Wikipedia:Reference desk/Archives/Mathematics/2007 November 27

= November 27 =

Probability notation: P or Pr?
When discussing probability, is it more common/is it the convention to use the notation P or Pr? Larry V (talk &#124; e-mail) 01:37, 27 November 2007 (UTC)


 * P is the convention that I have always seen.130.166.172.130 (talk) 01:51, 27 November 2007 (UTC)


 * We used Pr in my induction class, but that was in a philosophy class.--droptone (talk) 02:13, 27 November 2007 (UTC)

See the article on probability using P. The notation (A|B) for conditional probability allows you to omit the P alltogether. Bo Jacoby (talk) 07:35, 27 November 2007 (UTC).
 * I don't think writing (A|B) instead of P(A|B) is common. But I agree that the more common convention is P. -- Meni Rosenfeld (talk) 08:13, 27 November 2007 (UTC)

No, it might not be very common, but it is a straightforward generalization of the useful Iverson bracket. Bo Jacoby (talk) 16:55, 27 November 2007 (UTC).
 * $$\mathbb P$$ is also in common use, at least in Cambridge. I don't think I've ever seen Pr used. Algebraist 14:49, 27 November 2007 (UTC)
 * The book Randomized Algorithms by Motwani and Raghavan (ISBN 0521474655) uses the notation Pr[X]. Phaunt (talk) 15:22, 27 November 2007 (UTC)
 * Hmm, I've only ever seen $$\mathbb P$$ to refer to a projective space. Usually the blackboard bold letters indicate sets. Donald Hosek (talk) 16:01, 27 November 2007 (UTC)
 * I've found the blackboard bold letters (or handwritten versions of them) used a lot for statistical operators, e.g. $$\mathbb E$$ for expectation, $$\mathbb V$$ for variance, etc. It seems to be a personal preference though, rather than the common usage. Confusing Manifestation (Say hi!) 22:11, 27 November 2007 (UTC)

I have seen both notations. I have the impression that Pr is the usual convention in econometrics (well, maybe Economics in general). But it's not big deal if you make the proper disclaimer with the convention you have followed, I guess. By the way, $$\mathbb P$$ could be confused with the power set operator. And sometimes, the notation P(A) is used to affirm that the element A possesses the property P. Pallida Mors  22:38, 27 November 2007 (UTC)
 * Again in cambridge, powerset is normally $$\mathcal{P}$$ or some slightly curlier version, which is hard to confuse with $$\mathbb{P}$$ (and a lot easier to write than the $$\mathfrak{p}$$ our German lecturer insists on using for prime ideals!). Algebraist 00:00, 28 November 2007 (UTC)

I've seen P(X), Pr(X), p(X) (usually for continuous distributions) and even $$\mu(X)$$. I'd say that P is most common, at least in the field of Artificial Intelligence. risk (talk) 14:15, 28 November 2007 (UTC)


 * Using μ for a probability measure could be a sign that your lecturer is coming from an analysis background rather than a statistical or a combinatorial one; it could also be a sign that you're about to dive head-first into stuff like singular distributions and σ-algebras. One convention I've sometimes seen is to use μ for general measures and p for probability measures that are required to be normalized.  —Ilmari Karonen (talk) 15:43, 28 November 2007 (UTC)


 * I have seen and used both P and Pr at the same time, even in the same formula! Usually I used Pr for the most 'outer' probability space, and P for some smaller probability space. I believe it is quite common in bootstrapping (statistics)(Igny (talk) 04:35, 29 November 2007 (UTC))

I've seen most of the forms mentioned above in mathematics. My take is that there will never be a single accepted form there because of symbol conflicts. For example, I do too much that involves the symbol "P" to be comfortable with using it for probability. I even had problems with Pr once, ending up using "Prob" though not in published work. I think the tradition is to define the usage and then use it consistently through the work. In certain fields like statistics or econometrics, the usage may be standard enough to avoid this, I don't know. But there's too much variety in regular mathematics to standardize on this. -- 68.189.79.81 00:26, 3 December 2007 (UTC)

Ordinals
I don't have a solid understanding of what I'm talking about, so please bear with me if I'm not entirely clear.

The article on ordinals says that $$\omega^\omega$$ is countable. But $$2^{\aleph_0}$$ is the cardinality of an uncountable set, even though $$|\omega|=\aleph_0$$. I assume this is because of different meanings of the exponentiation notation and the use of $$2^{\aleph_0}$$ as the cardinaly of the power set of the natural numbers, whereas $$\omega^\omega$$ isn't defined in the same sense. This seems a bit counterintuitive as (in a handwavy way) $$ 2 < \aleph_0$$ so you would expect $$\omega^\omega$$ to be uncoutable as $$2^{\aleph_0}$$ is the cardinality of an uncountable set.

(The problem only gets worse with $$\omega^{\omega^\omega}$$ being countable, and so on, even including $$\varepsilon_0$$.)

What is the fundamental property I'm missing out on here ? I'm not entirely clear with the exponentiation of infinite ordinals either (I believe the article treats ordinals in a much too formal way).

--Xedi (talk) 02:29, 27 November 2007 (UTC)


 * Two different sorts of exponentiation, unfortunately represented by the same notation. $$\omega^\omega$$ is a countable ordinal, when you interpret it as ordinal exponentiation, but an uncountable cardinal, interpreted as cardinal exponentiation. --Trovatore (talk) 02:39, 27 November 2007 (UTC)
 * I'll add that unless specifed otherwise, the sort of exponentiation intended can be deduced from the notation used. Although $$\omega=\aleph_0$$ (using the convention that cardinals are a special case of ordinals), the sybmol &omega; is usually used when it is treated as an ordinal, so $$\omega^{\omega}$$ is ordinal exponentiation; $$\aleph_0$$ is usually used when it is treated as a cardinal, so $$2^{\aleph_0}$$ is cardinal exponentiation. -- Meni Rosenfeld (talk) 08:18, 27 November 2007 (UTC)


 * Each ordinal has a cardinality, but different ordinals may have the same cardinality, and I'm not aware of a convention that cardinals are a special case of ordinals – although it is possible to associate a canonical "initial ordinal" with a cardinal. The operation of taking the cardinality does not distribute over exponentiation, so it is not in general the case that |αβ| is the same as |α|undefined. This should not be entirely unfamiliar: |2−3| ≠ |2|undefined. --Lambiam 09:37, 27 November 2007 (UTC)
 * How do you construct cardinalities as a concrete, formally defined object? The only way I am aware of, which also appears to be found in our cardinal number, is "the cardinality of a set X is the least ordinal α such that there is a bijection between X and α". In particular it follows that every cardinal number is an ordinal number, and that an ordinal number is a cardinal number iff there is no bijection between it and a smaller ordinal. So &omega; is a cardinal number, but &omega;+1 is not - the cardinality of &omega;+1 is &omega;. -- Meni Rosenfeld (talk) 10:02, 27 November 2007 (UTC)
 * OK. So then the initial ordinal of a cardinal is that cardinal itself. It reminds me of the statement that the ordered pair (n, n+1) is a topological space, since n+1 is a set of subsets of n that contains the empty set and n itself, and is closed under union and finite intersection. --Lambiam 13:54, 27 November 2007 (UTC)
 * Yes, and $$2 \in (0,1)$$. On a serious note: Lambiam, how do you define cardinalities (in ZFC, so classes are out)? Meni, there is of course another way of constructing cardinalities, which works in ZF (it's given in our article). It's very silly.
 * To the OP: the problem is with the definition of ordinal exponentiation. To define (for example) the ordinal $$\omega^\omega$$, we would like it to be the order-type of some well-ordering of the set $$\omega^\omega$$ of functions from $$\omega$$ to $$\omega$$. Unfortunately, in the absence of choice this set might not be well-orderable, and even with choice there's no 'canonical' way of well-ordering it. Thus we are forced to slim down our definition so that $$\omega^\omega$$ is the order-type of a (lexicographic) well-ordering of a set of functions from $$\omega$$ to $$\omega$$ which we can well-order, namely the functions which are zero for all but finitely many arguments. There are only countably many such functions, so we end up with a countable ordinal $$\omega^\omega$$.Algebraist 14:46, 27 November 2007 (UTC)


 * All right, understood. Thanks everyone. --Xedi (talk) 18:46, 27 November 2007 (UTC)

Just a side note on the notation issue: Meni's right that the notation $$\omega^\omega$$ will ordinarily be understood as ordinal exponentiation, whereas $$\aleph_0^{\aleph_0}$$ will be understood as cardinal exponentiation. But you can't always count on this generalizing -- lots of times I've seen $$2^\omega$$ used for the cardinality of the continuum. Mostly on the blackboard, not in published papers. Presumably this is because it's easier to write an omega on the blackboard than it is to make a calligraphic aleph. I think there's such a thing as a non-calligraphic aleph, but those of us who don't speak Hebrew don't recognize it as an aleph, so it doesn't help. (I'm curious -- in Israel, is the non-calligraphic one used on the blackboard in a set theory context?) --Trovatore (talk) 20:09, 27 November 2007 (UTC)
 * I don't know if "calligraphic" is the correct term here. There is a handwritten aleph, which I have no idea how to reproduce here but it looks somewhat like $$|\!\!\subset$$, and there is a typed aleph, א, of which $$\aleph$$ is a slightly stylized version. I think in a Hebrew lecture the typed aleph would be used, but typically not stylized - just a diagonal line with two vertical segments attached. -- Meni Rosenfeld (talk) 21:46, 27 November 2007 (UTC)
 * That's certainly what was done in my (non-Hebrew) set theory lectures. It's not very difficult, so I'm not sure why one would avoid it in favour of incorrect use of $$\omega$$. Algebraist 23:55, 27 November 2007 (UTC)