How to calculate $\mathbb{E}((B_3-B_2)(B_4-B_{\pi}) \mid B_1)$ for a Brownian motion $(B_t)_{t \geq 0}$ I want to find the best predictor of $(B_3-B_2)(B_4-B_{\pi})$ given an observation of $B_1$
Where $B_t$ is brownian motion for time $t \geq 0$.
I am not sure how to approach this.
I know it will be the result of the conditional expecation: $\mathbb{E}[(B_3-B_2)(B_4-B_{\pi})|B_1]$ But have no idea how to compute this result.
Would $(B_3-B_2)$ be independent of $(B_4-B_\pi)$ ?
I know the property of an independence of increments in brownian motion exists, but was not sure if that could be applied here.

For any $0 = t_0 <  \ldots < t_n$, we know that
$$B_{t_n}-B_{t_{n-1}}, B_{t_{n-1}}-B_{t_{n-2}},\dots,B_{t_1}-\underbrace{B_{t_0}}_{0}$$
are independent random variables. Choosing $t_0 = 0$, $t_1 = 1$, $t_2 = 2$, $t_3 = 3$, $t_4 = \pi$, $t_5 = 4$, we get that
$$B_4-B_{\pi},B_{\pi}-B_3, B_3-B_2,B_2-B_1,B_1 \tag{1}$$
are independent. This implies in particular that  $(B_3-B_2) \cdot (B_4-B_{\pi})$ and $B_1$ are independent. Hence,
$$\mathbb{E}( (B_3-B_2) \cdot (B_4-B_{\pi}) \mid B_1) = \mathbb{E}((B_3-B_2) \cdot (B_4-B_{\pi})).$$
Now use that, by $(1)$, $B_4-B_{\pi}$ and $B_3-B_2$ are independent to conclude that the expression equals $0$.
