# 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 $X(t) = X(s) + (X(t) - X(s))$.
@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
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.