From: Dan Fabulich (dfabulich@warpmail.net)
Date: Thu Jun 19 2003 - 03:48:07 MDT
Eliezer S. Yudkowsky wrote:
> Dan Fabulich wrote:
> >
> > I think you're getting yourself confused with the language of
> > "unconstrained by territory". I also note that you simply snipped my
> > Bayesian argument, which I took to be the meat of my point.
> >
> > Given a theory T and our background theory B, we regard our theory T to
> > *more* likely if P(T|~B) is higher, *regardless* of P(B), all else being
> > equal. Do you contest this for some reason?
>
> It's a Bayesian answer to a most un-Bayesian question. We do not want the
> probability of our theory P(T) to be as high as possible. Seriously. We
> don't. There is no "our" theory in rational thinking. We want P(T) to be
> whatever P(T) really should be. We do not want to argue *for* P(T), we
> want to equably assemble all the evidence for and against T. This being
> the case, I as a human find it quite highly suspicious when people try to
> have their cake and eat it too.
This completely missed the point. The point is that we want to pick the T
with the highest probability. I'm not trying to MAKE P(T|~B) higher so as
to make P(T) turn out better for my favorite T, I'm suggesting that we
ought to select T over T', under which P(T|~B) is low; that, furthermore,
if we can find a T* that's even *more* likely, we should switch to that
instead.
> It moreover follows from Bayes' Theorem that if B is evidence for T,
> then ~B *must* be evidence against T - you *cannot* have your cake and
> eat it too.
The degree of confirmation that B is for T is given by the log of the
likelihood ratio
(lR) log (P(T|B)/P(T|~B)).
You're right, it can't be both positive and negative. But what of that?
I'm precisely proposing a case where T should be (largely) independent
from B, or at least more independent from B than T'. (Note that T is
fully independent from B only when lR is 0; but we can say that T is
*more* independent from B than T' if the lR of T is closer to 0 than the
lR of T'.)
> > Let's suppose I agree with you that any theory T for which P(T|B) and
> > P(T|~B) are both high is less "constrained by the territory" than an
> > alternate theory T', under which P(T'|B) = P(T|B), but P(T'|~B) is very
> > low. T' is constrained by the territory. T is less constrained by the
> > territory. Are you trying to tell me, against Bayes, that we should hold
> > T' to be more likely than T, because T is "mere philosophy" whereas T' is
> > "constrained by the facts"?
>
> Nope! Reality has no preference one way or the other. Therefore
> neither should our theories. If T and T' start out being mostly equal,
> then in the situation you name, we should slightly prefer T to T'.
Well, that's my whole point! You smear T as "mere philosophy", but my
whole point is that we should slightly *prefer* it for its modularity, not
reject it.
> *However* this is because the prior probability of T' starts out lower,
> and then, if we observe B, this will be evidence about T' which raises
> it to the around the level already occupied by T; however we haven't
> observed B yet.
Sure.
> You cannot say that both B and ~B are arguments for T - that is not
> possible.
Of course not. I'm saying that T is more independent of B than T', and
therefore, all else being equal, preferable.
> The most you can say is that T is about evenly compatible with B and ~B
> such that B does not interact much with T and is not evidence about T one
> way or the other. Interpreting P(T|B) and P(T|~B) as "high" does not mean
> that both B and ~B are strong arguments for T; this is *impossible*. It
> can only mean that the prior probability of T is high and that B is not
> much additional evidence one way or the other.
Yup. That's good ol' modularity for you. You have a problem with that?
> > But that's not at all the case in my T & T' example, by construction, and
> > it doesn't apply to our "Why believe the truth" argument either.
> > Nobody's saying that we shouldn't consider the probability of a
> > Singularity [P(T|B)] in our calculations. But we ARE saying that a theory
> > that applies well to Singularitarians and non-Singularitarians alike is a
> > better theory, more likely to be true, than a theory that applies equally
> > well to Singularitarians but not at all to non-Singularitarians, ceteris
> > paribus.
>
> The key word is that "ceteris paribus". It never is. Ceteris paribus,
> what you are saying is true, but ceteris is only paribus if you just
> happened to stumble across two theories T and T' and you're wondering
> vaguely which of them is more accurate. This business of trying to
> deliberately assemble evidence *for* a theory is never Bayesian. You
> are just trying to sum up all the support and anti-support you run
> across. If it happens to support the theory, great, but you can never
> *try* to make a theory stronger by *trying* to show it is compatible
> with ~B if that is not a natural way for the theory to work.
So, here's where I think you've got me confused. You see me as saying:
"Hmm. I wanna believe claim X. It's certainly probable given S. How
could I make it more probable given ~S?"
But I'm not saying/doing this at all. I'm saying: "You've got an argument
A that entails X. A is certainly sound given S, but it depends on S; if
~S, A is probably not sound. We should keep an eye out for some argument
A' (which might also entail X) that's independent of S." And, indeed, I
take myself to have offered a few such arguments, arguments which are
independent of Singularitarianism.
These independent arguments are better arguments for X; how much better
simply depends on how likely a near-future Singularity may be.
-Dan
-unless you love someone-
-nothing else makes any sense-
e.e. cummings
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