Re: Is this how you would have done it?
On Sat, 22 Dec 2001, Ralph Noble asked: How would you have done this? A local newspaper asked its readers to rank the year's Top 10 news stories by completing a ballot form. There were 10 choices on all but one ballot (i.e. local news, sports news, business news, etc.), and you had to rank those from 1 to 10 without duplicating any of your choices. One was their top pick, 10 their lowest. Only one ballot had more than 10 choices, because of the large number of local news stories you could choose from. I would have thought if you only had 10 choices and had to rank from 1 to 10, then you'd count up all the stories that got the readers' Number One vote and which ever story got the most Number One votes would have been declared the winner. That is certainly one way of determining a winner. But if one were going to do this in the end, there is not much point to asking for ranks other than 1, because that information is not going to be used at all. (Unless, of course, one uses a variant of this method for the breaking of ties, or for obtaining a majority of votes cast for the winner.) Not so in the case of this newspaper. So maybe I do not understand statistics. Non sequitur. You are not discussing statistics, you are discussing the choice of methods of counting votes. The newspaper told the readers there were several ways it could have tallied the rankings. This is true. Several may be an understatement. The newspaper decided to weight everybody's responses and gave each first place vote a value of 10, each second place nine, each third place eight, and so on. They then added together the values for each story and then ranked the stories by point totals. So is this an accurate way to have tallied the votes? Why not, assuming they didn't err in their arithmetic? In what sense do you want to mean accurate? I would use the word to describe the care with which the chosen method was carried out, not the choice of method, as you appear to mean. Accurate ordinarily refers, at least by implication, to how closely some standard or other is being met: what standard did you have in mind? And why weight them since the pool in all but one category only had 10 items to choose from? One answer is, precisely because all categories (but the one, and you haven't quite described what happened to the one, but I'll assume that only the ranks 1 to 10 were used in that case) had 10 items. If you add up all the 1st, 2nd, 3rd, etc. votes _without_ weighting them (that is to say, weighting them equally instead of unequally), you get the same total for each item, and have no way of declaring a winner. (This may not be true for the one category, since there are more than 10 items but only 10 ranks to be apportioned among them.) One could, of course, have weighted them according to their ranks (1st = 1, 2nd = 2, etc.) and chosen the one with the _lowest_ point total. (This of course is equivalent to what the newspaper actually did: this point total equals 11 minus the newspaper's point total, and you get the same winners this way.) Or according to the reciprocal of their ranks (1st = 1, 2nd = 1/2, 3rd = 1/3, etc.) and added those up, and taken the highest score. This is not equivalent to the method actually used, although sometimes the results are not different. Etc. If you conclude from all this that the choice of counting method for tallying votes is an arbitrary one, you are quite right. It is. -- DFB. Donald F. Burrill [EMAIL PROTECTED] 184 Nashua Road, Bedford, NH 03110 603-471-7128 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Is this how you would have done it?
In article [EMAIL PROTECTED], Donald Burrill [EMAIL PROTECTED] wrote: On Sat, 22 Dec 2001, Ralph Noble asked: How would you have done this? A local newspaper asked its readers to rank the year's Top 10 news stories by completing a ballot form. There were 10 choices on all but one ballot (i.e. local news, sports news, business news, etc.), and you had to rank those from 1 to 10 without duplicating any of your choices. One was their top pick, 10 their lowest. Only one ballot had more than 10 choices, because of the large number of local news stories you could choose from. I would have thought if you only had 10 choices and had to rank from 1 to 10, then you'd count up all the stories that got the readers' Number One vote and which ever story got the most Number One votes would have been declared the winner. That is certainly one way of determining a winner. But if one were going to do this in the end, there is not much point to asking for ranks other than 1, because that information is not going to be used at all. (Unless, of course, one uses a variant of this method for the breaking of ties, or for obtaining a majority of votes cast for the winner.) Not so in the case of this newspaper. So maybe I do not understand statistics. Non sequitur. You are not discussing statistics, you are discussing the choice of methods of counting votes. The newspaper told the readers there were several ways it could have tallied the rankings. This is true. Several may be an understatement. This problem is the type of problem which results in Arrow's Paradox. The problem is, given the opinions of all individuals, to form the opinion of the group. Arrow showed in his thesis that if certain reasonable conditions are satisfied, there is no way of doing this other than dictatorial or conventional. Now Arrow did not consider consistency under randomization. With this included, the proof becomes very short, and can be found as a comment in my paper showing that self-consistent behavior must be Bayesian. Briefly, the argument is that the group evaluation must be a positive linear combination of those of the individuals. However, how do we compare the scales? Any method to do this leads to paradoxes. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 [EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: How ro perform Runs Test??
Glen [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... [EMAIL PROTECTED] (Chia C Chong) wrote in message news:[EMAIL PROTECTED]... I am using nonlinear regression method to find the best parameters for my data. I came across a term called runs test from the Internet. It mentioned that this is to determines whether my data is differ significantly from the equation model I select for the nonlinear regression. Can someone please let me know how should I perform the run tests?? You need to use a runs test that's adjusted for the dependence in the residuals. The usual runs test in the texts won't apply. Glen I always understood that the runs test was designed to detect systematic departures from the fitted line because some other curve fitted the data better. In this context, it is a test for dependence of residuals. There is a discussion of this at http://216.46.227.18/curvefit/systematic_deviation.htm Any elementary text in Non-parametric Methods in statistics will give an example. Hope this helps Jim Snow [EMAIL PROTECTED] = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: How to prove absence of dependence between arguments of function?
- Original Message - From: Estimator [EMAIL PROTECTED] Newsgroups: sci.stat.edu Sent: Saturday, 22 December 2001 2:47 Subject: How to prove absence of dependence between arguments of function? *** post for FREE via your newsreader at post.newsfeeds.com *** I've got linear function Y=f(x1;x2;x3;x4) of theoretical distribution, in addition x3=g1(x1), x4=g2(x1;x2) Also I've got empirical sample consists of N sets of these values (magnitudes) Y1 x11 x12 x13 x14 .. Yn xn1 xn2 xn3 xn4 as since x3 x4 are dependent of x1 x2 it's reasonable to evaluate x3 x4 by x1 x2 accordingly and analyse Y only from x1 x2. If x3= g1(x1) x3 and x1 can only be independent if the function g1 is a constant, or at least degenerate wrt x1 and similarly for Y. But I've got a strong believe that in fact all of the arguments are independent or dependence is insignificant. How to prove this mathematically using empitical observations? You could plot x1 against x3 to convince yourself that there was no tendency for the points to depart from a random scatter. Similarly plot x1 vs x4 and x2 vs x4. But this would not give you objective grounds to include or reject x3 and x4 from the analysis. In fact, even if x3 and x4 are uncorrelated with x1 and x2 ,your best course would be to retain them in the analysis. Then you can formally test to see whether they contribute any explanatory power wrt Y. If the explanatory power of the model is not significantly improved by including x3 and x4, you have objective evidence to exclude them from the model. Is any sense in making correlation matrix 4x4 (Pearson) and proving insignificance of coefficient of correlation (Student t-criterion for example) between arguments? No. One reason is that you would have to conduct six significance tests and the chance of the 1 in 20 level of the test being exceeded by chance in one of them is too high. In any event correlation is only a measure of LINEAR dependence and frequently data has more complicated dependencies. There is no way to prove independence from a data set, in the same sense that scientific theories are never proved, only disproved. In particular, failure to reject a null hypothesis is not proof of its correctness. Dependence does not consist of only linear relationships and lack of correlation does not imply independence. For example if X is symetrically distributed on (-1,1) and Y = X^2 , then X and Y are uncorrelated although functionally related. Hope this helps Jim Snow [EMAIL PROTECTED] = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Is this how you would have done it?
Ralph Noble wrote: How would you have done this? A local newspaper asked its readers to rank the year's Top 10 news stories by completing a ballot form. There were 10 choices on all but one ballot (i.e. local news, sports news, business news, etc.), and you had to rank those from 1 to 10 without duplicating any of your choices. One was their top pick, 10 their lowest. Only one ballot had more than 10 choices, because of the large number of local news stories you could choose from. I would have thought if you only had 10 choices and had to rank from 1 to 10, then you'd count up all the stories that got the readers' Number One vote and which ever story got the most Number One votes would have been declared the winner. If you only count the #1 votes, then the other positions do not count for anything, and do not matter. You loose the opportunity to discover that lots of people felt a single item was #2. You might, in fact, elect Spiro Agnew as Gov. of Maryland. Depending on your politics, this may be a good thing or not. Not so in the case of this newspaper . so maybe I do not understand statistics. The newspaper told the readers there were several ways it could have tallied the rankings. The newspaper decided to weight everybody's responses and gave each first place vote a value of 10, each second place nine, each third place eight, and so on. They then added together the values for each story and then ranked the stories by point totals. this ranking system assumes that 2 second place votes are worth 1.8 times as much as 1 first place vote. And two 5th place votes are worth as much as one first place vote. My gut suspicion is that this puts more emphasis on the 3rd 4th place finishers than most people would use, if they could award the points of the scale as they choose. Maybe not. But this does give credit for the rank of a vote in the list. Even a last place finisher gets some credit. So is this an accurate way to have tallied the votes? I think 'accurate' as a term here, does not compute. there are many ways to consider preference for items in a list of more than 2 items. the complexity increases with the length of the list. the newspaper made up the rules; so long as they followed them, it was 'accurate.' And why weight them since the pool in all but one category only had 10 items to choose from? If they simply added the rank positions, 1, 2, 3... 10, then the 'winner' would be the one with the lowest score. Unless you play a lot of golf, this does not make much intuitive sense to many people. A simple first reaction is to reverse the rank 'position' number, as they did. Another way might be to weight the ranks by an exponential decay model, so that the difference in weight between a #1 and #2 would be greater than the difference between a #8 and #9. Like I said, the newspaper can make up whatever rules they like. Thanks! Ralph Noble [EMAIL PROTECTED] Cheers lots of diverse Holidays, Jay -- Jay Warner Principal Scientist Warner Consulting, Inc. North Green Bay Road Racine, WI 53404-1216 USA Ph: (262) 634-9100 FAX: (262) 681-1133 email: [EMAIL PROTECTED] web: http://www.a2q.com The A2Q Method (tm) -- What do you want to improve today? = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Maximum Likelihood Question
To all, Thanks so much for all your ideas and insights thus far. To those who have suggested a Baysean approach, I am interested, but I am weeks away from understanding it well enough to figure out if I can use it. Also, I think I am close to developing a usable technique along my current line. The only constrain on my parameters is that they remain positive. Occassionally one will approach zero, not often. I am reposting because I have another focused question stemming from the same problem. MY SITUATION: I am studying a time-dependent stochastic Markov process. The conventional method involved fitting data to exponential decay equations and using the F-test to determine the number of components required. The problem (as I am sure you all see) is that the F-test assumes the data is iid, and conflicting results are often observed. As a first step, I have been attempting to fit similar (simulated) data directly to Markov models using the Q-matrix and maximum likelihood methods. The likelihood function is: L= (1/Sqrt( | CV-Matrix |))*exp((-1/2)*(O-E).(CV-Matrix^-1).(O-E)) Where | CV-Matrix | is the determinant of the Covariance matrix, (O) is the vector of observed values in time order and (E) is the vector of the values predicted by the Markov model for the corresponding times. The Covariance matrix is generated by the Markov model. My two objectives are to determine the number of free parameters, and to estimate the values of the parameters. Because the data is simulated I know what the number of parameters and their values are. MY PROBLEM: I have been using the Log(Likelihood) method to compare the results of fitting to the correct model and to a simpler sub-hypothesis (H0). I am getting very small Log(Likelihood ratio)?¡¥s when I know the more complex model is correct (i.e. H0 should be rejected). When I first observed this I tried increasing the N values, and found a decrease rather than an increase in the Log(Likelihood ratio). When I look at the likelihood function, the weighted Sum of Squares factor : ( (O-E).CV^-1.(O-E) ) is very different between the two hypotheses (i.e. favoring rejection of H0), but difference in the determinant portion ( (1/Sqrt( | CV-Matrix |)) ) is in the opposite direction. As a result, the Log(Likelihood ratio) is below that needed to reject H0. I asked about just fitting (O-E).CV^-1.(O-E) and was reminded that without the determinant factor, the likelihood would be maximized by simply increasing the variance. This appears to be true in practice. In learning about the quadratic form, I read in several places that, for the distribution to approach a chi square distribution, the Covariance Matrix must be idempotent (CV^2 = CV). I am almost certain this is not the case. I am hoping to get feedback on this idea: THE QUESTION: Following maximization of the full likelihood function ( (1/Sqrt( | CV-Matrix |))*exp((-1/2)*(O-E).(CV-Matrix^-1).(O-E)) ) for both models, can I use the F-test to compare the weighted Sum of Squares (i.e. (O-E).CV^-1.(O-E) ) of the two models, rather than the likelihood ratio test. In other words, does correcting each (O-E) for its variance and covariance legitimize the F-test? Any insight is greatly appreciated. Thanks for your patience and consideration. James Celentano = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
