# Brownian motion

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!

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 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

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