# Proving Conditional Independence

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 $$U\perp X\mid V$$.

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.
• Many Thanks, Losif, for the reply. First, as I can understand, to show the desired conditional independence, one has to show the following: \begin{equation*}\nonumber E(f(U)g(Z+V)|V)\overset{\text{(?)}}=E(f(U)|V)\,E(g(Z+V)|V) \end{equation*} almost surely (a.s.) for all nonnegative Borel-measurable functions $f,g$. Immediately after the above you write, that is, \begin{equation*} Ef(U)g(Z+V)h(V)\overset{\text{(?)}}=EE(f(U)|V)E(g(Z+V)|V)h(V); \tag{0} \end{equation*} here and in what follows, $f,g,h$ are any nonnegative Borel-measurable functions." – user_akt Feb 24 at 23:37
• I do not understand why is $h(V)$ is required here. Secondly, it would easier for me to understand if there are parenthesis along with the expectation operator in (0). Also, which equation is equation (1), in which I have to assume $f =1$. I shall be grateful, if you could reply. – user_akt Feb 24 at 23:37
• The equivalence of formulas (-1) and (0) follows right by the definition of the conditional expectation; cf. e.g. the 1st display in section "Conditional expectation with respect to a random variable" at en.wikipedia.org/wiki/…; a condition guaranteeing that both integrals in that display exist is missing there (to that end, it was required in my answer that $f,g,h$ be nonnegative). – Iosif Pinelis Feb 25 at 1:05
• Previous comment continued: The roles of $X$, $Y$, $f$, and $g(Y)$ on that Wikipedia page are played, respectively, by $f(U)g(Z+V)$, $V$, $h$, and the right-hand side of formula (-1) in the transition from (-1) to (0) in my answer. I have also added some parentheses and/or brackets after the expectations sgins. Please let me know if something is still unclear. As for the missing tag (1), I guess it was due to my mistaken use of MathJax. Now this tag is restored. – Iosif Pinelis Feb 25 at 1:08