" If the occurrence of B makes A more likely, does the occurrence of A make B more likely?
I guess yes. But I could not figure out the reason. Thanks~~
Hello, shiningstarpxx!
An interesting question . . .
The answer is "Yes".If the occurrence of $\displaystyle B$ makes $\displaystyle A$ more likely,
does the occurrence of $\displaystyle A$ make $\displaystyle B$ more likely?
We need Bayes' Theorem: .$\displaystyle P(X\,|\,Y) \;=\;\frac{P(X \wedge\,Y)}{P(Y)}$
We are told that, if $\displaystyle B$ happens, $\displaystyle A$ is more likely to happen.
. . That is, the probability of $\displaystyle A$, given $\displaystyle B$, is greater than the probability of $\displaystyle A$.
In symbols: .$\displaystyle P(A\,|\,B) \:> \:P(A)$
This means: .$\displaystyle \frac{P(A \wedge B)}{P(B)} \:> \:P(A) \quad\Rightarrow\quad P(A\wedge B) \:>\:P(A)\!\cdot\!P(B)$
Hence, we have: .$\displaystyle \frac{P(B \wedge A)}{P(A)} \:>\:P(B) \quad\Rightarrow\quad P(B\,|\,A) \:>\:P(B)$
Therefore, the probability of $\displaystyle B$, given $\displaystyle A$, is greater than the probability of $\displaystyle B.$
. . The occurence of $\displaystyle A$ makes $\displaystyle B$ more likely.
What you're asking is the following:
Does Pr(A | B) > Pr(A) => Pr(B | A) > Pr(B) ?
The answer is yes. Consider:
$\displaystyle \Pr(A | B) \, \Pr(B) = \Pr(B | A) \, \Pr(A) \Rightarrow \frac{\Pr(A | B)}{\Pr(A)} = \frac{\Pr(B | A)}{\Pr(B)}$.
But it's given that $\displaystyle \Pr(A | B) > \Pr(A) \Rightarrow \frac{\Pr(A | B)}{\Pr(A)} > 1$.
Therefore $\displaystyle \frac{\Pr(B | A)}{\Pr(B)} > 1$ and the implication follows.
Edit: Too fast for me this time, Soroban. But I like to think our replies ...... complement each other lol!