Pricing an Asian option by Monte Carlo¶

Course exercise 180 · Part III

Problem¶

Price a six-month Asian call and put, whose payoff depends on the average price of the underlying over the life of the option rather than on its final price. There is no simple closed form: simulate one million paths of 100 steps each.

Method¶

Under the risk-neutral measure the underlying follows a geometric Brownian motion with drift \(r\). For each simulated path the arithmetic average of the \(m\) monitored prices is

and the expectations are replaced by averages over the paths. Data: \(S_0 = 36.64\), \(K = 37\), \(r = 1.75\%\), \(\sigma = 20\%\), \(T = 0.5\), \(m = 100\).

Code¶

clear all, clc

%% 1

n = 1e6; 
m = 100;
S_0 = 36.64; 
K = 37; 
r = 0.0175; 
T = 0.5; 
sigma = 0.2;

%% 2

Dt   = T / m;                                       
Z    = randn(n, m);                                 
dX   = (r - 0.5*sigma^2)*Dt + sigma*sqrt(Dt)*Z;    
S_tk = S_0 * exp(cumsum(dX, 2));                    

%% 3

S_bar = mean(S_tk, 2);                              

%% 4

C = exp(-r*T) * mean( max(S_bar - K, 0) )          
P = exp(-r*T) * mean( max(K - S_bar, 0) )         

import numpy as np

# %% 1
n = 10**6                               # MATLAB: 1e6   (10**6 is an integer, needed as a size)
m = 100
S_0 = 36.64
K = 37
r = 0.0175
T = 0.5
sigma = 0.2

# %% 2
Dt = T / m
Z = np.random.randn(n, m)
dX = (r - 0.5 * sigma**2) * Dt + sigma * np.sqrt(Dt) * Z
S_tk = S_0 * np.exp(np.cumsum(dX, axis=1))

# %% 3
S_bar = np.mean(S_tk, axis=1)           # average price of each path   (MATLAB: mean(S_tk, 2))

# %% 4
C = np.exp(-r * T) * np.mean(np.maximum(S_bar - K, 0))
P = np.exp(-r * T) * np.mean(np.maximum(K - S_bar, 0))
print("C =", C)
print("P =", P)

Results¶

One simulated run, random seed 0.

Takeaways¶

• The Asian options cost roughly half of the European ones: an average moves less than a single final price, so the payoff is less volatile and the option is worth less.
• Monte Carlo handles path-dependent payoffs with no extra effort: the paths are already simulated, only the payoff formula changes.
• The monitoring convention is part of the contract: this version averages the 100 prices after time zero; an earlier version of the same exercise averaged over a grid that included \(S_0\) itself, which gives a slightly different price.
• One million paths of 100 steps means three \(10^6 \times 100\) matrices in memory, several gigabytes: the price of vectorising everything.