ONT Re: Logic Of Relatives
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LOR. Note 65
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Let me try to sum up as succinctly as possible the lesson
that we ought to take away from Peirce's last "number of"
example, since I know that the account that I have given
of it so far may appear to have wandered rather widely.
| So if men are just as apt to be black as things in general:
|
| [m,][b] = [m,b]
|
| where the difference between [m] and [m,] must not be overlooked.
|
| C.S. Peirce, CP 3.76
In different lights the formula [m,b] = [m,][b] presents itself
as an "aimed arrow", "fair sample", or "independence" condition.
I had taken the tack of illustrating this polymorphous theme in
bas relief, that is, via detour through a universe of discourse
where it fails. Here's a brief reminder of the Othello example:
B C D E I J O
o o o o o o o 1
| | | |
| | | | m,
| | | |
o o o o o o o 1
|
| b,
|
o o o o o o o 1
B C D E I J O
The condition, "men are just as apt to be black as things in general",
is expressible in terms of conditional probabilities as P(b|m) = P(b),
written out, the probability of the event Black given the event Male
is exactly equal to the unconditional probability of the event Black.
Thus, for example, it is sufficient to observe in the Othello setting
that P(b|m) = 1/4 while P(b) = 1/7 in order to cognize the dependency,
and thereby to tell that the ostensible arrow is anaclinically biased.
This reduction of a conditional probability to an absolute probability,
in the form P(A|Z) = P(A), is a familiar disguise, and yet in practice
one of the ways that we most commonly come to recognize the condition
of independence P(AZ) = P(A)P(Z), via the definition of a conditional
probability according to the rule P(A|Z) = P(AZ)/P(Z). To recall the
familiar consequences, the definition of conditional probability plus
the independence condition yields P(A|Z) = P(AZ)/P(Z) = P(A)P(Z)/P(Z),
to wit, P(A|Z) = P(A).
As Hamlet learned, there's a lot to be learned from turning a crank.
Jon Awbrey
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