# is P( X | Y, Z ) proportional P(Z | X, Y) P(X | Y)

By the bayes theorem is true the following statement?

$P(S | X ,U) \propto P(X | S, U)P(S|U)$

Thanks

We can explicitly calculate it with a conditional variant of Bayes rule, with $X$ and $Z$ conditional on $Y$, or equivalently, expanding the expression $P(X, Z | Y)$ in two different ways:
$$P(X | Z,Y) P(Z | Y) = P(X, Z | Y) = P(Z | X, Y) P(X | Y)$$
So $P(Z | Y)$ (or its inverse) is the constant of proportionality, but this is different for different distributions.