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Recently, I see some properties from conditional independence wiki page https://en.wikipedia.org/wiki/Conditional_independence

I don't quite understand the properties of "Rules of conditional independence" in the wiki page.

Definition: $X\perp A$ means random variables $A$ and $X$ are independent from each other. $X\perp A\ |\ B$ means random variables $A$ and $X$ are conditionally independent given random variable $B$.

Question 1: how to prove "Contraction-weak-union-decomposition" property as follows: \begin{equation*} \left. \begin{array}{rl} X\perp A\ |\ B\\ X\perp B \end{array} \right\} and\ \ \ \ \ \ \Leftrightarrow \ \ \ \ \ \ X\perp A,B \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ and\left\{ \begin{array}{ll} X\perp A\ |\ B\\ X\perp B\\ X\perp B\ |\ A\\ X\perp A \end{array} \right. \end{equation*}

Question 2: how to prove the "Intersection" rule as follows: \begin{equation*} \left. \begin{array}{rl} X\perp A\ |\ B\\ X\perp B\ |\ A \end{array} \right\} and\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ X\perp A,B \end{equation*}

Question 3: Is the inverse proposition of "Decomposition" rule still true? how to comprehend it, please give me an example.

Original proposition: \begin{equation*} X\perp A,B\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ and \left\{ \begin{array}{rl} X\perp A\\ X\perp B \end{array} \right. \end{equation*} Inverse proposition: \begin{equation*} \left. \begin{array}{rl} X\perp A\\ X\perp B \end{array} \right\} and\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ X\perp A,B \end{equation*} Thanks very much!

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    $\begingroup$ Please make your question self-contained, one should not have to visit an offsite page to understand it. $\endgroup$ – Did Aug 28 '15 at 10:07
  • $\begingroup$ These are direct consequences of the definitions, for example $X\perp A\mid B$ and $X\perp B$ means that $P(X=x,A=a\mid B=b)=P(X=x\mid B=b)P(A=a\mid B=b)$ and that $P(X=x,B=b)=P(X=x)P(B=b)$ for every $(x,a,b)$, hence... $\endgroup$ – Did Aug 28 '15 at 14:21
  • $\begingroup$ I understand how to prove Question 1, but what about Question 2 and Question 3 $\endgroup$ – Johnny Ji Aug 28 '15 at 16:40
  • $\begingroup$ Please show your solution to 1 and your tries to 2 and 3. $\endgroup$ – Did Aug 28 '15 at 17:18
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Here is the proof of Question 1:

first arrow, left to right: \begin{eqnarray} P(X, A, B) & = & P(X, A\ |\ B)P(B)\\ & = & P(X\ |\ B)P(A\ |\ B)P(B)\ \ \ \ \ because\ X\ and\ A\ are\ conditionally\ independent\ on\ B\\ & = & P(X\ |\ B)P(AB)\\ & = & P(X)P(AB)\ \ \ \ \ because\ X\ and\ B\ are\ independent \end{eqnarray} first arrow, right to left: \begin{eqnarray} P(X, A, B) & = & P(X)P(A,B)\ \ \ \ \ because\ X\ and\ A,B\ are\ indepentdent\\ \int_BP(X, A, B) & = & \int_BP(X)P(A,B)\\ P(X,A) & = & P(X)P(A) \end{eqnarray} Hence, $X\perp A$, similarly $X\perp B$. \begin{eqnarray} P(X, A\ |\ B) & = & \frac{P(X,A,B)}{P(B)}\\ &=&\frac{P(X)P(AB)}{P(B)}\ \ \ \ \ because\ X\ and\ A,B\ are\ indepentdent\\ &=&P(X)P(A\ |\ B)\\ &=&P(X\ |\ B)P(A\ |\ B)\ \ \ \ \ because\ X\ and\ B\ are\ indepentdent \end{eqnarray} Hence, $X\perp A\ |\ B$

second arrow: we can obtain that from the first arraw.

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  • $\begingroup$ It seems that a similar approach would solve Q2, wouldn't it? Re Q3, you may want to look for a counterexample. $\endgroup$ – Did Aug 29 '15 at 9:51
  • $\begingroup$ I just can't prove Q2. For Q3, yes, I want to look for a counterexample. $\endgroup$ – Johnny Ji Aug 29 '15 at 10:41
  • $\begingroup$ Re Q2, you should be able to reach the identity $$P(X=x\mid A=a,B=b)=P(X=x\mid A=a)=P(X=x\mid B=b)$$ for every $(x,a,b)$. Hence, $$P(X=x)=\sum_aP(A=a)P(X=x\mid A=a)=\sum_aP(A=a)P(X=x\mid B=b)=P(X=x\mid B=b).$$ By symmetry, $P(X=x)=P(X=x\mid A=a)$, QED. $\endgroup$ – Did Aug 29 '15 at 11:35
  • $\begingroup$ Sorry, I think what you prove is about Q2. You prove that \begin{equation*} \left. \begin{array}{rl} X\perp A\ |\ B\\ X\perp B\ |\ A \end{array} \right\} and\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ (X\perp A)\ and\ (X\perp B) \end{equation*} not $X\perp A,B$ $\endgroup$ – Johnny Ji Aug 29 '15 at 12:49
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    $\begingroup$ "Sorry, I think what you prove is about Q2." ?? Yes, which is why the comment starts by "Re Q2". "You prove that ... not X⊥A,B" ?? One gets P(X=x∣A=a,B=b)=P(X=x∣A=a)=P(X=x∣B=b)=P(X=x). If this is not "X⊥A,B", then what is? $\endgroup$ – Did Aug 29 '15 at 14:22

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