My Math Forum The Cantor Ternary Set the "Vitali Way" - Uniform Distribution QA

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 May 6th, 2017, 07:21 PM #1 Senior Member   Joined: Jun 2014 From: USA Posts: 246 Thanks: 12 The Cantor Ternary Set the "Vitali Way" - Uniform Distribution QA A few weeks ago I was using Vitali sets as a means of considering a uniform distribution over $\mathbb{N}$. The intro to my previous work is a good place to start for this thread (the full original work is provided below for posterity): A Consideration of the Fabled Uniform Distribution Over the Naturals Introduction: It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$. Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers. Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well. Question(s) for This Thread: As opposed to partitioning $[ \,0, 1) \,$ as described above using non-measurable Vitali sets, we can instead partition the Cantor ternary set $\mathcal{C}$ into countable subsets* that are then bijected with $\mathbb{N}$. Unlike the non-measurable Vitali sets I used previously (see full paper below), each subset of $\mathcal{C}$ has a measure of 0 because $\mathcal{C}$ itself has a measure of 0. Selecting an element of $\mathcal{C}$ uniformly at random then results in the selection of a natural number uniformly at random as well. Only this time, the result is that the probability of selecting any given natural number $p$ becomes 0 because the measure of the subset of $\mathcal{C}$ containing all elements that map to $p$ is 0. We are left with: 1) $P(p = n) = 0$ for all $n \in \mathbb{N}$. 2) Much like selecting a real uniformly at random from $[ \,0, 1) \,$, we still paradoxically get $P( \,p=1) \,+P( \,p=2) \,+P( \,p=3) \,+... = 1$. 3) And finally, $P( \,p=1) \,=P( \,p=2) \,=P( \,p=3) \,=...$. I believe this is a true uniform distribution over the natural numbers in the classical sense??? * My method of partitioning $\mathcal{C}$ is to partition it into a countable number of uncountable sets that contain one and only one element from each equivalence class contained in a collection of equivalence classes that also form a partition of $\mathcal{C}$ (similar to how we can partition $[ \,0, 1) \,$ into a countable number of uncountable Vitali sets that contain one and only one element from each Vitali equivalence class where the collection of Vitali equivalence classes also partitions $[ \,0, 1) \,$ when restricted to $[ \,0, 1) \,$). I definitely hope for some feedback, please and thank you!!! ... Original Work Referenced Above: A Consideration of the Fabled Uniform Distribution Over the Naturals Introduction: It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$. Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers. Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well. Definitions: Let $V^{( \,0.5, 1) \,}$ 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. For any real number $r$, let $d(r)$ equal the one and only one element $v \in V^{( \,0.5, 1) \,}$ such that $r - v \in \mathbb{Q}$. Let $h : \mathbb{N} \longmapsto \mathbb{Q} [ \, 0, 1 ) \,$ be bijective. Let $k : [ \, 0, 1 ) \, \longmapsto \mathbb{N}$ be surjective: $$k(x) = \begin{cases} h^{-1}(x - d(x) + 0.5), && x \geq d(x) - 0.5 \\ h^{-1}((x + 1) - d(x) + 0.5), && x < d(x) - 0.5 \end{cases}$$ Let $V^n = \{ x \in [ \,0,1) \, : k(x) = n \}$ for each $n \in \mathbb{N}$. We then have $V^{( \,0.5, 1) \,} = V^{h^{-1}(0.5)}$, for example. Each $V^{n}$ will be a Vitali set by definition with the collection $\{ V^{n} : n \in \mathbb{N} \}$ forming a partition of $[ \,0, 1) \,$. Let $x$ be an element of $[ \,0, 1) \,$ selected uniformly at random. Let $p = k(x)$. By definition, $p$ has now been selected uniformly from $\mathbb{N}$. Comments on Uniformity: A uniform distribution is a concept of translation invariance. For example, if $S$ is a measurable set, we may want the probability of $S$ to be the same as the probability of $\{y : y = z + n, z \in S \}$ for each natural number $n$. In the case of function $k$ over the domain $[ \,0, 1) \,$, however, we end up mapping each element of each non-measurable Vitali set $V^{n}$ to a distinct natural number $n$. Where $a, b \in \mathbb{N}$, it is easy to see that the probability of $x$ falling within $V^{a}$ is equal to the probability of $x$ falling in $V^{b}$ (thus the probability that $p = a$ equals the probability that $p = b$), but we cannot rely on a Lebesgue measure as a means of establishing probability or creating any sort of cumulative distribution function on $\mathbb{N}$. The probability of selecting any given natural remains undefined. B.J.K. April 19, 2017
May 7th, 2017, 12:10 AM   #2
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Disclaimer: Apologies if this seems crabby. I only posted it to get it out of my system. Your post irritated me. Better if you don't read it.

Quote:
 Originally Posted by AplanisTophet A few weeks ago I was using Vitali sets as a means of considering a uniform distribution over $\mathbb{N}$. The intro to my previous work is a good place to start for this thread (the full original work is provided below for posterity):
Posterity? Why? Your idea was thoroughly debunked by multiple people on three separate websites.

Quote:
 Originally Posted by AplanisTophet A Consideration of the Fabled Uniform Distribution Over the Naturals
Is that phrasing some kind of joke? I don't get it. You never had a uniform probability distribution and at one point you seemed to understand that.

Quote:
 Originally Posted by AplanisTophet Introduction: It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$.
Why are you telling us all this? It seems very self-indulgent, since it's irrelevant to your argument.

Quote:
 Originally Posted by AplanisTophet Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers.
I assume you mean a countable collection of uncountable sets, yes?

Quote:
 Originally Posted by AplanisTophet Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well.
Have you devolved into complete crankery? You know that it does no such thing, since the Vitali set and all its rational translates are nonmeasurable. Why are you pretending to have forgotten that you understood this point? This entire paragraph seems very disrespectful to those readers (myself in particular) who have put in a lot of time to explain the problem with your idea.

Quote:
 Originally Posted by AplanisTophet Question(s) for This Thread: As opposed to partitioning $[ \,0, 1) \,$ as described above using non-measurable Vitali sets, we can instead partition the Cantor ternary set $\mathcal{C}$ into countable subsets* that are then bijected with $\mathbb{N}$.
But you're not going to bother to tell the readers what the Cantor set is? For anyone who cares it's on Wikipedia, but its key feature for your purposes is obviously that it's an uncountable set of measure zero. Won't help you though.

Quote:
 Originally Posted by AplanisTophet Unlike the non-measurable Vitali sets I used previously (see full paper below), each subset of $\mathcal{C}$ has a measure of 0 because $\mathcal{C}$ itself has a measure of 0.
True.

Quote:
 Originally Posted by AplanisTophet Selecting an element of $\mathcal{C}$ uniformly at random then results in the selection of a natural number uniformly at random as well. Only this time, the result is that the probability of selecting any given natural number $p$ becomes 0 because the measure of the subset of $\mathcal{C}$ containing all elements that map to $p$ is 0.
True. But we noted weeks ago that there's no great trick in putting a uniform distribution on the naturals by assigning 0 to each natural. What's the point of that?

Quote:
 Originally Posted by AplanisTophet We are left with: 1) $P(p = n) = 0$ for all $n \in \mathbb{N}$.
Yes.

Quote:
 Originally Posted by AplanisTophet 2) Much like selecting a real uniformly at random from $[ \,0, 1) \,$, we still paradoxically get $P( \,p=1) \,+P( \,p=2) \,+P( \,p=3) \,+... = 1$.
Nonsense. First, you give not a shred of justification for such an obviously false claim. Secondly, it's obviousy false because of countable additivity. You just told us that the measure of each natural is 0. I can't even imagine what you're thinking here.

Quote:
 Originally Posted by AplanisTophet 3) And finally, $P( \,p=1) \,=P( \,p=2) \,=P( \,p=3) \,=...$.
Yes, they're all zero, as is the sum of their probabilities. You've got nothing. Literally.

Quote:
 Originally Posted by AplanisTophet I believe this is a true uniform distribution over the natural numbers in the classical sense???
At this moment I say you've officially gone full crank. You've been repeatedly told by multiple people on multiple sites, and at various times you yourself have understood, that there is no such thing unless it's trivially 0. Which is what you've got here, despite your unsupported false claim to the contrary.

Quote:
 Originally Posted by AplanisTophet * My method of partitioning $\mathcal{C}$ is to partition it into a countable number of uncountable sets that contain one and only one element from each equivalence class contained in a collection of equivalence classes that also form a partition of $\mathcal{C}$ (similar to how we can partition $[ \,0, 1) \,$ into a countable number of uncountable Vitali sets that contain one and only one element from each Vitali equivalence class where the collection of Vitali equivalence classes also partitions $[ \,0, 1) \,$ when restricted to $[ \,0, 1) \,$).
That's a heck of an incoherent run-on sentence. Did you even read it?

But you are going to have a problem. I imagine (since you don't bother to say) that you are applying the same equivalence relation to the Cantor set that we did to the reals when we formed the Vitali set. But in that case, the difference of two reals is real hence either rational or not, so you have a well-defined binary relation that partitions the reals.

But what makes you think this is a well-defined relation on the Cantor set? Specifically, given two members of the Cantor set, what makes you think their difference is in the Cantor set? Offhand I don't think it is. The Cantor set consists of reals whose ternary expansion only contains 0's and 2's. If one number has a 0 in ternary position n, and the other number has a 2, the difference will have a 1 in that place. It does not appear that the Cantor set is closed under subtraction.

So you can't even get your construction off the ground; and worse, you have an incoherent run-on sentence where you needed a proof. You just waved your hands at this when you needed to do some math.

Quote:
 Originally Posted by AplanisTophet I definitely hope for some feedback, please and thank you!!!
This post irritated me. I only responded so I didn't have to pick it apart in my mind. You waved your hands at the places that needed you to do some math: one, your obviously false claim that the sum of the probabilities is one (how on earth could that be?) and two, you didn't show that you have a valid binary relation on the Cantor set. (And I don't think you do).

Quote:
 Originally Posted by AplanisTophet Original Work Referenced Above: A Consideration of the Fabled Uniform Distribution Over the Naturals Introduction: It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$. Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers. Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well.
Is there some reason you printed this twice? Oh now I remember. For posterity. Sorry if my annoyance is showing. Let me try to be more supportive. Wouldn't it have been more effective to shore up the parts of your exposition where the math is absent, rather than repeating an exposition that's not relevant to your new argument?

Quote:
 Originally Posted by AplanisTophet Definitions: Let $V^{( \,0.5, 1) \,}$ 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.
I already pointed out to you that the .5 business is an irrelevancy that adds nothing but confusion. Since you didn't even bother to fix this problem, I was strongly de-motivated from spending any more time helping you. As I say I am only posting this to get the bad taste out of my mind. It's really annoying to see you paying so little attention to not only me, but people on other sites who took the time to help you sort through your many conceptual errors.

Quote:
 Originally Posted by AplanisTophet For any real number $r$, let $d(r)$ equal the one and only one element $v \in V^{( \,0.5, 1) \,}$ such that $r - v \in \mathbb{Q}$. Let $h : \mathbb{N} \longmapsto \mathbb{Q} [ \, 0, 1 ) \,$ be bijective. Let $k : [ \, 0, 1 ) \, \longmapsto \mathbb{N}$ be surjective: $$k(x) = \begin{cases} h^{-1}(x - d(x) + 0.5), && x \geq d(x) - 0.5 \\ h^{-1}((x + 1) - d(x) + 0.5), && x < d(x) - 0.5 \end{cases}$$ Let $V^n = \{ x \in [ \,0,1) \, : k(x) = n \}$ for each $n \in \mathbb{N}$. We then have $V^{( \,0.5, 1) \,} = V^{h^{-1}(0.5)}$, for example. Each $V^{n}$ will be a Vitali set by definition with the collection $\{ V^{n} : n \in \mathbb{N} \}$ forming a partition of $[ \,0, 1) \,$. Let $x$ be an element of $[ \,0, 1) \,$ selected uniformly at random. Let $p = k(x)$. By definition, $p$ has now been selected uniformly from $\mathbb{N}$.
But you've already been shown several times that the translates are non-measurable so that you have done no such thing. Maybe this is petty or delusional of me but I take this as personally disrespectful to the time I've spent working with you.

Quote:
 Originally Posted by AplanisTophet Comments on Uniformity: A uniform distribution is a concept of translation invariance. For example, if $S$ is a measurable set, we may want the probability of $S$ to be the same as the probability of $\{y : y = z + n, z \in S \}$ for each natural number $n$. In the case of function $k$ over the domain $[ \,0, 1) \,$, however, we end up mapping each element of each non-measurable Vitali set $V^{n}$ to a distinct natural number $n$. Where $a, b \in \mathbb{N}$, it is easy to see that the probability of $x$ falling within $V^{a}$ is equal to the probability of $x$ falling in $V^{b}$ (thus the probability that $p = a$ equals the probability that $p = b$), but we cannot rely on a Lebesgue measure as a means of establishing probability or creating any sort of cumulative distribution function on $\mathbb{N}$. The probability of selecting any given natural remains undefined.
Yet you keep acting as if you think otherwise.

There's a real disconnect. You do understand that the translates are non-measurable, so that your idea is interesting but ultimately flawed. So why do you keep repeating things that you know are false? Why not write an exposition that respects the understanding you've gained and thereby respects those who have helped you toward this understanding?

Quote:
 Originally Posted by AplanisTophet B.J.K. April 19, 2017
Grandiosity, the sign of a crank.

You've gone over to the dark side, my friend. And frankly I have too. My own post here was very self-indulgent. I could have just kept it to myself. So take what's of value and leave the rest.

Three constructive criticisms to provide some redeeming social value:

1) Leave out the irrelevant exposition, especially the parts that claim things you know to be false, such as fining a uniform distribution.

2) Talk a little about the Cantor set and exactly, in detail, how you are partitioning it. The traditional equivalence relation doesn't work because the Cantor set does not appear to be closed under subtraction.

3) You need to justify or explain your claim that the sum of the probabilities is 1. It's obviously false but it would be helpful to know what you're thinking.

Ok now I won't have this running around my brain all night.

Last edited by Maschke; May 7th, 2017 at 12:50 AM.

May 7th, 2017, 05:10 AM   #3
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Quote:
1) My original idea was simply to devise a theoretical way to select a natural number from $\mathbb{N}$ with all naturals having an equal chance of being selected. According to you, I succeeded. What is this "debunked" bs you are referring to?

2) Ahhhh... I never claimed to have a uniform distribution over the naturals, I never directly tried to, and I've always understood that. The "Fabled" is an acknowledgement that such a distribution is non-existent.

3) It's the same thing I'm looking into doing with the Cantor ternary set as opposed to [0, 1]. You said before that you like that intro as a way of introducing conceptually what I was doing. I'm not trying to be self-indulgent...?

4) No, I said exactly what I meant. Yes, we also end up with a countable collection of uncountable sets (the rational translates) in addition to the uncountable collection of countable sets (the equivalence classes).

5) It does allow for the selection of a natural uniformly at random, which you admitted. It does not give a uniform distribution over the naturals or a cumulative distribution function because the probability of selecting each natural is undefined due the non-measurability of the Vitali sets. Originally, I kept trying to tell you that I didn't care about measure theory because I wasn't trying to do the latter (have a uniform distribution) and only do the former (devise a method to select a natural with no one natural being favored over any other). You kept pounding your fist on the measure theory stuff. It was irritating.

6) Getting somewhere, as in why it won't help me.

7) How can you say that? I thought a uniform distribution over the naturals was impossible. Now it's easy if we allow each natural to have probability 0 (noting one natural is guaranteed to be selected anyways)?

8- It's the old "we're sure to hit the dartboard but the chances of hitting any real on the dartboard are 0" paradox. See here, starting at 3:37 -

9) How can asking a question make me a full crank...?! Cranks assert they're right when they are wrong. I'm just asking. We are guaranteed to select a natural and all naturals have equal, definable, probability. I thought that was impossible, but now you're saying it's not if we let the probability trivially be 0. Ok, that answers my question I suppose.

10) It's actually not a run-on sentence... Anyways, you are misunderstanding. I have a method of partitioning the Cantor ternary set into countable sets so that I can biject each with the naturals. You should see that is easy enough. I have no intention of trying to partition the Cantor ternary set using the Vitali equivalence classes as you indicate.

11) I agree that dart board paradox is weird. I don't see how the probabilities can add to one either, but we're sure to hit the dart board, so yeah... Anyways, the hand waivery as to showing my actual partition of the Cantor ternary set is simply because I don't care to get into it if it doesn't do any good. It's not hard though either, as surely we can partition the Cantor ternary set into countable sets. That part is easy.

12) Yes. If I was considering doing the same thing, only with the Cantor ternary set instead of [0,1), then it seemed relevant to display the short process and conclusion as to what happens when Vitali sets were used.

13) You said the 0.5 thing was not a problem in the other thread. I could have used a MOD 1 method, but I like my function. There is nothing wrong with that, so wtf?

14) The method is to select a natural uniformly from all naturals, not develop a meaningful cumulative distribution function. Why is that so hard to understand? I succeeded in doing what I was trying to do. I already knew, and didn't care about, the fact that Vitali sets have an undefined measure when I started. You keep asserting I'm trying to do something I'm not.

15) No, I don't. wtf?

16) What is false in what I wrote? I acknowledged all the additional crap that you wanted me to in my comments on uniformity, did I not?

17) I think it's an interesting paper and it's my original work. So sue me for initialing it.

18- Your constructive criticisms: Thanks. So, assuming I can do with the Cantor ternary set what I said I could, you'd say it still doesn't matter anyways because apparently trivial 0 uniform distributions over that naturals don't matter.

Your whole problem with my method of selecting a natural uniformly at random (which I succeeded in doing) was that the probabilities assigned to each natural were undefined (like, duh) because Vitali sets are not measurable (again, like, duh). So now that I have a method of selecting that natural uniformly at random where I assign 0 probabilities based on measure theory to each natural, you say it doesn't matter. Explain yourself. As someone who has a little common sense even if not a regular mathematician, what you're saying now sounds conflicting.

Last edited by AplanisTophet; May 7th, 2017 at 05:22 AM.

May 7th, 2017, 08:56 AM   #4
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Morning. What set me off was the large amount of irrelevant fluff combined with the lack of specifics where they were needed. I had a nice night's sleep and I'm much less crabby today.

You can partition the Cantor set by intersecting it with the Vitali translates so no problem there, although you needed to make that explicit.

We discussed assigning 0 to everything weeks ago. If I manage to find it I'll post the reference.

You did claim the sum of the probabilities is 1, and that can't be. You say you're sure to hit something so the prob must be 1. No, the prob is still 0 because that's the measure of the Cantor set! To have a probability measure you'd have to assign 1 to the whole set and you haven't done that.

The bottom line is that your entire idea is a shaggy dog story. It boils down to: For each natural n, assign it probability 0. You found an incredibly convoluted way to do that, but it's not necessary.

It's like if we say, let $f(x) = 0$. It's the zero function because we say so. We are not required to produce a mechanism to make the assignment.

So your idea comes down to assigning 0 to each natural. Every word beyond that is irrelevant. You are fixated on the wrong things. And like I say, we did already discuss that possibility a long time ago.

Suppose we say, Ok, let $f : \mathbb N \to \mathbb R$ be the zero function $f(n) = 0$. So $f$ assigns $0$ to each natural number $n$.

Now suppose we randomly pick a natural number. We must pick SOME number so the probability of picking one is $1$. Hence we have a uniform probability measure on the naturals.

What's wrong with this argument? It's that you can't randomly pick a natural until you tell me what the probability distribution is; and there is no uniform probability distribution on the naturals. That's because the definition of probability distribution is that the measure of the entire set must be $1$. But you haven't got one of those, and you could never have one. Countable additivity again.

pps -- We had this conversation in December. See Set Theory Question

Quote:
 Originally Posted by Maschke So: Yes, you have a uniform measure on the naturals that assigns probability $0$ to every natural. However you don't have a probability measure, by the very definition of probability measure. I've made this point several times already and you don't seem to be engaging with it.
So you see you still have the exact same problem. You haven't got a probability measure AND you are not engaging with this argument.

I didn't look at the video but Numberphile is pretty good and I'd be very surprised if they said anything different.

Last edited by Maschke; May 7th, 2017 at 09:32 AM.

 May 7th, 2017, 06:01 PM #6 Senior Member   Joined: Jun 2014 From: USA Posts: 246 Thanks: 12 Ok, so let me sum this up: 1) With the Vitali method, we have that the probabilities must sum to 1 because the measure of [0, 1) is 1, but the individual probabilities themselves are undefined. 2) With the Cantor set method, we have that the probabilities do not sum to 1 because the measure of the Cantor set is 0, but the individual probabilities themselves are defined and equal to 0. That's the uniform distribution 'stuff' that you wanted me to make clear, or at least acknowledge, when I was trying to select a natural uniformly from $\mathbb{N}$, correct? I think we're all well and good on that. My confusion, or more specifically what I was asking for clarification on, was that when we are guaranteed to 'hit the dart board,' how can we not assert that the probabilities sum to 1? Numberphile says we don't have a good answer yet, or at least not a satisfying one, in the video I cited. It's short and interesting if you want to take a gander.
May 7th, 2017, 06:32 PM   #7
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Quote:
 Originally Posted by AplanisTophet 1) With the Vitali method, we have that the probabilities must sum to 1 because the measure of [0, 1) is 1, but the individual probabilities themselves are undefined.
The probabilities must sum to one because that's the definition of a probability measure. We can put a probability measure on the unit interval BECAUSE it already has measure 1. So your cause and effect are reversed here I think.

Quote:
 Originally Posted by AplanisTophet 2) With the Cantor set method, we have that the probabilities do not sum to 1 because the measure of the Cantor set is 0, but the individual probabilities themselves are defined and equal to 0.
Yes agreed. But as interesting as the "Cantor method" is, it's really not essential to the argument. We can just assign the probability 0 to every natural number. We don't need to define a specific method to do that.

Quote:
 Originally Posted by AplanisTophet That's the uniform distribution 'stuff' that you wanted me to make clear, or at least acknowledge, when I was trying to select a natural uniformly from $\mathbb{N}$, correct? I think we're all well and good on that.
Uniform distributions are easy, just assign everything the value 0. It's uniform probability distributions that are hard, because those require (by definition) that the measure of the whole set is 1.

Quote:
 Originally Posted by AplanisTophet My confusion, or more specifically what I was asking for clarification on, was that when we are guaranteed to 'hit the dart board,' how can we not assert that the probabilities sum to 1? Numberphile says we don't have a good answer yet, or at least not a satisfying one, in the video I cited. It's short and interesting if you want to take a gander.
I didn't watch the video. I don't watch videos. They take too long to get to the point. If you can summarize that would be great.

[Purely a personal quirk. I don't watch the wonderful Numberphile vids. I don't even usually watch the incredibly good Mathologer vids. I can't stand Vsauce. Etc. Don't take this personally. I'm not not watching YOUR video. I don't watch ANY videos.]

But a circle in the plane, for example, can always have its area normalized. If we have the unit circle, its area is $\pi r^2 = \pi$. So the circle whose radius is $\frac{1}{\sqrt{\pi}}$ has area $1$ so we can regard it as a probability space. Then the chance of hitting any particular point is $0$ but the chance of hitting SOME point is $1$. It's exactly like the unit interval.

So we can NEVER put a uniform probability measure on a countable set because singletons must have measure zero and then countable additivity forces the whole set to have measure zero.

But if a set is uncountable, we can sometimes put a probability measure on it by normalizing the measure, that is, scaling it so it's $1$.

That's the problem with the Cantor set. It's uncountable but its measure is $0$ so we can't normalize it no matter what we do.

Last edited by Maschke; May 7th, 2017 at 06:53 PM.

May 11th, 2017, 07:31 AM   #8
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Quote:
 Originally Posted by Maschke I didn't watch the video. I don't watch videos. They take too long to get to the point. If you can summarize that would be great.
It addresses the selection of a real on the interval [0, 1) uniformly at random and states that we don't have an answer in the infinite case where we assume:

$$\sum_{x \in [ \,0,1) \,} Pr(\ \,x) \, = \text{1 or 0?}$$

Quote:
 Originally Posted by Maschke But a circle in the plane, for example, can always have its area normalized. If we have the unit circle, its area is $\pi r^2 = \pi$. So the circle whose radius is $\frac{1}{\sqrt{\pi}}$ has area $1$ so we can regard it as a probability space. Then the chance of hitting any particular point is $0$ but the chance of hitting SOME point is $1$. It's exactly like the unit interval.
If I set aside trying to be a proper mathematician for a second and use only common sense and my intuition, then it seems to me that the sum of the probabilities must equal 1 because we are sure to hit the dart board. I'm no Kolmogorov, but if I was trying to create some probability axioms, that would be my first axiom (somewhat humorously):

"Who cares what your sample space actually is when, knowing only that it exists, we can assert that the sum of the probabilities assigned to each element in the sample space adds to 1."

I don't care to challenge any kind of standard math, so if we say that the probability of selecting any given natural is 0 using the Cantor set method, ok. If we further say that the sum of those probabilities is 0, ok. I do appreciate the clarification.

May 11th, 2017, 09:42 AM   #9
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Quote:
 Originally Posted by AplanisTophet It addresses the selection of a real on the interval [0, 1) uniformly at random and states that we don't have an answer in the infinite case where we assume: $$\sum_{x \in [ \,0,1) \,} Pr(\ \,x) \, = \text{1 or 0?}$$
It's $0$. It's an uncountable sum of terms, each of which is $0$. No matter what definition of uncountable sum you come up with, the sum will be $0$.

This does not contradict the fact that $\int_0^1 dx = 1$.

This is a philosophical mystery but it is not a mathematical mystery. I can't imagine why Numberphile would be so deliberately disingenuous unless you misinterpreted what they're saying. And even if they did get confused on this point, that only shows that even the great Numberphile makes mistakes.

But wait, let me offer a proof of my claim. Say we call the sum $1$. Now what is

$$\sum_{x \in [ \,0,2) \,} Pr(x)$$

Well by your logic, is it $1$? Or $2$? And what if I take the sum over the entire real line? Is the sum $1$? Or $\infty$?

If you think about this carefully you will see that the only consistent answer is that the sum of an uncountable number of $0$'s is $0$.

Quote:
 Originally Posted by AplanisTophet If I set aside trying to be a proper mathematician for a second and use only common sense and my intuition, then it seems to me that the sum of the probabilities must equal 1
If I put aside trying to be a physicist and use only common sense and intuition, stuff falls down because it's heavy. Or as Aristotle put it, everything goes to its natural place. Bowling balls fall to earth, and gases fly up into the air. Makes perfect sense.

What would be the point of this exercise in pretending to adopt 2000 year old physics? We have far more sophisticated theories of gravity and gases these days.

[Pro spelling tip: The plural of the noun gas is gases. The verb form is gasses. Gases or Gasses: What’s the Difference? - Writing Explained]

Since we have had this conversation numerous times and you clearly understand the point, what kind of response are you looking for here? Yes, intuitively heavy things fall faster than light things. Turns out they don't. Yeah science!

But, just for old time's sake, let's imagine this dialog:

You: Say we put all the natural numbers in a big bag. After all Cantor put them into a set, and that was just as strange an idea at the time and frankly it still is a strange idea if you think about it. Now we close our eyes, reach into the bag, and pull out a number at random. We must get SOME natural number. It's certain. So the probability must be 1.

Kolmogorov: Какого хрена? When you say that you pull out a number "at random," what is your probability distribution?

You: I haven't got one.

Kolmogorov: Aha!

Last edited by Maschke; May 11th, 2017 at 10:40 AM.

May 11th, 2017, 10:39 AM   #10
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Quote:
 Originally Posted by Maschke You: Say we put all the natural numbers in a big bag. After all Cantor put them into a set, and that was just as strange an idea at the time and frankly it still is a strange idea if you think about it. Now we close our eyes, reach into the bag, and pull out a number at random. We must get SOME natural number. It's certain. So the probability must be 1. Kolmogorov: Какого хрена? When you say that you pull out a number "at random," what is your probability distribution? You: I haven't got one. Kolmogorov: Aha!
To fix the bolded part:

Lol, me: Sure P(1) + P(2) + P(3) + ... = 1, but each P(n) is undefined. (i.e., I am perfectly not bothered by this fact. Jokingly, screw the Archimedean Property!!! I am even willing to look beyond the traditional definition of the reals to assign a positive value to each P(n) because such a value exists despite being undefined in the traditional sense... All I cared is that it exists).

Kolmogorov: I hate you!

...

Couldn't resist.

 Tags cantor, distribution, set, ternary, uniform, vitali way

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