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Maximum of a Brownian motion?
If S follows a Brownian motion, and we know that at time T, its value is B. Now we start at time 0.What is the probability of the process having value never higher than A during time 0 to T?i.e. what is the probability that the maximum value of the process in time 0 to T is less than A? Assume constants A > B.Thanks a million!!
Maximum of a Brownian motion?
By the way, we assume the brownian motion has zero drift and constant variance, i.e. its volatility at time t is sigma*square_root_of(t), and sigma is a constant.
Maximum of a Brownian motion?
The SDE for the brownian motion conditioned on its ending value isdx = (B-x)/(T-t) dt + sigma dzSo you can write the prob P of x(t) > X for any 0<t<T as the solution to the PDEsig^2/2 d^2P/dx^2 + (B-x)/(T-t) dP/dx + dP/dt = 0with boundary condition P(X,t) = 1 and P(x,T) = 0 for x<X. Then the prob you want is P(0,0).The solution of that PDE is left as an exercise for the reader.
Maximum of a Brownian motion?
Thanks!Is it possible to derive the result through Bayesian probability method?