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I have four events A,B,C and D and I have these conditional independencies: $ A \coprod B | C \\ A\coprod C |D $ The question is do these conditions imply $ A \coprod B | D $ I guess that this is not true in general, therefore I tried to construct a counterexample. However, I was not able to do it. Can you explain me how to construct a counterexample to show that given these conditions A and B are not conditionally independent given D.

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Consider $\Omega=\{1,2,3,4,5,6,7,8,9\}$ endowed with the sigma-algebra $2^\Omega$ and the uniform distribution $\mathbb P$, and $A=\{1,2,3\}$, $B=\{1,2,4,5\}$, $C=\{2,3,4,5,6,7\}$, $D=\Omega$.

Then $A$ has probability $\frac13$, $B$ has probability $\frac49$, $C$ has probability $\frac23$ and $D$ has probability $1$. Likewise, $A\cap B=\{1,2\}$ has probability $\frac29$, $B\cap C=\{2,4,5\}$ has probability $\frac13$, $A\cap C=\{2,3\}$ has probability $\frac29$, and $A\cap B\cap C=\{2\}$ has probability $\frac19$.

Thus, $\mathbb P(A\cap C)=\frac2{9}=\mathbb P(A)\mathbb P(C)$ and $\mathbb P(A\cap B\cap C)\mathbb P(C)=\frac2{27}=\mathbb P(A\cap C)\mathbb P(B\cap C)$, but $\mathbb P(A\cap B)=\frac2{9}\ne\frac4{27}=\mathbb P(A)\mathbb P(B)$.

One sees that the conditions $A\perp B\mid C$ and $A\perp C$ do not imply that $A\perp B$.

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    i accepted your answer, thanks. i wonder is there a way to prove this without using counterexamples?2012-10-28
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If I tell you that Bob thinks that when he buys kippers it will not rain ($D$), then you know that there's a linkage between Bob buying kippers ($A$) and Bob thinking that it will not rain ($B$).

However, if I tell you that Bob thinks that when he buys kippers it will not rain ($D$), that doesn't tell you anything about a linkage between Bob buying kippers ($A$) and an earthquake occurring in Guatemala ($C$), and if I tell you that an earthquake just occurred in Guatemala ($C$), that doesn't tell you anything about a linkage between Bob buying kippers ($A$) and Bob thinking that it will not rain ($B$).