Reputation: 55
I have to prepare an algorithm for my thesis to cross check a theoretical result which is that the binomial model for N periods converges to lognormal distribution for N\to \infty. For those of you not familiar with the concept i have to create an algorithm that takes a starter value and multiplies it with an up-multiplier and a down multiplier and continues to do so for every value for N steps. The algorithm should return a vector whose elements are in the form of StarterValueu^id^{N-i} i=0,\dots,N the simple algorithm i proposed is
rata<-function(N,r,u,d,S){
length(x)<-N
for(i in 0:N){
x[i]<-S*u^{i}*d^{N-i}
}
return(x)
}
N is the number of periods and the rest are just nonimportant values (u is for the up d for down etc)
In order to extract my results i need to make a histogram of the produced vector's logarithm to prove that they are normally distributed. However for a N=100000( i need an great number of steps to prove convergence) when i type hist(x)
i get the error :(invalid number of breaks)
Can anyone help?? thanks in advance.
An example
taf<-rata(100000,1,1.1,0.9,1)
taf1<-log(taf)
hist(taf1,xlim=c(-400,400))
Upvotes: 0
Views: 1321
Reputation: 132999
First I fix your function:
rata<-function(N,r,u,d,S){
x <- numeric(N+1)
for(i in 0:N){
x[i]<-S*u^{i}*d^{N-i}
}
return(x)
}
Or relying on vectorization:
rata<-function(N,r,u,d,S){
x<-S*u^{0:N}*d^{N-(0:N)}
return(x)
}
taf<-rata(100000,1,1.1,0.9,1)
Looking at the result, we notice that it contains NaN
values:
taf[7440 + 7:8]
#[1] 0 NaN
What happened? Apparently the multiplication became NaN
:
1.1^7448*0.9^(1e5-7448)
#[1] NaN
1.1^7448
#[1] Inf
0.9^(1e5-7448)
#[1] 0
Inf * 0
#[1] NaN
Why does an Inf
value occur? Well, because of double overflow (read help("double")
):
1.1^(7440 + 7:8)
#[1] 1.783719e+308 Inf
You have a similar problem with floating point precision when a multiplicant gets close to 0 (read help(".Machine")
).
You may need to use arbitrary precision numbers.
Upvotes: 3