Exercise: Monte Carlo Option Price with a Running Confidence Interval

Prerequisites: Law of Large Numbers, Central Limit Theorem, Geometric Brownian Motion

Problem

Price a European call with spot $S_0 = 100$ , strike $K = 110$ , risk-free rate $r = 0.02$ , volatility $\sigma = 0.25$ , and maturity $T = 1$ year, using Monte Carlo under the risk-neutral log-normal model

S_T = S_0\exp\!\left((r - \tfrac{1}{2}\sigma^2)T + \sigma\sqrt{T}\,Z\right),\qquad Z \sim \mathcal{N}(0, 1).

Simulate $N = 10^5$ i.i.d. terminal stock prices and compute the Monte Carlo estimator $\hat C_N = e^{-rT}\cdot \frac{1}{N}\sum_{i=1}^N \max(S_T^{(i)} - K, 0)$ . Report $\hat C_N$ , the sample standard deviation $\hat\sigma_Y$ of the discounted payoffs, and the 95% confidence interval $\hat C_N \pm 1.96\hat\sigma_Y/\sqrt N$ .
Compute and print the running estimator and its running 95%-CI at $N \in \{10^2, 10^3, 10^4, 10^5\}$ . Verify that the CI shrinks as $1/\sqrt N$ — specifically, that moving from $N = 10^2$ to $N = 10^4$ (a 100x increase) shrinks the CI by roughly a factor of $10$ .
The Black-Scholes closed-form price for these inputs is approximately $5.60$ . Check whether your 95% CI at $N = 10^5$ contains this value. Comment on whether containment is guaranteed or merely likely.
Extension. Implement antithetic variates: for each draw $Z_i$ , also use $-Z_i$ , and average the two payoffs before taking the overall sample mean. Compare the CI width at $N = 10^5$ with and without antithetics. Which variance-reduction ratio do you observe?

Hint

For part 1, the discounted payoff variance is what you divide by

N

in the CI — not the payoff variance before discounting. For part 4, the point of antithetic variates is that

\max(S_0 e^{\sigma\sqrt T Z} - K, 0)

and

\max(S_0 e^{-\sigma\sqrt T Z} - K, 0)

are negatively correlated, which reduces the variance of their sum.

Jump to the solution when you're ready.