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Let A, B and C are independent events. How am I supposed to prove that:

  1. A′, B′ and C′ are independent.

  2. A, B′ and C′ , are independent.

  3. A, B and C' are independend.

This is my approach:

for Nr 3.

$P(ABC') = P(A)P(B)P(C').$ But $P(AB)=P(ABC)+P(ABC')$ and using independence $P(A)P(B) = P(A)P(B)P(C)+ P(ABC')$, therefore $P(A)P(B)(1-P(C))=P(ABC')$, $P(ABC') = P(A)P(B)P(C')$.

for Nr 2.

$P(AB'C') = P(A)P(B')P(C')$. But $P(AC')=P(ABC')+P(ABC')$ and using independence $P(A)P(C') = P(A)P(B)P(C')+ P(A B' C')$, therefore $P(A)P(C')(1-P(B))=P(AB' C')$, $P(AB'C') = P(A)P(B')P(C')$.

And for Nr 1.

$P(A'B'C') = P(A')P(B')P(C')$. But $P(A'B')=P(A'B' C )+P(A'B'C')$ and using independence $P(A')P(B') = P(A')P(B')P(C)+ P(A'B'C')$, therefore $P(A')P(B')(1-P(C))=P(A'B'C')$, $P(ABC') = P(A')P(B')P(C')$.

What do you think people? is this way of proving right?

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Read the definition of independence carefully and look through this answer! –  user21436 Mar 27 '12 at 11:34
Hmmm, i see ! Thanks for your answer. what troubles me is that there are 3 events and not 2 like the most cases! Well i ll look into it now! thanks –  Adrian Carter Mar 27 '12 at 12:00
Hint: Try to prove first that $A$, $B$ and $C'$ are inpedendent, than 2. and finally 1 (using what you've done in step one). –  martini Mar 27 '12 at 12:26
Hey martini!! thanks thanks! –  Adrian Carter Mar 27 '12 at 13:43
I've merged the previous incarnation of the question with this one. –  Willie Wong Mar 29 '12 at 21:44

2 Answers 2

Your answer/proof is incomplete. In order to assert that three events $D$, $E$, $F$ are mutually independent, you have to verify that four equations hold: $$\begin{align*} P(DEF) &= P(D)P(E)P(F)\\ P(DE) &= P(D)P(E)\\ P(DF) &= P(D)P(F)\\ P(EF) &= P(E)P(F)\\ \end{align*}$$ Taking $D=A$, $E=B$, $F=C^\prime$, you have verified the first of the four equations above. Now you need to say that $P(AB)=P(A)P(B)$ follows from the independence of $A$ and $B$, and either prove that $P(AC^\prime) = P(A)P(C^\prime)$ and $P(BC^\prime) = P(B)P(C^\prime)$, or assert that these follow from the independence of $A$ and $C$, and $B$ and $C$ respectively if you have done these kinds of calculations previously.

Another definition of independence of $n$ events $A_i$ is that all $2^n$ equations $$P(A_1^*A_2^*\cdots A_n^*) = P(A_1^*)P(A_2^*)\cdots P(A_n^*)$$ must hold, where each $A_i^*$ stands for either $A_i$ or $A_i^\prime$, the same on both sides of the equation. With this definition, the statements to be proved in the OP's problem are true by definition and there is nothing to prove.

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^ quite enlightening post there, this is what i missed, finally , thanks, it seems that i totally forgot pairwise independence. Thats why, i had a weird feeling that my proof was just not right... Thanks! –  Adrian Carter Mar 29 '12 at 13:59

You have shown this, but for clarity, using independence you have $$\Pr(ABC)=\Pr(A)\Pr(B)\Pr(C)$$ and $$\Pr(AB)=\Pr(A)\Pr(B)$$ so $$\Pr(ABC')=\Pr(AB)-\Pr(ABC) $$ $$=\Pr(A)\Pr(B)- \Pr(A)\Pr(B)\Pr(C) $$ $$= \Pr(A)\Pr(B)(1-\Pr(C)) $$ $$=\Pr(A)\Pr(B)\Pr(C')$$

so $A$, $B$ and $C'$ are independent events.

Since this is all commutative, as martini says, you can then derive (2) and (1).

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^ Very nice! actually it makes much more sense to me now, clarity, thats exactly what i needed! Thanks! –  Adrian Carter Mar 28 '12 at 23:23
But i still can understand this --> WHY Pr(A)Pr(B)−Pr(A)Pr(B)Pr(C) =Pr(A)Pr(B)(1−Pr(C)) ? –  Adrian Carter Mar 28 '12 at 23:34
@Adrian: In general $x\times y - x\times y \times z = x\times y \times (1-z)$ because of the common factor of $x \times y$ –  Henry Mar 28 '12 at 23:36
Actually, depending on what one assumes, proving that $\Pr(ABC^\prime)=\Pr(A)\Pr(B)\Pr(C^\prime)$ is not quite sufficient to prove that $A$, $B$, $C^\prime$ are independent. It is also necessary to assert (or remind the reader) that the other conditons $\Pr(AB)=\Pr(A)\Pr(B)$, $\Pr(AC^\prime)=\Pr(A)\Pr(C^\prime)$, $\Pr(BC^\prime)=\Pr(B)\Pr(C^\prime)$ follow from the results on pairwise independence of $A, B, C$. –  Dilip Sarwate Mar 29 '12 at 3:32

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