tttjjj/elt_bench huggingface dataset:
Orange's features page says 'under construction' and they don't list screenshots like what I am doing. weka has no features list at all. They may be able to do what I want, but if they don't promote the feature, how can I tell. I am more convinced by the other choices.
Yes, it shares the data snooping properties of stepwise selection. In fact, stepwise selection is an efficient approximation to full subset selection.
EnsembleSVM may be of interest to you. It is made for exactly such situations. Unfortunately there is no R interface yet. http://www.esat.kuleuven.be/sista/ensemblesvm/
Glen_b -- I agree. I think what OP must have meant was that $ y_i - \\bar{y} = y_i - \\hat{y}_i + \\hat{y}_i - \\bar{y} $
Suppose you have $n$ observations. LOO-CV takes 1 observation out of your data set and uses all the others to train your model. Measures how well it predicted on the left out(LO) observation. Then it takes another observation wich is not the previous one and uses all the others to train your model. Measures how well this model predicts on this LO observation. Does this $n$ times. The error would be the average of all these measures. 10-fold separates your observations into 10 groups $|g1|+|g2|+...+|g10|=n$. Uses the groups 1 to 9 to train, tests on $g10$. Does this 10 times. A shorter process.
Could you please explain the sense in which your `x` and `y` values define a single "distribution"? Usually, in statistics a discrete distribution will be specified by a collection of probabilities, one per possible value, and the probabilities must be non-negative and sum to unity. It is puzzling that your numbers do not have these properties.
Fairly sure that, on the internet, there's a higher ratio of trolls to sane people than there is in the real world. That might bias your results. Readers of large blogs, such as Pharyngula or Andrew Bolt's blog, have been known to crash polls by overloading them, and/or totally bias the results. It's pretty common that someone will put up a poll relating to the atheist/creationist flamewar, and within a day it will go from 100:20 in favour to 300:2000 against, due to Pharyngula readers. Be prepared.
"I would take a bet this is homework." Check out the OP's questions on Math.SE where many people have stopped answering the questions perhaps because of the OP's complete lack of any response to answers or comments asking for clarifications and the 0% acceptance rate.
(+1) I have (re)read your paper, which looks pretty interesting, although I found amazing the use of a Rasch model in clusters of genes. Did you compare your results with a sparse PLS-DA approach?
From what I understood, in the second case you correlate the annotator score with the model score. So you treat equally an annotation from ann1, ann2 and ann3; the info on the annotator number is lost. I think this is valid but not insightful, because different annotators can have different inclinations to vote one way or another. If that is the case, this be revealed by different slopes in a multivariate regression.
+1 re the *a priori* constitution of $Y$. It's an important point and I should've mentioned that there is no *a priori* number of samples in any of the scenarios I describe, and $Y$ does follow Bernoulli. In favor of LR, it could be argued that LR offers a multivariate analysis. Any ideas on the notion of using both consecutively?
If your goal is specifically to estimate the conditional 99th percentile, I'd vote for nonlinear quantile regression (of some sort--I don't know the R packages well), since it doesn't sound like you know the true functional form. I still wasn't clear to me from your previous question what the actual goal is, though, so I would reiterate the comment on your previous question from Spacedman Jan 4 at 17:01
which tutorials and books have you looked at? This one are great at an introductory level http://cs229.stanford.edu/notes/cs229-notes3.pdf. This is quite more technical, but insightful http://research.microsoft.com/pubs/67119/svmtutorial.pdf.
Right on the money. Thanks, chl.
R is the best but some prefer Python.
As to graphing those effects, you can try `interaction.plot(income, education, transactions)`, but you'd get better results by cutting income in classes, e.g. income = medium/low/high.
A short proof of the stated limit goes as follows: If $X$ is standard normal, then $x \\mathbb P(X>x)/\\varphi(x)\\to 1$ as $x\\to \\infty$, so $\\log\\mathbb P(X>x)−\\log \\varphi(x) + \\log x \\to 0$. But $2 \\log \\varphi(x) \\sim −x^2$ and so the result follows. A rough sketch for the case of the Poisson seems to indicate that the given limit is $\\lambda$, but I didn't check that too closely.
I expanded my answer to adress your comment. I think the explanation in the link is better than what I can cook up right now.
It seems to me like your question does not make much sense as it stands now. Perhaps you'd benefit from reading introductory chapters/papers to the Bayesian approach.
Another way to go is to forget about using a finite dimensional family of distributions like the ones you KS-checked, and use a non-parametric approach. The details will depend on what exactly you're trying to do. Classical kernel density estimators (http://en.wikipedia.org/wiki/Kernel_density_estimation) may give you some answers. A Bayesian analysis may use some distibution constructed with the help of a Dirichlet process, like this one: http://www.math.ucla.edu/~tom/papers/densityestimation.pdf
Did you share patients between devices? If you didn't then there's got to be an additional assumption that the patients in two groups are _similar_ in a broad sense.
OK, now I understand.  Great question.  I'm changing my answer accordingly.
Good references! I think this one is also worth reading: Blair & Higgins, "Test of hypotheses for unbalanced factorial designs under various regression/coding method combinations" (1977) http://epm.sagepub.com/content/38/3/621.full.pdf+html Like Venables, they stress that type III model comparisons give different results for different coding methods. For correct type III SS, one needs a sum-to-zero coding scheme like effect coding or Helmert. Dummy coding (default in R) gives the wrong results.
@Dan Thanks - I think its a nice idea to have a small range around the desired sum.
As of today there are 154 articles on PubMed mentioning [credibility interval](http://www.ncbi.nlm.nih.gov/pubmed/?term=%22credibility+interval%22) and 489 mentioning [credible interval](http://www.ncbi.nlm.nih.gov/pubmed/?term=credible+interval). They are not as common as confidence interval (179811 articles and counting), but it's just due to frequentist approach being the dominating method. And yes, credible interval sounds awesome but it's only true if the prior distribution is correctly specified. The devils are all in the assumptions.
I am no time series expert, by any means, but there are some good examples in the SAS documentation (especially helpful if you are using SAS, of course, but pretty nice even if you aren't).
hi you post seems interesting but i'm confused by the presentation. It seems you have a model and a sequential sampling scheme of the data. What is you need to estimate? Could write things mathematically?
+1, although it's worth noting that there are a few ways to work with larger-than-memory datasets in R ([bigmemory](http://cran.r-project.org/web/packages/bigmemory/index.html), [ff](http://cran.r-project.org/web/packages/ff/index.html), chunking data from a database, a wide range of distributed computing options).  But all of that takes setup; SAS will indeed just chug through whatever you throw at it, which is a real advantage.
If the data below the 10th or above the 95th percentile can be arbitrary, your estimate may be arbitrarily bad. On the other hand, if the distribution is sufficiently nice, you might get a very good estimate.
@COOLS Yes, [it is possible to convert back](http://stats.stackexchange.com/questions/21735/what-are-the-mean-and-variance-of-the-ratio-of-two-lognormal-variables/21740#21740) (and easy if the logarithms have Normal distributions). But the point of my original comment was that it's probably not even useful to transform back to the linear scale in the first place.
@kwak, the spatial effect you suggest is what is referred to as local Geary's C, Global formula, http://en.wikipedia.org/wiki/Geary's_C, (or here is a link for the local version  http://www.passagesoftware.net/webhelp/Introduction.htm#Local_Geary_s_c.htm ) you could also consider local Moran's I, http://en.wikipedia.org/wiki/Indicators_of_spatial_association
0.03 is the standardized or unstandardized? If standardized, that's very small.  By hierarchical, do you mean 2nd order? A 2nd order CFA with 2 (or 3, for that matter) first order factors is exactly equivalent to a regular CFA with a correlation.
This question appears to be off-topic because it is about statistics and belongs on Cross Validated.
Confidence intervals make no sense any more after stepwise variable selection procedures (nor do p-values), unless you calculate them based on bootstrapping the entire procedure, including the stepping. There is an extensive literature on this, which I heartily recommend that you delve into. ORs may make sense, on the other hand (but you will be too sure of them after stepwise model building).
For a general overview, have a look at page 3 of [this paper](http://www.math.wm.edu/~leemis/2008amstat.pdf). It contains a chart depicting the relationships between many distributions. Quite neat.
+1 for the "Empirical Bayes" mentioning. :)
thanks for answering, why I should run the Mann-Whitney-Wilcoxon test, in this case?
You can run Stan through R (see [RStan](http://mc-stan.org/rstan.html)).
Ah, I was wondering why you linked to this queston, in this question... :)
Do your classifiers return continuous valued "probabilities" between 0/1 like a risk prediction or do they return binary Yes/No predictions?
Strange that you used model fit indices for a confirmatory analysis while you used parallel analysis (usually found with exploratory analysis) at the same time.
If you have census data, then you have population means (and all other properties of the distribution) for variables (at that point in time, at least), so there is no estimation problem - for example, the null hypothesis of no difference in means is only true when the observed means are exactly the same. For statistical analysis of census data to make sense, you have to be conceiving of your data as realizations from some hypothetical infinite population whose parameter values you want to estimate.
As Macro said, Section 2.11.1 explains that when the data sets are different, you cannot compare the AIC's. I think that is the reason why the criteria for the summed model are significantly different. Gung, don’t you think the one who asked the question should do a little bit of research him/herself?
Thanks for confirming part of my understanding. The MASS lda function uses normalized scores.  In this simplest of cases, is there a hand-calculation way to adjust the groups grand mean (weighted by different n) for the within-groups covariance matrix? I will look for such a formula.
My experience is that fitting q correlated DVs together in one model can obtain more accurate estimates than fitting them separately. It is easy to distinguish the difference by simulations.
Have you considered doing a logistic regression, this may yield nearly identical results but may run MUCH faster. Also, you can try stochastic gradient descent, this is very helpful is datasets with many rows and columns. You should really try to retain all the data you can when training, limiting it will only make a lesser model.
@John I absolutely don't know. The question is unanswered unfortunatelly...
Apologies, I didn't read your question carefully enough. I believe your maximization step to compute $Q_\\Theta^2$ is only valid if you allow any distribution, that is, if you only make the factorization assumption. But you additionally assumed that $Q_\\Theta^2$ is a delta distribution. Try to explicitly maximize the lower bound with respect to $\\Theta^2$, the parameter of $Q_\\Theta^2(\\Theta) = \\delta_{\\Theta^2}(\\Theta)$.
Could you clarify why you can't include both predictors in the model? In your case, `X1` and `X2` would probably be correlated, as the brown spots probably increase with increasing time lying on the table.
I would vote this as an answer (solutions are provided in the linked document) - but I can't because it is 'just' a comment.
A logistic curve isn't the only curve that 'levels off'. Indeed a multiple of any continuous cdf would satisfy that requirement.
You can obtain the median using an [online algorithm](http://stats.stackexchange.com/questions/346/what-is-a-good-algorithm-for-estimating-the-median-of-a-huge-read-once-data-set).  If that's not good enough, search our site for [additional approaches](http://stats.stackexchange.com/search?q=%2Bmedian+%2Bonline).  Note that finding a median is a $O(N)$ algorithm and typically all the data need to be read at most twice, so $10^8$ is scarcely large: the median can be computed reliably in the amount of time it takes to read the data.
@Q2: You can change your data to wide format easily using R. Look for package `reshape2` or look [here](http://stackoverflow.com/search?q=[r]+reshape+long+wide).
ASIDE continued: However, in psychophysics we have a system that starts off not working at all, gradually starts being able to process the information provided to it, and then asymptotes near 100% when the information is clearly available.  Here I may have presented stimuli that are too hard to discriminate (beyond its range on the lower bound) or presented stimuli that are too easy to discriminate (beyond its range on the upper bound).  Here it is reasonable ask what the system will do at those extreme values.
Given the number of votes to close, you should probably try to reorient your question toward some kind of a statistical aspect rather than ask for technicalities that indeed are of the resort of the Help Center or another SE site.
Thanks! +1 and accetped
My three predictors are binary, but my covariate is continuous. In a moderated regression, I would enter it as a continuous predictor. Does this make things clearer?
I think it is critical whether $X$ and $Y$ are dependent or independent. It is not clear from the question. Are they independently distributed?
@Philipp The author is using average to refer to the fact that the characteristic function of a random variable $X$ can be expressed as the expectation of $\\exp \\left( i X \\right)$, i.e. mathematically, $\\varphi \\left( X \\right) = \\int_{\\Omega_X} \\exp \\left( i X \\right) dF_x $ You also did not answer fg nu's question.
@AndreSilva Thanks for your response, I've edited my question with a little more information. Hopefully that is clearer
1. $X$ is symmetric without assumptions, yes. However, $X = C^{-1} - I$ is correct but misleading: $X$ is not defined that way, $C^{-1}$ is (because of its structure). Consider the simple case where $C^{-1}$ is a diagonal matrix, where each element on the main diagonal is of the form: $1 + x_{ii}$, with $x_{ii} > 1$. You would write that matrix as $C^{-1} = I + X$. 2. Unfortunately, no matter how unusual it seems, neither assumption holds in my problem. However, if you can solve the problem with the last column of $H$ being all 1s, please do so, it would still be an important result.
Martin, there's no question or ambiguity about the quartiles here, because they would comprise the first and second smallest, the third and fourth, and so on, giving ({1,1}, {4,4}, {4,4}, {8,8}), from lowest to highest. When the amount of data is not a multiple of four, however, then (somewhat) arbitrary choices have to be made. But what are the definitions you have read? Can you cite some of them?
To make progress, you need to come up with a statistical model of land cover change. This is important because the very strong spatial correlations will cause crucial assumptions of the standard tests (such as the Student t test) to fail. The model, whatever it may be, will suggest an appropriate statistic to measure the change. In the present case, difference of total area is not a good choice, because it effectively allows loss of land cover in one location to cancel gain of land cover in another location. Making a formal statement of your study question would be a good start.
@B_Miner You are right I have edited my answer.
Awesome, thanks. Using my example above, the and qbeta being my quantile function: lower bound = qbeta(0.025, 1001, 5001-1000) will give me the lower bound for a 95% confidence interval.Thanks for your help and thank you for providing references for me to review. I appreciate your help. This is what I was looking for.
community wiki please. Your question does not have an 'objective best' answer.
Why does $$P(C|A\\ \\text{and}\\ B)=P(A\\ \\text{and}\\ B\\ \\text{and}\\ C)\\times P(A\\ \\text{and}\\ B)??$$ I thought that $$P(C|A\\cap B)=\\frac{P(C \\cap (A \\cap B))}{P(A \\cap B)}=\\frac{P(A\\cap B\\cap C)}{P(A\\cap B)}$$ using the definition of conditional probability.
I do not believe that a closed-form expression exists for arbitrary distributions, though for particular cases, the computation would be relatively easy and one could write down a formula for future reference, e.g. $a+bX$ is uniformly distributed on $[\\alpha, \\beta]$ where $0 < \\alpha < \\beta$.
The aim of visualization is not to show the data, but to show a claim that this data supports -- so first try to focus on what you want to show, to make this question answerable. Or, if you just want a pool of visualization ideas, make it a community wiki.
Because you explicitly stipulated $k\\lt n$.
(+1) For the function rawToChar which I did not know.
Thanks! Could that also be worded the other way around? i.e. The effect of increasing the gene by 1 on cad is modulated by glucose levels? Thereby saying that glucose modulates the effect of the gene?
I see what you mean about the J components. I'm not using them so yes it's a constant model. Thanks for clarifying. As for your second comment, I have plotted the smoothed and filtered level and seasonal components and they look as expected. It's just that when I use forecasting, the level component stays constant throughout the forecast (as expected) but the periodic component also stays constant.
Great answer. I would only add to make sure that the syntax used to clean the variables is retained in documentation, with comments if not descriptive passages about why things were changed. :)
@andreas-h I wouldn't do that; the STL trend is over fitted. Fit the GAM with the AR() structure and interpret the trend. That will give a proper regression model which will be far more useful to you.
@Sneha Because it can occur that two of the Half-Gaussians are the two largest values.
Thanks Scott. Yes, I noticed it was 0.07 and started wondering if it is significant and I realized its about the scale and units or the impressions. For the sake of visual representation, can you tell me how I can scale the coef plot, so my intercept can be shown on y-axis?
It's always difficult when you're not even sure what you want to know about or how to ask, but can you say a little bit more about what you want? E.g., what would it mean to determine "the number of features one should select for the classification process"?
@Andrey The formulas in (2) and the second constraint in (4) make no sense: what do the hats mean in (4)?  What is $S$?  (You have only defined $S_j$, not $S$.)  Is the expression in (4) a sum of three *products* or something else?
There have been great many of similar questions on this site about clustering of mixed-type data. Just one recent: http://stats.stackexchange.com/q/99983/3277.
I don't know what that is or why it is approapriate.
It means the assumptions of Pearson correlation. There are five assumptions that are made with respect to Pearson's correlation:

1. The variables must be either interval or ratio measurements 
2. The variables must be approximately normally distributed.    
3. There is a linear relationship between the two variables.
4. Outliers are either kept to a minimum or are removed entirely.     
5. There is homoscedasticity of the data. 

If any of these assumption violate then I would recommend Kendall correlation.
@Khader, the pairs are defined by the matching. For each case there should be one control with same sex and age. The table format for this setup could be (for hyptertension, htn): row1, col1: number of pairs having case and control no htn; row1, col2: number of pairs having case no htn, control with htn; row2, col1: number of pairs having case with htn control no htn and finally row2, col2; number of pairs having case and control with htn.
Does this page : http://scikit-learn.org/stable/modules/hmm.html suit what you need ?
Let me reiterate my suggestion to [register](http://stats.stackexchange.com/faq#login) your account, michelle. I would also suggest to provide as much background as possible when asking such question: What is the context of your homework, what did you learn about multiple regression in your previous courses, what is the purpose of identifying significant predictors (BTW, is this for explanatory or predictive purpose)? Thank you.
Sorry, I should have been more clear.  I meant to say that I ran a two-sample Kolmogorov-Smirnov test to see if the two samples could be considered drawn from the same distribution.  Wouldn't that mean that the distribution and its parameters would be the same?  I may not be entirely understanding the test.
My first specific question to you would be, how did you define best? This is important. If you are thinking that you will only use a subset of possible trees, determining what best is, and what bad is, can make a big difference. Are you thinking that best will be something like mean squared prediction error? Or, do you care more about speed (generate a prediction but just do it quickly ... accepting that it may not be optimal). I ask because this can impact the direction one takes to solve the problem.
What you want sounds more like analysis of variance. Running a regression on case and batch identifiers needs to be recast using indicator variables. That is still regression, but doing it as analysis of variance is likely to be a little easier for you.
Great If you like my answers to your questions you can check them and upvote.
(+1) Here is another paper, using simulation and real datasets, that suggests that the N/p rule-of-thumb does not perform very well in practice, and that provides sample sizes required to obtain stable and accurate solution in EFA--controlling for various quality criteria--as a function of the number of factors and the number of items (and optionally the half-width of Cronbach's alpha 95% CI, based on Feldt's formula) in a psychiatric scale: [Sample size requirements for the internal validation of psychiatric scales](http://1.usa.gov/14w0ezL) Int J Methods Psychiatr Res. 2011 Dec;20(4):235-49.
I think this will do it. I really don't know f, g or h, but if I use Spearman's rcc then I can proceed. Thanks for your answer!
Thanks for the confirmation, Stefan. I was hoping there was some magic wand method I didn't know about :) A flat curve below 16 is definitely wrong, but at least I know in which direction it is wrong (overestimating growth)
@Biorelated As can be seen in my response, you'll need to compute SD or SE or 95% CI yourself, and add them to your data frame (or use another one), then call `geom_errorbar()`. If your question is about how to compute those estimates, then you may consider updating your question and provide more information on your dataset.
As for the community, no, there really isn't one. The statistics site here is probably the closest, with occasional forays into StackOverflow etc. when you need it. By coincidence however, I am trying to start one...http://area51.stackexchange.com/proposals/34565/public-health-epidemiology
Nice plots often beat blah blah blah :)
Answers without a single word of explanation, a bare code doing what only God knows, are hardly acceptable.