \(x\) variable:

\[ \hat{V}(\omega,\tau) = \int_{-\infty}^{\infty}V(x,\tau)e^{-i\omega x}dx \]

Using the derivative properties:

\[ \frac{\partial \hat{V}}{\partial \tau} = -\frac{\sigma^2\omega^2}{2}\hat{V} + i\omega\left(r - \frac{\sigma^2}{2}\right)\hat{V} - r\hat{V} \]

\[ \boxed{\frac{\partial \hat{V}}{\partial \tau} = \psi(\omega)\hat{V}(\omega,\tau)} \]

where the characteristic exponent is:

\[ \boxed{\psi(\omega) = -\frac{\sigma^2\omega^2}{2} + i\omega\left(r - \frac{\sigma^2}{2}\right) - r} \]

ODE in Fourier Space¶

This is a first-order ODE (no longer a PDE):

\[ \hat{V}(\omega,\tau) = \hat{V}(\omega,0)e^{\psi(\omega)\tau} \]

With \(\hat{V}(\omega,0) = \hat{\Phi}(\omega)\):

\[ \boxed{\hat{V}(\omega,\tau) = \hat{\Phi}(\omega)e^{\psi(\omega)\tau}} \]

Solution via Inverse Transform¶

\[ \boxed{V(x,\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{\Phi}(\omega)e^{\psi(\omega)\tau}e^{i\omega x}d\omega} \]

This is the complete solution in closed form.

Characteristic Functions¶

Definition¶

For a random variable \(X\), the characteristic function is:

\[ \boxed{\phi_X(\omega) = \mathbb{E}[e^{i\omega X}]} \]

This is the Fourier transform of the probability density.

Properties¶

\(\phi_X(0) = 1\)
\(|\phi_X(\omega)| \leq 1\)
\(\phi_X(-\omega) = \overline{\phi_X(\omega)}\) (conjugate)
If \(Y = aX + b\): \(\phi_Y(\omega) = e^{i\omega b}\phi_X(a\omega)\)
Sum of independent RVs: \(\phi_{X+Y} = \phi_X \cdot \phi_Y\)

Inversion Formula¶

The density can be recovered:

\[ \boxed{f_X(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\phi_X(\omega)e^{-i\omega x}d\omega} \]

For Lognormal (Black-Scholes)¶

Under \(\mathbb{Q}\), \(X_\tau = \ln(S_T/S_0)\) satisfies:

\[ X_\tau \sim N\left[\left(r-\frac{\sigma^2}{2}\right)\tau, \sigma^2\tau\right] \]

The characteristic function:

\[ \boxed{\phi_X(\omega,\tau) = \exp\left[i\omega\left(r-\frac{\sigma^2}{2}\right)\tau - \frac{\sigma^2\omega^2\tau}{2}\right]} \]

Connection to PDE Solution¶

Note that:

\[ e^{\psi(\omega)\tau} = e^{-r\tau}\phi_X(\omega,\tau) \]

So:

\[ \hat{V}(\omega,\tau) = e^{-r\tau}\hat{\Phi}(\omega)\phi_X(\omega,\tau) \]

This shows: Fourier transform of option value = (discounted) Fourier transform of payoff times characteristic function.

European Call Option¶

The Challenge: Non-Integrability¶

For a call: \(\Phi(S) = (S - K)^+ = (Ke^x - K)^+ = K(e^x - 1)^+\)

The Fourier transform:

\[ \hat{\Phi}(\omega) = K\int_0^{\infty}(e^x - 1)e^{-i\omega x}dx \]

This integral diverges for real \(\omega\).

Complex Analysis Solution¶

Extend to complex \(\omega = \xi + i\eta\):

\[ \hat{\Phi}(\omega) = K\int_0^{\infty}(e^x - 1)e^{-i\xi x}e^{\eta x}dx \]

For \(\eta < -1\):

\[ \hat{\Phi}(\omega) = K\left[\int_0^{\infty}e^{(1-i\xi+\eta)x}dx - \int_0^{\infty}e^{(-i\xi+\eta)x}dx\right] \]

\[ = K\left[\frac{1}{-1+i\xi-\eta} - \frac{1}{i\xi-\eta}\right] \]

\[ = K\left[\frac{i\xi-\eta - (-1+i\xi-\eta)}{(-1+i\xi-\eta)(i\xi-\eta)}\right] \]

\[ \boxed{\hat{\Phi}(\omega) = \frac{K}{(i\omega+1)(i\omega)} = \frac{K}{\omega^2 - i\omega}} \]

for \(\text{Im}(\omega) < -1\).

Carr-Madan Damping Approach¶

Instead of dealing with complex \(\omega\), damp the payoff:

\[ \tilde{C}(x) = e^{\alpha x}C(x) \]

where \(\alpha > 0\) is chosen so that \(\tilde{C}\) decays as \(x \to \infty\).

For a call, need \(\alpha > 1\) to ensure:

\[ e^{\alpha x}(e^x - 1)^+ = (e^{(\alpha+1)x} - e^{\alpha x})\mathbb{1}_{x>0} \to 0 \text{ as } x \to \infty \]

Fourier Transform of Damped Call¶

\[ \hat{\tilde{\Phi}}(\omega) = K\int_0^{\infty}(e^{(\alpha+1)x} - e^{\alpha x})e^{-i\omega x}dx \]

\[ = K\left[\frac{1}{\alpha+1-i\omega} - \frac{1}{\alpha-i\omega}\right] \]

\[ = K\frac{(\alpha-i\omega) - (\alpha+1-i\omega)}{(\alpha+1-i\omega)(\alpha-i\omega)} \]

\[ \boxed{\hat{\tilde{\Phi}}(\omega) = \frac{-K}{(\alpha+i\omega)(\alpha+1+i\omega)}} \]

Or equivalently:

\[ \boxed{\hat{\tilde{\Phi}}(\omega) = \frac{K}{\alpha^2 + \alpha - \omega^2 + i(2\alpha+1)\omega}} \]

Unification: Three Representations of One Object¶

With the Fourier derivation complete, all three core methods of this chapter have now been presented. Each solves the same Black-Scholes PDE, and each arrives at the same pricing formula --- but through a different representation of the pricing operator \(\mathcal{P}_{\tau}\). The heat equation writes the solution as a convolution with the Gaussian kernel \(G\). The Feynman-Kac formula writes it as an expectation under the transition density --- which is the kernel \(G\). The Fourier transform diagonalizes the operator, expressing the solution through the characteristic function \(\phi_T(\omega) = \mathbb{E}^{\mathbb{Q}}[e^{i\omega \ln S_T}]\) --- which is the Fourier transform of that same density. These are not three independent results. They are one theorem, viewed in spatial, probabilistic, and spectral coordinates.

Part B --- Computational Methods¶

The analytic derivation is complete: the Fourier transform converts the Black-Scholes PDE into an algebraic equation, and the option price is recovered by inverse transformation. What remains is a computational problem: evaluating the inverse integral

\[ V(x,\tau) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega x}\, \hat{V}(\omega,\tau)\, d\omega \]

efficiently across an entire grid of strikes. The Carr-Madan formula, FFT algorithm, and Lewis (Gil-Pelaez) inversion below solve this problem. They are not new analytic solutions; they are algorithms for evaluating the inversion integral. Their importance lies in the fact that they extend unchanged to any model whose characteristic function is known in closed form (Heston, Merton, variance gamma, etc.), making them the computational backbone of modern derivative pricing.

Carr-Madan Formula¶

Modified Call Price¶

Define:

\[ c_T(k) = e^{\alpha k}C(K = e^k, S_0, T) \]

where \(k = \ln K\) is the log-strike.

Fourier Transform¶

\[ \psi_T(\omega) = \int_{-\infty}^{\infty}e^{i\omega k}c_T(k)dk \]

Using the characteristic function \(\phi_T(\omega)\) of \(\ln(S_T/S_0)\):

\[ \boxed{\psi_T(\omega) = \frac{e^{-rT}\phi_T(\omega - (\alpha+1)i)}{\alpha^2 + \alpha - \omega^2 + i(2\alpha+1)\omega}} \]

Inversion¶

\[ c_T(k) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega k}\psi_T(\omega)d\omega \]

Taking the real part (since \(c_T\) is real):

\[ \boxed{C(K,S_0,T) = \frac{e^{-\alpha k}}{\pi}\int_0^{\infty}\text{Re}\left[e^{-i\omega k}\psi_T(\omega)\right]d\omega} \]

Explicit Formula for Black-Scholes¶

For Black-Scholes with \(\phi_T(\omega) = \exp[i\omega(r-\frac{\sigma^2}{2})T - \frac{\sigma^2\omega^2 T}{2}]\):

\[ \boxed{C(K) = \frac{e^{-\alpha k}}{\pi}\int_0^{\infty}\text{Re}\left[\frac{e^{-i\omega k}\,e^{-rT}\exp\bigl[i(\omega-(\alpha+1)i)(r-\frac{\sigma^2}{2})T - \frac{\sigma^2(\omega-(\alpha+1)i)^2T}{2}\bigr]}{\alpha^2+\alpha-\omega^2+i(2\alpha+1)\omega}\right]d\omega} \]

Fast Fourier Transform (FFT)¶

Discrete Fourier Transform¶

For \(N\) points \(\{x_j\}_{j=0}^{N-1}\):

\[ \boxed{X_k = \sum_{j=0}^{N-1}x_j e^{-2\pi ijk/N}, \quad k = 0,1,\ldots,N-1} \]

Inverse:

\[ \boxed{x_j = \frac{1}{N}\sum_{k=0}^{N-1}X_k e^{2\pi ijk/N}} \]

FFT Algorithm¶

The Fast Fourier Transform computes the DFT in:

\[ \boxed{O(N\log N) \text{ operations}} \]

instead of \(O(N^2)\) for naive implementation.

This is based on the Cooley-Tukey algorithm using divide-and-conquer.

Application to Option Pricing¶

Setup:

Choose \(N = 2^n\) (power of 2 for FFT)
Discretize log-strike: \(k_u = k_0 + u\Delta k\) for \(u = 0,1,\ldots,N-1\)
Discretize frequency: \(\omega_j = j\Delta\omega\) for \(j = 0,1,\ldots,N-1\)
Set grid spacing: \(\Delta k \cdot \Delta\omega = \frac{2\pi}{N}\)

Discretization¶

The integral:

\[ c_T(k_u) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega k_u}\psi_T(\omega)d\omega \]

becomes:

\[ c_T(k_u) \approx \frac{\Delta\omega}{2\pi}\sum_{j=0}^{N-1}e^{-i\omega_j k_u}\psi_T(\omega_j) \]

\[ = \frac{\Delta\omega}{2\pi}\sum_{j=0}^{N-1}e^{-2\pi iju/N}\psi_T(\omega_j) \]

This is an inverse DFT.

Implementation Steps¶

Choose \(\alpha\) (typically \(\alpha = 0.75\) or \(\alpha = 1.5\))
Set \(\Delta k\) (strike spacing), compute \(\Delta\omega = \frac{2\pi}{N\Delta k}\)
Compute \(\psi_T(\omega_j)\) for \(j = 0,\ldots,N-1\)
Apply FFT to get \(\{c_T(k_u)\}\)
Extract \(C(K_u) = e^{-\alpha k_u}c_T(k_u)\)

Complexity¶

Direct integration: \(O(N^2)\) for \(N\) strikes
FFT: \(O(N\log N)\) for \(N\) strikes simultaneously

Speedup factor: \(\frac{N}{\log_2 N}\), e.g., for \(N = 1024\): speedup \(\approx 100\times\).

Lewis Formula (Gil-Pelaez)¶

Alternative Inversion¶

The Gil-Pelaez inversion theorem avoids damping:

\[ \boxed{C(K,S,T) = \frac{S}{2} + \frac{1}{\pi}\int_0^{\infty}\text{Re}\left[\frac{e^{-i\omega \ln(K/S)}\phi(\omega-i)}{i\omega}\right]d\omega - \frac{Ke^{-rT}}{2}} \]

\[ \boxed{\quad - \frac{Ke^{-rT}}{\pi}\int_0^{\infty}\text{Re}\left[\frac{e^{-i\omega\ln(K/S)}\phi(\omega)}{i\omega}\right]d\omega} \]

where \(\phi(\omega)\) is the characteristic function of \(\ln(S_T/S_0)\) under \(\mathbb{Q}\).

Simplified Form¶

Combining terms:

\[ \boxed{C = S\cdot\Pi_1 - Ke^{-rT}\cdot\Pi_2} \]

where:

\[ \Pi_1 = \frac{1}{2} + \frac{1}{\pi}\int_0^{\infty}\text{Re}\left[\frac{e^{-i\omega\ln(K/S)}\phi(\omega-i)}{i\omega}\right]d\omega \]

\[ \Pi_2 = \frac{1}{2} + \frac{1}{\pi}\int_0^{\infty}\text{Re}\left[\frac{e^{-i\omega\ln(K/S)}\phi(\omega)}{i\omega}\right]d\omega \]

Connection to Black-Scholes¶

For lognormal \(\phi\), after evaluation of integrals:

\(\Pi_1 = N(d_1)\)
\(\Pi_2 = N(d_2)\)

Recovering the Black-Scholes formula. \(\square\)

Advantages over Carr-Madan¶

No damping parameter \(\alpha\) to choose
Direct formula without modification
Works for wider class of characteristic functions
Numerically stable for most models

Greeks via Fourier Methods¶

Since the option price is expressed as a Fourier integral, Greeks follow by differentiating under the integral sign.

Delta. With \(x = \ln S\):

\[ \boxed{\Delta = \frac{1}{S}\cdot\frac{1}{2\pi}\int_{-\infty}^{\infty}i\omega\,\hat{C}(\omega,\tau)\,e^{i\omega x}\,d\omega} \]

Gamma.

\[ \boxed{\Gamma = \frac{1}{S^2}\cdot\frac{1}{2\pi}\int_{-\infty}^{\infty}(-\omega^2 - i\omega)\,\hat{C}(\omega,\tau)\,e^{i\omega x}\,d\omega} \]

Theta.

\[ \boxed{\Theta = -\frac{1}{2\pi}\int_{-\infty}^{\infty}\psi(\omega)\,\hat{\Phi}(\omega)\,e^{\psi(\omega)\tau}\,e^{i\omega x}\,d\omega} \]

Computational efficiency. All Greeks are computed from the same FFT output: evaluate \(\hat{C}(\omega_j,\tau)\) once, multiply by \(i\omega_j\) (Delta), \(-\omega_j^2 - i\omega_j\) (Gamma), or \(\psi(\omega_j)\) (Theta), and apply the inverse FFT. The incremental cost per Greek is \(O(N\log N)\).

Comparison with Other Methods¶

Fourier methods are most effective for European options across many strikes simultaneously when the characteristic function is available in closed form. The FFT prices an entire strike grid in \(O(N\log N)\) operations, compared with \(O(N^2)\) for pointwise quadrature or one PDE solve per strike. Greeks come essentially for free from the same transform.

The principal limitations are: (i) the payoff must be Fourier-representable, which requires damping for calls and puts; (ii) American options and other free-boundary problems are not directly accessible; (iii) non-smooth payoffs (e.g., digitals) introduce Gibbs oscillations; and (iv) the method suffers from the curse of dimensionality for \(d > 3\) assets.

For single-strike pricing, direct PDE or closed-form evaluation may be simpler. For path-dependent or early-exercise problems, finite-difference and Monte Carlo methods are generally preferable.

Summary¶

The Fourier approach to Black-Scholes pricing proceeds in four steps:

Option Payoff Phi(S) | Transform to log-price: Phi(e^x) | Fourier Transform: hat{Phi}(omega) | Multiply by e^{psi(omega) tau} (characteristic function) | Inverse Fourier Transform | Option Value V(x, tau)

Key formulas. The PDE in Fourier space reduces to the ODE

\[ \frac{\partial\hat{V}}{\partial\tau} = \psi(\omega)\hat{V} \]

with solution

\[ \hat{V}(\omega,\tau) = \hat{\Phi}(\omega)e^{\psi(\omega)\tau} \]

and inversion

\[ V(x,\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{\Phi}(\omega)e^{\psi(\omega)\tau}e^{i\omega x}d\omega \]

The identity \(e^{\psi(\omega)\tau} = e^{-r\tau}\phi_X(\omega,\tau)\) connects the PDE characteristic exponent to the probabilistic characteristic function of the log-return. All three inversion routes -- direct quadrature, Carr-Madan with FFT, and Gil-Pelaez -- produce the Black-Scholes formula \(C = SN(d_1) - Ke^{-rT}N(d_2)\) as a special case.

Exercises¶

Exercise 1. Verify the characteristic exponent for the Black-Scholes model. Starting from the log-price formulation of the PDE, apply the Fourier transform to both sides and show that the ODE in Fourier space has the characteristic exponent \(\psi(\omega) = -\frac{\sigma^2\omega^2}{2} + i\omega(r - \frac{\sigma^2}{2}) - r\).

Solution to Exercise 1

The log-price PDE is:

\[ \frac{\partial V}{\partial \tau} = \frac{\sigma^2}{2}\frac{\partial^2 V}{\partial x^2} + \left(r - \frac{\sigma^2}{2}\right)\frac{\partial V}{\partial x} - rV \]

Apply the Fourier transform \(\hat{V}(\omega,\tau) = \int_{-\infty}^{\infty} V(x,\tau) e^{-i\omega x} dx\) to both sides.

Left side: \(\mathcal{F}\left[\frac{\partial V}{\partial \tau}\right] = \frac{\partial \hat{V}}{\partial \tau}\)

Right side, term by term:

\(\frac{\sigma^2}{2}\mathcal{F}\left[\frac{\partial^2 V}{\partial x^2}\right] = \frac{\sigma^2}{2}(-\omega^2)\hat{V} = -\frac{\sigma^2\omega^2}{2}\hat{V}\)
\(\left(r - \frac{\sigma^2}{2}\right)\mathcal{F}\left[\frac{\partial V}{\partial x}\right] = \left(r - \frac{\sigma^2}{2}\right)(i\omega)\hat{V}\)
\(-r\mathcal{F}[V] = -r\hat{V}\)

Combining:

\[ \frac{\partial \hat{V}}{\partial \tau} = \left[-\frac{\sigma^2\omega^2}{2} + i\omega\left(r - \frac{\sigma^2}{2}\right) - r\right]\hat{V} \]

This is a first-order ODE \(\frac{\partial \hat{V}}{\partial \tau} = \psi(\omega)\hat{V}\) with the characteristic exponent:

\[ \psi(\omega) = -\frac{\sigma^2\omega^2}{2} + i\omega\left(r - \frac{\sigma^2}{2}\right) - r \]

\(\square\)

Exercise 2. The Fourier transform of the call payoff \((e^x - 1)^+\) diverges for real \(\omega\). Show explicitly that the integral \(\int_0^{\infty}(e^x - 1)e^{-i\omega x}dx\) diverges by analyzing the behavior of the integrand as \(x \to \infty\). Then verify that introducing the damping factor \(e^{-\alpha x}\) with \(\alpha > 1\) makes the integral convergent and derive the resulting transform.

Solution to Exercise 2

Consider the integral \(I(\omega) = \int_0^{\infty}(e^x - 1)e^{-i\omega x}dx\) for real \(\omega\).

Divergence: For \(x \to \infty\), the integrand behaves as \(e^x \cdot e^{-i\omega x} = e^{(1-i\omega)x}\). Since \(|e^{(1-i\omega)x}| = e^x \to \infty\) as \(x \to \infty\), the integral diverges.

More precisely, \(|e^x e^{-i\omega x}| = e^x\) which is not integrable on \([0,\infty)\).

Damping factor: Now consider \(\int_0^{\infty}(e^x - 1)e^{-\alpha x}e^{-i\omega x}dx\) with \(\alpha > 0\). The integrand magnitude is:

\[ |(e^x - 1)e^{-\alpha x}e^{-i\omega x}| \leq e^{(1-\alpha)x} + e^{-\alpha x} \]

For the first term, \(e^{(1-\alpha)x}\) is integrable on \([0,\infty)\) if and only if \(\alpha > 1\). The second term \(e^{-\alpha x}\) is integrable for \(\alpha > 0\). Therefore, the damped integral converges for \(\alpha > 1\):

\[ \int_0^{\infty}(e^x - 1)e^{-(\alpha + i\omega)x}dx = \frac{1}{\alpha + i\omega - 1} - \frac{1}{\alpha + i\omega} \]

\[ = \frac{1}{(\alpha - 1 + i\omega)(\alpha + i\omega)} \]

\(\square\)

Exercise 3. Using the Carr-Madan formula with \(\alpha = 1.5\), \(S_0 = 100\), \(K = 100\), \(r = 5\%\), \(\sigma = 20\%\), and \(T = 1\), set up the numerical integration for the call price. Write the integrand explicitly and verify that evaluating the integral (e.g., via Simpson's rule with a fine grid) recovers the standard Black-Scholes call price to at least 4 decimal places.

Solution to Exercise 3

With \(\alpha = 1.5\), \(S_0 = 100\), \(K = 100\), \(r = 0.05\), \(\sigma = 0.20\), \(T = 1\), the log-strike is \(k = \ln K = \ln 100\).

The Carr-Madan integrand is:

\[ C(K) = \frac{e^{-\alpha k}}{\pi}\int_0^{\infty}\text{Re}\left[e^{-i\omega k}\psi_T(\omega)\right]d\omega \]

where:

\[ \psi_T(\omega) = \frac{e^{-rT}\phi_T(\omega - (\alpha+1)i)}{\alpha^2 + \alpha - \omega^2 + i(2\alpha+1)\omega} \]

The characteristic function for Black-Scholes is:

\[ \phi_T(u) = \exp\left[iu\left(r - \frac{\sigma^2}{2}\right)T - \frac{\sigma^2 u^2 T}{2}\right] \]

For \(u = \omega - (\alpha+1)i = \omega - 2.5i\):

\[ \phi_T(\omega - 2.5i) = \exp\left[i(\omega - 2.5i)(0.05 - 0.02)(1) - \frac{0.04(\omega - 2.5i)^2}{2}\right] \]

\[ = \exp\left[(0.03i\omega + 0.075) - 0.02(\omega^2 - 5i\omega - 6.25)\right] \]

\[ = \exp\left[0.03i\omega + 0.075 - 0.02\omega^2 + 0.1i\omega + 0.125\right] \]

\[ = \exp\left[-0.02\omega^2 + 0.13i\omega + 0.2\right] \]

The denominator is \(\alpha^2 + \alpha - \omega^2 + i(2\alpha+1)\omega = 3.75 - \omega^2 + 4i\omega\).

Setting up the numerical integration with Simpson's rule on a fine grid (e.g., \(\omega \in [0, 50]\) with \(\Delta\omega = 0.01\)), and computing \(\text{Re}[e^{-i\omega k}\psi_T(\omega)]\) at each point, the integral can be evaluated. The Black-Scholes price for these parameters is \(C \approx 10.4506\), which the numerical integration recovers to at least 4 decimal places. \(\square\)

Exercise 4. Explain the relationship \(e^{\psi(\omega)\tau} = e^{-r\tau}\phi_X(\omega, \tau)\) between the characteristic exponent of the PDE and the characteristic function of the log-return. Starting from the explicit forms of \(\psi(\omega)\) and \(\phi_X(\omega,\tau)\), verify the identity algebraically and explain why it is central to the Fourier pricing framework.

Solution to Exercise 4

The identity \(e^{\psi(\omega)\tau} = e^{-r\tau}\phi_X(\omega,\tau)\) connects the PDE characteristic exponent to the probabilistic characteristic function.

Derivation. The PDE characteristic exponent is:

\[ \psi(\omega) = -\frac{\sigma^2\omega^2}{2} + i\omega\left(r - \frac{\sigma^2}{2}\right) - r \]

The characteristic function of \(X_\tau = \ln(S_T/S_0)\) under \(\mathbb{Q}\) is:

\[ \phi_X(\omega,\tau) = \exp\left[i\omega\left(r - \frac{\sigma^2}{2}\right)\tau - \frac{\sigma^2\omega^2\tau}{2}\right] \]

Therefore:

\[ e^{-r\tau}\phi_X(\omega,\tau) = \exp\left[-r\tau + i\omega\left(r - \frac{\sigma^2}{2}\right)\tau - \frac{\sigma^2\omega^2\tau}{2}\right] = e^{\psi(\omega)\tau} \]

Why this is central. This identity means the Fourier-space solution \(\hat{V}(\omega,\tau) = \hat{\Phi}(\omega)e^{\psi(\omega)\tau}\) can be rewritten as \(\hat{V}(\omega,\tau) = e^{-r\tau}\hat{\Phi}(\omega)\phi_X(\omega,\tau)\). In other words, the option price in Fourier space equals the discounted payoff transform times the characteristic function of the log-return. This directly encodes risk-neutral pricing: the transform of the discounted expected payoff factorizes into a payoff piece and a distributional piece. \(\square\)

Exercise 5. The Gil-Pelaez formula writes the call price as \(C = S\Pi_1 - Ke^{-rT}\Pi_2\) where \(\Pi_1\) and \(\Pi_2\) are given by integrals involving the characteristic function. For the Black-Scholes model, show that \(\Pi_1 = N(d_1)\) and \(\Pi_2 = N(d_2)\) by substituting the lognormal characteristic function and evaluating the resulting Gaussian integrals.

Solution to Exercise 5

The Gil-Pelaez formula gives:

\[ \Pi_2 = \frac{1}{2} + \frac{1}{\pi}\int_0^{\infty}\text{Re}\left[\frac{e^{-i\omega\ln(K/S)}\phi(\omega)}{i\omega}\right]d\omega \]

With the Black-Scholes characteristic function \(\phi(\omega) = \exp[i\omega(r - \sigma^2/2)T - \sigma^2\omega^2 T/2]\), the integrand becomes:

\[ \frac{e^{-i\omega\ln(K/S)}\exp[i\omega(r-\sigma^2/2)T - \sigma^2\omega^2 T/2]}{i\omega} \]

Write \(\ln(K/S) = -\ln(S/K)\) and combine the exponentials:

\[ = \frac{\exp[i\omega(\ln(S/K) + (r-\sigma^2/2)T) - \sigma^2\omega^2 T/2]}{i\omega} \]

The exponent in the numerator is \(i\omega\, d_2\sigma\sqrt{T} - \sigma^2\omega^2 T/2\) where \(d_2 = \frac{\ln(S/K) + (r-\sigma^2/2)T}{\sigma\sqrt{T}}\). Substituting \(u = \omega\sigma\sqrt{T}\), the integral reduces to the standard result:

\[ \frac{1}{2} + \frac{1}{\pi}\int_0^{\infty}\frac{\sin(d_2 u)}{u}e^{-u^2/2}du = N(d_2) \]

using the known identity \(\frac{1}{2} + \frac{1}{\pi}\int_0^{\infty}\frac{\sin(ax)}{x}e^{-x^2/2}dx = N(a)\).

For \(\Pi_1\), the argument \(\phi(\omega - i)\) shifts the characteristic function. The shift replaces \(\omega\) by \(\omega - i\) in \(\phi\), which after simplification introduces an extra factor \(e^{rT}S/S\) and shifts \(d_2\) to \(d_1 = d_2 + \sigma\sqrt{T}\). The same Gaussian-integral identity then gives \(\Pi_1 = N(d_1)\). \(\square\)