[R-SIG-Finance] sufficient n for a binomial option pricing model

[Smith, Dale]

One way to terminate is to look at the consecutive differences between
the averages and terminate if the difference is less than your
tolerance. However, you should guard against the case where the
consecutive differences are never less than the tolerance. In this case,
just put in a maximum number of steps n and log the last average, the
number of steps, and a message. This allows the user to determine
whether they want to accept or reject the result.

Thanks,
Dale Smith, Ph.D.
Senior Financial Quantitative Analyst
Risk & Compliance
Fiserv.
107 Technology Park
Norcross, GA 30092
Office: 678-375-5315
Mail: dale.smith at fiserv.com
www.fiserv.com

-----Original Message-----
From: r-sig-finance-bounces at r-project.org
[mailto:r-sig-finance-bounces at r-project.org] On Behalf Of J Toll
Sent: Thursday, September 06, 2012 10:25 AM
To: r-sig-finance at r-project.org
Subject: [R-SIG-Finance] sufficient n for a binomial option pricing
model

Hi,

I have a question regarding the selection of n, the number of time
steps, in a binomial option pricing model.  I suppose my question is not
strictly related to R.  As larger values should be more accurate, what
I've read on the subject simply suggests that you use a sufficiently
large value for your purposes.  So I've been trying to evaluate what is
a sufficiently large value of n for my purposes.  Is there any rule of
thumb regarding the value of n?

When using the fOptions package CRRBinomialTreeOption function, with
varying n, the price oscillates back and forth converging on a price.
This can be clearly seen through plotting.

require(fOptions)

x <- function(n) {

  CRRBinomialTreeOption(TypeFlag = "ca",
                        S = 50,
                        X = 50,
                        Time = 1/12,
                        r = 0.02,
                        b = 0.02,
                        sigma = 0.18,
                        n = n)@price
}

y <- sapply(1:100, x)               # mean(y) == 1.079693
plot(y)

Given this oscillation, my question is whether it would be "better" to
compute two prices using two smaller, consecutive values of n rather
than one large value?  Or is there some other better way?

For example, using n =1000 or 1001, the option prices are within 5
hundredths of a cent, but the calculation is extremely slow for either.

x(1000)                                    # 1.077408
x(1001)                                    # 1.077926

mean(sapply(1000:1001, x))      # 1.077667

Comparatively, taking the mean of n= 40 and 41 yields a value very close
to the middle of the range, yet is much faster.

mean(sapply(40:41, x))             # 1.0776

It seems like averaging two smaller, consecutive values of n is
basically as accurate and far faster than using large values of n.  I
was hoping someone might have some insight into why this might or might
not be a valid approach.  Thanks.


James

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