Brownian Motion conditional distribution Let $\{X(u),u\geq0\}$ be a standard Brownian motion. What is the conditional distribution of $X(t)$ given $\{X(t_{1}),\dots,X(t_{n})\}$, where $0<t_{1}<\cdots<t_{n}<t_{n+1}=t$?
--So far, I have derived the joint pdf of $X(t_{n+1})$ and $X(t_{1}),\dots,X(t_{n})$ using the fact that each increment $X(t_{i})-X(t_{i-1})$ is independent and normally distributed. The pdf if (I think) given by
$$
\begin{align*}
   f(x_{1},x_{2},\dots,x_{n},x_{n+1})&=f_{t_{1}}(x_{1})f_{t_{2}-t_{1}}(x_{2}-x_{1})\cdots f_{t_{n}-t_{n-1}}(x_{n}-x_{n-1})f_{t_{n+1}-t_{n}}(x_{n+1}-x_{n})\\
   &=\frac{\exp\left\{-\frac{1}{2}\left[\frac{x_{1}^{2}}{t_{1}}+\frac{(x_{2}-x_{1})^{2}}{t_{2}-t_{1}}+\cdots+\frac{(x_{n}-x_{n-1})^{2}}{s-t_{n-1}}+\frac{(x_{n+1}-x_{n})^{2}}{t_{n+1}-t_{n}}\right]\right\}}{(2\pi)^{(n+1)/2}[t_{1}(t_{2}-t_{1})\cdots(s-t_{n-1})(t-s)]^{1/2}}.
  \end{align*}
$$
So the $X(t_{i})$s are jointly normal. Now to find the conditional distribution, I believe we need to use multivariate distribution theory. I.e., if $\mathbf{X}$ and $\mathbf{Y}$ have a joint normal distribution, then the conditional distribution of $\mathbf{Y}$ given $\mathbf{X}$ is also normally distributed such that
$$\mathbf{Y}|\mathbf{X}\sim\operatorname{Multivariate\ Normal}(\boldsymbol\mu^{*},\boldsymbol\Sigma),$$
where $\boldsymbol\mu^{*}=\boldsymbol\mu_{y}+\boldsymbol\Sigma_{yx}\boldsymbol\Sigma_{xx}^{-1}(\mathbf{x}-\boldsymbol\mu_{x})$ and $\boldsymbol\Sigma^{*}=\boldsymbol\Sigma_{yy}-\boldsymbol\Sigma_{yx}\boldsymbol\Sigma_{xx}^{-1}\boldsymbol\Sigma_{xy}$. How would I go about finding each of these components for $X(t)|X(t_{1}),\dots,X(t_{n})$? Thank you. 

One knows that the marginal distributions of Brownian motion are normal and that $X(t)-X(t_n)$ is independent of $\sigma(X(s);s\leqslant t_n)$. Hence, the conditional distribution of $X(t)$ conditionally on $\sigma(X(t_k);1\leqslant k\leqslant t_n)$, for every $t_k\leqslant t_n$ (or, conditionally on $X(t_n)$ only) is normal with mean $X(t_n)$ and variance $t-t_n$.
