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Another way to see it: there's a one-to-one correspondence between pairs of nonadjacent seats chosen from a row of N and arbitrary pairs of seats chosen from a row of N−1 (just add/remove a seat between the chosen seats). So there are 14·13 choices for the two most talkative students and then 13! for
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I'm interested in seeing whether or not the annual volatility changes over time. I was thinking that once you had found the confidence interval of the annual variance, one could split the data set into smaller intervals, calculate the annual variance of each interval (by multiplying the variance of
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Instead you might want to consider the way to seat the two most talkative together. So making them a block of 2 which can be arrange 2 (or 2! ways) you have 14 items, 13 children and the two most talkitive together. So these are arranged 14!*2 ways. Then you can just minus this from the total 15!
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In a row, things are more complicated - there are two cases to consider. First suppose that the most talkative student doesn't sit at the end (13 options). Then there are again 12 possible places for #2 to sit, and then we arrange the other 13 students as we like, so 13.12.13!. If we sit the most
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Well, if we considered it as a circle, then we place the most talkative student first (15 options). That leaves 12 places for the 2nd most talkative (can't use the seat taken, or the 2 on either side). After that you have 13 seats available which can be used in any order, so here the answer is
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variance, how would you change the formula? Could you just replace σ^2 with 365*σ^2, and N=3652 with N=(3652/365)? In which case the N would be the number of years (10)? Because when I do that I end up with an awfully wide interval. Or would you let N=365, the number of days in one year?
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In how many ways can 15 students be seated in a row such that the 2 most talkative children never sit together? I've been given the answer 14! * 13, but I don't compute this. Seems like it should be 13! ways for every two positions that are not adjacent.
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This is the lovely thing about combinatorics - no matter how clever you think you're being, there's always someone else who can find an easier way. That and it deals with proper numbers. But then, I like complex analysis so who am I to talk?
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talkative student at the end (2 options), then there are 13 places for #2, and then 13! ways to arrange the rest, so a tital of 2.13.13! Putting this together, we have 13!(13.12 + 2.13) = 14.13.13! = 13.14! as required. Nice problem. -
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I understand that you could find the yearly variance by multiplying the daily variance by 365 (and the annual standard deviation or 'volatility' by multiplying the daily standard deviation by sqrt(365).
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Given that it is normally distributed, I understand that once you calculate the daily variance over N days (σ^2_N, that is sigma squared subscript N), you can find its 95% confidence interval: σ^2 ∈
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each interval by 365), and see if 95% of the resulting set of annual variances falls within the confidence interval. Would this be a valid method of tracking changes in annual volatility?
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These are outgoing links, not incoming links (since
Knowledge (XXG) is a directed graph). Among articles with at least one link, the probability that an article has exactly one link is
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Well, let's gather some data. I clicked "random article" 20 times, and got articles with the following number of blue links (not counting boiler-plate text):
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The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the
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I know that if you consider a network model of
Knowledge (XXG) (in which nodes are pages, edges are links), then that network is (or is close to being)
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is less than 2 -- probably very close to 1. Someone with more statistics knowledge (or more data) could probably conjure up a reasonable estimate for
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Say you have a set of data that shows daily price movement in some commodity over ten years, with 3652 or so prices (one for each day).
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where N is the number of days (which in this data set would be 3652), and σ^2 is the daily variance.
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ways or arranging. I think this is a more simple way of getting your 13*14! answer. Hope this helps
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But if you wanted to find the 95% confidence interval for the
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