# [SOLVED] Covariance of Brownian Motion

• Mar 9th 2009, 08:33 PM
horan
[SOLVED] Covariance of Brownian Motion
Just a little confused about the following. Given that $\displaystyle W$ denotes a one-dimensional standard Brownian motion, find $\displaystyle Cov(W_t,W_s)$

Step 1: Assume that $\displaystyle s \leq t$, then
$\displaystyle Cov(W_t,W_s)=E&#91;W_tW_s&#93;=E&#91;(W_t-W_s+W_s)W_s&#93;=Var(W_s)=s$

Step 2: Then, in general:
$\displaystyle Cov(W_t,W_s)=\min \{s,t\}$

Could someone explain each step and properties used? Thanks.
• Mar 9th 2009, 09:18 PM
zigzag20
Fix 0 ≤ s ≤ t.
Step 1:
Cov(W(t), W(s)) = E[W(t)W(s)] − E[W(t)]E[W(s)] = E[W(t) * W(s)] as these are standard brownian motion. Hence, E(W(t)) = E(W(s)) = 0.

E[W(t)W(s)] = E[(W(s) + W(t) − W(s))W(s)]
= E[W^2(s)] + E[(W(t) − W(s))W(s)] = s

Why? Since E[W(t)] = 0 for all t and since by definition of the standard Brownian motion we have E[W^2(s)] = s.

And by independent increments property we have E[(W(t) − W(s))W(s)] = E[(W(t) − W(s))(W(s) − W(0))] = E[(W(t) − W(s)]E[(W(s) − W(0)] = 0, since
increments have a zero mean Gaussian distribution.

Step 2: Repeat the same for t < s. Hence, you shall see min(t,s) to be the general answer.

Quote:

Originally Posted by horan
Just a little confused about the following. Given that $\displaystyle W$ denotes a one-dimensional standard Brownian motion, find $\displaystyle Cov(W_t,W_s)$

Step 1: Assume that $\displaystyle s \leq t$, then
$\displaystyle Cov(W_t,W_s)=E[W_tW_s]=E[(W_t-W_s+W_s)W_s]=Var(W_s)=s$

Step 2: Then, in general:
$\displaystyle Cov(W_t,W_s)=\min \{s,t\}$

Could someone explain each step and properties used? Thanks.

• Mar 23rd 2009, 06:45 AM
sonyrobocup
I am reading the solution, and wondering in W^2(s), what does the ^ symbol stands for.