Mixing
Independent events in the context of Simpson's Paradox
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I came across this problem in the book "Introduction to Probability" by Dr. Joseph K. Blitzstein and Dr. Jessica Hwang. This problem deals with the concept of independent events in the context of the Simpson's Paradox. I seem to be a little stuck in the third part of the problem. Please let me know if my approach to the rest of the question is correct. Thank You!
The problem goes as follows:
Simpson’s paradox says that it is possible to have events
$A, B, C$ such that $P(A | B,C) < P(A | B^c, C)$ and $P(A | B,C^c) < P(A | B^c, C^c)$, yet $P(A | B) > P(A | B^c)$.
(a) Can Simpson’s paradox occur if $A$ and $B$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(b) Can Simpson’s paradox occur if $A$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(c) Can Simpson’s paradox occur if $B$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
Answers:
(a)
$P(A cap B) = P(A) P(B)$
By this equation, $P(A|B,C) = P(A|C)$ and $P(A|B^c,C) = P(A|C)$. Hence, these two expressions are equal. So Simpson's Paradox does not apply here.
(b)
$P(A cap C) = P(A) P(C)$
By the above equation, $P(A|B,C) = P(A|B)$ and $P(A|B^c,C) = P(A|B^c)$. By the inequality $P(A|B,C^c) < P(A|B^c, C^c)$, $P(A|B)$ should be less than $P(A|B^c)$. This then contradicts the final result that $P(A|B) > P(A|B^c)$.
(c)
Here I am not sure how to proceed. I am not sure how $B$ and $C$ being independent would affect the occurrence of $A$.
probability paradoxes
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
asked Jul 15 at 15:00
adhok
235
add a comment |Â
up vote
0
down vote
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I came across this problem in the book "Introduction to Probability" by Dr. Joseph K. Blitzstein and Dr. Jessica Hwang. This problem deals with the concept of independent events in the context of the Simpson's Paradox. I seem to be a little stuck in the third part of the problem. Please let me know if my approach to the rest of the question is correct. Thank You!
The problem goes as follows:
Simpson’s paradox says that it is possible to have events
$A, B, C$ such that $P(A | B,C) < P(A | B^c, C)$ and $P(A | B,C^c) < P(A | B^c, C^c)$, yet $P(A | B) > P(A | B^c)$.
(a) Can Simpson’s paradox occur if $A$ and $B$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(b) Can Simpson’s paradox occur if $A$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(c) Can Simpson’s paradox occur if $B$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
Answers:
(a)
$P(A cap B) = P(A) P(B)$
By this equation, $P(A|B,C) = P(A|C)$ and $P(A|B^c,C) = P(A|C)$. Hence, these two expressions are equal. So Simpson's Paradox does not apply here.
(b)
$P(A cap C) = P(A) P(C)$
By the above equation, $P(A|B,C) = P(A|B)$ and $P(A|B^c,C) = P(A|B^c)$. By the inequality $P(A|B,C^c) < P(A|B^c, C^c)$, $P(A|B)$ should be less than $P(A|B^c)$. This then contradicts the final result that $P(A|B) > P(A|B^c)$.
(c)
Here I am not sure how to proceed. I am not sure how $B$ and $C$ being independent would affect the occurrence of $A$.
probability paradoxes
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
asked Jul 15 at 15:00
adhok
235
Dumb question: what doesP(A,B)mean in your notation?
– barrycarter
Jul 15 at 15:46
$P(A | B,C)$ means Probability of A given B given C.
– adhok
Jul 16 at 1:38
SoP(A|(B|C))? OrP((A|B)|C). Or is 'given' associative (my math is a little rusty)
– barrycarter
Jul 16 at 14:59
I think they are both equal. The results would be the same no matter the order.
– adhok
Jul 17 at 3:29
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I came across this problem in the book "Introduction to Probability" by Dr. Joseph K. Blitzstein and Dr. Jessica Hwang. This problem deals with the concept of independent events in the context of the Simpson's Paradox. I seem to be a little stuck in the third part of the problem. Please let me know if my approach to the rest of the question is correct. Thank You!
The problem goes as follows:
Simpson’s paradox says that it is possible to have events
$A, B, C$ such that $P(A | B,C) < P(A | B^c, C)$ and $P(A | B,C^c) < P(A | B^c, C^c)$, yet $P(A | B) > P(A | B^c)$.
(a) Can Simpson’s paradox occur if $A$ and $B$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(b) Can Simpson’s paradox occur if $A$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(c) Can Simpson’s paradox occur if $B$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
Answers:
(a)
$P(A cap B) = P(A) P(B)$
By this equation, $P(A|B,C) = P(A|C)$ and $P(A|B^c,C) = P(A|C)$. Hence, these two expressions are equal. So Simpson's Paradox does not apply here.
(b)
$P(A cap C) = P(A) P(C)$
By the above equation, $P(A|B,C) = P(A|B)$ and $P(A|B^c,C) = P(A|B^c)$. By the inequality $P(A|B,C^c) < P(A|B^c, C^c)$, $P(A|B)$ should be less than $P(A|B^c)$. This then contradicts the final result that $P(A|B) > P(A|B^c)$.
(c)
Here I am not sure how to proceed. I am not sure how $B$ and $C$ being independent would affect the occurrence of $A$.
probability paradoxes
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
asked Jul 15 at 15:00
adhok
235
I came across this problem in the book "Introduction to Probability" by Dr. Joseph K. Blitzstein and Dr. Jessica Hwang. This problem deals with the concept of independent events in the context of the Simpson's Paradox. I seem to be a little stuck in the third part of the problem. Please let me know if my approach to the rest of the question is correct. Thank You!
The problem goes as follows:
Simpson’s paradox says that it is possible to have events
$A, B, C$ such that $P(A | B,C) < P(A | B^c, C)$ and $P(A | B,C^c) < P(A | B^c, C^c)$, yet $P(A | B) > P(A | B^c)$.
(a) Can Simpson’s paradox occur if $A$ and $B$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(b) Can Simpson’s paradox occur if $A$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(c) Can Simpson’s paradox occur if $B$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
Answers:
(a)
$P(A cap B) = P(A) P(B)$
By this equation, $P(A|B,C) = P(A|C)$ and $P(A|B^c,C) = P(A|C)$. Hence, these two expressions are equal. So Simpson's Paradox does not apply here.
(b)
$P(A cap C) = P(A) P(C)$
By the above equation, $P(A|B,C) = P(A|B)$ and $P(A|B^c,C) = P(A|B^c)$. By the inequality $P(A|B,C^c) < P(A|B^c, C^c)$, $P(A|B)$ should be less than $P(A|B^c)$. This then contradicts the final result that $P(A|B) > P(A|B^c)$.
(c)
Here I am not sure how to proceed. I am not sure how $B$ and $C$ being independent would affect the occurrence of $A$.
probability paradoxes
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
asked Jul 15 at 15:00
adhok
235
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
edited Jul 15 at 16:17
Jendrik Stelzner
7,55211037
7,55211037
asked Jul 15 at 15:00
adhok
235
asked Jul 15 at 15:00
adhok
235
asked Jul 15 at 15:00
adhok
235
235
Dumb question: what doesP(A,B)mean in your notation?
– barrycarter
Jul 15 at 15:46
$P(A | B,C)$ means Probability of A given B given C.
– adhok
Jul 16 at 1:38
SoP(A|(B|C))? OrP((A|B)|C). Or is 'given' associative (my math is a little rusty)
– barrycarter
Jul 16 at 14:59
I think they are both equal. The results would be the same no matter the order.
– adhok
Jul 17 at 3:29
add a comment |Â
Dumb question: what doesP(A,B)mean in your notation?
– barrycarter
Jul 15 at 15:46
$P(A | B,C)$ means Probability of A given B given C.
– adhok
Jul 16 at 1:38
SoP(A|(B|C))? OrP((A|B)|C). Or is 'given' associative (my math is a little rusty)
– barrycarter
Jul 16 at 14:59
I think they are both equal. The results would be the same no matter the order.
– adhok
Jul 17 at 3:29
Dumb question: what does
P(A,B) mean in your notation?– barrycarter
Jul 15 at 15:46
Dumb question: what does
P(A,B) mean in your notation?– barrycarter
Jul 15 at 15:46
$P(A | B,C)$ means Probability of A given B given C.
– adhok
Jul 16 at 1:38
$P(A | B,C)$ means Probability of A given B given C.
– adhok
Jul 16 at 1:38
So
P(A|(B|C))? Or P((A|B)|C). Or is 'given' associative (my math is a little rusty)– barrycarter
Jul 16 at 14:59
So
P(A|(B|C))? Or P((A|B)|C). Or is 'given' associative (my math is a little rusty)– barrycarter
Jul 16 at 14:59
I think they are both equal. The results would be the same no matter the order.
– adhok
Jul 17 at 3:29
I think they are both equal. The results would be the same no matter the order.
– adhok
Jul 17 at 3:29
add a comment |Â
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Dumb question: what does
P(A,B)mean in your notation?– barrycarter
Jul 15 at 15:46
$P(A | B,C)$ means Probability of A given B given C.
– adhok
Jul 16 at 1:38
So
P(A|(B|C))? OrP((A|B)|C). Or is 'given' associative (my math is a little rusty)– barrycarter
Jul 16 at 14:59
I think they are both equal. The results would be the same no matter the order.
– adhok
Jul 17 at 3:29