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This question is regarding an answer to the question below:

Expectation regarding Brownian Motion

This is a formula regarding getting expectation under the topic of Brownian Motion. $$ \begin{align*} E[W(s)W(t)] &=E[W(s)(W(t)−W(s))+W(s)^2]\\ &=E[W(s)]E[W(t)−W(s)]+E[W(s)^2]\\ &=0+s =\min(s,t). \end{align*} $$

One of Michael Hardy's comment is: "The step that says $$E[W(s)(W(t)−W(s))]=E[W(s)]E[W(t)−W(s)]$$ depends on an assumption that $t>s$."

So, finally, my question is how does the assumption $t>s$ play out in the $$E[W(s)(W(t)−W(s))]=E[W(s)]E[W(t)−W(s)]?$$

What if $t\leq s$?

Thanks a lot! Love the smart math stack exchange crowd!

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Which definition of Brownian motion have you been exposed to? –  Did Nov 30 '12 at 1:29
    
this question is under the brownian motion as Gaussian process. thx –  user1486802 Nov 30 '12 at 6:38
    
?? Sorry but this is quite insufficient to characterize Brownian motion... Surely you know more than this on the subject? –  Did Nov 30 '12 at 11:31

1 Answer 1

up vote 1 down vote accepted

For $t>s$ you know from the definition of a Brownian Motion that $W_t-W_s$ is independent of $W_s$ and this implies

$$\mathbb{E}(W_s \cdot (W_t-W_s)) = \mathbb{E}(W_s) \cdot \mathbb{E}(W_t-W_s)$$

If $t \leq s$ that's not true, but you can use

$$\mathbb{E}(W_s \cdot W_t) = \mathbb{E}((W_s-W_t) \cdot W_t +W_t^2)$$

instead (where $W_s-W_t$ is now independent of $W_t$) and do the same calculations as above.

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