Brownian Motion Covariance: max instead of min

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.