The h Bar

Secret

16:02

Yes exactly, basically, the idea of that fraction is so independent from the number of repeats that everytime you try to do it, you get a different fraction and it never converges to any nice pattern

It's as if each trial is an independent event, with a value of probability that is not fixed

barrycarter

@DavidZ I think it's a bit more than that. Your task is to prepare a bag full of plusses and minuses in such a way that no one, not even you, knows what the percentage of plusses is.

The problem is: no matter how deep you nest your random processes, there is always a best guess for the probability of pluses.

@Secret Now... if accept the negation of the Axiom of Choice, you can theoretically do this, but not in practice.

Secret

make sense

barrycarter

Ooooh, wait.

Suppose you use the results of your previous selection to choose what the distribution is for the next pick?

ACuriousMind

The law of large numbers says that any average you get from samples will converge to the true average with repeated sampling. I still don't get what we're looking for.

barrycarter

@ACuriousMind Can you find a selection process where you can't predict what that number will be?

0celo7

16:06

@Slereah QFT has never raised the GDP.

barrycarter

Actually, what I just said might work, but it's easier in the non binary case. Every time you select a number, the next number will be selected from a probability distribution based on the number you just selected. However, I think that's still wrong somehow.

ACuriousMind

@barrycarter Well...e.g. the Cauchy distribution doesn't have a mean.

barrycarter

@ACuriousMind Yeah, and the alpha and beta distributions don't have SDs, but I'm not sure they count. Let me take a quick look at Cauchy

@ACuriousMind OK, so if you kept picking numbers from Cauchy and averaged them, the average would run off to +Infinity or -Infinity (depending on your pre-chosen fixed parameters?)

0celo7

@ACuriousMind How

I don't believe that for a second.

barrycarter

@0celo7 Sort of like the martingale I'm guessing.

@0celo7 No finite mean.

0celo7

16:13

Can you not integrate xP(x)?

ACuriousMind

@barrycarter No, I think it will actually converge to the median...

@0celo7 Exactly

barrycarter

@ACuriousMind If that's true, than that's the mean.

ACuriousMind

@barrycarter No, the mean doesn't exist, the integral is undefined.

barrycarter

@0celo7 Just because you can integrate P(x) and get a finite result, doesn't mean you can integrate x P(x) and get a finite result.

@ACuriousMind OK, the integral is undefined, but if you average enough samples, you get the median? (cough cough)

0celo7

@barrycarter Proof?

ACuriousMind

16:15

@barrycarter It is just a hunch. I don't actually know what happens to sampling from the Cauchy distribution

barrycarter

@0celo7 WP the Cauchy distribution, although the Martingale is another example. 2^-i probability of getting i.

ACuriousMind

Perhaps it also just dances all over the place and never converges.

0celo7

WP?

ACuriousMind

@0celo7 Wikipedia.

0celo7

@ACuriousMind Stop personifying mathematics

barrycarter

16:16

@ACuriousMind Hmmm.... you might be making sense... but your argument would mean the Cauchy distribution is symmetric in values?

0celo7

@ACuriousMind how do you know that

barrycarter

@ACuriousMind If what you're saying about dancing is true, you've found @Secret's answer. However, I believe that's literally impossible.

I think that "no mean" just means there's no finite mean.

ACuriousMind

@0celo7 Through a combination of intelligence and intuition.

barrycarter

I am this close to doing some actual mathematics and seeing what happens when you integrate the Cauchy PDF.

Secret

ACuriousMind

16:18

@barrycarter No, the integral for the mean is also not divergent in the sense that it would actually be infinte. you just can't assign a finite value to it (or rather, depending on your regularization, you can assign any value to it)

I.e. the integral does not converge

barrycarter

@ACuriousMind For a given fixed set of parameters, you're saying the integral does not converge?

ACuriousMind

Even if you would allow it to be $\infty$ or $-\infty$.

@barrycarter Yes

Secret

If what you guys said is true, then this really surprise me, because this cauchy distribution looks really ordinary in terms of shape

barrycarter

@ACuriousMind I must call shenanigans.

Secret

I mean the first thing that came to mind when I look at it is it looks like any bell curve lookign symmetric distribution

Guess I should not be fooled by deceptively simple looking shapes like these when questions like these came up in the future....

0celo7

16:21

@ACuriousMind Pleasse give me some of that

ACuriousMind

@barrycarter Well, if you can integrate it, be my guest, but I'm pretty sure $xf(x)$ for $f$ the Cauchy distribution is just non-integrable.

Just choose the median at 0, then $f(x)$ is symmetric.

Hence $xf(x)$ is odd.

barrycarter

@ACuriousMind I call special case!

ACuriousMind

If the integral of $xf(x)$ existed, it would be zero.

barrycarter

I think that's a special case though.

ACuriousMind

@barrycarter You can always shift it inside the integral.

barrycarter

16:22

There are multiple parameters I think.

0celo7

@ACuriousMind Tue distribution is even

So why isn't the integral zero

ACuriousMind

58 secs ago, by ACuriousMind

If the integral of $xf(x)$ existed, it would be zero.

barrycarter

I realize there are functions like sin(x) which have divergent non-infinite integrals, I'm just saying it can't happen for a probability distribution.

0celo7

Does the integral of X over the real line exist?

Well it's zero!

So why shouldn't it exist

barrycarter

We're talking about a definite integral here. A symmetric bound function's integral .. no, that doesn't quite work.

ACuriousMind

16:24

@0celo7 Yes. The integrals of $x$ and $f(x)$ exist, separatey. But that doesn't mean the integral of $xf(x)$ exists.

0celo7

@ACuriousMind what the hell

That makes zero sense

ACuriousMind

The integrable functions are not closed under multiplication, and the Cauchy distribution is an example of that.

barrycarter

@0celo7 He's right about that part.

But the integral of x doesn't exist.

0celo7

I disagree, the integral is obviously zero

ACuriousMind

@0celo7 Well, prove it.

barrycarter

16:25

@0celo7 Only if the x itself is symmetrical.

0celo7

@ACuriousMind It's odd!

barrycarter

@0celo7 And other conditions apply.

@0celo7 What's the integral of sin(x)

ACuriousMind

Being integrable is like being guilty: Things are non-integrable (innocent) until proven otherwise

barrycarter

It's odd too

ACuriousMind

@0celo7 The theorem you're trying to use has "the integral exists" as a prerequisite.

barrycarter

16:26

Yes, you can re-arrange how you integrate if you know the integral converges.

Then the positives and the negatives would cancel out.

ACuriousMind

It's like that resummation theorem - if a series converges absolutely, you can reorder the summands however you like. However, if the series doesn't converge, then different orderings will give different "results".

Secret

classica example is the 1+2-3+4-5 ...

0celo7

@barrycarter zero

barrycarter

Cauchy distribution

The Cauchy distribution, named after Augustin Cauchy, is a continuous probability distribution. It is also known, especially among physicists, as the Lorentz distribution (after Hendrik Lorentz), Cauchy–Lorentz distribution, Lorentz(ian) function, or Breit–Wigner distribution. The Cauchy distribution is often used in statistics as the canonical example of a "pathological" distribution since both its mean and its variance are undefined. (But see the section Explanation of undefined moments below.) The Cauchy distribution does not have finite moments of order greater than or equal to one; onl...

I was going for a subsection link.

0celo7

@ACuriousMind No. Do the symmetric integral from -L to L and take the limit.

Secret

16:29

Mind is still blown It still puzzles me why this thing has such a nice shape

barrycarter

@0celo7 Do that for sin(x)

ACuriousMind

@0celo7 Ah, but that is not how the improper integral at both ends is defined!

barrycarter

@Secret No, there are other examples where like 1 + -1 + 1 + -1 + 1 where combining pairs of terms converges, but the series as a whole does not.

0celo7

@barrycarter 0

ACuriousMind

You have $\int_{-\infty}^\infty = \lim_{a\to -\infty}\lim_{b\to\infty}\int_a^b$, not $\int_{-\infty}^\infty = \lim_{a\to \infty}\int_{-a}^a$.

0celo7

16:30

@ACuriousMind do you have to take the limits separately

barrycarter

@0celo7 And do you actually believe the definite integral of sin(x) from -Infinity to +Infinity really is 0?

0celo7

@barrycarter yes

ACuriousMind

@0celo7 Yes, and it has to be independent of order.

0celo7

@ACuriousMind well that's a shitty rule

barrycarter

@0celo7 But it's not. It jumps between -1 and +1

0celo7

16:31

@barrycarter Proof?

Secret

@barrycarter This is why I like pathological mathematical objects, it really TEST one's understanding of the subject by stretching one's limit way way beyond the confines of intuition, (and I sort of believe that helps out of the box thinking)

barrycarter

@0celo7 Well, integrate sin(x) from 0 to L as L goes to infinity.

Secret

the problem is that there is no algorithm that can list them all out, which is why I often have trouble trying to study them because I don't even know their names

ACuriousMind

@0celo7 No, it's a necessary rule, otherwise you do not have that a function that is integrable on $X$ is integrable on every subset. E.g. the Cauchy distribution would be integrable on $\mathbb{R}$, but not on $\mathbb{R}_{\geq 0}$.

barrycarter

@Secret I like them because I'm pathological myself :)

0celo7

16:32

@ACuriousMind what

Oh

Well I see nothing wrong with that.

ACuriousMind

really? :P

0celo7

?

barrycarter

You could argue that the Cauchy distribution doesn't quite work because you can still predict its median and mode.

Secret

That's a possible extension, but my original question have not though that far. It is nice to know about that
which in that case, it will be either David's answer or your suggestion on the negation of axiom of choice

0celo7

@ACuriousMind I see nothing wrong with it.

barrycarter

16:35

Look at en.wikipedia.org/wiki/…

ACuriousMind

@0celo7 You see nothing wrong with a function that is integrable on $(a,b)$ not being integrable on $(0,b)$, where $a<0$?

barrycarter

(intentionally didn't put that link by itself so I can get a subsection link)

ACuriousMind

You started this because you wanted to use a nice property of integration (integrals of odd functions are zero). But this would destroy an even more basic property, namely $\int^a_b = \int_b^c + \int_c^a$.

Secret

@barrycarter (thoughts) Lp spaces are weird, where simple looking functions get pathological very quickly

barrycarter

Sadly, I don't know what an Lp space is.

0celo7

16:37

@ACuriousMind Of course, but that's not what I'm suggesting.

ACuriousMind

@0celo7 Okay, what are you suggesting?

0celo7

I'm talking about integrals over infinite sets

You're talking about integrals over sets with compact closure.

I agree that's a good rule for such integrals.

barrycarter

@0celo7 If I want to integrate sin(x) can I integrate from 0,Pi and then Pi,2*Pi, and then 2*Pi,3*Pi and so on?

ACuriousMind

@0celo7 You have to define the improper integral in the same way for both. Saying "take the limits separately, except when the limit point is $\infty$, then take only one symmetric limit" is just inconcsistent.

0celo7

@barrycarter sine is odd and its integral is 0

End of discussion

Slereah

16:40

and back home

0celo7

@ACuriousMind Inconsistent? I think it's the only reasonable thing to do.

ACuriousMind

Are you serious?

Secret

@ slereah you missed a great detail of juicy pathologicalness, lol

0celo7

Yes.

I don't see where it goes wrong

ACuriousMind

$\lim_{a\to\infty} \int_{-a}^a = \lim_{a\to\infty} \left(\int_{-a}^b + \int_b^{-b} + \int_{-b}^a \right)$.

Secret

16:42

@Acuriousmind another question (should be much easier to understand) Wikipedia said that conservation of probability is an exact conservation law due to symmetry. Is the symmetry U(1)?

barrycarter

@0celo7 LOL. "end of discussion" means you're conceding you're wrong :)

Slereah

Well, I guess that now it's time to prove things about the reps of the Lorentz group

0celo7

@barrycarter No.

barrycarter

Someone should go back to the definition of an integral from -Infinity to +Infinity to show engineer boy the error of his ways :)

0celo7

It means I'm in class.

barrycarter

16:44

@0celo7 Oh, so those were two separate statements?

0celo7

@barrycarter no.

ACuriousMind

@Secret Yes

barrycarter

@0celo7 You seem very negative today.

Secret

ok

barrycarter

I'm still annoyed about the Cauchy distribution. It.. must... have.. a ... mean.

Secret

16:46

I am still puzzle why it has such a NICE shape, lol

Slereah

there's plenty of distributions with no means

ACuriousMind

@0celo7 Here's where it goes wrong: If you define $\int_{-\infty}^\infty$ by the symmetric limit, you cannot write $\int_{-\infty}^b + \int_b^{-b} + \int_{-b}^\infty$ for some $b$, yet $\int_{U\cup V} = \int_U + \int_V$ is, for disjoint $U,V$, a basic property of any integral.

barrycarter

Women have a nice shape too, and no one understands them.

Slereah

Just look at the Saint Petersburg distribution~

barrycarter

@Slereah But usually because the mean is infinite, right?

@Secret I see a flaw in your question. In order to choose numbers from a distribution, you must already have a method of choosing random numbers, right?

Also, even though the Cauchy distribution doesn't have a mean, it still has a PDF.

Secret

16:50

Well it's more like what David said: I do a series of experiments, and I get no long term % sucess that is a well defined number

I am not sure if that implies I have implicitly chosen random numbers in the process
and yes

0celo7

@barrycarter I'm having life problems

barrycarter

@0celo7 You know the solution to that.

0celo7

Mass shooting?

barrycarter

@0

@0celo7 Just requires one shooting cowboy :)

@Secret In order to choose a number from a distribution, I think you need an existing random number.

@Secret Also, the Cauchy thing doesn't work for a binary experiment.

I think @ACuriousMind deflected the issue. Cauchy still has a PDF.

ACuriousMind

@barrycarter Well, what the hell are you looking for, then? Probability theory is usually defined by PDFs, what is a "random process" if not one given by a probability distribution?

barrycarter

16:55

@ACuriousMind It may not exist, but: can you fill a bag in such a way that no one, not even you, can predict anything about what the output of choosing things from that bag will be?

ACuriousMind

Sure. Put on a blindfold.

barrycarter

@ACuriousMind With Cauchy, you can't predict the mean, but you can predict the shape of the PDF.

@ACuriousMind sdafaosj

@ACuriousMind Oh, you meant in order to choose the input to the bag.

0celo7

@barrycarter what

barrycarter

@0celo7 Not a MASS shooting... unless you're talking about your own mass...

0celo7

@ACuriousMind I disagree with your assertion. It's obvious what the mean of the Cauchy distribution is

barrycarter

16:56

@ACuriousMind Well, this gets a little philosophical then. Blindfold or not, how do you choose which things to put in the bag? Describe your process?

0celo7

So anything you say that disagrees must be wrong.

barrycarter

@0celo7 Sadly, I have to agree with @ACuriousMind .. and I hate doing that, so please stop being wrong so much.

ACuriousMind

@0celo7 You claimed you were serious, and now you are saying something like that?

barrycarter

That's not even a complete sentence!

ACuriousMind

@barrycarter Well, that thing with the blindfold wasn't a serious suggestion. I don't think what you're looking for exists.

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