I have several statements related to the probability of three events and am trying to determine if the statements are true or false. For some of the answers, I have intuition but I would prefer to learn a provable answer.

  1. I have 3 events $E_1, E_2, E_3$ in a probability space. If $E_3 \subset E_1$ and $E_1, E_2$ are pairwise independent, are $E_2, E_3$ pairwise independent?

  2. $E_1, E_2, E_3$ are three jointly independent events. The, the events $E_1 \cup E_2$ and $E_3$ are pairwise independent.

  3. $E_1, E_2, E_3$ be three pairwise independent events in a probability space. Then the events $E_1\cup E_2$ and $E_3$ are pairwise independent.

My solutions thus far.

  1. My thought is it depends on the amount of space $E_3$ occupies of $E_1$ because independence is defined as $P(AB)=P(A)P(B)$.

  2. True. Jointly independent events are always pairwise independent.

  3. I am unsure about how to tackle this one.


1 Answer 1


For (1), pick $E_2,E_3$ your favorite dependent events and let $E_1$ be the sample space itself, the sure event. $E_1$ trivially contains $E_3$ and further $E_1$ is trivially independent with any other event.

For (2), this is a fair bit of algebra but fairly straightforward to prove. Just remember how intersection distributes over union, apply inclusion-exclusion and properties of independence, and factor.

$P((E_1\cup E_2)\cap E_3) = P((E_1\cap E_3)\cup (E_2\cap E_3))$ $ = P(E_1\cap E_3) + P(E_2\cap E_3) - P(E_1\cap E_2\cap E_3) $ $= P(E_1)P(E_3) + P(E_2)P(E_3) - P(E_1)P(E_2)P(E_3)$ $=P(E_3)\times (\cdots)$

For (3), consider an example where the events are pairwise but not mutually independent. For example considering the uniform distribution over the sample space $\{1,2,3,4\}$ and letting the event $E_i = \{i,4\}$. You have $E_1,E_2,E_3$ are each pairwise independent but not mutually independent.

Here we have $P(E_1\cup E_2)=\frac{3}{4}$, $P(E_3)=\frac{1}{2}$, and $P((E_1\cup E_2)\cap E_3) = \frac{1}{4}\neq \frac{3}{4}\times\frac{1}{2}$

  • $\begingroup$ Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes? $\endgroup$
    – Avedis
    Dec 31, 2018 at 18:44
  • 1
    $\begingroup$ @Avedis (1) is false by using as E_1 as the sure event and E_2,E_3 as any two dependent events. If you want an explicit example consider flipping a coin. E_1 is the event the coin is head or tail, E_2 is the event the coin is head, E_3 is the event the coin is tail. Note that E_1,E_2 are independent and E_3 is a subevent of E_1 though E_2,E_3 are clearly dependent on one another. $\endgroup$
    – JMoravitz
    Dec 31, 2018 at 18:51

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