Hey BrownianMan.
What did you get for your distribution of the maximum of B(u) for 0<=u<=t?
First some notation.
B(t) is a standard Brownian Motion.
M(t) = max(B(u): 0<=u<=t)
T(x) = min(u=>0: B(u)=x)
phi(x) = standard normal cdf
I know that P(B(t) > x) = 1-phi(x/sqrt(t)) and P(M(t)=>x) = P(T(x)<=t) = 2*(1-phi(x/sqrt(t))).
I have a problem that asks for P(M(4)<=2). Wouldn't this be equal to 2*(phi(x/sqrt(t))) = 2*(phi(2/sqrt(4))) = 2*phi(1)?? However, the book says the answer is 0.6826, which is not equal to 2*phi(1). What am I doing wrong?