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In reading about events being pairwise independent but not independent, I come across this equation.

$$P\left(A\bigcap B\bigcap C\right)\:=\:\left(X\right)\ne \:\left(Y\right)\:=\:P\left(A\right)P\left(B\right)P\left(C\right)$$

(X and Y being different possiblities)

this answer explains it well, but I didn't quite understand this particular part from it.

But: $$P\left(A\bigcap B\bigcap C\right)\:=\:P\left(A\bigcap B\right)\:=\:\frac{1}{2}$$


My main question is, how can $P\left(A\bigcap B\bigcap C\right)$ and $P\left(A\right)P\left(B\right)P\left(C\right)$ yield different results? Aren't they suppose to be the same equation

Community
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lasec0203
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  • They yield different results precisely because they are not independent. If they were independent, we would have exactly $P(A \cap B \cap C) = P(A)P(B)P(C)$. – Brian Tung Jan 14 '16 at 00:31

2 Answers2

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The answer explains that $P(A) = P(B) =P(C) = \frac{1}{2}.$

If we want to consider $P(A\cap B\cap C)$, then notice that $$A\cap B\cap C = \{\text{First is $H$, second is $T$, one is $H$ and other $T$}\} = A\cap B.\tag{$\star$}$$ Thus, $$P(A\cap B\cap C) = P(A\cap B) = P(A)P(B) = \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}.$$ Clearly, this does not equal $$P(A)P(B)P(C) = \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{8}.$$

Therefore $A,B,C$ are not independent.

Em.
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  • P(A∩B∩C)=P(A∩B)?? what happens to C? – lasec0203 Jan 14 '16 at 00:36
  • @lasec0203 Refer to $(\star)$. Notice that $A\cap B \subset C$. – Em. Jan 14 '16 at 00:38
  • @lasec0203 That's the point. $A$ says first flip is $H$, $B$ says that the second is $T$, and $C$ says one is $H$ and $T$. Notice that $C$ is repetitive, since $A\cap B$ satisfies condition $C$. So $A\cap B\cap C = A\cap B$. – Em. Jan 14 '16 at 00:41
  • thanks, but this is still confusing me. – lasec0203 Jan 14 '16 at 00:48
  • @lasec0203 Notice that $A = {HT, HH}, B = {HT, TT}$, and $C = {HT, TH}$. Then the intersection $A\cap B\cap C = {HT}$, since this is the only outcome they have in common is $HT$. Thus $$P(A\cap B\cap C) = P(A\cap B) = P(HT) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}.$$ – Em. Jan 14 '16 at 01:03
  • @lasec0203 Great, good luck. – Em. Jan 14 '16 at 01:10
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Your first question: From the linked answer, the events $A$,$B$, and $C$ are defined so that $(A \cap B) \subset C$, so necessarily $A \cap B \cap C = A \cap B$. (This is just set theory.)

Your main question: that is precisely the significance of independence. In general, $P(A \cap B \cap C) \ne P(A) P(B) P(C)$, as the example shows. However, if $A$, $B$, and $C$ are independent, then you do have $P(A \cap B \cap C) = P(A) P(B) P(C)$, by definition. They are not the "same equation" unless you have the assumption of independence.

[Similarly, $P(A \cap B) \ne P(A) P(B)$ in general, unless $A$ and $B$ are independent.]

angryavian
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