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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
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up vote 4 down vote accepted

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

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@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
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