Each of the scalar random variables, $ Y $, $ X $, $ U $, and $ V $, is continuous and possibly has $ \mathbb{R} $ as its support. The random variable, $Z$, could be vector valued, but continuous. 


I have the following 
\begin{align}\nonumber
Y = \beta X + U\\\nonumber
X = \pi Z + V,
\end{align}
where $ Z \perp (U,V) $ ( $ \perp $ denotes independence). But $ U\not\perp X$. In the second equation, given $ Z $, there is one-to-one mapping between $ X $ and $ V $. 


Given the set-up, I need to know whether the following conditional independence holds: $ U\perp X\mid V $. 


My question is: since, given $ Z $, there is one-to-one mapping between $ X $ and $ V $, is it true that $ \sigma(X, Z) = \sigma(V, Z) $, where $ \sigma(X, Z) $ is the sigma algebra generated by $ (X,Z) $ and $ \sigma(V, Z) $ that generated by $ (V,Z) $?


If $ \sigma(X, Z) = \sigma(V, Z) $, then I could write $ U\mid X,Z \sim U\mid V,Z $. But since $ Z \perp (U,V) $, I could write $ U\mid V,Z \sim U\mid V $. From this could we deduce that $ U\perp X\mid V $. 



Proving Conditional Independence
The answer is yes. Indeed, $Z\perp(U,V)$ implies $\pi Z\perp(U,V)$. So, without loss of generality $\pi Z=Z$ and $X=Z+V$ (the condition $Y=\beta X+U$ is irrelevant and not needed here). So, the desired conditional independence $(U\perp X)|V$ can be rewritten as $(U\perp Z+V)|V$, which means that 
\begin{equation}
 E\Big(\big(f(U)g(Z+V)\big)|V\Big)\overset{\text{(?)}}=E(f(U)|V)\times E\big(g(Z+V)|V\big) \tag{-1}
\end{equation}
almost surely (a.s.) for all nonnegative Borel-measurable functions $f,g$, that is, 
\begin{equation\*}
 E[f(U)g(Z+V)h(V)]\overset{\text{(?)}}=E[E\big(f(U)|V\big)\,E\big(g(Z+V)|V\big)\,h(V)]; \tag{0}
\end{equation\*} 
here and in what follows, $f,g,h$ are any nonnegative Borel-measurable functions. 


By the condition $Z\perp(U,V)$, 
\begin{equation}
 \begin{aligned}
 E[f(U)g(Z+V)h(V)]&=\iint P(U\in du,V\in dv)f(u)h(v)\tilde g(v) \\
 &=E[f(U)h(V)\tilde g(V)], 
\end{aligned}
\tag{1}
\end{equation} 
where 
\begin{equation\*}
\tilde g(v):=\int P(Z\in dz)Eg(z+v)=Eg(Z+v). 
\end{equation\*}
In particular, (1) with $f=1$ implies that 
\begin{equation\*}
 E\big(g(Z+V)|V\big)=E\tilde g(V) 
\end{equation\*}
a.s. 
So, by the definition of the conditional expectation, the right-hand side of (0) equals the last expression in (1). Thus, (0) is proved. 




---


If the random variables (r.v.'s) $U,V,Z$ are discrete, then the proof can be written in a simpler and more transparent way, as follows. Let us use notations such as $p\_V$ to denote the probability mass function (pmf) of the r.v. $V$; $p\_{U,V}$ to denote the joint pmf of the r.v.'s $U,V$; $p\_{U,X|V}$ to denote the joint conditional pmf of the r.v.'s $U,X$ given $V$; etc. Then the desired conditional independence $(U\perp X)|V$ can be rewritten as $p\_{U,X|V}=p\_{U|V}p\_{X|V}$ or, equivalently, as 
\begin{equation\*}
 \frac{p\_{U,X,V}}{p\_V}\overset{\text{(?)}}=\frac{p\_{U,V}}{p\_V}\frac{p\_{X,V}}{p\_V}. \tag{0'}
\end{equation\*}
Since $X=Z+V$ and $Z\perp(U,V)$, we have $p\_{U,X,V}(u,x,v)=p\_{U,V}(u,v)p\_Z(x-v)$ and 
$p\_{X,V}(x,v)=p\_V(v)p\_Z(x-v)$ for all $u,x,v$, whence (0') follows.


