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December 7[edit]

Measure Theory[edit]

If a function is bounded and measurable, is it automatically integrable? We are using Royden's book and I have not seen this as a theorem any where, but it could be here and I just don't understand. I know for a bounded measurable function, they define the Lebesgue integral but never what integrable means specifically for such functions. Does this mean automatically that they are integrable? I don't think this is implied by the definition because they also define the integral for nonnegative functions but they are only integrable if the integral is finite. I could use the definition of a general function. It is integrable if the positive and negative parts are integrable. Are the positive and negative parts automatically measurable if the function is? If so, then they are both integrable because the integral of a bounded function on a set of finite measure is definitely finite.

Any help would be appreciated. Thanks StatisticsMan (talk) 01:02, 7 December 2008 (UTC)[reply]

A measurable function is integrable if its absolute value has finite integral. This happens iff the positive and negative parts both have finite integral. A function is measurable iff its positive and negative parts are both measurable. A bounded measurable function on a set of finite measure is certainly integrable. If f is not defined on a set of finite measure, the conclusion may fail, e.g. the constant 1 function on R. Algebraist 01:27, 7 December 2008 (UTC)[reply]

More tersely: the answer is "yes" if the measure of the domain is finite. Michael Hardy (talk) 06:08, 7 December 2008 (UTC)[reply]

Finding root-mean-square error for various methods[edit]

I understand the rms error of the regression method is (1 - r²)^1/2 but how would one find the rms error for other methods, such as in the following?

Final average is twice the midterm average, so someone proposes to simply guess twice the midterm average to predict final score SD(midterm)= 8 SD(final)= 20 r= 0.60 —Preceding unsigned comment added by 169.229.75.140 (talk) 03:33, 7 December 2008 (UTC)[reply]

Could it be that you mean that since the final average is twice the midterm average, someone proposes to simply guess twice the midterm SCORE to predict the final score? (I'll return to your question tomorrow and think about the details.) Michael Hardy (talk) 06:06, 7 December 2008 (UTC)[reply]

ya that's what i meant sorry169.229.75.140 (talk) 05:34, 8 December 2008 (UTC)[reply]

Topology question[edit]

If I have a topological space $X=\prod _{\alpha \in J}X_{\alpha }$ , is the set $S=\prod _{\alpha \in J}S_{\alpha }$ closed, where $S_{\alpha }=x_{\alpha }$ (that is just a point of $X_{\alpha }$ ) for all $\alpha \neq \beta$ and $S_{\beta }=X_{\beta }$ ?

Thanks StatisticsMan (talk) 05:39, 7 December 2008 (UTC)[reply]

Not in general. All the X_\alpha could be 2-point trivial spaces, say. Algebraist 06:02, 7 December 2008 (UTC)[reply]

The conclusion is true if the spaces appearing in the product are all Hausdorff. 67.150.254.10 (talk) 06:24, 7 December 2008 (UTC)[reply]

Or just T1. Algebraist 06:31, 7 December 2008 (UTC)[reply]

We can be more direct about what is going on. First, the entire product can be broken up as

\left(\prod _{i\not =\beta }X_{i}\right)\times X_{\beta }

Doing this, we can investigate the set S in terms of its two factors. The right hand factor of S, which corresponds to coordinate β, is automatically closed in X_β. If the rest of S were closed in

\prod _{i\not =\beta }X_{i}

that would be sufficient to show that S is closed. But S is just a single point in that product, so it's enough to prove that the product is T1. In turn this means it is sufficient for each factor in that product to be T1. — Carl (CBM · talk) 13:15, 7 December 2008 (UTC)[reply]

Thanks every one. That is very good reasoning Carl, simple and good. Someone else told me another way! It's just based on the fact that the product of the closures is equal to the closure of the product in the product topology. So, if each set is closed, the whole product must be. I remembered after I posted this last night that the fact that each

X_{\alpha }

is Hausdorff would be important. I'm assuming the space as a whole is normal and want to show each

X_{\alpha }

is normal. I already proved that if the whole space is Hausdorff, then each

X_{\alpha }

is so we have that each is Hausdorff from that. So, construct this subspace

S

and it is a closed subspace of a normal space and thus normal by another problem, which I will do. StatisticsMan (talk) 14:29, 7 December 2008 (UTC)[reply]

The same fact holds in the box topology (your first problem). In general, it is helpful to use the projection map when proving facts about factor spaces in products (not in this case though).

An interesting (but not so trivial) problem would be to consider whether the countably infinite box product of normal spaces is normal (what conditions must be imposed on the factor spaces?). First of all, have a go when the factor spaces are all equal to the space of all real numbers (consider the countably infinite product). It is in fact true in this special case but I will leave you to work out why.

Topology Expert (talk) 16:04, 7 December 2008 (UTC)[reply]

Asking the gods of science for some help.[edit]

A member of my family is going to attent a game show of this format http://en.wikipedia.org/wiki/Deal_or_No_Deal

There are 22 boxes with progressively larger ammonts of cash in each box. I would like to ask if there are any factors which can give an advantage (something similar to monthy hall problem), something to increase the probability of winning?. —Preceding unsigned comment added by 77.243.73.133 (talk) 14:13, 7 December 2008 (UTC)[reply]

The probability of winning what? The probability of winning at least n is, of course, maximized by the simple strategy "keep opening boxes until you get an offer of at least n, and then stop". If, instead, you're after maximizing the expected value of the prize, you'd probably be better off not accepting any deals (because they tend to be lower than the average of the remaining boxes), but as you're only allowed to play once, this means taking an extremely large risk. In conclusion, the math/probability in this game is quite simple, but it's all about what you want to risk. It's like choosing between investing your money in an uncertain stock which may well give a huge pay-off and investing it with a safe but uninspiring 3% interest. No mathematic theory can really tell you what is the best option for you there. -- Jao (talk) 14:31, 7 December 2008 (UTC)[reply]

(edit conflict) Opening the boxes is completely random, you might as well do it in numerical order, it makes no difference. The skill comes in deciding when to deal. The obvious way is to take the average of all the values still available and deal if the offer is more than that, but chances are that won't give the best chances due to something called the "diminishing marginal utility of money". Basically, the difference in utility (usefulness for achieving whatever it is you value) between £100 and £200 is much more than the difference in utility between £10,000 and £10,100, even though the amount of money is the same. So, you need to work out how useful each amount of money would be and take the average of that and compare it to how useful the amount offered would be. For example, if you are down to two boxes remaining, the £100 and the £250,000 (I'm assuming the UK version here, it doesn't really matter), and the banker offers you £100,000 it is probably a good idea to take the deal. The £100 is pretty useless, it's a meal in a nice restaurant with your family and that's about it. £250,000 and £100,000 are much the same in terms of usefulness (sure, £250k is better, but not much better - neither is anywhere near enough to retire on, both are enough to pay off a sizeable chunk of a mortgage), so it's better to take the £100k than risk getting essentially nothing, even though £100k is less than the numerical average (which is £125,050). Exactly how you value each numerical amount is going to depend on your circumstances - if you're already a millionaire you might as well just stick to the numerical averages, if you're going to struggle to pay your rent for the next few months and need at least £10,000 to do so, then deal as soon as you are offered more than £10,000. Oh, and if you are offered the swap, it makes absolutely no difference if you take it or not. (PS. Now I've seen Jao's answer I realise I forgot to mention risk! A certain £100,000 is worth more than a gamble with an expected value of £100,000, even without the diminishing utility thing - to take a risk the expected value needs to be higher than the safe option so that you have a risk premium, how much higher depends on how risk averse you are.) --Tango (talk) 14:39, 7 December 2008 (UTC)[reply]

I too first thought that there should be a parallell to the monty hall problem since you get to switch the suitcase you've chosen. But in the monty hall problem the gameshow host reveals one of the goats, while in Deal or No Deal, opening a suitcase could reveal the grand prize as well. EverGreg (talk) 15:20, 7 December 2008 (UTC)[reply]

You could work out the expectation I guess. Just sum up the total number of cash on the board (means add up the money for each case to get the total) and then divide by the total number of cases. Generally the expectation should be about $8,000 or so.

I agree that there is no strategy (if there were, the show would go bankrupt) but in general, since the expectation is about $10,000, don't aim for money higher than $30,000 unless you are in a very good position because the chances are very low that you are going to keep a particular good position. I have watched this show a lot and in general, don't go for cases in a pattern (such as case 22, then case 21 etc...) because it is likely that you will knock out as many high valued cases as low valued ones (I think that the people who set the cases don't make them completely random although I may be wrong).

Hope this helps (I think I summarized some of what Tango said; apologies for that).

Topology Expert (talk) 15:56, 7 December 2008 (UTC)[reply]

They're supposed to be properly random (at least on the UK one), I would be surprised if they didn't use some kind of fairly reliable random number generator. --Tango (talk) 16:22, 7 December 2008 (UTC)[reply]

The reason why I suspect it is because I choose (a long time ago) on a ten day period to write down all the money values associated to each particular case on each day. I studied the sample over the ten days and noted that a particular case (number 5) seemed to get in the top 3 money ranges 8 out of 10 times, and the high money values seemed to alternate between case 5, then case 8, then case 19 (I think it was 19 but it could have been something else) and in general, if case n had a low value, then case n+1 would have a high value (this was exactly true on 5 or 6 occassions and mostly true on the other occassions). Maybe this was a low probably event that somehow occurred (and maybe I should have choose a larger sample) but I still suspect it...

Topology Expert (talk) 16:38, 7 December 2008 (UTC)[reply]

Or maybe that just isn't that low a probability event. For example, the chance of there being a box that contained a top 3 value at least 8 out of 10 times is probably higher than you think. Anyone care to calculate it? (I might try doing it by Monte Carlo means if no-one does it precisely - I can't see an easy way to do it.) --Tango (talk) 17:05, 7 December 2008 (UTC)[reply]

Yes. It seems it should be a low probably event but then somehow it has a high probability!

A sample size of 10 is not exactly representative... --140.247.10.42 (talk) 19:30, 7 December 2008 (UTC)[reply]

An interesting point: I read (sometime ago) that a type of lottery at a pub (one of those were it takes 2 seconds to play each time) had an expectation of +$2 for every time you played and some students in probability realized this. Guess what? They earned 1 million dollars (played for a few days or so). You never know when the people who make a lottery have done something terribly wrong so its always a good idea to find the expectation.

Topology Expert (talk) 19:32, 7 December 2008 (UTC)[reply]

The mechanics of certain versions of Deal have been worked out - see [1]. However, they do differ from country to country. This article demonstrates a formula for the deals, which you could theoretically use to work out the expected bank offer from your remaining cases, but in the Australian version of the show it seems fairly likely the formula isn't used, or isn't used consistently (I recently caught part of an episode where the contestant had said he was aiming to walk out with $14,500 and the bank offer was $14,499 which strongly suggests the "banker" is willing to throw out probability in favour of dramatics). Confusing Manifestation(Say hi!) 23:07, 7 December 2008 (UTC)[reply]

I'm pretty sure there is no formula for the UK version (there may be a formula they use as a guide, but that's it). The banker's offers are done in such a way as to make good TV. --Tango (talk) 23:19, 7 December 2008 (UTC)[reply]

If you search the net some places lists the percentage of total values left that the bank overs (i couldn't find a link today but i have searched it before). For example most TV shows offer 35% instead of 50% for the majority of the first few rounds. In the later rounds they may offer 55% to 65%. This makes the best play simply to keep going until the bank offers you above 50%. If they never offer above 50% keep going. If they randomly make overs (ie at least 2 offers always over 50%) and you know a higher than 50% offer is still to come, you should always keep going.--Dacium (talk) 02:43, 8 December 2008 (UTC)[reply]

Thanks a lot guysBastard Soap (talk) 09:11, 8 December 2008 (UTC)[reply]

How many groups are there of a particular order?[edit]

I don't know whether this question is obvious because I haven't thought about it much but it does not seem trivial. Is there a way, given a postive integer n, to find how many groups exist of order n? There is a analagous problem in topology: given a set of cardinality n, find the number of topologies that exist on that set. I am not a fan of counting but is there a rigorous mathematical way of solving this problem (without using a computer)?

Any help would be much appreciated (its strange that such a problem uses basic language and yet is difficult compared to 'real topology' or 'real group theory').

Topology Expert (talk) 16:16, 7 December 2008 (UTC)[reply]

If you restrict it to finitely generated abelian groups it's pretty easy (using the standard classification for them). I don't know of a general method (but I haven't really thought about it). --Tango (talk) 16:20, 7 December 2008 (UTC)[reply]

This is a hard problem. There's a bit about it in finite group#Number of groups of a given order. It's A000001 at Sloane's. Algebraist 16:23, 7 December 2008 (UTC) Algebraist 16:21, 7 December 2008 (UTC)[reply]

Thanks to both users; I will have a look at that link (and of course I should have said 'up to an isomorphism'). Yes, specific cases are (generally) easy (such as finitely generated abelian groups) but what about the topological analogue? It should be harder in my opinion but it is similar in the sense that for special cases it's easy (like if you restrict to Hausdorff spaces, you obviously get '1' for any cardinality). I wonder what would happen in the case of sigma algebras or in general, does category theory give us an algorithm?

Topology Expert (talk) 16:30, 7 December 2008 (UTC)[reply]

A sigma-algbera on a finite set is the same thing as a boolean algebra on a finite set, so the only question is what size the atoms are. Thus the number of sigma-algebras (up to isomorphism) on a set of size n is the number of partitions of n. Algebraist 16:37, 7 December 2008 (UTC)[reply]

But there is no (or does not seem to be an) easy way of determining the number of partitions of a given integer. I will think about whether there is a category theoretical way of answering this problem in general...

Topology Expert (talk) 16:44, 7 December 2008 (UTC)[reply]

It's a lot easier to compute than the number of abelian groups of order n (no prime factorization needed). Algebraist 16:49, 7 December 2008 (UTC)[reply]

The topological question is also hard. It is A000798 or A001930 depending on whether you want 'up to homeomorphism' or not. Algebraist 16:53, 7 December 2008 (UTC)[reply]

Thankyou for the links but I am really more interested in a general formulae (if there exists one) rather than numerical results. Anyway, I will think about the problem myself.

Topology Expert (talk) 19:22, 7 December 2008 (UTC)[reply]

There are p^{k³2/27 + O(k^5/2)} groups of order p^k. In general the number of groups of order n depends both on the size of n and on the highest power k appearing in the prime factorization of n: there are at most n^{k²2/27 + O(k^3/2)} groups of order n, and this is sharp for perfect powers. If k is 1 (n is square-free), then the number of groups of order n can be computed exactly for reasonable n through various number theoretic functions by work of Hölder. If n is a product of three primes, then there are only a few cases, and they can be distinguished easily. If k is more than 15, then it is unlikely that we will know the exact number of groups of order n in our lifetime. If k is more than 10, then it is not known now. JackSchmidt (talk) 14:44, 8 December 2008 (UTC)[reply]

Thankyou! I ill think about it (hen I type 'double u' my computer log out; in fact mot of my key don't ork...).

Topology Expert (talk) 17:41, 8 December 2008 (UTC)[reply]

There's an extensive table answering this question in a recent issue of the Mathematical Intelligencer. I'll see if I can find it. What leaps out at you is that for numbers with a large number of small prime factors there is a large number of groups. Michael Hardy (talk) 01:12, 10 December 2008 (UTC)[reply]

The true version of this is covered by the asymptotics above. Note there is exactly one group of order 3*5*17*23*29*53*83*89*113*149*173*197*257*263*269*293*317*353*359*383*389*419*449*467*479. Indeed, for any square-free number n, the number of groups of order n is at most n. What matters is not how many small factors the number has, but how many times a prime factor is repeated. This is the point of k mentioned above. JackSchmidt (talk) 06:27, 10 December 2008 (UTC)[reply]

Graph Sketching[edit]

Hi. Given that I've sketched $y=x^{5}-x^{3}$ , what transformations do I apply to this graph in order to sketch $y^{2}=x^{5}-x^{3}$ . Or perhaps more generally, if I have sketched $y=f(x)$ , what transformations do I apply to sketch $y^{2}=f(x)$ ? I have graphed both fuctions on Grapher but I do not understand how to get from one to the other. Thanks 92.2.197.63 (talk) 17:59, 7 December 2008 (UTC)[reply]

There is no simple transformation, but basically you reflect anything below the x-axis in the x-axis so everything is positive and then and then you need to scale all the heights. 1 stays were it is, anything less than one moves towards zero, the nearer to zero is starts the more it moves (proportionally speaking), anything more than 1 moves up, the higher it starts, the more it moves. --Tango (talk) 18:12, 7 December 2008 (UTC)[reply]

No, you delete everything below the x-axis and reflect everything above the x-axis in the x-axis, while keeping the original copy. Algebraist 18:18, 7 December 2008 (UTC)[reply]

Yeah, as usual, just ignore me... it's pretty much the exact opposite of everything I said... --Tango (talk) 19:06, 7 December 2008 (UTC)[reply]

One part of Tango's response got struck out which is still important: the original graph does need to be scaled in a non-uniform way so "sketch" is pretty important if you don't handle the scaling. Consider the simpler case of y=x versus yy=x. The reflection and deletion parts are needed, but so is the non-uniform scaling that changes a straight (half-)line into a (half-)parabola. JackSchmidt (talk) 06:48, 10 December 2008 (UTC)[reply]