>
> On Wed, 20 Oct 1999, Russell Standish wrote:
> > The measure of Jack Mallah is irrelevant to this situation. The
> > probability of Jack Mallah seeing Joe Schmoe with a large age is
> > proportional to Joe Schmoe's measure - because - Joe Schmoe is
> > independent of Jack Mallah. However, Jack Mallah is clearly not
> > independent of Jack Mallah, and predictions of the probability of Jack
> > Mallah seeing a Jack Mallah with large age cannot be made with the
> > existing assumptions of ASSA. The claim is that RSSA has the
> > additional assumptions required.
>
> That's total BS.
> I'll review, although I've said it so many times, how effective
> probabilities work in the ASSA. You can take this as a definition of
> ASSA, so you can NOT deny that this is how things would work if the ASSA
> is true. The only thing you could try, is to claim that the ASSA is
> false.
> The effective probability of an observation with characteristic
> 'X' is (measure of observations with 'X') / (total measure).
> The conditional effective probability that an observation has
> characteristic Y, given that it has characteristic X, is
> p(Y|X) = (measure of observations with X and with Y) / (measure with X).
> OK, these definitions are true in general. Let's apply them to
> the situation in question.
> 'X' = being Jack Mallah and seeing an age for Joe Shmoe and for
> Jack Mallah, and seeing that Joe also sees both ages and sees that Jack
> sees both ages.
I shall take X = being Jack Mallah. The rest is irrelevant.
> Suppose that objectively (e.g. to a 3rd party) Jack and Joe have
> their ages drawn from the same type of distribution. (i.e. they are the
> same species).
> Case 1: 'Y1' = the age seen for Joe is large.
> Case 2: 'Y2' = the age seen for Jack is large.
> Clearly P(Y1|X) = P(Y2|X).
Sorry, not so clear. It is true by symmetry that p(Y1)=p(Y2).
p(Y1|X) = p(Y1&X)/p(X)
p(Y2|X) = p(Y2&X)/p(X)
Why do you assume p(Y1&X) = p(Y2&X)? I can see no reason. They
certainly aren't symmetrical. About all one can say from symmetry is
p(Y1&X) = p(Y2&Z), where Z = being Joe Schmoe.
Incidently, if you took X to be being anyone (pretty much what you do
by assuming the long clause you gave above) then clearly
p(Y1|X)=p(Y1)=p(Y2)=p(Y2|X). As I said before, though, this has no
relevance to the QTI issue.
>
> - - - - - - -
> Jacques Mallah (jqm1584.domain.name.hidden)
> Graduate Student / Many Worlder / Devil's Advocate
> "I know what no one else knows" - 'Runaway Train', Soul Asylum
> My URL: http://pages.nyu.edu/~jqm1584/
>
>
>
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Dr. Russell Standish Director
High Performance Computing Support Unit,
University of NSW Phone 9385 6967
Sydney 2052 Fax 9385 6965
Australia R.Standish.domain.name.hidden
Room 2075, Red Centre
http://parallel.hpc.unsw.edu.au/rks
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Received on Mon Oct 25 1999 - 18:02:07 PDT