shn

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Dec 5	comment	Aggregating votes from different users with some constraints and probabilities @whuber the difference with the knapsack problem is that in our case, we need to maximize a sort of expected correctness for the overall answer (after aggregating all the received answers), while for the knapsack problem we do our best to put items in the sack until we can no more add items to it. There is no relation between items, they should not be aggregated to give a final "good item", while in our case the answers are aggregated to get the best final answer.
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities @Jonathan Christensen But you didn't really answered the last question: How can you decide to stop asking users? i.e., how can you check that the current better overall answer can probably not be improved by asking a new user ? I mean algorithmically speaking you have a sorted vector where each entry i is the ENCAD of the i'th user; you start with the first answer (of the 1st user having the highest ENCAD), you still have budget, you ask the 2nd user and get its answer, now you aggregate the 2 answers that you have, how can you check if you should or not ask another user ? What do you compute?
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities @whuber Yes reversing the answer of users having a low probability may be a good suggestion when we consider this probability to be a "probability of answering the question correctly", thank you. I can also consider the probability to be about the "reliability" of the user according to his domain knowledge about the question, in this case users with a low probability (having nearly no knowledge about the domain) may answer randomly, in this case it is better to consider their "probability of answering the question correctly" to be 0.5.
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities This is also a part of my question, and of your approximate algorithm: Using the information on how reliable each user is, how can you check that every additional answer you get from a user, will help you estimate a better overall answer ?
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities (Note that I'm talking just about one question to ask to users). If for example I have a very big budget (say infinit) which is sufficient to ask all the users, then if I ask all the users, the final answer that I aggregate/deduce may be worse (have less chance to be correct) than the case where I just ask only the first user which have an ENCAD equals to 0.98 (very hight) ... Your approximate algorithm is not considering such cases.
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities @whuber What is called "vote" is the classical aggregation of the answers, by taking the most common answer among the received ones (if there is more "Yes" answers then ok I will consider the final result as "Yes), but in my real example I have some constraints: probabilities associated to users, ammount of money requested by each user to answer my question, and the total amount of money that I have, so I need to also consider these constraints. If you want, forget about the word "vote" in the question, just tell me who to ask and how to aggregate the answers that I get. Is it clear ?
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities @whuber What I want is to maximize the fact of getting the good final answer by aggregating the obtained answers (want to know that the answer to my question is "Yes" 10/2=5). For this, which users should I ask (given my budget, and given that each user request a given amount of money to accept to answer my question), and how to aggregate their answers in order to have the correct one.
Dec 5	comment	Aggregating votes from different users with some constraints and probabilities @whuber "what does it mean for a vote to be correct": if the question is "does 10/2 equals 5?" then "Yes" is a correct answer to this question; now if the probability associated to the user is 1, that means that the user will answer this question correctly because it have a very good knowledge of the domain of the question (mathematics); if the probability associated to the user is 0, this means that the user will answer this question wrongly; if the probability associated to the user is 0.5, then his answer may be correct ("Yes 10/2=5") with a probability of 0.5, or maybe wrong wt proba 1-0.5
Dec 4	comment	Measuring correlation or dependence between two data sets @Glen_b What I was looking for is the second case, not the first one, and it seems that the canonical correlation analysis is what I was looking for.
Dec 3	comment	Measuring correlation or dependence between two data sets @Glen_b I don't have the I don't have the data-points on my computer, they are generated by a program on another computer. The data-points of the two datasets have only two attributes (x, y). Can you please tell me what are the statistical test or measure to evaluation the degree of correlation or dependence between two sets of data-points ?
Nov 29	comment	Combining results of two classifiers to better classify a data-point We know if LA1 = LB1 or not by just comparing LA1 with LB1 ! I've never said that we dont know whether or not two labels are equals !
Nov 29	comment	Combining results of two classifiers to better classify a data-point @jerad LA and LB are just notations, that means Label given by classifier A, or Label given by classifier B. I'm just saying that for a given point x, it will be classified in a given class by classifier A, and in another or same class by classifier B. Example: A classify x in class 1 (i.e. LA1) with probability 0.8, and in class 2 (i.e. LA2) with probability 0.1. B classify x in class 1 (i.e. LB1) with probability 0.6 and in class 3 (i.e. LB2) with probability 0.2. Here LA1 = LB1 = class 1.
Nov 9	comment	An incremental Gaussian mixture model 1) SUM_l Plj is the sum per object not per cluster, isn't it ?! 2) I don't know exactly how to update the covariance matrixes, how each cell sj[.][.] of each matrix sj should be updated ?!
Nov 9	comment	An incremental Gaussian mixture model I added a pseudo code to my question to see if I understand the concept. Can you check if I understand it the right way ? Also, I can update the mean vector for each cluster j (as I did in this code), but I don't see how to update the covariance matrix of each cluster, can you please explain it algorithmically speaking ?
Nov 8	comment	How to define a posterior probability of y given x when the model is not probabilistic? @Anony-Mousse I want to have some confidence about how probable the instance x is of a class y. I want to use this probability to let the learning algorithm be able to actively queries the data-points about which it is least certain how to label (i.e. ask for there labels). This is often strainghtforward for probabilistic learning models. For example when using a probabilistic model for binary classification, it will simply queries the instances whose posterior probability of being positive is nearest 0.5. Now how to define P(y\|x) when the model is not probabilistic (e.g. with the online KM) ?
Nov 4	comment	Classification error is lower when I don't do any learning on the dataset? Ok, the problem came from the duplicated points. By removing them, some classifiers achieve a slightly better recognition rate than the NN strategy. However, I didn't noticed that there was too much duplicated points, I removed them and I end-up with a much smaller dataset, the number of instances is not really sufficient to perform an online learning. Do you know any available labelled dataset about document classification which is ready to use (i.e. that I can use without doing preprocessing and much stuff ...) ? There is a great bag-of-words dataset on UCI repo, but labels are not provided.
Nov 4	comment	Classification error is lower when I don't do any learning on the dataset? I'm directly using the training set to do the classification of the test set points. There is no training phase that was performed on the training set. I didn't learnt nothing, I just classified my test points. I don't know why you call this "learning" just because the train set is used. Nonetheless, I just checked the dataset and you are right, there are some duplicated data-points, sometimes the same data-point is in the train and test set, this is not the case for all data-points, but I'll try to fix remove the duplicates and see if the problem is fixed.
Nov 4	comment	Classification error is lower when I don't do any learning on the dataset? @DouglasZare I've tested with many classifiers from Weka, and others that I implemented myself. For many datasets, using a classifier with a training step (i.e. case(2)) will give better results than case(1), however, for the three datasets of bag of words that I'm currently testing on, I get better results in case(1). I don't think that learning/classifier is not working well, since I've tested with many classifiers.
Nov 2	comment	Choosing which data-point to label (active learning) The objective is classification. I didn't yet read the review that you proposed, maybe I can find an answer in it, but let me ask this question: we should query the points in which we are the least certain of their label; Ok but how do we know that we are not confident about the label assignment for a given point ? For SVM for example it is the points near the margin; but generally speaking I guess that it is points that are in an overlapping area (between classes), so should we detect them based on the assumption that such a points are generally in a low-density area of data ?
Nov 2	comment	bag of words in an online configuration, for classification / clustering Another problem is that we can have may words which are frequent but do not relate to the topic of the document such as "the", "of", "to" etc ... and I hopefully prefer the method to be independent from the language.