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I want to prove that $P(A\cap B)>P(A)P(B)$ given:

  1. $P(A\cap B|C)=P(A|C)P(B|C)$
  2. $P(A\cap B|C')=P(A|C')P(B|C')$
  3. $P(A|C)>P(A|C')$
  4. $P(B|C)>P(B|C')$
  5. $P(C)\neq 0,1$

Where $C'$ is the compliment of $C$.

Jarle Tufto
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Professor Reichenbach
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1 Answers1

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Using the law of total covariance applied to indicator variables for the different events it follows that \begin{align} P(A\cap B)-P(A)P(B) &= E(I_{A \cap B})-E(I_A)E(I_B) \\ &= E(I_A I_B)-E(I_A)E(I_B) \\ &= \operatorname{Cov}(I_A,I_B) \\&= E \operatorname{Cov}(I_A,I_B|I_C)+\operatorname{Cov}(E(I_A|I_C),E(I_B|I_C)). \end{align} The first term in the final expression is zero because of conditional independence (1. and 2.). The second term is positive because of 3. and 4. which creates a positive association between the two conditional expectations.

An analogous treatment of the amount of linkage disequilibrium generated by mixing genetically differentiated populations can be found in Joe Felsenstein's population genetics textbook, pp. 177-178.

Jarle Tufto
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    I don't understand your second point, would it be possible for you to explain how 3. and 4. create a positive association between the two conditional expectations ? – Camille Gontier Dec 08 '20 at 18:10
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    @CamilleGontier The conditional expectations, being functions of the indicator $I_C$, are themselves random variables with a joint bivariate discrete distribution. They take the values $P(A|C),P(B|C)$ with probability $P(C)$ and the values $P(A|C'),P(B|C')$ with probability $P(C')$. So given 3. and 4. they clearly have positive covariance. – Jarle Tufto Dec 08 '20 at 18:49
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    @CamilleGontier Also, see https://stats.stackexchange.com/a/440597/77222 – Jarle Tufto Dec 08 '20 at 18:55