Does Penalized Maximum Liklihood Estimation in Logistic Regression produce similar estimates to exact logistic regression.  Can be univariate model or multivariate, can be all dichotomous, all continuous, or a mixture.
212767
1. Go to Google at http://www.google.com/\n\n2. Enter into the search field:\n\nPenalized Maximum Likelihood Estimation in Logistic Regression\n\nor even\n\nFirth algorithm Penalized Maximum Likelihood Estimation in Logistic Regression\n\n3. Wow!\n\nThere are resources such as this from the Oxford journal Biometrika:\n\nhttp://biomet.oxfordjournals.org/cgi/content/abstract/69/1/123\n\nPenalized maximum likelihood estimation in logistic regression and discrimination \nJ. A. ANDERSON and V. BLAIR \n\nDepartment of Statistics, University of Newcastle upon Tyne Manchester\nDepartment of Statistical Unit, Statistical Unit, Christie Hospital Manchester \n\nMaximum likelihood estimation of the parameters of the binary logistic regression model for pr(H|x) is discussed with separate discussion of sampling from (i) the conditional distribution of H given x, (ii) the joint distribution of H and x, and (iii) the conditional distribution of x given H. Difficulties associated with continuous x in the latter sampling scheme are discussed. To avoid these, penalized maximum likelihood estimates are introduced, which give estimates of the logistic parameters and a nonparametric spline estimate of the marginal distribution of x. Extensions to multinomial logistic regression are outlined. \n\nKey Words: Binary logistic regression • Discrimination • Maximum likelihood estimation • Multinomial logistic regression • Nonparametric density estimation • Penalized maximum likelihood estimation • Spline function\n\nThere should be enough there to give you a sense of the answer.
173076
You use regression when you want to fit a model to predict some value or outcome. Multinomial regression is where you have more than one input to predict one output. \n\nHere is an example of regression:\n\nSuppose you had the following data:\n\nHeight Weight Age for 100 people. \n\nYou might want to see if you can predict someone's age if you are given their height and weight. You call age the dependent variable because it depends on height and weight, and you can height and weight the independent variables. \n\nWhen you fit your model, you get something like this:\n\nAge = Intercept + A * Height * B * Weight\n\nThe model gives you the values of A and B and Intercept, so you can predict the Age of someone if you know their Height and Weight.\n\nThis is multinomial regression where the dependent variable is continuous, that is, it can be any value within reason and it varies.\n\nLogistic regression is used when you want a binary outcome, for example 1 meaning success and 0 failure.\n\nUsing the same example, you might want to predict someone's sex based on their height, weight and age. In this model, you would have three independent variables making it multinomial and your outcome would be 1 = male, 0 = not male, that is female. \n\nThe logistic regression model gives you the probability that, given someone's age, weight and height, that they are male (in this example, the higher the probability, the greater the chance that they person is male).\n\nAgain, you get a number of coefficients, but this time you use them to calculate a LOGIT:\n\nLOGIT = Intercept + A * Height + B * Weight * C * Age\n\nThe LOGIT is also written as:\n\nLOGIT(P) = Log [( P )/ ( 1 - P )]\n\nSo if you fit your model and get your values for A, B, C and the INTERCEPT, you can calculate the LOGIT value from some data and with this, calculate the probability P, that they are male. This probability P is given by:\n\nP = 1 / ( 1 + e**-(LOGIT))\n\nwhere ** is to the power of.\n\nThis is a much better way to model categories or classications, which are often used for understanding risk in banks and marketing, where you want to know the likelihood that someone will buy a product or default on a loan payment.
112827
One cool way that I have seen it derived is based on Jayne's Formalism.\n\nAssume we sample process, and measure the mean and standard devation of the samples. There are many possible distributions that will produce those exact mean and standard deviation. \n\nJayne's Formalism chooses the distribution which maximizes the entropy of the distribuiton. This is viewed mathematically as the distribution that makes the minimum number of assumptions regarding the distribution.\n\nI always thought that was kinda cool.
35666
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
557016
Something like that.......\nFor high school - average estimation + exam from language and from mathematics.......After few days you have point scoring finished and know can you go in school you wish or must try a differed school (obviously worst)......\nFor college (faculty) - average estimation from high school + 2-3 exams (deepens what faculty or college you want to sign  in).....About few days you have a list (like in the high school) and you see - do you en roll in faculty (college)
546215
= ln x / x^5\nln x -> inf\nx^5 -> inf\ndifferentiate both sides (Lagrange)\n\n=> 1/x  / 5x^4\n= 1/ (5x^5)\n=> 0 \n\nOr ln x increases much slower as x gets bigger, whereas x^5 increases faster, (exponentially) as it gets bigger. polynomial wins over logarithm. \n\nThe limit is 0.
285956
I think  you are asking about normalization.\nIf so please be more precise.
72436
As with any sample, you generally discard the highest and lowest extremes to reduce the rate of error. Millionaires can be found in any field and are atypical of the population anyways, unless you are talking about Wall Street :)
319911
Not sure if you can say if there is an equivalent between the two.  Runout is typically with respect to a datum AXIS.  The concentricity is with respect to a datum FEATURE.
522804
Yes, but they remain equal only if you differentiate both sides with respect to the same variable(in this case x). You can never only do one side.
330375
Prior to Billy Bean, estimates were made by experts with a lot of experience. Now, many methods can produce results that are similar to his. The best method would be a combination of experience and statistics. This is the heart of professional handicapper in any sport. In fact, it makes up the core skills of a arbitrageur in financial markets as well.\n\nAn example of a simple method is linear regression. For exmple, go to yahoo sports and get the statistics of each baseball player on a team. For each game, create a matrix such that each row is the vector of that player's stats (prior to the game) concatenated with a vector of the pitchers stats.\n\nFor example, the first column could be number of games, then at bats, then runs, homers, .... The second part of that row  would be the pitchers stats. The next row would be the next player with the same pitcher and so on. \n\nThere are many ways to construct this matrix. You could do Principal Components Analysis afterwords, you could include variance and covariance, you could include some weighted sum of relief pitchers, you could do different sets for AL or NL, the possibilities are endless. \n\nNext create a vector (called the  response vector) that for each row has the points scored. You'll have to devise a statistic that is some combination of runs and rbi. For simplicity, you could just sum the two. \n\nThis is the player based method. The team based method would construct the matrix as a weighted sum of the entire team's stats as oppossed to each player. Then the response vector would simply be the points the entire teamed score. If you want to be really thorough about it, do both and compare the difference.\n\nOnce you have the matrix and the response vector, you simply perform linear regression to obtain set of weights. If the matrix is 'A' and the points are 'b', then solve for 'x' in the equation in \n\n            Ax = b\n\nMatlab does this pretty well or you could use Mathematria. If you like free stuff, check out a copy of numerical recipies from the school library. There are actually some Perl modules that do this. Some calculators do this as well. Be careful about a method that uses Grahhm-Schmidt or Gaussian Elimination. For very large data sets like the one you should have, you'll have impossible memory requirements, the program will be really slow and you will have bad round off errors.\n\nWhen you have the weights, which is simply a vector, for any game that you want to handicap, you simply construct the vector or set of vectors (in the 'by player' case) that represent the game and then multiply the matrix representation of the game by the weights to obtain the estimate for the score. \n\nSometimes you may want to know the varianc of this estimate, because you believe your model may inaccurate for a particular game, or the estimate seems way off. So if the variance of this estimate is too high, you want to 'blend' it with a prior estimate which is simply the average total score of all games. A simple way to estimate error is to use the residuals of the regression estimate. You can also compute the the square of the score. Then the variance is the expectcted square of the score minus the square of the expected score.\n\nIn general, with enough data and enough stats (if you use some method to control for the curse of dimensionality), you can actually get really close to what bookies do, just using  Yahoo data.
96226
See comment above.\n\nYou do need to do several trials at various concentrations though. You can then determine it through a linear regression.\n\nI also suggest that you consult a book on Analytical chemistry, because Beer's law has limitations. For example, it loses linearity at higher concentrations.
479987
The bigger the sample size, the more accurate the estimate.\n\nSay I was estimating the percentage of red m&m's in a pack.\n\nIf I sampled 10 m&m's, I might I might find that the sample contains 1 red one. This could lead me to think that a pack contained 10% red m&m's.\n\nHowever, if I sampled 100 m&m's, I might find that the sample contains 15 red m&m's, leading me to think that a pack contains 15% red m&m's. This answer is impossible using the sample size of 10, as you can't get 1.5 m&m's of a particular colour.\n\nIf I sampled 1000 m&m's I might find that I have 143 red ones, leading me to think that the pack contains 14.3% red m&m's.\n\nSo, as you can see, the bigger the sample size, the closer the estimation comes to the actual answer.
495480
Brute force though it may be, an x-dimensional linear regression is still your best bet for optimization.  If there is a single optimal process, the method will yield it, but you are more likely ro arrive at a number of processes, each optimized, that will yield a maximum profit.\n\nYour other alternative is to optimize each process independently, and then choose the most profitable, but you are likely to discard combinations which may yield a better end result, and you will still need  n-dimensional regressions of some sort.\n\nIn lieu of dedicated software for the problem, a spreadsheet can be used quite effectively.  For x > 7, however, you may need to apply techniques to keep numbers within your computer's ability, although this is not as problematic as it is with polynomial regressions.
26748
do you mean things as far as max and mins, asymptotes and things like that?\n\nif you are a little more specific i can help you better, or e-mail me, it would be a little easier  matttlocke@yahoo.com
320595
You might be looking for Regression analysis.\nThis is the statistical technique used to look for correlations between variables.
145265
If both the dependent variable (Y) and the independent variable (X) are standardized, then both their means are zero and standard deviations are one.  The correlation coefficient is not effected by a change in scale, meaning that the correlation doesn't change if you add or subtract the same number from each X or Y or if you multiply X or Y by the same positive number.  An example of this is if the correlation between height and weight when height is in inches and weight is in pounds is 0.6, then the correlation if height is in centimeters and weight is in kilograms is also 0.6.  We would hope that the relationship between the two variables doesn't change if there is a change in how the variables were measured.\n\nThis is the case when standardizing, since standardizing has you subtract the mean from each value and then divide by the standard deviation.  So the correlation coefficient for the standardized values of X and Y are the same as the correlation coefficient for the original values of X and Y.\n\nNow, the formula for the slope coefficient is r(sy)/(sx), where r is the correlation coefficient and sx and sy are the standard deviations of X and Y respectively.  Since both X and Y are standardized, then both of their standard deviations are equal to one, so putting that back into the expression above, the slope coefficient is equal to r.\n\nTo summarize, if X and Y are standardized, then the slope coefficient is the same as the correlation coefficient.