Q. 3.8 - A First Course in Probability

Question

Show that 

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

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 as it is when $H$ is true, which hypothesis is more likely after the evidence has been observed?

Step-by-Step Solution

Answer

We use the definition of condition probability, $P\left(H\right|E)>P(G\left|E\right)$ this follows and proven in the equation.

Step1: Definition of Condition probability

The possibility of a happening or outcome occurring supported by the occurrence of a preceding event or outcome is understood as a contingent probability. 

The updated probability of the next or conditional, event is multiplied by the probability of the preceding, or conditional, event to get introduce contingent probability.

Step2: Prove the statement

Prove:

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

$P\left(B\right)$ isn't capable $0$ for occurrences $A$and $B$

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

Begin with the LHS and apply the definition:

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

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

and on the RHS, using the identical definition:

$\frac{P\left(G\right)}{P\left(H\right)}\times \frac{P\left(E\right|H)}{P\left(E\right|G)}=\frac{P\left(H\right)}{P\left(G\right)}\times \frac{{\displaystyle \frac{P(H\cap E)}{P\left(H\right)}}}{{\displaystyle \frac{P(G\cap E)}{P\left(G\right)}}}$

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

As a result, it's been established.

Step3: To search out P ( H ∣ E ) > P ( G ∣ E ) this follows and proven within the equation.

The following statements are converted into mathematical equations.

$P\left(H\right)=3P\left(G\right)$

$P\left(E\right|G)=2P(E\left|H\right)$

Using the formula,

$$\begin{gathered} \frac{P\left(H\right|E)}{P\left(G\right|E)}=\frac{P\left(H\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{P\left(E\right|G)} \\ =\frac{P\left(H\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{P\left(E\right|G)} \end{gathered}$$

Replace the given facts together with your own.

$$\begin{gathered} \frac{P\left(H\right|E)}{P\left(G\right|E)}=\frac{3P\left(G\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{2P\left(E\right|H)} \\ =3\times \frac{1}{2} \\ =1.5 \end{gathered}$$

multiplying by $P\left(G\right|E)$ and equating the primary and last expressions:

$P\left(H\right|E)=1.5\times P(G\left|E\right)$$\Rightarrow P(H\mid E)>P(G\mid E)$