Bayes Formula for more than 2 events

Question

I have the following result used in one of the paper that I'm reading. I can't seem to get around how it is derived. Please help.

$$ P(M\mid D,R) = \dfrac{P(M\mid R)\,P(D\mid M)}{P(M\mid R)\,P(D\mid M)+P(\overline M\mid R)\,P(D\mid \overline M)} $$

Answer 1

The formula doesn't hold in general. It only holds if $\ D\ $ is independent of $\ R\ $ given $\ M\ $, and independent of $\ R\ $ given $\ \overline M\ $. The correct general formula is $$P(M\left\vert D,R\right.) = \frac{P(M\left\vert R\right.)P(D\left\vert M,R\right.)}{P(M\left\vert R\right.)P(D\left\vert M,R\right.)+P(\overline M\left\vert R\right.)P(D\vert\overline M,R)}\ ,$$ which reduces to the one you've quoted whenever $\ P(D\left\vert M,R\right.)=P(D\left\vert M\right.)\ $ and $\ P(D\vert\overline M,R)=P(D\vert\overline M)\ $.

To show that the formula you have quoted does not hold in general, consider a single roll of a fair die, and take $\ D=\{1,2,3,4,5\}\ $, $\ M=\{1,2,3,6\}\ $, $\ R=\{4,5,6\}\ $. Here we have $$ P(M\left\vert D,R\right.) = 0\ ,$$ because $\ M\cap D\cap R = \emptyset\ $.

But $$ \frac{P(M\left\vert R\right.)P(D\left\vert M\right.)}{P(M\left\vert R\right.)P(D\left\vert M\right.)+P(\overline M\left\vert R\right.)P(D\vert\overline M)}= \frac{\frac{1}{3}\frac{3}{4}}{\frac{1}{3}\frac{3}{4}+\frac{2}{3}1}=\frac{3}{11}\ .$$

To prove the correct general formula, take the terms in the denominator of the left expression, \begin{eqnarray} P(M\left\vert R\right.)P(D\left\vert M, R\right.) &=& \frac{P(M\cap R)P(D\cap M\cap R)}{P(R)P(M\cap R)}\\ &=& \frac{P(D\cap M\cap R)}{P(R)}\ \ \ \ \ \mbox{and}\\ P(\overline M\left\vert R\right.)P(D\vert \overline M, R ) &=&\frac{P(D\cap \overline M\cap R)}{P(R)}\ , \end{eqnarray} and add them together to get \begin{eqnarray} P(M\left\vert R\right.)P(D\left\vert M,R\right.)+P(\overline M\left\vert R\right.)P(D\vert\overline M,R)&=&\frac{P(D\cap M\cap R)}{P(R)}\\ &&+\frac{P(D\cap \overline M\cap R)}{P(R)}\\ &=&\frac{P(D\cap R)}{P(R)}\ . \end{eqnarray} It then follows that \begin{eqnarray} \frac{P(M\left\vert R\right.)P(D\left\vert M,R\right.)}{P(M\left\vert R\right.)P(D\left\vert M,R\right.)+P(\overline M\left\vert R\right.)P(D\vert\overline M,R)}&=&\frac{P(D\cap M\cap R)}{P(D\cap R)}\\ &=& P(M\left\vert D,R\right.)\ . \end{eqnarray}