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Let $\Omega, F, P$ be a probability space and $A, B, D \in F$. If $P( A \mid D ) \ge P(A)$ and $P( B \mid D) \ge P(B)$. Show that if $A \cap B = \emptyset$ then $P(A^c \cap B^c \mid D) \le P(A^c \cap B^c)$.

I know that given that $A \cap B = \emptyset$ then they are mutually exclusives, this means that $P(A \cap B) = 0$

So, $P( A \cup B) = P(A) + P(B)$, and since $P( A \cup B) = 1 - P(A^c \cap B^c)$, this gives:

$P(A^c \cap B^c) = 1 - P(A) - P(B)$

I got an expression for $P(A^c \cap B^c)$ but I don't know how to develop the proof from here, $P(A^c \cap B^c \mid D)$ can be expanded using the Bayes theorem, but this lead's me to nowhere, Im stuck.

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  • $\begingroup$ Can you check if the condition is $P(D|A) \geq P(A)$ or $P(A|D) \geq P(A)$? $\endgroup$
    – Zhanxiong
    Dec 3, 2023 at 5:53
  • $\begingroup$ Sorry @Zhanxiong, the condition is $P(A \mid D) \ge P(A)$, im editing the original question. $\endgroup$
    – Occhima
    Dec 3, 2023 at 14:20

1 Answer 1

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Using the observation that you have made (i.e., the De Morgan law and the complementary law) shows that it is equivalent to prove \begin{align*} P(A \cup B) \leq P(A \cup B|D). \tag{1}\label{1} \end{align*} The condition $A \cap B = \varnothing$ implies that $(A \cap D) \cap (B \cap D) = \varnothing$, whence \begin{align*} P(A \cup B|D) &= \frac{P((A \cup B)\cap D)}{P(D)} \\ &= \frac{P(A \cap D) + P(B \cap D)}{P(D)} \\ &= P(A|D) + P(B|D) \\ & \geq P(A) + P(B) = P(A \cup B). \end{align*} This completes the proof.

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