Let P(C)P(C^c) 0. Then the inequalities P(A | C) P(B | C) and P(A | C^c) P(B | C^c)

Question

Let [latex]P(C)P(C^{c})\gt 0.[/latex] Then the inequalities [latex]P(A\mid C)\gt P(B\mid C)[/latex] and [latex]P(A\,|\,C^{c})\gt P(B\mid C^{c})[/latex] imply [latex]P(A)\gt P(B).[/latex]

Accepted Answer

The inequalities [latex]P(A\mid C)\gt P(B\mid C)[/latex] and [latex]P(A\mid C^{c})\gt P(B\mid C^{c})[/latex] are equivalent to [latex]P(A\cap C)\gt P(B\cap C)[/latex] and [latex]P(A\cap C^{c})\gt P(B\cap C^{c}).[/latex] Adding up these inequalities, we obtain [latex]P(A\cap C)+P(A\cap C^{c})\gt P(B\cap C)+ P(B\cap C^{c})\ {\mathrm{or}}\ P(A)\gt P(B),[/latex] since [latex] A=(A\cap C)\cup(A\cap C^{c})[/latex] and [latex]B=\left(B\cap C\right)\cup (B\cap C^{c}).[/latex]

REMARK 4 Once again, that the inequalities of the two conditional probabilities should imply the same inequality for the unconditional probabilities is quite obvious on intuitive grounds. The justification given above simply makes it rigorous.

Question 2.26: Let P(C)P(C^c) > 0. Then the inequalities P(A | C) > P......

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