Conditional Expectation of one dimensional Brownian Motion Consider the one dimensional Brownian motion $B=(B_t)_{t \ge 0}$.
How to prove that for conditional expectation holds $\mathbb{E}[B_t \vert B_1]= tB_1$ where $t \in [0,1]$?
 A: Set $W_t = \sqrt{t} X$ and $W_1-W_t = \sqrt{1-t} Y$. Then $X,Y$ are standard and independent Gaussian random variables.
You are asked to compute
$$
\mathbb{E}[W_t |W_1] = \mathbb{E}[W_t |W_t + (W_1-W_t)] = \mathbb{E}[\sqrt{t} X |\sqrt{t} X + \sqrt{1-t} Y] 
$$
To make things simpler, assume that we condition on $\sqrt{t} X + \sqrt{1-t} Y = z$ for some $z \in \mathbb{R}$.
Define 
$$
a = \mathbb{E}[\sqrt{t} X |\sqrt{t} X + \sqrt{1-t} Y = z]
$$
$$
b = \mathbb{E}[\sqrt{1-t} Y |\sqrt{t} X + \sqrt{1-t} Y = z]
$$
Obviously $a+b = z$.
Besides it is easy to find a value $\alpha$ such that $\sqrt{t}X + \alpha \sqrt{1-t}Y$ is uncorrelated from $\sqrt{t} X + \sqrt{1-t} Y$, and hence independent because $X,Y$ are independent Gaussians (so is any linear combination of those).
A quick computation yields $$\alpha = -\frac{t}{1-t}$$
This yields that 
$$
a + \alpha b = \mathbb{E}[\sqrt{t} X + \alpha \sqrt{1-t} Y |\sqrt{t} X + \sqrt{1-t} Y = z] = \mathbb{E}[\sqrt{t} X + \alpha \sqrt{1-t} Y] = 0
$$
Now, it is just a matter of solving for the system
$$
a+b=z, \quad a + \alpha b = 0
$$
which gives 
$$
a = \frac{\alpha}{\alpha - 1} z = t z
$$
Hence
$$
\mathbb{E}[W_t |W_1 = z] = t z, \quad \mathbb{E}[W_t |W_1] = t W_1
$$
