# If $A,B,C$ are events, can we show that $A\perp\!\!\!\perp_CB$ if and only if $A\perp\!\!\!\perp_{\sigma(C)}B$?

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space and $$A,B,C\in\mathcal A$$. Write $$A\perp\!\!\!\perp_CB:\Leftrightarrow\operatorname P\left[A\cap B\mid C\right]=\operatorname P\left[A\mid C\right]\operatorname P\left[B\mid C\right]\tag1$$ and $$A\perp\!\!\!\perp_{\sigma(C)}B:\Leftrightarrow\operatorname P\left[A\cap B\mid\sigma(C)\right]=\operatorname P\left[A\mid\sigma(C)\right]\operatorname P\left[B\mid\sigma(C)\right]\text{ almost surely}.\tag2$$ The crucial difference between $$(1)$$ and $$(2)$$ is that $$\operatorname P\left[A\cap B\mid C\right]$$ in $$(1)$$ is define in the elementary sense, i.e. $$\operatorname P\left[A\cap B\mid C\right]=\begin{cases}\displaystyle\frac{\operatorname P\left[A\cap B\cap C\right]}{\operatorname P\left[C\right]}&\text{, if }\operatorname P\left[C\right]>0\\0&\text{, otherwise}\end{cases},$$ while $$\operatorname P\left[A\cap B\mid\sigma(C)\right]$$ in $$(2)$$ is defined in terms of the conditional expectation.

Are we able to show that $$A\perp\!\!\!\perp_CB$$ if and only if $$A\perp\!\!\!\perp_{\sigma(C)}B$$?

For example, in the "$$\Rightarrow$$" direction, we would need to show that $$\operatorname P\left[A\cap B\cap\tilde C\right]=\operatorname E\left[1_{\tilde C}\operatorname P\left[A\mid\sigma(C)\right]\operatorname P\left[B\mid\sigma(C)\right]\right]\tag3$$ for all $$\tilde C\in\sigma(C)=\left\{\emptyset,C,C^c,\Omega\right\}$$. This is trivial for $$\tilde C=\emptyset$$, but for $$\tilde C=\Omega$$ this is not clear to me.

This is not true. Consider uniform distribution on $$\Omega =\{1,2,3\}$$. If $$A=\{1,2\}, B=\{1,3\}$$ and $$C=\{1\}$$. Then the $$A$$ and $$B$$ are conditionally independent given $$C$$ but the $$P(A\cap B \cap C^{c})=0$$ and you can see from this that $$A$$ and $$B$$ are not conditionally independent given $$C^{c}$$.
• But it should follow from $(1)$ that $(2)$ holds on $C$, right? – 0xbadf00d Jul 27 '19 at 5:36
• Yes, of course. 2) holds when you condition on $C$ it may not hold when you condition on $C^{c}$. – Kavi Rama Murthy Jul 27 '19 at 5:43