Given a Brownian motion $W$, and $k \in (a,b)$, I'm trying to find the distribution of $W(k)$ in terms of $W(b)$, $W(a)$, and $k$ I'm trying to perform this "interpolation" because I ultimately am trying to write a small library to simulate stochastic processes. I realized I might need to figure out what is the distribution of $W(k)$ when I already know the values of $W(a)$ and $W(b)$ during a sample run.  
Here's what I got so far
$$
W(k) = \frac{1}{2} (W(k)-W(a)) + \frac{1}{2}W(a) + \frac{1}{2}W(b) 
- \frac{1}{2}(W(b)-W(k))
$$
I know the first and last summands are independent normal r.v.s
$$
W(k) = \frac{1}{2} (W(b)+W(a))+ \frac{1}{2} ( N(0,k-a)-N(0,b-k)) \\
$$
which results in 
$$ 
W(k) = \frac{W(b)+W(a)}{2} + \frac{1}{2} N(0,b-a)
$$
The distribution of $W(k)$ doesn't seem to depend on $k$ at all, which is very counter-intuitive. For example, if during a sample I see $W(0)=0$ and $W(1)=1$, I refuse to believe that $W(0.1)$ has the same distribution as$W(0.99)$--namely, $N(0.5,0.25)$. I must be doing something wrong, but can't see what. 
The formal manipulations seem correct. Am I committing some assumption of independence that isn't true. 
 A: To see what's happening, set $X = \frac{W(b) + W(a)}{2}$ and $Y = W(k) -X$.  I agree that your computation shows $Y \sim \frac{1}{2} N(0,b-a)$, regardless of the value of $k$.  And the distribution of $X$ certainly doesn't depend on $k$.
But this does not imply that the distribution of $W(k) = X+Y$ is the same for all $k$, because the covariance of $X$ and $Y$ does depend on $k$.  (I will leave you to compute it.)
Put another way, to conclude that the distribution of $W(k)$ was the same for all $k$, you'd need to show that the joint distribution of $(X,Y)$ was the same for all $k$.  But you've only shown that the marginal distributions are the same for all $k$.
A: Turns out I was making my life more difficult by trying to be more clever than I actually am. 
Using the conditional distribution of a multivariate normal random variable (https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions) makes the calculation simple if a bit long. 
Using Wikipedia's notation (mostly):
$$
x = \begin{bmatrix} W(k) \\ W(a) \\ W(b) \end{bmatrix}
$$
with 
$
x_1 = \begin{bmatrix} W(K) \end{bmatrix}
$
and
$ x_2 = \begin{bmatrix} W(a) \\ W(b) \end{bmatrix}$
and
$$
\Sigma = \begin{bmatrix} k & a & k \\ a & a & a \\ k & a & b \end{bmatrix}
= \begin{bmatrix}
\begin{bmatrix} a \end{bmatrix} &
\begin{bmatrix} a & k \end{bmatrix} \\
\begin{bmatrix} a \\ k\end{bmatrix} &
\begin{bmatrix} a & a \\ a & b \end{bmatrix} 
\end{bmatrix}
= \begin{bmatrix}
\Sigma_{11} &
\Sigma_{12} \\
\Sigma_{21} &
\Sigma_{22}
\end{bmatrix}
$$
So
$ x \sim N(0,\Sigma) $, since the mean of any $W(.)$ is 0. 
So the distribution of of this multivariate random variable when we're given 
$x_2 = u  = \begin{bmatrix} u_a \\ u_b \end{bmatrix}$ will be a normal random variable (usually multivariate normal, but here it's univariate since $x_1$ has dimension 1).
The mean of this new random variable is
$$
\mathbb{E}(x_1) + \Sigma_{12} \Sigma_{22}^{-1} (u-\mathbb{E}(x_2))
= \Sigma_{12} \Sigma_{22}^{-1} u
= \frac{b-k}{b-a}u_a + \frac{k-a}{b-a}u_b
$$ 
and the variance is
$$
\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21} = \frac{(b-k)(k-a)}{b-a}
$$
This
$$
W(k) \sim N\left(\frac{b-k}{b-a}u_a + \frac{k-a}{b-a}u_b
,\frac{(b-k)(k-a)}{b-a}\right)
$$
Which "feels" like it should be true. 
EDIT:
I double checked the process by going through the same calculations but with $a<b<k$, and got the right result $W(k) \sim N(u_b,k-b)$.
