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April 15th, 2017, 08:16 PM   #1
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Selecting a Natural and a Real Uniformly at Random

First Statement to Prove:

Given

1) the axiom of choice,

2) a randomly selected infinite binary sequence $S = s_1, s_2, s_3,$... (created via the theoretical flipping of a coin infinitely many times where H = 1 and T = 0), and

3) a definition of "select an element of an infinite set uniformly at random" that simply means, in layman's terms, that one and only one element of an infinite set will be selected using a process where all elements of the set have an equal chance of being selected,

then it is possible to select a natural number uniformly at random.


Definitions:

Let $V$ be a set containing one and only one element from each Vitali equivalence class on the interval $[ \, 0.5, 1 ] \,$ (Vitali equivalence classes are equivalence classes of the real numbers that partition $\mathbb{R}$ under the relation $x \equiv y \iff ( \, \exists q \in \mathbb{Q} ) \, ( \, x - y = q ) \,$). The axiom of choice allows for such a selection.

Let $D = \{ r \in [ \, 0, 1 ) \, : r$ is a dyadic rational $\}$ (dyadic rationals are rational numbers whose denominator is a power of two and, as a result, whose binary expansion is finite).

Let $E = \{ r \in \mathbb{Q} ( \, 0, 1 ) \, : r \notin D \}$.

Let $f : \mathbb{N} \longmapsto D$ (here $\longmapsto$ denotes a bijection).

Let $g : \mathbb{N} \longmapsto E$.

Let $h : \mathbb{N} \longmapsto \mathbb{Q} [ \, 0, 1 ) \,$.

Let $x$ be a real number in $[ \, 0, 1 ] \,$ selected uniformly at random via the binary sequence from (2) above such that $x = 0.s_1s_2s_3$... .

Let $n \in \mathbb{N}$ denote the natural number that is selected uniformly at random as a result of the following process.


Process for Selecting $n$ Uniformly at Random:

There are three possibilities for $x$. It may be a dyadic rational, a non-dyadic rational, or an irrational.

1) If $x \in D$ or $x = 1$:

Each dyadic rational not equal to 0 or 1 has two binary expansions: one finite and one infinite (e.g. 0.1 = 0.0111…). The randomly selected number $x$ may take the form of either expansion. Let $n$ be such that $f(n) = x$ if $0 < x < 1$ and $f(n) = 0$ if $x = 0$ or $x = 1$. At this point, there are two and only two possible ways of selecting each possible natural number.


2) If $x \in E$:

Let $n$ be such that $g(n) = x$. At this point, there are now three and only three ways of selecting each possible natural number.


3) If $x \in [ \, 0, 1 ] \, \setminus \mathbb{Q}$:

There will exist one and only one element $v \in V$ such that $v - x \in \mathbb{Q}$. For each rational $q$ on the interval $[ \, 0, 1 ) \,$, there will be one and only one corresponding irrational $i$ on the interval $[ \, v - 0.5, v + 0.5 ) \,$ such that $q - 0.5 = i - v$.

There are two possibilities for $x$. It will either fall in the interval $[ \, v - 0.5, 1 ) \,$ or in the interval $( \, 0, v - 0.5 ) \,$. For each irrational $y$ on the interval $( \, 0, v - 0.5 ) \,$, there will be one and only one irrational $z$ on the interval $( \, 1, v + 0.5 ) \,$ where $y + 1 = z$. In this fashion, we can biject all possible values for $x$ on the interval $[ \, 0, 1 ) \,$ that are within the Vitali equivalence class containing $v$ with all possible irrationals on the interval $[ \, v - 0.5, v + 0.5 ) \,$ that are also in the Vitali equivalence class containing $v$ (i.e., if $x \in [ \, v - 0.5, 1 ) \,$, then utilize $x$ itself whereas, if $x < v - 0.5$, utilize $x + 1$ instead). Select a suitable $n$ as follows:

If $x \in [ \, v - 0.5, 1 ) \,$, let $n$ be such that $h(n) - 0.5 = x - v$.

If $x \in ( \, 0, v - 0.5 ) \,$, let $n$ be such that $h(n) - 0.5 = (x + 1) - v$.

At this point, all possible values for $x$ will lead to the selection of a distinct value for $n$ and all $n \in \mathbb{N}$ have an equal chance of being selected.

End proof.


Second Statement to Prove:

It is possible to select an element of $\mathbb{R}$ uniformly at random given the first proof above.

Definitions:

Let $j : \mathbb{N} \longmapsto \mathbb{Z}$.

Let $n$ be selected uniformly at random via the above process.

Let $x$ be a real number in $[ \, 0, 1 ) \,$ selected uniformly at random. Given the selection process used in the first proof, this requires two additional steps. First, since each dyadic rational has two binary expansions, each of the expansions must be related to a distinct and separate real number (this is trivial in that the dyadics may be listed along with their possible expansions, allowing for a mapping by indexes). Second, it is important that $x \neq 1$, which is again accomplished trivially by bijecting $[ \, 0, 1 ] \,$ with $[ \, 0, 1 ) \,$.

Let $r \in \mathbb{R}$ denote the real number that is selected uniformly at random as a result of the following process.

Process for Selecting $r$ Uniformly at Random:

$r = j(n) + x$.

End proof.
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April 15th, 2017, 08:38 PM   #2
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Before I dive into this, I would like to ask if you are willing to spend a little time talking about the fact that there's no uniform, countably additive measure on the natural numbers, or for that matter on any countable set, as long as you want the total probability to be $1$.

Now I don't care if we end up agreeing or if you really think I'm just being annoying about this.

Before I do you the courtesy of spending my time finding the error in your idea, would you do me the courtesy of at least

* Acknowledging that I keep badgering you with this; and

* Tell me whether you understand it.

* Tell me if you agree with it.

* Or tell me you really don't want to hear it anymore and you're not interested. That's a perfectly valid response. I like to explain math to people who would like me to explain math to them. Like that paradoxical barber. I explain math to exactly those people who don't explain math to themselves.

So ok just engage with me long enough to feel like we are having a conversation.

ps -- I have skimmed your proof and it is subject to exactly the point I made at the end of the other thread. There are uncountably many equivalence classes, it's no use to locate your $x$ in one of them, they all have measure zero. They're each countable sets! Countable sets have measure zero.

And since there are uncountably many of them, you can't map them to the natural numbers in any useful fashion.

Instead you have to locate your $x$ in one of the countably many rational translates of the Vitali set; and each of those translates is nonmeasurable. There are countably many uncountable and nonmeasurable translates. It's true you can associate a unique natural number with a translate; but you can not possibly associate a probability to it.

tl;dr: You think you can read off the probability of each rational translate of $V$ and transfer it to its corresponding natural number.

But what you've really proved is that those translates must not be measurable! Because there are countably many of them and by uniformity they all have the same measure. Which can't be zero, else the unit interval would be too small; and can't be nonzero, else the unit interval would be too big.

So they can't be anything. They are sets of reals that can not possibly be assigned a sensible measure or probability.

Last edited by Maschke; April 15th, 2017 at 09:36 PM.
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April 15th, 2017, 11:19 PM   #3
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Before I go through and address all your points here, I suggest you do the proper thing which is find the faulty line of the proof and don’t just skim it assuming it’s wrong.

Quote:
Originally Posted by Maschke View Post
Before I dive into this, I would like to ask if you are willing to spend a little time talking about the fact that there's no uniform, countably additive measure on the natural numbers, or for that matter on any countable set, as long as you want the total probability to be $1$.

Now I don't care if we end up agreeing or if you really think I'm just being annoying about this.

Before I do you the courtesy of spending my time finding the error in your idea, would you do me the courtesy of at least

* Acknowledging that I keep badgering you with this; and

* Tell me whether you understand it.

* Tell me if you agree with it.

* Or tell me you really don't want to hear it anymore and you're not interested. That's a perfectly valid response. I like to explain math to people who would like me to explain math to them. Like that paradoxical barber. I explain math to exactly those people who don't explain math to themselves.

So ok just engage with me long enough to feel like we are having a conversation.
Ok. The process will result in a distinct natural $n$ given the real $x$. The probability of selecting some natural $n$ is therefore 1. The question is whether each $n$ has an equal probability of being selected given my method.

Yes, I understand that the probability or selecting each given $n$ must be greater than 0. This has always been easy to do. It’s getting the probability of selecting each possible $n$ to be equal, or uniform, that has been considered impossible because such uniformity leaves you wondering whether the sum is 0 or 1. I get that, no need to keep bugging me. I have noted that $\frac{1}{\infty}$ is undefined.


Quote:
Originally Posted by Maschke View Post
ps -- I have skimmed your proof and it is subject to exactly the point I made at the end of the other thread. There are uncountably many equivalence classes, it's no use to locate your $x$ in one of them, they all have measure zero. They're each countable sets! Countable sets have measure zero.
The fact that each equivalence class has a measure of 0 as you say is of no concern to the proof. We aren’t dealing with a Lebesgue measure.

Quote:
Originally Posted by Maschke View Post
And since there are uncountably many of them, you can't map them to the natural numbers in any useful fashion.
I believe I did map each equivalence class to the naturals in a useful fashion, so we disagree on that. You’ll have to state precisely why what I did is not “useful” and in what sense.

Quote:
Originally Posted by Maschke View Post
Instead you have to locate your $x$ in one of the countably many rational translates of the Vitali set; and each of those translates is nonmeasurable. There are countably many uncountable and nonmeasurable translates. It's true you can associate a unique natural number with a translate; but you can not possibly associate a probability to it.
I’m not trying to associate a given probability to any natural as such a probability, $\frac{1}{\infty}$, is undefined.

Quote:
Originally Posted by Maschke View Post
tl;dr: You think you can read off the probability of each rational translate of $V$ and transfer it to its corresponding natural number.
In no way do I try to come up with a probability of each “rational translate” (I believe you mean equivalence class).

Quote:
Originally Posted by Maschke View Post
But what you've really proved is that those translates must not be measurable! Because there are countably many of them and by uniformity they all have the same measure. Which can't be zero, else the unit interval would be too small; and can't be nonzero, else the unit interval would be too big.
You just said those translates have a measure of 0, but now you say they aren’t measurable (a Vitali set is not measurable, I’ll give you that), but again, it’s irrelevant because I’m not concerned with a Lebesgue measure.

Quote:
Originally Posted by Maschke View Post
So they can't be anything. They are sets of reals that can not possibly be assigned a sensible measure or probability.
Simple question: using my method, can you show that any given $n$ has a probability of getting selected that differs from some other natural number, thus showing it’s not uniform? If yes, do so. If not, then I think we have our answer.
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April 16th, 2017, 08:54 AM   #4
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Quote:
Originally Posted by AplanisTophet View Post
Before I go through and address all your points here, I suggest you do the proper thing which is find the faulty line of the proof and don’t just skim it assuming it’s wrong.
I'm disappointed that you won't engage on the math. You are the one claiming to violate a well known and accepted mathematical fact. The burden's on you engage with what I'm saying to you. I have no obligation here.

I say again. Do you understand that there can be no countably additive, uniform measure on the natural numbers that assigns 1 to the entire set?

I'm not asking you to agree or even to understand the details. I'm asking you to acknowledge reading the words, and make a response.

If you will not even acknowledge reading those words, then you are not willing to engage in a mathematical conversation. I'd be foolish to continue.

Last edited by Maschke; April 16th, 2017 at 09:42 AM.
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April 16th, 2017, 12:59 PM   #5
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Quote:
Originally Posted by Maschke View Post
I say again. Do you understand that there can be no countably additive, uniform measure on the natural numbers that assigns 1 to the entire set?

I'm not asking you to agree or even to understand the details. I'm asking you to acknowledge reading the words, and make a response.
I did answer you. See bold.

Quote:
Originally Posted by AplanisTophet View Post
Yes, I understand that the probability or selecting each given $n$ must be greater than 0. This has always been easy to do. It’s getting the probability of selecting each possible $n$ to be equal, or uniform, that has been considered impossible because such uniformity leaves you wondering whether the sum is 0 or 1. I get that, no need to keep bugging me. I have noted that $\frac{1}{\infty}$ is undefined.
Now can you answer my simple question?: Using my method, can you show that any given $n$ has a probability of getting selected that differs from some other natural number, thus showing it’s not uniform? If yes, do so. If not, then I think we have our answer.
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April 16th, 2017, 01:49 PM   #6
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Quote:
Originally Posted by AplanisTophet View Post
Now can you answer my simple question?: Using my method, can you show that any given $n$ has a probability of getting selected that differs from some other natural number, thus showing it’s not uniform? If yes, do so. If not, then I think we have our answer.
CR Greathouse patiently and meticulously debunked and falsified your arguments three years ago here. Probabilities and Infinite Sets. Any readers who think I'm being unfair to you should read through that 15 page thread.

In his first response to you (out of many) he said:

"... there is no uniform distribution over $\mathbb N$".

That was three years ago. What could I possibly add? You didn't get the point then, you don't get it now. It's five minutes worth of work. If you aren't interested, well there's no crime in not being interested in infinitary probability theory. But you're the one trying to make an argument in a field you refuse to learn the basics of.

I haven't the patience of CR Greathouse. But you did him the courtesy of supplying far more detail than you did with me. When you make statements like "I'm not talking about Lebesgue measure," or, "I'm not assigning probabilities to natural numbers," you leave me baffled.

In that other thread you had the courtesy to state your argument in far more detail. You think probabilities are inverses of the Aleph numbers. That's wrong but at least it's an articulated idea that can be refuted. In your current exposition you're simply tossing out contradictions and expecting that I'm going to clean up the mess.

I have no obligation to do so.

As far as your specific argument, it's incoherent and I haven't the patience to follow it. But I will do you the service of calling out a few of your more obvious misunderstandings.

Quote:
Originally Posted by AplanisTophet View Post
...no need to keep bugging me. I have noted that $\frac{1}{\infty}$ is undefined.
You must think we're in that other thread, which I'm not a participant in. You truly don't seem willing to simply engage me on a specific point that I regard as one of the keys to the issue. If so, why should I do you the favor of spending my time finding your error?

Quote:
Originally Posted by AplanisTophet View Post
The fact that each equivalence class has a measure of 0 as you say is of no concern to the proof. We aren’t dealing with a Lebesgue measure.
LOL Then what the heck are you talking about? If you're still on the inverse Aleph idea, say so. If you have some other idea, then say so. Stop playing coy.

Quote:
Originally Posted by AplanisTophet View Post
I believe I did map each equivalence class to the naturals in a useful fashion, so we disagree on that. You’ll have to state precisely why what I did is not “useful” and in what sense.
Do you understand that there are uncountably many equivalence classes? How are you mapping them back to natural numbers? And not only that, but uniformly? Personally I haven't got that much imagination. And you haven't explained your idea with sufficient clarity to make me want to read it, since it contradicts standard math.

Quote:
Originally Posted by AplanisTophet View Post
I’m not trying to associate a given probability to any natural as such a probability, $\frac{1}{\infty}$, is undefined.
If you're not doing that, can you please me more clear about what you are doing? And please, stop writing $\frac{1}{\infty}$. This is not that other thread. You're getting confused.

Quote:
Originally Posted by AplanisTophet View Post
In no way do I try to come up with a probability of each “rational translate” (I believe you mean equivalence class).
I mean rational translate. If you're not trying to come up with probabilities, what are you doing? You are claiming to do exactly that.

Quote:
Originally Posted by AplanisTophet View Post
You just said those translates have a measure of 0, but now you say they aren’t measurable
The equivalence classes are each countable, hence have measure zero. The rational translates of $V$ are nonmeasurable. The fact that you're unclear on what I'm talking about should be a clue that you are missing a huge part of your own argument.

Quote:
Originally Posted by AplanisTophet View Post
... I’m not concerned with a Lebesgue measure.
Then what on earth are you talking about? Is your measure countably additive? Is it a measure at all? What is it? You can't just make up words without defining them.

Quote:
Originally Posted by AplanisTophet View Post
Simple question: using my method, can you show that any given $n$ has a probability of getting selected that differs from some other natural number, thus showing it’s not uniform?
Your exposition was not sufficiently coherent for me to do that. But please do tell me, do you understand that there are uncountably many equivalence classes?

Quote:
Originally Posted by AplanisTophet View Post
If yes, do so. If not, then I think we have our answer.
Now that I've discovered the prior 15 page thread from three years ago, I see little point in playing this game. You won't respond directly to my mathematical arguments, but you expect me to sort out your errors. Actually I have, several times over. You just won't engage with the points I've made.

I read through that 15 page thread and there is nothing I could possibly say that CR Greathouse didn't already tell you three years ago.

Last edited by Maschke; April 16th, 2017 at 02:17 PM.
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April 16th, 2017, 02:27 PM   #7
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But look, nevermind all that. To bring focus: Do you understand that there are uncountably many equivalence classes? How can you hope to map them uniformly to the natural numbers?

ps -- LOL I see you're having a go at the Physics forums. https://www.physicsforums.com/thread...random.911544/

Good luck. They have a very low tolerance over there.

Last edited by Maschke; April 16th, 2017 at 03:12 PM.
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April 16th, 2017, 03:24 PM   #8
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Quote:
Originally Posted by Maschke View Post
But look, nevermind all that. To bring focus: Do you understand that there are uncountably many equivalence classes? How can you hope to map them uniformly to the natural numbers?

ps -- LOL I see you're having a go at the Physics forums. https://www.physicsforums.com/thread...random.911544/

Good luck. They have a very low tolerance over there.
Yes, I mapped the uncountably many equivalence classes back to the naturals uniformly.

Remember not long ago when I assumed that we couldn't partition $\mathbb{R}$ into an uncountable number of countable sets, because if we could, then I said we could select a natural uniformly at random. I assumed then we couldn't select a natural uniformly at random for all the reasons you say now and that CRGreathouse discussed with me long ago. You showed me we could partition $\mathbb{R}$ though, so now that I've followed through on my end, you refuse to even read it?

Do you realize that in the time you've spent digging up that old thread you could have read the OP?! I'm baffled...

What I did long ago with CRGreathouse is not comparable to what I've done here. I've learned a lot since then.

Occasionally I buzz over to the physics forums when I just want a straight answer because, yes, they don't mess around over there.

I gave you a very clean OP. You've said multiple times now that you haven't read it. Why waste my time? As you said, you are not obligated. Yes, I'm sure I'll get a straight answer at the physics forums. I only go there once in a great while because they don't mess around like here.
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April 16th, 2017, 04:02 PM   #9
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PS - that "15 page thread" was mainly between CRGreathouse and Rob Osborn with very few posts by me.
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April 17th, 2017, 12:44 PM   #10
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Interesting that you should bring in Physics here because random is one of those terms that the Physicist and the Mathematician interpret differently.

The Mathematician is mostly interesting in the result (the kolmogorov definition) whereas the Physicist is mostly interested in the journey.
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