Re: Calculus XOR Probability

On Tue, 21 Mar 2006 16:08:29 +0100, Han de Bruijn
<Han.deBruijn@xxxxxxxxxxxxxx> wrote:

Robert Low wrote:

Han de Bruijn wrote:

What !? I've resorted to the hyperreals in order to _have_ the
probability of choosing a finite integer. And that probability
turns out to be an infinitesimal in the hyperreals *R.

No, it doesn't. If N is a hyperfinite natural, and you associate
the probability 1/N with each natural 1 <= n <= N, then for any
*finite* n, the probability of choosing a natural less than
or equal to n is infinitesimal.

Yeah. And that's fine, isn't it? So we are finished. But, but, but ...

But the probability of choosing
*some* finite natural would be the sum over all finite naturals
of 1/N. Unfortunately, the finite naturals don't comprise an internal
set, and this sum does not exist as a nonstandard real.

Sorry. I don't understand even a little bit of what you are saying here.

I believe I do. (If you understood the material on that page
that you sent me to to read about the transfer principle
you'd understand what he was saying as well.)

Let's try looking at it this way. Consider three statements:

(i) The (standard) probability that a (standard) integer is
even is 1/2. Or just "can be defined using non-standard
analysis, and then it turns out to be infinitely close to 1/2.")

(ii) The hyper-probability that a hyper-integer is even is
1/2. (Or maybe just infinitely close to 1/2, whatever).

(iii) The hyper-probability that a hyper-integer <= N is even
is (infinitely close to) 1/2.

In (iii) N is that hyper-real you defined above.

Now, possibly you've proved (iii). I'd want to be much more
careful about definitions and axioms, but let's assume
_for the sake of argument_ that you've proved (iii).
It does not follow that you've proved (i). The reason is
that the standard integers are just a tiny subset of the
hyper-integers <= N. (For example, every standard integer
is also <= N/2.) Supposing that you _have_ (or could)
prove (iii), that's not much more interesting than the fact
that the probability that an integer <= 4 is even is 1/2.

On the other hand, it's conceievable that if you proved (ii)
then (i) would follow, or some version of (i). But you have
definitely not proved (ii). (Before saying that (ii) does
imply (i) I'd definitely want to see more details. But never
mind that - in fact, assuming that we _do_ extend probability
theory to *R then there's simply no such thing, in *R or
anywhere else, as the probability that a random hyperinteger
is even. The reason there's no such thing is the same as the
reason for R and the standard integers.)

Or maybe you don't buy that. Whether you buy my explanation
of why (ii) is simply meaningless or not, you certainly
have not proved (ii), and (ii) is the only thing that
could possibly give (i) via the transfer principle.

******************

That's not so much an explanation of exactly what Robert
said as an explanation of why mumbling about non-standard
analysis is simply not going to do what you want. A
explanation of what Robert said:

Ok, so now we have this hyper-probability defined on
hyper-naturals <= N. Say it's P. So if n is a random
hyper-natural <= N and n_0 is a specific such hypernatural
then P(n = n_0) = 1/N.

Now then, we can define the probability that a _finite_
natural is even like so:

P(n is finite) / P(n is finite and even).

And that turns out to be infinitely close to 1/2.

Q: What's wrong with that?

A: It doesn't work. For the P that we've defined, there's no
such thing as P(n is finite). The reason is that the finite
naturals are not an internal set; P(n is natural) would be
the sum of 1/N, once for each finite natural, but there is
no such sum in *R.

Q: Why is there no such sum again?

A: Well, you tell me why there _is_ such a thing.
You can't use the transfer principle to say what it
is - for one thing right now we're not "transferring"
anything, we're just talking about *R. And in any
case the transfer principle transfers some statements
about *{standard naturals}, not about the hypernaturals
<= N. If we're talking about the sum of 1/N, once for
each finite natural, we're talking about an infinite sum.
So it doesn't obviously exist. Since *R is not complete
it's not true that any infinite sum with partial
sums bounded above must exist. Why _does_ that sum exist?

The only _definition_ for the sum I can imagine, since
there's nothing in R that we can "transfer", is
the supremum of the finite partial sums. But the
finite partial sums do not _have_ a supremum,
for more or less the same reason as the set of
infinitesmals has no supremum: If s is the sup
of the finite partial sums then it's easy to see
that s = s + s, which cannot happen for s > 0 since
*R is an ordered field.

So, no transfer trick, the obvious definition
doesn't work. What do you even _mean_ by

sum_{n a finite natural} 1/N ?

Han de Bruijn


************************

David C. Ullrich
.

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