I'm trying to calculate the conditionals within a Bayesian network as shown in the picture, e.g. $P(X_t|X_{t-1}, S)$, which resorts to the calculation of the marginals $P(X_{t-1}, S)$ and $P(X_t, X_{t-1}, S)$ because $$P(X_{t-1}|X_t, S)=\frac{P(X_t, X_{t-1},S)}{P(X_{t-1},S)}$$
In the Bayesian network, the joint probability can be written as $$P(X_{t-2}, X_{t-1}, X_t, X_{t+1}, S) = P(X_{t-2})P(X_{t-1}|X_{t-2})P(X_t|X_{t-1},)P(X_{t+1}|X_t) P(S|X_{t-2},X_{t-1},X_t)$$ Therefore the marginals can be calculated by integrating the unrelated variables: $$P(X_t, X_{t-1},S) = \int{P(X_{t-2}, X_{t-1}, X_t, X_{t+1}, S)} dX_{t-2}dX_{t+1} = \int{\int{P(X_{t-2})P(X_{t-1}|X_{t-2})P(X_t|X_{t-1},)P(X_{t+1}|X_t) P(S|X_{t-2},X_{t-1},X_t)} dX_{t-2}} dX_{t+1} =\int{P(X_{t-2})P(X_{t-1}|X_{t-2})P(X_t|X_{t-1},)P(S|X_{t-2},X_{t-1},X_t)} dX_{t-2} $$
$$P(X_{t-1},S) = \int{P(X_{t-1}, X_t, S)} dX_t = \int{\int{P(X_{t-2})P(X_{t-1}|X_{t-2})P(X_t|X_{t-1},)P(S|X_{t-2},X_{t-1},X_t)} dX_{t-2}}dX_t $$ such that $$P(X_{t-1}|X_t, S)=\frac{P(X_t, X_{t-1},S)}{P(X_{t-1},S)}=\frac{\int{P(X_{t-2})P(X_{t-1}|X_{t-2})P(X_t|X_{t-1},)P(S|X_{t-2},X_{t-1},X_t)} dX_{t-2} }{ \int{\int{P(X_{t-2})P(X_{t-1}|X_{t-2})P(X_t|X_{t-1},)P(S|X_{t-2},X_{t-1},X_t)} dX_{t-2}}dX_t}$$
My question is: such expressions are too heavy and not very useful for further analysis. Is there any way to simplify the integrals in the marginals in this non-parametric setting? Otherwise, is there any means if $P(S|X_{t-2},X_{t-1},X_t)$ can be written explicitly?
Great thanks!!!
I would recommend looking into d-separation and directed acyclic graphs (DAGs). For example, number 7 on page 2 here and the examples here. Since S has an edge coming to it from those three nodes, it is not conditionally independent of any of them given any of the others.
However, $x_{t+1}$ is d-separated from the rest and thus conditionally independent given $x_t$ from the other three nodes.