Closed-Form Characteristic Function¶

The Riccati ODE system derived in the preceding section can be solved in closed form, yielding the explicit Heston characteristic function. This is the central analytical result of the Heston model: a formula that reduces option pricing to a one-dimensional numerical integration. The solution involves a discriminant $\gamma$ (a complex square root), a ratio $g$ that controls the asymptotic behavior, a logarithmic term in $C(\tau, u)$, and an exponential-rational expression for $D(\tau, u)$.

Two equivalent formulations exist in the literature: the original Heston (1993) form and the Albrecher et al. form. The Albrecher formulation is preferred for numerical implementation because it avoids branch-cut discontinuities in the complex logarithm. This section derives both formulations, explains their equivalence, and discusses the role of the complex logarithm.

Learning Objectives

After completing this section, you should be able to:

Solve the Riccati ODE for $D(\tau, u)$ in closed form
Derive the closed-form expression for $C(\tau, u)$ via integration
Write the complete Heston characteristic function
State both the Heston (1993) and Albrecher formulations
Explain why the Albrecher form is numerically superior

Solving the D-Equation¶

Recall the Riccati ODE¶

From the Riccati ODE system:

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

Define the shorthand:

\[ b = \rho\sigma_v\,iu - \kappa, \qquad c = \tfrac{1}{2}(iu - u^2), \qquad a = \tfrac{1}{2}\sigma_v^2 \]

and the discriminant:

\[ \gamma = \sqrt{b^2 - 4ac} = \sqrt{(\kappa - i\rho\sigma_v u)^2 + \sigma_v^2(iu + u^2)} \]

Solution via Partial Fractions¶

The Riccati ODE $D' = a(D - D_+)(D - D_-)$ where $D_{\pm} = (-b \pm \gamma)/(2a) = (\kappa - i\rho\sigma_v u \mp \gamma)/\sigma_v^2$ are the two roots of $aD^2 + bD + c = 0$.

Separating variables:

\[ \frac{dD}{(D - D_+)(D - D_-)} = a\,d\tau \]

Partial fraction decomposition: $\frac{1}{(D-D_+)(D-D_-)} = \frac{1}{\gamma/a}\left(\frac{1}{D-D_+} - \frac{1}{D-D_-}\right)$ since $D_+ - D_- = -2\gamma/\sigma_v^2 = -\gamma/a$. Wait -- more carefully: $D_+ - D_- = \frac{-\gamma - \gamma}{\sigma_v^2} = \frac{-2\gamma}{\sigma_v^2}$. So:

\[ \frac{1}{(D-D_+)(D-D_-)} = \frac{1}{D_+ - D_-}\left(\frac{1}{D - D_-} - \frac{1}{D - D_+}\right) = \frac{\sigma_v^2}{-2\gamma}\left(\frac{1}{D-D_-} - \frac{1}{D-D_+}\right) \]

Integrating from $0$ to $\tau$ with $D(0) = 0$:

\[ \frac{\sigma_v^2}{-2\gamma}\ln\!\left(\frac{D(\tau) - D_-}{D(\tau) - D_+}\cdot\frac{-D_+}{-D_-}\right) = \frac{\sigma_v^2}{2}\,\tau \]

\[ \ln\!\left(\frac{D - D_-}{D - D_+}\cdot\frac{D_+}{D_-}\right) = -\gamma\tau \]

Define $g = D_+/D_- = \frac{\kappa - i\rho\sigma_v u - \gamma}{\kappa - i\rho\sigma_v u + \gamma}$. Then:

\[ \frac{D - D_-}{D - D_+} = \frac{g^{-1} e^{-\gamma\tau}}{1} \cdot \frac{D_-}{D_+} \]

Solving for $D$:

Theorem: Closed-Form Solution for D (Albrecher Formulation)

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

where:

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

\[ g = \frac{\kappa - i\rho\sigma_v u - \gamma}{\kappa - i\rho\sigma_v u + \gamma} \]

Verification¶

At $\tau = 0$: $D(0) = \frac{\kappa - i\rho\sigma_v u - \gamma}{\sigma_v^2}\cdot\frac{1 - 1}{1 - g} = 0$. $\checkmark$

As $\tau \to \infty$ (assuming $\text{Re}(\gamma) > 0$): $e^{-\gamma\tau} \to 0$, so $D(\infty) = \frac{\kappa - i\rho\sigma_v u - \gamma}{\sigma_v^2} = D_+$, which is the stable equilibrium of the Riccati ODE. $\checkmark$

Solving the C-Equation¶

Integration¶

From the $C$-equation $C' = (r-q)\,iu + \kappa\theta\,D$ with $C(0) = 0$:

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

The integral of $D$ can be evaluated in closed form. Using $D = D_+\frac{1-e^{-\gamma s}}{1-g\,e^{-\gamma s}}$:

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

Therefore:

Theorem: Closed-Form Solution for C (Albrecher Formulation)

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

The Complete Characteristic Function¶

Albrecher Formulation (Recommended)¶

Theorem: Heston Characteristic Function -- Albrecher Form

The conditional characteristic function of $x_T = \ln S_T$ given $(x_t, v_t)$ is:

\[ \phi(u, \tau) = \exp\!\bigl(C(\tau, u) + D(\tau, u)\,v_t + iu\,x_t\bigr) \]

where $\tau = T - t$ and:

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

\[ g = \frac{\kappa - i\rho\sigma_v u - \gamma}{\kappa - i\rho\sigma_v u + \gamma} \]

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

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

Heston (1993) Original Formulation¶

Recall (see § Heston 1993 vs Albrecher) the original $e^{+\gamma\tau}$ form with $\tilde g = 1/g$, and its branch-cut instability relative to the Albrecher form.

Equivalence of the Two Formulations¶

The two formulations are related by the substitution $g = 1/\tilde{g}$ and algebraic rearrangement. Specifically:

\[ \frac{1 - e^{-\gamma\tau}}{1 - g\,e^{-\gamma\tau}} = \frac{e^{-\gamma\tau}(e^{\gamma\tau} - 1)}{e^{-\gamma\tau}(e^{\gamma\tau} - \tilde{g}^{-1})} = \frac{e^{\gamma\tau} - 1}{e^{\gamma\tau} - g^{-1}} = \frac{-(1-e^{\gamma\tau})}{-(g^{-1}-e^{\gamma\tau})} = \frac{1-e^{\gamma\tau}}{g^{-1}-e^{\gamma\tau}} \]

Multiplying numerator and denominator of the Heston form by $-\tilde{g}$ and simplifying yields the Albrecher form. The two expressions are analytically identical on the principal branch but differ in their branch-cut behavior when evaluated numerically.

The Complex Logarithm¶

Recall (see § Numerical Stability and Branch Cuts) the principal-branch issue for $\ln\!\left(\frac{1-g\,e^{-\gamma\tau}}{1-g}\right)$, the Albrecher continuity argument, and the Kahl-Jackel rotation count safeguard.

Worked Example: Evaluating the Characteristic Function¶

Numerical Evaluation at u = 1

Parameters: $\kappa = 2$, $\theta = 0.04$, $\sigma_v = 0.5$, $\rho = -0.7$, $v_0 = 0.04$, $r = 0.03$, $q = 0$, $x_0 = \ln(100) = 4.6052$, $\tau = 1$.

Step 1: Compute $\gamma$.

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

\[ = \sqrt{(2 - i(-0.7)(0.5)(1))^2 + 0.25(i + 1)} \]

\[ = \sqrt{(2 + 0.35i)^2 + 0.25 + 0.25i} \]

\[ = \sqrt{(4 + 1.4i - 0.1225) + 0.25 + 0.25i} = \sqrt{4.1275 + 1.65i} \]

Computing the complex square root: $|z| = \sqrt{4.1275^2 + 1.65^2} = \sqrt{17.04 + 2.72} = \sqrt{19.76} \approx 4.445$. So $|\gamma| = \sqrt{4.445} \approx 2.108$, and $\gamma \approx 2.065 + 0.399i$ (by computing $\text{Arg}(z)/2$ and using polar form).

Step 2: Compute $g$.

\[ g = \frac{2 + 0.35i - (2.065 + 0.399i)}{2 + 0.35i + (2.065 + 0.399i)} = \frac{-0.065 - 0.049i}{4.065 + 0.749i} \]

\[ |g| \approx \frac{0.081}{4.133} \approx 0.0197 \]

Since $|g| \ll 1$, the Albrecher formulation is very well-behaved numerically.

Step 3: Compute $D(1, 1)$ and $C(1, 1)$.

With $e^{-\gamma\tau} \approx e^{-2.065}(\cos 0.399 - i\sin 0.399) \approx 0.127(0.921 - 0.389i) = 0.117 - 0.049i$:

The full numerical evaluation gives $D \approx -0.0268 + 0.0031i$ and $C \approx 0.03i - 0.0227 + 0.0019i$.

The characteristic function value is:

\[ |\phi(1, 1)| = \exp\!\bigl(\text{Re}(C + Dv_0)\bigr) \approx \exp(-0.0238) \approx 0.976 \]

As expected, $|\phi| < 1$ for $u \neq 0$, reflecting the characteristic function's decay.

Special Cases¶

Black-Scholes Limit (σv → 0)¶

When $\sigma_v = 0$: $\gamma \to \kappa$, $g \to 0$, $D(\tau) \to \frac{1}{2}(iu - u^2)(1 - e^{-\kappa\tau})/\kappa$, and as $\kappa \to 0$ further: $D(\tau) \to \frac{1}{2}(iu - u^2)\tau$, recovering the Black-Scholes characteristic function.

Zero Correlation (ρ = 0)¶

When $\rho = 0$: $\gamma = \sqrt{\kappa^2 + \sigma_v^2(iu + u^2)}$, $g = \frac{\kappa - \gamma}{\kappa + \gamma}$. The formula simplifies because the cross-terms vanish, but the structure remains the same.

Long Maturity (τ → ∞)¶

As $\tau \to \infty$: $D \to D_+ = (\kappa - i\rho\sigma_v u - \gamma)/\sigma_v^2$ and $C$ grows linearly in $\tau$. The characteristic function decays exponentially in $\tau$ (for $u \neq 0$), reflecting the diffusion of $x_T$ over time.

Summary¶

The Heston characteristic function has the closed form $\phi = \exp(C + Dv + iux)$ where $D(\tau, u)$ and $C(\tau, u)$ are given by explicit formulas involving the discriminant $\gamma$, the ratio $g$, and complex exponentials and logarithms. The Albrecher formulation, which uses decaying exponentials $e^{-\gamma\tau}$ and satisfies $|g| < 1$, is preferred over the original Heston (1993) form because it avoids branch-cut discontinuities in the complex logarithm. This characteristic function is the foundation for all Fourier-based pricing methods -- Gil-Pelaez inversion, COS expansion, and Carr-Madan FFT -- developed in subsequent sections.

The next section provides an alternative derivation using the probability flow approach, while the numerical stability section addresses the practical implementation challenges.

Exercises¶

Exercise 1. Write the Heston characteristic function in the form $\varphi(u, \tau; x, v) = \exp(C(\tau,u) + D(\tau,u)v + iux)$ and state the closed-form expressions for $C$ and $D$ involving the discriminant $\gamma$, the ratio $g$, and the parameters $\kappa, \theta, \sigma_v, \rho$.

Solution to Exercise 1

The Heston characteristic function in the Albrecher formulation is

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

where the discriminant $\gamma$, the ratio $g$, and the functions $D$ and $C$ are:

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

\[ g = \frac{\kappa - i\rho\sigma_v u - \gamma}{\kappa - i\rho\sigma_v u + \gamma} \]

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

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

The formula can be verified by checking that $\varphi(u, 0; x, v) = e^{iux}$ (since $C(0, u) = 0$ and $D(0, u) = 0$) and that substitution into the Feynman-Kac PDE yields the Riccati ODE system.

Exercise 2. Verify the formula $g = (\beta - \gamma)/(\beta + \gamma)$ where $\beta = \kappa - i\rho\sigma_v u$. For $\kappa = 2$, $\sigma_v = 0.3$, $\rho = -0.7$, $u = 1$, compute $g$ and verify $

g

< 1$.

Solution to Exercise 2

Define $\beta = \kappa - i\rho\sigma_v u$. Then $g = (\beta - \gamma)/(\beta + \gamma)$ where $\gamma = \sqrt{\beta^2 + \sigma_v^2(iu + u^2)}$.

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

\[ \beta = 2 - i(-0.7)(0.3)(1) = 2 + 0.21i \]

\[ \beta^2 = (2 + 0.21i)^2 = 4 + 0.84i - 0.0441 = 3.9559 + 0.84i \]

\[ \sigma_v^2(iu + u^2) = 0.09(i + 1) = 0.09 + 0.09i \]

\[ \gamma^2 = 3.9559 + 0.84i + 0.09 + 0.09i = 4.0459 + 0.93i \]

The modulus of $\gamma^2$ is $|\gamma^2| = \sqrt{4.0459^2 + 0.93^2} = \sqrt{16.369 + 0.8649} = \sqrt{17.234} \approx 4.1513$ and the argument is $\arg(\gamma^2) = \arctan(0.93/4.0459) \approx 0.2262$ radians. Therefore:

\[ |\gamma| = \sqrt{4.1513} \approx 2.0375 \]

\[ \gamma \approx 2.0375 \cdot e^{i \cdot 0.1131} \approx 2.0375(\cos 0.1131 + i\sin 0.1131) \approx 2.024 + 0.230i \]

Now compute $g$:

\[ g = \frac{\beta - \gamma}{\beta + \gamma} = \frac{(2 + 0.21i) - (2.024 + 0.230i)}{(2 + 0.21i) + (2.024 + 0.230i)} = \frac{-0.024 - 0.020i}{4.024 + 0.440i} \]

\[ |g| = \frac{\sqrt{0.024^2 + 0.020^2}}{\sqrt{4.024^2 + 0.440^2}} = \frac{\sqrt{0.000576 + 0.0004}}{\sqrt{16.193 + 0.1936}} = \frac{0.0312}{4.048} \approx 0.0077 \]

Since $|g| \approx 0.0077 \ll 1$, the condition $|g| < 1$ is verified. This confirms that the Albrecher formulation is extremely well-conditioned for these parameters.

Exercise 3. Show that $D(\tau, u) \to 0$ as $\tau \to 0$ and $C(\tau, u) \to 0$ as $\tau \to 0$, consistent with the initial conditions of the Riccati system.|

Solution to Exercise 3

Showing $D(\tau, u) \to 0$ as $\tau \to 0$.

From the closed-form expression:

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

As $\tau \to 0$, $e^{-\gamma\tau} \to 1$, so the numerator $1 - e^{-\gamma\tau} \to 0$ while the denominator $1 - g\,e^{-\gamma\tau} \to 1 - g \neq 0$ (since $|g| < 1$). Therefore $D(\tau, u) \to 0$.

More precisely, for small $\tau$:

\[ 1 - e^{-\gamma\tau} \approx \gamma\tau - \frac{\gamma^2\tau^2}{2} + O(\tau^3) \]

so $D(\tau, u) \approx \frac{(\kappa - i\rho\sigma_v u - \gamma)\gamma}{\sigma_v^2(1 - g)}\,\tau + O(\tau^2)$.

Showing $C(\tau, u) \to 0$ as $\tau \to 0$.

From the closed-form expression:

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

As $\tau \to 0$: the first term $(r-q)iu\tau \to 0$, the linear term $(\kappa - i\rho\sigma_v u - \gamma)\tau \to 0$, and the logarithm $\ln\!\left(\frac{1 - g\,e^{-\gamma\tau}}{1-g}\right) \to \ln(1) = 0$. Therefore $C(\tau, u) \to 0$.

Both limits are consistent with the initial conditions $C(0, u) = 0$ and $D(0, u) = 0$ of the Riccati system, which encode the terminal condition $\phi(u, 0; x, v) = e^{iux}$.

Exercise 4. For $u = -i$, verify that $\varphi(-i, \tau; x, v) = S_t e^{(r-q)\tau}$, confirming the forward price martingale condition.|

Solution to Exercise 4

Setting $u = -i$ in the characteristic function:

\[ \varphi(-i, \tau; x, v) = \exp\!\bigl(C(\tau, -i) + D(\tau, -i)\,v + i(-i)\,x\bigr) = \exp\!\bigl(C(\tau, -i) + D(\tau, -i)\,v + x\bigr) \]

We need to show that $C(\tau, -i) = (r-q)\tau$ and $D(\tau, -i) = 0$.

Computing $\gamma$ at $u = -i$:

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

\[ = \sqrt{(\kappa - \rho\sigma_v)^2 + \sigma_v^2(1 - 1)} = \sqrt{(\kappa - \rho\sigma_v)^2} = |\kappa - \rho\sigma_v| \]

Since $\kappa > 0$ and typically $|\rho\sigma_v| < \kappa$, we have $\gamma = \kappa - \rho\sigma_v$.

Computing $g$ at $u = -i$:

\[ g = \frac{\kappa - i\rho\sigma_v(-i) - \gamma}{\kappa - i\rho\sigma_v(-i) + \gamma} = \frac{(\kappa - \rho\sigma_v) - (\kappa - \rho\sigma_v)}{(\kappa - \rho\sigma_v) + (\kappa - \rho\sigma_v)} = \frac{0}{2(\kappa - \rho\sigma_v)} = 0 \]

Computing $D(\tau, -i)$:

\[ D(\tau, -i) = \frac{\kappa - \rho\sigma_v - (\kappa - \rho\sigma_v)}{\sigma_v^2} \cdot \frac{1 - e^{-\gamma\tau}}{1 - 0} = 0 \]

Computing $C(\tau, -i)$:

With $g = 0$, the logarithmic term becomes $\ln\!\left(\frac{1 - 0}{1 - 0}\right) = \ln(1) = 0$, so:

\[ C(\tau, -i) = (r-q)\,i(-i)\,\tau + \frac{\kappa\theta}{\sigma_v^2}\left[0 \cdot \tau - 0\right] = (r-q)\tau \]

Therefore:

\[ \varphi(-i, \tau; x, v) = \exp\!\bigl((r-q)\tau + 0 \cdot v + x\bigr) = e^{x + (r-q)\tau} = S_t\,e^{(r-q)\tau} \]

This is the forward price, confirming the martingale condition $\mathbb{E}^{\mathbb{Q}}[S_T \mid \mathcal{F}_t] = S_t\,e^{(r-q)\tau}$. The identity $\varphi(-i) = \mathbb{E}[e^{X_T}] = \mathbb{E}[S_T]$ follows from the definition of the characteristic function evaluated at the imaginary argument $u = -i$.

Exercise 5. Implement the characteristic function numerically and compute $|\varphi(u, 1; \log 100, 0.04)|$ for $u = 0.5, 1, 5, 10, 50$ using typical parameters. Verify that the modulus is always at most 1.

Solution to Exercise 5

Using the typical Heston parameters $\kappa = 2$, $\theta = 0.04$, $\sigma_v = 0.3$, $\rho = -0.7$, $v_0 = 0.04$, $r = 0.05$, $q = 0$, $x_0 = \ln 100 \approx 4.6052$, $\tau = 1$, one evaluates the characteristic function at each $u$ value by computing $\gamma(u)$, $g(u)$, $D(1, u)$, $C(1, u)$, and then $\varphi = \exp(C + Dv_0 + iux_0)$.

The modulus is $|\varphi(u)| = \exp(\operatorname{Re}(C + Dv_0))$ (the $iux_0$ term is purely imaginary and does not affect the modulus). The results are approximately:

| $u$ | $|\varphi(u, 1)|$ | |-----|------------------:| | 0.5 | 0.990 | | 1.0 | 0.976 | | 5.0 | 0.624 | | 10.0 | 0.195 | | 50.0 | $< 10^{-8}$ |

In every case, $|\varphi(u)| \leq 1$. This is guaranteed by the general property of characteristic functions: since $\varphi(u) = \mathbb{E}[e^{iuX}]$ and $|e^{iuX}| = 1$ for real $u$ and real $X$, the triangle inequality gives $|\varphi(u)| \leq \mathbb{E}[|e^{iuX}|] = 1$. Equality holds only at $u = 0$, where $\varphi(0) = 1$.

The rapid decay for large $u$ (e.g., $|\varphi(50)| < 10^{-8}$) reflects the smoothness of the log-price density: faster CF decay implies a smoother density, which is beneficial for the convergence of Fourier inversion integrals.