I am trying to simulate Geometric Brownian Motion in Python, to price a European Call Option through Monte-Carlo simulation. I am relatively new to Python, and I am receiving an answer that I believe to be wrong, as it is nowhere near to converging to the BS price, and the iterations seem to be negatively trending for some reason. Any help would be appreciated.
import numpy as np
from matplotlib import pyplot as plt
S0 = 100 #initial stock price
K = 100 #strike price
r = 0.05 #risk-free interest rate
sigma = 0.50 #volatility in market
T = 1 #time in years
N = 100 #number of steps within each simulation
deltat = T/N #time step
i = 1000 #number of simulations
discount_factor = np.exp(-r*T) #discount factor
S = np.zeros([i,N])
t = range(0,N,1)
for y in range(0,i-1):
S[y,0]=S0
for x in range(0,N-1):
S[y,x+1] = S[y,x]*(np.exp((r-(sigma**2)/2)*deltat + sigma*deltat*np.random.normal(0,1)))
plt.plot(t,S[y])
plt.title('Simulations %d Steps %d Sigma %.2f r %.2f S0 %.2f' % (i, N, sigma, r, S0))
plt.xlabel('Steps')
plt.ylabel('Stock Price')
plt.show()
C = np.zeros((i-1,1), dtype=np.float16)
for y in range(0,i-1):
C[y]=np.maximum(S[y,N-1]-K,0)
CallPayoffAverage = np.average(C)
CallPayoff = discount_factor*CallPayoffAverage
print(CallPayoff)
Monte-Carlo Simulation Example (Stock Price Simulation)
I am currently using Python 3.6.1.
Thank you in advance for the help.
Here's a bit of re-writing of code that may make the notation of S
more intuitive and will allow you to inspect your answer for reasonableness.
Initial points:
deltat
should be replaced by np.sqrt(deltat)
. Source here (yes, I know it's not the most official, but results below should be reassuring).First, here is a GBM-path generating function from Yves Hilpisch - Python for Finance, chapter 11. The parameters are explained in the link but the setup is very similar to yours.
def gen_paths(S0, r, sigma, T, M, I):
dt = float(T) / M
paths = np.zeros((M + 1, I), np.float64)
paths[0] = S0
for t in range(1, M + 1):
rand = np.random.standard_normal(I)
paths[t] = paths[t - 1] * np.exp((r - 0.5 * sigma ** 2) * dt +
sigma * np.sqrt(dt) * rand)
return paths
Setting your initial values (but using N=252
, number of trading days in 1 year, as the number of time increments):
S0 = 100.
K = 100.
r = 0.05
sigma = 0.50
T = 1
N = 252
deltat = T / N
i = 1000
discount_factor = np.exp(-r * T)
Then generate the paths:
np.random.seed(123)
paths = gen_paths(S0, r, sigma, T, N, i)
Now, to inspect: paths[-1]
gets you the ending St
values, at expiration:
np.average(paths[-1])
Out[44]: 104.47389541107971
The payoff, as you have now, will be the max of (St - K, 0
):
CallPayoffAverage = np.average(np.maximum(0, paths[-1] - K))
CallPayoff = discount_factor * CallPayoffAverage
print(CallPayoff)
20.9973601515
If you plot these paths (easy to just use pd.DataFrame(paths).plot()
, you'll see that they're no longer downward-trending but that the St
s are approximately log-normally distributed.
Lastly, here's a sanity check through BSM:
class Option(object):
"""Compute European option value, greeks, and implied volatility.
Parameters
==========
S0 : int or float
initial asset value
K : int or float
strike
T : int or float
time to expiration as a fraction of one year
r : int or float
continuously compounded risk free rate, annualized
sigma : int or float
continuously compounded standard deviation of returns
kind : str, {'call', 'put'}, default 'call'
type of option
Resources
=========
http://www.thomasho.com/mainpages/?download=&act=model&file=256
"""
def __init__(self, S0, K, T, r, sigma, kind='call'):
if kind.istitle():
kind = kind.lower()
if kind not in ['call', 'put']:
raise ValueError('Option type must be \'call\' or \'put\'')
self.kind = kind
self.S0 = S0
self.K = K
self.T = T
self.r = r
self.sigma = sigma
self.d1 = ((np.log(self.S0 / self.K)
+ (self.r + 0.5 * self.sigma ** 2) * self.T)
/ (self.sigma * np.sqrt(self.T)))
self.d2 = ((np.log(self.S0 / self.K)
+ (self.r - 0.5 * self.sigma ** 2) * self.T)
/ (self.sigma * np.sqrt(self.T)))
# Several greeks use negated terms dependent on option type
# For example, delta of call is N(d1) and delta put is N(d1) - 1
self.sub = {'call' : [0, 1, -1], 'put' : [-1, -1, 1]}
def value(self):
"""Compute option value."""
return (self.sub[self.kind][1] * self.S0
* norm.cdf(self.sub[self.kind][1] * self.d1, 0.0, 1.0)
+ self.sub[self.kind][2] * self.K * np.exp(-self.r * self.T)
* norm.cdf(self.sub[self.kind][1] * self.d2, 0.0, 1.0))
option.value()
Out[58]: 21.792604212866848
Using higher values for i
in your GBM setup should cause closer convergence.