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The wikipedia article on conditional independence states that the following two statements are equivalent, but I can't see how one implies the other and vice versa:

$$\text{Pr}(R\cap{B}\rvert{Y})=\text{Pr}(R\rvert{Y})\;\text{Pr}(B\rvert{Y})\tag{1}$$

$$\text{Pr}(R\rvert{B\cap{Y}})=\text{Pr}(R\rvert{Y})\tag{2}$$

Using Bayes's rule, I can rewrite equation 1 as follows:

$$\text{Pr}(R\cap{B}\rvert{Y})=\frac{\text{Pr}(Y\rvert{R\cap{B}})\;\text{Pr}(R\cap{B})}{\text{Pr}(Y)}$$

but that doesn't help me prove equation 2 -- I don't see a pathway to getting a $B\cap{Y}$ and we have no information whether $R$ and $B$ are independent, so I can't rewrite $\text{Pr}(R\cap{B})$ as $\text{Pr}(R)\;\text{Pr}(B)$. Likewise, expanding equation 2 via Bayes's doesn't help me prove equation 1.

I'm sure it's something obvious, but I just don't see it.

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Assuming $\Pr(B \mid Y) \not = 0$, the following statements are equivalent:

$$\Pr(R\cap{B}\mid {Y})=\Pr(R\mid{Y})\;\Pr(B\mid {Y})$$

$$\Pr(R\mid {B \cap Y})\;\Pr(B \mid Y) =\Pr(R\mid{Y})\;\Pr(B\mid {Y})$$

$$\Pr(R\mid {B \cap Y})=\Pr(R\mid{Y})$$

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  • $\begingroup$ I'm afraid I don't see where your second line comes from. Doesn't the definition of conditional probability say $\text{Pr}(R\mid{B\cap{Y}})=\text{Pr}(R\cap{B}\cap{Y})/\text{Pr}(B\cap{Y})$? Even so, expanding out all the other conditional probabilities, I can't see how to get your second line. $\endgroup$
    – Reb.Cabin
    Feb 16, 2018 at 18:52
  • $\begingroup$ @Reb.Cabin Are you happy with saying $\Pr(R\cap B) =\Pr(R\mid B)\;\Pr(B )$? Then $\Pr(R\cap B \mid Y) =\Pr(R\mid B\cap Y)\;\Pr(B\mid Y )$ should be seen as the same. If you insist, $\Pr(R\cap B \mid Y) = \frac{\Pr(R\cap B \cap Y)}{\Pr(Y)} = \frac{\Pr(R\cap B \cap Y)}{\Pr(B \cap Y)} \times \frac{\Pr(B \cap Y)}{\Pr( Y)} = \Pr(R\mid B\cap Y)\;\Pr(B\mid Y )$ $\endgroup$
    – Henry
    Feb 16, 2018 at 18:59
  • $\begingroup$ I see it now. Marking the question as answered. $\endgroup$
    – Reb.Cabin
    Feb 16, 2018 at 19:55

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