Lookback option in C++

[ckwright]

Here's a question for the coding community here . . .


If I'm building a model for a lookback option, the number of possible paths at period 'n', should be equal to 2^n.

THe problem is once you try to extend the lattice to certain point, the array's dimension won't accomodate for the number of possible option price at that node.

Anyone have any tips for extending the lattice past, lets just say, 60 periods.


I know of other methods, such as monte carlo, however I was specifically worried for the 'lattice' method now.


ckwright

 

[msj]

use the auxiliary variable method.

 

[ckwright]

use the auxiliary variable method.

Is there a good academic paper out there on this?

 

[SteveTownsend]

A Google search on lookback "auxiliary variable" could be helpful. I found a lot of relevant stuff quickly.

 

[msj]

well it's covered in textbooks, eg Wilmott or my book

 

[bob]

Here's a question for the coding community here . . .


If I'm building a model for a lookback option, the number of possible paths at period 'n', should be equal to 2^n.

THe problem is once you try to extend the lattice to certain point, the array's dimension won't accomodate for the number of possible option price at that node.

Anyone have any tips for extending the lattice past, lets just say, 60 periods.


I know of other methods, such as monte carlo, however I was specifically worried for the 'lattice' method now.


ckwright

I haven't ever done it, and haven't checked out the auxiliary variables method msj cites, but from a quick look at the problem, it seems that as long as u = 1/d, binomial trees to price lookbacks do "recombine" in some sense.

For example, consider the paths udu and duu. Both end at the same spot (u), and both have the same running maximum (u). So you can add the probabilities of the two paths and treat them as the same node. I haven't worked it all the way out, but the number of nodes under this scheme seems to increase far less quickly than 2^n. What you'd wind up with at the terminal time step is a probability associated with each ordered pair (S(T), running max), at which point you could take expected values and be done with it.

I feel sure that there are more elegant solutions around, but this is the first one that occurred to me.