Does this independence property hold?

Let $$x \sim N(\mu_x,\Sigma_x)$$ and $$v \sim N(0,\Sigma_v)$$ be independent multivariate Gaussian random vectors, and let $$y = Ax + v$$ for some square matrix $$A$$ such that $$y \sim N(A\mu_x, A\Sigma_xA^T + \Sigma_v)$$. Now let $$Z$$ be a Bernoulli random variable with parameter $$p_Z$$, where $$Z$$ is marginally independent $$x$$ and marginally independent of $$y$$, such that the joint CDF of $$x$$ and $$Z$$ is equal to the product of the CDF of $$x$$ and the CDF of $$Z$$, and the joint CDF of $$y$$ and $$Z$$ is equal to the product of the CDF of $$y$$ and the CDF of $$Z$$. Let $$F_{Z\mid x,y}(z \mid x,y)$$ be the CDF of $$Z$$ conditioned on $$x$$ and $$y$$, and let $$F_Z(z)$$ be the CDF of $$Z$$. Is it true that, $$F_{Z\mid x,y}(z \mid x,y) = F_Z(z)$$ In the following, it seems to be true as long as $$x$$ is conditionally independent of $$y$$ given $$Z$$: \begin{align} F_{Z\mid x,y}(z \mid x,y) &= \frac{F_{x,y,Z}(x,y,z)}{F_{x,y}(x,y)} \\ &= \frac{F_{x \mid y,Z}(x \mid y,z) \cdot F_{y \mid Z}(y \mid z) \cdot F_Z(z)}{F_{x \mid y}(x \mid y) \cdot F_{y}(y)} \\ &= \frac{F_{x \mid y,Z}(x \mid y,z) \cdot F_{y}(y) \cdot F_Z(z)}{F_{x \mid y}(x \mid y) \cdot F_{y}(y)} \\ &= \frac{F_{x \mid y,Z}(x \mid y,z) \cdot F_Z(z)}{F_{x \mid y}(x \mid y)} \\ \end{align} However, I don’t think I can reduce this any further.

• It is equivalent to prove $Z \perp (X, Y)$ given $Z \perp X$ and $Z \perp Y$. While this is obviously not true for general $X, Y, Z$, it seems hard to construct a counterexample for this specific setup (though I am still inclined it is not true). Commented Feb 10, 2023 at 2:08
• As for your derivation, the last "if" is clearly unwarranted with provided conditions. Basically, to evaluate $F(x, y, z)$, we need to know the joint distribution of $(X, Y, Z)$, however, all the conditions are merely about the distributions of $(X, Y), (Z, X), (Z, Y)$, which are not sufficient to determine the distribution of $(X, Y, Z)$. Commented Feb 10, 2023 at 2:13
• This answer may provide a basis to construct a counterexample for disproving your conjecture. Commented Feb 10, 2023 at 4:50

In abstract, your question is "do $$Z \perp X$$ and $$Z \perp Y$$ imply $$Z \perp (X, Y)$$"? This is in general not true (pairwise independence does not guarantee joint independence). Your question re-examined this relationship after imposing more conditions on the joint distribution of $$(X, Y)$$. For simplicity, consider the univariate case (the logic is that, if this conjecture cannot hold for the univariate case, then it won't hold for multivariate case):
\begin{align} Y = X + V, \quad X \sim N(\mu, \sigma_x^2), V \sim N(0, \sigma_v^2), X \perp V. \tag{1} \end{align} $$(1)$$ implies that the joint distribution of $$(X, Y)$$ is $$N_2((\mu, \mu)', \Sigma)$$, where $$\Sigma = \begin{bmatrix} \sigma_x^2 & \sigma_x^2 \\ \sigma_x^2 & \sigma_y^2 \end{bmatrix}$$ and $$\sigma_y^2 = \sigma_x^2 + \sigma_v^2$$. So the problem reduces to if $$Z$$ is independent of every component random variable of a Gaussian random vector, is $$Z$$ also independent of the random vector itself?
Unfortunately, this is still not true. To disprove, first note that \begin{align} & Z \perp X \iff Z \perp \sigma_x^{-1}(X - \mu), \\ & Z \perp Y \iff Z \perp \sigma_y^{-1}(Y - \mu), \\ & Z \perp (X, Y) \iff Z \perp \Sigma^{-1/2} \left(\begin{bmatrix}X \\ Y \end{bmatrix} - \begin{bmatrix}\mu \\ \mu \end{bmatrix}\right). \end{align} Therefore, for the counterexample, it is sufficient ($$\dagger$$) to construct a random vector $$(X, Y, Z)$$ such that $$(X, Y), (X, Z), (Y, Z)$$ are bivariate $$N_2(0, I_{(2)})$$ random vectors, but $$X, Y, Z$$ are not jointly independent. The clever example provided by Dilip Sarwate in this answer then perfectly fits in.
$$\dagger$$: Whether $$Z$$ is binary is immaterial: to make $$Z$$ binary, you can use indicator functions after the Gaussian $$Z$$ has been constructed.