# Bayes’ formula conditions

I was just wondering if given Bayes’ formula, $P(A|B) = \frac{ P(B|A) P(A)}{P(B)}$

can one always claim that $0 < \frac{P(B|A)}{P(B)} < 1$?

Can a proof be given?

Note: A and B are different events.

Thanks.

• Are you sure you're not asking whether $P(B|A)P(A)/P(B)\in (0,1)$? That is, there is a $P(A)$ missing in the numerator? – Herr K. Feb 8 '18 at 23:01
• Yes, I’m sure. I’m asking specifically about that ratio – Thomas Moore Feb 8 '18 at 23:04
• Then @ConMan's answer is it. See the last sentence in his answer, where he covered the possibility that A and B are different events. – Herr K. Feb 8 '18 at 23:17

No, because it's not true.

Suppose I flip a coin. Let A = "the coin is heads". Let B be the same event. Then $P(B) = 0.5$, but $P(B | A) = 1$, because in the event the coin is heads, the coin is definitely still heads. Thus, $\frac{P(B | A)}{P(B)} = 2$.

Similarly, if B = "the coin is tails", then $P(B | A) = 0$, so $\frac{P(B | A)}{P(B)} = 0$.

• I picked a deliberately obvious counterexample, but Graham shows the more general case. I could also point out that if A and B are independent, then P(B|A) = P(B) and so the ratio is exactly 1. – ConMan Feb 9 '18 at 3:17

Can a proof be given?

No, it is not always so.   Many counterexamples exist.

We just need any events $A,B$ where outcomes for event $A$ are underrepresented within event $B$ than within the sample space.

Let $\Omega =\{1,2,3,4,5,6\}$ be the outcome set for a sample space with no bias between outcomes.   Let our events of interest be: $B=\{1,2,3\}$ and $A=\{2,4,6\}$.   Ie: roll a fair six sided dice, and call $B$ the event of rolling $3$ or less and $A$ the event of rolling an even result.$$\begin{split}\mathsf P(A)&=1/2\\\mathsf P(A\mid B)&=1/3\\\mathsf P(B)&=1/2\\\mathsf P(B\mid A)&=1/3\\[2ex]\frac{\mathsf P(B\mid A)}{\mathsf P(B)}&=\frac 32\end{split}$$