# How to prove if P(B|A) > P(B) then P(B|A') < P(B)

I hate proof questions, I never know how to start. My question says:

"Let A and B be two events in a sample space such that 0 < P(A) < 1. Let A' denote the compliment event of A. Show that if P(B|A) > P(B) then P(B|A') < P(B)."

How do I go about answering this question? I don't understand why P(B) is between 1 and 0 but not P(B)?

Has it got something to do with using that Bayesian Theorem?

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It has to do with relating $P(B|A)$ to $P(A|B)$, if that's what you mean. –  Gerry Myerson Oct 4 '12 at 13:25

We are told that $\Pr(B|A)\gt \Pr(B)$. So knowing that $A$ holds makes $B$ more likely. It stands to reason that knowing that $A'$ holds makes $B$ less likely. "It stands to reason" is presumably not enough for a proof, so we give full details.

Of course we will use $\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}$. For this we need $\Pr(A)\ne 0$. From $\Pr(B|A)\gt \Pr(B)$ we get $$\Pr(A\cap B)\gt \Pr(A)\Pr(B).\tag{1}$$

Note that $$\Pr(A\cap B)+\Pr(A'\cap B)=\Pr(B).$$ Substituting in Equation $(1)$ we get $$\Pr(B)-\Pr(A'\cap B)\gt \Pr(A)\Pr(B),$$ which can be rewritten as $$\Pr(A'\cap B)\lt \Pr(B)-\Pr(A)\Pr(B)=\Pr(B)(1-\Pr(A))=\Pr(B)\Pr(A').$$ If $\Pr(A')\ne 0$, we can divide, obtaining $$\frac{\Pr(A'\cap B)}{\Pr(A')}\lt \Pr(B).$$ This says exactly what we want, since the left side is just $\Pr(B|A')$.

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Here's one method. It uses Bayes theorem, which says that:

$$P(A|B) = \frac{P(A)}{P(B)} P(B|A)$$

for any $A$ and $B$.

Proof: You know that

$$P(B|A)>P(B)$$

Use Bayes rule, then cancel and rearrange:

$$\frac{P(B)}{P(A)} P(A|B) > P(B) \quad\Rightarrow\quad P(A|B) > P(A)$$

Since $A'$ is the complement of $A$ you have

$$1-P(A|B) < 1-P(A) \quad\Rightarrow\quad P(A'|B) < P(A')$$

Now use Bayes rule again, and cancel and rearrange one more time:

$$\frac{P(A')}{P(B)}P(B|A') < P(A') \quad\Rightarrow\quad P(B|A') < P(B)$$

Make sure you understand all of the steps and can go through them yourself.

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Hint: $P(B)=P(B|A) P(A) + P(B|A') P(A')$, hence $P(B)$ is a convex combination of $P(B|A)$ and $P(B|A')$.

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No need for Bayes.

Observe that by the definition of conditional probability:

$$P(B|A)P(A) = P(A \text{ and } B)$$

and that

$$P(A \text{ and }B) + P(A' \text{ and } B) = P((A \text{ or }A') \text{ and } B) = P(B)$$

So we have

$$P(B) = P(B|A) P(A) + P(B|A') P(A') \tag{%}$$

If $P(B|A)> P(B)$ and $P(B|A') > P(B)$, equation (%) implies that

$$P(B) > P(B) P(A) + P(B) P(A') = P(B)\left[ P(A) + P(A')\right] = P(B)$$

Intuitively this is the statement that "the weighted average of two numbers must lie between the two numbers". $P(B)$, the probability of $B$ happening over the whole sample space, is the weighted average of $P(B|A)$, the probability of $B$ happening over the sample space for which $A$ happens, and $P(B|A')$, the probability of $B$ happening over the remaining part of the sample space. Hence we must have that if $P(B|A) > P(B)$, the left over bit satisfies $P(B) > P(B|A')$.