# Probability for a countable sample space

• Jun 18th 2010, 04:16 AM
aman_cc
Probability for a countable sample space
Cosider sample space, S, defined as

S = {1,2,3,4,5,6,7,8,9,................}

Experiment is to pick a number randomly from this set (with equal probability).

Question: How do you define probability measure for this?

Confusion I have -

Consider sets A1 = {1}, A2 = {2}, A3 = {3}, A4 = {4}, ..., An = {n},.... so on

Now let P(An) = x

P(S) = P(A1)+P(A2)+P(A3)+....+P(An)+..... (under the axiom of countable additivity)

LHS is 1 but RHS = 0 or infinite, so where is that I'm going wrong?
• Jun 18th 2010, 10:09 AM
Plato
Quote:

Originally Posted by aman_cc
Cosider sample space, S, defined as
S = {1,2,3,4,5,6,7,8,9,................}
Experiment is to pick a number randomly from this set (with equal probability).
Question: How do you define probability measure for this?

Each event has the probability say p>0.
It well known that the sum of countable infinite positive constants is inifinte.
So?
• Jun 18th 2010, 11:19 AM
undefined
I think the setup to the problem is flawed; there is no uniform probability distribution function for this set. I'm not entirely sure though. Here is a reference.
• Jun 18th 2010, 12:49 PM
SpringFan25
My guess:

You require:
$\sum_x^\infty {P(x)} = 1$

$P(x) = P(y) ~~~~ \forall x,y >0$

It seems clear (as you have reasoned) that there is no solution. Can you define an appropriate limit and find that? (You will get that in the limit, P(.) -> 0)
• Jun 18th 2010, 07:16 PM
aman_cc
@Plato - I'm sorry not able to take your argument fwd.

@undefined - Thanks for your reference. Yes even I feel it is flawed - but where exactly?
• Jun 18th 2010, 07:42 PM
undefined
Quote:

Originally Posted by aman_cc
@undefined - Thanks for your reference. Yes even I feel it is flawed - but where exactly?

The flaw in my opinion is in assuming that it is possible to "randomly" choose an element of an infinitely countable set such that any element is equally likely to be chosen as any other. The arguments presented show that this assumption is false. Maybe in terms of measure theory there is a better way to say this; I have not studied measure theory.

I also found this thread on another forum; post #16, which seems to be by the same HallsofIvy who posts on this forum, might be helpful to you.
• Jun 20th 2010, 09:41 PM
aman_cc
Thanks. But haven't been able to make much sense. So am I correct in saying - you cannot define a probabiity measure for the set S, which conforms to our intution of 'equally likely'?
• Jun 20th 2010, 09:44 PM
matheagle
the probabilities must go to zero, since the sum is finite

$\sum_{k=1}^{\infty}p_k<\infty$