I have gone through the chain rule of probability Which specified
$P(A1,A2…An)=P(A1)P(A2|A1)…P(An|A1,A2…An−1)$
But when I see the derivation of Bayes rule for three events I came across the formula to be
Source:
$P(ABC) \;= P(A|BC)P(BC)\\ \qquad\quad\quad= P(B|AC)P(AC)\\ \qquad\quad\quad= P(C|AB)P(AB)$
But if we see according to the chain rule $P(ABC) = P(A)P(B|A)P(C|A,B)$
So why the two are different is there a mistake in my understanding
Kindly enlighten me
Regards, Siddartha C.S
On
They are all true, so there is no inconsistency.
More generally by symmetry
$\quad P(A,B,C) \\ = P(A\mid B,C)P(B,C) = P(A\mid B,C)P(B\mid C)P(C)= P(A\mid B,C)P(C\mid B)P(B)\\ = P(B\mid A,C)P(A,C) = P(B\mid A,C)P(A\mid C)P(C)= P(B\mid A,C)P(C\mid A)P(A)\\ = P(C\mid A,B)P(A,B) = P(C\mid A,B)P(A\mid B)P(B)= P(C\mid A,B)P(B\mid A)P(A)\\ $
where you can see your first block of equalities in the left column and your chain rule equalities at the start
On
They are equivalent. It is just varying the degree to which is applied the definition of conditional probability.
$\def\P{\operatorname{\mathsf P}}\P(A_1,A_2,A_3) ~{= \P(A_1)\P(A_2,A_3\mid A_1)\\= \P(A_1)\P(A_2\mid A_1)\P(A_3\mid A_1,A_2) \\ = \P(A_1,A_2)\P(A_3\mid A_1,A_2) \\ ~\vdots~\textit{et cetera}}$
Using the "product" notation
$\P(ABC) ~{= \P(A)\P(BC\mid A) \\ = \P(A)\P(B\mid A)\P(C\mid AB) \\ = \P(AB)\P(C\mid AB) \\ ~\vdots~\textit{et cetera}}$
$$P(A|BC)P(BC) = \frac{P(ABC)}{P(BC)}P(BC)=P(ABC)$$
$$P(B|AC)P(AC) = \frac{P(ABC)}{P(AC)}P(AC)=P(ABC)$$
$$P(C|AB)P(AB) = \frac{P(ABC)}{P(AB)}P(AB)=P(ABC)$$