$B_{i} $ is a standard Brownian motion. $$E[\prod_{i=1}^3 B_{i}]$$
I know how to find $E[B_{1}B_{2}]$, but how do I find the expectation of this one?
$E[B_{1}(B_{2}-B_{1}+B_{1})(B_{3}-B{2}+B_{2})]=E[B_{1}(B_{2}-B_{1})(B_{3}-B_{2})+B_{1}B_{2}(B_{2}-B_{1})+B_{1}^2B_{2}+B_{1}^2(B_{3}-B_{2})]=E[B_{1}B_{2}^2]+E[B_{1}^2(B_{3}-B_{2})]$
This is my workings so far.
We write $\Delta_{ij}$ for the increment $B_j-B_i$, $j>i$, and have soon products of only increments: $$ \begin{aligned} B_3 &= (B_1-B_0) + (B_2-B_1)+ (B_3-B_2)\\ &= \Delta_{01}+\Delta_{12}+\Delta_{23}\ ,\\ B_2 &= (B_1-B_0) + (B_2-B_1)\\ &= \Delta_{01}+\Delta_{12}\ ,\\ B_1 &= (B_1-B_0)\\ &= \Delta_{01}\ .\\[3mm] &\qquad\text{This implies:}\\[3mm] \Bbb E[\ B_3B_2B_1\ ] &= \Bbb E\left[\ \Big(\ \Delta_{01}+\Delta_{12}+\Delta_{23}\Big) \Big(\ \Delta_{01}+\Delta_{12}\Big) \Delta_{01} \ \right] \\ &= \Bbb E\left[\ \Big(\ \Delta_{01}+\Delta_{12}+\Delta_{23}\Big) \Big(\ \Delta_{01}^2+\Delta_{12}\Delta_{01}\Big) \ \right] \\[3mm] &= \Bbb E\left[\ \Delta_{01}^3+\Delta_{01}^2\Delta_{12}+\Delta_{01}^2\Delta_{23} \ \right] \\ &\qquad\qquad + \Bbb E\left[\ \Delta_{01}^2\Delta_{12}+\Delta_{01}\Delta_{12}^2 +\Delta_{01}\Delta_{12}\Delta_{23} \ \right] \\[3mm] &= \Bbb E\left[\Delta_{01}^3\right] + \Bbb E\left[\Delta_{01}^2\Delta_{12}\right] + \Bbb E\left[\Delta_{01}^2\Delta_{23}\right] \\ &\qquad\qquad + \Bbb E\left[ \Delta_{01}^2\Delta_{12}\right] + \Bbb E\left[\Delta_{01}\Delta_{12}^2\right] + \Bbb E\left[\Delta_{01}\Delta_{12}\Delta_{23}\right] \\[3mm] &= \Bbb E\left[\Delta_{01}^3\right] + \Bbb E\left[\Delta_{01}^2\right]\; \Bbb E\left[\Delta_{12}\right] + \Bbb E\left[\Delta_{01}^2\right]\; \Bbb E\left[\Delta_{23}\right] \\ &\qquad\qquad + \Bbb E\left[ \Delta_{01}^2\right]\; \Bbb E\left[\Delta_{12}\right] + \Bbb E\left[\Delta_{01}\right]\; \Bbb E\left[\Delta_{12}^2\right] + \Bbb E\left[\Delta_{01}\right]\; \Bbb E\left[\Delta_{12}\right]\; \Bbb E\left[\Delta_{23}\right] \\[3mm] \end{aligned} $$ Among the many products some are vanishing... i hope it is clear now how to proceed...
$$E(B_1B_2B_3)=E[E(B_1B_2B_3|B_1,B_2)]$$ $$=EB_1B_2E(B_3|B_1,B_2)$$ and $$E(B_3|B_1,B_2)$$ $$=E(B_3-B_2|B_1,B_2)+E(B_2|B_1,B_2)=B_2.$$ Hence, $E(B_1B_2B_3)=EB_1B_2^{2}$. Now write $B_2$ as $(B_2-B_1) +B_1$. Can you finish?
