Bayesian formula,
\n[P(A|B)=\\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\\bar{A})P(\\bar{A})}]
\nis conceptually straightforward, but amazingly useful in statistics. It turns calculation of (P(A|B)) into finding out (P(B|A)) by simply making use of the rule of total probability,
\n[P(A\\cap B) = P(A|B)P(B) = P(B|A)P(A)]
\nand
\n[P(A) = P(A\\cap B) + P(A\\cap \\bar{B}). ]<\/p>\n
This seems rather trivial to me. Here comes the surprising part. Let us now add another set (C) in the following way,
\n[P\\left(A|B\\right) = P\\left(\\left(A|C\\right)|B\\right)P\\left(C|B\\right) + P\\left(\\left(A|\\bar{C}\\right)|B\\right)P\\left(\\bar{C}|B\\right), \\hspace{2cm} (1)]
\nor in this way,
\n[P\\left(A|B\\right) = P\\left(\\left(A|B\\right)|C\\right)P\\left(C\\right) + P\\left(\\left(A|B\\right)|\\bar{C}\\right)P\\left(\\bar{C}\\right). \\hspace{2cm} (2)]<\/p>\n
Now let us ask which one of the above two formulae is the proper one, or both, or none?<\/p>\n
It is easy to verify the first one: Assuming
\n[P\\left(\\left(A|C\\right)|B\\right) = P\\left(A|\\left(C,B\\right)\\right) = \\frac{A\\cap B \\cap C}{B \\cap C}, \\hspace{2cm} (3)]
\nthen right-hand side of Equ(1) becomes
\n[\\frac{A\\cap B \\cap C}{B \\cap C}\\frac{B\\cap C}{B} + \\frac{A\\cap B \\cap \\bar{C}}{B \\cap \\bar{C}}\\frac{B\\cap \\bar{C}}{B} = \\frac{A\\cap B \\cap C}{B} + \\frac{A\\cap B \\cap \\bar{C}}{B} = \\frac{A\\cap B}{B}, \\hspace{1cm} (4)]
\nwhich is exactly the left-hand side of Equ(1).<\/p>\n
Verifying Equ(2) is however not easy. If the assumption in Equ(3) is right, then Equ(2) becomes
\n[\\frac{A\\cap B \\cap C}{B \\cap C}\\frac{C}{\\Omega} + \\frac{A\\cap B \\cap \\bar{C}}{B \\cap \\bar{C}}\\frac{\\bar{C}}{\\Omega}. \\hspace{1cm} (4)]
\nI can see no clue that this expression should be (\\frac{A\\cap B}{B}).<\/p>\n
However, if (P\\left(A|B\\right)) is the probability of a set of events, then the second one should be correct too. So what is the problem? It seems to me that when discussing (P\\left(A|B\\right)), we have implicitly limited the whole set, which originally is (\\Omega), to be (B), therefore, all the expressions derived from there should have carried the condition (B) forever. So lesson one: Keeping the condition (B) as the condition for all other events. Therefore, Equ(1), not Equ(2), should be used in our case.<\/p>\n
Another lesson learned is that conditional probability is a tricky concept and one has to deal it with extra attention.<\/p>\n","protected":false},"excerpt":{"rendered":"