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?
Brownian motion has independent increments and $X(t) \sim N(0,t)$.
If $t < s$ then
$$0 = E[X(t)(X(s)-X(t))] = E[X(t)X(s)]-E[X(t)^2]=E[X(t)X(s)]-t.$$
Hence,
$$E[X(t)X(s)] = t = \min(t,s)$$
Similarly if $s < t$, then $E[X(t)X(s)] = s = \min(t,s)$