Brownian Motion Covariance: max instead of min

Question

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.

Answer 1

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}

Answer 2

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 $\max(s,t)$.