Re: Probability in an infinite sample space


"Rob Johnson" <rob@xxxxxxxxxxxxxx> wrote in message
news:20060321.002431@xxxxxxxxxxx
In article <441f8549$0$20114$afc38c87@xxxxxxxxxxxxxxxxxxxx>,
"Peter Webb" <webbfamily-diespamdie@xxxxxxxxxxxxxxx> wrote:
"Dave Seaman" <dseaman@xxxxxxxxxxxx> wrote in message
news:dvmg28$8la$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Tue, 21 Mar 2006 00:14:00 +1100, Peter Webb wrote:
Consider the set of reals [0,1] that has a terminating decimal
representation. Consider the same probability distribution over this as
for all Reals [0,1] uniformly. Isn't this a uniform distribution over a
countably infinite set?

It isn't a probability distribution. The total weight is 0, not 1.

Probability distributions are required to be countably additive, but
they
are not *uncountably* additive.

Sorry about the questions, but this is still bugging me.

Lets call the set of Reals of [0,1] as R, and the set of non-terminating
decimals within [0,1] as D.

I looked for the definitions of a probability distribution and found this:

http://en.wikipedia.org/wiki/Kolmogorov_axioms

My probability distribution fails condition 3, because the sum of the
probabilities of an infinite number number of discrete elements is always
zero.

Why does Axiom 3 single out countable sums but not uncountable sums? Put
slightly differently, what would "break" if the definition dropped the
requirement that the sum be countable, but instead imposed a requirement
that the probabilities being summed over were non-zero? This slightly
changed definition appears to work in exactly the same manner, except that
it also allows distributions over countable sets (and is hence more
powerful).

How do you take a sum of an uncountable set? You can't, unless all
but countably many are 0. Since we cannot make a sequence of partial
sums with uncountably many elements, there can be no conditional
convergence. It only makes sense to speak of absolute summability of
an uncountable set. So we can assume that the uncountable set we are
summing is non-negative. If we sum a mixed uncountable set, we must
simply sum the positive and negative terms and add those sums. In
the case of sums in a probability measure, all terms are non-negative
anyway.

Suppose we have a set of non-negative numbers X = { x_a }, whose sum,
S, is finite. Let X_n be the set { x_a in X : x_a > 1/n }. There
are at most nS elements in X_n since the sum of the elements in X_n
is less than S and each element of X_n is greater than 1/n. The set
of positive elements of X is the union of the X_n, for n = 1, 2, ...
(each postitive element must be greater than 1/n for some n). So the
set of positive elements in X is a countable union of finite sets,
therefore countable. So in a set of non-negative numbers whose sum
is finite, all but countably many must be zero.

Axiom 3 is generous; it allows us to take countable sums. It is not
restricting us from taking uncountable sums, since we can only sum an
uncountable set of non-negative numbers if all but countably many are
zero.

Rob Johnson <rob@xxxxxxxxxxxxxx>

I understand that if a sum is finite, only a countable number of terms can
be non-zero. However, I don't believe this is relevant to the wording of the
Axiom.

I believe the restriction in Axiom 3 that the sum be over a countable set
is for a completely different reason, to eliminate a contradiction that
arises when every term in the sum is zero. The contradiction is that we can
work out the probability that a number selected at random as being in the
interval [0,1] by two different methods and get two different answers.

Firstly, if we restrict ourselves to a Real randomly selected in [0,1], then
the probability that it lies in [0,1] is 1, by definition.

Alternatively, for any x element of [0,1], the probability of this x being
selected at random is zero. By adding probabilities, the probability that a
number picked at random lies between [0,1] is also zero, as it is the sum of
an uncountable number of zeroes (there is zero probability it is 0.5, zero
probability it is 1/7, zero probability it is pi-3, etc etc).

So we get for the same question (what is the chance that a randomly selected
number between 0 and 1 is between 0 and 1) a probability of 1 by the first
technique, and a probability of zero by the second.

Axiom 3 bypasses this inconsistency by not requiring that probabilities be
additive for uncountable sets. This is all fine, and I can see why the
additive rule for probabilities needs to be modified for infinite cases.
What I cannot understand is why the Axiom makes an exception only for
"uncountable" sets. Exactly the same issues exist for countably infinite
sets. However, because the Axiom excludes uncountable sets but not countable
infinite sets, my proposed probability distribution fails to meet the
Probability Axioms. As I said before, this strikes me as strange because I
cannot see the motivation. The uniform probability distribution I proposed
over the countable set of Reals [0,1] with terminating decimal
representations seems to be an absolutely well behaved and reasonable
function. To me, it only fails to be a probability distribution because
Axiom 3 seemingly treats countable infinite sets differently from
uncountable sets. The only effect that I can see of this somewhat arbitrary
wording of Axiom 3 is to eliminate uniform probability distributions over
countable sets, which is exactly what I want to create. I am sure that Mr
Kolmogorov didn't devise this wording purely to eliminate some otherwise
quite legitimate distributions; there must be some other reason, some
motivation.

I still cannot see any underlying problem with a distribution over the set
of terminating decimals in [0,1] defined by
Prob(x is in [a,b]) = b -a.

This strikes me as a uniform probability distribution over a countable set;
its only problem is that it fails to be additive (Axiom 3). Uncountably
infinite sets also fail to be additive, but Axiom 3 excludes those. It would
seem with very minor re-wording Axiom 3 could eliminate the requirement to
treat countable and uncountable sums differently, and suddenly my
distribution becomes legal.

That this hasn't occurred suggests a deeper reason for why we cannot have
uniform distributions over countable sets (deeper than the fact that Axiom 3
arbitrarily excludes it). What is it?


.

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