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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.

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1  
Presumably you have some concrete problem in mind. Without further information, $P(B|A)$ could be $0$, or it could be $1$, or anywhere in between. The ratio you wrote down could be huge negative. It is never greater than $1$. If there are additional details that you know, perhaps a more explicit answer could be supplied. –  André Nicolas Oct 4 '11 at 3:03
    
That is all to the question. the value of P(B|A) can be anything. So the upper bound of the equation is 1. How would one solve to a possible lower bound. Thank you. –  Jimmy Oct 4 '11 at 3:06
    
I will give an answer to the lower bound issue, because it is unpleasant to type it as a series of comments. –  André Nicolas Oct 4 '11 at 3:10
    
When does the probability function get its minimum and maximum values. –  Jimmy Oct 4 '11 at 3:17
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I am concerned about your use of the term "probability function." For a probability function, you would need first of all a sample space. And as has been observed in my comments and answer, the expression you gave can in some situations be negative. Then certainly it cannot be interpreted as a probability. –  André Nicolas Oct 4 '11 at 3:48

2 Answers 2

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.

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Thank you for that great analysis. I was wondering what is A -> B? –  Jimmy Oct 4 '11 at 3:28
    
Your meaning is not clear to me. One can make sense of $A \implies B$ in a probabilistic context, but you would have to frame a specific question. –  André Nicolas Oct 4 '11 at 3:41

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.

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