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Question on BM.
QuoteOriginally posted by: NormalAt the risk of sounding a little pedantic, I think there should be a square root inside the Arctan The way I drew the picture, I think that tan theta = -(t_2-t_1)/t_1, so I think the ratio in the arctan should be flipped around. But what is the reason for the square root?
Question on BM.
Conditioned on B_t1 > 0, the probability of staying above zero on [t1, t2] is 2 \Phi( B_t1/(Sqrt[t2-t1]) ) - 1. Integrating over B_t1 ~ N(0, t_1) gives (1/Pi) Arctan(Sqrt(t1/t2-t1). I guess that's probably what you did, but you might have missed the "sqrt" changing variables.
Last edited by Normal on September 12th, 2006, 10:00 pm, edited 1 time in total.
Question on BM.
Okay, I see---I was thinking of B_t as having standard deviation t when in fact it has variance t
Last edited by gentinex on September 13th, 2006, 10:00 pm, edited 1 time in total.
Question on BM.
QuoteOriginally posted by: FedorEgood to know! will have a look.P.S. sometimes it's useful not just to have a book, but actually read it LOLas far as I am concerned, the important thing is to buy it, all else is optional