Conditional independence of random variables under isomorphism of measurable spaces

Question

I am looking for a proof of what should be a trivial fact:

Let $(\Omega, \mathcal F, P)$ be a probability space and $\tilde x, \tilde y, \tilde z$ be random variables on the measurable space $(X, \mathcal A)$. Let another measurable space $(Y, \mathcal B)$. Suppose that $(X, \mathcal A)$ and $(Y, \mathcal B)$ are isomorphic as measurable spaces, and let $X \xrightarrow{\pi}Y$ be the isomorphism with $E=\pi^{-1}$. Define the random variables $(x,y,z)=\pi(\tilde x, \tilde y, \tilde z)$ on $(Y, \mathcal B)$. If $\tilde x, \tilde y$ are conditionally independent given $\tilde z$, then $x, y$ are conditionally independent given $z$.

My attempt so far, given $C,D \in \mathcal B$:

$$P(x\in C, y\in D \mid z) = \mathbb E [1_C(x)1_D(y) \mid z]= \mathbb E [1_{\pi^{-1}C}(\tilde x)1_{\pi^{-1}D}(\tilde y) \mid \pi \tilde z] =\mathbb E [ \mathbb E [1_{\pi^{-1}C}(\tilde x)1_{\pi^{-1}D}(\tilde y) \mid \tilde z]\mid \pi \tilde z]=\mathbb E [ \mathbb E [1_{\pi^{-1}C}(\tilde x)\mid \tilde z]\mathbb E [1_{\pi^{-1}D}(\tilde y) \mid \tilde z]\mid \pi \tilde z]=\mathbb E [ \mathbb E [1_{C}(x)\mid E z]\mathbb E [1_{D}( y) \mid E z]\mid z]=\mathbb E [1_{C}(x)\mid E z]\mathbb E [1_{D}( y) \mid E z]=P(x\in C\mid \tilde z) P( y\in D \mid \tilde z)$$

And I cannot replace $\tilde z$ by $z$ in the latter.

Any help or pointers are welcome.

Answer 1

Conditioning with respect to a random variable is defined in terms of conditioning with respect to the $\sigma$-algebra it generates. What you want to show then is that under these conditions the $\sigma$-algebra generated by $z$ and $\tilde{z}$ are the same. For this you will need that $\pi$ is measurable-- i.e. $\pi$ is a Borel-isomorphism. https://en.wikipedia.org/wiki/Borel_isomorphism

Answer 2

Here is another perspective. The isomorphism $\pi :(\mathcal{X},\mathscr{A})\to (\mathcal{Y},\mathscr{B})$ is usually defined s.t. $\pi(\mathscr{A})=\mathscr{B},\pi^{-1}(\mathscr{B})=\mathscr{A}$. Note that for all $B \in \mathscr{B}$

$$Z^{-1}(B)=(\pi \circ \tilde{Z})^{-1}(B)=\tilde{Z}^{-1}(\underbrace{\pi^{-1}(B)}_{\in \mathscr{A}})$$

Recall: equalities will now be in the a.s. sense. So for $f$ Borel and bounded we get $$\begin{aligned}E[\mathbf{1}_{Z^{-1}(B)}f(\tilde{X},\tilde{Y})]&=E[\mathbf{1}_{\tilde{Z}^{-1}(\pi^{-1}(B))}f(\tilde{X},\tilde{Y})]=\\ &=E[\mathbf{1}_{\tilde{Z}^{-1}(\pi^{-1}(B))}E[f(\tilde{X},\tilde{Y})|\tilde{Z}]]=\\ &=E[\mathbf{1}_{Z^{-1}(B)}E[f(\tilde{X},\tilde{Y})|\tilde{Z}]]\end{aligned}$$

We also know that for some Borel $\phi:\mathcal{X}\to \mathbb{R}$ we have $$E[f(\tilde{X},\tilde{Y})|\tilde{Z}]=\phi(\tilde{Z})$$ and for any $E$ Borel, there exists $F\in \mathscr{B}$ s.t. $\phi^{-1}(E)=\pi^{-1}(F)$ because $\phi^{-1}(E) \in \mathscr{A}$ so $$\begin{aligned}(\phi \circ \tilde{Z})^{-1}(E)&=\tilde{Z}^{-1}(\phi^{-1}(E))=\\ &=\tilde{Z}^{-1}(\pi^{-1}(F))=\\ &=(\pi\circ \tilde{Z})^{-1}(F)=\\ &=Z^{-1}(F)\end{aligned}$$ so $E[f(\tilde{X},\tilde{Y})|\tilde{Z}]$ is $\sigma(Z)$-measurable and since $B$ was arbitrary we get $$E[f(\tilde{X},\tilde{Y})|\tilde{Z}]=E[f(\tilde{X},\tilde{Y})|Z]$$ So now we conclude: for any $C,D\in \mathscr{B}$ $$\begin{aligned}P(X \in C,Y \in D|Z)&=P(X^{-1}(C)\cap Y^{-1} (D)|Z)=\\ &=P(\tilde{X}^{-1}(\pi^{-1}(C))\cap \tilde{Y}^{-1} (\pi^{-1}(D))|Z)=\\ &=P(\tilde{X}^{-1}(\pi^{-1}(C))\cap \tilde{Y}^{-1} (\pi^{-1}(D))|\tilde{Z})=\\ &=P(\tilde{X}^{-1}(\pi^{-1}(C))|\tilde{Z})P(\tilde{Y}^{-1} (\pi^{-1}(D))|\tilde{Z})=\\ &=P(\tilde{X}^{-1}(\pi^{-1}(C))|Z)P(\tilde{Y}^{-1} (\pi^{-1}(D))|Z)=\\ &=P(X^{-1}(C)|Z)P(Y^{-1}(D)|Z)=\\ &=P(X\in C|Z)P(Y \in D|Z) \end{aligned}$$