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.