*4.3: 12.13: news.*

**math**/probability function over naturals:**askamathematician.com 2010:**

. 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)?

askamathematician states:

". 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..infinity) f(x)

= integration(-infinity..c) f(x)

+ integration(c..infinity) f(x);

so, integration(-infinity..infinity)

= 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 .