# Properties of independence and conditional independence

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!

• Please make your question self-contained, one should not have to visit an offsite page to understand it. – Did Aug 28 '15 at 10:07
• 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... – Did Aug 28 '15 at 14:21
• I understand how to prove Question 1, but what about Question 2 and Question 3 – Johnny Ji Aug 28 '15 at 16:40
• Please show your solution to 1 and your tries to 2 and 3. – Did Aug 28 '15 at 17:18

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, $X\perp A\ |\ B$
• 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. – Did Aug 29 '15 at 11:35
• 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 $X\perp A,B$ – Johnny Ji Aug 29 '15 at 12:49