Wikipedia:Reference desk/Archives/Mathematics/2009 August 14

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August 14

Infinity--Question/comment/request for comments

Gauss quote from Actual infinity: "I protest against the use of infinite magnitude as something completed, which is never permissible in mathematics. Infinity is merely a way of speaking, the true meaning being a limit which certain ratios approach indefinitely close, while others are permitted to increase without restriction." In Kenneth Kunen's book on forcing, I recall he said roughly proper classes are not defined in the most popular theory ZFC but are frequently talked about in an informal way and used as an aid to intuition. That is pretty akin to being not permissible, to being a manner of speaking, as Gauss described actual infinity. That makes me suspect that although mathematicians have genuinely tackled infinity and have some really big infinite sets like the reals and more, Aren't we confronting the same issue Gauss had? That is, we've tried to deal with his issue by grappling with infinite sets but ran into it again in proper classes?Rich (talk) 06:47, 14 August 2009 (UTC)

I'm not sure what the question is. What are you hoping for? Some comment that's better than what's already at that article? The reference desk isn't really a discussion board though sometimes it might seem otherwise. Dmcq (talk) 13:19, 14 August 2009 (UTC)

ZFC doesn't axiomatize classes but other set theories like NBG do. ZFC itself postulates the existence of certain collections of objects (called "sets") that obey a certain bunch of axioms (the ZFC axioms). It turns out that any model of ZFC must necessarily also contain some collections of objects (e.g. the ordinals in the model) that can't follow the axioms without leading to contradiction. This was considered paradoxical for a while (Burali-Forti paradox) but it just means that not every collection is a set. 70.90.174.101 (talk) 01:24, 15 August 2009 (UTC)

Well, actually, from a realist point of view, it means a bit more than that. The class of all ordinals is not a completed totality at all (if it were, it would have to be a set). So when we speak of "the class of all ordinals", what we're actually referring to, via a metalingustic circumlocution, is the predicate "is an ordinal", rather than a collection of any sort. --Trovatore (talk) 01:43, 15 August 2009 (UTC)

Responding to the original question: It seems to me that what you are talking about is the notion of the absolute infinite. As I understand it (and this is sort of a reconstruction from a modern point of view, one for which I don't exactly have clear references to point you to), you can look at it like this: Gauss says, "you can't treat infinite collections as being actual". Cantor says "you're right, but many of the things that were formerly thought of as being infinite, say the set of all natural numbers, are not "infinite" in that sense. They are transfinite; that is, beyond some limit, but not infinite or entirely without limit.".

See also limitation of size. Our article on that could use some (or a lot of) work. I started it but never got around to doing the literature search to really bring it up to snuff. --Trovatore (talk) 01:51, 15 August 2009 (UTC)

• Maybe Gauss had a notion of absolute infinite, since he was very smart and didn't publish many of his brilliant ideas? If he did maybe that means his quote was a denial of what he really thought, to avoid controversy, sort of like what some historian claimed was Gauss's lack of courage for not publishing his thoughts on noneuclidean geometry, though I've never heard that he actually REJECTED noneuclidean geometry.(I don't know if the claim of lack of courage is fair, since Gauss was a perfectionist and didn't publish things he hadn't had time to polish.) But if Gauss didn't have a notion of absolute infinity, then I think his quote must mean he rejected all quantities that weren't finite, that no infinity could be actual, really exist, like a zfcer would say some classes could not exist as sets, and only as collections"in a manner of speaking." Also, do we know if Gauss thought in terms of "cardinality of sets" for measuring quantity? Thanks to both of you for your thoughtful answers.Rich (talk) 15:21, 16 August 2009 (UTC)`

Regifting Robin

http://www.regiftable.com/regiftingrobinpopup.html

What's the secret behind this little game? --Halcatalyst (talk) 13:54, 14 August 2009 (UTC)

The same as it was last time. Algebraist 13:57, 14 August 2009 (UTC)

Sometimes, though, it comes up with something other than "board game." --Halcatalyst (talk) 16:29, 14 August 2009 (UTC)

When you take a 2-digit number and subtract both digits, you always get a number divisible by 9. So all that game has to do is label all the numbers divisible by 9 with the same item. --COVIZAPIBETEFOKY (talk) 17:58, 14 August 2009 (UTC)

Take a number x, it is made up of digits a, and b. in other words x = a * 10 + b.

now, take x and subtract a and b to get y: y = x - a - b.

Substitute values for x: y = (a * 10 + b) - a - b. y = a * 10 - a + b - b. y = a * 9.

if the possible values for a are (0-9) then the possible values for y are 0,9,18,27,36,45,54,63,72,81. If you look, you'll see that all of those positions have the same item. 148.134.37.3 (talk) 18:42, 20 August 2009 (UTC)

Showcase Showdown strategy

When should contestants spin again or stay to have the best chance of winning? I guess it would be different depending on whether you go first or second, and whoever goes third doesn't have to worry about it since they are just trying to beat the best score. Recury (talk) 17:29, 14 August 2009 (UTC)

What's the difference?

My statistics text book says this:

Suppose we calculate from one sample in our battery example the following confidence interval and confidence level: "We are 95 percent confident that the mean battery life of the population lies within 30 and 42 months' This statement does not mean that the chance is 0.95 that the mean life of all our batteries falls within the interval established from this one sample. Instead, it means that if we select many random samples of the same size and calculate a confidence interval for each of these samples, then in about 95 percent of these cases, the population mean will within that interval.

My question is - don't the two statements that are bolded mean the same thing? That is, doesn't one imply the other? What's the difference between the two? I have been scratching my head for a long time over this but can't figure it out and I'm feeling extremely stupid now =/ --ReluctantPhilosopher (talk) 20:47, 14 August 2009 (UTC)

I think that "We are 95 percent confident that the mean battery life of the population lies within 30 and 42 months" is talking about the accuracy of the calculation of the mean value. The statement "...the chance is 0.95 that the mean life of all our batteries falls within the interval established" is talking about the distrabution of the data. For example, let's say you tested one million batteries, and you found that 500,000 batteries had lives of 1 month and that 500,000 had lives of 71 months. In this case the mean battery life would be exactly 36, but none of the batteries would have a battery life within 30 and 42 months. ~~ Dr Dec (Talk) ~~ 22:19, 14 August 2009 (UTC)

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95% of the time the confidence interval will contain the population mean. That means if you have independent repetitions of the experiment with the mean remaining the same throughtout, in 95% of cases that will happen. But you're looking at just one case, where you got the interval from 30 to 42. The conclusion that they're saying is not justified is that you can be 95% sure in that one case. The difference is that being 95% sure in that one case—that one repetition of the experiment—is not a statement that says in 95% of all repetitions of the experiment, a specific thing happens.

In fact, sometimes you may even find something in your data that tells you that the one specific repetition of the experiment is one of the other 5% of repetitions, where the specified method of finding an interval gives you an interval that fails to include the population mean. And sometimes you might find information in your data that doesn't tell you for sure that you've got one of the other 5%, but makes it probable. That only happens when you've got a badly designed method of finding confidence intervals, but nonetheless the 95% confidence level is correctly calculated. Ronald Fisher's technique of "conditioning on an ancillary statistic" was intended to remedy that problem.

The statement that in one particular repetition of the experiment, which gave you the interval from 30 to 42, the population mean has a 95% chance of being in that interval, is a statement about 95% of possible values of the population mean, not about 95% of repetitions of the experiment.

The argument that one should be 95% sure, conditional on the outcome of that one particular repetition of the experiment may actually be reasonable in cases where all of the information in the data was taken into account in forming the interval, but it's not actually backed up by the math involved. Something other than mathematics, not as well understood, is involved. Michael Hardy (talk) 22:33, 14 August 2009 (UTC)

Summary: One statement is about 95% of all independent repetions of the experiment. The other is about 95% of all equally (epistemically) probable values of the population mean, given the outcome of one particular repetition of the experiment. "95% of repetitions" is a relative frequency, not an epistemic probability. Michael Hardy (talk) 22:36, 14 August 2009 (UTC)

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So, Michael, please tell us: what's the Wikipedia convention that made you put a line (like the one above) before your answer? ~~ Dr Dec (Talk) ~~ 22:53, 14 August 2009 (UTC)

You should not be feeling stupid. Your professor should. The concept of confidence interval is low quality science. The dispute between frequentist and bayesian statistics is behind this sad state of affairs. Bo Jacoby (talk) 11:56, 15 August 2009 (UTC).

I didn't want to indent at a different level from the previous comment but I want the boundary between the previous comment and mine to be clear. Michael Hardy (talk) 19:29, 15 August 2009 (UTC)

One of the problems you're facing is that the population mean is not a random variable. The mean lifetime of your batteries is a fixed number. We don't know what it is, but in repeated experiments it will never change. This raises issues with the the frequency view of probability. You can't really assign a (frequency based) probability to the population mean, as it's always exactly the same, no matter how you conduct your sample. Any discussion of probability with respect to the population mean would refer, rather, to our state of knowledge (or lack thereof) about the mean - a Bayesian or epistemic probability. The catch is that the standard confidence intervals were derived from frequency-based statistics. Bayesian statistics has it's own related measure, the credible interval, but the two are not necessarily equivalent. So formally, the "95%" has to refer to a frequency-based probability for a random variable: "If you carry out random sampling multiple times, 95% of the time the calculated confidence interval (the endpoints are random variables) will enclose the population mean." You can't say: "If you carry out random sampling multiple times, 95% of the time the population mean (NOT a random variable) will be within 30 and 42 months", because the population mean either is in that interval or it isn't. It doesn't jump around with repeated sampling. -- 128.104.112.102 (talk) 19:30, 15 August 2009 (UTC)

Lipschitz, absolute continuity

I'm working through problems in Royden today. Probablem 5.20b says:

Show that an absolutely continuous function f satisfies a Lipschitz condition if and only if |f'| is bounded.

This isn't true is it? I mean, |x| is Lipschitz but the derivative does not exist at x = 0 so it is not bounded. Something that doesn't exist can not be bounded. But, whenever it does exist, it is bounded. Would it be correct if it were changed to "if and only if |f'| is bounded whenever it exists."??? StatisticsMan (talk) 23:45, 14 August 2009 (UTC)

Yes; more precisely, recall that an absolutely continuous function f on an interval I is differentiable a.e. in I, so that f' is defined a.e.; then, f is Lipschitz if and only if f' is essentially bounded. If I is a bounded interval youu may also rephrase it saying that Lip(I) coincides with the Sobolev space W^1,∞(I). --pma (talk) 07:54, 15 August 2009 (UTC)

Okay, thanks. So, say f is absolutely continuous and Lipschitz. Then, if the derivative exists at a point a, it is $\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$. By the Lipschitz condition, $-M\leq {\frac {f(x)-f(a)}{x-a}}\leq M$, so the limit is also within these bounds, when it exists. Now, assume f is absolutely continuous and the derivative is bounded, whenever it exists. I'll have to think about this one. Thanks! StatisticsMan (talk) 13:01, 15 August 2009 (UTC)

And, as to the first implication, remember that a Lipschitz function is in particular absolutely continuous. For the other implication, remember the generalization of the fundamental theorem of calculus in the setting of absolutely continuous functions. --pma (talk) 13:57, 15 August 2009 (UTC)

Okay, so this is pretty simple too. Assume |f'| <= M whenever it exists, which is almost everywhere since f is absolutely continuous on [a, b]. Also, the derivative is measurable and f is equal to the antiderivative of its derivative, $f(x)=f(a)+\int _{a}^{x}f'(t)\,dt$. Then, for any x < y in [a, b], we have

$|f(y)-f(x)|\leq \int _{x}^{y}|f'(t)|\,dt\leq \int _{x}^{y}M\,dt=M|y-x|$. StatisticsMan (talk) 15:08, 15 August 2009 (UTC)

Measurable functions/Derivatives

Let f : [0, 1] to R be a measurable function and E a subset of {x : f'(x) exists}. If m(E) = 0, show that m(f(E)) = 0.

This is a qual problem from the past that I have not seen a solution for. Any ideas? Thanks StatisticsMan (talk) 23:46, 14 August 2009 (UTC)

Some hints for a completely elementary proof.

For any positive integer k consider

$\scriptstyle E_{k}:=\{x\in E\ :\ |f'(x)|<k\}$.

Since by assumption $\scriptstyle \mathrm {m} (E_{k})=0$, there exists a relatively open nbd $\scriptstyle U\subset [0,1]$ of $\scriptstyle E_{k}$, such that $\scriptstyle \mathrm {m} (U)<1/k^{2}$.

Consider the collection $\scriptstyle {\mathcal {J}}$ of all those (relatively) open intervals $\scriptstyle J\subset U$ such that $\scriptstyle \mathrm {diam} (f(J))<k\mathrm {diam} (J)$, that is, such that $\scriptstyle f(J)$ is contained in an interval of length less than k times the length of $\scriptstyle J$.

Consider the union $\scriptstyle V$ of these intervals. The relevant facts that you can check are: $\scriptstyle V$ is a relatively open neighborhood of $\scriptstyle E_{k}$, and $\scriptstyle V\subset U$; moreover (key point) each connected component of $\scriptstyle V$ is an interval $\scriptstyle J$ that belongs to the class $\scriptstyle {\mathcal {J}}$.

This implies that $\scriptstyle \mathrm {m} (f(E_{k}))<1/k$, whence $\scriptstyle \mathrm {m} (f(E))=0$ since $\scriptstyle E$ is the increasing countable union of the $\scriptstyle E_{k}$. --pma (talk) 09:53, 15 August 2009 (UTC)