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Exercise 2.58

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(a)

A

and

B

are independent, then

P (A | B, C) = P (A | B^{c}, C) = P (A | C) .

P (A | B, C^{c}) = P (A | B^{c}, C^{c}) = P (A | C^{c}) .

Thus, Simpson’s Paradox does not hold.

(b)

A

and

C

are independent, then

P (A | B, C) < P (A | B^{c}, C) ⟹ P (A | B) < P (A | B^{c}) .

Thus, Simpson’s Paradox does not hold.

(c)

B

and

C

are independent, then

P (A | B) = P (C) P (A | B, C) + P (C^{c}) P (A | B, C^{c}) .

P (A | B^{c}) = P (C) P (A | B^{c}, C) + P (C^{c}) P (A | B^{c}, C^{c}) .

Since $P (A | B, C) > P (A | B^{c}, C)$ and $P (A | B, C^{c}) > P (A | B^{c}, C^{c})$ , $P (A | B) > P (A | B^{c})$ , so Simpson’s Paradox does not hold.

fifthist

2021-12-05 00:00

Exercise 2.58

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