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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

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  • $\begingroup$ In what exactly they are different? They are actually the same. See for instance the third equality and expand $P(AB)$ through the chain rule. Consider also that $P(AB)=P(A,B)$ (in both case they indicate the probability of both $A$ and $B$ happening). $\endgroup$
    – nicola
    Commented Jul 14, 2017 at 9:43

3 Answers 3

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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

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$$P(A|BC)P(BC) = \frac{P(ABC)}{P(BC)}P(BC)=P(ABC)$$

$$P(B|AC)P(AC) = \frac{P(ABC)}{P(AC)}P(AC)=P(ABC)$$

$$P(C|AB)P(AB) = \frac{P(ABC)}{P(AB)}P(AB)=P(ABC)$$

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  • $\begingroup$ Sorry I did not understand, can you please add more details? $\endgroup$
    – Siddarth
    Commented Jul 14, 2017 at 9:31
  • $\begingroup$ @Siddarth Define $D=BC$ i.e. $D$ is the intersection of events $B$ and $C$. Now, use the chain rule to get $P(AD) = P(D)P(A|D)$ and then replace back $D=BC$. $\endgroup$
    – Dhruv Kohli
    Commented Jul 14, 2017 at 10:14
  • $\begingroup$ Just thinking can this be also true $P(A|BC) = P(A)*P(B|A)*P(C|A)\\P(A)*P(BC|A))+P(!A)*P(BC|!A)$ for the above equation? $\endgroup$
    – Siddarth
    Commented Jul 17, 2017 at 14:55
  • $\begingroup$ No. Just use $P(X|Y) = \frac{P(X,Y)}{P(Y)}$ to see for yourself. $\endgroup$
    – Dhruv Kohli
    Commented Jul 17, 2017 at 15:14
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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

$\P(ABC) ~{= \P(A)\P(BC\mid A) \\ = \P(A)\P(B\mid A)\P(C\mid AB) \\ = \P(AB)\P(C\mid AB) \\ ~\vdots~\textit{et cetera}}$

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  • $\begingroup$ :Can i write this as $P(ABC) = P(A)P(B|A)P(C|A) = P(AB)P(C|A)?$ $\endgroup$
    – Siddarth
    Commented Jul 19, 2017 at 1:40
  • $\begingroup$ No, not generally. Only in the special case where $B,C$ are conditionally independent for given $A$. $\endgroup$
    – Graham Kemp
    Commented Jul 19, 2017 at 1:46

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