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I have this given problem, how can I calculate this mixed moment?

$$E[|B(t)|* B(1-t)]$$

I know that the expected value of the Brownian motion is $0$, so this is a covariance, but the absolute value make me insecure on how calculate the result.

Gerardo Santonicola
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1 Answers1

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$B$ has the same distribution as $-B$, so $$J:=E[|B(t)|* B(1-t)]= E[|-B(t)|* (-B(1-t))]=-J.$$ Therefore $J=0$.

Yuval Peres
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