conditional dependence, strictly monotone transformation

Question

Suppose $Y \perp\!\!\!\perp X\mid Z$. Let $g(\cdot)$ be a strictly increasing or decreasing function (i.e. one-to-one, invertible). Then, is the following true?

$Y \perp\!\!\!\perp X\mid Z \iff g(Y) \perp\!\!\!\perp X\mid Z$

Answer 1

Suppose we talk about the mass distribution of discrete random variables.

Since $g$ is a bijection we have:

$$\begin{align} \mathsf P(g(Y) =w\cap X=x\mid Z=z) & = \mathsf P(Y=g^{-1}(w)\cap X=x\mid Z=z) \\[1ex] & = \mathsf P(Y=g^{-1}(w)\mid Z=z)\;\mathsf P(X=x\mid Z=z) \\[1ex] & = \mathsf P(g(Y)=w\mid Z=z)\;\mathsf P(X=x\mid Z=z) \\[4ex] \therefore \big[Y\perp\!\!\!\perp X\mid Z\big] & \iff \big[g(Y)\perp\!\!\!\perp X\mid Z\big] \end{align}$$

Clearly this can be extended to density distributions of continuous RV (or any sigma algebra).