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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.

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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 $C^{c}$.

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  • $\begingroup$ But it should follow from $(1)$ that $(2)$ holds on $C$, right? $\endgroup$ – 0xbadf00d Jul 27 '19 at 5:36
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    $\begingroup$ Yes, of course. 2) holds when you condition on $C$ it may not hold when you condition on $C^{c}$. $\endgroup$ – Kavi Rama Murthy Jul 27 '19 at 5:43
  • $\begingroup$ Thanks. Could you take a look at this question: math.stackexchange.com/q/3304535/47771? I'm struggling to see how I need to phrase the conditional independence correctly there. $\endgroup$ – 0xbadf00d Jul 27 '19 at 5:45

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