Affine Structure and Characteristic Function¶
A key advantage of the Heston model is its affine structure, which enables semi-closed-form pricing of European options via characteristic functions. This section derives the characteristic function explicitly and explains its use in option pricing.
Affine Processes¶
Definition¶
A Markov process \((X_t, V_t)\) is affine if its conditional characteristic function has exponential-affine form:
where \(\tau = T - t\) and the functions \(A, B\) satisfy ordinary differential equations (Riccati equations).
Why Affine Structure Matters¶
- Semi-closed form: Characteristic function given by solving ODEs
- Fast evaluation: No Monte Carlo needed for European options
- Gradient availability: Derivatives for calibration are tractable
- Fourier pricing: Efficient methods via FFT
The Heston model belongs to the affine class, making it computationally tractable.
Characteristic Function of Log-Price¶
Setup¶
Let \(X_t = \log S_t\). Under the risk-neutral measure \(\mathbb{Q}\):
with \(\langle W^S, W^V \rangle_t = \rho t\).
The Characteristic Function¶
The conditional characteristic function of \(X_T\) is:
where \(\tau = T - t\).
Derivation of Riccati Equations¶
Feynman–Kac Approach¶
Define \(f(t, x, v) = \mathbb{E}[e^{iuX_T} | X_t = x, V_t = v]\).
By Feynman–Kac, \(f\) satisfies the PDE:
with terminal condition \(f(T, x, v) = e^{iux}\), where \(\mathcal{L}\) is the infinitesimal generator.
Generator Calculation¶
The generator is:
Affine Ansatz¶
Substitute the ansatz \(f = \exp(C + Dv + iux)\):
Collecting Terms¶
Substituting into the PDE and dividing by \(f\):
Separating terms constant in \(v\) and linear in \(v\):
Constant terms:
Linear in \(v\):
Solving the Riccati Equations¶
The D Equation (Riccati)¶
with terminal condition \(D(0, u) = 0\).
This is a Riccati ODE with solution:
where:
The C Equation (Linear)¶
with terminal condition \(C(0, u) = 0\).
Integrating:
Complete Characteristic Function¶
Final Formula¶
with:
Alternative Formulation (Lord–Kahl)¶
For numerical stability, use the alternative form:
This avoids numerical issues when \(|g| > 1\).
Numerical Stability Issues¶
The "Little Heston Trap"¶
The original Heston formulation can suffer from: 1. Branch cut issues: Complex logarithm discontinuities 2. Cancellation errors: Near-zero denominators 3. Overflow: Large exponentials
Rotation Count Method¶
Track the branch of the complex logarithm by counting rotations:
where \(n\) is chosen for continuity as \(u\) varies.
Practical Recommendations¶
- Use the Lord–Kahl formulation
- Implement rotation counting for \(\ln(1 - ge^{d\tau})\)
- Test against known limits (Black–Scholes when \(\xi \to 0\))
- Validate with Monte Carlo for extreme parameters
Moments from Characteristic Function¶
Moment Generating Property¶
The moments of \(X_T = \log S_T\) are obtained from:
First Two Moments¶
Mean of log-price:
where \(\mathbb{E}[\int_0^T V_s\,ds] = \theta T + (V_0 - \theta)\frac{1 - e^{-\kappa T}}{\kappa}\).
Variance of log-price:
The full expression involves the covariance between \(X_T\) and \(\int_0^T V_s\,ds\).
Option Pricing via Fourier Methods¶
Density Recovery¶
The risk-neutral density of \(X_T\) is:
Gil-Pelaez Inversion¶
For the distribution function:
Call Price Formula (Heston's Original)¶
where:
and \(\varphi_1\), \(\varphi_2\) are modified characteristic functions.
Carr–Madan Formula¶
See Section 9.4 for the complete Fourier pricing methodology.
Key Takeaways¶
- Affine structure yields exponential-affine characteristic functions
- Heston CF is given by explicit formulas involving Riccati solutions
- Numerical stability requires careful implementation (Lord–Kahl formulation)
- The characteristic function enables Fourier-based option pricing
- Moments and Greeks can be derived from CF derivatives
Further Reading¶
- Heston, S. (1993). A closed-form solution for options with stochastic volatility. Review of Financial Studies.
- Duffie, D., Pan, J., & Singleton, K. (2000). Transform analysis and asset pricing for affine jump-diffusions. Econometrica.
- Lord, R. & Kahl, C. (2010). Complex logarithms in Heston-like models. Mathematical Finance.
- Albrecher, H., Mayer, P., Schoutens, W., & Tistaert, J. (2007). The little Heston trap. Wilmott Magazine.
- Carr, P. & Madan, D. (1999). Option valuation using the fast Fourier transform. Journal of Computational Finance.
Exercises¶
Exercise 1. Verify the affine structure of the Heston model by showing that the conditional characteristic function \(\mathbb{E}[e^{iuX_T} | X_t, V_t]\) has the exponential-affine form \(\exp(C(\tau, u) + D(\tau, u)V_t + iuX_t)\). Start from the Feynman-Kac PDE and substitute the affine ansatz. Identify the two ODEs that result from separating terms constant in \(V\) and linear in \(V\).
Solution to Exercise 1
We start from the Feynman--Kac PDE. Define \(f(t, x, v) = \mathbb{E}[e^{iuX_T} | X_t = x, V_t = v]\). This satisfies:
with terminal condition \(f(T, x, v) = e^{iux}\).
Substitute the affine ansatz \(f = \exp(C(\tau, u) + D(\tau, u)v + iux)\) where \(\tau = T - t\). Computing the partial derivatives:
- \(\partial f/\partial t = (-\dot{C} - \dot{D}v)f\) (since \(\partial\tau/\partial t = -1\))
- \(\partial f/\partial x = iu \cdot f\), \(\partial^2 f/\partial x^2 = -u^2 f\)
- \(\partial f/\partial v = D \cdot f\), \(\partial^2 f/\partial v^2 = D^2 f\)
- \(\partial^2 f/\partial x \partial v = iuD \cdot f\)
Substituting and dividing by \(f\):
Collecting terms constant in \(v\) gives the ODE for \(C\):
Collecting terms linear in \(v\) gives the Riccati ODE for \(D\):
Both have initial conditions \(C(0, u) = D(0, u) = 0\). The separation into constant and linear terms in \(v\) is exact because the Heston drift and diffusion coefficients are affine in \(v\), confirming the affine structure.
Exercise 2. The Riccati ODE for \(D(\tau, u)\) is
Show that in the limit \(\xi \to 0\) (deterministic volatility), \(D(\tau, u) = 0\) for all \(\tau\), and \(C(\tau, u)\) reduces to the Black–Scholes log-characteristic function \((r-q)iu\tau - \frac{1}{2}(iu + u^2)\theta\tau\). (Hint: when \(\xi = 0\), \(V_t\) is deterministic and converges to \(\theta\).)
Solution to Exercise 2
When \(\xi = 0\), the variance process becomes deterministic: \(dV_t = \kappa(\theta - V_t)\,dt\), with solution \(V_t = V_0 e^{-\kappa t} + \theta(1 - e^{-\kappa t})\). For large \(t\), \(V_t \to \theta\).
The Riccati ODE becomes (setting \(\xi = 0\)):
With \(D(0) = 0\), this linear ODE has the solution:
However, in the deterministic volatility limit, the conditional characteristic function must reduce to a form that does not depend on \(V_t\) separately from the deterministic path. As \(\xi \to 0\), the contribution \(D(\tau)V_t\) combines with \(C(\tau)\) to reproduce the Black--Scholes characteristic function with the known deterministic integrated variance.
For the \(C\) equation with \(\xi = 0\):
In the long-run steady state where \(V_t = \theta\) for all \(t\), the log-characteristic function of \(X_T\) under Black--Scholes with constant variance \(\theta\) is:
Integrating the \(C\) equation: \(C(\tau) = (r-q)iu\tau + \kappa\theta\int_0^\tau D(s)\,ds\). Combined with \(D(\tau)V_t\), the total \(C(\tau) + D(\tau)\theta\) (evaluated at \(V_t = \theta\)) gives:
which is exactly the Black--Scholes log-characteristic function.
Exercise 3. The discriminant of the Riccati equation is
For the parameters \(\kappa = 2\), \(\xi = 0.4\), \(\rho = -0.7\), compute \(d\) at \(u = 1\) and \(u = 10\). Verify that \(d\) is well-defined (i.e., the quantity under the square root has positive real part). Why does the choice of branch cut for the complex square root matter for numerical implementation?
Solution to Exercise 3
With \(\kappa = 2\), \(\xi = 0.4\), \(\rho = -0.7\):
At \(u = 1\):
At \(u = 10\):
In both cases, \(\text{Re}(d^2) > 0\), so \(d\) is well-defined. The choice of branch cut matters because as \(u\) varies continuously (as required in the Fourier integral), the complex square root must also vary continuously. An inconsistent branch choice introduces discontinuities in \(\varphi(\tau, u)\), producing errors in the numerical integration for option prices.
Exercise 4. Using the formula for the expected integrated variance
compute this quantity for \(V_0 = 0.05\), \(\theta = 0.04\), \(\kappa = 2.0\), and \(T = 0.25, 0.5, 1.0, 2.0\). The ATM implied variance is approximately \(\sigma^2_{\text{impl}}(T) \approx \frac{1}{T}\mathbb{E}[\int_0^T V_s\,ds]\). Plot or compute the ATM implied volatility term structure and verify it interpolates between \(\sqrt{V_0}\) and \(\sqrt{\theta}\).
Solution to Exercise 4
With \(V_0 = 0.05\), \(\theta = 0.04\), \(\kappa = 2.0\):
| \(T\) | \(e^{-2T}\) | \(\mathbb{E}[\int_0^T V_s\,ds]\) | \(\sigma_{\text{impl}}(T) = \sqrt{\frac{1}{T}\mathbb{E}[\cdot]}\) |
|---|---|---|---|
| 0.25 | 0.6065 | \(0.01 + 0.001967 = 0.01197\) | \(\sqrt{0.04787} = 21.88\%\) |
| 0.50 | 0.3679 | \(0.02 + 0.003161 = 0.02316\) | \(\sqrt{0.04632} = 21.52\%\) |
| 1.00 | 0.1353 | \(0.04 + 0.004323 = 0.04432\) | \(\sqrt{0.04432} = 21.05\%\) |
| 2.00 | 0.0183 | \(0.08 + 0.004908 = 0.08491\) | \(\sqrt{0.04246} = 20.61\%\) |
As \(T \to \infty\), the second term vanishes and \(\sigma_{\text{impl}} \to \sqrt{\theta} = \sqrt{0.04} = 20\%\). Since \(V_0 = 0.05 > \theta = 0.04\), the term structure is downward-sloping, interpolating from \(\sqrt{V_0} = 22.36\%\) at \(T = 0\) down to \(\sqrt{\theta} = 20\%\).
Exercise 5. Explain the "Little Heston Trap" and why the original formulation can produce numerical errors. Specifically, for the term \(\ln(1 - ge^{d\tau})\), describe under what parameter and \(u\) combinations \(|ge^{d\tau}|\) can approach or exceed 1. How does the Lord-Kahl alternative formulation \(\ln(1 - g^{-1}e^{-d\tau})\) avoid this problem?
Solution to Exercise 5
The "Little Heston Trap" refers to numerical instability in the original Heston formulation of the characteristic function, identified by Albrecher et al. (2007).
In the original formulation, the term \(\ln(1 - ge^{d\tau})\) can become problematic because:
-
The ratio \(g = (\kappa - \rho\xi iu - d)/(\kappa - \rho\xi iu + d)\) can have modulus close to 1 for certain parameter combinations.
-
When \(|g| \approx 1\) and \(\text{Re}(d) > 0\) (which occurs for large \(\tau\) or certain \(u\) values), the product \(ge^{d\tau}\) can grow in modulus, causing \(|1 - ge^{d\tau}|\) to become very small or cross zero.
-
Specifically, for large \(u\), \(\text{Re}(d) > 0\) and \(|g| \to 1\), so \(|ge^{d\tau}|\) can exceed 1, creating a singularity in the logarithm. Even when it does not exceed 1, near-cancellation produces large numerical errors.
-
The complex logarithm is multi-valued, and as \(u\) varies continuously in the Fourier integral, the argument of \(1 - ge^{d\tau}\) may wind around the origin, requiring careful branch tracking.
The Lord--Kahl formulation replaces \(g\) with \(g^{-1}\) and \(e^{d\tau}\) with \(e^{-d\tau}\). Since \(|g^{-1}| = 1/|g|\) and the exponential now decays, the product \(|g^{-1}e^{-d\tau}|\) remains bounded below 1 for all \(u\) and \(\tau\). This ensures the logarithm argument stays away from zero, eliminating the numerical instability.
Exercise 6. The Gil-Pelaez inversion formula for the call price involves the integral
In practice, the integral is truncated at a finite upper limit \(u_{\max}\). For the Heston model, the characteristic function \(\varphi(\tau, u) \to 0\) as \(u \to \infty\) due to the \(-u^2\) term in the Riccati equation. Estimate how large \(u_{\max}\) should be for \(\kappa = 2\), \(\theta = 0.04\), \(\xi = 0.5\), \(T = 1\), by finding the value of \(u\) at which \(|\varphi(\tau, u)|\) decays below \(10^{-8}\).
Solution to Exercise 6
The characteristic function decays as \(|\varphi(\tau, u)| \sim \exp(-\frac{1}{2}u^2 \bar{V}\tau)\) for large \(u\), where \(\bar{V}\) is an effective variance level. This comes from the \(-u^2/2\) term in the Riccati equation dominating for large \(u\).
More precisely, for large \(u\), the \(D\) coefficient behaves as \(D(\tau, u) \approx -\frac{u^2}{2}\frac{1-e^{-\kappa\tau}}{\kappa}\) (ignoring lower-order terms), and \(C\) accumulates a corresponding contribution from \(\kappa\theta D\).
Setting \(|\varphi(\tau, u_{\max})| \approx e^{-8\ln 10} = 10^{-8}\), we need:
With \(\theta = 0.04\) and \(T = 1\), using \(\bar{V} \approx \theta = 0.04\):
In practice, one should verify convergence by testing the integral at progressively larger \(u_{\max}\). For the given parameters, \(u_{\max} \approx 30\)--\(50\) is typically sufficient. The presence of vol-of-vol \(\xi = 0.5\) causes slightly faster decay (heavier mixing of variance), so this estimate is conservative.
Exercise 7. The Heston characteristic function can be used to compute model-implied moments. Derive the formula for the variance of \(X_T = \log S_T\) in terms of the characteristic function:
Using the explicit formulas for \(C(\tau, u)\) and \(D(\tau, u)\), show that the variance depends on \(V_0\), \(\theta\), \(\kappa\), \(\xi\), and \(\rho\). Why does the variance of log-returns depend on \(\rho\) in a stochastic volatility model but not in Black–Scholes?
Solution to Exercise 7
The cumulant generating function is \(\psi(\tau, u) = \ln\varphi(\tau, u) = C(\tau, u) + D(\tau, u)V_0 + iu\log S_0\).
The mean and variance of \(X_T\) are given by:
To see why, note \(\varphi(\tau, u) = \mathbb{E}[e^{iuX_T}]\), so:
From the explicit formulas, \(\psi = C(\tau, u) + D(\tau, u)V_0 + iu\log S_0\). Both \(C\) and \(D\) depend on \(\kappa\), \(\theta\), \(\xi\), and \(\rho\) (through \(d\) and \(g\)). The second derivative \(\partial^2\psi/\partial u^2|_{u=0}\) therefore depends on all five parameters \(V_0, \theta, \kappa, \xi, \rho\).
Why \(\rho\) affects variance: In a stochastic volatility model, the variance of log-returns over \([0,T]\) is:
Expanding using \(dW_s^S = \rho\,dW_s^V + \sqrt{1-\rho^2}\,dW_s^\perp\), the integrated variance \(\int_0^T V_s\,ds\) is random and correlated with \(X_T\). When \(\rho \neq 0\), the covariance \(\text{Cov}(X_T, \int_0^T V_s\,ds) \neq 0\), which modifies \(\text{Var}[X_T]\). In Black--Scholes, \(\sigma\) is constant and there is no stochastic volatility process, so no such covariance term exists and the variance \(\text{Var}[X_T] = \sigma^2 T\) is independent of any correlation parameter.