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Topic: the probability theory has holes!
Replies: 31   Last Post: Dec 3, 2009 7:55 PM

 Messages: [ Previous | Next ]
 Jim Burns Posts: 1,200 Registered: 12/6/04
Re: the probability theory has holes!
Posted: Dec 3, 2009 11:53 AM

Ostap S. B. M. Bender Jr. wrote:
> On Dec 1, 11:39 am, James Burns <burns...@osu.edu> wrote:
>

>>One of the more pleasant surprises on the Internet
>>is finding someone who will admit that they were wrong
>>(when they finally understand that they were wrong).
>>
>>I wonder, are you going to surprise us?
>>

> I have a surprise for him: the uniform measure on Z+
> (or any other infinite countable set) doesn't exist,
> so there can be no valid probability arguments
> of this sort.

I suppose I understand what you're saying here:
There is no function f: Z+ -> R such that both
f(m) = f(n) for all m,n in Z+
and
Sum_{n in Z+} f(n) = 1
That's easy enough to prove. Certainly, I am not
disputing that.

However, it looks to me as though a good argument
very much like the quick-and-dirty arguments
being used in this thread is possible with
an expanded definition of a measure on Z+,
a definition much like that for generalized
functions, with the Dirac delta function as a
particular example.

As I understand it, the Dirac delta is described
as a function that is 0 everywhere except one
point, but, when it is integrated across that
point, we get 1. This is obviously (and provably)
impossible, as it is stated. The work-around that
I am familiar with uses a sequence of
functions < d_n > such that
d_n(x) -> 0, as n -> infinity, for x <> 0,
and
INT{-inf, +inf} d_n(x) dx = 1, all n
There are a lot of choices for such a sequence.

A rigorous argument involving Dirac deltas
would solve some problem using a convenient
choice for the d_n, get an answer involving
n, and then take the limit as n -> infinity
as the "true" answer. The operations behind
what we call the Dirac delta are completely
unexceptional, but they have the same effect
as the provably impossible function.

My question is: why can't something similar
be done to create a uniform measure on Z+,
or, at least, something that would give us
the answers a uniform measure would, if such
a thing existed?

For concreteness, suppose
f_n : { m in Z+ | m =< n } -> R
where
f_n(m) = 1/n

Suppose we ask what the probability that a
"random number" is not divisible by 2, 3, or 5
(for example), and get a constant plus some
bounds that -> 0 as n -> infinity. Without
doing very much in the way of calculation, I
feel confident that the constant would be
(1/2)(2/3)(4/5).

What is wrong with this argument?

Jim Burns

Date Subject Author
12/1/09 eestath
12/1/09 Kook Spotter
12/1/09 Bart Goddard
12/1/09 eestath
12/1/09 Bart Goddard
12/1/09 eestath
12/1/09 Jim Burns
12/1/09 Bart Goddard
12/1/09 Pubkeybreaker
12/1/09 Bart Goddard
12/1/09 Gerry Myerson
12/1/09 Bart Goddard
12/1/09 eestath
12/2/09 Dik T. Winter
12/2/09 Bart Goddard
12/2/09 eestath
12/2/09 Bart Goddard
12/2/09 eestath
12/3/09 Bart Goddard
12/2/09 eestath
12/1/09 Nick
12/2/09 Dik T. Winter
12/3/09 ostap_bender_1900@hotmail.com
12/3/09 Jim Burns
12/3/09 FredJeffries@gmail.com
12/2/09 Dik T. Winter
12/1/09 Henry
12/1/09 eestath
12/1/09 eestath
12/2/09 Richard Tobin
12/1/09 eestath
12/3/09 ostap_bender_1900@hotmail.com