tttjjj/elt_bench huggingface dataset:
@martino Fair enough. Step-down methods would also be less conservative than Bonferroni adjustment. The reason I gave the latter is that if the customer "insists on using hypothesis testing to identify significant segments/clusters" I would not expect he is looking for complex adjustment to mass bivariate screening of candidate predictors; hence my suggestion for a 'simple' method of correction for multiple comparisons, if any should be given.
Could you be a little more explicit as to what you mean by "sparse" in this context and how this differs from the Box-Jenkins approach?
@caracal: yes, each observer tried all the stimuli in 2 conditions (EXP1 and EXP2). What I am interested in understanding if the differences between the responses of the males are significantly higher than those of the females. Do you think that is correct to do what you suggested? I.e. : > aov3 = aov(response ~ stimulus*condition*sex + Error(subject/(stimulus*condition)), data=scrd)
> summary(aov3)
I tested difference from known optimal solution using approximately equal number of scoring function calculations.
I think I see. So essentially, we can split the parameters $\\theta, \\theta'$ into those that those for which $X|\\theta'$ depends and for which $X|Y,\\theta$ depends conditionally. Furthermore we can assume independence of these parameters $\\theta, \\theta'$. Finally, one particular case of this is $\\theta'$ empty, so that $X$ has no dependence on parameters. Am I interpreting your report correctly?
@COOLSerdash Sure, it would look something like this:
`library(nlme); mod4 <- gls(y.new~x.new*dummy.var, weights=varIdent(form= ~1 | dummy.var))`
Why not? Binary [Phi](http://stats.stackexchange.com/a/26111/3277) is a symmetric measure. Odds ratio is asymmetric, either (ad)/(bc) or (bc)/(ad). That is, either 0.56 or 1.80.
@grautur: That's four excellent questions (plus many more, because your response applies to every answer in this thread).  They all deserve elaborate answers, but obviously there's no space for that here: one question at a time, please!
As with just about any basic introductory statistics question (i.e. an issue that is covered any introductory statistics class), this has been answered multiple times on this site before - e.g. [What is a complete list of the usual assumptions for linear regression?](http://stats.stackexchange.com/questions/16381/what-is-a-complete-list-of-the-usual-assumptions-for-linear-regression). Please consider taking a minute to search the site before asking/answering questions like this (finding this duplicate took approx 30 seconds), as cluttering the site with duplicates does no one any good.
What about having insignificant variables in a model makes you uncomfortable?  If you are doing an ANOVA on 5 groups do you collapse groups that are not "significant"?
Right, but this is not a discussion about dogmatic truths but about words. You have asked about entropy, so I answered about entropy. Because now I see that you indeed need an answer about time series descriptors, write a question about time series descriptors, only then you'll get an useful answer.
I think if the terms of the statement are not clear, it's not really appropriate to stand on its own in an encyclopedia.

That said, my interpretation (based on "within decision theory") was that a decision function based on the interval must minimize the expected loss with respect to an arbitrary loss function... which is why I found it dubious.
@TheAugust  There! Done now. Sorry for the neglect.
Perhaps I took the wording a little too literally, but it struck me as slightly odd; if I share (only) 50% of my genes with my fraternal twin, it is quite likely one of the two of us is not human.
No setting is needed: Select the range of cells in the two rows, invoke the Chart Wizard, specify "XY (Scatter)", and you have it.
Thanks so much for your response @Aniko. I was thinking of a similar solution (i.e. calculate AvgTimeBetween), but could not figure out how to incorporate the badgeless periods. In this situation Pathway 5 and 4 will have a badgeless period of 1. Do you suggest I should add that 1 to the AvgTimeBetween to calculate the AvgWaitingTime? Thanks.
@Nambari For this to work you have to start at a minimum of `t=8` because you need at least 2 periods to set the season index or it will not work.
$k$ random variables are mutually independent if $\\displaystyle P\\left( \\cap_i R_i \\right) = \\Pi_i P(R_i)$
$x_{ij}$ is the number of item j. As you can see, this is a 0-1 problem which restricts the number of copies of each item to 1.
@Glen_b:Thanks! "A random effects model has a "different" parameter or parameters (say, a different intercept) for each subject, drawn from a population distribution of parameters." I don't understand what is "drawn from a population distribution of parameters" and how that is like? Could you explain more specifically? (I think math formulation will make it clear)
Of course you can reduce "extra columns" that won't change the result. I just wanted to say that the book itself doesn't say anything about that important trick.
@Zen I already wrote in the question that the distribution is discontinuous. Also the RHS of $f$ makes it obvious that this $f$ stands for a density in $[a_i,k_i)$, but for a probability for $k_i$ -and I do prefer compact notation.
The mistake is that the MGF of $a_iX_i$ is $exp(\\lambda_i (exp(a_i t)-1))$ and not $exp(a_i\\lambda_i (exp(t)-1))$
Yes, I think you are correct. In fact I expect to use several different predictors, each with it's small contribuition.
@Mark If I understand correctly, then, what you are asking for is the MLE when you assume that certain entries in the covariance matrix are necessarily zero.  But isn't the answer to that question routine (even trivial)?  What am I missing?
How do you intend to interpret this average density?
Regarding your update. 1) I can see no point in doing this. Arima() will produce the fitted values and forecasts. Why should I produce additional R code to do the same thing as Arima() already does? 2) Yes, but differencing X-hat does not give you Y-hat. So undifferencing Y-hat does not give you X-hat.
@susan: I am not a big expert on multiple correlation, but I think that this stuff is pretty standard so you don't need to describe it in any particular detail. You would want to mention the value of your correlation coefficient, the F statistic and the p-value, e.g. "multiple correlation between this and that was $R^2=0.9$ ($F_{3,33}=10.22$, $p<0.0001$)".
Here you use the left-continuous version of the Kaplan-Meier estimate (option right=FALSE), so the result is correct.
I'm just confused how it fits in with the question. For example, the words "prediction", "decision theory" or "unsupervised" don't appear in the accepted answer
A notational problem still remains: your description suggests that the $x_{i,miss}$ are vectors of *numbers*, in which case the integral makes no sense. If they are to be considered random vectors, we can make sense of the integral by assuming you mean "$dx_{i,miss}$" to refer to the distributions of those vectors, but this is still an abuse of notation and it is ambiguous, because some sense of a *joint* distribution is needed. Evidently that comes from a context and additional assumptions adopted implicitly both by you and the OP. Could you state those assumptions and clarify the notation?
If $a=80$%, say, do you want the reader to conclude that $1.5\\times 80$% = $120$% is a *probability*? Really?
A good starting point is the list of earliest known uses of mathematical terms: http://jeff560.tripod.com/c.html which gives a reference to 1949 (but with covariate meaning specifically the "x" variable in a regression.)
You've put it really nicely. To update on the topic, right now I am trying both approaches on different multivariate time series data sets  and see which one performs better, Although this is not an exact way to deal with this. It is the best I could come up with 

Another doubt : What is the minimum no. of records you need to fit a VAR model  on a time series with n variables
PCR=Principle Component Regression not analysis. The basic of PCR is a bilinear model: a relation between a set of p-dimensional predictorsX=(x_1,...,x_n)and a response variable y=(y{1,...,y_n) through the latent component matrix T=(t_1,...,t_k). we assume that : X=X^bar+Pt_i+e_i, y_i=y^bar+qt_i+z_i. and then the bilinear model in terms of original predictors is: y_i=beta_0+beta^T+ebsilon_i.and then beta=Pq
Welcome to the site, @B_D_Dubbya. I took the liberty of formatting your answer, I hope you don't mind. You can find more info about editing your answers on CV [here](http://stats.stackexchange.com/editing-help).
$d(A,B)$ as defined above is a metric. When two objects have no features in common it yields the maximum distance between two normalized vectors, which is $\\sqrt{2}$ (since $cos = 0$)
Don't assume a normal distribution, assume a biniomial distribution.  Maybe a logistic curve.  Your "standard deviation" is how far to the left/right you go on the s-curve, which maps to probability between zero and one.
Sympathies that you are confused, but adding extra questions in the comments to an already broad question is not the way to go. http://stats.stackexchange.com/questions/2691/making-sense-of-principal-component-analysis-eigenvectors-eigenvalues offers various perspectives on eigenvalues and eigenvectors, which seems to be what you want to understand first.
+1 Very good @Zen.  This explains in detail the idea that I was getting at in my very brief answer.  I agree that the chi-square test for homogeneity and alternatives are exactly what the OP needs to solve his problem.  Your answer is a nice little tutorial on this that should be helpful to the OP. The table in your example is exactly how I was suggesting that the OP structure his data and the chi square test you set up and found the critical value for Will work very well to answer the OPs question.  He really justs needs to compare his test statistic to this critical value to get his answer.
Are items in S similar to each other?
It would be normal to avoid naming a variable "c" since there's a function by that name already. It's not that you *can't* do it - you can - it's just that it's likely to be less confusing if you avoid using standard function names as variable names, especially given that functions are variables.
+1, you might note that all of the $\\chi^2$s will have $1$ degree of freedom.
+1 Elegant approach.
The question seems slightly ambiguous.  For example, if the customer in question orders *first* then the probability of getting a steak is 1. If she orders *last* then there's a 3.5% chance she gets no steak. The answers you link to presuppose that orders are made simultaneously and that steaks are then randomised over customers who order them.
Based on how I posed my question, I implied that the population parameter is a constant rather than a random variable, so this answer is good.
A GLS model allows the errors to be correlated and/or have unequal variances. If you do *not* specify such a correlation or difference of residual variance with the options `correlation` or `weights` within the `gls` function, the results from GLS are equal to those from `lm`.
+1. Out of the box solutions rarely work for a specific business problem of your organisation. You need to tailor and improve, and for that you need to understand what is under the hood.
"Hierarchical linear models demonstrated that" is not a description of the model, it is a description of a (broad) category of regression methods. But the way the OP present the stats from the paper already made that clear: these are regression model outputs. Eoin's answer provides an example of the model.
Cross-correlation is for two time-series.
One of them would be a leading variable.
SPSS has the option to compute cross-correlation.
@Glen_b: I admit the notation is a bit garbled. I shall fix it.
I don't have a worked out answer, but I suggest this book: [Extending the Cox model](http://www.powells.com/biblio/61-9781441931610-1)
A friend pointed out that in this wikipedia article (http://en.wikipedia.org/wiki/Beta_distribution#Effect_of_different_prior_probability_choices_on_the_posterior_beta_distribution) there is a good discussion about the priors. Apparently, one should use Beta(0,0)=uniform when one does not know anything, Beta(1,1) when we know that both extremes of probability are possible and Beta(1/2,1/2)=Jeffrey's somewhere in the middle (? Maybe sth close to 1/2?). I have the impression that there is much more rhetorics than straightforward mathematics in this solution.
Depending on the available explanatory variables, you might think of using [Poisson regression](http://en.wikipedia.org/wiki/Poisson_regression).
Bootstrapping can be done in R with the `boot` library: `boot.ci(boot(y,R=1000,statistic=function(y,ind) { mean(y[ind]) }))`
Thanks for clarifying that. I have limited knowledge of statistics, so that was helpful. On the survey, there was a question asked twice. The means don't match up for the question, there's a difference of 3-5%. If the difference is greater than the variance, should the data be adjusted in any way? And yes, I'm comparing experimental subjects to choices and also the response on one choice to another.
It would be helpful if you described how you are doing this.
@DWin, not so sure that Hotelling's $T^2$ is really applicable here. At least from formulas from wikipedia page it is not immediately clear that $T$ in OP question can be represented as $T^2$. Can you please elaborate on this more?
Not sure if I understand this, can you provide me with some more details?
(+1) It seems to me this advice is complementary rather than countervailing.  After all, using Student's t matters when the degrees of freedom are small and that's usually where you can't detect heteroscedasticity anyway.  So with lots of data you can check assumptions and protect yourself somewhat with robust SEs and, furthermore, you won't really care whether you're using t tables or z tables.
I corrected 'saturated' to 'full', thinking that must be what you meant. See [here](http://stats.stackexchange.com/questions/283/what-is-a-saturated-model) for an explanation of what a saturated model is.
Given that even the *input* to the $\\mathcal{N}_p(\\mu,\\Sigma)$ problem is $O(p^2)$ and the [conversion from Cartesian to spherical coordinates](http://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates) is $O(p)$, the time needed for conversion is inconsequential except for small $p$.
Great answer. If I had another vote, I would vote for this answer again.
@cardinal Please, answer. The team is later complaining that we are not fighting with low answered ratio.
Yes: if the model is $y_i= x_i \\beta+ \\varepsilon_i$ with the $\\varepsilon_i$ normally distributed with mean zero and a constant variance, then what you have done looks reaonable, though of course the numbers are overprecise.  `plot(Sepal.Length~Petal.Length, data=iris); abline(h=5)` gives you a view of the information you actually have.
sorry but there seems to be a misunderstanding: I'm familiar with nonlinear optimisation; here I would like to know what smoothing techniques I can use on such data when fitting at every wavelength is not reliable because of the noise in both dimensions. Granted, my dummy example seems pretty workable, but if I'd added more noise it would have been harder to visualise. I like the fda approach suggested earlier, as it encompasses both the fitting part and the smoothing in one methodology.
Maybe http://gis.stackexchange.com/ would be more appropriate in your case. Anyway, you might think of updating the title of your question because as it stands it doesn't really reflect your aims (the use of "classifier" is misleading here).
(+1) The R code looks promising! I will look further into it when I have time.
Thanks for clarifying the relation and difference between those two concepts (bernoulli trial and normal distribution). I was actually looking for something along that line (didn't know how to express that accurately).
First, because it looks trivial it exaggerates focal effects on "magic" numbers 5, 7, 10 - you always encounter this 3-modality in the distribution. Second, looking at "intrapsychic scale" or usual respondent of satisfaction question, 9/10 almost indistinquishable, but stand apart from 5-7 or 5-8 range, and this is in turn apart from 1-3 or 1-4 range. Third, 10 points is just too much for a usual human differentiation capacity. 6 or 7 Likert scales are optimal for an adult, and 3 to 4, for a child.
Maybe even more general: one-class classification is useful, if the "out" cases do not have a better description than "not the class" (outliers, processes out of specification, a patient not having a particular disease). You can even have several "in" classes in a one-class classification, and a case can belonging to none, one or more of the defined ("in") classes (imagine a patient with some tumor and hepatitis).
"Want to assume spatial stationarity" appears directly to contradict the OP's assumption that "the intensities taper off radially in all directions."
Thank you Zhang. One tiny nit: if you wanted to use the Normal distribution for your prior over the probability of heads, I'd say you should truncate it so that p lies in [0, 1].
Adjustment is necessary when pre-specified millions of tests;ie on genetic studies
I believe you cannot fit the model that you have described: to my knowledge in JAGS you are not able to scale a distribution. What distribution $\\phi$ is supposed to follow? If it is simple enough, you can derive the distribution of $x$ directly and sample from it.
(+1) Welcome to our site and thank you for contributing such a well-developed answer!
Hi Patrick, thanks for your answer. Yes, I have tried to use bootstrap when estimating with DWLS, but the coefficients se estimation are wrong (se=0). You're right, the bayesian approach has been a bit hard so far. I had a try in WinBugs and I have specified my model, syntactically correct, but I can compile it and I getting a Trap windows with 21 preconditions violated, which are foreign language for me.
Agresti and Finlay mention that generalizing Fisher's exact test also makes sense when cell counts fall below 5 in any contingency table.
Your question strikes me as oddly worded. You don't "choose" residuals. Those are given / implied by a procedure / analysis. The underlying model implies certain (abstract) error terms, and the analysis produces residuals. In that sense you choose the *analysis* not the *residuals*. Maybe rephrasing to "what type of post-fit analysis of residuals do you use" may be better? You seem to be leaning that way asking the question in the context of a OLS model. And in that case, looking at influential observations ("hat matrix") is a good method.
Pleased that you got what you wanted, but I am not sure what long-term value this question has for the forum.
You don't make clear whether all your variables are Likert scales; if not, please explain what your dependent variable(s) is or are. Testing Likert scales for normality is futile; a variable defined to have a small number of values could not possibly be normally distributed. Conversely, the underlying distribution has no relevance to Kruskal-Wallis, which only uses ranks. You should worry about whether ties frustrate Kruskal-Wallis P-value calculations. You need to review which assumptions are associated with which procedures, as you appear seriously confused.
It is rare for that vector to be the first principal component, but it is possible. If your "no" is your response to my question, then that means you are missing out on the good information in the link I gave: please go read it.
If you are able to answer some of the questions in my answer that request clarification, I may be able to add some detail.
I had a look at this, and it's perfectly fine but has a bit too much of the feel of an undergraduate text book for my tastes. Not that there's anything wrong with that...
Re (1): It is impossible for a uniform variate and a normal variate to have a correlation of $1$.
Have you tried expliciting calculating $$\\text{Cor}(X,Y)=\\text{Cor}(A-B,C-B)=...$$?
I have updated the question with specifics, answering some of your questions.
If your wish is the first one I said about, i.e. only the fact that n<p bothers you and only this fact forces you to apply PCA, - then you should retain 28 (all but one last) components in PCA and do LDA on those 28 components, leaving there few strongest discrimimants. Each of those is the linear combination of 28 components, each of which is in turn the linear combination of 100 variables.
So far, I like this solution. The only question I have would be how one would calculate each bin. For example, what would the boundaries be between each bin? If I use outliers to calculate that, it might influence the result.
@rafalotufo Sorry, I don't know that one exists as what I laid out is state-of-the-art. If I can help with some examples I would certainly do that. You can post 1 of your sequences and I will analyze it ( not with R ) and post the results.
I still don't understand what you want to do, but maybe you could use simulations to compare what happens if you simulate using a t distribution compared with using a normal distribution.
@whuber, you are right, I have to provide more information. Feature f can assume a discrete number of values (in practise many of them), so if we consider the probability of a particular feature occurring in case of Background, p(f1|B), it should be a low number and all p(fi|B) should sum up to 1. The same happens for the foreground.  
p(f|F) = 1-p(f|B) is impossible. Assumption I-C5 states: “Pixels corresponding to visitors have a color distribution distinct from the background distribution”
I would really like to thank everyone, every answer has improved my understanding of regression.. I was mislead by the high p-value values, which, as i now discover, are not that significant (since translating -2000 years solves the problem of p-values, but of course the regression it's exactly the same)
@Wayne, I was thinking more of a "weight" graph, I don't know the actual name for it. In Github, they have it named a Punchcard.  Both would show the same thing though, so I think I'll do both if possible.  As for examples, it'd just be a set of numbers, ie: [23, 9, 187, 2, 222, 23, 75, 250], although it can be rather large.
How would you go to test whether those differences are significant or merely due to chance? I guess you could do some simulation by randomly shuffling the data a large number of times and have some sort of 95% confidence bands of the differences. Would that make sense?
@david w: The only question I had is the elbow method shows increasing values and your plot shows decreasing. Is this because you are using the distortion values directly from kmeans? How can I convert this to look like the Wikipedia's elbow plot? And as a last question, would you happen to know how to do this for kmeans2 instead of kmeans?
@ kjetil b halvorsen: Thanks for the suggestions! I've looked into Relative Distributions and QQ-plots, but unless I'm misunderstanding them, I don't think they help me to answer the problem I'm after. In both cases I'm visually interpreting a plot, but don't have a way of testing the difference between the distributions at every position along the curve? Is that correct?
If your time series is exactly $x_1, x_2, \\ldots, x_n$ with all other $x_i$, $i < 1$ or $i >n$ being unknown, then the sum must necessarily stop at $t=n-h$ when $x_{t+h}=x_n$ occurs in the sum: the next term (for $t=n-h+1$) that would be included in the sum would have $x_{n-h+1+h}=x_{n+1}$ in it, and $x_{n+1}$ is not part of the sample.
Are you asking for multivariate hypothesis tests? For your first question, if you're looking for a multivariate equivalent to a t-test, you might consider Hotelling's T-Squared. For the second question, are you trying to compare k groups? Is MANOVA appropriate?