This post is a continuation of my previous one, but can be read as being stand-alone. I am trying to understand the concept of conditional independence of two random variables given a third one.

Following is the definition that I think is the right one:

Two random variables $X$ and $Y$ are conditionally independent given a random variable $Z$ if they are independent given $\sigma(Z)$: the $\sigma$-algebra generated by $Z$.

Two events $R$ and $B$ are conditionally independent given a $\sigma$-algebra $\Sigma$ if $$\Pr(R \cap B \mid \Sigma) = \Pr(R \mid \Sigma)\Pr(B \mid \Sigma)\ a.s.$$ where $\Pr(A \mid \Sigma)$ denotes the conditional expectation of the indicator function of the event $A$, given the sigma algebra $\Sigma$.

Two random variables $X$ and $Y$ are conditionally independent given a $\sigma$-algebra $\Sigma$ if the above equation holds for all $R$ in $\sigma(X)$ and $B$ in $\sigma(Y)$.

My questions are

  1. Is the definition equivalent to that $X$ and $Y$ are conditional independent given $Z$ iff $E(X \times Y \mid Z) = E(X\mid Z) \times E(Y \mid Z)$? If yes, can this be proved from the definition, by that any measurable function (here random variable) can be pointwise approximated by a sequence of simple functions (a simple function is the finite linear combination of indicator functions), just as used in definition of Lebesgue integral, and that conditional expectation and taking pointwise limit can be exchangeable?
  2. Since both $E(X\mid Z)$ and $E(Y \mid Z)$ are random variables, is conditional independence between $X$ and $Y$ given $Z$ equivalent to that $E(X\mid Z)$ and $E(Y \mid Z)$ are independent random variables w.r.t. the underlying probability measure $P$ of the underlying sample space $(\Omega, \mathcal{F}, P)$? Or are the two related somehow?

Thanks and regards!


1 Answer 1


For (1), the forward direction is true and can be proved as you suggest, provided $X,Y$ are integrable. The converse is false even for ordinary independence (where $\Sigma$ is the trivial $\sigma$-field $\{\Omega, \emptyset\}$ or $Z = c$); we have $E[XY] = E[X] E[Y]$ iff $X,Y$ are uncorrelated, which is weaker than independence. (Wikipedia gives as an example $X \sim U(-1,1), Y=X^2$).

What is true is that $X,Y$ are conditionally independent given $\Sigma$ (where $\Sigma = \sigma(Z)$ is a special case) iff $$E[f(X) g(Y) | \Sigma] = E[f(X) | \Sigma] E[g(Y)|\Sigma]$$ for all $f,g$ in some suitably large class of measurable functions from $\mathbb{R}$ to $\mathbb{R}$. For instance, one could take:

  • all bounded measurable functions
  • indicator functions of measurable sets
  • bounded continuous functions
  • smooth and compactly supported functions
  • indicators of open/closed/half-open intervals or half-lines.

For (2), this is false. You can get counterexamples to both directions by taking $X,Y$ to be not (unconditionally) independent and considering the extreme cases where $\Sigma$ is trivial or $\Sigma = \mathcal{F}$.

  • $\begingroup$ I tried to prove the forward direction 1) for indicator functions, that is $E(\mathbb{I}_{A}\mathbb{I}_{D} \mathbb{I}_{G})=E(E(\mathbb{I}_{A}|\mathbb{G}) E(\mathbb{I}_{D}|\mathbb{G})\mathbb{I}_{G})$ for all $G$ in the sigma-algebra $\mathbb{G}$, with $X=\mathbb{I}_{A}, Y=\mathbb{I}_{D}$ for measurable sets $A$ and $D$. But I couldn't do it, how would one prove this? $\endgroup$
    – Joogs
    Jan 10, 2018 at 10:30

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