**From:** Lee Corbin (*lcorbin@tsoft.com*)

**Date:** Tue Jul 08 2003 - 19:13:28 MDT

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I wrote

*> I can't answer your question, but did you check out problem
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*> number three on my list? Apparently Bayesian statisticians
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*> can solve it whereas those like me trained in classical
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*> mathematics, do not have enough information. I'm still in
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*> shock that not in all instances does P(A & B) = P(A|B)*P(B)
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*> according to the Bayesians.
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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
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*> Rosetta Stone that does not require talking of distributions
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*> to see Bayesianism rear its head. (I do have one smaller
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*> and much easier problem that lies bare the differences, but
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*> this here might be much hotter stuff.)
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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
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*> > traceable to a sickness of its mother during pregnancy is 1%.
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*> > If a child is born healthy and normal, the probability that
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*> > the mother had rubella during her pregnancy is 10%. If a
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*> > child is born with deformities that can be traced to a
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*> > sickness of the mother, the probability that the mother had
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*> > rubella during her pregnancy is 50%. What is the probability
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*> > that a child will be born with deformities if its mother had
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*> > rubella during her pregnancy?
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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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