From: Joseph W. <jo...@co...> - 2004-12-31 07:16:39
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I think the basic problem is that when you turn x = log(S) into an operator, dx = dS / S and you start getting extra terms in the difference equation that aren't being put into BSMOperator. One could go and work out what the operator should be with a logarithmic grid, but I suspect that the second derivative is going to add a lot of nasty terms. |