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?

Short 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.

Step-by-step solution

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.

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.

To search outP(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.

localid="1649658537441" $$\begin{gathered} \frac{P\left(H\right|E)}{P\left(G\right|E)}=\frac{3P\left(G\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{2P\left(E\right|H)} \\ =3\times \frac{1}{2} \\ =1.5 \end{gathered}$$

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)$