tttjjj/elt_bench huggingface dataset:
@Jonathan, keep the holdout as holdouts and train the second classifier on the training data again. That is, Node 0 is just your training data. It may or may not give improved performance, but that is how you should proceed.
@Macro In the univariate case I think this method yields very larger confidence intervals than those based on tolerance intervals
Could you give us a better explanation of the context of your question?
thanks for taking a look, I really appreciate it. the help shows definitions for N, A and M and I wasn't sure what Md meant. multiplicative with dampening maybe? I will down sample this data set before it's used. thanks again.
Verify that the residuals of the model are uncorrelated and homoskedastic. All time-series auto-regressive models incorproate lags. It's also not clear that the series you are using is stationary.
So your numbers, 1304 133 6095 3404, are counts of the population inside your experiments? If you assume there hasn't been a significant change in your data over the years, you could segregate the population by year and you'll have 5 replicates per combination.
Thanks for the reference.  It's always fun to see early work (even when it was published in one's own lifetime, LOL!). The results are simple because they combine two simple results: (1) we know the pdf of the Normal distribution (which allows us to give a name to its cdf, the Gaussian integral,which has no closed form evaluation).  (2) the standard expressions for distributions of the order statistics of any pdf.  Thus you do obtain "exact" expressions for the order statistics, but in they end they must be polynomials in the Gaussian integral. (Computing them in 1961 wasn't easy.)
(+1) I gave a link to Robert Pruzek's article, but I didn't know it was available in R.
Are you perhaps trying to see if other variables in your data are related to the probability of someone having a job?
John, I've merged your 4 accounts (thanks to @gung who noticed that). Please, [register](http://stats.stackexchange.com/faq#login) your account once and for all so that you get all benefits from Stack Exchange system-wide facilities.
Since the OP indicates there is no censoring, the K-M estimator corresponds exactly to subtracting the empirical cdf from 1.
Great! So, as I understand, if I have some prior knowledge about $a$ and $\\mathbf{B}$ (maybe obtained with a previous sample), I can sample new points trying to optimize the Informational gain? Thanks a lot!
I don't know too much about it, but the way you describe the problem makes me think that you need [shrinkage](http://en.wikipedia.org/wiki/Shrinkage_estimator). Or to put it another way, you want a different [bias-variance tradeoff](http://scott.fortmann-roe.com/docs/BiasVariance.html) than the one you're getting from your current modelling approach. [Multilevel models](http://bit.ly/YilFES) provide a principled way to trade off larger bias for smaller variance, hopefully in a way that reduces overall mean squared error. Maybe [this paper](http://bit.ly/14WeapO) will help you.
The main objective is to see how variable the data around these anchors are. And if this variance is roughly consistent across various datasets. The choice of normals is just for convenience, I do not aim to have a super precise model or anything, more to have a tool to quickly describe my datasets in a couple numbers.
I added a more complete explanation at (http://stats.stackexchange.com/questions/28474/how-can-adding-a-2nd-iv-make-the-1st-iv-significant). What I'm suggesting is that your professor was addressing the modeling issue, not the mathematical issue. The mathematical issue is a "how", while the modeling issue is a "why". Look at the analogy I provided at the link.
When you can't write down a plausible model. Then the question becomes which systems are not amenable to statistical modeling?
@whuber I have added some stats. Is this what you had in mind ? I am sorry if I am misunderstanding you.
Really?? I've never seen any automated model selection algorithm that would return a non-invertible model.
Are you aware that PCA (*i.e.*, `prcomp`) applies to the correlation or covariance matrix of its input, which means in either case that *first the column means are removed?*  Thus, to make a valid comparison, you need to apply `svd` not to `x` but instead to `sweep(x, 2, apply(x, 2, mean))`.
@user2385 You got the idea, but don't forget that the 1% are distributed in both tails so that you have to work with .005 ($Z$ will be $<0$, $X$ will be $<68$) and .995 ($Z$ will be $>0$, $X$ will be $>68$), for the total area exceeding the limit equals 1%. No need to substract to the mean, you already know that $X=Z\\mu +\\sigma$. Just use the correct $\\sigma$ to go back from the std normal to the sampling distribution of your empirical mean.
Even assuming that 0/0 =0 would not be sufficient to assume that X/Y is Poisson.  The Poisson distribution only takes on integer values, so the ratio clearly cannot be Poisson.  What exactly do you mean by an empirical analysis?
Do you mean "Wallace's generator for normal variates (see e.g. TAoCP V2) and for uniforms on a sphere" are non-uniform random number generators not relying on uniform random number generators?
@Alecos Papadopoulos.I edit the question
But the underlying theory is in terms of covariances.  I don't do this manually, so it's harder to advise on that.  As for getting sums of squares, you should be able to call `anova(model)` to get that.
Have you considered natural splines rather than higher order polynomials?
You might be interested in this post, [How to transform ordinal data from questionnaire into proper interval data?](http://stats.stackexchange.com/q/27927/930), where I gave a brief overview of the use of latent variable models for ordered response categories.
No that should be enough; among economists that would qualify as "asymptotic" ;). But again, check first if you need robust SEs at all.
Have you got as far as reading the data into R to make a data table?
Added a massive amount of information for you, including the solution to an analogous problem. Good luck!
In a well-defined sense "most" distributions do not even have a PDF.  In another well-defined sense "most" distributions that have a PDF do not have a unique PDF (they can differ on sets of measure zero).  Thus this question will have interesting answers only for those who focus exclusively on distributions that have everywhere continuous PDFs, in which case there may be some room for debate.
Cyrus, it may be time to review some of your past questions and see whether any answer helped you to solve your problem. A good way to indicate the ones that were useful is to vote on them, and eventually mark them as accepted. This will help future users identify how to solve a similar problem. To do so, you will need to [register your account](http://stats.stackexchange.com/faq#login), though.
That's a sensible approach.
Actually tlo and thi might be more the CI taking the variance of the random effect into account...
Of *course* the transformation is not a bijection, because the components of $Y$ are not dependent: they must sum to zero and their squares must sum to $N-1$. But that does not imply the problem has no solution! Indeed, once you observe the $Y_i$ are identically distributed, you ought to be able to deduce a tremendous amount about them (because you can easily compute their first few moments).
What is your definition of a useful question? I am not sure if I can extrapolate your definition by the examples you gave.
Why not use maximum likelihood instead of GMM?
This is very helpful!  Thank you.
@Actually, I got the expression above. I am not sure this is what I am supposed to get. I don't think it will give me the target expression
Have you seen [this](http://rss.acs.unt.edu/Rdoc/library/TTR/html/MovingAverages.html)?
Thanks for your reply, I would rather stick to this comparison pre and post 2008 (with 08 included in pre as pointed out) ... what would you recommend? Thanks a lot.
Look for Googles word2vec which is probably the most successful attempt at that so far. But beware, you need a LOT (as in Google scale) of data for this to be really good.
Read SPSS Help, Tutorial and especially Case Studies.
I just wanted to be careful not to give more approval than I can; bootstrapping is complex. There's no code or details in the post, so ... I prefaced my remarks with that qualifier.
Thanks for your analysis. If my described accuracy in the graph also happens for _validation set $=$ training set_, where can the problem be?
Good idea wolfgang, I didn't think of it.
Can you be more specific about what you mean by "is bounded"? Has a finite expectation? Almost surely? What can the bound depend on? ($X$? the distribution of $X$?)  The difficulty of the question depends strongly on the type of bound you are looking for.
Nice explanation, it would be great if you could also add the frequentist approach to this problem so that it would be possible to compare the two :)
@user2370334 Good catch.  The argument is `xreg`.  Type `?forecast.Arima` for more info.
Yes, that makes sense. If the denominator is always the same, though, you could re-create the numerator from the proportion. Then you could do regular logistic. But ordinal may be just as good or better - depending on how you are thinking of the proportions.
This should be migrated to stackoverflow.
Thank you whuber
Look at the `bty` argument in `?par`, for example, `bty="n"`. You can pass this argument in your call to `plot`.
$\\sigma^2$ is the error variance. For purposes of estimation it is replaced by $s^2$, the mean squared error. That when the $t_{n-2}$ distribution comes in.
@soakley The question came to me when I read about conditional forecasting in my uni script.(its mostly about time series in this case, but also about regression)
Is $f_{z_i}$ the same as $\\phi(z_i)$? If so, may be clearer to just use $\\phi$ throughout.
See my answer to your cross-post: http://stackoverflow.com/a/18822701/1939754
It would help if you'd define the terms in your equations. Also, is $\\widehat{{\\bf B}}$ a maximum likelihood estimator (if so, under what assumptions) or..?
Do you mean that the healthy control has not baseline and follow-up? What is the hypothesis that is being tested?
Thank you for the detail explanation and the links. I'm still at the learning stage. Few months ago, I tried to understand Bayes factor but no avail in how to put the theory into pymc code(practical). So, the idea of using Bayes factor was gave up.
Just a history remark, the table originally comes from Sewell, Grandville, and P. J. Rousseau. "Finding groups in data: An introduction to cluster analysis." (1990). http://img546.imageshack.us/img546/4523/cnfg.png (maybe published earlier in one of the authors' articles)
Doesn't http://en.wikipedia.org/wiki/Kalman_filter#Unscented_Kalman_filter answer this question?
Hint: think about t-test's assumpion of independent observations.
@user12694 The test assumes the binomial which really falls out by assuming that the presence of the elements are IID Bernoulli with probability of occurrence 0<p<1.  Sometimes the normal approximation to the binomial is used but that is a valid approximation based on the central limit theorem and not an assumption of normality.  However the exact binomial test can be used (or alternatively Fisher's exact test).
I agree, this question needs another exchange site :)
+1 for thinking about this _before_ collecting data. Good on you! I already answered that MANOVA (http://en.wikipedia.org/wiki/Multivariate_analysis_of_variance) might be the way to go, but this presupposes normally distributed residuals, which your Likert scales will not provide, so I deleted my non-answer right away...
Origin was randomized, but all had the same orientation.
Love this one -- a wonderful bonus of being a statistician.
@Daniel: Hmm, your actual situation is a little unclear.  Maybe you could edit the question to say more about what your doing.  Anyway, I'm operating on the assumption that the references to normality may be a bit of a red herring.
Thank you very much for joining us to answer this question!  We are grateful to have such authoritative advice.
Your questions cannot be answered with the information given: you need to supply a formula expressing how $E[Y]$ depends on $X$.
Do $P_1$, $P_2$, $T_1$ and $T_2$ all refer to vectors of *exchangeable* values (exchangeable within a vector)? Or is there, for example, some sensible a priori pairing between the values in $P_1$ and $T_1$, so that the ordering of the values in each vector is *not* arbitrary?
To elaborate: we should also consider the difference between the pre-hoc and post-hoc probability. If you were to tell me in advance that you would roll 10101010, and you did, I would certainly be astonished. However, if you were to tell me that you did roll <any> k-tuple, and you did, I most certainly would not, as the chance of this event under the null is 1.
@Misakov, probably.
In what sense are you assuming a multivariate response distribution? Typically we think there is a univariate normal response distribution that has been categorized by replacing the value of the latent variable with an ordinal category based on where the value is relative to the thresholds.
@JohnK you don't need it. We have shown the density is equal $x^{\\alpha - 1}(1-x)^{\\beta-1}$ times some constant. Because we know that $g(x)$ integrates to 1 we can recover the constant $\\Gamma(\\alpha+\\beta)/[\\Gamma(\\alpha)\\Gamma(\\beta)]$. You can actually redo the whole thing keeping track of all the constants if you want, it might be a good exercise - you can get the denominator in closed form easily enough, i.e. That isn't the hard part.
@Néstor Which "that" are you asking about? 1. The significance of the difference between the P values is irrelevant (and might indicate that you are conflating significance tests and Neyman-Pearson hypothesis tests). 2. Yes, you do have a null hypothesis, but it may be false. Why does that matter to comparing P values? 3. No, I don't think that you can perform significance testing on P values, but you can make the relevant likelihood functions in many cases. Comparing P values does not need significance testing or hypothesis testing, and the original question does not ask for it.
I have replied to mpiktas, before Breusch-Pagan I check the series with: Phillips-Perron and KPSS unit root tests. Unfortunately, the series pass these tests. I also check the cointegration with Johansen and the series pass again(cointegrated). What could I do?
Thanks! However, AFAIK, for the ANOVA I would need a continuous response, which I don't... my response variable consists of ceroes and ones (binomial).
Nope. To do this all at once (take into account the professional group and calculate the association between your 2 other variables), you will need to use linear regression.
@LCialdella Alright Ill add a link.
At least one of your variables is ordered; you may get more power if you account for ordering. What do you mean by "which one"? Are you wanting to get 2x2 multiple comparisons? Or do you want Pearson residuals or contributions to chi-square (which kind of tells you where the main contributions to the large statistic come from but isn't a formal hypothesis test)
His example is more extreme than it needs to be.  It's possible for you to have Boolean variables $A, B$, and $C$ such that $A$ and $B$ are causes of $C$, and $C = A + B$ (mod 2).  Then, absent of knowledge of $B$, $A$ and $C$ have no mutual information.  $B$ is an undiscovered confounder — what you are calling "gremlins" even though it is something very common.
+1 for (whether it deserves it or not) :)
You could perhaps accept your own answer to show that the question did receive an answer.
Thanks for the answer. I have implemented the suggestion by @rbatt to introduce the random features and I will now explore how to do the same using a SARIMA approach.
@KennyPeanuts: Ah.  There are various conventions.  The most common by far is that $10\\% = (\\delta r) / r$, but multiples of this would not be surprising.  With three samples, you can use the sample standard deviation to get an estimate for $\\delta m$, though it's not quite as simple as the sample standard deviation.  http://en.wikipedia.org/wiki/Standard_error_(statistics)#Standard_error_of_the_mean
What you describe appears "reasonable" as a first approach but you need to need to be careful on how will you generalize your findings. Check the responses in the following link concerning Feature selection and Cross-Validation, I think you'll find them quite helpful: http://stats.stackexchange.com/questions/27750/feature-selection-and-cross-validation/
I believe the Prais-Winston autoregression assumes errors are AR(1). For SPSS command line syntax see the AREG command. Depending on the extent of the serial-correlation, ARIMA models might need to be considered.
My actual dataset is about 20,000 and i would break it up into less than 100 subsets, so that gives me a bessel-corrected std deviation of >0.9975.
Should I edit my question to reflect this?
@Mustafa. I am glad to have helped. If you liked it, you can accept it.
Waiting eagerly for the day when one can "test causality" in a cookbook fashion.
The RTs were behavioural observations using a video recorder.  Sometimes the behaviour occurred slightly off-camera and therefore an RT value cannot be made. In terms of the MRI, the design is like this.. 10 seconds prep, 12 seconds 'action', 8 seconds rest.  This is repeated 10 times in a block.  The RT eats into the 12 seconds 'action' - thus lengthening the prep time and lessening the 'action' time.  I want to do a comparison which represents this.  So the analysis will be looking at functional data from the new 'prep+RT' and the new 'Action-RT' times.
+1 I was about to post the same answer!
Maybe a duplicate: http://stats.stackexchange.com/questions/94845/how-to-estimate-confidence-level-for-svm-or-random-forest/94894#94894
$\\max \\sigma_Z^2 = \\sigma_X^2 + \\sigma_Y^2 + 2\\sigma_X\\sigma_Y \\leq 2(\\sigma_X^2 + \\sigma_Y^2)$ is true if and only if $\\sigma_X^2 + \\sigma_Y^2 \\geq 2\\sigma_X\\sigma_Y$. I think there might be something in my answer that can be used to show that the latter assertion does always hold, and I will leave it to you to figure out what it might be.
Facebook, Microsoft and Google (and many others large companies I suppose) employ a lot of visiting researchers, and full-time or part-time academics. I think it is these people (lured in from academia) and the virtually limitless resources of these corporations that primarily drive progress in the industry.
Thanks @Frank. To make sure I understand your last sentence: do you mean "ultra-low-precision accuracy" because, based on my example, using V1 and V2 only would collect ~50% of tha variance? I put that example just to explain myself based on some specific data, but not to actually do it in this case. Thanks for your comments.
... and similarly you can drop $U$ in your problem. Thus the question is about the existence of the joint distribution with specified "full conditional distributions" in the language of Gibbs sampling theory.
Thanks! If I would like to collect iid sample, those points in $X_1$ are not independent from each other. So here is how I understand: Suppose from a MC, I get $X_1, X_2, \\dots, X_N$ from each iteration of a single sample path of the chain (where the sample path length $N$ is sufficiently large), and the burn in period is $t_b$, and the sampling period is $t_s$. Is it correct to collect $X_{t_b}, X_{2*t_b}, \\dots, X_{t_s*t_b}$ as iid sample of the stationary distribution? So does the sample period $t_s$ mean the sample size for the sample that I need to collect?
Singularity means that one of your predictor variable is redundant.  How exactly you need to recode data, depends on what are you trying to achieve. I suggest asking a new question with more details about what your analysis is about.
I just finished this course and it is awesome! Also, it gave me a great start to understanding books on machine learning.
Thank you!  I had been trying drop, but missed the extra comma.
Pre-testing (as suggested here) can invalidate the Type I error rate of the overall process; one should take into account the fact that a pre-test was done when interpreting the results of whichever test it selected. More generally, hypothesis tests should be kept for testing null hypothesis one actually cares about, i.e. that there is no association between variables. The null hypothesis that the data is exactly Normal doesn't fall into this category.
The question itself really begs the question that you need to ask yourself which is why am I building this regression model? What will you use the model for? What do you hope to learn from the model? These are huge questions that will surely guide the steps you take in building the model.
@DrWho: The model for STATE1 is a Transfer Function Model because it includes causal variables ( # of cases plus the Interventions that were Detected ) and an ARIMA component (1,0,0). The model for STATE2 was similar but had an ARIMA component (0,0,0) which means it was a simple multiple regression model (OLS) . Regression(OLS) is a particular case of a Transfer Function.
Do your offers expire? That is, can you accept offer $X_1$ on a later day $n \\ne 1$, so that your profit is actually $X_1 - C_n$ instead?
Can you give us some information on your design? Usually, a prediction interval does in fact give predictions for all coming observations (with respect to its error probability, of course) - however, if it is the right approach depends on your problem and design.
If an authority tells you what to do then you have little choice. People, however, very rarely think exactly the same. It's more that the 1-2-3-4-5 scale did not have the resolution to capture the variation among the opinions. You will get a confidence interval of zero width, understating the inaccuracy. If I had to follow the standardized calculation I would put a remark into my report about this methodical weakness.
I used restitution because a direct translation of what we use in Portuguese. It seems that paid loss is more common in the literature, so I'll use with it from now on.
Also note that the statement of [Chebyshev's Inequality](http://en.wikipedia.org/wiki/Chebyshev's_inequality#Statement) gives you those probabilities, in general.
Gung,the paper is "Model Selection Criteria Using Likelihood Functions and Out-Of-Sample Performance", by Norwood, Ferrier, and Lusk. The url is http://ageconsearch.umn.edu/bitstream/18947/1/cp01no01.pdf
It looks to me like $A$ is a $1\\times 5$ matrix.
@Glen_b I would just calculate quantiles at each timepoint and make a timeseries out of those rather than use boxplots, I expect to see complex distributions though.
@DilipSarwate, with all due respect, Proof by authority does not impress me. The supposition that "hence" is "authoritative" is a "Straw-man" that I'd rather ignore. The level of detail in any given statement is commensurate with the inclination and/or the pedagogical significance of the same in a given context. I'd assume that a person who is asking "What is a standard deviation, how is it ....so forth?" may not wish to be burdened with rigorous mathematical definitions of the same. The simplification is deliberate and, let me assure you, not a result of not being aware.
[This](http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables#Second_moment) might be of some use
What constitutes 'bell-shaped' exactly?
Basically supervised LDA on a sentence-level, i.e. labeling sentences. The first approach shows that a response variable is created for each sentence conditioned on the label set. The second shows exactly the same model as described in the paper but the topic proportions are restricted according to the label set via the Dirichlet alpha.
@Joris Other links that might be useful: http://j.mp/bzetkQ, http://j.mp/ac20UQ, http://j.mp/a1UNRb. HTH