Probability Range? How do you find the range ( min and max) for a probability function such as $$\frac{P (B|A) − P (B)} {1−P (B)}\;?$$
What I tried was to use Venn diagrams, but I couldn't find a solution as the circles must overlap completely to max P (B|A). But that causes P(B) to decrease.
Any thoughts?
Thank you.
 A: No lower bound: We give an example to show that the expression 
$$\frac{P(B|A)-P(B)}{1-P(B)}$$
can be arbitrarily large negative. 
Imagine tossing a very unfair coin which has probability of head equal to $10^{-6}$, and therefore probability of tail equal to $1-10^{-6}$.
Let $B$ be the event "tail" and let $A$ the event "head."  It is clear that $P(B|A)=0$.
Thus our expression is equal to 
$$\frac{0 -(1-10^{-6})}{1-(1-10^{-6})}.$$
This simplifies to $-999999$.
By choosing $10^{-66}$ instead of $10^{-6}$, we can make our expression inconceivably huge negative.  So there is no universal lower bound for our expression.  (If suitable restrictions are put on $B$ and $A$, there may be a lower bound.)
An upper bound: Choose any $A$ such that $P(B|A)=1$, for example choose $A=B$.  Then our expression is equal to $1$.  It cannot ever be larger than $1$, since for given $P(B)$, the numerator is maximized if $A$ is such that $P(B|A)=1$.  Thus $1$ is an upper bound for our expression. This upper bound can be attained, so there is no cheaper universal upper bound.
A: As another way of looking at André Nicolas's solution, note that your expression is the same as  $$1-\dfrac{\Pr(\text{not }B|A)}{\Pr(\text{not }B)}$$  where the quotient can take any non-negative value and so the whole expression can take any value from 1 downwards including any negative value.  
As he says in a comment, this may be a function of probabilities but it is not a probability itself. 
