Conditional expectation of Wiener process I want to calculate $E(W_t | W_1)$, $E(W_t^2 | W_1)$ and $E(W_t^2 | W_1, W_2)$, where $(W_t)_{t\geq0}$ is a Wiener process. 
For the first one I used the conditional distribution formula for the multivariate normal distribution and I got: 
$E(W_t | W_1)=EW_t+\sum_{12}\sum_{22}^{-1}(W_1-EW_1)=0+  (1\wedge t)\cdot 1\cdot(W_1-0)=(1 \wedge t) W_1$.
For the second I did:
$E(W_t^2 | W_1)=VAR(W_t |W_1)+E^2(W_t|W_1)=t-(t\wedge 1)^2+ ((1\wedge t )W_1)^2$
How can I do the third one?
 A: You can extend your first two answers readily to the third. Consider that for $t\in[0,1]$ we know that $\mathbb{E}(W_t|W_1) = t\, W_1$ but this is really $\mathbb{E}(W_t|W_1,W_0) = t\,W_1 + (1-t)W_0$ just that $W_0=0$. In general we have
$$\mathbb{E}(W_t|W_r,W_s) = \frac{t-r}{s-r} W_s + \frac{s-t}{s-r} W_r \qquad\text{for }\,\, r\leq t\leq s$$
This is the general result for a Brownian bridge. We can combine the $\min$ and $\max$ functions ($\wedge$ and $\vee$ respectively) to produce an arbitrary continuous piecewise linear function. Specifically in our case we get
$$\mathbb{E}(W_t|W_2,W_1,W_0=0) = (t\wedge1)W_1 + (0\vee(1\wedge t))(W_2-W_1)$$
For the Brownian bridge we also have that
$$\mathrm{Var}(W_t|W_r,W_s) = \frac{(s-t)(t-r)}{s-r} \qquad\text{for }\,\, r\leq t\leq s$$
Thus we have
$$\begin{align}\mathrm{Var}(W_t|W_1,W_0) &= t(1-t) \qquad&\text{for }\,\, 0 \leq t < 1\\
\mathrm{Var}(W_t|W_2,W_1)&= (2-t)(t-1) \qquad&\text{for }\,\, 1 \leq t < 2 \end{align}$$
These are parabolas, which are negative outside the intervals $[0,1]$ and $[1,2]$ respectively thus we can combine them easily with the $\max$ function as well as accounting for the linearly increasing variance for $t>2$ giving
$$\mathrm{Var}(W_t|W_2,W_1,W_0=0) = \big(t(1-t)\big)\vee\big((2-t)(t-1)\big)\vee(t-2)$$
Now, exactly as you did for your second answer, we combine the variance and square of the expectation through
$$\mathbb{E}(W_t^2|W_2,W_1,W_0=0) = \mathrm{Var}(W_t|W_2,W_1,W_0=0) + \mathbb{E}(W_t|W_2,W_1,W_0=0)^2,$$
to get the final answer. (I haven't actually written out in full but you should be able to see what it looks like and if it simplifies.)
