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.


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.


If $0<s<t$ then $$ \operatorname{cov}(B_t,B_s) = \operatorname{cov}(B_s + (B_t-B_s),B_s) = \operatorname{cov}(B_s,B_s) + \operatorname{cov}(B_t-B_s,B_s) = \operatorname{var}(B_s) + 0 = s. $$

  • $\begingroup$ thanks! but how would you represent this backward-brownian-motion in terms of Ito process or Ito integral? $\endgroup$ – Troy McClure Dec 28 '12 at 1:09
  • $\begingroup$ @MichaelHardy The covariance is never max(s,t). $\endgroup$ – Did Dec 28 '12 at 22:19
  • $\begingroup$ @did : Let $C_t=B_t-B_1$. Then for $0<s<t<1$, you've got $\operatorname{cov}(C_s,C_t) = t$, i.e. it's the maximum. $\endgroup$ – Michael Hardy Dec 28 '12 at 23:37
  • $\begingroup$ @MichaelHardy Not at all. For the reason why this example fails and why any other would fail as well, see my now deleted answer. $\endgroup$ – Did Dec 28 '12 at 23:39
  • $\begingroup$ @did : OK, I see: it's the distance from the point where the process is fixed (in the example, $1$) and the nearest of the two indices to that point. $\endgroup$ – Michael Hardy Dec 28 '12 at 23:47

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