Re: Probability in an infinite sample space


"Dave Seaman" <dseaman@xxxxxxxxxxxx> wrote in message
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On Tue, 21 Mar 2006 00:14:00 +1100, Peter Webb wrote:

"Dave Seaman" <dseaman@xxxxxxxxxxxx> wrote in message
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On 19 Mar 2006 19:04:41 -0800, mikeh106@xxxxxxxxxxx wrote:
If you choose a natural number at random, that it will be a multiple of
3 is there a 1/3, 1/2, or undefined chance?

That's a meaningless question, because you have not identified a
probability distribution.

It is often assumed that if no distribution is specified, then a uniform
distribution is intended. That can't be the case here, because there is
no such thing as a uniform probability distribution on a countably
infinite sample space.

Does it mean something to select "at random" from an infinite number of
objects?

Certainly, provided a probability distribution is specified. For
example, there is a uniform distribution on the unit interval [0,1],
which is an uncountably infinite space. You can also have a probability
distribution on a countably infinite space, but it can't be uniform.
For
example, the probability might be P(n) = 2^(-n) for n = 1, 2, 3, ....


--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>

You and another poster strengthened "uniform distribution over N" to
"uniform distribution over countably infinite", so I realise that my
following remarks are almost certainly wrong.

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.



--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>

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.

However, this strikes me as a definition which is explicitly designed to
make distributions like mine invalid, but allowing similar distributions
over uncountable sets.

For example, if we removed the word "countable" from the definition, there
would be no uniform distributions over R. Alternatively, if we constrained
the intervals to have non-zero probabilities, then my distribution over D
would comply.

So I accept that my distribution does not meet the definition, but this
definition seems quite arbitrary to me, as its only effect that I can see is
to eliminate distributions of the type I propose.

Putting aside this definition, my distribution seems just as well behaved as
the standard uniform distribution over R. The probability that an element of
D is in [0.1, 0.2] is 0.1, just as it is for the Real case, indeed all
probabilities work out to be exactly the same, in both cases the probability
of a discrete event is zero, and the probability of a number X being in the
range [a,b] is b-a.

Sure, I can't form the sum of the probability of a randomly selected element
of D being less than 0.5 by forming the sum of every discrete event where
X<0.5 and adding them up, but nor can we for R - its just that the
definition (Axiom 3) gives uncountable sets a get-out-of-jail free card,
whilst explicitly denying this for countable sets.

My distribution seems to walk like a duck, quack like a duck and swim like a
duck.

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).





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