Exercise: Antithetic for Asian options

For an arithmetic Asian call paying $\big(\bar S - K\big)^+$ where $\bar S = \frac{1}{M}\sum_{k=1}^M S_{t_k}$:

Tasks

1. Implement antithetic Monte Carlo for the Asian call. The path generation uses $M$ independent normal draws per path; the antithetic twin negates all $M$ of them simultaneously.

2. With $S_0 = K = 100, T = 1, r = 0.05, \sigma = 0.2, M = 50, N = 10{,}000$ pairs, estimate $\rho$ between the two arms of each pair and the variance reduction factor.

3. Component-wise antithetic. Instead of negating all $M$ draws, negate only half of them. Does this perform better or worse than the full antithetic? Why?
4. Comparison with European. For the European call (same parameters, only $S_T$ matters): antithetic gives $\sim 9\times$ variance reduction. For the Asian, expect roughly $\sim 2\times$ — verify this numerically. Argue why path-dependent payoffs see less benefit.