Probability measure: finitely additive but not countably additive
vv
exclusive events, then P[A + B] = P[A] + P[B], and induction leads to
finite additivity. What will be an example of probability measure
that is finitely additive but not countably additive? Thanks.
-vv
Mariano Suárez-Alvarez
Let X be a countable set, endow it with the
algebra A of sets consisting of all its subsets,
and consider the measure mu : A --> R such that
mu(G) = 0 for all *proper* subsets G of X, and
mu(X) = 1.
-- m
G. A. Edgar
<d35ffcbc-a758-42ee...@l42g2000hsc.googlegroups.com>,
Mariano Suárez-Alvarez <mariano.su...@gmail.com> wrote:
Not good enough: X is the disjoint union of two proper subsets.
Here is another attempt...
Let X be a countable set, endow it with the albebra A of
all finite sets and all cofinite sets (= complements of finite sets).
Assign P[A] = 0 if A is countable, P[A] = 1 if A is cocountable.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
brieucs
And if X is the set of natural numbers, and the
set-function is the *density* (natural one) of
the set, when it makes sense ? maybe those sets
with natural density do not buid an algebra ?
Robert Israel
They are not an algebra. Consider sets such that for every n,
exactly one of 2 n and 2 n + 1 is in the set. Such a set has natural
density 1/2. But the intersection of two such sets could be e.g. any
set of odd numbers.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Robert Israel
> <d35ffcbc-a758-42ee...@l42g2000hsc.goog
Of course, he meant
P[A] = 0 if A is finite, P[A] = 1 if A is cofinite.
brieucs
this is fine ! thanks !
Are there ways to extend such finitely-additive-function
to sigma-additive-funtion of subsets ?
Shall we extend the X base space, or the algebra of sets ?
Herman Rubin
brieucs <brieuc@orange> wrote:
>G. A. Edgar a crit :
>> In article
>> <d35ffcbc-a758-42ee...@l42g2000hsc.googlegroups.com>,
>> Mariano Surez-Alvarez <mariano.su...@gmail.com> wrote:
>>> -- m
They do not. One can produce an integral which is of
this type, the Cesaro sum, but not a measure.
As an example, suppose the integers are divided into
groups of four, and each set has two elements in each
group. These sets have density 1/2. But we can choose
the pairs in each foursome in an arbitrary manner, so
that the first pair has one number for the intersection,
the second another, the third 4 = 3!-2! another, the
fourth 18 = 4!-3! another, etc. As the n-th group
has all except 1/n of the total to that point, the
intersection does not have a limiting density.
One can use sets which are periodic modulo m for some m;
these form a field on which we can define a finitely
additive but not countably additive probability measure.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
vv
PG) EE level, where students are not exposed to even real analysis,
leave alone measure theory. Since posting my query, I came across,
"Countable Additivity for Probability Measures" by David T. Price, The
American Mathematical Monthly, Vol. 81, No. 8 (Oct., 1974), pp.
886-889. There the author's motivation was to provide "an example,
related to the beginning coursework in finite probability, where more
than finite additivity occurred." This is related but not directly
the query I had asked. Would I be correct in concluding that there
may not be a simple engineering-students-understandable example to
illustrate finitely additive does not imply countably so?
Link to Price's paper: http://www.jstor.org/pss/2319450
-vv
Robert Israel
I think the closest would be Gerald Edgar's example of finite and cofinite
sets, which might be restated this way for an audience of engineering
students.
Suppose you wanted to assign probabilities to sets of integers, such that
P(A) is the probability that an integer "chosen at random" is in the set A.
You might start with:
1) If A is finite, then P(A) = 0
2) If the complement of A is finite, then P(A) = 1.
It is possible (but not elementary, and I can't define an explicit example)
to have P defined on all sets of integers and still have finite additivity.
But already (1) and (2) imply that countable additivity is impossible here,
because the probability of the union of the sets {j} for all integers j
is 1, while the sum of their probabilities is 0.
--
Bill Taylor
> It is possible (but not elementary,
> and I can't define an explicit example)
> to have P defined on all sets of integers
> and still have finite additivity.
It is certainly possible, by the theorem that all filters
can be extended to ultrafilters. But whether an "explicit"
example can be found is highly doubtful.
The ultrafilter theorem is weaker than full AC,
but some sort of Choice seems unavoidable.
One way to make a pseudo-explicit extension to all
*definable* sets of integers would be as folows,
(which also depends on one's extension of the idea of definable.)
List all definable subsets according to some Godel-type coding,
(this is "explicit in principle" - what a ghastly concept!),
then go through the list successively adding each new
undetermined subset into the sets of measure zero, then
taking all unions/intersections/differences with the previous
determined sets and assigning measure to them, consistently.
This eventually gets through the whole list, but in a rather
non-constructive fashion, as it's not likely that every set
on one's list of definable sets is going to be amenable
to such (explicit!) treatment.
Still, it's a start.
----------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
----------------------------------------------------------------
The concept of definability is necessarily undefinable.
----------------------------------------------------------------
Jonathan
This example leaves something to be desired, since the finite and cofinite subsets of (say) the naturals do not form a sigma-algebra. For example, the countable union of the singletons containing evens is not cofinite.
I'm told the standard example of an additive-but-not-countably-additive probability measure comes from Bruno de Finetti's "Theory of Probability", (1972). I think the way it works is: let P_n be the uniform distribution over the set {1, ..., n}. Then let P(S) be the limit of the sequence P_1(S), P_2(S),...
I'm afraid I don't know how to prove that this limit exists for each subset of N, and that the resulting P is countably additive (though it is obviously not countably additive). Hopefully somebody else can verify this?
Herman Rubin
>> <d35ffcbc-a758-42ee...@l42g2000hsc.goog
>> > legroups.com>,
>> > Mariano Surez-Alvarez
>> > <mariano.su...@gmail.com> wrote:
>> > > -- m
There are problems with this distribution even if the set is
uncountable, such as the reals. To get useful results, one
needs locally sigma-finite measures with countable additivity.
>I'm told the standard example of an additive-but-not-countably-additive probability measure comes from Bruno de Finetti's "Theory of Probability", (1972). I think the way it works is: let P_n be the uniform distribution over the set {1, ..., n}. Then let P(S) be the limit of the sequence P_1(S), P_2(S),...
>I'm afraid I don't know how to prove that this limit exists for each subset of N, and that the resulting P is countably additive (though it is obviously not countably additive). Hopefully somebody else can verify this?
Not only does the limit not exist for each subset of N, but the
class of sets for which it does exist is not even closed under
finite union. However, the integral corresponding to it is
finitely additive.
One can get a finitely additive measure on the class of
periodic sets, in the obvious manner. This class is closed
with respect to finite union.
The problem with finite additivity is that there is no
known way of getting any reasonable Radon-Nikodym theorem.
This is what is needed for statistical inference.
There is a type of Fubini theorem for finitely additive
integrals, similar to the Moore-Osgood theorem for
double limits and iterated limits. Limit is a
finitely additive integral.