There are 3 events $A, B, D$ such that $D$ makes $A$ more likely and $D$ makes $B$ more likely. Does this mean that $D$ makes it more likely that both $A$ and $B$ occur? How can you prove this using mathematical formalism to prove it if it’s true or what counter example makes it false?
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$\begingroup$ I helped you did formatting with MathJax this time. Going forward, please make sure your math expressions are formatted properly. $\endgroup$– ZhanxiongSep 18 at 15:21
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5$\begingroup$ Draw a Venn diagram and attempt to construct a counterexample from what it shows you. Consider a case where $A\cap B \cap D=0.$ $\endgroup$– whuber ♦Sep 18 at 15:27
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1 Answer
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If inequalities are strict, then counter examples are easy to make. For example, consider $\Omega = \{1, 2, 3, 4, 5\}$ where $P(\{1\}) = P(\{2\}) = P(\{3\}) = P(\{4\}) = P(\{5\}) = 1/5$. Let $A = \{1, 2\}, B = \{4, 5\}$, $D = \{2, 4\}$. Then $P(A|D) = 1/2 > P(A) = 2/5$, $P(B|D) = 1/2 > P(B) = 2/5$, while $P(A \cap B|D) = 0 = P(A \cap B)$, because $A \cap B = \varnothing$.
A more interesting question is thus whether $P(A \cap B|D) \geq P(A \cap B)$ holds given $P(A|D) \geq P(A)$ and $P(B|D) \geq P(B)$. This is still false, and we can tune the first counter example by making $A \cap B \cap D = \varnothing$ but pairwise intersection events $A \cap D$ and $B \cap D$ are not. This can be done by setting $A = \{1, 3\}$, $B = \{3, 5\}$, and $D = \{1, 5\}$, under which $P(A|D) = 1/2 \geq P(A) = 2/5$, $P(B|D) = 1/2 \geq P(B) = 2/5$, but $P(A \cap B|D) = 0 < P(A \cap B) = 1/5$.
