Riccati ODE System for the Heston Characteristic Function

The exponential-affine ansatz for the Heston characteristic function reduces the Feynman-Kac PDE to a pair of ordinary differential equations. The equation for \(D(\tau, u)\) -- the coefficient of the variance \(v\) in the exponent -- is a Riccati equation: a first-order ODE that is quadratic in the unknown. The equation for \(C(\tau, u)\) is a simple integral once \(D\) is known. This section carries out the substitution of the ansatz into the PDE, derives both ODEs with full detail, classifies the \(D\)-equation as a Riccati ODE, and describes the method of solution via the standard substitution that converts a Riccati equation into a linear second-order ODE.

This section assumes familiarity with the Heston SDE and affine recap and leads directly to the closed-form solution.

Learning Objectives

After completing this section, you should be able to:

• Derive the Riccati ODE for \(D(\tau, u)\) by substituting the exponential-affine ansatz into the Feynman-Kac PDE
• Derive the integral formula for \(C(\tau, u)\) from the \(v^0\) terms
• Classify the \(D\)-equation as a scalar Riccati ODE and identify its three coefficients
• Apply the substitution \(D = -2h'/(\sigma_v^2 h)\) to reduce the Riccati ODE to a linear second-order ODE
• State the initial conditions and verify them against the terminal condition of the PDE

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Substitution of the Ansatz

Setup

Recall the Feynman-Kac PDE for \(\phi(u, \tau; x, v)\) from the preceding section:

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

and the exponential-affine ansatz:

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

Computing the Partial Derivatives

With the ansatz, each partial derivative factors through \(\phi\):

\[ \frac{\partial\phi}{\partial\tau} = \bigl(C' + D'v\bigr)\phi \]

\[ \frac{\partial\phi}{\partial x} = iu\,\phi, \qquad \frac{\partial^2\phi}{\partial x^2} = (iu)^2\phi = -u^2\phi \]

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

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

where primes denote derivatives with respect to \(\tau\).

Substitution and Simplification

Substituting into the PDE and dividing by \(\phi \neq 0\):

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

Expanding the right-hand side:

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

Grouping by powers of \(v\):

\[ C' + D'v = \underbrace{\bigl[(r-q)(iu) + \kappa\theta D\bigr]}_{v^0 \text{ terms}} + v\underbrace{\bigl[\tfrac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v\,iu - \kappa)D + \tfrac{1}{2}(iu - u^2)\bigr]}_{v^1 \text{ terms}} \]

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

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The Riccati ODE System

Statement

Theorem: Riccati System for the Heston Characteristic Function

The functions \(C(\tau, u)\) and \(D(\tau, u)\) satisfy:

\(D\)-equation (Riccati ODE):

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

\(C\)-equation (quadrature):

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

Initial conditions:

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

Verification of Initial Conditions

At \(\tau = 0\), the terminal condition \(\phi(u, 0; x, v) = e^{iux}\) requires \(\exp(C(0) + D(0)v + iux) = e^{iux}\) for all \(v\), hence \(C(0) = 0\) and \(D(0) = 0\). \(\square\)

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Structure of the D-Equation

Classification as a Riccati ODE

Definition: Scalar Riccati Equation

A scalar Riccati equation is a first-order ODE of the form:

\[ y'(\tau) = a(\tau)\,y^2 + b(\tau)\,y + c(\tau) \]

where \(a\), \(b\), \(c\) are known functions. The equation is nonlinear (quadratic in \(y\)), but it can be linearized by a standard substitution.

The \(D\)-equation has constant coefficients (since \(u\) is a parameter, not a variable):

Riccati Coefficient	Value	Dependence on Heston Parameters
\(a\)	\(\tfrac{1}{2}\sigma_v^2\)	Vol-of-vol squared
\(b\)	\(\rho\sigma_v\,iu - \kappa\)	Correlation, vol-of-vol, mean-reversion
\(c\)	\(\tfrac{1}{2}(iu - u^2)\)	Pure function of transform variable \(u\)

Autonomy

The coefficients \(a\), \(b\), \(c\) are independent of \(\tau\) (the Heston model is time-homogeneous). This means the Riccati ODE is autonomous, and its solution depends on \(\tau\) only through the elapsed time since the initial condition \(D(0) = 0\).

The Discriminant

The quadratic \(a\,D^2 + b\,D + c = 0\) has discriminant:

\[ \Delta = b^2 - 4ac = (\rho\sigma_v\,iu - \kappa)^2 - \sigma_v^2(iu - u^2) \]

\[ = (\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(u^2 - iu) \]

\[ = (\kappa - i\rho\sigma_v u)^2 + \sigma_v^2 u(u - i) \]

Define the discriminant square root:

\[ \gamma = \sqrt{(\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2)} \]

Complex Square Root

For complex \(u\), the square root \(\gamma\) is complex-valued, and its branch must be chosen carefully to ensure continuity. This branch-cut issue is addressed in the numerical stability section and is one of the main practical challenges in implementing the Heston characteristic function.

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Solving the Riccati Equation

Linearization via Substitution

The standard method for solving a constant-coefficient Riccati ODE is to substitute \(D = -\frac{2}{\sigma_v^2}\frac{h'}{h}\), which converts the quadratic ODE into a linear second-order ODE.

Proposition: Linearization of the D-Equation

The substitution \(D(\tau) = -\frac{2}{\sigma_v^2}\frac{h'(\tau)}{h(\tau)}\) transforms the Riccati ODE for \(D\) into the linear second-order ODE:

\[ h''(\tau) - \frac{b}{1}\,h'(\tau) + \frac{ac}{1}\cdot\frac{\sigma_v^2}{1}\,h(\tau) = 0 \]

More explicitly:

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

This is a constant-coefficient linear ODE whose characteristic equation is:

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

with roots:

\[ \lambda_{\pm} = \frac{-(\kappa - i\rho\sigma_v u) \pm \gamma}{2} \]

Proof. Substitute \(D = -\frac{2}{\sigma_v^2}\frac{h'}{h}\) into the Riccati ODE. Using \(D' = -\frac{2}{\sigma_v^2}\left(\frac{h''}{h} - \frac{(h')^2}{h^2}\right)\) and \(D^2 = \frac{4}{\sigma_v^4}\frac{(h')^2}{h^2}\):

\[ -\frac{2}{\sigma_v^2}\left(\frac{h''}{h} - \frac{(h')^2}{h^2}\right) = \frac{1}{2}\sigma_v^2 \cdot \frac{4}{\sigma_v^4}\frac{(h')^2}{h^2} + (\rho\sigma_v iu - \kappa)\left(-\frac{2}{\sigma_v^2}\frac{h'}{h}\right) + \frac{1}{2}(iu - u^2) \]

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

The \((h')^2/h^2\) terms cancel, giving:

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

Multiplying by \(-\sigma_v^2 h/2\):

\[ h'' + (\kappa - i\rho\sigma_v u)\,h' + \frac{\sigma_v^2}{4}(iu - u^2)\,h = 0 \]

Wait -- let us redo this more carefully. The Riccati ODE is \(D' = \frac{1}{2}\sigma_v^2 D^2 + bD + c\) where \(b = \rho\sigma_v iu - \kappa\) and \(c = \frac{1}{2}(iu - u^2)\). Substituting \(D = -\frac{2}{\sigma_v^2}\frac{h'}{h}\) and simplifying leads to \(h'' - b\,h' + \frac{\sigma_v^2 c}{2}\,h = 0\), i.e.:

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

\[ h'' + (\kappa - \rho\sigma_v iu)\,h' + \frac{1}{4}\sigma_v^2(iu-u^2)\,h = 0 \]

The characteristic roots of \(\mu^2 + (\kappa - \rho\sigma_v iu)\mu + \frac{1}{4}\sigma_v^2(iu-u^2) = 0\) are:

\[ \mu_\pm = \frac{-(\kappa - \rho\sigma_v iu) \pm \sqrt{(\kappa-\rho\sigma_v iu)^2 - \sigma_v^2(iu-u^2)}}{2} = \frac{-(\kappa - i\rho\sigma_v u) \pm \gamma}{2} \]

\(\square\)

General Solution

The general solution of the linear ODE is \(h(\tau) = A\,e^{\mu_+\tau} + B\,e^{\mu_-\tau}\). The constants \(A\) and \(B\) are determined by the initial conditions.

From \(D(0) = 0\), we need \(h'(0)/h(0) = 0\), hence \(h'(0) = 0\). This gives:

\[ A\mu_+ + B\mu_- = 0 \qquad \Longrightarrow \qquad \frac{A}{B} = -\frac{\mu_-}{\mu_+} \]

Substituting back into \(D = -\frac{2}{\sigma_v^2}\frac{h'}{h}\) yields the closed-form expression for \(D(\tau, u)\), which is derived in the next section.

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The C-Equation as Quadrature

Once \(D(\tau, u)\) is known, \(C(\tau, u)\) is obtained by direct integration:

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

Structure of the Solution

The term \((r-q)\,iu\,\tau\) is the drift contribution from the log-price. The integral \(\kappa\theta\int_0^\tau D(s)\,ds\) is the contribution from the mean-reversion drift of the variance process. Because \(D\) has an explicit form (involving exponentials and \(\gamma\)), this integral can also be evaluated in closed form, producing a logarithmic term.

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Worked Example: Riccati Coefficients at Small Time

Taylor Expansion Near tau = 0

At \(\tau = 0\), \(D(0) = 0\). The Riccati ODE gives:

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

Differentiating the Riccati ODE:

\[ D'' = \sigma_v^2 D\,D' + (\rho\sigma_v iu - \kappa)\,D' \]

At \(\tau = 0\): \(D''(0) = 0 + (\rho\sigma_v iu - \kappa)\cdot\frac{1}{2}(iu - u^2)\).

So the Taylor expansion is:

\[ D(\tau) \approx \tfrac{1}{2}(iu - u^2)\,\tau + \tfrac{1}{2}(\rho\sigma_v iu - \kappa)(iu - u^2)\,\frac{\tau^2}{2} + O(\tau^3) \]

For \(C\): \(C(\tau) \approx (r-q)iu\,\tau + \kappa\theta\cdot\frac{1}{2}(iu-u^2)\frac{\tau^2}{2} + O(\tau^3)\).

To zeroth order in \(\sigma_v\) (i.e., deterministic volatility), \(D(\tau) \approx \frac{1}{2}(iu - u^2)\tau\) and the characteristic function becomes:

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

This is exactly the Black-Scholes characteristic function with constant variance \(v\), confirming that the Heston model reduces to Black-Scholes when \(\sigma_v = 0\).

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Summary

Substituting the exponential-affine ansatz into the Feynman-Kac PDE yields a scalar Riccati ODE for \(D(\tau, u)\) and a simple quadrature for \(C(\tau, u)\). The \(D\)-equation has constant complex coefficients (parametrized by \(u\)) with discriminant \(\gamma^2 = (\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2)\). The standard Riccati-to-linear substitution \(D = -\frac{2}{\sigma_v^2}h'/h\) reduces the ODE to a constant-coefficient linear ODE with characteristic roots involving \(\gamma\). The initial condition \(D(0) = 0\) determines the integration constants. The closed-form solution for \(D\) and \(C\) -- the Heston characteristic function itself -- is presented in the next section.

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Exercises

Exercise 1. Write the \(D\)-equation: \(D' = \frac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v iu - \kappa)D + \frac{1}{2}(iu - u^2)\). Identify the coefficients \(\alpha\), \(\beta\), \(\gamma\) in the standard Riccati form \(D' = \alpha + \beta D + \frac{1}{2}\gamma D^2\) and compute the discriminant \(\gamma^2 = \beta^2 - 2\alpha\gamma\).

Solution to Exercise 1

The \(D\)-equation is:

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

To match the standard Riccati form \(D' = \alpha + \beta D + \frac{1}{2}\gamma_R D^2\) (using \(\gamma_R\) to distinguish from the discriminant), the coefficients are:

\[ \alpha = \frac{1}{2}(iu - u^2), \qquad \beta = \rho\sigma_v iu - \kappa, \qquad \gamma_R = \sigma_v^2 \]

Note the factor of \(\frac{1}{2}\) in front of \(\gamma_R D^2\), consistent with the original equation having \(\frac{1}{2}\sigma_v^2 D^2\).

The discriminant of the quadratic \(\frac{1}{2}\gamma_R D^2 + \beta D + \alpha = 0\) is:

\[ \Delta = \beta^2 - 2\alpha\gamma_R = (\rho\sigma_v iu - \kappa)^2 - 2 \cdot \frac{1}{2}(iu - u^2) \cdot \sigma_v^2 \]

\[ = (\kappa - i\rho\sigma_v u)^2 - \sigma_v^2(iu - u^2) \]

\[ = (\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(u^2 - iu) \]

\[ = (\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2) \]

(where the last step uses \(u^2 - iu = -(iu - u^2)\), and noting the sign: \(-\sigma_v^2(iu - u^2) = \sigma_v^2(u^2 - iu)\); the standard convention writes this as \(\sigma_v^2(iu + u^2)\) which equals \(\sigma_v^2(u^2 + iu)\).) The discriminant square root is:

\[ \gamma = \sqrt{\Delta} = \sqrt{(\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2)} \]

This is the key quantity appearing in the closed-form Heston characteristic function.

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Exercise 2. Apply the substitution \(D = -\frac{2}{\sigma_v^2}\frac{h'}{h}\) to transform the Riccati ODE into a second-order linear ODE for \(h(\tau)\). Solve \(h(\tau) = Ae^{r_+\tau} + Be^{r_-\tau}\) where \(r_\pm\) are the characteristic roots.

Solution to Exercise 2

Starting from the Riccati ODE \(D' = \frac{1}{2}\sigma_v^2 D^2 + bD + c\) where \(b = \rho\sigma_v iu - \kappa\) and \(c = \frac{1}{2}(iu - u^2)\), apply the substitution:

\[ D(\tau) = -\frac{2}{\sigma_v^2}\frac{h'(\tau)}{h(\tau)} \]

Computing \(D'\):

\[ D' = -\frac{2}{\sigma_v^2}\left(\frac{h''}{h} - \frac{(h')^2}{h^2}\right) \]

Computing \(D^2\):

\[ D^2 = \frac{4}{\sigma_v^4}\frac{(h')^2}{h^2} \]

Substituting into the Riccati ODE:

\[ -\frac{2}{\sigma_v^2}\frac{h''}{h} + \frac{2}{\sigma_v^2}\frac{(h')^2}{h^2} = \frac{1}{2}\sigma_v^2 \cdot \frac{4}{\sigma_v^4}\frac{(h')^2}{h^2} + b\left(-\frac{2}{\sigma_v^2}\frac{h'}{h}\right) + c \]

The \((h')^2/h^2\) terms cancel on both sides, leaving:

\[ -\frac{2}{\sigma_v^2}\frac{h''}{h} = -\frac{2b}{\sigma_v^2}\frac{h'}{h} + c \]

Multiplying by \(-\frac{\sigma_v^2}{2}h\):

\[ h'' - b\,h' + \frac{\sigma_v^2 c}{2}\,h = 0 \]

Substituting \(b = \rho\sigma_v iu - \kappa\) and \(c = \frac{1}{2}(iu - u^2)\):

\[ h'' + (\kappa - \rho\sigma_v iu)\,h' + \frac{\sigma_v^2}{4}(iu - u^2)\,h = 0 \]

This is a constant-coefficient second-order linear ODE with characteristic equation:

\[ r^2 + (\kappa - \rho\sigma_v iu)\,r + \frac{\sigma_v^2}{4}(iu - u^2) = 0 \]

By the quadratic formula:

\[ r_\pm = \frac{-(\kappa - i\rho\sigma_v u) \pm \sqrt{(\kappa - i\rho\sigma_v u)^2 - \sigma_v^2(iu - u^2)}}{2} = \frac{-(\kappa - i\rho\sigma_v u) \pm \gamma}{2} \]

The general solution is \(h(\tau) = Ae^{r_+\tau} + Be^{r_-\tau}\).

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Exercise 3. From the initial condition \(D(0) = 0\), determine the ratio \(A/B\) in terms of \(r_+\) and \(r_-\). Substitute back to obtain \(D(\tau)\) in closed form.

Solution to Exercise 3

From \(D(0) = 0\) and \(D = -\frac{2}{\sigma_v^2}\frac{h'}{h}\), we need \(h'(0) = 0\) (with \(h(0) \neq 0\)).

With \(h(\tau) = Ae^{r_+\tau} + Be^{r_-\tau}\):

\[ h'(\tau) = Ar_+e^{r_+\tau} + Br_-e^{r_-\tau} \]

At \(\tau = 0\): \(h'(0) = Ar_+ + Br_- = 0\), so:

\[ \frac{A}{B} = -\frac{r_-}{r_+} \]

Now compute \(D(\tau)\):

\[ D = -\frac{2}{\sigma_v^2}\frac{Ar_+e^{r_+\tau} + Br_-e^{r_-\tau}}{Ae^{r_+\tau} + Be^{r_-\tau}} \]

Substitute \(A = -\frac{r_-}{r_+}B\) and divide numerator and denominator by \(Be^{r_-\tau}\):

\[ D = -\frac{2}{\sigma_v^2}\frac{-\frac{r_-}{r_+}\cdot r_+ \cdot e^{(r_+-r_-)\tau} + r_-}{-\frac{r_-}{r_+}\cdot e^{(r_+-r_-)\tau} + 1} \]

\[ = -\frac{2}{\sigma_v^2}\frac{r_-(-e^{(r_+-r_-)\tau} + 1)}{-\frac{r_-}{r_+}e^{(r_+-r_-)\tau} + 1} \]

\[ = -\frac{2}{\sigma_v^2}\frac{r_-(1 - e^{(r_+-r_-)\tau})}{1 - \frac{r_-}{r_+}e^{(r_+-r_-)\tau}} \]

Note that \(r_+ - r_- = \gamma\) and define \(g = r_-/r_+\). Then:

\[ D(\tau) = -\frac{2r_-}{\sigma_v^2}\frac{1 - e^{\gamma\tau}}{1 - g\,e^{\gamma\tau}} \]

Since \(r_- = \frac{-(\kappa - i\rho\sigma_v u) - \gamma}{2}\), we have \(-\frac{2r_-}{\sigma_v^2} = \frac{\kappa - i\rho\sigma_v u + \gamma}{\sigma_v^2}\), giving the Heston 1993 form. Alternatively, multiplying numerator and denominator by \(e^{-\gamma\tau}\) and using \(g_{\text{Alb}} = r_+/r_-\):

\[ D(\tau) = \frac{\kappa - i\rho\sigma_v u - \gamma}{\sigma_v^2}\cdot\frac{1 - e^{-\gamma\tau}}{1 - g_{\text{Alb}}\,e^{-\gamma\tau}} \]

which is the Albrecher form with \(g_{\text{Alb}} = \frac{\kappa - i\rho\sigma_v u - \gamma}{\kappa - i\rho\sigma_v u + \gamma}\).

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Exercise 4. The \(C\)-equation is \(C'(\tau) = (r-q)iu + \kappa\theta D(\tau)\). Given \(D(\tau)\) in closed form, perform the integration to obtain \(C(\tau)\) and verify that \(C(0) = 0\).

Solution to Exercise 4

The \(C\)-equation is \(C'(\tau) = (r-q)iu + \kappa\theta D(\tau)\) with \(C(0) = 0\). Integrating:

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

Using the Albrecher form \(D(s) = D_+ \cdot \frac{1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}}\) where \(D_+ = \frac{\kappa - i\rho\sigma_v u - \gamma}{\sigma_v^2}\):

\[ \int_0^\tau \frac{1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}}\,ds \]

Split: \(\frac{1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}} = 1 + \frac{(1-g)e^{-\gamma s} - 1 + 1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}} = 1 - \frac{(1-g)}{1-g\,e^{-\gamma s}} + \frac{1-g}{1-g\,e^{-\gamma s}}\). More directly, use the substitution \(w = e^{-\gamma s}\), \(dw = -\gamma e^{-\gamma s}\,ds\):

\[ \int_0^\tau \frac{1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}}\,ds = \tau + \frac{1}{\gamma}\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1 - g}\right) \]

This can be verified by differentiating with respect to \(\tau\): \(\frac{d}{d\tau}\left[\tau + \frac{1}{\gamma}\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right)\right] = 1 + \frac{1}{\gamma}\cdot\frac{g\gamma e^{-\gamma\tau}}{1 - g\,e^{-\gamma\tau}} = \frac{1 - g\,e^{-\gamma\tau} + g\,e^{-\gamma\tau}}{1 - g\,e^{-\gamma\tau}} = \frac{1}{1 - g\,e^{-\gamma\tau}}\). Hmm, this gives \(\frac{1}{1 - g\,e^{-\gamma\tau}}\), not \(\frac{1 - e^{-\gamma\tau}}{1 - g\,e^{-\gamma\tau}}\), so the integral formula needs refinement.

The correct evaluation proceeds by writing:

\[ \frac{1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}} = \frac{1}{1 - g\,e^{-\gamma s}} - \frac{e^{-\gamma s}}{1 - g\,e^{-\gamma s}} \]

For the second term, \(\int_0^\tau \frac{e^{-\gamma s}}{1 - g\,e^{-\gamma s}}\,ds = -\frac{1}{\gamma}\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1 - g}\right)\) (by the substitution \(w = g\,e^{-\gamma s}\)).

For the first term, \(\int_0^\tau \frac{ds}{1 - g\,e^{-\gamma s}}\), substitute \(w = e^{-\gamma s}\):

\[ = -\frac{1}{\gamma}\int_1^{e^{-\gamma\tau}} \frac{dw}{w(1 - gw)} = -\frac{1}{\gamma}\int_1^{e^{-\gamma\tau}}\left(\frac{1}{w} + \frac{g}{1-gw}\right)dw \]

\[ = -\frac{1}{\gamma}\left[-\gamma\tau - \ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right)\right] = \tau + \frac{1}{\gamma}\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right) \]

Combining: \(\int_0^\tau D_+\frac{1 - e^{-\gamma s}}{1 - g\,e^{-\gamma s}}\,ds = D_+\left[\tau + \frac{2}{\gamma}\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right)\right]\). Wait --- the two integrals give \(\left[\tau + \frac{1}{\gamma}\ln(\cdots)\right] - \left[-\frac{1}{\gamma}\ln(\cdots)\right] = \tau + \frac{2}{\gamma}\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right)\).

Therefore:

\[ C(\tau) = (r-q)iu\,\tau + \frac{\kappa\theta}{\sigma_v^2}\left[(\kappa - i\rho\sigma_v u - \gamma)\tau - 2\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right)\right] \]

Verification that \(C(0) = 0\): At \(\tau = 0\), the linear term gives \(0\) and the logarithm gives \(\ln(1) = 0\), so \(C(0) = 0\). \(\checkmark\)

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Exercise 5. For parameters \(\kappa = 2\), \(\sigma_v = 0.3\), \(\rho = -0.7\), compute the discriminant \(\gamma\) at \(u = 1\) and verify \(\operatorname{Re}(\gamma) > 0\).

Solution to Exercise 5

For \(\kappa = 2\), \(\sigma_v = 0.3\), \(\rho = -0.7\), \(u = 1\):

\[ \gamma^2 = (\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2) \]

\[ = (2 - i(-0.7)(0.3)(1))^2 + (0.3)^2(i + 1) \]

\[ = (2 + 0.21i)^2 + 0.09(1 + i) \]

\[ = (4 + 0.84i - 0.0441) + (0.09 + 0.09i) \]

\[ = 4.0459 + 0.93i \]

To compute \(\gamma = \sqrt{4.0459 + 0.93i}\), use polar form. The modulus is \(|\gamma^2| = \sqrt{4.0459^2 + 0.93^2} = \sqrt{16.369 + 0.865} = \sqrt{17.234} \approx 4.151\), and the argument is \(\theta = \arctan(0.93/4.0459) \approx 0.2262\) radians.

Therefore \(|\gamma| = \sqrt{4.151} \approx 2.037\) and \(\arg(\gamma) = 0.2262/2 \approx 0.1131\) radians. So:

\[ \gamma \approx 2.037(\cos 0.1131 + i\sin 0.1131) \approx 2.037(0.9936 + 0.1129i) \approx 2.024 + 0.230i \]

Since \(\operatorname{Re}(\gamma) \approx 2.024 > 0\), the condition \(\operatorname{Re}(\gamma) > 0\) is verified. This ensures that the Albrecher formulation uses decaying exponentials \(e^{-\gamma\tau}\) and that \(|g| < 1\).

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Exercise 6. Explain why the \(D\)-equation is autonomous in \(D\) (it does not involve \(C\)), while the \(C\)-equation depends on \(D\). How does this hierarchical structure simplify the solution process?

Solution to Exercise 6

The hierarchical (triangular) structure of the Riccati system arises from the affine structure of the Heston model and the form of the exponential-affine ansatz.

Why \(D\) is autonomous: The \(D\)-equation \(D' = \frac{1}{2}\sigma_v^2 D^2 + (\rho\sigma_v iu - \kappa)D + \frac{1}{2}(iu - u^2)\) collects all terms proportional to \(v\) in the PDE. Since the PDE coefficients that multiply \(v\) (namely, the drift \(-\frac{1}{2}v\) and \(-\kappa v\), and the diffusion terms \(\frac{1}{2}v\), \(\rho\sigma_v v\), \(\frac{1}{2}\sigma_v^2 v\)) depend only on the model parameters and \(D\) (through the ansatz), but not on \(C\), the resulting ODE for \(D\) is self-contained.

Mathematically, \(C\) enters the ansatz only through the \(v\)-independent part of the exponent. When collecting \(v^1\) terms in the PDE, the \(C\)-dependent terms (which are all \(v^0\)) do not contribute. Hence \(D' = f(D, u)\) with no \(C\) dependence.

Why \(C\) depends on \(D\): The \(C\)-equation \(C' = (r-q)iu + \kappa\theta D\) collects the \(v^0\) terms. The term \(\kappa\theta D\) arises from the mean-reversion drift \(\kappa\theta\,\partial_v\varphi = \kappa\theta\,D\,\varphi\), which contributes to the \(v^0\) coefficient. Thus \(C'\) depends on \(D\) through this coupling.

Simplification: The hierarchical structure means the system can be solved sequentially rather than simultaneously:

1. First, solve the \(D\)-equation as a standalone scalar Riccati ODE. This is the harder step (nonlinear ODE), but it is a single equation in one unknown.
2. Second, with \(D(\tau)\) known, compute \(C(\tau)\) by direct integration (quadrature): \(C(\tau) = (r-q)iu\tau + \kappa\theta\int_0^\tau D(s)\,ds\). This is a trivial integration step --- no ODE-solving is required.

Without this hierarchical structure, we would face a coupled system of two nonlinear ODEs, which would be considerably harder to solve in closed form. The decoupling is a direct consequence of the affine property: the state variable \(v\) enters the drift and covariance linearly, ensuring that the \(v^1\) and \(v^0\) coefficient equations separate cleanly.