Probability question.

Originally posted by mtthw
Not if there is no P! That's the point.

(Palynka - if I've misunderstood what you're getting at and that isn't the point, let me know and I'll butt out 🙂)

You can find such a measure and can call it P, so that P(X) = 0. The error is thinking this is a statement about a probability, which it isn't, because P will not have the properties required to be a probability measure.

Edit - No need to butt out, we're just having a chat not competing (although measures are involved! 😵)

Originally posted by adam warlock
But this is not how things work when you apply measure theory to probability... 😕

http://en.wikipedia.org/wiki/Probability_theory#Measure-theoretic_probability_theory
http://en.wikipedia.org/wiki/Probability_space
http://en.wikipedia.org/wiki/Lebesgue_measure
http://en.wikipedia.org/wiki/Measure_theory

Let me quote myself:
[quote]When one is it is obvious that I'm assuming that all steps are valid in the context of probability theory.

It's right there, man!

http://en.wikipedia.org/wiki/Probability_space#Definition

a probability space is a triple \scriptstyle (\Omega,\; \mathcal{F},\; P) consisting of:

* the sample space \Omega — an arbitrary non-empty set,
* the \sigma-algebra \mathcal{F} ⊆ 2Ω (also called \sigma-field) — a set of subsets of \Omega, called events, [...]
* the probability measure P: \mathcal{F}→[0,1] — a function on \mathcal{F} such that:
o P is countably additive: [...]
o the measure of entire sample space is equal to one: P(\Omega) = 1.

Originally posted by Palynka
It's right there, man!

http://en.wikipedia.org/wiki/Probability_space#Definition

a probability space is a triple \scriptstyle (\Omega,\; \mathcal{F},\; P) consisting of:

* the sample space \Omega — an arbitrary non-empty set,
* the \sigma-algebra \mathcal{F} ⊆ 2Ω (also called \sigma-field) — a set of subsets ...[text shortened]... o [b]the measure of entire sample space is equal to one: P(\Omega) = 1.[/b]

http://en.wikipedia.org/wiki/Measure_theory#Definition

A probability measure is a measure with total measure one (i.e., μ(X) = 1); a probability space is a measure space with a probability measure.

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

The difference between a probability measure and the more general notion of measure (which includes concepts like area or volume) is that a probability measure must assign 1 to the entire probability space.

Originally posted by adam warlock
http://en.wikipedia.org/wiki/Measure_theory#Definition

A probability measure is a measure with total measure one (i.e., μ(X) = 1); a probability space is a measure space with a probability measure.

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

The difference between a probability measure and the more general notion ...[text shortened]... ume) is that a probability measure [b]must assign 1 to the entire probability space.

[/b]

LOL, did you miss the first line?

A probability measure is a real-valued function defined on a set of events in a probability space that satisfies measure properties such as countable additivity.

Originally posted by Palynka
LOL, did you miss the first line?

A probability measure is a real-valued function defined on a set of events in a probability space that satisfies measure properties such as [b]countable additivity.[/b]

Third property on http://en.wikipedia.org/wiki/Measure_theory#Definition

Second property on http://en.wikipedia.org/wiki/Probability_measure#Definition

Originally posted by adam warlock
Third property on http://en.wikipedia.org/wiki/Measure_theory#Definition

Second property on http://en.wikipedia.org/wiki/Probability_measure#Definition

Yes, and?

Are you going to address this or just pretend you know more by quoting some wiki links that actually prove my point?

"The proof is simple. If m a measure over the real line then to be a probability measure m(omega)=1 and m needs to be countably additive (the measure of a union of countable sets is equal to the sum of the measures). But under a potential uniform over the real line then any finite interval has measure 0.

Since omega={...U [-1,0[ U [0,1[ U [1,2[ U ...} then m(omega) = m({...U [-1,0[ U [0,1[ U [1,2[ U ...})=1. Yet the sum of the measures of those intervals is 0 as every one of them has measure 0.

The measure is then not a probability measure and P(X) is not a statement about probability."

Originally posted by Palynka
Are you going to address this or just pretend you know more by quoting some wiki links that actually prove my point?

"The proof is simple. If m a measure over the real line then to be a probability measure m(omega)=1 and m needs to be countably additive (the measure of a union of countable sets is equal to the sum of the measures). But under a potential u ...[text shortened]...
The measure is then not a probability measure and P(X) is not a statement about probability."

Why do I need to address this? When one uses measure theory in order to support probability theory one isn't restricted to a potential uniform over the real line.

Originally posted by adam warlock
Why do I need to address this? When one uses measure theory in order to support probability theory one isn't restricted to a potential uniform over the real line.

Certainly. I didn't say it was.

All I said was that there was no uniform over the real line, that one could find non-probability measures that assign equal measure to any point in the real line but these will not sum up to one. So such a measure would not be a probability measure. And if I call this measure P, then P(X) = 0 where X are the rationals but P would not be a probability measure.

Do you then agree with this?

Originally posted by Palynka
All I said was that there was no uniform over the real line, that one could find non-probability measures that assign equal measure to any point in the real line but these will not sum up to one. So such a measure would not be a probability measure. And if I call this measure P, then P(X) = 0 where X are the rationals but P would not be a probability measure.

Do you then agree with this?

Yes I do. And I still don't understand why I need to address this.

Originally posted by adam warlock
Yes I do. And I still don't understand why I need to address this.

Because before you called it nonsensical. Perhaps you should read before you started disagreeing and we would avoid these little misunderstandings.

Originally posted by Palynka
Because before you called it nonsensical. Perhaps you should read before you started disagreeing and we would avoid these little misunderstandings.

The example you gave is an example of measure theory not supporting probability theory which is outside of everything I've said so far.

Perhaps you should read before you started disagreeing and we would avoid these little misunderstandings.

Originally posted by adam warlock
The example you gave is an example of measure theory not supporting probability theory which is outside of everything I've said so far.

Perhaps you should read before you started disagreeing and we would avoid these little misunderstandings.

Where did I disagree with you before this...exactly? Quotes please.

Originally posted by Palynka
Where did I disagree with you before this...exactly? Quotes please.

The bigger point is that your example fell of the context of my posts and this is what you should be addressing instead of moving the goalposts (again).

As for the quotes:

he error is thinking this is a statement about a probability, which it isn't

Yes, that sentence is incorrect/imprecise because, first, he doesn't specify a distribution (if you want me to define what I mean by distribution please say so) although for people used to probability when you don't specify a distribution it usually means the uniform

Originally posted by adam warlock
The bigger point is that your example fell of the context of my posts and this is what you should be addressing instead of moving the goalposts (again).

As for the quotes:

he error is thinking this is a statement about a probability, which it isn't

[b]Yes, that sentence is incorrect/imprecise because, first, he doesn't specify a di ...[text shortened]... sed to probability when you don't specify a distribution it usually means the uniform

[/b]

I agreed with your main point (do you need quotes?) but said you were either imprecise or incorrect because you do need to specify that this is only always true for continuous distributions. Your insistence in that you didn't need to specify anything is, mildly put, weird. As it is also weird that before you replied "this is not how things work when you apply measure theory to probability" to a post of mine and a few posts later agreed with it when I re-posted the same thing. Who's moving the goalposts again? At least I know I've been coherent throughout.

As for the uniform over the real line, that was more directed at twhitehead's comments about what you meant by random draw (as he thought it was synonymous with a draw from a uniform distribution). Although it's common for people to say "a random number from 1 to 10" to mean drawing from a uniform, that could not have been the case here. If that's what you meant then you would have been wrong, and if not then it would be imprecise because you neglected the possibility of discontinuous distributions assigning positive probability to certain sets of rationals. It turned out that you were simply imprecise and not incorrect, I just don't understand why you get so worked up about it.

mtthw understood what I meant from the start, so perhaps the problem is not me. twhitehead also seemed to understand what I meant.