I am confused about the following identities of probability densities $p$ which lead to an apparent contradiction. Assume there are two random variables $X$ and $W$ and $Y=f(X,W)$.
$$p(Y|X) = \int p(Y|X,W)p(W|X)dW$$
$p(Y|X,W)$ is the conditional density, but does it have positive density anywhere? If not, what happens to the integral if we multiply $p(W|X)$ by zero?
I switch to lower case letters as arguments of the density functions, since upper case letters are usually reserved, in my experience, for the stochastic variables.
As discussed in the comments if Y is a deterministic function of X and W than:
$p(y|x,w)=\delta(y-f(x,w))$
Therefore the integral becomes:
$p(y|x)=\int dw \delta(y-f(x,w)) p(w|x)$
Calling $w_i(x,y)$ the $w$ s.t. $f(x,w_i)=y$ this gets expanded as:
$p(y|x)=\sum_i \frac{p(w_i|x)}{|(\partial_w f)(x,w_i)|}$
I do not know if this can be made more explicit without fixing $f$. I stress the dependence of the $w_i$ on $x,y$ even if to lighten the notation I did not indicate it...