$\frac{P(H\mid E)}{P(G\mid E)}=\frac{P\left(H\right)}{P\left(G\right)}\frac{P(E\mid H)}{P(E\mid G)}$

Suppose that, before new evidence is observed, the hypothesis $H$is three times as likely to be true as is the hypothesis $G$. If the new evidence is twice as likely when $G$is true than it is when $H$is true, which hypothesis is more likely after the evidence has been observed?

To prove:$\frac{P(H\mid E)}{P(G\mid E)}=\frac{P\left(H\right)}{P\left(G\right)}\frac{P(E\mid H)}{P(E\mid G)}$.

For events $A$and $B$such that $P\left(B\right)\ne 0$(so that the fraction is defined):

$P(A\mid B):=\frac{P(A\cap B)}{P\left(B\right)}$

Start from the left hand side, apply this definition:

$\frac{P(H\mid E)}{P(G\mid E)}=\frac{\frac{P(H\cap E)}{P\left(E\right)}}{\frac{P(G\cap E)}{P\left(E\right)}}=\frac{P(H\cap E)}{P(G\cap E)}$

And by using the same definition on the right hand side:

$\frac{P\left(H\right)}{P\left(G\right)}·\frac{P(E\mid H)}{P(E\mid G)}=\frac{P\left(H\right)}{P\left(G\right)}·\frac{\frac{P(H\cap E)}{P\left(H\right)}}{\frac{P\left(G\right)E)}{P\left(G\right)}}=\frac{P(H\cap E)}{P(G\cap E)}$

As both sides equal to the same expression the properties of equality prove the given statement.

Given statements are transferred into mathematical equations:$P\left(H\right)=3P\left(G\right)$

$P(E\mid G)=2P(E\mid H)$

Using the equation from part I):

$\frac{P(H\mid E)}{P(G\mid E)}=\frac{P\left(H\right)}{P\left(G\right)}·\frac{P(E\mid H)}{P(E\mid G)}=\frac{P\left(H\right)}{P\left(G\right)}·\frac{P(E\mid H)}{P(E\mid G)}$

Substitute the given facts:

Equating the first and the last expression, and multiplying by $P(G\mid E)$ :

To prove the identity use the definition of conditional probability

$P(H\mid E)>P(G\mid E)$ this follows from the given statements and the proven equation.