Why is this wrong - conditional expectation of brownian motion: $\mathbb{E}[B_1 | B_2]$ Trying to find $$ \mathbb{E}[B_1 | B_2]$$
$$\mathbb{E}[(B_1 - B_2 + B_2) | B_1] = \mathbb{E}[(B_1-B_2)|B_2] + \mathbb{E}[B_2 | B_2] $$
$$\mathbb{E}[ -(B_2 - B_1)| B_2] + B_2$$
Since $$ -(B_2  - B_1) \sim N(0, (2+1))$$
$$\mathbb{E}[ -(B_2 - B_1)| B_2] + B_2 = 0 + B_2$$
The answers I have is it is $$\frac{1}{2}B_2$$
and they use the formula: $$\mathbb{E}[B_1|B_2] = \mathbb{E}[B_1] + \frac{Cov(B_1,B_2)}{Var(B_2)} \dots$$
Why is my method not applicable here? Where is the mathematical error?
 A: Well first keep in mind that the variance for $B_2-B_1$ is incorrect (you can only add variances of normals if they are independent, and the only thing we know about Brownian motion is that it has independent increments). 
Also Graham is correct in that the conditional expectation is not zero. Regarding your response to him, the last equality follows because we can show $B_2-B_1$ is independent from $B_1$:
$$Cov(B_2-B_1,B_1)=Cov(B_2,B_1)-Cov(B_1,B_1)=\min(2,1)-\min(1,1)=0$$
Thus we can write $\mathbb{E}[B_2-B_1|B_1]=\mathbb{E}[B_2-B_1]=0$.
That formula for conditional normals that you gave is the most convenient way to solve the problem, but if you wanted to approach finding the distribution of $B_1|B_2=b$ without systematically calculating the expectation and variance, you could try:
Time Reversal- $\overline{B}_t=tB_{1/t}$. $\{\overline{B}_t\}$ is standard Brownian motion iff $\{B_t\}$ is standard Brownian motion. After doing this, you can make the problem equivalent to finding the distribution of $\overline{B}_1|\overline{B}_t=\overline{b}$ where $t \in (0,1)$. Then you can just shift the time and position based on the conditioning.
A: The distribution of the difference being normal does not mean that the conditional distribution of the difference will also be normal.
$(X-Y)\sim N(0,\sigma^2) \quad\;\;\!\!\!\implies\quad \mathsf E(X-Y) =0$
$(X-Y)\sim N(0,\sigma^2) \quad{\;\;\;\not\!\!\!\implies\;}\quad \mathsf E(X-Y\mid Y) =0$
