tttjjj/elt_bench huggingface dataset:
@DT: Again, someone might say that those cases don't really matter, since your system will spend a lot more time in production than in development. But if you're the developer, it makes at least a psychological difference.
Could you please explain in what sense one might "use" a *distribution* as an *estimator*? In the conventional technical sense of these words, such a phrase would not be meaningful.
Step-wise model selection methods are not recommended for either continuous or categorical variables. To understand why, you may find it helpful to read my answer here: [algorithms-for-automatic-model-selection](http://stats.stackexchange.com/questions/20836//20856#20856).
Some tricks, Michael: (1) http://www.math.harvard.edu/texman/ lets you quickly look up some common expressions. (2) Start $\\TeX$ expressions with both the enclosing "$\\$$" characters, so that as you type you can see the expression previewed below. (3) For using $\\TeX$ in comments, open a separate window and start to *ask a new question* on this site. It will give you a preview. Cut and paste it into your comment, then abandon the new question. (4) You can right-click on any $\\TeX$ on a page to see the original markup: learn from it (and use copy-and-paste judiciously).
Maybe the level that was removed had a mean very similar to the others so, by removing it, there was more heterogeneity in the means?
Although not an answer, it seems like all your examples are all ordinal.  Pass and fail are not just names, but measures of ability on a scale (ordinal). Grade as A/B/C... is ordinal.  Your example of interval (percent grades) is ordinal as well, as the difference in ability between a 90-80 range is probably not the same difference in ability between 40-30. It also has no meaningful zero point so it is not ratio. Percent grades or scores are really not ratio or interval, but ordinal.  Although we treat them routinely as interval, we should feel awkward about it.
Unless you can do 1.5 or 3.78 activities its probably best to make it a factor variable since the activity variable really isn't continuous.
I am not a time series expert by any means, but it sure looks wrong to me.
Is this question *only* about how to get something done in R / WinBUGS? If so, it would be off-topic for CV (see our [FAQ](http://stats.stackexchange.com/faq)), but on topic for [Stack Overflow](http://stackoverflow.com/). If you have a substantive statistical question, please edit to clarify it, if not, you can flag your Q for migration (*please don't cross-post, though*).
Hmmm I've tried tuning the weights manually though what the SVM predicts is or all 1s or all 0s. No in-between. (Btw I've added confusion matrices to the question for clarification, as you can see by using SMOTE, the unbalance is gone when using a test set which is also composed of synthetic data but there is still in unbalance when a test is run on the validationset.)
It really depends on what you mean by 'trust'. If the assumptions hold, the chance of a type I error is just your significance level - it's not a function of sample size (*power* sure is - but that's not an issue for you). Your power analysis shows that you don't have a high power ... but nevertheless you did reject. Small sample size doesn't increase the chance that you'll wrongly reject - on average it decreases it.
What kinds of analyses will you be performing with these data?
Can you at least say something about what kind of statistical problem you are dealing with, which kind of model you are using and which numerical method you are planning to use? Bayesian inference is a very broad topic - at the moment you are basically asking for a checklist for all statistical problems that we can think of!
Your example L-matrix (L1,L2,L3) is the _repeated contrasts_ whereby each category is compared to the following category. Neither these contrast predictors are orthogonal nor they are binary (coded as 0-1). In fact, their values are .75 vs -.25 (1st variable), .5 vs -.5 (2nd variable), .25 vs -.75 (3rd variable)
(1) What you have drawn isn't really a histogram (in histograms, contigous bars touch); it'a a barplot, which is more appropriate for non-contiguous categorical variables. (2) You could compare ecdfs. If you can unbin, you might compare boxplots, or kernel density estimates (or, combining the two, violin plots). The suggestion of ttnphns to plot them back-to-back makes sense.
Without a lot more information, this is unanswerable. It's unclear what you did and what you wanted to find out. My blog post [how to ask a statistics question](http://www.statisticalanalysisconsulting.com/how-to-ask-a-statistics-question/) may help.
Alternatively, if you have an exponential function;y=exp(xb)+e, were x is matrix of explanatory variables, how would one estimate this when one of the x's is endogenous and one has two good instruments for it? Using the MLE framework.
A link to a CV answer or wiki that nicely enumerates these important points would give a nice polish to your answer (if not actually enumerating them here). :)
@Noah As Seth asked below, why are you drawing random betas as well?
Don't know much about Naive Bayes. However, looking at the graph, Y1 and Y2 are only conditionally independent given X. Think of them as points uniformly distributed on a frisbee, and X as the co-ordinates of where the frisbee lands
One idea to explore might be the inclusion of Latitude in your model.  Another might be to look at Longitude, though if the mountains are in west then this could be highly correlated with altitude.
In general the definition is $Var(Y)=E[(Y-E[Y])^2]$ if $E[Y]$ exists and is finite.  That can then also be written as  $E[Y^2] - (E[Y])^2$ by expanding the square and simplifying.  You just have a special case of this.
@sesqu Hmm...  You ought to rethink that last statement, because it seems you have rejected out of hand some significant advances in statistical concepts and methodology.  The first statement seems to misunderstand what the ecdf is; it perhaps is related to an unusual conception of what a distribution function is.  Are you using a standard definition such as that given at http://mathworld.wolfram.com/DistributionFunction.html ?
For R it's in the `car` package as `leveneTest`
Almost the entire reply was taken wholesale from two sources without any attribution.  (I have edited it accordingly.)  The lack of attribution is plagiarism, no matter in what medium it occurs.  Do not do that again on this site.
A quick idea coming is to set all within-class distances ("class" is points with the same label) in the distance matrix to 0. So that they won't be ever separated. But this idea may be incompatible for some hierarchical methods. One should think out it.
@Michael: I interpreted it that way when I read the original question as well, but decided to write my answer after reading the comment. Otherwise, I don't think that there'd be much to add to your (+1) answer (other than the small remarks I made in a comment there).
Thank you for an extensive and very informative explanation!

To answered your question, my vss, with varimax rotation, suggests 1 factor at complexity 1, 8  factors at complexity 1, and The Velicer MAP criterion achieves a minimum of 1 factor. With the oblimin rotation the vss suggest 1 factor at complexity 1, 3 factors at complexity 1, and The Velicer MAP criterion achieves a minimum of 1 factor. Finall, Ruscio and Roche's CD suggest 1 factor. It might be that my data is not that good.
@deps_stats, yes, thanks; changed the title and first line. @Dmitrij, 1) yes little-l is strictly speaking correct, but big-L is common and understandable. 2) yes one can find an optimal p for a given problem, but what's your first choice, and why ?
+1 Did you mean to write "correlation" instead of "covariance" in the last sentence?
Determining what is an [outlier](http://en.wikipedia.org/wiki/Outlier) depends greatly on the source of error. This question lacks information sufficient to even guess which method you should use.
you can get predictions using predict.
dat[,"y0"]<-predict(res,newdata=data.frame(x1=dat[,"x1"],x2=0))
Saves a bit of typing.
@DikranMarsupial +1 I really like your answer and also your followup comment.  I think the war between the Bayesian's and frequentists is over and nobody won.  A lot of statisticians including Brad Efron are adopting the position you describe in last sentence of the statement above.
FWIW, any Box-Cox transformation of power greater than 1 will reduce negative skew.  In light of Rob Hyndman's response that's not the first thing to try, though.
Frankly, it will be hard to do actual research without a supervisor. At least try to pin down a subject or specific learning task first. If you don't know what interests you, both current answers give useful tips. Be aware of the fact that, without either a supervisor or deep domain knowledge, you're prone to reinventing the wheel because you don't know the field slang/state-of-the-art unless you read textbooks about literally *everything*. You can find your way by yourself once you are familiar with the domain, but starting will be difficult. Kudos for being willing to try, though!
Instead of "orthogonality", the concept you are describing is usually referred to as *linear (in)dependence*.
@Glen_b In that case, http://stats.stackexchange.com/questions/14483/intuitive-explanation-for-density-of-transformed-variable is extremely closely related.
*Nelsen's* book? Did you try [contacting the author](http://legacy.lclark.edu/~mathsci/nelsen.html) directly?
I could do an Confidence interval but i'm somehow missing the standard deviation which would be needed to derive theorethically? I assume it to be close to zero for 500 values for the skewness, respectively to 3 for the kurtosis..?
I evaluate the performance by checking training and test error. I just wanted to try out the concept with neural networks. Actually I wanted to reproduce the time series prediction result from the book (Machine Learning: An Algorithmic Perspective) and couldn't achieve comparable results so I wanted to find the main cause of bad performance.
You probably misunderstood the documentation. When you extracted factors by PC method (that is, factors are pr. components) then regression method found under Scores button will save exact component scores for you - same scores you might obtain if calculate yourself by matrix algebra.
+1, great answer. At first, I didn't see how you have arrived at the formula for the approximation of the $F$ distribution using the Wilson-Hilferty transformations for $\\chi^{2}$ random variables. I have found the explanations at the beginning of [this paper](http://comjnl.oxfordjournals.org/content/11/2/209.full.pdf) quite useful. I wonder if you have used the same reasoning (using Fiellers' result) or a completely different derivation?
@amoeba I think it solved my problem although i'm not sure how to correctly interpret the meaning of the plots, need to do more reading I guess. some previous questions helped a bit like this one: http://stats.stackexchange.com/questions/62034/are-there-examples-of-more-informative-pca-plots
@SteveS, thanks! I have started working on collecting more data. Since this is not an easy task, I wanted to check if there was anything else that I could do on my current data set (that I have not already tried) before proceeding with data collection.
OK, thanks @PeterFlom.  I can see how you use a mixed effects model in longitudinal panel data to create a group effect for each case with multiple observations, and maybe even each time point with multiple cases but it seems to me that if the issue is that volatility in $t-1$ impacts on volatility in $t$ nesting or grouping won't help and you do need time series methods (perhaps in conjunction with a mixed effects model).
"...reaching the memory limit on my computer to use" - I presume that you are thinning as you go, rather than post processing your samples?
Hmm, picture is missing... If you can't attach it, put it on some picture hosting site and paste the URL, I'll fix it then.
yeah but that is only for split nodes.... I want to compute it for all features at the root level :/
Mostly one of my 20-year-old college textbooks.  I don't have it in front of me at the moment, but with some help from Amazon I think it's either Karlin & Taylor "Second Course in Stochastic Processes" or Hoel/Port/Stone's "Intro to Stochastic Processes".  I've been trying to supplement that with online searches for more papers or class notes on the topic.
@EricPeterson Do you mean, use LOESS *instead* of piecewise regression? Or is there a option of R LOESS that does piecewise regression?
@NickCox: we were just too busy writing answers :-)
Although it is not correct that standardization changes collinearity in a purely mathematical sense, it can improve the numerical stability of algorithms to solve linear systems.  That might be the source of the confusion in this reply.
I think what you are after is the idea of an [uninformative prior](http://en.wikipedia.org/wiki/Prior_probability#Uninformative_priors).
What plotting techniques do you know? Let's be specific: part (a) asks you to graph the function $1 e^{-1 x} = e^{-x}$ for $x \\ge 0$. Where exactly are you having trouble doing that?
I used SAS enterprise miner. The error message is "no support vector found".
If he will continue to sample again infinitely, & the null hypothesis is true, his type I error rate will be $1$. If you wanted to know how many samples would be required, on average, to get a type I error, that can be analyzed w/ a parametric survival analysis using an exponential hazard.
@whuber ThePawn below suggests that it is C. Do you have more detailed concerns?
Can you define exactly what you mean by these two approaches?
(a) [All models are wrong but some are useful](http://stats.stackexchange.com/questions/57407/what-is-the-meaning-of-all-models-are-wrong-but-some-are-useful), and (b) the requirements of a model depend on its intended use;  both (a) and (b) contribute to the murkiness of my question.  I guess I'm looking for insights that help to see through that fog.
A one-class SVM with Gaussian kernel (for example) generally works well even if the data is not normally distributed. Typically, these are not the issues that cause a lot of headaches.
Can you tell us what you have so far? Where are you stuck?
The original inputs take values >1, by transformation, I put them in an interval of 0 and 1 , and for the output activation function I use the Sigmoid ==> results between 0 and 1
I'd like to know if I randomly sampled a car from say, 2003 would it be cheaper than if I got a car from 2005.
Can you clarify what you mean by "hierarchical multiple regression"? It might also help if you can provide some context (eg, your field); it is not my impression that path analysis is 1 of the "most common analytic methods".
Not quite sure what you're asking - are you asking why your 24-sized sample statistics don't match the theoretical values? One check would be to increase the number of samples and see that improves the sample statistics. If with $n=1000$ you still get a dramatic difference between realisations and theoretical, then there's probably something wrong with your code - I suggest posting the code :)
Just to underline that your picture of seasonality is based on one annual cycle, pretty much, so don't expect too much.
Note that the Cauchy takes negative values, so in effect almost all fractional moments are ruled out immediately - as they are for the normal, say (assuming you expect real-valued answers).
I tried plot(res.pca, choix="ind") but it just plotted x and y axis nothing else. What should I do now. Waiting for your positive reply
@Glen_b, your answer is amazingly helpful. Error term added. $t$ is known for each event. I have one to four observations per subject, so 0-4 events per subject. Obviously, I haven't presented the above equation as accounting for the nesting, but I wanted to get a non-nested idea down first.
Thanks Gung for your reply. My problem in understanding this issue is how come "correlation" between 2 independent variable lead to change in coefficients. I guess Yang to great extend help me understand this issue. But I'm not confident yet .
[tag:[tag:_tag_content_here_]] You can do it in comments too, but (as you see $\\leftarrow$ here) this just produces a hyperlink, so it's not as cool. I find the in-answer tags helpful for providing easy access to the wiki excerpts, but I'm a little frustrated by the disruption of line spacing...
Ty, fixed error in formula. My answer is for the second part of question regarding the proportion of variance explained by the regression line.
@calbear You start by doing five separate hypothesis tests. Each one is much more variable than a test based on the mean of the five  samples (by the factor of √5 that I mentioned).  So if the mean is actually <0 the p-values for the individual test should all be similar and a lot higher than for the mean of the five.  Your so-called decision rule allows five distinct cutoff points.  But if the samples are iid there would be no reason not to use the same cutoff point rahter than using the observed value as the cutoff.
This looks good too. Will try it and see how it compares to the algo above.
"New cases" is what correspondence analysis calls "supplementary" or "passive" categories. Most implementations of CA includes the option of such categories.
How does this change now that `Stan` is out?
Aha! This totally cleared up the "larger than" bit. Thanks!
Do you also count zeros, i.e., e-commerce for which no reply is posted? If so you could do some kind of poisson regression where the depended variable count the number of replys on e-commerce. Here the, though this can be relaxed, the variance changes if the mean will change.
I have mentioned what the abbreviations stand for with links to papers. Thank you
@NickStauner, check out some visually weighted regression plots in R ([1](http://is-r.tumblr.com/post/32256789836/visually-weighted-regression-plots-with-zelig),[2](http://www.nicebread.de/visually-weighted-watercolor-plots-new-variants-please-vote/)). Originally via [Solomon Hsiang](http://www.fight-entropy.com/2012/08/watercolor-regression.html). For a start on the lit. of viz. uncertainty I would recommend the paper [Spiegelhalter, David, Mike Pearson, and Ian Short. 2011. Visualizing uncertainty about the future. Science 333: 1393–1400.](http://goo.gl/RAv8wY)
I believe that the MARS/Earth algorithm used in the earth package does not let the user set the number of splines because the whole point of the [algorithm](http://en.wikipedia.org/wiki/Multivariate_adaptive_regression_splines) is to choose the number of splines endogenously. You do have control over [`nk` and `nprune`](http://cran.r-project.org/web/packages/earth/vignettes/earth-notes.pdf) though.
Part of the problem may be the notation, which I haven't been able to completely sort out yet.  I figured out you are using $\\phi$ in two different senses, but I haven't yet grasped what $X_1$ is.  But I do get the main point now, which is a good one but becomes somewhat obscured by the context: when continuous distributions can be reparameterized arbitrarily, their modes are not well defined.  (Indeed, by a suitable choice of parameterization you can place the mode *anywhere* within their support.)  Providing that cautionary note is worth an upvote IMO (+1).
Classification or regression over the B elements inside an A event, depending of the more efficient — both may be relevant.  I would like to consider each B element independently from other A events, so even though it may not be the case in reality, in practical the number of Bs is larger than the number of As and elements from the test set will differ from the training set.  And there are only a fairly small number of B elements per A event.
>A probability is not used to estimate models. 

This is obviously true. One of two things is happening here, (1) I'm am not being clear enough in my responses or (2) you are entertaining yourself with your pedantism. Either way, I shall be leaving our conversation. All the best.
@amoeba, off the top of my head, I'm not sure what corresponds to what I would call "repeated measures" in the engineering perspective on DoE. I suppose you could say "Montgomery's 'repeated measurements' can be interpreted as repeated measures with respect to furnaces (fair enough), but furnaces are not the object of study in this quote; wafers are", but M's point is that the repeated measurements are information about "differences among the wafers and other sources of variability within that particular furnace run". Identifying sources of variability is the point of DoE in engineering.
What do you mean with "statistical a-priori method"? Because there are certainly ways to identify causality from purely observational data even under the absence of time. See the book linked by Peter Flom.
Ignoring the linearity assumption is an excellent way to create biased estimates.
Although I cannot reproduce the "correct" answer, I can get close ($2.008$) by using a Student t distribution with $60-1$ degrees of freedom instead of a Normal distribution. Is the Student t also covered in this study material?
I believe you can seriously consider http://www.amazon.com/Introduction-Mathematical-Analysis-Economic-Econometrics/dp/0691118671
$-K_1^2$, being the negative of a square, is always negative when applied to nonzero arguments and by assumption $K_2$ is always positive when applied to nonzero arguments, QED.
Regarding 1) there are a bunch of different names for different formulations of purity. These get a bit complex to write, but are covered in any good book on classification/regression trees or recursive partitioning. Regarding 2) I'm talking about self-similar and other dis-similar
-0.5 is not much skewness; I don't think this should be a concern unless there are other problems (e.g outliers).
Hint: There are _four_ outcomes of the experiment. _Two_ of the 4 outcomes result in $X+Y=1$.  Can you identify what the values of $(X,Y)$ are? Let these be $(x_0, y_0)$ and $(x_1,y_1)$.  Then, conditioned on $X+Y=1$, $X^Y$ takes on values $x_0^{y_0}$ and $x_1^{y_1}$.  Can you figure out these values? Can you figure out the probabilities? Can you put it all together?
@Tim, the $F$-tests used in ANOVA are equivalent to likelihood ratio tests based on the normal error distribution.
+1 Cool!  (BTW, the links to the MaplePrimes site are returning a 404.)
Thanks for the update. All I can say at this point is that if a procedure to produce confidence ellipses contains the true point only 60% of the time, then it is an (approximate) 60% confidence region, *not* a 95% region. It might be fine in application, but *calling* it a 95% region is a mistake no matter what. I do not see the sense in which these triangulation estimates only assess precision instead of accuracy: the purpose is to *find the location* and any variation in its estimate ineluctably is a combination of inaccuracy and imprecision, not just one or the other.
Both answers are great and help me with other questions in which we do have matched samples.
How is $z$ a random variable?
Another question: do R and Minitab give similar outputs after a Shapiro test?
Only 5th set does not have a tie-breaker in Wimbledon, Australian Open and French Open. The first 4 sets are played with tie-breakers.
Thanks for the update - much better question.  Now though, I think the issue is that the correlation coefficient is not the best way to describe the relationship between these two variables.  A better way would be a regression of some sort - then you could report the slope coefficient and have a confidence interval for that instead.  You will also have a problem that income and savings variables are usually quite skewed distributions. The confidence interval for your correlation coefficient probably assumes they (or one of them) is normally distributed and hence will be incorrectly calculated.
Have you seen [Diehard](http://www.stat.fsu.edu/pub/diehard/)?