Pricing of Asian Option using R - r

Question

The pricing of the Asian option is approximated, using Monte Carlo simulation, by:
delta <- 1/12
T <- 2
S0 <- 100
sigma <- 0.20
K <- 100
r <- 0.01
n <- 10^4
m <- T/delta
S <- S0
for(i in 1:n) {
for(j in 1:m) {
W <- rnorm(1)
Si <- S[length(S)]*exp((r-0.5*sigma^2)*delta + sigma*sqrt(delta)*W)
S <- c(S, Si)
}
Si.bar <- mean(S[-1])
Ci <- exp(-r*T)*max(Si.bar - K, 0)
}
mean(Ci)
The for(j in 1:m) for loop runs perfectly, I think... But when I run it n times, using for(i in 1:n) S gets smaller and smaller by n. It decreases to almost zero when n grows. This leads to a mean (Si.bar <- mean(S[-1]) well below the strike price, K= 100.
I can't figure out what is wrong with the two last lines of codes. I'm getting a value on the Asian call option of 0, due to the payoff function. The correct solution to this option is a value of approximately 7 (mean(Ci))

Answer 1

There's a couple of issues with your code. Firstly, it's inefficient in R to build a vector by repeated concatenation. Instead, you should allocate the vector up front and then assign to its members.
Secondly, as I understand it, the aim is to repeat the inner loop n times and store the output into members of a vector C before taking the mean. That's not what you're doing at the moment - each iteration of the outer loop makes S longer and overwrites Ci such that the last statement, mean(Ci) is meaningless.
Here's an amended version of the code. I've used plyr partly to make the code neater, and partly for its progress bar functionality.
library(plyr)
delta <- 1/12
T <- 2
S0 <- 100
sigma <- 0.20
K <- 100
r <- 0.01
n <- 10^4
m <- T/delta
S <- numeric(m + 1)
S[1] <- S0
asian_price <- function() {
for(j in 1:m) {
W <- rnorm(1)
S[j + 1] <- S[j] * exp((r - 0.5 * sigma^2) * delta + sigma * sqrt(delta) * W)
}
Si.bar <- mean(S[-1])
exp(-r * T) * max(Si.bar - K, 0)
}
C <- raply(n, asian_price(), .progress = "text")
mean(C)
# [1] 7.03392