tttjjj/elt_bench huggingface dataset:
A good answer to this question would consider the trade-off between the (presumably) additional costs of designing to 16 compared to 12, against the (presumably) additional costs of having to handle more than 12 items for a machine designed only for 12.  If this observation is not enough to help you answer your question, then please let us know something about these costs and the frequencies with which you expect them to be incurred. There are other issues, too: the future will *not* be like the past, but we can use statistical methods to estimate how much different it might be.
Try auto.arima(data.s.g, stepwise=FALSE)
Hi D Greenwood, welcome. These three posts from the site could be helpful: [First](http://stats.stackexchange.com/questions/55501/test-a-significant-difference-between-two-slope-values), [second](http://stats.stackexchange.com/questions/33013/what-test-can-i-use-to-compare-slopes-from-two-or-more-regression-models?rq=1), [third](http://stats.stackexchange.com/questions/59327/how-to-compare-two-regression-slopes-for-one-predictor-on-two-different-outcomes/59331#59331). Since you have repeated measurements, you'll probabily need to fit a linear mixed effect model (e.g. with `lme` in `R`).
I believe there are many forms of matrix factorization. What one do you mean? I'd also ask you to unpack your question, make it more specific. As far as you're studying the subject you might have faced concrete issues.
If you look at e.g. my answer below, you have the number of votes probably top left of that (for now that's a 2 - it has an arrow up and down around it that look like normal distributions). If I'm not mistaken, just below that, there should be an empty tick right below that. As you are the one who asked the question, you are the only one who can say which of the answers was the one that helped you most (or more technically, which answer was most to the point). You do that by clicking in the empty tick, which 'accepts' that answer (and fills the tick).
@NickStauner It would be for the entire population.. .I've updated my question as well.
since it is not log-log regression, without constant you have not the elasticity interpretation as you wrote. Why you believe that log transform is needed? Is your $x_1$ log-normally distributed? Is it a time series or just cross-sectional data you work with? What is the nature of your data?
'Non-independent' is a bit vague; there are many causes and models for dependence (clustered data, hierarchical/multilevel models, time series, ...). Could you be a bit more specific on your area of interest?
@Arun, take $n=1$ in formula (32).
Boris, it is not necessary for *either* variable to have a normal distribution and therefore a test of normality would be of little use. What matters most is that on a scatterplot the data appear to trend along a line and there are no extraordinary ("outlying") excursions from that line. A secondary issue is that the scatter of the data around the line should be approximately cigar-shaped or (American) football-shaped. All these can be checked simply by looking at the scatterplot.
Check your calculation again, $\\widehat{a}=\\frac{X_1+X_2}{2(X_1+X_2+X_3)}$.
@Gary - the observation set must indeed scale like you say, but note that $K=2d+1$ and this is enough to satisfy the semi-symmetric condition. Pick any set of $d$ vectors, add their negations, add zero vector and you have a semi-symmetric set of means (yet it is probably not optimal)
Could you indicate: (1) what your test statistic is, and (2) what hypothesis test you are interested in. I anticipate you want to test the null hypothesis of zero correlation (of $x$ with $y_a$ and $y_b$), but in that case your test statistics and DF look quite strange.
@darkcminor: the `xtest` that I have, and the sample data you provide are drawn from $\\mathcal{N}(0,1)$. The classifier is always going to prefer the $\\mathcal{N}(5,1)$ class to the $\\mathcal{N}(6,1)$ class in this case! Instead, why not try comparing to two classes, one with mean $-1$ and the other with mean $1$.
whuber, based on your contributions to this site, I am convinced your students (if you are still teaching) are very lucky.
I'll edit the above to give these forms and references.
I expanded my answer to include a link to a review paper discussing (among other things) multivariate time series. As to the current trends in the use of NN and related techniques, try and find a more recent review paper.
(I presume a sort of ad hoc engineering approach here) Then I would suggest dividing by an $N$ every week (or day, hour, etc..). I.e., (1) above. This will discount observations last week by $N$, observations from the week before that by $N^2$ and so on. What you are in effect doing is a weighted average where you give more weight to more recent observations.
This is not a clustering problem.  In statistics, "clusters" consist of sets of objects that are mutually *closer* to one another than other objects.  Closeness does not capture circularity: that's why neither K-means nor any other clustering algorithm will work.  For this reason, this question probably fits better in the image processing or GIS sites, where you might find some experts on this issue.
Linear regression with a binary dependent variable? That is asking for trouble in so many ways beyond the constancy of variance. Can you elaborate why you would want to do this instead of (e.g.) logistic regression?
+1 For 579, I really enjoyed the videos (I added descriptions for some of the videos)
@Glen_b Thank you. Do you have in mind some specific count functions?
I'd even say [box-counting dimension](http://en.wikipedia.org/wiki/Box-counting_dimension) for a more practical statistic.
love = favorite?  you're not posting degrees of loving.  if you love it, it is your favorite.  clarify if needed.
I am not sure; I mostly use SAS for this sort of thing. Others here are expert in R
@Simon: I don't disagree with your comment on the odds ratio, but I have seen people fail to take this back to consequences for the population correctly after stratified sampling when they had done so in other cases.  For example, if you find that people for which factor A is true have twice the odds of being "bad" as those without factor A, this should not change with stratified sampling, but if you want to know what proportion of the population will be unnecessarily affected if you target those with factor A, then you need to carefully weight the information from your samples.
Why not just sort the values and assign 0's to the first half of them and 1's to the rest? That's not random (because the sort probably does not randomly shuffle tied values) but it's easy, reproducible, and does what you need.
Stepwise regression can result in misleading insights; especially if you are using $p$-values. Check this thread (http://stats.stackexchange.com/questions/20836/algorithms-for-automatic-model-selection) and Zach's post in particular. If your adamant about using direct model comparisons only use at least an information criterion like AIC or BIC instead of "raw" $p$-values, it will give similar but slightly more coherent results.
I think your problem is very solvable, but we need to understand your data better. Let's see if I understood you correctly: i. there are three factors (trt) of interest -- *development*, *region*, and *genotype*. ii. there is a *batch* effect also, about which you don't care. iii. I still don't understand the design (confusing to me at least). Could you tell us in plain English what is your experiment, how do you go about measuring the variables of interest, and how does batch effect creep in. Given this info, modeling will follow.
@ttnphns : thanks, good point. I looked a bit more and found a source for an alternative to explained variance that is more appropriate for FA.
@Masato: +1 for your answer to this badly formulated question.
I think the term you are looking for is [Association rule learning](http://en.wikipedia.org/wiki/Association_rule_learning).
Maybe [this post](http://stats.stackexchange.com/a/60083/21054) or [this one](http://stats.stackexchange.com/q/56066/21054) can help.
You have to learn the basic principles of hypothesis testing ! If you want to prove that the odds ratio is significantly greater than 1 then you must use "greater" for the alternative hypothesis.
In a comment I've since removed, I had provided the output as [IrishStat](http://stats.stackexchange.com/users/3382/irishstat) requested, in hopes it could help him answer my question. However, this resulted in attention to the specific application of the question instead of the question itself. Also, the pastebin postings have since expired. These two reasons are why I have removed the comment.
This is an amazing answer; I'm much obliged! I've been playing with this, and the results look very promising. I'll need a bit more time to fully understand the reasoning, though :)
Also: could I contact you through your website for one additional (private) question, regarding acknowledgements?
Well that IS my assumption, but I don't think the censors will be impressed by that argument :)
Perhaps outliers..a good place to start. No one said anything about logistic regression though, these are normally boosted decision tree models.
Tests, p-values, & confidence intervals all assume a pre-specified model, & don't take into account that you're picking the "best" model out of several in training. Even the point estimates from training will have a bias towards larger absolute values. That's the point of re-fitting the "best" model on the test set - now it *is* pre-specified. See @gung's answer [here](http://stats.stackexchange.com/questions/20836/) for an intuitive explanation.
There seems to be a bit of confusion here because dividing the range into five equal parts has nothing to do with the standard deviation.  You might enjoy researching the [tag:clustering] questions on this site, with especial attention to "K-means" solutions (which do involve standard deviations).  On the Web you could also research "Jenks' method" or "natural breaks classification," which is the one-dimensional version of K-means rediscovered by cartographers to group data into classes for mapping purposes.  (The 5-class solution is (1,1,3), (7,9), (15,22), (99,100), (240).)
Did you google "homoscedasticity"? The first link, not surprisingly, is Wikipedia: http://en.wikipedia.org/wiki/Homoscedasticity. You will find that "homoscedasticity" and "homogeneity of variance" are the same thing.
Thanks for your reply, and I am very sorry that my question initially was so off-the-cuff and unclear. I have tried to clarify it now. 

The link from Bernoulli to Poisson is interesting, but what I am looking for are examples of situations in continuous time that are not suitable to be modeled by a Poisson distribution, the simpler the better.
http://en.wikipedia.org/wiki/Mahalanobis_distance and http://stat.ethz.ch/R-manual/R-patched/library/stats/html/mahalanobis.html are good starts for the Mahalanobis distance, which is a good measure for multivariate outliers. Of course you can still check for univariate outliers!
Very nice, clear answer.
I think the paper, "[Testing for the Markov Property in Time Series](http://www.wise.xmu.edu.cn/Activity/XMU-HUB2008/paper/HONG.pdf)" contains useful insight and literature review.
If you are running 3 chains, you should provide initial values; otherwise, JAGS (rjags) would use the same generated initials for all 3 chains.
I am working on inference problems in dynamical systems which involve models like above mentioned model. The motivation is: since these models can be represented as a generative model, inference algorithms can be immediately applied, therefore they can use for inference problems. To make concrete, suppose simpler model, $$x_{t+1} = x_t + w_t$$ where $w_t$ is a zero mean, unit variance Gaussian random variable and $x, w \\in \\mathbb{R}$. Then, this model will correspond to the following generative model:$$p(x_{t+1}|x_t) = \\mathcal{N}(x_t,1)$$ or $$x_{t+1}|x_t \\sim \\mathcal{N}(x_t,1)$$
@Aaron it works pretty well now thanks! but I still don't know how to fit the extra line. can someone help? also what am I doing wrong with this? colors<-c("#000000","#F8766D","#7CAE00")
> d+scale_fill_manual(values=colors)
+1 for the reference: it's a lengthy treatise on practical die testing.  Halfway down the author suggests using a KS test and then goes into ways to identify specific forms of deviation from fairness.  He's also well aware that chi-square is an approximation for small numbers of rolls per face (e.g., for 100 rolls of a 20-sided die), that power varies, etc., etc.  In short, anything the OP might like to know is clearly laid out.
@macro Excellent answer, but I would stress (for those new to this subject) that R^2 can be very low even with a strong relationship, if the relationship is nonlinear, and particularly if it is nonmonotonic. My favorite example of this is the relationship between stress and exam score; very low stress and very high stress tend to be worse than moderate stress.
No problem, I can't really advise on time-series as I don't have much experience with that.  I seem to recall that Johan Suykens wrote a good paper on time series with the least-squares support vector machine (essentially kernelised ridge regression) for time series, IIRC in the late 90s/early 2000s, might be a good place to look (Johan's group has done lots of interesting things, so well worth checking out anyway!)
Change-of-variable formula (with some thought regarding the support of X)?... but $P(X=x)$ is non-zero only for discrete r.v.'s. Is $X$ a discrete r.v.?
If you have 0's then, of course, a pure log transform will not work. Are there any actual 0's? If so, what are you doing with  them?
Does the number of reactions depend on all $X_1,\\ldots,X_n$ or is there a different number of reactions measured for each of these $n$ positions? It might even help if you edited the question to contain your complete model.
Ah, I see your point. I think I'll write above that I'm assuming the RVs are independent otherwise this cannot be done. I don't think the professor would be trying to throw us for a loop like this. Thank you!
No my comment was not to concentrate on residual plots but rather to make the model fit "up front".  Peter the proportion classified "correctly" has multiple problems, including the fact that it is the lowest precision estimator of predictive performance, is arbitrary, and will lead to selection of the wrong model.  I have an example where adding a very strong predictor (on the basis of likelihood ratio $\\chi^2$ or $c$-index (ROC area) makes the proportion classified correctly **decrease** significantly.
In case you're looking for further reading, this is all covered in detail with examples in Gelman and Hill's textbook on mixed effects modeling. It's a very, very good reference.
@Junier Drawing a picture often helps :-).
(+1) I'm glad you turned this into an answer. Note that, strictly speaking, the uncorrelated case is a *subcase* of $\\mathbf K_{X+Y} = \\mathbf K_X + \\mathbf K_Y$. (Counterexamples, e.g., in the bivariate situation, are easy to construct.)
Edit: I was able to use the lmtest package with the coeftest function where you can just choose which VCV-Matrix you would like to perform the Coefficient tests.
@suncoolsu In terms of "Google fu", a good approach seems to be to start with broad searches and then narrow them in response to the hits you are getting.  Starting with an extremely narrow search risks overlooking good stuff.  In this case even my search perhaps was too aggressive.  When you consider, though, that every word in that quote can have many meanings unrelated to your intended topic, using the entire quotation seemed to be a reasonable starting point.
I already was aware that Cohen's d was a biased statistic.  I appreciate the details regarding the reasons why it is biased.  Nevertheless, I'm a little skeptical that it is biased to the degree I'm observing.  The Wikipedia article doesn't define 'a' in the referenced equation.  In addition the referenced equation and yours appear to reference the two sample version of Cohen's d.  So, I'm unsure what magnitude of bias I should be expecting in this case and whether your answer covers the difference I'm seeing.
You should answer your own question for people to see in the future! You can even select it as correct.
In addition to fg nu's answer, the variance process in GARCH is time-varying. However, there is a trick here is that given a time-series of log-return of SP500, then to obtain the volatility process what should we do? Some people say that we need using the ARMA model to withdraw the residual series, then plug this residual series into the GARCH model to obtain the conditional variance process? Or directly plug the log-return plug the log-return process of SP500 into the GARCH model to obtain the conditional variance?
It is - as discussed for example [here](http://stats.stackexchange.com/q/79343/805), [here](http://stats.stackexchange.com/q/9610/805), [here](http://stats.stackexchange.com/q/56844/805), and [here](http://stats.stackexchange.com/q/48615/805) ... but I don't think that's a suitable way to approach modelling the shape of what's clearly a time-series. For starters, you have dependence over time that isn't accounted for by the approach you're taking here.
Actually it optimizes $\\frac{wx_i+b}{|w|}$ to $\\frac{1}{|w|}$. But in mathematical view, the numerator and denominator both have $w$. So I can't understand
I don't think there will be a threshold; certainly not a simple one. Different errors will be of different severity: "Slightly messy" vs. "crashes the computer".
@whuber we just have 8 numerical data for 100% and 4 entries for 50% and 75% each. The question is the second part of a question about ANOVA analysis if that helps. I'm just not sure how to make one confidence interval for two differences. or is it asking for multiple confidence intervals?
For reference: http://en.wikipedia.org/wiki/File:Vector_Video_Standards2.svg I want to plot the line which acts as the boundary for yellow on the lower right (>1M pixels)
This question appears to be off-topic because it is about data analysis methods -- migrate to CrossValidated?
@Frank Harrell Efron and Gong described optimism bootstrap for classification problem and 0-1 loss. Though it seems straightforward for regression and quadratic loss, somehow I could not find it in the literature. Can you suggest a reference? Thank you.
Isn't this solution kind of lame?  It doesn't take into account saturation effects at all.
Ask a question. It is not clear what it is that you want to do. You cannot push analytics any further than your last equation (which you would find in most introductory econometrics textbooks when discussing endogeneity and introducing instrumental variables... which is something you may want to look up).
On the last, yes indeed. For a start, consider units of measurement here. Cumulative probabilities are unit-free; densities aren't. Also, cumulative probabilities are easy to get from data. Densities (now) are too, except that there are lots of ways to do it and they won't agree in general.
Considering your problem from a purely statistical perspective, my recommendations would be to use autoregressive (p) model to see if we could say the future value of $y$ depends on the previous value. The space between each peak and trough seem like there is some pattern in that.

Alternatively, consider estimation of first differences to see if there are any relationships between them.
@Glen_b: I mean since solving the problem involves the computation of Hessian (typical), which is N^2 in size, so the evaluation of functions (N) is not likely to be the bottleneck of the performance when Hessian is computed analytically.
It sounds like you're looking at two variables where there might really not be a relationship but trying to discover a relationship anyway. This way lies madness and papers which turn out not to be replicable.
Michael, I've merged all your replies into a single response. Unless response are really different, posting several replies is not encouraged, as per our [FAQ](http://stats.stackexchange.com/faq).
Polynomials are *far* more sensitive to anomalies within the data than splines are. An outlier anywhere in the data set has a massive *global* effect, while in splines the effect is local.
Your $\\hat y$'s should presumably all have subscripts
This should *NOT* have been closed as off topic - the underlying issue is clearly a statistical misunderstanding, but the OP is perhaps too confused to realize that. Instead this should be closed as a *duplicate*.
@Abhishak.Unfortunately your seasonal indices assume a specific weighting structure i.e.the # of periods to use to compute the "weights" and that there are no Pulses or Level Shifts or Time Trends. In terms of seasonality changing over time you are not quite correct. Adapting to change requires a model form. For example if you had 6 years of data and thee was no June effect for the the first 3 years . The correct seasonal factor would reflect the June impact for the last three years. There is no adaptivity here in the sense you are using the word. Your approach has too many presumptive steps.
wow. thank you Macro, i wish i could give more points.
@SimonHayward: yes, \\overline{y}.it is the denominator of y_i. I want to get the catch-weighted cumulative curve, so what I did was to get the frac{y_i}{y} and multiply with the corresponding x_i (temperature measurement at day i) and tried ecdf in R (wrong result). What's the right approach in evaluating this expression. thanks much.
Need context here. In marketing, this would be a chart which would indicate the percent sales increase expected from various marketing activities, but you probably have a different context in mind.
What does your model look like? (can you specify the canonical equations at least?) Is it discrete or continuous? Is your noise additive?
It is noteworthy that Batschelet's approach appears to rule out certain kinds of confidence intervals that give thoughtful readers pause, such as CIs that can be empty. Such a CI would scarcely capture the idea of "indications of deviations caused by chance fluctuations." This hints that perhaps the standard definition of confidence interval does not quite accomplish what is intended. Regardless, in the absence of any clear indication of what "confidence" means in question (5), we have to discount any conclusions drawn by the authors based on the answers they got to that question.
@whuber: I guess I was thinking if I knew 30% of the observations came from $f$ and 70% came from $g$ in $s_1(i)$, and lets say 50-50 for sample $s_2(i)$, then I could set up an equation system like $s_1(i) = 0.2f(i)+0.8g(i)$ and $s_2(i) = 0.2f(i)+0.8g(i)$. Solve for $f$ and $g$. In this case, then $f(i)=-2.5s_1(i)  + 3.5s_1(i) $ and  $g(i)=2.5s_1(i)  - 1.5s_2(i)$.
The two variables need not be independent. Consider $(X_0,X_1)$ uniform on $\\{(1,0), (0,1), (-1,0), (0,-1)\\}$. Though I have not checked this carefully (yet), more generally, if $U_1$ and $U_2$ are iid symmetric random variables, I believe $X_0 = U_1 + U_2$ and $X_1 = U_1 - U_2$ should work.
AIC/BIC values are often used for model selection. If that would be your goal, then yes, this is one approach. If your goal is to determine the relative importance of the variables, then you could use Akaike weights. A very nice and clear presentation on that idea can be found here: http://bit.ly/pDQ7eq
Holy crap, I think that SAX page is the ugliest web page I have ever seen!
I think the difference to MNIST is scaling up the number of classes. Unicode does have a lot of confusables. I was just googling and [found this table of Unicode confusables](http://www.unicode.org/Public/security/latest/confusables.txt). I could possibly use this list to create the categories for supervised learning. But maybe, as you suggest, it is not a big problem in the end. I will have to try it out. Regarding training data, I'm creating it automatically. I want to solve OCR of screen dumps, and it is possible to auto-generate a lot of training/validation data.