From: Lee Corbin (lcorbin@tsoft.com)
Date: Tue Jul 08 2003 - 19:13:28 MDT
I wrote
> I can't answer your question, but did you check out problem
> number three on my list? Apparently Bayesian statisticians
> can solve it whereas those like me trained in classical
> mathematics, do not have enough information. I'm still in
> shock that not in all instances does P(A & B) = P(A|B)*P(B)
> according to the Bayesians.
Those remarks, as well as the problem below, come from a link
http://www.apa.org/journals/xge/press_releases/september_2001/xge1303380.html
in an email written by Damien Sullivan on January 26, in the
thread entitled "Teaching Bayesian Inference in Two Hours".
I continued:
> Anyway, problem number three is hopefully my long sought
> Rosetta Stone that does not require talking of distributions
> to see Bayesianism rear its head. (I do have one smaller
> and much easier problem that lies bare the differences, but
> this here might be much hotter stuff.)
Well, now it seems to me that in this particular problem, it's
easy to think, wrongly, that an answer is possible. So I apologize
for putting it on the list (at the time, I thought that I just
wasn't quick enough.) I still maintain that there is not enough
information in it. But now I believe that that goes for
Bayesians as well as for everyone else.
> > 3. The probability that a newborn will have deformities
> > traceable to a sickness of its mother during pregnancy is 1%.
> > If a child is born healthy and normal, the probability that
> > the mother had rubella during her pregnancy is 10%. If a
> > child is born with deformities that can be traced to a
> > sickness of the mother, the probability that the mother had
> > rubella during her pregnancy is 50%. What is the probability
> > that a child will be born with deformities if its mother had
> > rubella during her pregnancy?
In this stupid problem, the authors badly misstate at least one
of the premises, in my opinion. If you draw a picture of the
problem, perhaps, you may be less likely to misread it than if
you plug the numbers into a formula.
And the heresy P(AB) is not equal to P(A|B)*P(B) appears in this
quote from the above paper:
In Germany, every expectant mother must have an
obligatory test for rubella infection because
children born to women who have rubella while
pregnant are often born with terrible deformities.
The following information is at your disposal:
The probability that a newborn will have deformities
traceable to a sickness of its mother during pregnancy is 1%.
If a child is born healthy and normal, the probability
that the mother had rubella during her pregnancy is 10%.
If a child is born with deformities and it can be traced
to some sickness of the mother, the probability that the
mother had rubella during her pregnancy is 50%.
What is the probability that a child will be born with
deformities if its mother had rubella during her pregnancy?
The Bayesian solution p (H | D) is .048. But participants
who use one of two non-Bayesian algorithms, computing
p (H&D) = .005 or picking p (H) = .01, will produce
estimates that lie in the interval of ±5 percentage points
around the Bayesian solution.
But why shouldn't p(H&D) --- that is, p(defect & rubella) be equal
to .005? What other conclusion can one come to from the first
statement together with the third statement?
Unless, they fail to believe that p(H&D) = p(H|D)*p(D). Sure
enough, go back a couple of paragraphs in that paper, and find
The most frequent non-Bayesian algorithms they identified
include computing p (H&D) by multiplying p (H) and p (D | H);
No wonder I can't grasp Bayesianism. ;-)
Of course, it is a libel to say that Bayesians would be guilty of
believing that p(H&D) was not equal to p(H|D)*p(D). But I was
hoping that it would turn out to be the case that in some problems
they do not---now I simply believe that the authors are full of shit.
And I think that their study is deeply flawed too, as a predictable
result of inflicting this problem on a lot of innocent victims.
Lee
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