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If $A$ is independent of $(B,C)$ does it mean $A$ is independent of $B$ and also independent of $C$? What does jointly independent mean?

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Hereby $\perp$ means independence. So in your case: A $\perp$(B,C)

This yields $P(A, (B,C))=P(A)\times P(B,C)$ In general you can write $P(B,C)$ as $P(B|C)\times P(C)$ or $P(C|B)\times P(B)$.

You can check whether actually $A \perp B$ and $A \perp C$. in those cases you should be able to write: $P(A,B) = P(A)\times P(B)$

So let's see:

$P(A,B,C)=P(A)\times P(B)\times P(C|B)$

$P(A,B)=\sum_{c\in C}P(A,B,C)=\sum_{c\in C}P(A)\times P(B)\times P(C|B)$

$=P(A)\times P(B)\times \sum_{c\in C}P(C|B)$ , where $\sum_{c\in C}P(C|B)$ in general is 1. so Actually $P(A,B)=P(A)\times P(B)$

But beware that the reverse does not necessarily hold; so you cannot infer independence of $A \perp (B,C)$ from $A \perp B$ and $A \perp C$.

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You know how independence between $A$ and $B$ means that $P(AB)=P(A)\cdot P(B)$? Well, you might think that if $A$ and $B$ are independent and $A$ and $C$ are independent according to this definition, then that we also have that $A$ and $BC$ are independent. So we would have $P(ABC)=P(A)\cdot P(BC)$. However this isn't the case. We can have the first without the second. Wikipedia has a nice example on this: https://en.wikipedia.org/wiki/Independence_(probability_theory)#Pairwise_and_mutual_independence Therefor we have a seperate name for this: mutual independence.

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  • $\begingroup$ your answer is conflicting and opposite of what Yazdiha proved..interms of factoring $P(ABC)$... $\endgroup$ – user251385 Oct 25 '16 at 15:51

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