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This is what I tried to do.

$$E(B_2 B_3B_5)=E(E[(B_2B_3B_5)|\mathcal{F}_2]))=E(B_2(E[(B_3B_5|\mathcal{F}_2]))\\=E(B_2(E[(B_3-B_2+B_2)(B_5-B_2+B_2)|\mathcal{F}_2])).$$

Please help me !

1 Answers 1

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The answer is zero by symmetry. ($B$ and $-B$ are statistically the same.)

To see this via direct calculation: $$ E(B_2B_3B_5) = E(B_2E(B_3 E(B_5|\mathcal{F}_3)|\mathcal{F}_2)) =E(B_2E(B_3^2|\mathcal{F_2}))\\=E(B_2(B_2^2+1))=E(B_2^3+B_2) =0$$ where we used the fact that conditional on $\mathcal{F_t},$ $B_{t'}$ for $t'>t$ is distributed as $N(B_t,t'-t).$

Note that the symmetry argument would work for any odd number of $B$'s but wouldn't for an even number. And for an even number it's generally not zero. For instance $$E(B_2B_5) = E(B_2E(B_5|\mathcal{F}_2)) = E(B_2^2) = 2. $$

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    But , E(B3^2|F2) = 2 , right ? Because B2 is distributed N (0,2) – 2017-02-06
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    @carla $B_3^2|\mathcal{F}_2$ is conditionally distributed $N(B_2, 1)$. – 2017-02-06
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    Yes , you're right . Thank you very much ! – 2017-02-06