From: Eliezer S. Yudkowsky (sentience@pobox.com)
Date: Thu Sep 08 2005 - 10:22:55 MDT
Ben Goertzel wrote:
>
> I am curious how you propose to handle Hempel's paradox of confirmation
> using probabilistic semantics alone.
>
> There are many solutions to this problem, but I'm curious which one you
> advocate...
Like I said, this challenge was meant for Loosemore alone; I don't have
enough time to respond if everyone piles on...
Googling on Hempel + Bayes turns up some standard responses, for example
in http://plato.stanford.edu/entries/epistemology-bayesian/
"Hempel first pointed out that we typically expect the hypothesis that
all ravens are black to be confirmed to some degree by the observation
of a black raven, but not by the observation of a non-black, non-raven.
Let H be the hypothesis that all ravens are black. Let E1 describe the
observation of a non-black, non-raven. Let E2 describe the observation
of a black raven. Bayesian Confirmation Theory actually holds that both
E1 and E2 may provide some confirmation for H. Recall that E1 supports H
just in case Pi(E1/H)/Pi(E1) > 1. It is plausible to think that this
ratio is ever so slightly greater than one. On the other hand, E2 would
seem to provide much greater confirmation to H, because, in this
example, it would be expected that Pi(E2/H)/Pi(E2) >> Pi(E1/H)/Pi(E1)."
A fine answer so far as it goes, though really it is only half the
solution. The two main points to bear in mind are that, first, a
hypothesis tells us which experiences to anticipate, not which facts to
believe. Second, evidence never confirms or disconfirms a hypothesis by
itself, it only confirms or disconfirms one hypothesis relative to an
alternative hypothesis. So I need to know what the alternative
hypothesis is to "All ravens are black", and I need to know what
probability both hypotheses assign to seeing a red lampshade given the
local sampling method. Then I'll tell you whether a red lampshade
confirms "All ravens are black" over the alternative hypothesis.
In the usual scenario, we suppose that objects are sampled randomly from
the set of all objects. "All ravens are black" does not tell us with
what probability we ought to observe a raven. It does not even imply
that any ravens exist at all. The strength of a hypothesis is what it
tells us *not* to expect. "All ravens are black" only excludes a very
tiny amount of probability mass, the possibility that we will observe a
raven-shaped object whose color is not black. Since, a priori, the
chance of seeing a raven-shaped object was 1/2 to the Kolmogorov
complexity of "raven-shaped", the hypothesis excludes only a tiny realm
of possibilities, and therefore only slightly concentrates its force
into other possibilities relative to the hypothesis of maximum entropy.
Therefore seeing a red lampshade is only infinitesimal evidence in
favor of the hypothesis that all ravens are black, as compared to the
null hypothesis. If, on the other hand, we have already seen at least
one raven, then this confirms other hypotheses which assign a
substantial probability to seeing more raven-shaped objects. And when
we do see another raven, its color will confirm subclasses of these
hypotheses which say "All ravens are black", over alternative hypotheses
that permit ravens to be more colors.
Since "All ravens are black" and "All ravens are white" exclude around
the same amount of probability mass from the maxentropy distribution
(thereby concentrating it into other possibilities), seeing a red
lampshade does not confirm one hypothesis over the over, and
infinitesimally confirms both over the null hypothesis.
But it really depends on the hypotheses that seem likely a priori, and
even more so on the sampling method. Suppose there are only two highly
probable hypotheses under consideration. The first hypothesis, that in
a group all the ravens are black and one-third of the lampshades are
red. The second hypothesis, that in a group not all ravens are black
and one-ninth of the lampshades are red. If the two hypotheses have
equal prior odds, seeing a red lampshade randomly sampled from the
subgroup of lampshades, makes it three times as probable that all ravens
are black as that not all ravens are black.
> In PTL (Novamente's probabilistic inference component) we handle
> this sort of thing via augmenting probability theory with
> other mathematics.
I shall now demonstrate the folly of adulterating Bayes with lesser wares.
Suppose that I know that, in a certain sample, there is at least one
black raven, and at least one blue teapot, and some number of other
ravens of unknown color. I now observe an item from the group that is
produced by the following sampling method: Someone looks over the
group, and if there are no non-black ravens, he tosses out a blue
teapot. If there are non-black ravens, he tosses out a black raven.
Now observing a black raven definitely shows that not all ravens are black.
How would Novamente's "augmented" probability theory handle that case, I
wonder?
From http://www.iep.utm.edu/h/hempel.htm:
"Hempel's paradoxes are a straightforward consequence of the following
apparently harmless principles:
1. the statement (x)(Rx --> Bx) is supported by the statement (Ra & Ba)
2. if P1 and P2 are logically equivalent statements and O1 confirms P1,
then O1 also supports P2."
As the above example demonstrates, to a Bayesian, the idea that "All
ravens are black" is supported by observing a black raven, is in general
false. Though it may oft be true in the particular that observing a
black raven confirms subclasses of hypotheses which hold "All ravens are
black" over their alternatives.
-- Eliezer S. Yudkowsky http://intelligence.org/ Research Fellow, Singularity Institute for Artificial Intelligence
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