Questions about the independence of three events
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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.
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?
$E_1, E_2, E_3$ are three jointly independent events. The, the events $E_1 cup E_2$ and $E_3$ are pairwise independent.
$E_1, E_2, E_3$ 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.
My thought is it depends on the amount of space $E_3$ occupies of $E_1$ because independence is defined as $P(AB)=P(A)P(B)$.
True. Jointly independent events are always pairwise independent.
I am unsure about how to tackle this one.
probability probability-theory
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add a comment |
$begingroup$
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.
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?
$E_1, E_2, E_3$ are three jointly independent events. The, the events $E_1 cup E_2$ and $E_3$ are pairwise independent.
$E_1, E_2, E_3$ 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.
My thought is it depends on the amount of space $E_3$ occupies of $E_1$ because independence is defined as $P(AB)=P(A)P(B)$.
True. Jointly independent events are always pairwise independent.
I am unsure about how to tackle this one.
probability probability-theory
$endgroup$
add a comment |
$begingroup$
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.
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?
$E_1, E_2, E_3$ are three jointly independent events. The, the events $E_1 cup E_2$ and $E_3$ are pairwise independent.
$E_1, E_2, E_3$ 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.
My thought is it depends on the amount of space $E_3$ occupies of $E_1$ because independence is defined as $P(AB)=P(A)P(B)$.
True. Jointly independent events are always pairwise independent.
I am unsure about how to tackle this one.
probability probability-theory
$endgroup$
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.
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?
$E_1, E_2, E_3$ are three jointly independent events. The, the events $E_1 cup E_2$ and $E_3$ are pairwise independent.
$E_1, E_2, E_3$ 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.
My thought is it depends on the amount of space $E_3$ occupies of $E_1$ because independence is defined as $P(AB)=P(A)P(B)$.
True. Jointly independent events are always pairwise independent.
I am unsure about how to tackle this one.
probability probability-theory
probability probability-theory
edited Dec 31 '18 at 18:43
Avedis
asked Dec 30 '18 at 23:20
AvedisAvedis
647
647
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1 Answer
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$begingroup$
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 $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$
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Can you elaborate on 1. Also, how can i word my questions better so I dont get downvotes?
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– 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
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$
$endgroup$
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 $P((E_1cup E_2)cap E_3) = frac{1}{4}neq frac{3}{4}timesfrac{1}{2}$
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
add a comment |
$begingroup$
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
add a comment |
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