Black-Scholes Monte Carlo Pricer

Worked Solution

How to Think About It: This is the bread-and-butter of quant interview take-homes. The interviewer is checking three things: (1) you know what risk-neutral pricing means and can apply Ito's lemma to get the terminal distribution, (2) you can translate a mathematical formula into a clean simulation, and (3) you understand that naive Monte Carlo is noisy and know practical tricks to fix it. Think of this as three mini-problems stacked together -- derivation, implementation, engineering.

Part (a): Risk-Neutral Distribution of $S_T$

Under the risk-neutral measure $Q$, the drift $\mu$ is replaced by $r$, so:

$$dS_t = r S_t \, dt + \sigma S_t \, dW_t^Q$$

Apply Ito's lemma to $\ln S_t$. Let $f(S) = \ln S$. Then $f'(S) = 1/S$ and $f''(S) = -1/S^2$, so:

$$d(\ln S_t) = \left(r - \frac{\sigma^2}{2}\right) dt + \sigma \, dW_t^Q$$

Integrating from $0$ to $T$:

$$\ln S_T = \ln S_0 + \left(r - \frac{\sigma^2}{2}\right) T + \sigma \sqrt{T} \, Z, \quad Z \sim N(0,1)$$

Equivalently:

$$S_T = S_0 \exp\!\left[\left(r - \frac{\sigma^2}{2}\right) T + \sigma \sqrt{T} \, Z\right]$$

So $\ln S_T$ is normally distributed with mean $\ln S_0 + (r - \sigma^2/2)T$ and variance $\sigma^2 T$, meaning $S_T$ is lognormal. The option price is:

$$C = e^{-rT} \, E^Q[(S_T - K)^+]$$

Part (b): Monte Carlo Algorithm and Code

The algorithm is direct sampling from the exact terminal distribution -- no need to discretize the path since we only care about $S_T$.

Algorithm: 1. For each path $i = 1, \ldots, M$: draw $Z_i \sim N(0,1)$ and compute $S_T^{(i)} = S_0 \exp[(r - \sigma^2/2)T + \sigma \sqrt{T} \, Z_i]$. 2. Compute the payoff $V_i = (S_T^{(i)} - K)^+$. 3. Estimate $\hat{C} = e^{-rT} \frac{1}{M} \sum_{i=1}^{M} V_i$.

```python import numpy as np from scipy.stats import norm

def bs_mc_call(S0, K, T, r, sigma, M, seed=42): rng = np.random.default_rng(seed) Z = rng.standard_normal(M) ST = S0 * np.exp((r - 0.5 * sigma**2) * T + sigma * np.sqrt(T) * Z) payoffs = np.maximum(ST - K, 0.0) price = np.exp(-r * T) * np.mean(payoffs) se = np.exp(-r * T) * np.std(payoffs, ddof=1) / np.sqrt(M) return price, se ```

Complexity: Each path requires $O(1)$ work (one random draw, one exp, one max). Summing over $M$ paths gives $O(M)$ total time and $O(M)$ space (or $O(1)$ space if you accumulate online). The standard error of the estimator decreases as $O(1/\sqrt{M})$, so getting one more decimal digit of accuracy requires $100\times$ more paths.

Part (c): Variance Reduction

Antithetic variates. For each $Z_i$, also compute the payoff using $-Z_i$. The pair $(S_T(Z_i), S_T(-Z_i))$ shares the same random draw but covers both tails. Your estimator becomes:

$$\hat{C}_{\text{AV}} = e^{-rT} \frac{1}{M} \sum_{i=1}^{M} \frac{V(Z_i) + V(-Z_i)}{2}$$

This works because the payoff is a monotone function of $Z$, so $V(Z)$ and $V(-Z)$ are negatively correlated, reducing variance. In practice this cuts variance by roughly 20-30% for vanilla options.

Control variates. Use $S_T$ itself as a control. We know $E^Q[S_T] = S_0 e^{rT}$ exactly. Define:

$$\hat{C}_{\text{CV}} = \hat{C} - \beta \left(\frac{1}{M}\sum_{i=1}^{M} S_T^{(i)} - S_0 e^{rT}\right)$$

The optimal $\beta$ is $\text{Cov}(V, S_T) / \text{Var}(S_T)$, estimated from the same sample. Since the call payoff and terminal stock price are highly correlated, this typically reduces variance by 50-80%.

Empirical convergence verification: - Benchmark against Black-Scholes closed form. For a European call, the BS formula gives the exact answer. Run your MC pricer for increasing $M$ (e.g., $10^3, 10^4, 10^5, 10^6$) and confirm the MC estimate converges to the BS price and that the standard error shrinks proportional to $1/\sqrt{M}$. - Confidence intervals. At each $M$, compute a 95% CI as $\hat{C} \pm 1.96 \cdot \text{SE}$. The BS price should fall inside the CI roughly 95% of the time across repeated runs. - Log-log plot. Plot $\log(\text{SE})$ vs. $\log(M)$. You should see a slope of $-1/2$. If the slope is flatter, something is wrong (e.g., correlated draws, a bug in the payoff).

Answer: The option price is $C = e^{-rT} E^Q[(S_T - K)^+]$ where $S_T = S_0 \exp[(r - \sigma^2/2)T + \sigma\sqrt{T} Z]$ with $Z \sim N(0,1)$. The Monte Carlo estimator averages $M$ i.i.d. discounted payoffs in $O(M)$ time, converging at rate $O(1/\sqrt{M})$. Antithetic variates exploit monotonicity to pair $Z$ and $-Z$ for negative correlation; control variates use the known mean of $S_T$ to correct the estimator. Both can be verified by benchmarking against the Black-Scholes closed-form price.