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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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.