Brownian Bridge™

1. The forward finite difference approximation is $f'(x) \approx [f(x+h) - f(x)]/h$ . What is its truncation error order, and what limits how small $h$ can be made?

O(h) truncation error; roundoff error O(εmach/h) limits h from below, giving an optimal h* ≈ √εmach ≈ 10⁻⁸O(h²) truncation error; memory constraints limit h from belowO(h) truncation error; no lower bound on h existsO(1/h) truncation error; machine epsilon sets the upper bound on h

2. Central finite differences use $f'(x) \approx [f(x+h) - f(x-h)]/(2h)$ . Compared to forward FD at their respective optimal step sizes:

Central FD is ~10,000× more accurate: O(ε²/³) vs O(ε¹/²) at optimal hForward FD is more accurate because it uses fewer function evaluationsBoth achieve the same accuracy at their optimal step sizesCentral FD is only more accurate for smooth functions; they are equivalent for non-smooth functions

3. The complex step method computes $f'(x) \approx \text{Im}[f(x+ih)]/h$ . Its key advantage over forward finite differences is:

No cancellation error: taking the imaginary part avoids subtracting nearly-equal numbersIt requires fewer function evaluations than forward FDIt is applicable to discontinuous functions where FD failsIt automatically computes higher-order derivatives

4. Reverse-mode AAD computes the full gradient $\nabla_\theta \mathcal{L} \in \mathbb{R}^p$ in $O(1)$ passes (independent of $p$ ). What is the key reason reverse mode is preferred over forward mode for a scalar calibration objective?

The objective ℒ(θ) is scalar: one backward pass computes all p partial derivatives simultaneouslyReverse mode avoids storing intermediate values (the tape), making it memory-efficientReverse mode is always faster than forward mode regardless of the number of outputsForward mode cannot handle non-linear functions

5. For a Heston calibration with $p = 5$ parameters and $N = 40$ instruments, how many pricing engine evaluations does the central FD Jacobian require per LM iteration?

10 evaluations: 2 perturbations × 5 parameters (all 40 instruments evaluated at each perturbed θ)80 evaluations: 2 × 40 (one perturbation per instrument)200 evaluations: 5 parameters × 40 instruments5 evaluations: one per parameter, instruments evaluated simultaneously

6. A developer implements the complex step method for Black-Scholes pricing but uses `abs(sigma)` inside the NormCdf computation to ensure a positive argument. What will happen to the derivative?

The derivative will be wrong: abs() is not analytic at zero, and its complex extension does not equal |Re(sigma)| for complex sigmaThe derivative will be correct because abs() returns the real part for complex inputsThe derivative will be correct but slower due to the complex arithmeticThe derivative will have O(h) error instead of O(h²)

7. You need the full Jacobian $J \in \mathbb{R}^{100 \times 50}$ for an LMM calibration (100 swaptions, 50 parameters). Which method requires the fewest pricing engine evaluations?

Forward-mode AAD (50 forward passes, one per parameter): fewer than central FD (100 passes)Central FD (100 passes: 2 per parameter × 50 parameters)Reverse-mode AAD (100 backward passes, one per swaption)Forward FD (51 passes: 50 + 1 base)

8. A pricing function contains the line `d1 = log(S/K) + 0.5*sigma**2*T`. You differentiate with respect to sigma using forward AAD. Which elementary operation in this line contributes a non-zero tangent?

0.5*sigma**2*T, with tangent ∂(0.5σ²T)/∂σ = σTlog(S/K), because log depends on sigma through the smileS/K, because the forward depends on the volAll terms contribute equally

Quiz: Jacobian Computation — Finite Difference and AAD

Quick Quiz