Let $B_t$ be brownian motion at time $ t \geq 0$.
Then I want to find the best predictor of $B_8 + 4$ given that there are observations of brownian motion up to time $t = 1$.
Approach:
Essentially, if $X = X_8 +4$, then $Y$ is our observation, we want to find $\hat X = \mathbb{E}(X|Y)$
So:
$$\mathbb{E}(B_8 + 4 | B_1) = \mathbb{E}(B_8|B_1) + 4$$ since $4$ is a constant.
$$\mathbb{E}(B_8 - B_1 +B_1 | B_1) + 4 = \mathbb{E}(B_8 - B_1|B_1) + \mathbb{E}(B_1|B_1)$$
$$\mathbb{E}(B_8-B_1|B_1) = 0$$ since $B_8-B_1$ is independent of $B_1$.
and $\mathbb{E}(B_1|B_1) = B_1$
Therefore the best predictor is $B_1 + 4$.
Is this correct?