0
$\begingroup$

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 $(1-\kappa)^{-1}$.

$\endgroup$
4
  • 1
    $\begingroup$ If you denote by $\cal F$ the sigma-field generated by the variables you condition on, then the conditional law is Gaussian. The conditional variance (which will determine an upper bound on the conditional density) can be computed and depends on the covariance kernel of your process (you need the inverse of the covariance kernel, viewed as an operator). $\endgroup$ Jun 12, 2015 at 11:08
  • $\begingroup$ Ok thanks for the answer. How can I make sure the covariance is not zero or infinite? Do you have a reference? I would imagine the covariance formula is the analogue of the discrete one, with the inverse of the matrix replaced by the inverse of the kernel. $\endgroup$ Jun 12, 2015 at 14:46
  • $\begingroup$ I wonder if there is not a need for an $\epsilon^{-1}$ in the partial answer I added above $\endgroup$ Jun 12, 2015 at 16:12
  • $\begingroup$ I had the luck to see a related paper on the arxiv yesterday and it seems that conditioning by a continuous Gaussian field is not trivial at all. arxiv.org/abs/1506.04208 I think I will discretize my way around the problem. I'm still interested if someone has a neat answer. $\endgroup$ Jun 17, 2015 at 0:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.