Python Code: Geometric Brownian Motion - what's wrong?

Question

I\'m pretty new to Python, but for a paper in University I need to apply some models, using preferably Python. I spent a couple of days with the code I attached, but I can\'t re

Answer 1

An additional implementation using the parametrization of the gaussian law though the normal fonction (instead of standard_normal), a bit shorter.

import numpy as np

T = 2
mu = 0.1
sigma = 0.01
S0 = 20
dt = 0.01
N = round(T/dt)
# reversely you can specify N and then compute dt, which is more common in financial litterature

X = np.random.normal(mu * dt, sigma* np.sqrt(dt), N)
X = np.cumsum(X)
S = S0 * np.exp(X)

Answer 2

According to Wikipedia,

So it appears that

X=(mu-0.5*sigma**2)*t+(sigma*W) ###geometric brownian motion#### 

rather than

X=(mu-0.5*sigma**2)*dt+(sigma*sqrt(dt)*W)

---

Since T represents the time horizon, I think t should be

t = np.linspace(0, T, N)

---

Now, according to these Matlab examples (here and here), it appears

W = np.random.standard_normal(size = N) 
W = np.cumsum(W)*np.sqrt(dt) ### standard brownian motion ###

not,

W=(standard_normal(size=Steps)+mu*t)

Please check the math, however, I could be wrong.

---

So, putting it all together:

import matplotlib.pyplot as plt
import numpy as np

T = 2
mu = 0.1
sigma = 0.01
S0 = 20
dt = 0.01
N = round(T/dt)
t = np.linspace(0, T, N)
W = np.random.standard_normal(size = N) 
W = np.cumsum(W)*np.sqrt(dt) ### standard brownian motion ###
X = (mu-0.5*sigma**2)*t + sigma*W 
S = S0*np.exp(X) ### geometric brownian motion ###
plt.plot(t, S)
plt.show()

yields