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June 22

Linear topological space

What is the exact definition of a linear topological space (the formal definition). I am unable to find it on the internet. Thanks. —Preceding unsigned comment added by 58.161.138.117 (talk) 04:42, 22 June 2009 (UTC)[reply]

The same thing as a topological vector space. Bo Jacoby (talk) 04:53, 22 June 2009 (UTC).[reply]

The gambler's fallacy

I am an EMT and far too often, I hear my co-workers say, "They didn't get a single [emergency] call during the last shift, we're going to get slammed now". Clearly, each 9-1-1 call we receive is completely independent of the next, making this statement an example of the Gambler's fallacy; however, at what point does the law of large numbers take over? We typically get 15 calls a day. Say, for example, suddenly we get 0 calls over 5 days. Those five days have no bearing on what the 6th day will bring (or the 7th, 8th, etc.). However, is it fair to say that it is probable that we will get a number of calls that "make up" for that five day stretch over the course of the rest of the year? Even this seems to imply a causation effect, but really it's just looking at statistical data and assigning probabilities to it, right? -- MacAddct1984 ^{(talk  contribs)} 05:09, 22 June 2009 (UTC)[reply]

It's important to note that the law of large numbers only provides stability of the mean, not the total. Assume you normally get 6 calls a day. If you get zero calls the first five days, your per-day average is 0. But you should still expect to get on average 5 calls a day for the rest of the year, bringing you to an average of (360*6)/365, which is 5.92. Over an even longer period, the expected average would be even closer to the normal 6, but the expected total will still be 30 calls short—it's just divided by a larger number of days. (This is all assuming that the probabilities really are independent.) —JAO • T • C 05:46, 22 June 2009 (UTC)[reply]

Ah, that makes more sense to me, thanks. -- MacAddct1984 ^{(talk  contribs)} 13:53, 22 June 2009 (UTC)[reply]

Why do you think the 911 calls are independent of each other? Say it's snowing heavily. Everyone stays indoors keeping warm, and you don't get any calls. Then the snow stops, those same people go outside to shovel their driveways or drive someplace, a lot of them have heart attacks or road accidents, and the 911 board lights up. That's hypothetical of course, but I don't have too much trouble with the idea that a flurry of calls is correlated with an earlier lull in them. 67.122.209.126 (talk) 06:55, 22 June 2009 (UTC)[reply]

Assuming independence, the number of 911 calls on a shift is likely to have a poisson distribution. (This is called the law of small numbers, by the way). Your description, that you 'typically get 15 calls a day. Say, for example, suddenly we get 0 calls over 5 days' is very unlikely for a poisson distribution, but unlikely events are remembered very well. You need better historical data. Don't you have a record of the total number of calls over a large number of shifts? Bo Jacoby (talk) 07:21, 22 June 2009 (UTC).[reply]

67.122, that seems true for the most part^{[original research]}, but even in that case the calls can still be considered independent of each other, no? The fact that Joe calls 911 on one side of town after a snowfall still has no bearing on whether Jane, another resident, will.

Bo, we do keep track of call statistics. We've typically seen a 5-10% increase in call volume each year. Assuming the sample size is large enough, is it possible/logical to make predictions based off of that data? -- MacAddct1984 ^{(talk  contribs)} 13:53, 22 June 2009 (UTC)[reply]

No, in the case described different people calling are not independent. The fact that Joe has just called makes it more likely that now is a time of higher-than-usual call density, which makes it more likely that Jane will call. Algebraist 13:57, 22 June 2009 (UTC)[reply]

Perhaps you are getting a tad confused about the meaning of 'independent'. It does not necessarily describe a cause-effect relationship. Obviously, Joe's calling 911 does not cause Jane to call 911 around the same time (at least, not usually), but as Algebraist explained, the knowledge that Joe called does affect the probability that Jane will call about the same time. And that is what it means for the two events to be 'dependent', as opposed to independent. --COVIZAPIBETEFOKY (talk) 14:25, 22 June 2009 (UTC)[reply]

Where the probability is a gauged value from the frame of reference of an observer (us), rather than a descriptor of Jane's actual propensity to make the call. —Anonymous Dissident^Talk 14:54, 22 June 2009 (UTC)[reply]

Have you checked against the TV listings? I wouldn't put it past people to wait till a soap is over to ring up about their heart attack. Dmcq (talk) 12:43, 22 June 2009 (UTC)[reply]

Let the number of calls on day number i be X_i. Compute the number of days S₀ = ΣX_i⁰, the number of calls S₁ = ΣX_i¹, and the square sum S₂ = ΣX_i². Compute the mean value μ = S₁/S₀ and the variance σ² = (S₂/S₀)−(S₁/S₀)². Now, if the number of calls are really independent of one another and of time, the distribution is poisson and μ = σ². If this is not the case, but for example μ < σ², then the distribution may be the negative binomial distribution. (See the article on cumulant for explanation). If the historical data fit reasonably well into a mathematical model, then it is reasonable to make predictions based on that model. Bo Jacoby (talk) 14:30, 22 June 2009 (UTC).[reply]

I think the key thing to remember here is that how many calls you got over the last 5 days isn't a random variable, it's a known value. It's meaningless to talk about the probability distribution for it because it doesn't have one - there is a 100% chance that you saw what you saw and a 0% chance that you didn't. If you expect 15 calls a day then at the start of the year you would expect 365*15 calls that year, but once you've experienced a few days you can improve that estimate by replacing the expected number of calls on days past with the actual number. --Tango (talk) 14:49, 22 June 2009 (UTC)[reply]

Not all calls will be independent. A house starts on fire, and 3 different neighbors could very well call 911 over the same event. Or are such incidents not counted? 65.121.141.34 (talk) 18:49, 22 June 2009 (UTC)[reply]

We're after that step, the calls have already been filtered out by the time we get them. Very rarely will we get a call for, say, a stabbing, and then 30 minutes later we'll get another call because they found the other party involved who's also injured. -- MacAddct1984 ^{(talk  contribs)} 13:28, 23 June 2009 (UTC)[reply]

question on stats

Assume I have a giant jar of jelly beans (say 500 individual beans) that is half red and half green. Without being able to see what I am doing, I randomly select one. I replace it and select another. I note if I have drawn 2 red, a red and a green, or two greens and replace the bean I just drew. Now if I get 50 pairs, Logically I should have 25 red/green, 12.5 red/red and 12.5 green/green (ok so 12 of one and 13 of the other). But I know that this exact outcome by itself is somewhat unlikely. How do I determine the probability of the various combinations, (22rg, 10rr, 18gg), (28rg, 11 rr, 11gg) etc?

Our article on the urn problem is a good starting point. You may also find the article on the binomial distribution helpful. Ray^Talk 21:54, 22 June 2009 (UTC)[reply]

The probability of the combination (22rg, 10rr, 18gg), is $\scriptstyle {\frac {50!\,0.5^{22}\,0.25^{10}\,0.25^{18}}{22!\,10!\,18!}}=0.00385358$ . Bo Jacoby (talk) 08:33, 23 June 2009 (UTC).[reply]

Door codes

In order to get into the building I live in, I need to type in a four-digit code, and today, I noticed something that leads to a problem that I don't know how to solve.

I had assumed that if I typed in an incorrect digit that I would have to wait five seconds or thereabouts so that the pad-thing would reset itself, and then try again. (Trust me, this leads to the problem). So, in this situation, in order to type in every possible combination, I would need 40,000 keystrokes, since there are 10,000 codes, from 0000 to 9999, and each code has four digits. Correct me if I'm wrong, but I believe that this is the minimum number of keystokes to type in every combination.

What I noticed today is that I don't need to wait for the pad-thing to reset itself; I can simply start over without waiting. That is, if the code is 1234, and I type 51234, the door will unlock. As far as I can tell, this means that, as far as the pad-thing is concerned, typing 51234 is essentially the same as entering two codes, to wit, 5123 and 1234. The problem that I cannot solve is: What is the minimum number of keystrokes needed to enter all 10,000 codes if they can be chained up like this. And, if you would continue to humor my curiosity, could you describe what (one of) the minimal patterns would be?

Thank you for your attention to my question. 84.97.254.29 (talk) 23:48, 22 June 2009 (UTC)[reply]

You obviously need at least 10,003 keystrokes, and in fact this is sufficient. See De Bruijn sequence. Algebraist 23:57, 22 June 2009 (UTC)[reply]

That was fast. Thanks. 84.97.254.29 (talk) 00:05, 23 June 2009 (UTC)[reply]

I disagree -- but I may not have understood the question correctly.

My understanding of these machines is that they only store (or pay attention to) the last four digits entered. When you were successful by entering 51234, you did NOT test 5123; rather, the 5 was simply discarded.

(Further, if you know that you've made a keying error before you push enter, you need not wait any time at all. Just continue entering digits; as long as the last four entered are correct, you're in!)

The devices are far less complex than you may be giving them credit for. --DaHorsesMouth (talk) 22:39, 23 June 2009 (UTC)[reply]

There isn't always an "enter" key - when I set or unset the alarm system in my house I just have to press the correct 4 digits. AndrewWTaylor (talk) 22:55, 23 June 2009 (UTC)[reply]

You're wrong, DaHorsesMouth. OP talks about system with no Enter key, so if he enters 51234, he tests 5123 (and the test fails), then enters 4 which causes dropping the first digit, appending 4 at the end and testing a new 1234 code. Consider it as a 4-position shift-register, filled with keystrokes, which tests itself against a pre-set value each time a new digit is entered. CiaPan (talk) 05:36, 24 June 2009 (UTC)[reply]