Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.