Re: More Hard Problems Using Bayes' Theorem, Please

From: Dan Fabulich (dfabulich@warpmail.net)
Date: Tue Jul 08 2003 - 22:35:37 MDT

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    Eliezer S. Yudkowsky wrote:

    > >>>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.
    >
    > No, Lee, it's perfectly straightforward to solve the problem from these
    > premises.
    >
    > Observe:
    >
    > p(deformity) = 0.01
    > p(~deformity) = 0.99
    > p(rubella|~deformity) = 0.1
    > p(rubella|deformity) = 0.5
    > p(rubella&~deformity) = p(r|~d)p(~d) = .1*.99 = .099
    > p(rubella&deformity) = p(r|d)p(d) = .5*.01 = .005
    > p(rubella) = p(r&d) + p(r&~d) = .104
    > p(deformity|rubella) = p(rubella&deformity)/p(rubella) = .005/.104 = .048

    Now, wait a minute.

    You begin with p(deformity) = 0.01. I presume that's your interpretation
    of the claim that "a newborn will have deformities traceable to a sickness
    of its mother during pregnancy", but that doesn't seem right to me at all.
    [Obviously p(deformity) should include cases where there are deformities
    that aren't traceable to a sickness of the mother.]

    Here's my interpretation of the problem.

    p(deformity&traceable) = 0.01
    p(rubella|~deformity) = 0.1
    p(rubella|deformity&traceable) = 0.5

    Where the question is: what's p(deformity|rubella)?

    Now, perhaps I can assume that deformity&rubella entails traceable, but I
    can't just assume that deformity->traceable. (And let's not even get into
    the possibility that the child may not be "born healthy and normal" but
    may have no "deformities".)

    But unless we know p(deformity&~traceable), an obviously non-zero number,
    we can't get p(deformity) or p(~deformity) at all, which we need to get
    p(rubella&~deformity).

    Of course, in light of what they think the answer is, it's obvious that
    they think trivial details like these are irrelevant to the problem;
    perhaps they think my point is a nitpick or something.

    But I'm on Lee's side here... this problem is very badly stated.

    > I would (as stated more briefly in my first reply) recommend that you read
    > through the "Intuitive Explanation" from start to finish without skipping
    > anything; and then, if you are still interested in understanding the
    > difference between Bayesians and frequentists (which is something the
    > intro does not address), I would recommend reading the E.T. Jaynes
    > lectures given in the "Further Reading" section at the end of the intro.
    > Jaynes gives specific examples of cases where frequentist methods are both
    > more complicated than and inferior to Bayesian methods.

    I note here that he has worked out a number of very clear frequentist
    accounts; accounts which don't suffer from as much vagueness as my own,
    but which suffer from their own kinds of problems.

    Note especially section 3.4 and Lecture 5 though his objections against
    frequentism are available througout.

    -Dan

          -unless you love someone-
        -nothing else makes any sense-
               e.e. cummings



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