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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.

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    $\begingroup$ 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). $\endgroup$
    – Zhanxiong
    Commented Feb 10, 2023 at 2:08
  • $\begingroup$ 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 $(X, Y, Z)$. $\endgroup$
    – Zhanxiong
    Commented Feb 10, 2023 at 2:13
  • $\begingroup$ This answer may provide a basis to construct a counterexample for disproving your conjecture. $\endgroup$
    – Zhanxiong
    Commented Feb 10, 2023 at 4:50

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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.

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