Is it true that $P(z\mid x,y)=P(z\mid x)P(z\mid y)$ if x and y are independent?

Question

Is it true that $P(z\mid x,y)=P(z\mid x)P(z\mid y)$ if $x$ and $y$ are independent? I'm struggling to prove it(or disprove it).

Answer 1

$$P(z|x,y)=\frac{P(z,x,y)}{P(x,y)}=\frac{P(z,x,y)}{P(x)P(y)}$$

while

$$P(z|x)P(z|y)=\frac{P(x,z)}{P(x)}\frac{P(y,z)}{P(y)}=\frac{P(x,z)P(y,z)}{P(x)P(y)}$$

we can see that $P(z|x,y)= P(z|x)P(z|y)$ is only possible if $P(x,y,z)=P(x,z)P(y,z)$. It should be fairly easy to find examples where this is not true. For example, as the comments already mention, if $x,y,z$ are independent, then $P(x,z)P(y,z)=P(x)P(y)P(z)^2$, while $P(x,y,z)=P(x)P(y)P(z)$, and if $P(x), P(y), P(z)\neq 0$, these two numbers are not the same.

For example, if $x,y,z$ are the results of three independent coin flips, then $$P(z=H|x=H,y=H)=\frac12$$ and $$P(z=H|x=H)P(z=H|y=H)=\frac12\cdot\frac12=\frac14\neq\frac12$$