tttjjj/elt_bench huggingface dataset:
I think it would be excellent if you could give more detail (see suncoolsu's answer). Furthermore, with design, do you mean how to analyze the data?
The nominal coverage probability is the "target" coverage probability: the one we try to attain when we derive a method providing a confidence interval. The actual coverage is the "true" coverage. Some people say that the confidence interval is *exact* when the actual coverage equals the nominal coverage. Scotchi and Unwisdom have mentionned that the confidence interval is never exact for discrete data. Another example is when we use an asymptotic confidence interval: it is exact only when $n \\to \\infty$. I totally understand your idea because "actual" is also a synonym of "present".
thanks for the clarification. Since this is all in a small file, it is one line anyway: `source('foo.r')` ;)
The OP requests a theoretical proof, not a thought experiment.
@Yuan I don't know, because simple "MLE" can mean different things... Try reading ?naivebayes, it is quite detail description what is what in table, maybe you could find something familiar there. Or try to specify what you need in more detailed way -- what is your desired use, for what you need it?
The difference between the parametric and nonparametric bootstrap is that the former generates its samples from the (assumed) distribution of the data, using the estimated parameter values, whereas the latter generates its samples by sampling with replacement from the observed data - no parametric model assumed.
I edited my description by adding more information.
Wonderful!! Thank you (both) for your time.
The $\\chi^2$-test for a 2x2-table is more or less based on the same normality assumptions than the z-test in your reference, so you can use either test (note it's not the data that has to be close to normal, it's the test statistic). The z-test has the advantage to provide straightforward approximate confidence interval for the true shift, so you might go with this version to make life easier.
+1 However, MSV, while for very large data sets this is probably the approach I'd suggest (taking into account @shujaa's clarification), one must understand that this approach nevertheless contains assumptions (such as independence of (x,y) pairs from other pairs, that the future will be like a random sample of the past). By the way, your histograms in your question should really have more bins.
1) Any data preparation (centering, various scalings) is done prior PCA. 2) The results of PCA strongly depend on those data transformations, as if it were results for another dataset. Transformations take you away from original dataset. 3) You may do any such transformation - whatever makes sense to you in your wish to obtain meaningful results. 4) Centering is usually done before scaling, not after it - this way is more meaningful.
The latest version of the package on github (https://github.com/robjhyndman/forecast) will now find the `xreg` info in the first example.
I thought about this more and corrected my question to focus on users not customers. For a freemium service the broader objective is to survey users most of which are unbilled. For paying customers, your answer is helpful. Thank you.
Since you are fitting the distributions with MLE, have you thought of using the maximum likelihood theory to assess the goodness of fit?
In order to get the most useful answers you may need to break down your question into more digestible parts.  I, for one, could use clarification on:  why do you need multiple groups?  what does "n" mean in this case, and how would it be determined?   what does "optimise the interval for controls" mean?  (the interval of what?)
I think Henry's answer below is clear but you can also check out this detailed [answer](http://stats.stackexchange.com/questions/47771/what-is-the-intuition-behind-beta-distribution/47782#47782) on how to apply conjugate priors.
Hi, thank you for the interest, here are the references: 
Dormann et al, 2007 "Methods to account for spatial autocorrelation in the analysis
of species distributional data: a review", Ecography 30: 609628 
Paper : http://www.yale.edu/jetz/pdf/Dormann%20et%20al%202007%20Ecography.pdf
Appendix: http://www.oikos.ekol.lu.se/appendixdown/E5171.pdf
Borcard et al., 2011, Numerical Ecology with R, p.275-277 (book)
Related: http://stats.stackexchange.com/questions/28178/what-is-empirical-entropy
+1. Large numbers of digits seem to generate irreverent responses: see the last few lines of [my reply to a similar question](http://gis.stackexchange.com/questions/8650/how-to-measure-the-accuracy-of-latitude-and-longitude/8674#8674) on another SE site.
This is a really open-ended question! If you're trying to spec out a project, it might work better to come up with a problem first and then go from there. I've added a few more examples, but if you're asking "what should I do with these scanned documents?", then only you (or your boss) can really give you a solid answer.
Both are continuous, they don't have spikes in either position. You could try to model it as a 0-inflated, 1-inflated beta (total of 4 parameters), which might fit pretty well as long as the distribution is continuous between the extremes.
@probabilityislogic Say I found component 3 unconvincing because the authors failed to report crucial information in the paper. I multiplied by ten just to make it look more like common likelihood functions I have seen.
Ah ok. So I have to use the NbClust with a distance matrix. I have to use another method.
You could check what predictions it gives to your training data. If loss goes down to zero, all the instances should be classified perfectly
I think power-law functions may be of interest. But also realize that if you define an objective function to optimize (some measure of error), then to get estimates you can use a numerical optimizer.
Very interesting suggestion!

Some remarks:
 1. if you assume C(i|t) instead of C(i|t-i), you still have a convolution, and there is no really good _a priori_ reason to suppose that this model will be worse.
 2. when doing *deconvolution* (that's the official name), one is likely to run into problems and has to do _regularization_. Spline regularization is an option (pretty advanced); normally people just add a small constant to F(D) for starters.
Thanks for your detailed and patient answer. I just find I actually haven't figured out such a question: if we solve $min \\quad \\phi(x)$ using coordinate descent with tiny step size, do we get the $\\ell$-1 regularization path for $min \\quad \\left(\\phi(x) + \\lambda||x||_1 \\right)$, where $\\phi(x)$ is smoothly convex function?
Agree with last para. The Gaussian kernel uses all the data points but other commonly-used kernels (Epanechnikov, biweight, cosine, Parzen, triangular, ...) are finite-width, i.e. they are defined to be zero outside a finite interval. R's density() function defaults to Gaussian but Stata's -kdensity- command defaults to Epanechnikov.
@probabilitylogic: Thank you for your answer. A previous discussion (http://stats.stackexchange.com/questions/577/is-there-any-reason-to-prefer-the-aic-or-bic-over-the-other) stated that "AIC is best for prediction as it is asymptotically equivalent to cross-validation." and that "AIC is equivalent to leave-one-out cross-validation." If so, can I use AIC alone (instead of  cross-validation) to fight against over-fitting in prediction?
You will probably like [Many Eyes](http://www-958.ibm.com/software/data/cognos/manyeyes/), [dataviz](http://www.improving-visualisation.org/visuals/), [datavisualization.ch](http://datavisualization.ch/), and [Ideas2evidence](http://www.ideas2evidence.com/showcase.html), to name a few.
+1 I would always opt for option B even for large datasets.  How can we be sure that the difference in measured performance is due to the difference in generalisation performance of the classifiers, rather than being due to the difference in cross-validation spits?  Even if the difference is likely to be negligible it is easier to avoid the question altogether by using the same splits.
Personally, I would just look at the distribution and use the old ocular test. If the data looks skewed, use the Johnson. When the data are skewed, then *one-sided* t tests can be badly off. I think that's less of an issue with *two-sided* tests, like you're doing here.
The 64 choose 8 calculation divides by $8!$ because order does not matter.  But the $8! \\times 8!$ assumes that order does matter.  You can either divide the restricted case by $8!$ to say order does not matter, or take the $8!$ out of the denominator of the choose to say that order does matter.  Either way should give the correct result.
+1 Nice applications.  They remind me strongly of [influence diagrams](http://sspp.proquest.com/archives/vol3iss1/0610-036.grosskurth-print.html) (and maybe they are meant to).
@whuber Thank you for the definition!
Your question is very hard to follow. Please edit your question to clearly express what you want to ask. In particular, (1) use capitalization where it belongs; (2) use complete sentences to express clear ideas, and clearly identified questions, not some stream of consciousness; (3) Lay out what you're saying more clearly, using blank lines and punctuation and so on; (4) take particular care with the first two lines of your second paragraph.
My point is that unless you *do* make it a matrix equation, your characterization of ANOVA is too limited to be useful: it is *identical* to your characterization of the t-test and so is more confusing than it is helpful. When you start introducing more groups, you suddenly change the equation, which may also be less than clear. Whether you want to use matrix notation is of course up to you, but in the interest of communicating well you should strive for consistency.
It is straightforward to *prove* that scaling does not alter the correct estimates. Noting that the two estimates for `y2` are not significantly different, one has to suspect that the nonlinear fitting procedure is finding its solutions with relatively poor precision. Providing better starting values is one way to improve such behavior. Consider estimating $a$ and $b$ with a preliminary *linear* model $\\log(y) = \\log(a) + b\\log(x)$ to obtain good starting values. This can also give a large set of diagnostics to assess other potential problems, such as a large VIF in the design matrix.
You are correct that I am pointing at a modeling approach requiring more technical effort. But, well... not to be too much of a brat, but one *might* respond by asking if unjustified assumptions of linear relationships and unjustified hemorrhage of statistical power are necessary. But I will instead say you're not wrong to raise this critique: there's a trade off, and "how little can we get away with knowing" is not a bad value (among others) to guide one's analytic approach.
Have you noted the nice symmetry of this distribution?  Because the height at $20-y$ is the same as the height at $20 + 2y$ for $0 \\le y \\le 20,$ then for small $dy$ the number of families with incomes between $20-y$ and $20-(y+dy)$ for $0\\lt y\\le 20$ is one-half the number of families with incomes between $20 + 2y$ and $20 + 2y + 2 dy.$  Thus the contribution from each of the latter families, which is proportional to $2 y t,$ must exactly balance the transfer to the former families, who receive an amount proportional to $y/2.$  That makes it obvious what $t$ should be.
Thanks for the response. Unfortunately, the data I have does not lend itself to imputing values.
Your proposal of age-indicators + shrinkage is essentially the same thing as a smoothing spline of order 0.
I have to say that without your blue/green colouring I don't think I'd have seen two clusters there. In fact I'd rather have described the data as a central clould with outliers (in the sense of "points further away") in two directions.
Is it possible to run this program with more than one categorical predictor?  On the upload page it looks like only the first column can be the "class" of data...  Perhaps I do not understand how the data is supposed to be formatted.
The quantity $(N(0,\\frac{\\sigma^2}{n}))^2$ is a scaled Chi-square. As a check, the variance of a Chi-square with $k$ dof is $2k$, which gives the same result you get.
Right; without additional scaling the beta-binomial won't converge to a proper probability distribution as $n\\rightarrow\\infty$ if $\\alpha, \\beta$ are fixed (I think). From his phrasing ("count distribution") I think the OP may be looking for the analogue of a binomial approaching a Poisson as $n\\rightarrow \\infty$ if $p\\rightarrow 0$ and $np$ is constant. Probably there is a similar result if you send $\\alpha/(\\alpha+\\beta)$ to zero at a proper rate or something, but I don't know it.
@AlexxHardt This is not quite correct (as I have mentioned in my first comment). Please, have a look at this document about [AIC Myths](http://warnercnr.colostate.edu/~anderson/PDF_files/AIC%20Myths%20and%20Misunderstandings.pdf), particularly the last paragraph of pp. 2. Other references by Brian Ripley: [1](http://www.stats.ox.ac.uk/~ripley/ModelChoice/Ox08_2.pdf), [2](http://www.stats.ox.ac.uk/~ripley/Nelder80.pdf), [3](http://www.stats.ox.ac.uk/~ripley/ModelChoice.pdf).
I see. However, if the examples aren't really typical, or contain details that don't really apply, it becomes even harder to see what the question is. The bottom line is that you seek a precise answer to a very fuzzy question. Do distributions on different domains differ? has to be firmed up in some way.
@Gilles: what could possibly be the reason to change `PCA` into `$\\text{PCA}$`?
@JuhoKokkala I think this is simply the pdf for x being nth order statistics times integrating over all possible values of x (the pdf).
(The assumption is that few doctors may not be able to follow all cases in ICU and may lead to some deaths). Anyway, that was just an example to simplify the question.  The problem is that we do not have the values of A and B obtained in the same order.
Do you mean you have several random effects and that you assume all of them to have the same variance?
You don't correct the mean!
@Jeromy, in response to your earlier question, for some of the nonparemtric tests the dependent variable is discrete ie. scores out of 20 on a test and for some it is continuous
What you consider positive is entirely semantic. It makes no difference to the underlying algorithms.
No help?? It's quite important to me to find a solution!
(+1) nice example. Mapping in excel is difficult, Jorge Camoes has some [nice examples](http://www.excelcharts.com/blog/making-maps-in-excel-cross-stitching-edition/), but visualizing detailed choropleth maps is quite impractical (it takes manually digitizing the shapes in excel). Point patterns aren't as hard though. A related question came up recently on the GIS site and suggested an [ESRI ArcMap](http://gis.stackexchange.com/a/27657/751) add-on for excel.
@Xi'an Thanks for pointing me the paper and book chapter! (though Amazon doesn't want me to read Section 10.4.3 as preview...)
The difference between a proportion and a count is the total that you scaled by; surely that can't be a problem; just move between unscaled and scaled counts as needed. If your Bernoulli's are zero-inflated, then as suggested there may be no issue. However, there are zero-inflated binomials in R (`VGAM::zibinomial`).
"*Q= i.A i.S i.Ec i.Pr i.xx i.dd#i.ii*" -- what is this? Did a cat walk on the keyboard? Is it a sequence of keystrokes for something? I don't understand.
Could it be that you were in a region with a local maximum and happened to move into a different region with a different local maximum?  It seems like that could explain the behavior.
<disclaimer> I'm someone with just enough stats background to be dangerous.  </disclaimer> So lets get dangerous....  My intuition agrees with yours that the Gaussian isn't the way to handle non-continuous data.  For continuous data one value on the number line has a different sort of relationship with all other values than does a number on an integer line, or a binary variable.  Binomial distributions describe binary variables.  Multinomial distributions describe multinomial variables.  Aren't these all exponential family members?
It means that the odds of whatever your DV is being 1 when the person is other Asian are 1.123 times those of Indians, but that this difference is not statistically significant.
The [Wilcoxon rank sum test](http://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U) does not evaluate a difference in *means*: it tests for stochastic inequality.  The proper measure of an effect size therefore is the probability that $B \\gt A$, minus the null value of $1/2$.
I don't see that R cannot be by-passed, it seems to only be called in the robust.cpp file for some R-specific data types. The R-scripts are in /R/ the .cpp files are the cplusplus codes.
@berbelein the images are from R.
Clarifying question: if rows in your matrix are "events" and columns are "trials," then what is an entry in the matrix?  A more typical design would have rows as "individuals" (say, people in a repeated experiment), and columns as "treatments" (a series of experimentally manipulated treatments), with entries in the matrix denoting "responses" (how each individual reacted to each treatment).  Is that right, or are you doing something special?
How does "making sense" differentiate this question from the duplicate? Having binary data makes no difference in either the application or the interpretation, so could you explain how that might matter?
Thanks for writing this much.
@David please see the edit...
I'm not an expert on clustering, but w/ so much data & just 1 dimension, I wonder if you could just make some [kernel density plots](http://en.wikipedia.org/wiki/Kernel_density_estimation) using different bandwidths and see how many modes / peaks you find, & whether the result seems like it would be useful for you.
Hope you don't mind but I added a link to "VECM". Feel free to correct.
What makes you think it depends upon the ratio between the number of subjects and features?
It's merely the most obvious and probably biggest issue. The space of ways other things might go wrong if things are sufficiently bad is quite big (e.g. if you have very large amounts of data or very large numbers, or very small numbers with fairly large numbers) -- pretty much any of those things - and many more - can be much more likely to cause problems with that formula. Rules of thumb aren't much good because there are so many other things it depends on (even the *order* of the data can matter - a lot - as to how much of a problem you have). ... ctd
@Procrastinator: yes, I know, but if you take $n\\to \\infty$, then it is just a Riemann integral, isn't it? (the bins are equal [I just asked the author of the paper - She's in the same office :)], and it does state that the skillscore is a measure of "the common area between two PDFs"). I mean, it's still going to be calculated numerically, but writing it in a continuous fashion doesn't restrict one to using a particular numerical method..
Your question is answered (in different ways!) at http://stats.stackexchange.com/questions/16230 and at http://stats.stackexchange.com/questions/16117. It is not possible to say which is "best" without knowing more about the kinds of errors you might incur and what decisions you will be making based on the resulting estimate.  I have not voted to close this thread, though, because you ask a novel follow-on question: how to estimate the error in the calculation.  I don't think that has been adequately addressed here before.  We could do a good job with some more info about your errors.
Generalization error is far from trivial. Unfortunately rules of thumb don't help much in this regard.
How many times do you plan on doing this?
It is answerable because accessible isn't the operative term.  :)  Instead, "how have you typically provided" is one operative phrase.  This is open ended, and I agree that CW is perfectly appropriate.  Thanks for making it so.
A book that goes "axioms-->theorems-->proofs,axioms-->theorems-->proofs,axioms-->theorems-->proofs‌​,axioms-->theorems-->proofs" might potentially lead to some understanding of probability (though, frankly, I've seen such approaches fail at imparting understanding), but will not really prepare you to understand statistics. In fact in my experience such a background on its own seems to prompt many mathematically able people to *think* they understand statistics[,](http://xkcd.com/793/) without necessarily doing so[.](http://xkcd.com/675/)
What is the number of dimensions of your dataset? What statistical model are you fitting? (an outlier is defined wrt to a model). Give more details. As formulated the question doesn't make sense.
is it always the same house, probably not? Is each house measured once? once per season?
Do the two realizations of $Z$ share the same $Y$ or is the case that $Z = X+Y$ and $Z^\\prime = X^\\prime + Y^\\prime$? Are $X$ and $Y$ _jointly_ normal? If so, what is their correlation coefficient?
Definitely! I don't have much more time to devote to this, but I'm sure you can find others to help.
Both since the standard error of estimate is √[p(1-p)/N] But if I can take just one it will be N because regardless of pthe standard error is bounded above by 1/[2√N]
+1, I'm a big fan of using simulations to help work through statistical ideas.
I cannot make any sense of these equations because you have not clearly distinguished variables from coefficients from random error terms. In fact, you seem to be referring to their *products* as "regressors," which makes little sense.  IMHO it would help immensely to drop the "IMR" and "U5MR" superscripts because they create a lot of visual clutter and detract from the apparent point of your question. You could also simplify these equations by reducing the number of variables that appear in each: it seems to be immaterial whether you have two groups, half a dozen, or more.
Well, ClassificationTree for example requires a matrix of numeric values. Information about conversion class-number is lost. Hence my question. But I just decided to use my option.
@Wolfgang, thank you very much for your answer, I'm actually using what you describe, in my work, so I need some references... can you helpme with a citation for the formulas?? thank you in advance
And what we see, of course, depends mainly on what we look for. --John Lubbock
Thank you for your help :)! That's exactly what I needed to know.
@Brandon Please, register your account and consider accepting this answer (green check mark).
I don't see a problem. This is just how it works. You just have to understand these facts when you view / interpret this plot.
If you need a command to do that, you better provide us at least with the language/package you use. Are we talking SAS, SPSS, Stata, SPlus, Minitab, R, ... ?
@MarcClaesen it does not seem like the same/similar question.
It's the relative *cost* of the two kinds of misclassification together with their probabilities that should determine the cut-off. If you just want to validate the probability model, calculate its AUC or Brier score when applied to the test set.
You should pool the $n$ imputations but more important part of MI is the number of imputations, $n$. If I remember correctly, you need to impute at least 40 times with 50% missing information to guarantee less than 1% power falloff compared to the comparable full-information maximum likelihood analysis. This [paper](http://www.stats.ox.ac.uk/~snijders/Graham2009.pdf) should be helpful.
Thanks a lot mzuba! What should I do if I have an unbalanced design between the trials in which efficiency is tested? Can I still use a two-way ANOVA, or do I have to use a regression analysis?
See [here](http://stats.stackexchange.com/questions/577/is-there-any-reason-to-prefer-the-aic-or-bic-over-the-other).  And note that none of the three information criteria is about assessing how well the model fits the training set - they all penalize over-fit.
(+1) Well done! -- it was not obvious where to look for that Wikipedia article; you had to think of quantile regression.