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Let's say that A, B, C takes on any integer values between 1 to 10.

Event I: A is greater than C. Event II: B is greater than C.

Why are Event I and Event II independent? The intuitive definition of independence is that one event gives NO information about the other (as in two independent coin flips). But, in this case, Event I prevents C from taking on 10 (since A is guaranteed to be greater than C) which impacts our thinking of Event II.

My question, in general, is how should be think about independence when "some information" is revealed.

Thanks!!

Statistics
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  • Your intuition is right: those events are not independent (assuming A,B,C are independent die rolls). – Karl Feb 04 '22 at 08:37
  • The intuitive definition of independence is that one event gives NO information about the other. Not quite; your characterisation is stronger than necessary: for independence of two events, it is sufficient that one event occurring affects the probability of the other occurring. I wrote more here. – ryang Feb 04 '22 at 10:18

1 Answers1

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Let $E_1$ and $E_2$ be the events that $A > C$ and $B > C$, respectively. Then $$\Pr[E_1] = \sum_{c=1}^{9} \Pr[A > C \mid C = c]\Pr[C = c] = \sum_{c=1}^9 \frac{10-c}{10} \cdot \frac{1}{10} = \frac{9}{20}.$$ Similarly, $\Pr[E_2] = 9/20$.

What is $\Pr[E_2 \mid E_1]$? That is to say, given that $A > C$, what is the probability that $B > C$ also? We have by definition of conditional probability,

$$\Pr[E_2 \mid E_1] = \frac{\Pr[E_1 \cap E_2]}{\Pr[E_1]},$$ and $$\Pr[E_1 \cap E_2] = \sum_{c=1}^9 \Pr[(A > C) \cap (B > C) \mid C = c]\Pr[C = c].$$ Now, because $A$ and $B$ are independent and identically distributed (note we are not saying $E_1$ and $E_2$ are independent!), we have for each $c$, $$\Pr[(A > c) \cap (B > c)] \overset{\text{ind}}{=} \Pr[A > c]\Pr[B > c] \overset{\text{id}}{=} \left( \frac{10 - c}{10} \right)^2.$$ Hence $$\Pr[E_1 \cap E_2] = \sum_{c=1}^9 \left(1 - \frac{c}{10}\right)^2 \frac{1}{10} = \frac{57}{200},$$ therefore $$\Pr[E_2 \mid E_1] = \frac{57/200}{9/20} = \frac{19}{30}.$$ But if $E_1$ and $E_2$ are independent events, then we should have $\Pr[E_2 \mid E_1] = \Pr[E_2]$. Since $19/30 \ne 9/20$, this shows they are not independent.

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