Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What would the covariance function be of $V(t) = (1-t) B[t/(1-t)]$ if $B(t)$ is standard Brownian motion. Also $t$ is between $0$ and $1$.

Thanks for the help!

EDIT:

Here is where I am stuck:

I believe that $Cov(V(t),V(s)) = E[V(t)V(s)]-E[V(t)][V(s)]$. We know that $E[V(t)]$ and $E[V(s)]$ are $0$ so the second term disappears. We now need to make $E[V(t)V(s)]$ independent so we can separate it out. We can do this by $E[(V(t))(V(s)-V(t)+V(t))]$ which is equal to $E[V(t)V(s-t)] - E[(V(t))^2]$. $E[V(t)V(s-t)]$ is $0$ and we are left with $E[(V(t))^2]$ which is simply the Variance of $V(t)$. Subbing back in $B(t)$ into this expression, we have $Var((1-t)\cdot B[t/(1-t)])$ and this is where I am stuck... Thanks for the help.

share|improve this question
1  
What have you tried? Do you know the covariance of $(B(t))$? What does it mean to say that the covariance is such and such? Can you apply the formula for $(B(t))$ to the process $(V(t))$? (Yes you can.) –  Did Apr 4 '11 at 5:26
    
@DidierPiau here is what I tried so far: I believe that Cov(V(t),V(s)) = E[V(t)V(s)]-E[V(t)][V(s)]. We know that E[V(t)] and E[V(s)] are 0 so the second term disappears. We now need to make E[V(t)V(s)] independent so we can separate it out. We can do this by E[(V(t))(V(s)-V(t)+V(t))] which is equal to E[V(t)V(s-t)] - E[(V(t))^2]. E[V(t)V(s-t)] is 0 and we are left with E[(V(t))^2] which is simply the Variance of V(t). Subbing back in B(t) into this expression, we have Var((1-t)* B[t/(1-t)]) and this is where I am stuck... Thanks for the help. –  icobes Apr 4 '11 at 5:32
    
So you want to compute $E(V(t)V(s))$ for fixed $t$ and $s$. Now, you could try to compute $E(a(t)B(c(t))a(s)B(c(s))$ for any fixed numbers $a(t)$, $a(s)$, $c(t)$ and $c(s)$. –  Did Apr 4 '11 at 5:53
    
@DidierPiau So what I have done is wrong then? E[V(t)V(s)] is not equal to Var(V(t)) as I showed through my calculations? –  icobes Apr 4 '11 at 5:55
    
Edit: To do that, you should assume that $c(t)\le c(s)$... –  Did Apr 4 '11 at 5:55
show 14 more comments

1 Answer

up vote 1 down vote accepted

Hint: The standard Brownian bridge, $X$, can be defined by $X(t) = B(t) - tB(1)$, $0 \leq t \leq 1$. Can you calculate the covariance function of $X$?

EDIT (more details).

Suppose that $Y$ is defined by $Y(t) = f(t)B(h(t))$, for $t \in I$. Then, for any $s,t \in I$ (say with $s \leq t$; see remark at the end), $$ {\rm Cov}(Y(s),Y(t)) = {\rm Cov}(f(s)B(h(s)),f(t)B(h(t))) = f(s)f(t) {\rm Cov}(B(h(s)),B(h(t))) . $$ Now, for any $u,v \geq 0$, ${\rm Cov}(B(u),B(v)) = \min \{u,v\}$. Hence, it is easy to calculate ${\rm Cov}(B(h(s)),B(h(t)))$ when $h$ is a monotone function. To check yourself, note that the covariance function of the process $V$ defined by $V(t)=(1-t)B(t/(1-t))$ is the same as that of the process $X$ defined by $X(t)=B(t)-tB(1)$, $t \in [0,1]$. (You can assume that $V(1)=0$.) Since the covariance function of a zero mean Gaussian process determines the law of the entire process, it would follow that $V$ is a standard Brownian bridge on $[0,1]$ (since $X$ is).

General remark: By symmetry, it suffices to calculate covariance functions for $s \leq t$ only.

share|improve this answer
    
Sure this will help the OP? See the succession of comments on the main post. Right now, we are struggling to know how to write $E((aX)(bY))$ as a function of $E(XY)$, so... –  Did Apr 4 '11 at 7:10
    
@Didier: I'll add more details. –  Shai Covo Apr 4 '11 at 7:15
    
Details added... –  Shai Covo Apr 4 '11 at 7:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.