Exercise: Pricing a geometric Asian option — Sobol vs Monte Carlo

A geometric-average Asian call on

S_0 = 100

with

K = 100

r = 0.05

\sigma = 0.2

T = 1

, sampled at

n = 32

equally spaced dates, has a closed-form analytic price under Black-Scholes (Kemna-Vorst 1990). This is a rare case: we can measure Monte Carlo error against ground truth.

The closed-form price under log-normal dynamics with geometric-average sampling is $C_{\text{geo}} = S_0 e^{(\mu^* - r)T}\Phi(d_1^*) - K e^{-rT}\Phi(d_2^*)$ with adjusted parameters:

\sigma^*{}^2 = \sigma^2 \frac{(n+1)(2n+1)}{6n^2}, \quad \mu^* = \frac{1}{2}\left[r - \frac{\sigma^2}{2} + \sigma^*{}^2\right],

d_1^* = \frac{\ln(S_0/K) + (\mu^* + \sigma^*{}^2/2)T}{\sigma^*\sqrt{T}}, \quad d_2^* = d_1^* - \sigma^*\sqrt{T}.

For the parameters above, $C_{\text{geo}} \approx 5.88$ (compute to verify).

Tasks

Compute the analytic price $C_{\text{geo}}$ to six decimals.
Implement pure Monte Carlo pricing with $N_{\text{paths}} \in \{2^8, 2^{10}, 2^{12}, 2^{14}\}$ . For each sample size, run $M = 100$ independent replicates and estimate the RMSE of your price estimator.
Implement Sobol-based QMC pricing (scrambled Sobol, inverse-CDF transform to Gaussian). For the same sample sizes, run $M = 100$ independent scrambles and estimate the RMSE.
Plot $\log(\text{RMSE})$ vs $\log N$ for both. Fit lines to each and report the empirical convergence rate (slope).
At what sample size does Sobol-QMC reach the accuracy that pure Monte Carlo achieves at $N = 2^{14}$ ?

Hint

Scramble Sobol by using scipy.stats.qmc.Sobol(d=32, scramble=True, seed=k) for

k = 0, 1, \ldots, M - 1