Brownian Bridge™

1. The Heston model specifies the instantaneous variance $V_t$ as: $dV_t = \kappa(\theta - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^V$ What are the roles of the three parameters $\kappa $,$ \theta $, and$ \xi$ in shaping the implied vol surface?

κ controls mean-reversion speed (term structure decay); θ sets the long-run variance level; ξ (vol of vol) controls smile curvature (butterfly).κ controls skew; θ controls curvature; ξ controls the overall vol level.κ controls the initial variance; θ controls how fast the smile decays; ξ controls the risk-reversal.κ and θ together determine the skew; ξ determines the term structure slope.

2. For the Heston model with $\kappa = 1.5$ , $\theta = 0.04$ , $\xi = 0.6$ , is the Feller condition satisfied? What are the consequences if it is not?

2κθ = 0.12 < ξ² = 0.36: Feller violated. The variance hits zero with positive probability; Monte Carlo schemes that ignore this introduce pricing bias.2κθ = 0.12 > ξ² = 0.36: Feller satisfied. The variance remains strictly positive almost surely.2κθ = 0.12 < ξ² = 0.36: Feller violated, but this only affects long-dated options; short-dated pricing is unaffected.2κθ = 0.12 < ξ² = 0.36: Feller violated, making the characteristic function undefined and semi-analytic pricing impossible.

3. The Heston characteristic function is $\phi_T(u) = \exp(A(u,T) + B(u,T)V_0)$ . Why is this exponential-affine form available in closed form, and what property of the Heston model enables it?

The Heston model is affine: drift and squared-diffusion coefficients of (X_t, V_t) are affine in the state. The Duffie-Pan-Singleton theorem guarantees an exponential-affine CF, with A and B satisfying Riccati ODEs with constant coefficients.The Heston model is Markovian, and all Markovian models have closed-form characteristic functions by the Feynman-Kac theorem.The closed form follows from the fact that V_t is lognormal; lognormal processes always have explicit characteristic functions.The affine form arises because ρ = 0, decoupling the spot and variance SDEs and allowing independent CF computation.

4. The Albrecher et al. (2007) reformulation of the Heston characteristic function is preferred in production code over the original Heston (1993) formula. What specific numerical problem does the Albrecher form fix?

The original formula has a branch-cut discontinuity in the complex logarithm: the imaginary part of ln(g·e^{−dT}) jumps by 2π for large u or large T, causing the integral to be evaluated incorrectly. The Albrecher form avoids this.The original formula requires computing a complex square root, which is numerically unstable for |ρ| close to 1. The Albrecher form replaces the square root with a real-valued expression.The original formula diverges at u = 0 (ATM), whereas the Albrecher form has a removable singularity handled by L'Hôpital's rule.The original formula requires matrix exponentiation for multi-dimensional extensions, whereas the Albrecher form is scalar.

5. Under Heston with baseline parameters ( $\kappa=2$ , $\theta=0.04$ , $\xi=0.5$ , $\rho=-0.7$ , $V_0=0.04$ ), what is the stationary variance of $V_t$ (as $t \to \infty$ )?

θξ²/(2κ) = 0.04 × 0.25 / 4 = 0.0025θ² = 0.04² = 0.0016ξ²/(2κ) = 0.25/4 = 0.06252κθ/ξ² = 0.32

6. A structuring desk prices a 2-year cliquet paying the sum of annually-reset at-the-money call returns. They calibrate a Heston model to the current vanilla surface and price the cliquet. Compared to stochastic vol models with explicit forward vol control (e.g., Bergomi), what systematic bias should they expect from Heston?

Heston will underprice the cliquet: the forward smile at year 1 is too flat (the Heston skew decays exponentially at rate κ), underestimating the value of forward-starting optionality.Heston will overprice the cliquet: stochastic vol inflates the forward smile relative to local vol, causing Heston to overestimate the cliquet premium.There is no systematic bias because Heston is calibrated to the current surface, which encodes the market's expectation of all future smiles.Heston will misprice the cliquet only if ρ ≠ 0; for ρ = 0, the forward smile is exact.

7. A quant observes that in a calibrated Heston model, the ATM implied vol at $T = 3$ months is $25\%$ but the long-dated ATM vol ( $T = 5$ years) is $20\%$ . What Heston parameters primarily drive this term structure, and in what direction?

V₀ > θ drives the downward term structure: V₀ = 0.0625 (25% vol) > θ = 0.04 (20% vol), and κ controls how quickly ATM vol decays from √V₀ to √θ.ξ > 0 drives the downward term structure by pushing variance upward at short maturities.ρ < 0 drives the downward term structure by inflating short-dated implied vols through the leverage effect.The term structure is flat in Heston; the difference is purely a bid-offer artefact.

8. A desk scenario: calibrated Heston parameters are $\kappa=3$ , $\theta=0.04$ , $\xi=0.9$ , $\rho=-0.8$ , $V_0=0.05$ . A risk manager asks whether the variance process can reach zero and what Monte Carlo scheme should be used. What is the correct answer?

Feller is violated (2κθ=0.24 < ξ²=0.81): V_t hits zero with positive probability. Use full-truncation Euler (clamp V_t at 0 in the diffusion term) or the Andersen QE scheme for accurate pricing.Feller is satisfied (2κθ=0.24 > ξ²=0.81): V_t is strictly positive; standard Euler-Maruyama is sufficient.Feller violation only affects the drift of V_t, not the diffusion; any scheme that uses V_t⁺ = max(V_t, 0) in the drift handles this correctly.Feller violation makes the Lewis formula invalid; Monte Carlo is the only viable pricing method when the condition is violated.

Quiz: The Heston Stochastic Volatility Model

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