Using R for loop to create a time series

Question

I'm forecasting a stock price 'St1' starting at 'Stt = 10' at time t. Dates are not important. I want to find the next 100 prices using the following formula:

St1 <- Stt*exp((r-0.5*sigma^2)*T/n+sigma*E*sqrt(T/n))

So I'm starting with 10 and then need the next 100 values and put it in a data.frame or whatever is easy. How do I do this? I'm stuck trying this...

Stt = 10
r = 0.15
sigma = 0.2
T = 1
n = 100
E = 0.15

St1 = Stt
for (i in 1:100)
{
    St1[i] <- Stt*exp((r-0.5*sigma^2)*T/n+sigma*E*sqrt(T/n))
}

Please add a reproducible example

Thierry
– Thierry

2016-03-18 09:55:45 +00:00
Commented Mar 18, 2016 at 9:55 — Thierry
– Thierry, Commented Mar 18, 2016 at 9:55

Karsten W. · Accepted Answer · 2016-03-18 11:31:05Z

1

One way is this

r = 0.15
sigma = 0.2
T = 1
n = 100
E = 0.15

Stt<- rep(NA,n) # preallocate
Stt[1] <- 10
for (i in 2:100)
{
    Stt[i] <- Stt[i-1]*exp((r-0.5*sigma^2)*T/n+sigma*E*sqrt(T/n))
}

Without for loop, and in order to use rnorm for E, this code could work:

# for fixed E of length one
Stt <- cumprod(c(10,rep(exp((r-0.5*sigma^2)*T/n+sigma*E*sqrt(T/n)), n-1)))

# for random vector E 
E <- rnorm(n-1, mean=0.15, sd=0.01)
Stt <- cumprod(c(10, exp((r-0.5*sigma^2)*T/n+sigma*E*sqrt(T/n))))

edited Mar 18, 2016 at 11:31

answered Mar 18, 2016 at 10:20

Karsten W.

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3 Comments

cJc Over a year ago

Works like a charm, tnx a lot!

cJc Over a year ago

What if I wanted to make E a normal random variabel? rnorm(100) - gives me error message " number of items to replace is not a multiple of replacement length"

jlesuffleur Over a year ago

In the formula, replace E by E[i] since E is a vector. By the way, I would recommend you to vectorize the whole formula.

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