I know that if $(X,Y)$ is a Gaussian vector, then $(X|Y=y)$ is a Gaussian vector which covariance matrix is explicit in function of the covariance matrix of $(X,Y)$, and does not depend on $y$.
What happens if $Y$ is a Gaussian field? For instance, let $(X(t),t\in [0,1])$ be a Gaussian process. Is the variable $(X(1)|X(t),t\in [0,\epsilon])$ well defined? Gaussian? My real question is: what is the conditional density? Or what is an upper bound on the conditional density?
$\epsilon$ can be chosen as small as we want, and we assume that $X(0)$ and $X(1)$ don't have correlation $1$.
EDIT: I think I figured out the answer thanks to Ofer's tip, but I still have holes to fill. If anyone can help that would be appreciated.
Call $\sigma_\epsilon(x,y)=E X(x)X(y)$ for $x,y\in [0,\epsilon]$, and also denote by $\sigma_\epsilon$ the functional operator acting on $L^2([0,\epsilon])$. It is a compact operator by Ascoli theorem. I would like to prove that, given the function $f(x)=E X(1)X(x)$ the mapping $$ H:\epsilon\mapsto \int_0^\varepsilon f(x)(\sigma_\epsilon^{-1}\cdot (f\cdot 1_{[0,\epsilon]}))(x)dx$$ is bounded by $\kappa<1$ in the neighbourhood of $0$ using the fact that $f(0)=E X(0)X(1)<1$. Then the conditional density I'm talking about would be bounded by $(1-\kappa)^{-1}$.