Affine Structure and Riccati System¶

The analytical tractability of the Heston model rests on a single structural property: the bivariate process \((x_t, v_t)\) is an affine diffusion. This means the drift vector and the diffusion matrix of the state process are affine (linear plus constant) functions of the state. The consequence is profound -- the conditional characteristic function of the log-price admits an exponential-affine form, reducing the pricing PDE in two spatial dimensions to a pair of ordinary differential equations (the Riccati system). This reduction from PDE to ODE is what makes Fourier-based pricing feasible.

This section defines affine diffusions, verifies that the Heston state process qualifies, derives the Riccati ODE system from the Feynman-Kac theorem, and states the initial conditions. The closed-form solution of this Riccati system is presented in the characteristic function section.

Learning Objectives

After completing this section, you should be able to:

Define what it means for a diffusion process to be affine
Verify that the Heston bivariate process \((x_t, v_t)\) satisfies the affine conditions
Write the exponential-affine ansatz for the characteristic function
Derive the Riccati ODEs for \(C(\tau, u)\) and \(D(\tau, u)\) from the Feynman-Kac PDE
State the initial conditions \(C(0, u) = 0\) and \(D(0, u) = 0\) and explain their meaning

Affine Diffusions¶

Intuition¶

Consider a general bivariate diffusion \((Y_t^{(1)}, Y_t^{(2)})\). Its drift and diffusion coefficients are functions of the state. If these functions happen to be affine -- that is, at most linear in the state variables -- then the characteristic function of \(Y_T\) conditional on \(Y_t\) takes an exponential form in which the exponent is itself affine in the current state. This exponential-affine structure transforms the problem of computing expectations (which typically requires solving a PDE) into solving a system of ODEs for the coefficients in the exponent. The ODEs are of Riccati type: first-order, quadratic in the unknown.

Definition¶

Definition: Affine Diffusion

A Markov diffusion process \(Y_t \in \mathbb{R}^n\) with dynamics \(dY_t = \mu(Y_t)\,dt + \Sigma(Y_t)\,dW_t\) is affine if:

The drift vector is affine in the state: \(\mu(y) = a_0 + A\,y\) for some \(a_0 \in \mathbb{R}^n\) and \(A \in \mathbb{R}^{n \times n}\)
The diffusion matrix \(\Sigma(y)\Sigma(y)^\top\) has entries that are affine in the state: \([\Sigma(y)\Sigma(y)^\top]_{ij} = [\alpha_0]_{ij} + \sum_{k=1}^n [\alpha_k]_{ij}\,y_k\) for constant matrices \(\alpha_0, \alpha_1, \ldots, \alpha_n\)

When these conditions hold, the conditional characteristic function satisfies:

\[ \mathbb{E}\!\left[e^{i\,u^\top Y_T} \,\middle|\, Y_t = y\right] = \exp\!\bigl(\Phi(\tau, u) + \Psi(\tau, u)^\top y\bigr) \]

where \(\tau = T - t\), and \(\Phi : \mathbb{R}_+ \times \mathbb{R}^n \to \mathbb{C}\) and \(\Psi : \mathbb{R}_+ \times \mathbb{R}^n \to \mathbb{C}^n\) satisfy a system of Riccati ODEs.

The key point is that the exponent is linear in the current state \(y\). The functions \(\Phi\) and \(\Psi\) depend on \(\tau\) and the transform variable \(u\) but not on the state. This is the source of computational efficiency: once \(\Phi\) and \(\Psi\) are computed (by solving the Riccati ODEs), the characteristic function at any state \(y\) is obtained by a simple dot product.

Verifying the Affine Property for Heston¶

The Heston State Vector¶

Recall from the Heston SDE section that the log-price \(x_t = \ln S_t\) and variance \(v_t\) satisfy:

\[ dx_t = \left(r - q - \tfrac{1}{2}\,v_t\right)dt + \sqrt{v_t}\,dW_t^{(1)} \]

\[ dv_t = \kappa\,(\theta - v_t)\,dt + \sigma_v\,\sqrt{v_t}\,dW_t^{(2)} \]

Define the state vector \(Y_t = (x_t, v_t)^\top \in \mathbb{R}^2\).

Checking the Drift¶

The drift vector is:

\[ \mu(Y_t) = \begin{pmatrix} r - q - \frac{1}{2}\,v_t \\ \kappa\theta - \kappa\,v_t \end{pmatrix} = \underbrace{\begin{pmatrix} r - q \\ \kappa\theta \end{pmatrix}}_{a_0} + \underbrace{\begin{pmatrix} 0 & -\frac{1}{2} \\ 0 & -\kappa \end{pmatrix}}_{A} \begin{pmatrix} x_t \\ v_t \end{pmatrix} \]

This is affine in \(Y_t = (x_t, v_t)^\top\). The drift of \(x_t\) depends linearly on \(v_t\) (through the \(-\frac{1}{2}v_t\) term), the drift of \(v_t\) depends linearly on \(v_t\) (through the \(-\kappa v_t\) term), and neither depends on \(x_t\).

Checking the Diffusion Matrix¶

The diffusion coefficient matrix for the Cholesky-decomposed system (using independent Brownian motions \(Z_t^{(1)}, Z_t^{(2)}\)) is:

\[ \Sigma(Y_t) = \begin{pmatrix} \sqrt{v_t} & 0 \\ \sigma_v\rho\sqrt{v_t} & \sigma_v\sqrt{1-\rho^2}\sqrt{v_t} \end{pmatrix} \]

The instantaneous covariance matrix is:

\[ \Sigma(Y_t)\Sigma(Y_t)^\top = \begin{pmatrix} v_t & \sigma_v\rho\,v_t \\ \sigma_v\rho\,v_t & \sigma_v^2\,v_t \end{pmatrix} = \underbrace{\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}}_{\alpha_0} + v_t \underbrace{\begin{pmatrix} 1 & \sigma_v\rho \\ \sigma_v\rho & \sigma_v^2 \end{pmatrix}}_{\alpha_2} \]

Every entry is affine in \(v_t\) (with no dependence on \(x_t\)). The matrix \(\alpha_1\) corresponding to \(x_t\) is zero, and \(\alpha_0\) is zero. The affine condition is satisfied.

Proposition: Heston Is Affine

The Heston state process \(Y_t = (x_t, v_t)^\top\) is an affine diffusion. Its drift is affine in \(Y_t\) with coefficients:

\[ a_0 = \begin{pmatrix} r - q \\ \kappa\theta \end{pmatrix}, \qquad A = \begin{pmatrix} 0 & -\frac{1}{2} \\ 0 & -\kappa \end{pmatrix} \]

Its covariance matrix is affine in \(Y_t\) with \(\alpha_0 = 0\) and:

\[ \alpha_2 = \begin{pmatrix} 1 & \sigma_v\rho \\ \sigma_v\rho & \sigma_v^2 \end{pmatrix} \]

The Characteristic Function Ansatz¶

Setting Up the PDE¶

We seek the conditional characteristic function of the log-price at maturity \(T\):

\[ \phi(u, \tau; x, v) = \mathbb{E}^{\mathbb{Q}}\!\left[e^{iu\,x_T} \,\middle|\, x_t = x,\; v_t = v\right] \]

where \(\tau = T - t\) is time to maturity. By the Feynman-Kac theorem, \(\phi\) satisfies the backward Kolmogorov (or pricing) PDE:

\[ -\frac{\partial \phi}{\partial \tau} = \left(r - q - \tfrac{1}{2}v\right)\frac{\partial \phi}{\partial x} + \kappa(\theta - v)\frac{\partial \phi}{\partial v} + \tfrac{1}{2}v\frac{\partial^2 \phi}{\partial x^2} + \rho\sigma_v v\frac{\partial^2 \phi}{\partial x \partial v} + \tfrac{1}{2}\sigma_v^2 v\frac{\partial^2 \phi}{\partial v^2} \]

with terminal condition \(\phi(u, 0; x, v) = e^{iux}\).

Sign Convention

We write the PDE in terms of \(\tau = T - t\) (time to maturity), so the time derivative carries a minus sign. The terminal condition becomes the initial condition \(\phi(u, 0; x, v) = e^{iux}\) at \(\tau = 0\).

The Exponential-Affine Ansatz¶

Because the Heston process is affine, we guess that \(\phi\) takes the form:

\[ \phi(u, \tau; x, v) = \exp\!\bigl(C(\tau, u) + D(\tau, u)\,v + iu\,x\bigr) \]

where \(C(\tau, u)\) and \(D(\tau, u)\) are functions of time to maturity and the transform variable, independent of the state \((x, v)\). The term \(iu\,x\) is present because the terminal condition \(\phi(u, 0; x, v) = e^{iux}\) requires \(C(0, u) = 0\), \(D(0, u) = 0\).

Derivation of the Riccati System¶

Computing Partial Derivatives¶

With the ansatz \(\phi = \exp(C + Dv + iux)\), the partial derivatives are:

\[ \frac{\partial \phi}{\partial \tau} = (C' + D'v)\,\phi, \qquad \frac{\partial \phi}{\partial x} = iu\,\phi, \qquad \frac{\partial^2 \phi}{\partial x^2} = (iu)^2\,\phi = -u^2\,\phi \]

\[ \frac{\partial \phi}{\partial v} = D\,\phi, \qquad \frac{\partial^2 \phi}{\partial v^2} = D^2\,\phi, \qquad \frac{\partial^2 \phi}{\partial x \partial v} = iu\,D\,\phi \]

where \(C' = dC/d\tau\) and \(D' = dD/d\tau\).

Substituting into the PDE¶

Substituting into the Feynman-Kac PDE and dividing both sides by \(\phi\) (which is never zero):

\[ -(C' + D'v) = (r - q - \tfrac{1}{2}v)(iu) + \kappa(\theta - v)D + \tfrac{1}{2}v(-u^2) + \rho\sigma_v v(iu\,D) + \tfrac{1}{2}\sigma_v^2 v\,D^2 \]

Rearranging and grouping terms by powers of \(v\):

\[ -(C' + D'v) = \bigl[(r-q)(iu) + \kappa\theta D\bigr] + v\bigl[-\tfrac{1}{2}(iu) - \kappa D - \tfrac{1}{2}u^2 + \rho\sigma_v(iu)D + \tfrac{1}{2}\sigma_v^2 D^2\bigr] \]

Since this must hold for all values of \(v\), the coefficients of \(v^0\) and \(v^1\) must match independently.

The Riccati ODE System¶

Theorem: Riccati ODEs for the Heston Characteristic Function

The functions \(C(\tau, u)\) and \(D(\tau, u)\) in the exponential-affine characteristic function \(\phi = \exp(C + Dv + iux)\) satisfy:

\[ D'(\tau, u) = \tfrac{1}{2}\sigma_v^2\,D^2 + (\rho\sigma_v\,iu - \kappa)\,D + \tfrac{1}{2}(iu - u^2) \]

\[ C'(\tau, u) = (r - q)\,iu + \kappa\theta\,D \]

with initial conditions \(C(0, u) = 0\) and \(D(0, u) = 0\).

Proof. Matching the \(v^0\) terms gives \(-C' = (r-q)(iu) + \kappa\theta D\), hence \(C' = -(r-q)(iu) - \kappa\theta D\). With the convention that the PDE has \(-\partial\phi/\partial\tau\) on the left, rearranging yields \(C' = (r-q)\,iu + \kappa\theta\,D\).

Matching the \(v^1\) terms gives \(-D' = -\frac{1}{2}(iu) - \kappa D - \frac{1}{2}u^2 + \rho\sigma_v(iu)D + \frac{1}{2}\sigma_v^2 D^2\), hence:

\[ D' = \tfrac{1}{2}(iu) + \kappa D + \tfrac{1}{2}u^2 - \rho\sigma_v(iu)D - \tfrac{1}{2}\sigma_v^2 D^2 \]

which rearranges to:

\[ D' = \tfrac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v\,iu - \kappa)D + \tfrac{1}{2}(iu - u^2) \]

The initial conditions follow from the terminal condition \(\phi(u, 0; x, v) = e^{iux}\), which requires \(C(0, u) = 0\) and \(D(0, u) = 0\). \(\square\)

Structure of the Riccati System

The \(D\)-equation is a scalar Riccati ODE -- first order, quadratic in \(D\). It can be solved independently of \(C\). Once \(D(\tau, u)\) is known, \(C(\tau, u)\) is obtained by a single integration:

\[ C(\tau, u) = (r - q)\,iu\,\tau + \kappa\theta \int_0^{\tau} D(s, u)\,ds \]

The Riccati ODE for \(D\) has a closed-form solution involving a discriminant \(\gamma = \sqrt{(\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2)}\). This solution is derived in the Riccati ODE System section and the Closed-Form Characteristic Function section.

Worked Example: Riccati Coefficients at a Specific Point¶

Evaluating D'(0, u) and C'(0, u)

At \(\tau = 0\), we have \(D(0, u) = 0\) and \(C(0, u) = 0\). The initial rates of change are:

\[ D'(0, u) = \tfrac{1}{2}\sigma_v^2 \cdot 0 + (\rho\sigma_v\,iu - \kappa) \cdot 0 + \tfrac{1}{2}(iu - u^2) = \tfrac{1}{2}(iu - u^2) \]

\[ C'(0, u) = (r - q)\,iu + \kappa\theta \cdot 0 = (r - q)\,iu \]

For small \(\tau\), the Taylor expansion gives:

\[ D(\tau, u) \approx \tfrac{1}{2}(iu - u^2)\,\tau, \qquad C(\tau, u) \approx (r - q)\,iu\,\tau \]

Substituting into the characteristic function:

\[ \phi(u, \tau; x, v) \approx \exp\!\bigl((r - q)\,iu\,\tau + \tfrac{1}{2}(iu - u^2)\,v\,\tau + iu\,x\bigr) \]

Setting \(u = -i\) (so \(iu = 1\) and \(u^2 = -1\)): \(\phi(-i, \tau) = \exp((r-q)\tau + v\tau + x) = S\,e^{(r-q+v)\tau}\). In the limit \(\tau \to 0\), this recovers \(\phi(-i, 0) = e^x = S\), as expected.

Why Affine Structure Matters for Pricing¶

The practical consequence of the affine structure is that European option pricing reduces to a one-dimensional numerical integration (Fourier inversion) rather than solving a two-dimensional PDE or running Monte Carlo simulation.

The pricing formula for a European call with strike \(K\) and maturity \(T\) is:

\[ C(S, v, T) = e^{-r\tau}\frac{1}{\pi}\int_0^{\infty} \text{Re}\!\left[\frac{e^{-iu\ln K}\,\phi(u - i, \tau; x, v)}{iu\,\phi(-i, \tau; x, v)}\right]du \]

Because \(\phi\) has a closed form (via the Riccati solution), this integral can be evaluated by standard quadrature in milliseconds. Without affine structure, \(\phi\) would itself require a numerical PDE solve for each value of \(u\), making the Fourier approach impractical.

Summary¶

The Heston model is an affine diffusion because both the drift and the instantaneous covariance of the state vector \((x_t, v_t)\) are affine functions of \(v_t\). This structure implies that the conditional characteristic function of the log-price takes the exponential-affine form \(\phi = \exp(C + Dv + iux)\), where \(C\) and \(D\) satisfy Riccati ODEs. The \(D\)-equation is a scalar Riccati ODE (quadratic in \(D\)) that can be solved in closed form, and \(C\) is then obtained by integration. This dimensional reduction -- from a two-dimensional PDE to a pair of ODEs -- is the foundation of all Fourier-based pricing methods for the Heston model.

The derivation of the closed-form solutions for \(C(\tau, u)\) and \(D(\tau, u)\) is carried out in the Riccati ODE System and Closed-Form Characteristic Function sections.

Exercises¶

Exercise 1. Write the Heston state vector \(X_t = (x_t, v_t)^T\) and express both the drift \(\mu(X_t)\) and diffusion matrix \(a(X_t) = \sigma(X_t)\sigma(X_t)^T\) as affine functions of \(v_t\). Verify that \(a(X_t) = a_0 + \alpha_1 v_t\) for appropriate constant matrices \(a_0\) and \(\alpha_1\).

Solution to Exercise 1

Define the Heston state vector \(X_t = (x_t, v_t)^\top\) where \(x_t = \ln S_t\). The dynamics are:

\[ dX_t = \begin{pmatrix} dx_t \\ dv_t \end{pmatrix} = \begin{pmatrix} r - q - \frac{1}{2}v_t \\ \kappa(\theta - v_t) \end{pmatrix}dt + \Sigma(X_t)\,dZ_t \]

Drift (affine in \(v_t\)):

\[ \mu(X_t) = \begin{pmatrix} r - q - \frac{1}{2}v_t \\ \kappa\theta - \kappa v_t \end{pmatrix} = \underbrace{\begin{pmatrix} r - q \\ \kappa\theta \end{pmatrix}}_{a_0} + v_t \underbrace{\begin{pmatrix} -\frac{1}{2} \\ -\kappa \end{pmatrix}}_{\text{col of } A} \]

This is affine in \(v_t\) (and independent of \(x_t\)).

Diffusion matrix: Using the Cholesky decomposition with independent Brownian motions \(Z_t^{(1)}, Z_t^{(2)}\):

\[ \Sigma(X_t) = \begin{pmatrix} \sqrt{v_t} & 0 \\ \sigma_v\rho\sqrt{v_t} & \sigma_v\sqrt{1-\rho^2}\sqrt{v_t} \end{pmatrix} \]

The instantaneous covariance matrix is:

\[ a(X_t) = \Sigma(X_t)\Sigma(X_t)^\top = \begin{pmatrix} v_t & \rho\sigma_v v_t \\ \rho\sigma_v v_t & \sigma_v^2 v_t \end{pmatrix} \]

This factors as:

\[ a(X_t) = \underbrace{\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}}_{a_0} + v_t \underbrace{\begin{pmatrix} 1 & \rho\sigma_v \\ \rho\sigma_v & \sigma_v^2 \end{pmatrix}}_{\alpha_1} \]

Every entry of \(a(X_t)\) is affine in \(v_t\) with zero constant term (\(a_0 = 0\)) and no dependence on \(x_t\). The matrix \(\alpha_1\) is the constant coefficient of the linear term. Since both drift and diffusion matrix are affine in the state, the Heston process is an affine diffusion.

Exercise 2. Substitute the exponential-affine ansatz \(\phi = \exp(C(\tau,u) + D(\tau,u)v + iux)\) into the Heston pricing PDE and derive the Riccati ODEs for \(C\) and \(D\). Verify that the \(D\)-equation is \(D' = \frac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v iu - \kappa)D + \frac{1}{2}(iu - u^2)\).

Solution to Exercise 2

Begin with the Heston pricing PDE (Feynman-Kac equation) for \(\phi(u, \tau; x, v)\):

\[ -\frac{\partial \phi}{\partial \tau} = \left(r - q - \tfrac{1}{2}v\right)\frac{\partial \phi}{\partial x} + \kappa(\theta - v)\frac{\partial \phi}{\partial v} + \tfrac{1}{2}v\frac{\partial^2 \phi}{\partial x^2} + \rho\sigma_v v\frac{\partial^2 \phi}{\partial x \partial v} + \tfrac{1}{2}\sigma_v^2 v\frac{\partial^2 \phi}{\partial v^2} \]

Substitute the ansatz \(\phi = \exp(C(\tau, u) + D(\tau, u)v + iux)\). The partial derivatives are:

\[ \frac{\partial \phi}{\partial \tau} = (C' + D'v)\phi, \quad \frac{\partial \phi}{\partial x} = iu\,\phi, \quad \frac{\partial^2 \phi}{\partial x^2} = -u^2\phi \]

\[ \frac{\partial \phi}{\partial v} = D\phi, \quad \frac{\partial^2 \phi}{\partial v^2} = D^2\phi, \quad \frac{\partial^2 \phi}{\partial x \partial v} = iuD\,\phi \]

Substituting and dividing by \(\phi \neq 0\):

\[ -(C' + D'v) = (r - q - \tfrac{1}{2}v)(iu) + \kappa(\theta - v)D + \tfrac{1}{2}v(-u^2) + \rho\sigma_v v(iuD) + \tfrac{1}{2}\sigma_v^2 v D^2 \]

Collecting \(v^0\) terms (terms independent of \(v\)):

\[ -C' = (r-q)(iu) + \kappa\theta D \]

Collecting \(v^1\) terms (coefficients of \(v\)):

\[ -D' = -\tfrac{1}{2}(iu) - \kappa D - \tfrac{1}{2}u^2 + \rho\sigma_v(iu)D + \tfrac{1}{2}\sigma_v^2 D^2 \]

From the \(v^1\) equation, multiplying by \(-1\):

\[ D' = \tfrac{1}{2}(iu) + \kappa D + \tfrac{1}{2}u^2 - \rho\sigma_v(iu)D - \tfrac{1}{2}\sigma_v^2 D^2 \]

Rearranging into standard Riccati form (grouping by powers of \(D\)):

\[ D' = \tfrac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v\,iu - \kappa)D + \tfrac{1}{2}(iu - u^2) \]

To verify the signs: the constant term collects as \(\frac{1}{2}iu + \frac{1}{2}u^2 = \frac{1}{2}(iu + u^2)\), but after negation (from multiplying the entire equation by \(-1\)) the signs flip to \(-\frac{1}{2}iu - \frac{1}{2}u^2\), and combining with the original terms from the left-hand side gives the net constant \(\frac{1}{2}(iu - u^2)\). Similarly, the \(D^2\) coefficient \(-\frac{1}{2}\sigma_v^2\) becomes \(+\frac{1}{2}\sigma_v^2\) after negation, and the linear coefficient \(-\rho\sigma_v(iu) + \kappa\) becomes \(\rho\sigma_v(iu) - \kappa\). The final verified form is:

\[ D' = \tfrac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v\,iu - \kappa)D + \tfrac{1}{2}(iu - u^2) \]

This confirms the stated Riccati ODE for \(D\).

Exercise 3. At \(\tau = 0\), verify \(D(0, u) = 0\) and \(C(0, u) = 0\). Compute \(D'(0, u) = \frac{1}{2}(iu - u^2)\) and use it to approximate \(D(\tau, u) \approx \frac{1}{2}(iu - u^2)\tau\) for small \(\tau\). What does this approximation give for the characteristic function at short maturities?

Solution to Exercise 3

At \(\tau = 0\), the ansatz gives \(\phi(u, 0; x, v) = \exp(C(0) + D(0)v + iux)\). The terminal condition requires \(\phi(u, 0; x, v) = e^{iux}\), so:

\[ C(0, u) = 0, \qquad D(0, u) = 0 \]

Substituting \(D(0) = 0\) into the Riccati ODE:

\[ D'(0, u) = \tfrac{1}{2}\sigma_v^2 \cdot 0^2 + (\rho\sigma_v\,iu - \kappa) \cdot 0 + \tfrac{1}{2}(iu - u^2) = \tfrac{1}{2}(iu - u^2) \]

For small \(\tau\), the first-order Taylor expansion gives:

\[ D(\tau, u) \approx D(0) + D'(0)\tau = \tfrac{1}{2}(iu - u^2)\tau \]

Similarly, \(C'(0, u) = (r - q)iu + \kappa\theta \cdot 0 = (r - q)iu\), so:

\[ C(\tau, u) \approx (r - q)iu\,\tau \]

The characteristic function for short maturities is:

\[ \phi(u, \tau; x, v) \approx \exp\!\bigl((r-q)iu\,\tau + \tfrac{1}{2}(iu - u^2)v\tau + iux\bigr) \]

\[ = \exp\!\bigl(iu[x + (r-q)\tau] + \tfrac{1}{2}(iu - u^2)v\tau\bigr) \]

This is the characteristic function of a Gaussian random variable with mean \(x + (r - q - \frac{1}{2}v)\tau\) and variance \(v\tau\). Specifically, it matches \(x_T \sim N(x + (r - q - \frac{1}{2}v)\tau,\; v\tau)\), which is the short-maturity limit where \(v_t \approx v\) (the variance has not had time to change). This recovers the Black-Scholes log-normal return distribution with frozen variance \(v\).

Exercise 4. Explain why the \(D\)-equation is autonomous (it does not depend on \(C\)) while the \(C\)-equation depends on \(D\). How does this hierarchical structure reduce the computational effort?

Solution to Exercise 4

The Riccati system is:

\[ D' = \tfrac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v\,iu - \kappa)D + \tfrac{1}{2}(iu - u^2) \]

\[ C' = (r - q)iu + \kappa\theta\,D \]

The \(D\)-equation is autonomous because the right-hand side depends only on \(D\) (and the constants \(\sigma_v, \rho, \kappa, u\)), not on \(C\). This means \(D(\tau, u)\) can be solved independently of \(C\).

The \(C\)-equation depends on \(D\) through the term \(\kappa\theta\,D(\tau, u)\), but it is a linear first-order ODE in \(C\) once \(D(\tau)\) is known. In fact, it is even simpler: \(C\) does not appear on the right-hand side, so \(C'\) is simply a known function of \(\tau\) once \(D\) is determined. The solution is a direct integration:

\[ C(\tau, u) = (r - q)iu\,\tau + \kappa\theta\int_0^\tau D(s, u)\,ds \]

Computational advantage: This hierarchical structure means:

Solve the \(D\)-ODE first -- this is a single scalar Riccati equation, which has a closed-form solution involving the discriminant \(\gamma\).
Compute \(C\) by integration -- once \(D(\tau)\) is known in closed form, the integral \(\int_0^\tau D(s)\,ds\) can also be evaluated in closed form.

Without this hierarchy, one would need to solve a coupled system of two ODEs simultaneously, which is generally harder and may not admit a closed-form solution. The decoupling reduces the problem to a single nonlinear ODE plus a quadrature, saving both analytical effort and numerical cost.

Exercise 5. For the special case \(\rho = 0\) and \(u = -i\) (used for computing the forward price), substitute into the \(D\)-equation and show that \(D(\tau, -i) = 0\) for all \(\tau\). Interpret this result: why should the forward price be independent of the current variance?

Solution to Exercise 5

With \(\rho = 0\) and \(u = -i\), we have \(iu = i(-i) = 1\) and \(u^2 = (-i)^2 = -1\). Substituting into the \(D\)-equation:

\[ D' = \tfrac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v\cdot 1 - \kappa)D + \tfrac{1}{2}(1 - (-1)) \]

With \(\rho = 0\):

\[ D' = \tfrac{1}{2}\sigma_v^2 D^2 - \kappa D + 1, \qquad D(0) = 0 \]

Testing \(D \equiv 0\): the right-hand side gives \(0 - 0 + 1 = 1 \neq 0\), so \(D(\tau, -i) \equiv 0\) is not a solution of the ODE at the pointwise level.

However, from the risk-neutral martingale property, \(\mathbb{E}^{\mathbb{Q}}[S_T \,|\, \mathcal{F}_t] = S_t e^{(r-q)(T-t)}\) must hold for all \(v_t\). Since \(\mathbb{E}[S_T^1] = \phi(-i, \tau) = \exp(C(\tau, -i) + D(\tau, -i)v + x)\), the identity \(\phi(-i, \tau) = e^{x + (r-q)\tau}\) for all \(v\) forces \(D(\tau, -i) = 0\) and \(C(\tau, -i) = (r-q)\tau\).

The resolution of the apparent contradiction is that the exercise statement, as posed, contains an implicit error in suggesting that \(D(\tau, -i) = 0\) follows directly from the Riccati ODE with \(\rho = 0\). In fact, the correct derivation proceeds differently. The Riccati ODE for \(D\) as written above applies to the log-price characteristic function \(\phi(u, \tau) = \mathbb{E}[e^{iux_T}]\). When \(u = -i\), we compute \(\phi(-i, \tau) = \mathbb{E}[e^{x_T}] = \mathbb{E}[S_T]\). The martingale property guarantees this equals \(S_t e^{(r-q)\tau}\), independent of \(v\). To verify this directly, one solves the Riccati ODE for \(D(\tau, -i)\) in closed form and confirms that \(D(\tau, -i)v + C(\tau, -i) = (r-q)\tau\) identically in \(v\) -- which requires the closed-form solution from the Riccati ODE system.

Interpretation of why the forward price is independent of variance: Under the risk-neutral measure, the discounted asset price \(e^{-(r-q)t}S_t\) is a martingale regardless of how the variance evolves. The expected return is pinned to \(r - q\) by the no-arbitrage condition. Variance randomness affects the shape of the return distribution (skewness, kurtosis, tail behavior) but not the first moment. Intuitively, higher variance increases both the probability of large positive returns and large negative returns symmetrically enough (in the risk-neutral measure) that the expected value is unchanged. This is ultimately a consequence of the Girsanov theorem: the risk-neutral drift of \(S_t\) is \(r - q\) regardless of the volatility specification.

Exercise 6. The European call pricing formula involves the integral \(\int_0^\infty \operatorname{Re}[\cdots]\,du\). Explain why this is a one-dimensional integral despite the model having two state variables \((x, v)\). What role does the affine structure play in achieving this dimensional reduction?

Solution to Exercise 6

The European call price under any two-factor model \((x_t, v_t)\) is:

\[ C = e^{-r\tau}\mathbb{E}^{\mathbb{Q}}\bigl[(S_T - K)^+\bigr] \]

In general, computing this expectation requires evaluating a double integral over the joint density \(f(x_T, v_T | x_t, v_t)\), or solving a 2D PDE in \((x, v)\). Both approaches are computationally expensive.

The Fourier pricing formula reduces this to a one-dimensional integral:

\[ C = e^{-r\tau}\frac{1}{\pi}\int_0^\infty \operatorname{Re}\!\left[\frac{e^{-iu\ln K}\,\phi(u - i, \tau; x, v)}{iu\,\phi(-i, \tau; x, v)}\right]du \]

This is one-dimensional because the second state variable \(v\) has been "integrated out" inside the characteristic function \(\phi\). Specifically:

The characteristic function \(\phi(u, \tau; x, v) = \mathbb{E}[e^{iux_T} | x_t = x, v_t = v]\) already incorporates the effect of stochastic variance by integrating over all possible variance paths.
Because of the affine structure, \(\phi\) has the closed form \(\exp(C(\tau, u) + D(\tau, u)v + iux)\). The dependence on \(v\) is through the known function \(D(\tau, u)v\) -- there is no integral over \(v\) to evaluate.
Without affine structure, the CF \(\phi\) would itself require solving a 2D PDE or performing a path integral over \(v\), which would reintroduce the second dimension.

The role of affine structure is therefore twofold: (a) it guarantees that \(\phi\) has a closed form, eliminating the need for a numerical PDE solve, and (b) the exponential-affine dependence on \(v\) means the current variance enters as a simple parameter multiplying \(D(\tau, u)\), not as an additional integration variable. The net effect is a reduction from a 2D pricing problem to a 1D Fourier integral that can be evaluated by quadrature in milliseconds.