Matching Geometric Brownian Motion and Lognormal Price Simulations

# Geometric Brownian Motion vs Lognormal (Monte Carlo)


# Geometric Brownian Motion vs Lognormal (Monte Carlo)












I am a bit confused. I thought that simulating stock prices via the GBM model is equivalent to drawing returns from a lognormal distribution.

In particular, consider the following Python code, which showcases both approaches I am referring to. It yields slightly different results. My questions are: Why is that so - what is the difference here?

EDIT: When I do not transform the standard deviation like I do in approach 2, both approaches yield a much more similar result. However, I was of the opinion that the lognormal standard deviation =/= the normal standard deviation and hence the transformation. What am I missing here? I appended this code below the original code (Implementation 2).

```
# IMPLEMENTATION 1
import numpy as np

# Parameters
mu = 0.05  # annual return
sigma = 0.2  # annual volatility
S0 = 100  # initial stock price
dt = 1/252  # daily time step
years = 5 # years
total_steps = 252 * years # total steps
M = 10000  # number of simulations

# Method 1: Geometric Brownian Motion
np.random.seed(42)
St1 = np.exp(
    (mu - sigma ** 2 / 2) * dt
    + sigma * np.random.normal(0, np.sqrt(dt), size=(total_steps, M))
)
final_prices1 = S0 * np.cumprod(St1, axis=0)

# Method 2: Direct Log-Normal
np.random.seed(42)
# Transforming the parameters to those of the underlying normally distributed log returns
norm_sigma = np.sqrt(np.log((sigma / (1 + mu)) ** 2 + 1))
norm_mu = np.log(1 + mu) - 0.5 * norm_sigma ** 2
returns2 = np.random.lognormal(
    mean=norm_mu * dt,
    sigma=norm_sigma * np.sqrt(dt),
    size=(total_steps, M)
) - 1
final_prices2 = S0 * np.cumprod(1 + returns2, axis=0)

# Compare statistics
print("Method 1 Final Price - Mean:", np.mean(final_prices1[-1]))
print("Method 2 Final Price - Mean:", np.mean(final_prices2[-1]))
print("Method 1 Final Price - Std:", np.std(final_prices1[-1]))
print("Method 2 Final Price - Std:", np.std(final_prices2[-1]))

# IMPLEMENTATION 2
import numpy as np

# Parameters
mu = 0.05  # annual return
sigma = 0.2  # annual volatility
S0 = 100  # initial stock price
dt = 1/252  # daily time step
years = 5 # years
total_steps = 252 * years # total steps
M = 10000  # number of simulations

# Method 1: Geometric Brownian Motion
np.random.seed(42)
St1 = np.exp(
    (mu - sigma** 2 / 2) * dt
    + sigma * np.random.normal(0, np.sqrt(dt), size=(total_steps, M))
)
final_prices1 = S0 * np.cumprod(St1, axis=0)

# Method 2: Direct Log-Normal
np.random.seed(42)
norm_mu = np.log(1 + mu) - 0.5 * sigma ** 2
returns2 = np.random.lognormal(
    mean=norm_mu * dt,
    sigma=sigma * np.sqrt(dt),
    size=(total_steps, M)
) - 1
final_prices2 = S0 * np.cumprod(1+returns2, axis=0)

# Compare statistics
print("Method 1 Final Price - Mean:", np.mean(final_prices1[-1]))
print("Method 2 Final Price - Mean:", np.mean(final_prices2[-1]))
print("Method 1 Final Price - Std:", np.std(final_prices1[-1]))
print("Method 2 Final Price - Std:", np.std(final_prices2[-1]))
```

## Answer by Ted Black (score 1)

https://quant.stackexchange.com/a/81281

Method 1 uses $$ S = S_0 \prod_{i=1}^N\exp\left( \left(\mu - \frac{1}{2}\sigma^2 \right)\Delta t + \sigma \sqrt{\Delta t} x_i \right) $$ where $x_i \in \texttt{np.random.normal}(0,\sqrt{\Delta t})$.

Method 2 should use $$ S = S_0 \prod_{i=1}^N z_i $$ where $z_i \in \texttt{np.random.lognormal}((\mu - 0.5 \sigma^2)\Delta t,\sigma \sqrt{\Delta t})$.

In your version of Method 2 you have $$ S = S_0 \prod_{i=1}^N (1+r_i) $$ where $r_i$ is what you call $\texttt{returns2}$. You model this as a lognormal variable; but clearly $z_i=1+r_i$ is a lognormal variable; and if $r_i$ is a lognormal variable, $z_i$ is not a lognormal variable.