tttjjj/elt_bench huggingface dataset:
For transparency reasons, I should note that I am using software that I have been involved with as a developer. Most available software will duplicate the findings that I have presented here.
@amoeba Your point and that of ssdecontrol are related. I want a confidence interval at every time point, which reflects the different uncertainty at each time point.
Ah, I understand the confusion. There was a typo in my answer. Fixed now.
@BabakP good catch, it should be $a, b < 0$ instead of $\\alpha, \\beta$.
+1 ... however, if the data is categorized (and it sounds like it may be) the K-S (which assumes continuous random variables) will be conservative. The same test statistic could still be used, but the standard calculation of p-values is wrong. Simulation of the distribution of the test statistic might be used to assess significance, however.
I don't have an answer to your question but I'm trying to do something similar. I've been looking into using Taken's Theorem of Time Lag and Embedded vectors. So far I'm pretty sure that you can construct a matrix of Taken's Embedded vectors and using then as you x in a f(x) = y formulation. But, I'm unsure what the y would be. My confusion mainly comes from how the embedded vectors are constructed. I know it's not much but just thought I'd try and help out. Good luck!
I'm sorry, but it is in no way clear what beta 1 hat is! Can you elaborate on the model in question?
Many thanks for your feedback - that is very helpful!
Because documents are often represented as vectors. Cosine similarity is a measure of whether these vectors are pointing in the same direction. In practice, this often works as a measure of whether documents are about the same thing.
Thank you very much Peter, really appreciated. What I like about your answers is also how straight to the point and short without going into complex details thank you again
@amcnabb Sounds like so. The paper your were reading used the notation p.e. on page 97 as well: "... we find that the joint p.e. (probability element) of Y and X1 is ... ".
It would be best to be slightly more specific.  Surely your question arises from some specific case you're thinking about.  Say at least a little bit about that specific case.
search "cross wavelet transform" , wavelet coherence ;-)
To clarify, is it correct that you are already able to get the required $n$, but you're looking for a general formula? I would be very surprised if there is a useful general formula.
Hello, Taimoor, and welcome to the site!  It would be easier to answer your question if you could add some more detail to it, for example (but not limited to) what you are trying to estimate, why you want to use a time series model, what you have tried so far, and so on.
@rdorlearn Linear regression is just a GLM with an identity link function.
Also - I am entirely serious - if this method works I am more than willing to mention you by name in the acknowledgement section of resulting refereed publication :)
I've added an example.
Oh!  I didn't know.  Crap.
A 2-second Google search would have saved you from a 15-second question typing effort: http://math.stackexchange.com/questions/337689/if-you-roll-5-standard-six-sided-dice-whats-the-probability-that-you-get-at
Downvoter, any particular reason you object to my answer?
No problem at all, I didn't realise either until you mentioned it. I edited the question a bit, for it to be more clear.
@MichaelBishop: I'll try to do that in a bit (issues with my FTP right now).  For now, just picture two beautiful curve fits on a log plot, with one roughly a constant amount above the other.
@kanzen_master: Sure you may observe decrease of performance with more training samples: there is a variance aspect as well as a systematic (bias) one to the learning curve. That's why Dikran talks about multiple random samples. Pls. see also my answer.
Related: http://stats.stackexchange.com/questions/90005/predict-probability-of-failure-in-r-using-survreg
Although this makes sense in general, it seems to me it is a little bit at odds with the usual practice of spatial statistics. There, stationarity is a *decision* made by the modeler, not an inherent property of a process or the data.  It refers not necessarily to the data but to *residuals* from an external trend, or "drift." In your rats example, the differences in expected numbers of rats with location could be modeled with a drift and the stochastic part of the model could be considered stationary.
"the book" is slightly ambiguous here. I'd consider Barnett and Lewis the leading monograph, but it's not the only book on outliers. http://www.amazon.com/Outlier-Analysis-Charu-C-Aggarwal/dp/1461463955/ is recent. There is also an older book by D.M. Hawkins.
Do you need the expected distance?  Its distribution?  The first few moments?
A [review](https://xianblog.wordpress.com/2012/05/02/whos-1/) of this book by @Xi'an.
I'm choosing this as the answer because I find it a bit easier to understand (my statistics is pretty basic). Thanks!
You are right with your grievance, but it is not Bonferroni correction's problem; it is SPSS options problem. Unfortunately, SPSS doesn't allow to specify user-defined number of comparisons by hand, e.g. via syntax. So, you should just produce the table with uncorrected p-values and then multiply it according to the number of comparisons you actually imply.
How's that? ;-)
@AdamO, I still don't understand. They are both `.265`, how are they not identical?
Ordinary least squares is an estimator. There are a variety of  algorithms for computing the estimate: usually some sort of orthogonal matrix decomposition, such as QR, is used. See http://en.wikipedia.org/wiki/Numerical_methods_for_linear_least_squares
May I ask what kind of data do you want to model and for what usage you want to do it?
Might help if you say *why* you need to find outliers. By eye, I wouldn't say there are any particularly severe outliers, considering the size of the data set. The only obvious anomaly is the little line of values with y=0 for x=1,2,3,4.. with a gap just above it.
A very closely related thread appears at http://stats.stackexchange.com/q/6022: it concerns the same problem where only three percentiles are known for data about amounts of government funding, which also has a distinctly skewed distribution.  One answer includes clearly-explained working `R` code.
(That question I link to in my previous comment might even be close enough to this one to count as a duplicate.)
Your question (and the handout) misses the connection between $n$ and $m$. Is this the number of iid observations of the $m$ coins? Or the total number of observations? If $m$ is fixed, this shows that the max is a $o_P(1)$, not a $O_P(1)$. If $m$ is not fixed, we need $m$ to increase slowly enough to get $o_P(1)$.
I posted [similar question here](http://stackoverflow.com/questions/7611169/intelligent-point-label-placement-in-r), have a look on the answers there..
@ user11852: as far as I know a Box-Cox transformation requires positive values. Of course, one can add the absolute value of the minimum to the original values before the Box-Cox transformation. But is that better?
You should not be using least squares for this regression: a binary response needs a generalized linear model (GLM) instead.
I don't think that maximizing pseudo-r-squared is a good strategy, although i agree that model goodness-of-fitting is important to assess the model...
I believe that you could characterize it as "a mess". This paper, http://www.jstor.org/stable/2236545 , examines the matter, with more scientific rigor.
I believe it represents 1 SE above the minimum possible cross-validated error.
I think this suggestion was already covered by Zeron's answer some hours earlier. The only piece of additional information ('it is free') would be better as a comment on that answer.
You need to explain the data structure more, otherwise it is not clear how would a chi-square test apply. What do the values represent? Are the rows meaningful?
Can you clarify your question? Are you wondering how to move from a confidence region for the entire parameter vector to a confidence interval for a *single* element of the vector?
@PeterFlom this paper has an example: http://www.ophi.org.uk/wp-content/uploads/Widaman-1993.pdf
I found an answer. I now use the package "glmADMB". In addition, I also changed quasipoisson to Negative Bionominal Distribution. I heard that quasipoisson is a "quick and dirty" method for overdispersion.
If you're counting objects you likely don't want a decomposition like this. That'd be called Kernel PCA. Maybe something like skimage or imagej?
Henry, that's very helpful, thanks. Yes, it does seem that some users rate a page negatively (and others just don't respond) simply because the page wasn't what they were looking for, rather than it being poor quality in itself. Other data such as search keywords bears this out.

So not only should I not exclude these pages; I should pay extra attention to them and why people might be erroneously landing on them.
@Zhang You are correct about the $F$ *distribution*, but a statistic is computed from data: it is not a distribution. For instance, to help clarify the distinction, we all know that a (squared) correlation coefficient is described by a non-negative distribution *when certain assumptions hold*, but nevertheless the $R^2$ statistic [*can* be negative.](http://stats.stackexchange.com/a/12929)  I wonder whether the present question might be of a similar nature.
Thanks @whuber, you help me to better formulate my question. I've just edited it again.
If you always output 0 as your estimate then your estimate will have zero entropy but is not necessarily a good estimate.
As Nick said, the Gaussian is *also* a special case of the t-distribution (a limiting case), though in fact the Cauchy isn't the extreme case in the other direction if one allows non-integer degrees of freedom; I've seen $t_\\frac{1}{2}$ used for example. I could write down a hundred differences between the two that one might discern (e.g. a common observation people make is that the Cauchy is heavier-tailed and more peaked). To list every difference you *might* mean could fill a book (and mean your question will be closed). What kind of things do you mean?
Note that the correct df to use is likely neither 19 nor 20, because I strongly doubt the parameters were estimated via MLE from the binned counts and I suspect the bin cutpoints may have been based on the data.  The $\\chi^2$ approximation is so exceptionally poor with so many zero bins it hardly merits being called "approximate"; *wrong* would be more accurate.
It is usually a bad idea to have regression without a constant term unless there is strong reason to do so.
You have to indent it with 4 spaces manually or by selecting and clicking `{}` button.
Yes, BIC is an approximation of the marginal likelihood.  However, it is only an approximation that converges to the "truth" when the sample size grows to infinity. It is therefore not directly Bayesian (does not use the prior, for one thing!) and completely unrelated to MCMC (where the approximation is of a Monte Carlo type: if I increase the number of simulations, the approximation improves). DIC is deemed to be more Bayesian by many (incl. B. Carlin and D. Spiegelhatler)
(+1) Nice plot! Do you maybe want to remove at least some of the code excerpts from your question to improve readability? It's up to you, of course.
If your data would be in a `data.frame` you could use basically the same call (with the added `data = your.df` argument)  to function `aov.car` from the `afex` package (on CRAN) to obtain a more informative output.
@BrashEquilibrium: It seems that the word *independent* is being used in two very different ways here. I think whuber, and most any other reader here, would interpret that as *statistical independence*. I suspect that you are confusing this with *linear independence*, which is a very different thing. The application of the FFT will provide *coordinates* of the input time series in an orthonormal basis, so to speak of "independence" of the real and imaginary components in the sense of linear independence makes no real sense.
That would answer the question 'to guarantee with absolute certainty to get every number at least once'. For the question that was asked, the answer is a random variable, the distribution of which can be well approximated.
A nice exploration of different issues.
As far as I am aware, randomized sampling is not used to reduce bias, nor in many circumstances can it validly be claimed that it does reduce bias.
I and most people I know also pronounce the rope thing a "LAS-so". Then again I've never even touched a real lasso in my life so what do I know.
In this study, subjects are exposed to different drug and excercise combinations over time and the outcome of interest is the presence of a certain respiratory condition (yes/no). Patients are measured repeatedly every 2 weeks over 6 months. In terms of software, I use SAS and R. The 9 IVs where choosen by the investigator due to their scientific importance.
An intuitive explanation (in the bivariate case) is given on the GIS site at http://gis.stackexchange.com/questions/1553/how-to-interpret-grass-v-kernel-results/1589#1589 (where "half-width" is synonymous with "bandwidth"). It would be no trouble to adapt this explanation to the univariate case.
I agree, but as should be clear this is a small dataset. Note that quantile regression (L_1 norm) yields a loosely similar but not identical line.
In (ii) it appears that you have specialized the setting to the case $\\mathbf{R} = \\mathbb{R},$ the set of Real numbers. Here are some hints. Evidently you will need to exploit the assumed continuity of $f$ and, almost as evidently, you really only need continuity in a neighborhood of $0$. Can you find a simple formula for $\\int_\\mathbf{R} f d\\delta_{x}$ in terms of $f$ and $x$?
+1 very good question! I was asking this some bayesian specialists and they only say that in some cases (weights according to categorical covariate), you can compute the posterior distribution of parameter(s) for each category and then combine them in a weighted mean. They did not give me a general solution though, so I would be really interested if it exists or not!
@AndreSilva MAP = [Maximum a posteriori](http://en.wikipedia.org/wiki/Maximum_a_posteriori_estimation) - the mode of the posterior
You might use linear regression if the assumptions are appropriate; you might use GLMs; or you might use nonlinear regression; or you might use some other models.
You refer to regression in the title & use that tag, however in the body of the question, you seem to be discussing a classifier. In machine learning, people often think of regression & classification as being somewhat distinct. Can you clarify what topic you are thinking about? Are you thinking about both at a greater level of abstraction?
I took the liberty of formatting your post using the $\\LaTeX$ markup our site affords. Please ensure it still says what you want it to.
It would be of great help to link or define each specific term you mention, especially the model you are asking for (I don't know what $y$, $x$, constant, unit dummies, and time dummies you are talking about). To emphasize variables put *$$* around them.
One possible, although incomplete answer is that the conditional distribution over the total indeed changes in light of different evidence. The number 7 is the only magical number which probability will be the same given any evidence.
@PeterFlom I don't care about the coefficients. I care about the residuals, which I then plan to use in a SEM using package 'sem' in R, which relies on polychoric/polyserial correlations for categorical variables.
Thanks you very much to u too.
Your second bullet point can be read in (at least) two possible ways. I'm wondering which you're interested in. (**1**) The probability that the $n$th string matches one of the previous strings or (**2**) The probability that by the time the $n$th string is selected there exists *some* duplicate within the collection of strings drawn so far. The answers to these two questions will be ***very*** different. :)
Thanks Andy for your quick response.
good link, it's a pitty that it is not available freely.
ANOVA *is* regression. 
The problem would come in if you want to include relationships between your two DVs.
Yes, of course it will; but as you're working with 10 observations per group, don't expect very much from this setting; check for yourself at http://statpages.org/#Power
Gibbs sampling is a special type of MCMC.
upvoted for the reference to the paper and the book, I'll read them
Note that the answers demonstrate the assertion is not quite right: the distribution of RSS is $\\sigma^2$ (not $n-p$) times a $\\chi^2(n-p)$ distribution where $\\sigma^2$ is the true variance of the errors.
Addendum: Ok, I think I understand what you mean: Put the datasets in the same grid (in their original spatial allignment with the original grid resolution), but with lots of space between the points for each profile, right? That is a very very good idea, I will put that into execution immediately.
@James I don't see that anywhere in the formula and I don't think that's true. From http://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm : "Although the optimal choice [between mean and median] depends on the underlying distribution, the definition based on the median is recommended as the choice that provides good robustness against many types of non-normal data while retaining good power. If you have knowledge of the underlying distribution of the data, this may indicate using one of the other choices."
Well, in many practical situations you more-less know the environment, so optimizing selection of a classifier makes sense and do help.
As far as I know, these are the ones people should be interested in for a repeated-measures design, because they are the residuals from the model that we are applying to our data. The df from the anova = 4, so we would expect 5 residuals (only one variable is being modelled in this case, that represents the difference between the two groups, i.e. the within-subjects effect). I am sure the values are somehow mathematically related to the ones you calculated above.
Also frequently abbreviated ATT
@Glen_b But what does "easy to spin mean" for a probability distribution.
It looks like a mixture model: 2 tight distributions centered around X1 and X2 respectively. That's all I can say, since I don't know where the numbers come from. It doesn't look as if knowing a Y value helps you predict X - or vice versa.
Have you tried using any of the standard statistical methods for gene expression analysis?  There are a lot of methods out there and a good place to start is by simply googling gene expression analysis.
I made some edits and hope it's okay now.
My main questions (the two in bold) cannot be addressed using `anova()`. I want to see whether the log-odds in each of the conditions is significantly different from 0.
I think it is straight forward. If K(x,y) is a valid dot product (it is \\gamma(x)\\cdot \\gamma(y) in some space), $K(x,y)/\\sqrt{K(x,x)K(y,y)}$ is the dot product in the following mapping \\theta(x)=\\gamma(x)/||\\gamma(x)||
@MichaelChernick Lastly: I would like to find an analytical solution since I hope this is a well-known problem and people proposed flexible enough analytical solution. But I agree with our suggestion that modelling the 'real dynamics' is more important than the computational cost in general. Sadly this is for big data and a slow algo will be useless :-(
That's what I wanted to do, but the problem is that the automatic linear modeling output never actually gives you the model itself, with all precise coefficients etc.
Good point, and interesting links.
As a general rule, _any_ statement that includes probabilities conditioned on different events should be assumed to be false unless you have solid proof that it is true.  One cannot "mix and match" conditional probabilities conditioned on different events. One proven case where mixing and matching works (the exception that proves the rule?) is the law of total probability which in its simplest form says that $$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$$ but otherwise, assume all such statements to be false and you will not be wrong in most cases.
Multiple axes are nearly always misleading, and changing the limits of one or the other can change the whole meaning of the plot.  I think this was illustrated in some of Howard Wainer's work.
"and only one robust regression gives a sensible result." A better way to say that is that only one of the robust functions was designed to work with the 'confint()' interval..."Is it a correct way to produce a confidence interval for the robust regression model?" yes. "May the same method be used for the quantile regression model?' just use summary on an object produced by 'rq' (quantreg).
Adjusted R-square is another measure you can enter into the mix.  For each variable the percentage of variance it explains given the other variables in the model would be an intuitive measure of the variable's importance.  This would be a partial R-square.
are you looking for code? Many of the packages that make ROC curves also compute the CI.
From what I understand, it is best to not do LR comparisons with models fit by REML. In R's lmer, for example, you should set REML=FALSE when doing LRT. (It defaults to TRUE, which is otherwise better.)
Thanks. I worked through the first three examples there, but I don't really have a problem with understanding the contrasts and their interpretation when doing a lm().

I'm more confused about the relation between the coding matrix and resulting contrast matrix (see footnote [1] in my question). Perhaps it's more the linear algebra that's eluding me?
Variance is simply a property of a set of numbers. The existence of a variance does not presuppose that the variable was stochastic. You can take a variance of a variable fixed at prespecified levels. What you might legitimately conclude from the variance may be another issue, though.
Thank you. It was a valuable read.
@Andy W I believe he means "eligibility" in the sense that college athletes can play for up to four years. To the OP, it seems to be your measure, which is essentially "Average playing years left" seems as decent a way to measure it as anything.
@Darren Young: Correct, you don't need to square the variances again; but you do need make sure you are using [unbiased estimates of the population variances](http://en.wikipedia.org/wiki/Variance#Population_variance_and_sample_variance).
Darn; that makes it harder.  You could think of the outcomes as mixture of some small values (due to the obstruction) and then a distribution of "real" values whose mean is the parameter you're looking for.  It would be easier if the "obstruction" component were strictly a point mass at 0.  I'll think some more.
A less intrusive tweak would be adding to only those variables with this problem?
Welcome to the site and it's an excellent question to ask.  You'll need to help us with what is meant by "95% of precision".  Most likely, this might mean something like "we created a 95% confidence interval that for the true parameter, which was the range [lower, upper]."  That would mean you actually have six numbers: either three upper and three lower limits, or three point estimates and three standard errors or relative margins of error (eg +/- X %).  So maybe edit your question to be a bit more precise on exactly what information you've got.
The model you've specified - `lmer(S~time + Direction + Summit + time:Direction+time:Summit+(time|plot)` makes time a fixed effect (in the main effects, interactions) and, for each `plot`, there is a random slope in time and a random intercept. There are no other random effects. I'm not sure if that is what you intended. In any case, I think this question is more at home on stackoverflow, unless there is a _statistical_ question.
Could you add a link to the support docs?
I mean the observation come with measurement error. The covariance matrix of each error is known.
The argmax of $\\cal{L(1)}$ and $\\cal{L(2)}$, by dint of the words "maximum" and "likelihood".