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Given that $A,B, C$ are independent events, I am trying to prove that $A \cup B$ and $B \cup C$ are independent.

$$P((A \cup B) \cap (B \cup C)) \\= P(B \cup (A \cap C)) \\= P(B) + P(A \cap C) - P(A \cap B \cap C) \\= P(B) + P(A)P(C) - P(A)P(B)P(C)\\ = P(B) + P(A)P(C) \left(1 - P(B) \right).$$

This clearly does not equal $P(A \cup B)P(B \cup C).$ How do I find an explicit counterexample to the claim that $A \cup B$ and $B \cup C$ are independent events? I am struggling because of the condition that $A,B,C$ must be independent.

ryang
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    Why do you say "clearly"? If it is clear that these are not the same, then you should be able to design a counterexample. Try the simplest example...let $A,B,C$ represent the outcome of the tosses of three independent fair coins. – lulu Mar 01 '23 at 14:19
  • @lulu could I use numbers instead, so like $\Omega =$ {$0,1$}, as this is easier for me ? –  Mar 01 '23 at 14:26
  • Absolutely. you have three ordered numbers, each of which is either $0$ or $1$ with equal probability (independent of the others). $A$ is the event the first one is $1$, and so on. – lulu Mar 01 '23 at 14:29
  • @lulu so I will let A be event $1$st is heads, B $2$nd heads, and C$3$rd heads. $P(A) = P(B) = P(C) = $\frac{1/2}$. Then each individual event is independent since for example P($A \cap B$) = P(HHH,HHT) = $\frac{1}{4}$ = $\frac{1}{2} \cdot \frac{1}{2}$. –  Mar 01 '23 at 14:36
  • Yes, those are three independent events. Good. Now you need to analyze $A\cup B$ and $B\cup C$. – lulu Mar 01 '23 at 14:37
  • @lulu Then for $P( (A \cup B) \cap (B \cup C)) = P((HHH, HHT, HTH, THH, THT)) = \frac{5}{8}$. Did I calculate this probability correctly? And then this is not equal to $(\frac{6}{8})^2 = \frac{9}{16}$ –  Mar 01 '23 at 14:40
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    Right. The only events not in that intersection are $HTT, TTH,TTT$. Your analysis looks good, and complete. If you are so inclined, I suggest posting your own solution below. The site prefers not to leave questions unanswered, and if you write it all out, users here can check it. – lulu Mar 01 '23 at 14:42
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    The simplest case is just to let $A$ and $C$ be events that never happen, so they're obviously independent of anything. – Zoe Allen Mar 01 '23 at 14:44
  • @lulu Can you please check my answer below ? –  Mar 01 '23 at 15:00
  • I'd have thought $A\cup B, B\cup C$ are always dependent as they both contain $B$ (except if $B=\emptyset$ or $B\subset A,C$). – JMP Mar 01 '23 at 15:26
  • @JMP But no-disjointedness does not imply dependence (in fact, it is disjointedness that implies dependence); for example, in Universe $U_3$ here, events $X$ and $Y$ have a common outcome and are independent. (In general, it is mutual exclusivity, not independence, that is immediately obvious from a Venn diagram.) – ryang Mar 01 '23 at 15:54
  • *non-disjointedness – ryang Mar 03 '23 at 18:30

3 Answers3

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Consider an experiment of tossing two coins where each outcome is uniform. We have $\Omega =$ {H,T}$^2$. Label events:

$A =$ Event that first coin toss is heads. $P(A) = \frac{1}{2}$

$B =$ Event that second coin toss is heads $P(B) = \frac{1}{2}$

$C = \emptyset$ $P(C) = 0$

Check that $A,B,C$ are independent.

$P(A \cap B) = P({HH}) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} $

$P(A \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $

$P(B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $

$P(A \cap B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot \frac{1}{2} \cdot 0 $

Now, check whether $A \cup B$ and $B \cup C$ are independent.

$P((A \cup B) \cap (B \cup C)) = P({HH,TH}) = \frac{1}{2}$. However, this is not equal to $P(A \cup B) \cdot P(B \cup C) = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$, so not independent.

  • You were right in the comments...$P(A\cup B)=\frac 34=P(B\cup C)$ so the product is $\left(\frac 34\right)^2=\frac 9{16}$. Same conclusion, of course. Besides that, it looks good. – lulu Mar 01 '23 at 15:02
  • @lulu $P(B \cup C) = \frac{3}{4}$ ? Are you sure ? The only outcomes for that are HH and TH –  Mar 01 '23 at 15:15
  • @lulu I think you are getting confused with my previous events in the comments. They have been labelled differently in my answer. –  Mar 01 '23 at 15:16
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    Ah, you are right. Personally, I prefer the other example..."degenerate" events always make independence look a bit artificial. After all, if $E$ has probability $0$ then $E$ is independent from itself, which seems...unintuitive. But, it's a perfectly good example nonetheless. – lulu Mar 01 '23 at 15:28
  • @lulu Okay thanks for your help. Should I accept my answer ? –  Mar 01 '23 at 15:42
  • Sure! Why not? $\quad$ – lulu Mar 01 '23 at 15:48
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Distilling the OP's counterexample in the self-answer: consider a single toss of a fair coin, and let $A,B$ and $C$ be the events of obtaining 2 Tails, a Tail, and 3 Tails, respectively.

Then $A \cup B=\{\text{tail}\}=B \cup C,$

and $P(A \cup B)P(B \cup C)=\frac14\ne\frac12=P\big((A \cup B)\cap(B \cup C)\big).$

ryang
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Let us assume that with a few inependent events A,B,C the equality is satisfied: $${P(B)+P(A)P(C)(1-P(B))=P(A\cup B)P(B\cup C) (I)}$$

From set theory knowing that: $${A \cup B = A \cup (B \setminus (A \cap B))}$$ $${P(A \cup B)=P(A \cup (B \setminus (A \cap B)))}$$

Since sets {${A}$} and {${B \setminus (A \cap B)}$} do not have any common elements, then we can use that ${P(X \cup Y)=P(X)+P(Y)}$, resulting in: $${P(A \cup B)=P(A \cup (B \setminus (A \cap B)))=P(A)+P(B)-P(A \cap B)}$$ while ${P(A \setminus B)=P(A)-P(B)}$ is also true while ${A \supset B}$.

Then rewriting the right side of the (I) equation we get: $${P(A\cup B)P(B\cup C)=(P(A)+P(B)-P(A \cap B))(P(B)+P(C)-P(B \cap C))}$$ $${=(P(A)+P(B)-P(A)P(B))(P(B)+P(C)-P(B)P(C))}$$ $${=P(A)P(C)[(1-2P(B)+P^2(B))]+P(B)[(P(A)+P(B)+P(C)-P(B)P(C)-P(A)P(B))] (II)}$$

Now we can find the equality between (I) equation's left side and (II): $${P(B): 1=P(A)+P(B)+P(C)-P(B)P(C)-P(A)P(B)}$$ $${P(A)P(C): 1-P(B)=1-2P(B)+P^2(B) (III)}$$

From the (III) equation we have: $${0=-P(B)+P^2(B)=P(B)(-1+P(B))}$$

Which means it is forced that P(B) is either 0 or 1 and it obviuosly satisfies (I) equation.

So the condition for your statement to be true, ${P(B)=0 \lor 1}$ have to be true.

Jokūbas Žitkevičius
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