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.
 A: 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.
