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Say I have discrete random variables $A$ and $B$. Is it true that $P(A|B)=1-P(\overline{A}|B)$?

My intuition is conflicting; it makes sense that if $P(A|B)=p$, then it is only possible that $P(\overline{A}|B)=1-p$ and vice versa, since the probability of not A given B and the probability of A given B must add to 1 since there are no other options. However, my intuition is usually wrong with probability, things are rarely as they seem. Can anyone give me insight into this?

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Yes this is true. If B is the condition that must occur in both cases, then it sort of becomes irrelevant, because cases where B does not occur can be ignored. In this case, we can look at the space of outcomes where B occurs, within which there are two options: A and not A. Therefore these probabilities must add to 1.

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  • $\begingroup$ thank you, that makes more sense now! $\endgroup$ Apr 20 '20 at 23:19
  • $\begingroup$ An intuitive way of looking at it: conditioning on $B$ means your "world" is smaller: you're only looking at the events where $B$ occurred. All the usual rules of probability still apply in the smaller world. $\endgroup$ Apr 21 '20 at 1:38
  • $\begingroup$ As a related question but for 3 variables, if $A$ and $B$ are conditionally independent given $C$, does this imply $P(A=a|B=b, C=c) = P(A=a|B=\overline{b}, C=c)$? $\endgroup$ Apr 21 '20 at 3:03
  • $\begingroup$ @ajax2112 Well based on such an approach of 'intuitive reasoning' it should also be true that $P(A|B) = 1 - P(A|\bar{B})$, no? Because you have only two options, either $B$ occurred or it did not occur... ;-) $\endgroup$ Apr 22 '20 at 7:25
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I like the other answer (+1) as it is simple and based on intuition. This can also be demonstrated using relatively simple probability rules.

\begin{align*} P(\bar A | B) &= \frac{P(\bar A\cap B)}{P(B)} \\[1.2ex] &= \frac{P(B) - P(A\cap B)}{P(B)} \\[1.2ex] &= 1 - \frac{P(A\cap B)}{P(B)} \\[1.2ex] &= 1 - P(A|B) \\[1.2ex] &\square \end{align*}


Update to address question in the comments.

Claim: If $A$ and $B$ are conditionally independent given $C$, then $P(A|B\cap C) = P(A|\bar{B}\cap C)$.

Again, this can be demonstrated using only simple probability rules. The idea is, once we know that $C$ occurred, the probability of $A$ "doesn't care" whether or not $B$ occurred. So we will demonstrate this by showing both the LHS and RHS are equal to $P(A|C)$.

\begin{align*} P(A|B\cap C) &= \frac{P(A\cap (B\cap C))}{P(B\cap C)} \\[1.2ex] &= \frac{P(A\cap B | C)P(C)}{P(B\cap C)} \\[1.2ex] &= \frac{P(A|C)P(B|C)P(C)}{P(B|C)P(C)} && \text{(conditional independence)} \\[1.2ex] &= P(A|C) \end{align*}

The RHS is slightly trickier. We use the fact $P(E\cap \bar F) = P(E) - P(E\cap F)$ several times.

\begin{align*} P(A|\bar B\cap C) &= \frac{P(A\cap (\bar B\cap C))}{P(\bar B\cap C)} \\[1.2ex] &= \frac{P(A\cap C) - P(A\cap B \cap C)}{P(C) - P(B\cap C)} \\[1.2ex] &= \frac{P(A|C)P(C) - P(A\cap B | C)P(C)}{P(C) - P(B|C)P(C)} \\[1.2ex] &= \frac{P(A|C) - P(A|C)P(B|C)}{1 - P(B|C)} && \text{(conditional independence)} \\[1.2ex] &= \frac{P(A|C)(1-P(B|C)}{(1-P(B|C)} \\[1.2ex] &= P(A|C) \end{align*}

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  • $\begingroup$ Thank you! As a related question but for 3 variables, if $A$ and $B$ are conditionally independent given $C$, does this imply $P(A=a|B=b, C=c) = P(A=a|B=\overline{b}, C=c)$? $\endgroup$ Apr 21 '20 at 3:03
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    $\begingroup$ @user123544, that statement is also true. I recommend adding this to your question, or making a new question. $\endgroup$
    – knrumsey
    Apr 21 '20 at 17:12

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