Basic Gaussian Conditioning Suppose $X_t$ satisfies the SDE $dX_t = a \; dt + dW_t$, where $a$ is a constant and $X_0 = 0$. What is the distribution of $X_3$ given that $X_1 = -1, X_4 = 2$ and $X_5 = 1$? Give the type of distribution and the parameters that describe the conditional distribution of $X_3$ completely.
OK I am a bit embarrassed that I do not remember how to do this. Here is what I have started on: one can see by integrating that $X_t$ is distributed as $N(at, t)$, so each $X_t$ is Gaussian. I know that $X_3$ conditioned on $X_1, X_4, X_5$ is a one dimensional Gaussian, but what is the easiest approach to finding the conditional mean and variance? I remember being able to do this given a covariance matrix, but I suspect that's not really necessary here. Any help would be greatly appreciated.
 A: Given that SDE $\mathrm{d} X_t = a \mathrm{d} t + \mathrm{d} W_t$ with the initial condition $X_0 = 0$, the process $X_t$ is Gaussian, meaning that the joint distribution of $(X_1, X_3, X_4, X_5)$ has the normal distribution with mean vector $(a, 3a, 4a, 5a)$, and the covariance matrix $\Sigma_{t,s} = \min(t,s)$, i.e.:
$$
   \Sigma = 
\begin{pmatrix}  1 & 1 & 1 & 1 \\
 1 & 3 & 3 & 3 \\
 1 & 3 & 4 & 4 \\
 1 & 3 & 4 & 5 \\
\end{pmatrix}
$$ 
The conditional normal of a multi-normal vector is also normal. Using notation's from the wikipedia page:
$$
   \hat{\Sigma}_{11} = \begin{pmatrix} 3 \end{pmatrix} \qquad
   \hat{\Sigma}_{22} = \begin{pmatrix} 1 & 1 & 1\\ 1 & 4 & 4\\ 1 & 4 & 5 \end{pmatrix} \quad
   \hat{\Sigma}_{12} = \begin{pmatrix} 1 & 3 & 3 \end{pmatrix}
$$
Notice that
$$
   \hat{\Sigma}_{22}^{-1} =\frac{1}{3} \left( \begin{array}{rrr}
 4 & -1 & 0 \\
 -1 & 4 & -3 \\
 0 & -3 & 3
 \end{array}\right)
$$
Conditional distribution of $X_3$ given $X_1 = x_1$, $X_4 = x_4$ and $X_5 = x_5$ is normal  with mean
$$ \begin{eqnarray}
    \mathbb{E}(X_3 &|& X_1=x_1, X_4=x_4, X_5=x_5) \\&=& \mathbb{E}(X_3)  + \hat{\Sigma}_{12} \hat{\Sigma}_{22}^{-1}  \begin{pmatrix} x_1 - \mathbb{E}(X_1)  & x_4 - \mathbb{E}(X_4) &
x_5 - \mathbb{E}(X_5)   \end{pmatrix}^t \\
   &=& 3 a + \left( -3 a + \frac{x_1}{3} + \frac{2 x_4}{3} \right) = \frac{x_1 + 2 x_4}{3}
\end{eqnarray}
$$
The conditional variance:
$$
  \mathbb{Var}(X_3 | X_1=x_1, X_4=x_4, X_5=x_5) = \hat{\Sigma}_{11} - \hat{\Sigma}_{12} \hat{\Sigma}_{22}^{-1} \hat{\Sigma}_{12}^t = \frac{2}{3}
$$
Thus the conditional distribution of $X_3$ results to be $\mathcal{N}\left( \frac{x_1 + 2 x_4}{3}, \sqrt{\frac{2}{3}} \right) = \mathcal{N}\left(1,\sqrt{\frac{2}{3}}\right)$. 
As you see, the conditional distribution of $X_3$ only depends on the conditional value of $X_4$, but not $X_5$. 
Indeed, the conditional Brownian motion $X_t$ for $1 \leqslant X_t \leqslant 4$, conditioned on $X_1 = x_1$ and $X_4 = x_4$ is known as the Brownian bridge process.

My original incorrect response:

Remember that $X_t$ is an independent increment process, such that $X_t - X_s \sim X_{t-s}$, that is $X_3$, given $X_1 = -1$ is equal in distribution to $X_3 = X_1 + (X_3-X_1) \sim -1 + X_2$, that is it is normal with variance $t$ and mean $-1 + 2 a$.
