On Tue, 26 Oct 1999, Russell Standish wrote:
[JM wrote] [&BTW I am getting tired of RS omitting the attribution]
> > 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.
I must disrespectfully disagree.
It is obvious that p(Y1&X) = p(Y1&Z), because in all instances in
which there is an observation with Y1 & X, there is observation by Joe
Shmoe with Y1 & Z, of equal measure. That's why I added the extra
conditions, to make it real obvious.
(Since there are no near-zombies in the ASSA. They are both there
(in that branch/universe), both with human brains, so they get the same
measure.)
So p(Y1&X) = p(Y2&Z) = p(Y1&Z). OK we have shown it for Joe,
Jack's case works the same way: p(Y1&X) = p(Y2&X).
- - - - - - -
Jacques Mallah ([EMAIL PROTECTED])
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/
