Exercise: Itô vs. Ordinary Chain Rule — Where They Diverge Numerically

Prerequisites: Itô's Lemma, Geometric Brownian Motion, Monte Carlo Pricing (Basic)

Problem

A trader simulates geometric Brownian motion with drift $\mu = 0.1$ , volatility $\sigma = 0.4$ , horizon $T = 1$ , and $N = 200{,}000$ paths. They want to confirm that $\mathbb{E}[\ln(S_T/S_0)] = (\mu - \tfrac{1}{2}\sigma^2)T$ by Monte Carlo.

Two code snippets compute the mean of $\ln(S_T/S_0)$ :

# Snippet A: naive (ordinary chain rule)
# "Since d(ln S) = dS/S, log-return has drift mu."
log_ret_mean_A = mu * T   # = 0.10

# Snippet B: Itô-correct
# "d(ln S) = (mu - 0.5*sigma^2) dt + sigma dW, so log-return has drift mu - 0.5*sigma^2."
log_ret_mean_B = (mu - 0.5 * sigma**2) * T   # = 0.02

Simulate the process yourself in Python or R and report the empirical mean of $\ln(S_T/S_0)$ . Which of snippet A or snippet B does the data support?
The naive prediction (A) would say that the arithmetic expected return $\mathbb{E}[S_T/S_0] - 1 = \mathbb{E}[e^{\ln S_T/S_0}] - 1 \approx 0.10$ . Compute the correct value of $\mathbb{E}[S_T/S_0] - 1$ from GBM theory and from the simulation. How far off is the naive prediction?
For what range of $\sigma$ (holding $\mu = 0.1$ ) is the naive log-return drift of A within 10% of the correct Itô log-return drift of B? Beyond that volatility, the naive calculation mis-prices options badly. Give a qualitative interpretation: why does the mistake grow with $\sigma$ ?
Optional extension. Repeat the simulation with $\sigma = 0.8$ . Plot histograms of $\ln(S_T/S_0)$ for $N = 10^5$ paths. Overlay both theoretical means (A and B). Which one lands in the histogram's centre?

Hint

Use rng = np.random.default_rng(0) to make your results reproducible. Part 2 needs

\mathbb{E}[S_T/S_0] = e^{\mu T}

from the MGF of a Gaussian — no Itô correction here because you are taking the expectation of an exponential directly.

Jump to the solution when you're ready.