stochastic processes & statistics

samyonez · December 8th, 2004, 8:04 am

standard brownian motion B(0) = 0, B(t)~N(0,t)0<t1<t2given the value of B(t2), what's the distribution of B(t1)?since i have two normally distributed variables B(t1) and B(t2) with cov(B(t1),B(t2)) = t1, in general the question is, in a bivariate normal distribution, what's the distribution of one variable given the value of the other?i guess it's normal with mean (t1/t2)*B(t2), but why exactly and what would be the variance?

epiccn · December 8th, 2004, 7:40 pm

your guess is correct. the point here isp{B(t1)=x; B(t2)=y} = p(t1,0,x) p(t2-t1,x,y) =p(t2,0,y) p(t1(t2-t1)/t2, yt1/t2,x)here p(t,a,b) is the transition density: travelling from a to b in time t