The following is taken from Selfsimilar Processes by Paul Embrechts:
I have only seen this before as: $\ E[B(t)B(s)]$ (without the ' prime next to the $\ B(s)$)
The technique I have seen before involves using the substitution:
$\ B(s) = B(t) + (B(s) - B(t))$
And after some rearranging, you pop out that:
$\ E[B(t)B(s)] = t$ if you arbitrarily picked $\ t<s$ or
$\ E[B(t)B(s)] = s$ if you arbitrarily picked $\ s<t$, (which is the same result as above, albeit less elegant.)
BUT: I can't see (a) what the prime is supposed to denote in Embrechts' proof and (b) how he gets the first line of his proof.
(a) I initially thought the prime was supposed to specify that $\ B(s)'$ is a different Brownian Motion to $\ B(s)$. But then in the first line of the proof, the prime is scoping over an entire bracket: $\ ((B(t) - B(s))'$. Is this meant to be the incremental noise process for the separate 'primed' Brownian Motion? What would be the motivation for distinguishing between 2 separate processes?
(b) The jump is just not clear to me.
Many thanks, all!
EDIT: the $\ I$ in Embrechts' proof is the identity matrix, since this is supposed to hold true of general n-dimensional Brownian motion.