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It's a bit weird to just throw up your hands just because you need to take a logarithm of a negative number. Plugging ina negative number might cause the soluti
by contravariant 7y ago
It's a bit weird to just throw up your hands just because you need to take a logarithm of a negative number. Plugging ina negative number might cause the solution to become complex but as far as I can tell it should remain a solution of the corresponding differential equation. So either the partial differential equation has no real solutions or the complex part cancels out or can be removed. Some information seems to be missing to truly determine Black Scholes to be broken.
- cm2187 7y agoI wonder how you transfer a complex amount of cash through the banking system...
- williamscales 7y agoPresumably the cash amplitude can be taken as positive.
- contravariant 7y agoWell, to illustrate my point the differential equation f'(x) = 1/x has the solution log(x) for positive x. If you let x become negative then that particular solution ends up becoming complex, however there is still a real solution namely log(-x) (which only differs by a constant). The Black Scholes equation is a tad too complicated to say whether something like that will happen but it's too simple to just conclude everything breaks because of one complex logarithm.