# covariance function for Brownian motion

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.

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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

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 $X$?
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.
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