For the graphical model $X\rightarrow Y\leftarrow Z$, I am given the distributions $p(y\mid x)$ and $p(y\mid z)$ (and also maybe $p(x)$ and $p(z)$ if needed). Is there any possible way for me to derive (or at least approximate) $p(y\mid x, z)$?
I was thinking of just multiplying the two maybe $p(y\mid x, z) \propto p(y\mid x)p(y\mid z)$ ? This was only from intuition and I have no solid reasoning why this should be the case.
If you knew that $X,Z$ are conditionally independent given $Y$ then you can compute $p(y|x,z)$. First notice $p(y)$ can be determined: $$p(y)=\int{p(y|x)p(x)}dx$$ Now using Baye's Theorem along with conditional independence, $$p(y|x,z)=\frac{p(x,z|y)p(y)}{\int p(x,z|y)p(y)dy}=\frac{p(x|y)p(z|y)p(y)}{\int p(x|y)p(z|y)p(y)dy}$$ Since we have $$p(x|y)=\frac{p(y|x)p(x)}{p(y)}$$ and also $$p(z|y)=\frac{p(y|z)p(z)}{p(y)}$$ we can say that $$p(y|x,z)=\frac{p(y|x)p(y|z)p(x)p(z)}{p(y)\int\Big({\frac{p(y|x)p(y|z)p(x)p(z)}{p(y)}}\Big)dy}$$ If you didn't know $X,Z$ are conditionally independent given $Y$ then you would have to know some information about the density $p(x,z|y)$ in order to compute $p(y|x,z)$.