For a fixed $t>0$, how to compute the conditional distribution of a Brownian motion at time $t$, $B(t)$, given that $B(2t)=x$?
My reasoning is that this is just equivalent to the distribution of $B(t)$ given $B(0)=x$. So I am guessing that $\mathcal{N}(x,t)$ is the answer, but I am not sure. What looks really odd in my reasoning is the reversed time.
Any thoughts on this? Any help is appreciated.
The joint distribution of $\begin{bmatrix} B(s) \\ B(t)\end{bmatrix} $ is a joint Gaussian distribution with mean $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$, and covariance matrix $\begin{bmatrix} s & s \wedge t \\ t \wedge s & t \end{bmatrix}$.
Note that the conditional distribution is still Gaussian with some well known mean and covariance (which you can calculate via Proposition 3.9 in these notes; Eq. 3.10 has a typo it should read $\hat{E}[X|Y=y] = E[X] + \ldots$)