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Heston's option price >(=) BS-price
Posted: August 14th, 2012, 8:00 pm
by japanstar
Thank you KnutSchnute!
Heston's option price >(=) BS-price
Posted: August 15th, 2012, 12:07 am
by Alan
QuoteOriginally posted by: japanstarHi Alan,Sorry for the probably banal question: would you mind showing how you arrived atC(sigma + d sigma) - 2 C(sigma) + C(sigma - d sigma) from d^2 C/ d sigma^2 ?Thank you very much!Ps: it is really great to have someone like you on this forum!Thanks.That deriv. has a standard centered difference approximation:f'+(x) = [f(x+dx) - f(x)]/dx (forward dif)f'_(x) = [f(x) - f(x-dx)]/dx (backwards dif)f''(x) = [f'+(x) - f'_(x)]/dx = [f(x+dx) - 2 f(x) + f(x-dx)]/ dx^2 (centered 2nd dif)Then I dropped the dx^2 because it is positive and doesn't affect the sense of the inequality I wrote.
Heston's option price >(=) BS-price
Posted: August 15th, 2012, 4:04 am
by japanstar
QuoteOriginally posted by: AlanQuoteOriginally posted by: japanstarHi Alan,Sorry for the probably banal question: would you mind showing how you arrived atC(sigma + d sigma) - 2 C(sigma) + C(sigma - d sigma) from d^2 C/ d sigma^2 ?Thank you very much!Ps: it is really great to have someone like you on this forum!Thanks.That deriv. has a standard centered difference approximation:f'+(x) = [f(x+dx) - f(x)]/dx (forward dif)f'_(x) = [f(x) - f(x-dx)]/dx (backwards dif)f''(x) = [f'+(x) - f'_(x)]/dx = [f(x+dx) - 2 f(x) + f(x-dx)]/ dx^2 (centered 2nd dif)Then I dropped the dx^2 because it is positive and doesn't affect the sense of the inequality I wrote.Great!