Given a Wiener process X, how do I prove this?

$R_x(s,t) = E[X(s)X(t)] = min(s,t)$

There seems to be a trick with dividing to two cases of $s<t$ and $s>t$, but I can't figure out how this would be helpful.

This is what I've got so far:

$R_x(s,t) = E[X(s)X(t)] = E[X(s)(X(t)-X(s)+X(s))]$ $=E[X(s)^2]+E[X(s)(X(t)-X(s)]=E[X(s)^2]+E[(X(s)-X(0))(X(t)-X(s)]$ $=E[X(s)^2]+E[(X(s)-X(0))]E[(X(t)-X(s)]$ $=E[X(s)^2]+E[(X(s))]E[(X(t)-X(s)] = E[X(s)^2] = Var[X(s)] = s$

But the same thing could be done with t instead of s... So what am I doing wrong?

  • $\begingroup$ Hello and welcome to MSE! Be sure to provide thorough background or your own work to avoid having your questions prematurely closed! $\endgroup$ – Adam Hughes Aug 1 '14 at 2:55
  • $\begingroup$ @Ana Mzmz: You edited with the proof as I was posting -- the argument is symmetric with respect to $s$ and $t$. $\endgroup$ – RRL Aug 1 '14 at 3:13

Brownian motion has independent increments and $X(t) \sim N(0,t)$.

If $t < s$ then

$$0 = E[X(t)(X(s)-X(t))] = E[X(t)X(s)]-E[X(t)^2]=E[X(t)X(s)]-t.$$


$$E[X(t)X(s)] = t = \min(t,s)$$

Similarly if $s < t$, then $E[X(t)X(s)] = s = \min(t,s)$

  • $\begingroup$ I still don't understand the importance of assuming $t<s$. What's wrong with the following, for example? Assume $s<t$. So $0=E[X(t)(X(s)−X(t))]=E[X(t)X(s)]−E[X(t))^2]$, hence $E[X(t)(X(s)]=t \neq min(t,s)$ $\endgroup$ – Ana M Aug 1 '14 at 3:14
  • $\begingroup$ That is incorrect because if $s < t$ then $t = \max(t,s)$. Furthermore $X(t)-X(0)$ and $X(s)-X(t)$ are not independent in this case -- the increments overlap. $\endgroup$ – RRL Aug 1 '14 at 3:18
  • $\begingroup$ Thank you! The overlapping explanation is what I was looking for. $\endgroup$ – Ana M Aug 1 '14 at 3:23
  • $\begingroup$ You're welcome. $\endgroup$ – RRL Aug 1 '14 at 3:23

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