4.3: 12.13: news.math/probability function over naturals:
. can we define a probability distribution on the set of integers
(rather than the real numbers between 0 and 1)
such that they each occur with equal probability
(i.e. does a uniform distribution on the integers exist)?
". it can not be the case that
each of the infinity of integers
has the same probability
and the probability density function integrates to 1"
. but intuitively you can see
that the value is non-zero:
it is the infinitesimal defined by 1/infinity .
. by using L'Hopital's Rules for Improper Integration
we can get a definite integral from
an infinity of infinitesimals .
[12.13: I doubt that applies to a constant function over
naturals or integers, but I have another argument ...]
. I think it should be obvious from
the definition of probability function and infinity
that integration(0..infinity) 1/infinity = 1
or even integration(-infinity...infinity) 1/infinity = 1
simply because that is the mathematical way of saying
the fact that the chance of picking
the same item from an infinite set
is infinitesimally unlikely but not impossible:
the probability = 1/infinity = infinitesimal .
. integration(0..infinity) 1/infinity
= infinity * 1/infinity = 1;
but, integration(-infinity...infinity) 1/infinity = 1?
. the basic integration rules have this to say:
= integration(-infinity..c) f(x)
+ integration(c..infinity) f(x);
= integration(negatives) + integration(naturals).
= infinity * 1/infinity + infinity * 1/infinity
= 1+1 =2, which is not what we want
for a probability function .
. if the cardinality of naturals is infinity
then cardinality of integers is 2*infinity
and cardinality of reals is 2*infinity*infinity,
because there an infinity of integers to choose from
and then an infinity of fractions also .