# brownian motion

• Feb 26th 2007, 03:00 AM
chogo
brownian motion
also for a brownian motion process

why is 2B(3) - B(2) - 2B(1) = N(0,1) + N(0,1) + N(0,4)

thanks alot
• Feb 26th 2007, 03:05 AM
topsquark
Quote:

Originally Posted by chogo
also for a brownian motion process

why is 2B(3) - B(2) - 2B(1) = N(0,1) + N(0,1) + N(0,4)

thanks alot

What??? :confused:

-Dan
• Feb 26th 2007, 03:30 AM
chogo
sorry

if X(t) = -1 for 0 <= t <= 1
X(t) = 1 for 1 < t <= 2
X(t) = 2 for 2 < t <= 3

then the ito integral int_0^3 X(t)dB(t) = 2B(3) - B(2) - 2B(1)

that i understand. But why does 2B(3) - B(2) - 2B(1) have a distribution Normal(0,6) or if put as sums N(0,1) + N(0,1) + N(0,4).

I know from the definition of wiener process that Bt - Bs = N(0,t-s) for 0<s<t
• Feb 26th 2007, 10:44 AM
chogo
ive learnt on this forum if no one responds its cos they dont understand the question. Is this the case or does no one know this?
• Feb 26th 2007, 05:53 PM
JakeD
Quote:

Originally Posted by chogo
sorry

if X(t) = -1 for 0 <= t <= 1
X(t) = 1 for 1 < t <= 2
X(t) = 2 for 2 < t <= 3

then the ito integral int_0^3 X(t)dB(t) = 2B(3) - B(2) - 2B(1)

that i understand. But why does 2B(3) - B(2) - 2B(1) have a distribution Normal(0,6) or if put as sums N(0,1) + N(0,1) + N(0,4).

I know from the definition of wiener process that Bt - Bs = N(0,t-s) for 0<s<t

I have only a casual knowledge of Brownian motion, but I think it goes like this. The integral is

int_0^3 X(t)dB(t) = 2(B(3)-B(2)) + (B(2)-B(1)) - (B(1)-B(0)).

That (B(t) - B(t-1)) ~ N(0,1) for t = 1,2,3 follows from (B(t) - B(s)) ~ N(0,t-s) when s = t-1 and B(0) = 0. Next, after the multiplication by 2, 2(B(3)-B(2)) ~ N(0,2^2) = N(0,4). Finally, use the sum of independent normal variables has a normal distribution with variance equal to the sum of the variances to get

2(B(3)-B(2)) + (B(2)-B(1)) - (B(1)-B(0)) ~ N(0,4+1+1) = N(0,6).
• Feb 27th 2007, 02:18 AM
chogo
that was a complete world of help THANK YOU SO MUCH