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In general, Simpson's Paradox occurs because situation such as following occurs for some arbitrary events $A,B,$ and $C$:

$P( A | B , C) < P(A| B^c,C) \tag{1}$

$P( A | B , C^c ) < P(A| B^c,C^c) \tag{2}$

Can someone show me a step-by-step way to arrive at $P( A|B) > P(A|B^c)$ from (1), (2)?

The Law of Total Probability

$P( A | B ) = P( A | B , C ) P( C | B) + P( A | B, C^c) P(C^c | B)$

appears somehow involved but I don't see how. Any help would be appreciated.

Jenna Maiz
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1 Answers1

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There is no derivation of the third equation from the first two. If there were, then it would be the case that Simpson's paradox occurs for $all$ $A,B,C$. This is clearly not true. The correct statement is that $there$ $exists$ $A,B,C$ such that those three conditions hold. The fact that this occurs highly depends on the events $A,B,C$

Stella Biderman
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  • I realize the question may not be the best. Maybe you can explain why the paradox occurs using the 4th equation I gave above ? – Jenna Maiz Dec 07 '15 at 23:02
  • All I can do is give you some numerical examples... There's no algebraic explanation, because it's not always true. – Stella Biderman Dec 07 '15 at 23:04
  • I have heard that $P( C | B )$ and $P(C^c|B)$ may be seen as weights in the 4th equation that contribute to the paradox but I am not 100% sure I understand that idea ? – Jenna Maiz Dec 07 '15 at 23:05
  • I would recommend you read the wiki page on Simpson's Paradox, it does multiple numerical examples. https://en.wikipedia.org/wiki/Simpson%27s_paradox – Stella Biderman Dec 08 '15 at 17:56