Questions about the independence of three events












1












$begingroup$


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 $E_1cup E_2$ 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.







probability probability-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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 $E_1cup E_2$ 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.







    probability probability-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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 $E_1cup E_2$ 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.







      probability probability-theory







      share|cite|improve this question











      $endgroup$




      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 $E_1cup E_2$ 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.







      probability probability-theory




      probability probability-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 18:43







      Avedis

















      asked Dec 30 '18 at 23:20









      AvedisAvedis

      647




      647






















          1 Answer
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          1












          $begingroup$

          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.




          $P((E_1cup E_2)cap E_3) = P((E_1cap E_3)cup (E_2cap E_3))$
          $ = P(E_1cap E_3) + P(E_2cap E_3) - P(E_1cap E_2cap 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 $P(E_1cup E_2)=frac{3}{4}$, $P(E_3)=frac{1}{2}$, and $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
            $endgroup$
            – Avedis
            Dec 31 '18 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 '18 at 18:51













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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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.




          $P((E_1cup E_2)cap E_3) = P((E_1cap E_3)cup (E_2cap E_3))$
          $ = P(E_1cap E_3) + P(E_2cap E_3) - P(E_1cap E_2cap 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 $P(E_1cup E_2)=frac{3}{4}$, $P(E_3)=frac{1}{2}$, and $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
            $endgroup$
            – Avedis
            Dec 31 '18 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 '18 at 18:51


















          1












          $begingroup$

          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.




          $P((E_1cup E_2)cap E_3) = P((E_1cap E_3)cup (E_2cap E_3))$
          $ = P(E_1cap E_3) + P(E_2cap E_3) - P(E_1cap E_2cap 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 $P(E_1cup E_2)=frac{3}{4}$, $P(E_3)=frac{1}{2}$, and $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
            $endgroup$
            – Avedis
            Dec 31 '18 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 '18 at 18:51
















          1












          1








          1





          $begingroup$

          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.




          $P((E_1cup E_2)cap E_3) = P((E_1cap E_3)cup (E_2cap E_3))$
          $ = P(E_1cap E_3) + P(E_2cap E_3) - P(E_1cap E_2cap 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 $P(E_1cup E_2)=frac{3}{4}$, $P(E_3)=frac{1}{2}$, and $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$







          share|cite|improve this answer









          $endgroup$



          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.




          $P((E_1cup E_2)cap E_3) = P((E_1cap E_3)cup (E_2cap E_3))$
          $ = P(E_1cap E_3) + P(E_2cap E_3) - P(E_1cap E_2cap 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 $P(E_1cup E_2)=frac{3}{4}$, $P(E_3)=frac{1}{2}$, and $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 23:51









          JMoravitzJMoravitz

          48k33886




          48k33886












          • $begingroup$
            Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
            $endgroup$
            – Avedis
            Dec 31 '18 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 '18 at 18:51




















          • $begingroup$
            Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
            $endgroup$
            – Avedis
            Dec 31 '18 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 '18 at 18:51


















          $begingroup$
          Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
          $endgroup$
          – Avedis
          Dec 31 '18 at 18:44




          $begingroup$
          Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
          $endgroup$
          – Avedis
          Dec 31 '18 at 18:44




          1




          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 '18 at 18:51






          $begingroup$
          @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 '18 at 18:51




















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