Why using the transformation x=Ln(s) is computationally more efficient ?

Seagull77 · August 3rd, 2005, 8:51 am

Hi,I am trying to figure out why using the transformation x= Ln(S) is computationally more efficient when using finite difference methods . Can anyone share with me why we use this transformation ? One reason that I can think of is that after the transformation, the differential equation will have a constant drift and variance and will cut down the computation time for the finite difference method. Anyone can help ? Thank you.

ppauper · August 3rd, 2005, 12:42 pm

another will be that there are fewer multiplications:instead of terms likeS dV/dS you get terms like dV/dx

spursfan · August 3rd, 2005, 1:18 pm

better to spend less time figuring and more time doing - if you're interested in computational efficiency, code it up both ways and compare speed and accuracy