1. 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

2. 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???

-Dan

3. 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

4. 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?

5. 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).

6. that was a complete world of help THANK YOU SO MUCH