Brownian Bridge™

1. The Feller condition for the Heston variance process is `2κθ > ξ²`. Which calibrated parameter set **violates** the Feller condition?

κ = 2.0, θ = 0.04, ξ = 0.3 → 2×2×0.04 = 0.16 > 0.09κ = 0.5, θ = 0.02, ξ = 0.5 → 2×0.5×0.02 = 0.02 < 0.25κ = 4.0, θ = 0.05, ξ = 0.4 → 2×4×0.05 = 0.40 > 0.16κ = 1.0, θ = 0.04, ξ = 0.2 → 2×1×0.04 = 0.08 > 0.04

2. In the full-truncation Euler scheme for the Heston model, the variance step is: ```cpp V = V + kappa*(theta - V_pos)*dt + xi*sqrt(V_pos*dt)*Z1; ``` where `V_pos = max(V, 0)`. What is the purpose of using `V_pos` in the drift term `kappa*(theta - V_pos)` rather than `V`?

It speeds up the computation by avoiding negative square rootsWhen V goes negative (below the Feller boundary), using V_pos in the drift prevents the mean-reversion term from pushing V further negative — it treats V=0 as the boundary and pulls the variance back up toward thetaIt implements the reflection boundary conditionIt is a typo; V_pos should only appear inside sqrt()

3. In the Heston model, when `ρ = 0` (no correlation between spot and variance), the implied volatility smile is symmetric: the implied vol at strike K and strike S₀²/K (equidistant in log-moneyness) is the same.

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4. The Lewis (2001) pricing formula integrates the characteristic function along the real axis: ``` C = S₀ - sqrt(S₀·K)·exp(-rT/2)/π · ∫₀^∞ Re[exp(iu·log(S₀/K)) · φ(u-i/2;T) / (u²+1/4)] du ``` Why does the integrand involve `φ(u - i/2; T)` rather than `φ(u; T)`?

It is a change of numeraire from the risk-neutral to the T-forward measureThe `- i/2` shift ensures the characteristic function is evaluated along a contour where the call payoff `max(S-K,0)` has a convergent Fourier transform — the `1/sqrt(S₀K)` factor in the Lewis formula requires integrating in the strip Im(u) ∈ (-1, 0)It is a numerical stabilisation to prevent the integrand from oscillating at high uThe i/2 shift converts the risk-neutral measure to the physical measure

5. You implement a control variate for the Heston MC pricer using the Black-Scholes price with vol = sqrt(V₀) as the control. The BS price is 10.45. Your MC gives a raw estimate of 10.52 ± 0.30. The control variate MC gives 10.43 ± 0.03. Which statement is correct?

The control variate estimate is biased because it subtracts a model priceThe control variate is unbiased: it adjusts the MC estimate by the difference between the simulated control payoff mean and the analytic BS price, leaving a residual that converges fasterThe BS control is invalid because Heston and BS are different modelsThe standard error reduction from 0.30 to 0.03 is impossible — it would require 100x more paths

6. The Heston model can reproduce any arbitrage-free implied volatility surface exactly, given the right choice of parameters (κ, θ, ξ, ρ, V₀).

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Quiz: Heston Stochastic Volatility: Theory and C++ Implementation

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