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

Verify that $E( X(t) X(s) | X(0)=0 ) = min (t, s)$, where $X(t)$ is standard Brownian motion. I don't know where to start. Thanks!

share|cite|improve this question
What is the definition of a Brownian motion in your textbook? (My question is motivated by your reaction to Nate's hint.) – Did Apr 25 '11 at 19:57
@Didier: My book is very difficult, and we don't use it too much. – user9636 Apr 28 '11 at 15:07
up vote 5 down vote accepted

Hint: if $t > s$, write $X(t) = X(s) + (X(t) - X(s))$.

share|cite|improve this answer
@Nate: Thanks. Can you show a few more steps? – user9636 Apr 25 '11 at 4:02
@user9636: If $t>s$ then $X(s)$ and $X(t)-X(s)$ are independent standard Brownian motions. This may be useful when you take the expectation of their product. – Henry Apr 25 '11 at 16:33
@Henry Sorry but $X(s)$ and $X(t)-X(s)$ are not Brownian motions. They are independent centered Gaussian random variables (with respective variance $s$ and $t-s$). – Did Apr 25 '11 at 19:56
@Didier: True enough, I was careless when abusing the language in the question. I was concentrating on independent and zero mean. – Henry Apr 25 '11 at 21:13
@Nate: Thanks for your hint. Finally, it was useful. I got it. – user9636 Apr 28 '11 at 15:06

Brownian motion has independent increments and $X(t)$ $\sim$ $N(0,t)$, given that it starts from 0. Independent increment also means that $X(t)-X(s)$ is independent of $X(s)$, given that t>s. Therefore you can write $X(t)=X(t)-X(s)+X(s)$ Which gives you the formula of expectation that $$E(X(t)-X(s)+X(s))(X(s))=E((X(t)-X(s))X(s))+E(X^2(s))$$ $$=E(X(t)-X(s))E(X(s))+E(X^2(s))=0+s=s.$$ And the case where $s>t$ is done in the same way.

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.