Brownian Bridge under observational error Suppose that $Z_t$ follows a simple discrete random walk $Z_t=Z_{t-1}+e_t$ , where $e_t$ are a bunch of uncorrelated normal variables with arbitrary variance sigma^2, and that there are observations of the series at t=a and t=b, with both observations having normal uncorrelated observational error with variance $O_a$ and $O_b$.
How can I find distribution for intermediate values between a and b?
 A: if you consider a Gaussian vector $V=(X,Y) \in \mathbb{R}^{d=m+n}$, you know how to find the conditional distribution of $X$ knowing the value of $Y=y$, right ? This is exactly the same thing here.
For example, let us suppose that $a=0, b=N+1$: 


*

*you have a noisy observation $Y=(y_1, y_2)=(O_a, O_b)$ with know covariance matrix $\Sigma_Y$

*the data you are looking for, $X=(z_1, \ldots, z_N) \in \mathbb{R}^N$, have a known covariance matrix $\Sigma_X$

*the covariance matrix $E[X Y^t] = \Sigma_{X,Y}$ is also known.


A quick way to find the conditional distribution of $X$ knowing $Y$ is to write
$$X = AU + BV$$
$$Y=CU$$
where $U,V$ are independent standard Gaussian random variable of size $2$ and $N$ respectively, while $A \in M_{N,2}(\mathbb{R})$ and $B \in M_{N,N}(\mathbb{R})$ and $C \in M_{2,2}(\mathbb{R})$. Because


*

*$CC^t = \Sigma_Y$ gives you $C=\Sigma_y^{\frac{1}{2}}$,

*$AC^t = \Sigma_{X,Y}$ then gives you $A=\Sigma_{X,Y}\Sigma_y^{-\frac{1}{2}}$,

*$AA^t + BB^t = \Sigma_{X}$ then gives you $B=(\Sigma_{X}-\Sigma_{X,Y}\Sigma_y^{-1}\Sigma_{X,Y})^{\frac{1}{2}}$,


the $3$ matrices are easily computable, and $C$ is invertible in the case you are considering. This shows that if you know that $Y=y$, the conditional law of $X | Y=y$ is given by
$$X = AC^{-1}y + BV,$$
which is a Gaussian vector with mean $AC^{-1}y = \Sigma_{X,Y}\Sigma_y^{-1}y$ and covariance $BB^t = \Sigma_{X}-\Sigma_{X,Y}\Sigma_y^{-1}\Sigma_{X,Y}$
