RE: Bayes' formula can yield P>1
May 4, 2014 at 12:18 am
(This post was last modified: May 4, 2014 at 12:22 am by Categories+Sheaves.)
Oh, cool. That's actually a pretty cute mistake--or feat of misdirection, not sure which. But still cool!
The problem comes up because the spread of values you assigned is inconsistent. IT should be straightforward to see that P(O and T) = P(O|T)*P(T) = (3/100)*(4/10) = 12/1000
But P(O) = 1/100 = 10/1000 < 12/1000, which means we've already made a contradiction. Having O happen is a strictly weaker condition than having O happen and having T be true, so P(O) should never be less than P(O and T). But that's a consequence of the values you assigned, so there can't exist a probability space with those properties in the first place. Since we're only supposed to use Bayes' theorem on well-defined probability measures, your example 2 isn't a problem for the theorem
You understand? Ja? Did I do good at making sense?
The problem comes up because the spread of values you assigned is inconsistent. IT should be straightforward to see that P(O and T) = P(O|T)*P(T) = (3/100)*(4/10) = 12/1000
But P(O) = 1/100 = 10/1000 < 12/1000, which means we've already made a contradiction. Having O happen is a strictly weaker condition than having O happen and having T be true, so P(O) should never be less than P(O and T). But that's a consequence of the values you assigned, so there can't exist a probability space with those properties in the first place. Since we're only supposed to use Bayes' theorem on well-defined probability measures, your example 2 isn't a problem for the theorem
You understand? Ja? Did I do good at making sense?
