Brownian Motion Covariance: max instead of min It is known that $\operatorname{Cov}(B_t,B_s)=\min(t,s)$ where $B$ is Brownian motion.
Can one think of an Ito process or integral (preferrably plain Gaussian process) $W$ such that $\operatorname{Cov}(W_t,W_s)=\max(t,s)$?
Let me ask it in another way: it is known that $k(x,y)=\min(x,y)$ is the reproducing kernel of the Cameron Martin RKHS. What is the RKHS (if any) of the kernel $k(x,y)=\max(x,y)$?
Thanks for your help!
EDIT: Please recall that
$$\operatorname{Cov}(B_s,B_t)=\operatorname{Cov}(sB_{1/s},tB_{1/t})$$
but I didn't manage to go further.
 A: So that this has a (non-deleted) answer:
No such process can exist.  If $\operatorname{Var}(W_0) = \max(0,0) = 0$ then $W_0$ is a constant.  In particular, $W_0$ and $W_1$ are independent, so we have $\operatorname{Cov}(W_0, W_1)= 0 \ne 1= \max(0,1)$.
More generally, Cauchy-Schwarz shows that for any process with finite variance, we have $\operatorname{Cov}(X_s, X_t) \le \sqrt{\operatorname{Var}(X_s) \operatorname{Var}(X_t)}$.  Thus the covariance function $k(s,t)$ must satisfy $k(s,t) \le \sqrt{k(s,s) k(t,t)}$.  The function $k(s,t) = \max(s,t)$ does not, so it cannot be a covariance function.
A: If $0<s<t$ then
\begin{align}
& \operatorname{cov}(B_t,B_s) \\[8pt]
= {} & \operatorname{cov}(B_s + (B_t-B_s),B_s) \\[8pt]
= {} & \operatorname{cov}(B_s,B_s) + \operatorname{cov}(B_t-B_s,B_s) \\[8pt]
= {} & \operatorname{var}(B_s) + 0 = s.
\end{align}
A: The function $k:(t,s)\mapsto\max(t,s)$ is not a covariance kernel since, for example, the matrix $\begin{pmatrix}1 & 2\\ 2 & 2\end{pmatrix}$ has a negative eigenvalue.
Edit: About the Edit in the question, note that
$$
\operatorname{Cov}(sB_{1/s},tB_{1/t})=st\cdot\operatorname{Cov}(B_{1/s},B_{1/t})=st\cdot\min(1/s,1/t)=\min(s,t),
$$
which is certainly not equal to $\max(s,t)$.
