Joint probability of two variables X and Y that is conditioned on a third variable Z, which is independent from X and independent from Y

Question

Assume that we have the random variables X, Y and Z.

Suppose that X and Z are independent. This means that $$P(X=x, Z=z) = P(X=x) \cdot P(Z=z).$$

Also, suppose that Y and Z are independent, so $$P(Y=y, Z=z) = P(Y=y) \cdot P(Z=z).$$

With this information, can we say that $$P(X=x, Y=y) = P(X=x, Y=y | Z=z)?$$

$$\rule{18cm}{0.1cm}$$

My attempt

If we start from the second part of the equation we have $$P(X=x, Y=y | Z=z) = P(X=x | Z=z) \cdot P(Y=y | X= x, Z=z)$$

Now, $P(X=x | Z=z) = P(X=x)$, since X and Z are independent.

Likewise, $P(Y=y | X=x, Z=z) = P(Y=y | X=x)$, since Y and Z are independent (this is the step I am not sure about).

We can conclude that

$P(X=x, Y=y | Z=z) = P(X=x) \cdot P(Y=y | X=x) = P(X=x, Y=y)$

Answer 1

As is well known there exist events $A,B,C$ such that any two of them are independent but $P(A\cap B\cap C) \neq P(A)P(B)P(C)$. Take $X=I_A,Y=I_B,Z=I_C$ and $x=y=z=1$ to get counterexample.

Answer 2

Think of two independent coin tosses where $X,Y$ both take values in $\{H,T\}$.

Further let $Z$ take value $1$ if the tosses give the same result and takes value $0$ otherwise.

Then $X,Z$ are independent and also $Y,Z$ are independent.

However $P(X=H\wedge Y=T\mid Z=1)=0\neq P(X=H)P(Y=T)$.