Orton's Arm Posted December 28, 2006 Author Posted December 28, 2006 I have provided specifics. And math. Regression is a function of the variance in a statistical distribution. Period. That's specific, and it's math. And it's right. And it's exactly what you haven't been saying. The one topic I've repeatedly asked for specifics on is one which you've consistently avoided. Suppose you have a population that's normally distributed with respect to I.Q., and an I.Q. test which has measurement error. The measurement error is normally distributed with a mean of zero. Suppose you were to give everyone in the population an I.Q. test; and were to retest those who obtained, say, a 140 the first time they took the test. Would people retaking the test tend to score closer to the population's mean upon being retested?
Bungee Jumper Posted December 28, 2006 Posted December 28, 2006 Give up already. You're wrong, and you'll continue to be wrong until you understand the math.
Orton's Arm Posted December 28, 2006 Author Posted December 28, 2006 Give up already. You're wrong, and you'll continue to be wrong until you understand the math. The math is actually fairly simple and straightforward. But I can see why you'd want to make it seem harder than it really is as an excuse to once again dodge my question.
Bungee Jumper Posted December 28, 2006 Posted December 28, 2006 The math is actually fairly simple and straightforward. Yes, it is. I've done it. You should too. But I can see why you'd want to make it seem harder than it really is as an excuse to once again dodge my question. I have answered it, several times. You're just too !@#$ing dumb to notice. Go back and look it up.
Orton's Arm Posted December 28, 2006 Author Posted December 28, 2006 Yes, it is. I've done it. You should too.I have answered it, several times. You're just too !@#$ing dumb to notice. Go back and look it up. Your post is wrong from start to finish. For one thing, I cannot recollect a single instance in this discussion when you've done any math. Not once. Secondly, I have done math, both by setting up my Monte Carlo simulation, as well as in the examples I've sometimes given. Third, you've given contradictory answers to the question I've asked, which is why I'm asking it again. When I first announced that people who get high I.Q. scores tend to score somewhat lower upon being retested, you ridiculed both the phenomenon itself, and me for proposing it. Later, you told Wraith that you weren't disputing the existence of this phenomenon. So I want you to come out and give a definitive statement about what you do or don't believe regarding this phenomenon. I've asked you for clarity on this issue before, only to have you dodge the question.
Bungee Jumper Posted December 28, 2006 Posted December 28, 2006 Your post is wrong from start to finish. For one thing, I cannot recollect a single instance in this discussion when you've done any math. Not once. Secondly, I have done math, both by setting up my Monte Carlo simulation, as well as in the examples I've sometimes given. Third, you've given contradictory answers to the question I've asked, which is why I'm asking it again. When I first announced that people who get high I.Q. scores tend to score somewhat lower upon being retested, you ridiculed both the phenomenon itself, and me for proposing it. Later, you told Wraith that you weren't disputing the existence of this phenomenon. So I want you to come out and give a definitive statement about what you do or don't believe regarding this phenomenon. I've asked you for clarity on this issue before, only to have you dodge the question. I'm not even reading this. Shut up and do the damn math already. Here's the same hint I've been giving you for weeks: "variance".
Orton's Arm Posted December 29, 2006 Author Posted December 29, 2006 I'm not even reading this. Shut up and do the damn math already. Here's the same hint I've been giving you for weeks: "variance". I've done the relevant math. You've promised plenty of math in the form of Monte Carlo simulations and the like, but you've yet to deliver on any of those promises. The people at Stanford, Berkeley, the University of Chicago, etc., know plenty of math, and guess what? they're saying the same things you ridiculed me for saying. You're dead wrong about this. Stop hiding behind the word "math."
Dibs Posted December 29, 2006 Posted December 29, 2006 Give up already. You're wrong, and you'll continue to be wrong until you understand the math. Why didn't you answer his question? The one before your response above. To this question here..... .......Suppose you have a population that's normally distributed with respect to I.Q., and an I.Q. test which has measurement error. The measurement error is normally distributed with a mean of zero. Suppose you were to give everyone in the population an I.Q. test; and were to retest those who obtained, say, a 140 the first time they took the test. Would people retaking the test tend to score closer to the population's mean upon being retested? It's a straight forward question. Why avoid answering it?
Bungee Jumper Posted December 29, 2006 Posted December 29, 2006 Why didn't you answer his question?The one before your response above. To this question here..... It's a straight forward question. Why avoid answering it? I did answer it. Eons ago. He's just too !@#$ing dumb to look it up.
/dev/null Posted December 29, 2006 Posted December 29, 2006 I did answer it. Eons ago. He's just too !@#$ing dumb to look it up. or maybe he doesn't want to read thru 23 pages and a couple merged threads of namecalling
Bungee Jumper Posted December 29, 2006 Posted December 29, 2006 or maybe he doesn't want to read thru 23 pages and a couple merged threads of namecalling Don't see why not. He wrote half of it.
/dev/null Posted December 29, 2006 Posted December 29, 2006 Don't see why not. He wrote half of it. just because he wrote half of it doesn't mean he actually read the other half
Bungee Jumper Posted December 29, 2006 Posted December 29, 2006 just because he wrote half of it doesn't mean he actually read the other half Very true. Also not my problem.
syhuang Posted December 29, 2006 Posted December 29, 2006 .......Suppose you have a population that's normally distributed with respect to I.Q., and an I.Q. test which has measurement error. The measurement error is normally distributed with a mean of zero. Suppose you were to give everyone in the population an I.Q. test; and were to retest those who obtained, say, a 140 the first time they took the test. Would people retaking the test tend to score closer to the population's mean upon being retested? It's a straight forward question. Why avoid answering it? Dibs, this question is actually not that straight forward. First, in such a group, some of them are likely to get scores closer to population mean upon retaking the test. Some of them are likely to get scores higher than 140. Furthermore, the first subgroup is larger than the second subgroup. Therefore, the more accurate statement is "When retaking the test, more people in this group are likely to get scores closer to population mean than not" or "The average score of the group is likely to be closer to population mean when retaking the test". However, when talking about "people tend to score closer to the population's mean", I think it's kind of ambiguous. One side can say "Yes, they are, becuase there're more people likely to score closer to population mean or the average score is likely to be closer to population mean". The other side may say "No, because not all people are likely to score closer to population mean". Therefore, it's kind of about "semantics", does "average score" or "majority of the group" represent the whole group? Some may agree, some may disagree. I can understand what both sides are talking about, but I really don't want to join this name-calling war. So, let me try it once to see if it helps the discussion. First, the phenomenon HA described exists, but the more accurate statement is "average score of the group" instead of "people in the group". However, does the average score regress toward the "population mean"? This is another ambiguous part. I think HA's answer is yes, becuase "the average score of the group" is likely to be closer to population mean upon retaking the test. BJ's answer is no, becuase the target (not a good statistics term, but you know what I mean) of the "average score" regressing toward to is not "population mean", it's "mean of error". What's the difference? Let's say in a normally distribute population with mean of 100, one of them has a real IQ of 120 (from other more accurate tests) and he scores 140 the first time. Of course, the assumption here is normally distributed error with mean of zero. When retaking the test, (1) is he likely to score lower than 140? Yes. (2) is he likely to score closer to population mean(100)? Yes. (3) is he likely to score closer to his real IQ (120)? Yes. So what's the target of his score regress toward to? 120 or 100? I think most of you will say 120, which is his real IQ. Ok, now, go back to the example with a group of people who score 140 the first time. The assumption is the same (normally distributed error with mean of error and normally distributed population. The population mean is 100), and I simplified the example. Let's say there're three people who score 140 the first time and we know from other more accurate tests that their real IQ are Person A: 150 Person B: 125 Person C: 115 The average of their real IQ is (150+125+115)/3=130. (again, this may not be a good statistics way. Hope this can show the idea) Thus, when retaking the test, A is likely to get a score higher than 140 and be closer to his real IQ. B and C are likely to get a score lower than 140. The average score of the group is likely to be lower than 140. When retaking the test, (1) is the average score likely to be lower than 140? Yes. (2) is the average score likely to be closer to population mean(100)? Yes. (3) is the average score likely to be closer to their mean of real IQ (130)? Yes. Like the example of one person earlier, what is the target of the average score regressing toward? The population mean(100) or mean of real IQ (130)? I think most will say mean of their real IQ. Thus, the argument here is again about "semantics". You can say the statement of "regression toward mean of population" is right because the average score is likely to be closer to population mean". You can also say the statement is not right becuase the target of the regression is not population mean, which happens to be in the same direction of mean of erorr. In short, the phenomenon exists, but can the statement be called "regression toward the population mean"? it depends on what you refer to, a direction or a target.
Bungee Jumper Posted December 29, 2006 Posted December 29, 2006 Thus, the argument here is again about "semantics". You can say the statement of "regression toward mean of population" is right because the average score is likely to be closer to population mean". You can also say the statement is not right becuase the target of the regression is not population mean, which happens to be in the same direction of mean of erorr. In short, the phenomenon exists, but can the statement be called "regression toward the population mean"? it depends on what you refer to, a direction or a target. True, EXCEPT... "Regression" has a very specific mathematical definition (again, having to do with that pesky "variance" idea that HA can't wrap his empty little head around). In the strict mathematical definition, what HA is talking about is not regression toward the mean (unless he's talking about regression of the error toward the mean error of zero - which he's obviously not, by virture of the way he arbitrarily constructs his example). Calling his example an example of regression toward the population mean of a population is, in fact, absolute idiocy: it can't even begin to be mathematically correct, as it attributes the variance of the error distribution to the population distribution, which is mathematically complete and utter nonsense. Yes, it's a semantic argument. That's precisely because HA refuses to make it a mathematical one - he's spent a thousand posts arguing vocabulary and calling it math.
Dibs Posted December 30, 2006 Posted December 30, 2006 .....Dibs, this question is actually not that straight forward............Thus, the argument here is again about "semantics"...... Thanks for that syhuang. Well explained. I figured as much myself but not being conversant with all the correct terminology I was unsure. To me it felt like a semantic argument due simply to the fact that I totally understood what HA meant.....& it seems a totally logical premise.....yet people continued to argue the point. There is somebody I know (in real life, not cyber life) who seems to feel that the people should only use the exact meaning of words & that anyone who has errors in this regards is a moron(sounds stupid, but he really does). I tend to view the English language as not only very pliable.....i.e. words have multiple meanings(which can cause confusion when using a general meaning for a word in a specific area....such as statistics....which has a very specialized & exact meaning for the word).....but....is designed for communication.....not miscommunication. IMO if person A understands what person B means.....even though person B has used words incorrectly(very obvious example is pacific/specific), person A is an arse to belabour the fact that what person B said was technically incorrect, if in fact what person B had meant was a correct. As an example..... "Regression" has a very specific mathematical definition is fairly irrelevant if you know what the person you are talking to means. Being a stickler for specific definitions is laudable when it is important for communication. In the cases where it isn't however, it is simple elitism. What it means is that people who are not fully conversant in a fields terminologies would be totally excluded from discussion in that field. Sometimes though, the boffins cannot seem to understand this.
Dibs Posted December 30, 2006 Posted December 30, 2006 Yes, it's a semantic argument. That's precisely because HA refuses to make it a mathematical one - he's spent a thousand posts arguing vocabulary and calling it math. It seems to me that it is a semantic argument because you refuse say to HA.... "Your premise is correct.....but you are using certain words incorrectly." Either you are.... blinded(fixated on the specific definitions), an idiot(in that you cannot see the logic behind HAs premise) or an arse(you see the logic but just wish to continually be a stickler for 'rules' because it's annoying to him). I'm figuring blinded.
Bungee Jumper Posted December 30, 2006 Posted December 30, 2006 It seems to me that it is a semantic argument because you refuse say to HA...."Your premise is correct.....but you are using certain words incorrectly." Either you are.... blinded(fixated on the specific definitions), an idiot(in that you cannot see the logic behind HAs premise) or an arse(you see the logic but just wish to continually be a stickler for 'rules' because it's annoying to him). I'm figuring blinded. Well...no. The crux of the argument is and has been that he's specifically manufacturing a specific effect and calling it something else, and pretending it's rigorous math. That fails on three points: 1) His example is completely manufactured. 2) He's grossly misusing terminology. To the point where his premise is not correct (specifically, he's confusing two very different things - error distribution and population variance). 3) He's pretending to be mathematically rigorous without being the least bit mathematically rigorous. By way of analogy, he's pointing at a dog and calling it a cat because it happens to be a domestic pet with fur and four legs, so what's the difference? It's a semantic argument because in this case the semantics are actually important. In this case, "regression" has a specific and definite mathematical meaning - either linear regression or multiple regression, take your pick in this context. The math gives the same result either way...regression is directly related to the correllation coefficient of pairs of variables, the random error HA's talking about by definition has a correllation coefficient w/r/t measurable IQ of precisely zero (again, by definition), which means that error CANNOT cause regression toward the mean as HA keep saying. QED. Though I recommend the Oxford Dictionary of Statistics for brevity and price, any decent statistics book covers the math in enough detail that anyone with college math skills can prove it (I'd post the math myself if the board format were conducive to equations). If you're semantically rigorous, that's easy to explain. If you're a complete dunce like HA, and keep using inexact and colloquial terms like "luck", no amount of explanation is going to help. He himself is too semantically challenged to understand the concepts, which is probably why his explanation has changed precisely four times and his only proof is "Stanford said so".
Dibs Posted December 30, 2006 Posted December 30, 2006 Well...no. The crux of the argument is and has been that he's specifically manufacturing a specific effect and calling it something else, and pretending it's rigorous math..... Are you sure about that? As far as I could tell, the actual crux of what HA initially was saying was the question he proposed in post #441 above. The fact that an argument ensued leads me to believe nobody actually gave the base concept any credence and instead pounced on....perhaps certain wordings....which then logically degenerated into tens of pages of argument over irrelevancies to the initial premise. Basically.....if you answer his question.....and the answer is "yes"(which it obviously is).....you acknowledge that the basics of what HA said is correct(regardless of his ability to prove it mathematically....or his ability to word the question in a totally 'correct' manner.). So....what's the answer to the question? Here it is again so you don't have to scroll up..... .......Suppose you have a population that's normally distributed with respect to I.Q., and an I.Q. test which has measurement error. The measurement error is normally distributed with a mean of zero. Suppose you were to give everyone in the population an I.Q. test; and were to retest those who obtained, say, a 140 the first time they took the test. Would people retaking the test tend to score closer to the population's mean upon being retested?
Bungee Jumper Posted December 30, 2006 Posted December 30, 2006 Are you sure about that? As far as I could tell, the actual crux of what HA initially was saying was the question he proposed in post #441 above. The fact that an argument ensued leads me to believe nobody actually gave the base concept any credence and instead pounced on....perhaps certain wordings....which then logically degenerated into tens of pages of argument over irrelevancies to the initial premise.Basically.....if you answer his question.....and the answer is "yes"(which it obviously is).....you acknowledge that the basics of what HA said is correct(regardless of his ability to prove it mathematically....or his ability to word the question in a totally 'correct' manner.). So....what's the answer to the question? Here it is again so you don't have to scroll up..... Yes, I'm sure about that. He's been calling that "regression toward the mean" for the past two months, and it's not. If you're just reading this thread, you don't have the whole absurd story.
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