Probability question.

Originally posted by adam warlock
There's no chooser. Probability isn't a fraction between favorable cases and totality of cases. Probability is a function and is calculated by evaluating an integral.

This statement is just a rephrasing of Cantor's diagonal argument. And I had already told you that. If you don't know what Cantor diagonal argument is about either you google it or you ask for a clarification.

I am trying to learn some more and understand what you are saying.
I looked up
http://en.wikipedia.org/wiki/Probability_theory#Discrete_probability_distributions

Is the definition under "Modern definition: " what you are talking about?

Suppose we are talking about the interval [0,1]
If so, can I define f(x) such that f(x) = 0 for all x not equal to 0.5 and f(x)=1 for x=0.5
What am I missing?

Originally posted by twhitehead
Suppose we are talking about the interval [0,1]
If so, can I define f(x) such that f(x) = 0 for all x not equal to 0.5 and f(x)=1 for x=0.5

No problem.

As for the definition, we are talking about distributions with continuous support so you need to look at that subsection in the wikipedia article.

Originally posted by Palynka
No problem.

As for the definition, we are talking about distributions with continuous support so you need to look at that subsection in the wikipedia article.

Is the case of a human being picking a number in the interval an example of a distribution with continuous support?

Originally posted by twhitehead
Is the case of a human being picking a number in the interval an example of a distribution with continuous support?

Yes, the interval is continuous and that is the support. They can pick any number in the interval. That many decide to pick rational numbers is a question of the distribution over that support.

Originally posted by Palynka
Yes, the interval is continuous and that is the support. They can pick any number in the interval. That many decide to pick rational numbers is a question of the distribution over that support.

So my proposed f(x) fails because there is supposedly a very small non-zero chance that a human would pick pi/5 ?

Further down the Wikipedia page I find

An example of such distributions could be a mix of discrete and continuous distributions, for example, a random variable which is 0 with probability 1/2, and takes a random value from a normal distribution with probability 1/2.

Are such distributions within what we are discussing or not? If they are, then surely the result claimed by adam would not be correct. So am on the wrong track again?

Originally posted by twhitehead
So my proposed f(x) fails because there is supposedly a very small non-zero chance that a human would pick pi/5 ?

Further down the Wikipedia page I find
[quote]An example of such distributions could be a mix of discrete and continuous distributions, for example, a random variable which is 0 with probability 1/2, and takes a random value from a normal ...[text shortened]... re, then surely the result claimed by adam would not be correct. So am on the wrong track again?

It could even be zero but the support is the same because the issue is about numbers between between 0 and 1. THAT is the support.

I'm not sure what result of adam you are talking about and what exactly you disagree on. I thought it was on events happening despite having probability zero, but you already agreed that that is true...

Originally posted by Palynka
I'm not sure what result of adam you are talking about and what exactly you disagree on. I thought it was on events happening despite having probability zero, but you already agreed that that is true...

I agree that events of probability zero happen. I disagree that a human choosing a rational number in the interval [0,1] has a probability of zero and thus is not an example of a probability zero event actually happening.

Originally posted by twhitehead
I agree that events of probability zero happen. I disagree that a human choosing a rational number in the interval [0,1] has a probability of zero and thus is not an example of a probability zero event actually happening.

Yes, I agree. I'm not sure that's what he meant, though. If he did then he would be wrong.

Originally posted by Palynka
Sure, but you need to use probability measure if you want to talk about probability. Only some measures are probability measures and the Lebesgue measure over the real line is not one of them. Measure 0? Yes. P(X)=0? No. There simply is no uniform over the real line.

When one is using the machinery of measure theory to support probability theory saying that a given set has zero measure is the same as saying that that set has a 0 probability outcome.

Thus saying this: Measure 0? Yes. P(X)=0? No. is just nonsensical.

Originally posted by twhitehead
I disagree that a human choosing a rational number in the interval [0,1] has a probability of zero

Are you still under the delusion that I deny that?

Originally posted by adam warlock
Thus saying this: Measure 0? Yes. P(X)=0? No. is just nonsensical.

Before you start arguing about things you actually agree on again (🙂), I think all Palynka is saying is that there is no uniform probability distribution over the real line. So P(X) is undefined in that case. So P(X) = 0 is not a true statement. He isn't saying P(X) = a where a != 0.

Originally posted by adam warlock
When one is using the machinery of measure theory to support probability theory saying that a given set has zero measure is the same as saying that that set has a 0 probability outcome.

Thus saying this: Measure 0? Yes. P(X)=0? No. is just nonsensical.

Why is it nonsensical? Not all measures are probability measures. Do you not agree? I don't imagine you would agree that all measures of entire spaces are finite (and they have to be 1 in case of probability measures)...

So, yes, something can have measure 0 under a measure that does not satisfy the basic properties of probability measures.

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 mtthw
So P(X) = 0 is not a true statement.

But in the case of X representing the set of rational numbers on the whole of the real line (it doesn't even have to be a finite interval) P(X)=0 is a true statement if you use measure theory.

Originally posted by Palynka
Why is it nonsensical? Not all measures are probability measures. Do you not agree? I don't imagine you would agree that all measures of entire spaces are finite (and they have to be 1 in case of probability measures)...

So, yes, something can have measure 0 under a measure that does not satisfy the basic properties of probability measures.

The proof i ...[text shortened]...

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

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:

When one is using the machinery of measure theory to support probability theory

If I'm talking about supporting probability theory it is obvious that I'm assuming that all steps are valid in the context of probability theory.

Originally posted by adam warlock
But in the case of X representing the set of rational numbers on the whole of the real line (it doesn't even have to be a finite interval) P(X)=0 is a true statement if you use measure theory.

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 🙂)