tttjjj/elt_bench huggingface dataset:
This approach would be appropriate to fix some of the coefficients of the ARIMA model (or of the external regressors). In this case, the parameter $\\delta$ is not the coefficient of the regressor, it is a parameter to generate the values of the regressor (the intervention variable). The question may mislead you because it mentions constrain $\\delta=1$. Nevertheless, I think it is usually more likely that we want to fix the coefficients of the ARIMA model rather than the external regressors. In the latter case we could simply fit a model for $y_t-\\kappa x_t$, being $\\kappa$ the fixed coeff.
Bayes theorem is introduced in 1.2, Bayesian linear regression in 3.3, graphical models (which are also Bayesian) in chapter 8, and sequential models (including particle filters in 13.3.4) in chapter 13.
I was troubled with something similar recently, but was able to run it to sample sizes up to a few thousands using Wallenius' Markov approximation. More info [here](http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2370823) and in his original PhD thesis and [Agner Fog's work](http://www.agner.org/random/nchyp1.pdf).
What would be "too high" a correlation so that it becomes "troublesome?" This seems subjective. Is it troublesome for the model itself? For just the interpretation?
A threshold-$\\eta$ rule using the likelihood ratio is a Bayes classification rule for prior $(\\pi_1, \\pi_2)$ where $\\eta = \\frac{\\pi_2}{\\pi_1} = \\frac{1-\\pi_1}{\\pi_1}$. Are there any admissible rules that are _not_ threshold tests on the likelihood ratio? If so, they would be a counter-example to what you are asked to prove.  On the other hand, if all admissible rules are threshold tests, then you are done.
I would recommend the great book [Hierarchical Linear Models: Applications and Data Analysis Methods](http://books.google.com/books?id=uyCV0CNGDLQC&printsec=frontcover#v=onepage&q&f=false) by Raudenbush and Bryk.
Thanks for this! Hamming distance in this case is just $L_1$ in $\\mathbb R^J$ restricted to the cube verticies; in that context, Hamming distance is acting isotropically. Getting away from that I guess complicates these things because I have more than $J$ different values for my distance measure? Any general comments on this?
I can't check right now (with SPSS but no SAS now, ironically), but the percentages you want come as default (e.g. [1](http://support.sas.com/documentation/cdl/en/statug/63347/HTML/default/viewer.htm#statug_factor_sect004.htm), [2](http://www.yorku.ca/pek/index_files/quickstart/FactorQuickStart.pdf)). My guess is your nfact and mineigen statements might have suppressed them. Start from a simpler command and see if that's true.
"ADF and PP seem to detrend the series before starting the tests". I don't want to sound brash, again, but indeed, when you carry the regression of y_t on f_{t-1} with y defined as above, yes that (removing trends by differentiation) is what you do --but not 'before starting the tests' as you write, no, this is part of the test. Again, i don't want to sound brash, but you should pick up a book on time series; i'm not sure that cargo cult is the best approach to statistics.
OR ANY IRT Model
An easy way of looking at it is: if gender is a confounder it will have a significant coefficient. Sometimes the gender effect is a proxy for other confounders such as age and then the gender coefficient might turn out not to be signinficant when you've added the age.
+1 for the link to the previous question on the subject in S.E. (thaks for the pointer to agrep()).
I believe this answer confuses *population* with *sample.*  The question concerns data about *entire populations* (the cities), not samples from those cities.  The point of the question is that when there is no association between safety and location, each city can be viewed *as if* it were a random sample from the entire population of drivers in all 200 cities.  The variance of the mean will be higher in smaller cities.
For Kullback–Leibler divergence I need a distrubiton per linear model right? How do i get this. I found ~AIC = exp((AIC min - AIC i)/2) on wiki, to see if a model has more info then the minimum AIC, but it doesn't say what 'number range' we look for in a model to select it. I have created 10 cross validated models using LASSO just for different 'fractions'. The link you provided explains I need to cross validate it again but I'm unsure how to implement this in R (assume all models are in linear model form). Infact reading the link again it seems I cannot pick a model?
My other advice is 'you must read more than one book'. Get a mix of books pitched at different levels and with different mixes of rigor and applied-level guidance.
Thanks, this was very helpful. The link you provided also was a big help.
Isin't Hotelling's T2 test used in such contexts?
AS an extreme example,with 4 evenly distributed classes of documents, { A,B,C,D }, one way that the classifier could be 70% accurate is with the following class accuracies: Class D - 100% correct, Class C 100% correct - Class B 40% correct - Class A 40% correct. The overall accuracy is 70%, the accuracy on class A is only 40%. The distribution of overall classifier errors amongst classes is important, as it directly affects the class accuracy.
Readers here might also be interested in reading this thread: [period-detection-of-a-generic-time-series](http://stats.stackexchange.com/questions/1207/).
...as for your second comment on Assumption A2, I guess it means that the whole maximum likelihood framework does not quite  meet the needs of your field, except perhaps when the distributions involved have log-concave densities.
You are approaching this the wrong way.  A graphical representation is not always the best way to look at things.  Variance is just the square of the standard deviation, which you already understand.  A better question is: why is the square interesting enough that it has its own name?  The answer to that is that variances are [additive](https://en.wikipedia.org/wiki/Variance#Sum_of_uncorrelated_variables_.28Bienaym.C3.A9_formula.29) (while standard deviation is not).
from summary.lm():         ```ans$adj.r.squared <- 1 - (1 - ans$r.squared) * ((n - df.int)/rdf)```, where ans$r.squared = R^2; n = n, rdf = residual df, df.int = intercept df (0 or 1).
Thanks for the answer! Do you have some good references in this context for me? A good book or a paper.
Sorry, yes .. and I did.. thanks
21 stocks,  Observations: The average price per trading session of each stock for 496 trading sessions (3 sessions a week), so 3 years (496 trading days). standard error = standard deviation / SQRT(N) so annualizing any of them would do the task to get the other,
You could always download R and look at the source code of `arima()` for inspiration.
Thank you Greg. I assume there's nothing wrong with the procedure per-se? I understand that I might not be asking the right question, but my concern is equally about the statistical test/procedure and understanding itself given two samples. 

I'll check if I am asking the right question and come back with questions, if any. Maybe if I explain the biological problem, it would help with more suggestions. Thanks again.
This lead me to relevel which is what I ended up going with
You really have to define what 'better' means for you. Besides which has the better mean, you might for example be interested in 'which is better more often' ($P(y_A<y_B)$) against a null that the probability is $\\frac{1}{2}$. I believe this corresponds to a sign test on the differences. But you may actually prefer some other measure; there's not enough detail to even really guess what 'better' means here.
I use R and I will check it out. Thank you for elaborating.
@Daniel: I suspect it doesn't matter if it's normal or not. If you want to know how normal it is, generate N numbers from a normal distribution, sort it, and plot against the numbers in your table. If your table is normal you'll get a straight line (QQ plot). If not, it will not be straight. Useful thing.
Might be some help in here: http://stats.stackexchange.com/questions/26016/sample-size-for-logistic-regression. If that still doesn't work, you could use a Bayesian model, which is typically more stable in small samples, or a regularized regression to improve the effective degrees of freedom (which is a special case of Bayesian regression anyway)
@chl I am not completely set on using a Welch's t-test. I'm trying to find out what the best test for this situation would be. On [this wikipedia page](http://en.wikipedia.org/wiki/T-test) it says if you have unequal sample sizes and unequal variance that you should use a Welch's t-test. Maybe I don't understand what constitutes unequal variance, all I've done is calculate the variance of the control and the treatment and they are "different" (7.12 vs 6.02). But maybe I'm going about that the wrong way.
I found someone who has addressed the Monty Hall problem in SAS and R here http://sas-and-r.blogspot.com/2010/01/example-723-monty-hall-problem.html -- enjoy
Could you please clarify the question you would like to obtain from the community, and give more details on the process you are trying to model (where binary outcomes come from, do you have a time series process (polynomial), covariates to regress with)?
Your algorithm tens to give more money to the first person than to any of the others.  There are other approaches which do not have this bias.
I have to admit I still haven't figured out how MCMC works, which is one of the reasons I opted for bootstrapping instead. While bootstrapping should be possible with random slopes, as you intimated, it may be that  pvals.fnc doesn't let you do CIs for models with random slopes because this is for some reason invalid, and further it may be that this invalidity extends to bootstrapping such models. I don't intuitively think there would be any problem with bootstrapping, but that may be a function of my limited expertise.
I did not affirm the conclusion: I merely observed that the two values drawn are *exchangeable*. If this is not perfectly obvious, imagine drawing the two members of the sample without looking at them, and then observing them in random order. A mathematically rigorous account of this is that every two-element sample $(x,y)$ from a set $S$ is in a unique one-to-one correspondence with the sample $(y,x)$. When you assume all samples are equally likely, then clearly the chance that $y$ is the second element drawn equals the chance that $y$ is the first element drawn.
Assuming a uniform prior, the answer is 848437653385740282130263/1167635946155025951609137 = about 72.66%.
This is in part asking us to elucidate a tutorial (please give a full reference or a link) and in part asking what PCA does. There are many threads here on PCA; have you looked at any of them? There is no sense in which PCA purports to eliminate the correlations between the original variables; it is designed to make use of them. Also, reduction of the number of variables is a user's choice, not what the method does automatically. Either you use some of the PCs or you use PC results to decide which variables should be used in some other method.
Thank you, I am so much grateful
(+1) Thanks for mentioning the BA plot. You might be interested in an earlier response I made on a similar topic, http://stats.stackexchange.com/questions/527/what-ways-are-there-to-show-two-analytical-methods-are-equivalent/2834.
Thanks for making an official summing-up of the comments (+1).
+1 Thanks for answering. We won't be able to spot problems like that if we can't see the form of the likelihood.
**Note** that the way you resolve the succesive line estimation results in different algorithms. Possible options: 1) the line estimation remains the same (from first 2 points), this is better if you know positively that the data fits a line very closely up to a point, or equivalently can be modelled as linear up to some value. 2) Re-estimating the line, taking into account the new data point (either by an average or least-squares), this may over-fit sometimes. 3) Always use last 2 points. 4) Always use 1st and the last point.
However, you could probably learn something by looking at a histogram of the second sample. For example, if the sample is roughly normally distributed, the delay is large relative to the variance, and a non-negligible proportion were subjected to the delay, then you should see a bimodal distribution. The relative heights of the two modes should give you a good idea of what proportion of the points were delayed. To actually estimate that proportion in a principled way, you'd probably need a mixture model.
Just to make sure we're on the same page : I mean linear thresholds in the *feature* space not the output, i.e. if you plot the data in the 5 dimensional space where the input vectors you are training on live (if I got you right), you want the different classes to be well seperated by linear thresholds
@ayush Yes, as far as I see from here (if the predictor/predictor information is always available). Does the meaning of the predictor does not indicate this ? 1. As I said, I recommend to talk to someone with domain language, if it is not instantly clear. 2. Although you hit 60%, you still miss 40%. So your precision is solid, but not excellent. Depends if this is enough for your business.
No it's not inappropriate, it just assumes a quadratic shape.  It requires 3 parameters whereas the most flexible mean model with 4 treatments requires the same number.
The need to obtain a test set using some form of randomization is basic, but pointing that out does not answer the question. It can be answered provided the classifier's results are tabulated. That is, for each pair $(i,j)$ of the possible categories you need to record the number of individuals in the test set who were in category $i$ and classified into category $j$. This gives sufficient information to estimate (and provide confidence limits for) the proportions of each category in the parent population as well as the likely range (and types) of misclassifications that would occur.
@Neptune, General answer to this is: toolboxes and modules for structure learning, parameter learning, and inference for BN's with continuous nodes is a very under-developed topic. The vast majority of BN applications discretise their continuous values.
@MilanAmrutJoshi I've rewritten the sum part also. Basically, a derivative of sum is sum of derivatives so usually you just "ignore" the summation writing it on the RHS and focus on differentiation of what's inside the sum.
When you use a function from a non-default package, please state which package you're using. Some function names occur in more than one package. Are you using the `perturb` package?
Proposition 7.9 of Bilodeau and Brenner (the pdf is freely available on the web) gives a promising result for the Wishart (perhaps it carries over for the inverse Wishart). If you partition $X$ in blocks as $X_{11},X_{12};X_{21},X_{22}$, then $X_{22}$ is Wishart, as is $X_{11} - X_{12}X_{22}^{-1}X_{21}$, and they are independent.
In the future, there are lots and lots of questions like this already asked on the stan-users mailing list. Might be worth checking that one out.
@varty: This is the alternative notation to ${2k \\choose k}$, used for instance in French textbooks.
Thank you. I think I can understand more about it now, I'll try more.
Sure, I've added an example of what the data looks like. For "growth" I have measurements from 16 plates per strain. For "virulence" I have measurement from 13 plants per strain. So I want to correlate strain growth and strain virulence but I need to take the within strain variance for both variables into account.
Polynomial regression seems like the obvious choice here, given that you are willing to consider you $x$ measurement error negligible.  A "quick and dirty" approach (the first that came to my mind) is to do ordinary least squares of $y$ on $(1,x,x^2,x^3)$.  Then just use standard prediction formula from that.  Perhaps I am missing something?
Honestly, I can't make it out how is your answer related to the OP's question. I see no connection between the adjusted R-sq and the question.
@probabilityislogic thanks so much for pointing that out! Yes, I think the correct expression should have been $p(y|\\theta) = \\int p(y|\\theta,\\alpha)p(\\alpha|\\theta)d\\alpha$ which contains an unknown
@IrishStat I would disagree if the residual model were stationary.  However, since exponential smoothing is nonstationary and the mean drifts the trend in the residual might have better been handled in the regression piece.
Can you say a little more about your situation? Are you trying to detect if A & B are the same actual person? Is this something like you have two sets of lists, but the names were left off, & some of the information is missing? Eg, do you know that A *doesn't* know Ryan, or is it that you don't know *whether or not* A knows Ryan?
Can you clarify what you mean by your last sentence? I cannot understand it.
Are you trying to model a structural change in the data? If so, then the package [strucchange](http://cran.r-project.org/web/packages/strucchange/index.html) might be appropriate.
I second that: I recommend moving to stats.SE.
@Stéphane You are correct--I misread that section. Unless the statement is changed to refer to the *laws* (as you point out) instead of the RVs, or to change "sum" to "mixture," then it is indeed wrong.
Thank you for the warning. I'll take that into consideration when I decide to use either the K-S test or the Chi-Squared
Nice job explaining some possibilities.  You had a difficult task with the research question(s) left undefined.
Unless "observation" has some order (which it usually does not) then you should not draw a line plot like this. It implies that "observation" isn't just a label.
All the answers to your questions are in the text above, I think you may have mis-read it.

I am doing 2 FAs; one with complete-cases data  (i.e. list wise deletion; 370 cases have complete data) and one with imputed data (370 cases have complete data, 280 have some imputed data for a total of 650 cases). I don't know what you mean by 'gross n' in my incomplete datasets.
'*I would have loved to refer to this process as "normalization"*' --- I think 'standardization' might be less prone to confusion in a statistics context and has some precedent.
If you give the money out one by one, there are many ways to distribute them evenly and many more to distribute them almost evenly (e.g. a distribution which is almost normal and with a mean of $50000$ and a standard deviation close to $224$)
Reporting possible negative number of visits per years in the descriptive statistics looks a bit odd to me "1.2 visits per year (± 1.5)".
You may want to read this answer to a seperate question and see why adjusting p-values in such a manner may not be the best solution, http://stats.stackexchange.com/questions/3200/is-adjusting-p-values-in-a-multiple-regression-for-multiple-comparisons-a-good-id/3317#3317
Found by myself: 'D' is a given function and it raise an exception. I changed the name and using a new function with different name a needed to create it in .GlobalEnv (I know that it's not optimal) and it runs.
Please tell us what you mean by the "order" of a statistic.  Do you mean its asymptotic behavior as $n$ increases?
(+1) .. good question. The Bayesian Central Limit Theorem is called the Laplace Approximation ie the posterior behaves "more or less" like a normal distribution. (formally posterior converges in distribution to a normal distribution)
@gung i updated the question
It would help if you describe your current mathematics standing. Three semesters of calculus? BSc in math? Any measure theory? Any matrix algebra? Any real analysis? Any programming experience?
While looking for other names in Wikipedia, I stumbled on the fact that Box was a son-in-law of Fisher's.
@Frank is right about ordinal logistic. About model selection, I don't think you should test all 255 models. If you wanna try all combinations, why stop here? Why not try transformations of variables (log, square root etc.)? Why not try interactions terms? And quadratic and cubic terms? Do you see that in the limit you will end with more models than degrees of freedom? Try to use theory to assess which variables to include and look at descriptive statistics to assess visually if it does make sense to include or not a variable.
I compute feature vectors (color, texture and shape) and obtain similarity scores for each, sum them up for a total similarity score, then sort descending. the top 10 image indices are the most relevant ones. I can obtain the class index from the image index since the images are ordered (20 gorillas, 20 giraffes etc.) I hope I made myself clear, since I don't fully understand the concepts classifier / descriptor etc.
Hi @zaynah and welcome to the site. I'm not sure if your question is *if* your data are compatible with a weibull distribution or what the parameters of a weibull distribution describing your data would be. If you assume that your data follow a weibull distribution and want to find the parameters, you can use `fitdistr(mydata, densfun="weibull")` in `R` to find the parameters via MLE. To make a graph, use the `qqPlot` function from the `car` package: `qqPlot(mydata, distribution="weibull", shape=, scale=)` with the shape and scale parameters that you've found with `fitdistr`.
Thanks to the m1 and m2 are nested to m3, likelihood-ratio test is highly recommended. AIC and BIC are better to compare models where they have same number of predictors. So, given the AIC result above, I'd think the LRT between m1 and m2 vs m3 may be not significant and then m1 < m2, and thus the m1 is the best model. That's something like parsimonious rule about.
Have you tried something like: 'r1 <- lm( I(volume)^(2/3) ~ deceased * sports * smoker, data=X)
summary(r1)' ? (To add interaction terms). Also, it would certainly be useful to build a linear mixed model (package 'nlme', function 'lme'). And be careful about the syntax for the power operator '^'.
Glen_b, can you elaborate? For what initial values this high BP measure of scale (like MAD, I take it?) is estimated for? In the section for fixed X matrix it is stated that they use the L1 estimator, then count the MAD of L1 residuals, and this value is inserted into the equation. Is that correct?
Does this mean that et = yt - u ? Does et need to come from a differenced series ? Or a stationary series ? is this talking about the change in yt or the actual value of yt ? If I make up a pretend series that follows a perfectly predictable pattern of 2,3,4,2,3,4,2,3,4 and use an order one model, should this method predict the series to continue ?  Apologies because this is probably confusing and not using the right terminology I'm just trying to understand the basics a bit better. Thanks.
I've been lookinhg for another program that works like ibm data generator... and i found ARTOOL [link](http://www.cs.umb.edu/~laur/ARtool/)... it's interesting and works. but i'm still trying to understand the IBM genetator...
@all -- can we take this post as an opportunity to discuss the interpretation of probit and CLL in some what detail..I'll state the CLL formula here and start it. log(-log(1-pi)).
https://github.com/jbryer/irutils/blob/master/R/likert.R
is the source for the stacked bar charts you want.
I considered it.  However there there are enough  other papers that claim to use the same statistic that it seemed more appropriate to discuss the point here if only to provide a web resource for those who have a similar question.
Many thanks, atiretoo. This raises another question. What's the difference between display() and summary()? It seems to me that the output from the former is just the output from the latter, less two columns, and rounded to 2 digits. It would appear so, from Gelman's post you've linked above.
Hi Adam, to clarify further, were you wanting to use the visualizations to show differences between the groups? If so, that is not a recommended method.
Thanks. This is an alternative way to an additive perturbation. In this solution there are no worries about negative values..
@cada A CFA with 45 subjects would be surrealist :-) I'll try to add my thoughts later.