tttjjj/elt_bench huggingface dataset:
This is a nice paper on MI using chained equations http://onlinelibrary.wiley.com/doi/10.1002/sim.4067/abstract
My first intuition is all things being equal, there is no effect.
The principal components can only be computed for the set of data that is included. How it would be affected by knowing the correct values for the missing data would depend on how the data became missing (i.e. MAR, MCAR or non-ingnorable).
I agree with @NickCox, it just depends if you decide the scale can be considered linear. If it can be, regression interpretation may be simpler. However if your purpose is purely test score prediction it is probably best to use dummy variables anyway
If you don't **show** us the results of your test--how you ran it and what it said--then how can you expect anyone to tell you how to interpret the results?
Thank you very much :).
Is 'proving it' related to some subject? For the purpose of private study? (By 'nothing to do with' above I mean it's a result that always holds, so you don't need to refer to the distribution.)
np. I was actually just inspecting y as you were writing your answer and thinking "I was not expecting to see 15 columns for m<-3..Should be choose(5,3) =10". But your edit fixed that. This is way cool! Thank you for very much!
(cont.) and many variants have a rich theory and plentiful applications. (3) As far as I can tell, I think all the commenters have an acute grasp of the terminology. What I think you'll find as you continue to contribute here (!!) is that one of the main challenges is to tease out the (actual) question of interest and the attendant level at which an answer is to be pitched. One nice thing about this site is that we get questions from a very wide audience; respondents are generally very conscientious about identifying and trying to help clarify points of confusion.
Welcome anna. Could you be more precise? For what parameter is the confidence interval? Usually, if the confidence interval does not include the null value, the result is said to be statistically significant. If the 95%-CI does not include the null, the corresponding $p$-value is < 0.05 and you have evidence that the corresponding parameter is different (in your case smaller) than the null value.
Do you assume that the player always declares the Joker to be of the suit for which she has the most cards? Do you know that such a policy is optimal for your game?
Our esteemed colleague Onestop has a pet peeve about all caps text; I have one about calling something a "Likert scale" when in fact it is only a survey item involving an ordinal rating.  I don't like the way Likert's substantial contribution gets overlooked, or changed into something trivial.  See http://stats.stackexchange.com/questions/10382/general-advice-on-forming-an-index-of-attitude-to-the-environment-from-a-set-of
@BJessop, I have tried to provide a more detailed explanation for the two group case.
Added a link to a histogram and a potential null hypothesis but without a specific level of significance
Are you talking about markov chain ? http://en.wikipedia.org/wiki/Markov_chain
From what I understand  you look for stationary distribution.
Thanks a lot. Now I assume something will change in my estimation process. As a first step, I should estimate the covariance $ \\mathbf{\\hat{\\Sigma}} $ with your procedure, then my distribution given the estimated hypothesis woulb be $ \\mathbf{P} (\\mathbf{X} | \\mu, \\mathbf{\\hat{\\Sigma}} )$ and since $\\mathbf{\\hat{\\Sigma}} $ is estimated and has its own distribution I am pretty sure this will somehow change my previous formula to compute $\\mathbf{\\hat{\\mu}_n} $ (as it happens on gaussian MLE when using the sample variance).
An alternative (and more easily accessible) link to the original paper should be [here](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aoms/1177730256). Note that the tables given there are quite extensive. The $L(z)$ values correspond to $(1-\\alpha)$ and the $z$ to $c(\\alpha)$.
The model was built on 80,000 individuals and is being used to predict an out-of-sample 20,000 individuals. I don't think sample size is a concern. I suspect it is because it is a very random process to model and the aggregated probabilities are just reflecting the (fairly stable) base rate.
NO! The syntax for formulas with `aov` is always `dependent.variable ~ independent.variable` where `independent.variable` needs to be a nominal factor (see `?factor`). And please edit your question and do not post questions as answers.
+1 That's a clever approach.  Note that it works in the example because the SD of [length] is a small fraction of the mean (about $22.3/188 \\approx 12\\%$): logarithms of values $1+x$ in the range $1 \\pm 2 SD \\approx (0.75, 1.25)$ are very close to $x$ itself, whence the logarithm is approximately a linear function within this range.  When any one of the independent variables exhibits appreciably more relative variation than this, all bets are off.
I remember it is implemented in a Perl module available in cpan. Otherwise i do not know any other. Nevertheless, if you have an implementation of nb, you can just add the correction for skew classes to your code
What's inappropriate about least squares estimation when the data are non-normal or heteroskedastic? It is not fully efficient, but it is still unbiased and consistent.
Feedback is used irrespectively, right or wrong.
Thanks! It is solved now, though with **hist** only.
As regards this code below which works well, may I ask you to exlain how it works? i.e. Does this compare one factor level (coded as "1") against the the other two factor levels (grouped by coding as "0")....  or does it instead run a signifcance test only on the slope classed as a "1" (and the "0" are ignored), so that in effect this code tests each slope independently (hence three slopes = 3 tests)?        Also, are the levels of factor organised alphabetically, so that the "1" in (1,0,0) refers to herbivore, while "1" in (0,0,1) refers to predator?
Everything hinges on what is meant by "best".  That, in turn, should depend on what use will be made of the estimated PDF.  Will you be testing a scientific theory, predicting a future outcome, using the PDF as input for additional analyses, advising people what action to take?  Do you need to reproduce the mean and variance accurately or perhaps (if you are doing risk assessment or failure analysis) do you need to characterize one of the tails and extrapolate with it?  Without this information, we can only guess what answer may be appropriate.
Actually, `choose(1000,10)` works just fine in R for me. If it hadn't, in R I'd just use the `lgamma` function, but more generally, [Stirling's approximation](http://en.wikipedia.org/wiki/Stirling%27s_approximation) would usually work okay for an approximate answer
Thanks for this piece of code @COOLSerdash. I had to get rid of some duplicates and NA's from my data set to get this to work, but it appears to be working now.
In addition, given the covariates are covarying with X it's best that's it's the response variable and not a predictor.
Well, as stated in the question, this is exactly the point: I know that the data comes from two distinct processes, so there is no mixture. You can think of it as a piece-wise pdf. A mixture model would be more like a joint pdf, where both pdfs are active through out the whole range. In a piece-wise model each data gets its likelihood strictly from one pdf only.
@whuber Quite right; and to think that I've spent a few decades emphasising precisely that in introductory courses, but omitted to mention it this time....
Ed Rigdon's page is http://www2.gsu.edu/~mkteer/. To make this answer even more complete, I would add that SEM with observed variables only have been first studied in econometrics in 1950s or so under the title of simultaneous equation models. Ken Bollen's excellent book (http://www.amazon.com/Structural-Equations-Latent-Variables-Kenneth/dp/0471011711) is actually one of the few books on SEM that covers it in enough detail to learn something useful.
I don't know the answer to this question, but simulation might give you at least an idea of which method is the "best" (whatever you mean by _best_).
A Binomial needs 2 quantities: the parameter representing the probability of a 'head' p, and the 'number of tosses' n.  Its 'response' is (equivalently) either a number k between 0 and n representing that k 'heads' appeared in n 'tosses', or a pair of numbers [k, n-k] indicating k heads and n-k tails. (see http://en.wikipedia.org/wiki/Binomial_distribution). When n=1 this is a Bernoulli distribution. You'd be saying: "in each of my cases there were n=iso1+iso2 'tosses' of which iso2 came up heads and the rest tails. Larger n makes me more sure about iso2/(iso2+iso1) as an estimate of p"
@Tal, I don't use `rpart` and I will leave the answer to more experienced members of this forum.
Can you include code to reproduce those toy examples here?  It would help in exploring this interesting behavior.
Thanks. But could you elaborate why? It seems to me that the 24 responses made by the subjects are correlated, and so their residuals should be correlated too.  Is this correlation somehow accounted for by the mixed effects model? How?
@image_doctor yes, I've adapted some ideas of "only-positive and unlabeled (PU) learning" to end up with a set of unknowns that I can pretty safely regard as "likely negatives".
(I've updated that part of the answer: the problem I refer to is already the variance. The Elements give a short discussion, but not about the finite test-sample-size for classification.) The Elements say in 7.10.3 that if you do the CV correctly, it is unbiased or slightly pessimistically biased: i.e. on average you are OK. They just mention the variance in one sentence. But that variance is exactly what you need to look at for your question: that variance means that indeed by chance you can observe high accuracy even if generalization power is low and you estimate is unbiased.
+1 Although there is more that can be said, as you intimate, this is a nice start. (I don't want to leave you hanging with that cryptic remark, so I will be more explicit. One direction this conversation could go in would be to point out that any randomized sampling procedure is typically subject to greater uncertainties ("unreliability") than the output of a well-done MC calculation. This has a bearing on question (2).)
Although the use of Feynman's trick is interesting, why not just integrate by parts to get $$\\int_x^\\infty t\\lambda e^{-\\lambda t}\\,dt = -t e^{-\\lambda t}\\biggr |_x^\\infty + \\int_x^\\infty e^{-\\lambda t}\\,dt = \\left(x + \\frac{1}{\\lambda}\\right)e^{-\\lambda x}?$$
A very poorly worded question. b) and c) are quite vague. The answer is actually e, I think. Although what b) is is not that clear.
I don't know, I posted it on theoretical CS site, I got voted down. And they refer to this site
Although I have no idea about how it will scale with such huge number of data points, you might want to take a look at [Cubism.js](http://square.github.com/cubism/).
Re weights: if feature A and B are both predictive, but strongly correlated they may have relative low weights (weight shared between them). Re drivers: from observations you can only get associations, not cause-effect relationships; however your knowledge on the mechanisms may help you decide what are causes, effects or just correlates among the associations.
@CJH I think we can reject Option #2 - standardizing the response variable doesn't make sense, only the predictor variables ought to be standardized. The aim is to reduce multicollinearities between the predictor variables due to scaling, not reduce collinearity between the predictors and response variable.
I think you have some big troubles on you data. What hypothesis are you trying to test?, what is your response variable? what does life cycle means? (I am a biologist but it is not clear in your data). You have many categories on the "variable" groups (some of them are non-sense), what about replicates?, as stated you have no replicates?. Please REVISE the WHOLE DESIGN of your study. It is not clear at all.
Let's consider p1 and p2 and imagine they are approximatively equal, but with n1 >> n2. How can I compare both subpopulations by using this information and, precisely, is there a minimum size of n2 to compute the proportion of success with confidence ? I would like to cut the node, for which the population size is so small that we can't trust the proportion of sucess. –
May be you could solve it for binomial first and then move to multinomial.
The quoted sentence is not something you can conclude from an Anderson-Darling test --- failure to reject normality *does not imply the data follows a normal distribution*. So making people understand an incorrect conclusion might be counterproductive.
Okay, I see. You cannot have both cov3 and cov4 (as you have defined them) in the model at once, because one way or another you are going to end up with one column of the design matrix being a perfect linear combination of the other columns... as you say, the model is overparameterized. You can verify this by attempting to do it and noting what happens (the model will fail with an obscure-looking error message). The solution is to remove cov3 and just use cov4, giving you essentially the set of codes that I posted, although with different column names.
The $best$ libraries can be highly subjective to your problem. The 'toy' mathematical programming problems you show here, should be easily solved by most off-the-shelf libraries.
You're right. I will have to think this problem through a lot more.
@Bruder I await your contact . You might check out my contact info. I don't necessarily see a role for intervention detection but rather incorporating possible canibilizers as predictor series in a Transfer Function. Annheuser-Busch used our software to relate sales of a product in a particular size as it was impacted by sales of the same product but in different sizes . They also included the price of the possible canibalizers in order to assess the effect.
Are you using many dummies in your model? The multicolinearity is harming your model. Being parsimonious would help. Best
I don't see any problem w/ resurrecting an old thread when it's appropriate, @Trevor. I think what you are looking for is something along the lines of this nice answer by conjugate prior: [does-an-unbalanced-sample-matter-when-doing-logistic-regression](http://stats.stackexchange.com/questions/6067//6086#6086).
There looks like an interesting question here, but "how good is this article" isn't it, because it's too vague and subjective. Could you please reformulate your query?
+1 Excellent post. "*how important personal responsibility for statistical scrutiny is*" -- I must applaud. Ultimately, this is where responsibility must lie, as onerous as that may be for someone already trying to get work done in an area of research to which they wish to apply statistics.
As long as both Y and $X_1$ have *any* relationship with time, any graph or equation that does not account for time will be misleading.
"... and when everyone's a super, no one is" - Syndrome, in the Incredibles.  (http://www.youtube.com/watch?v=A8I9pYCl9AQ)  There is not enough fame to go around.  If everyone has an equal share then there is not enough for the demand.  It is an economic conundrum that requires you to either become truly different and exceptional, to become a predator, or to fail.  "'Cause we all just wanna be big rockstars, And live in hilltop houses driving fifteen cars.."  Media does not have the answer - it is also peopled with predators and fails.  Exercise Caution.
I suggest backing up and saying much more about what you want to do and why. In any case, do you expect an answer independently of dataset? Is having _annual_ data intrinsic to your problem? (If so, please explain why. If not, please generalise it.)
You can find your aswer here: http://stats.stackexchange.com/questions/105501/understanding-roc-curveedited
@Dirkran Thanks for your explanation. Do you know any other source where i can get multilabel dataset other than http://www.csie.ntu.edu.tw/~cjlin/libsvmtools/datasets/multilabel.html and http://mulan.sourceforge.net/datasets.html
Point is in this case there'd be no sense in checking the normality of the residuals - they're clearly not normally distributed, nor should they be. It's only as the number of observations for each predictor pattern increases that the distribution of residuals (one residual being calculated per predictor pattern) tends to the normal. Similarly for a Poisson or negative binomial model - the counts need to be large-ish for the normal approximation to be good.
Are you training this periodic function with the x axis as the input, and the y axis as the label for x <= 245, and then testing for x > 245?  Or am I misinterpreting your plot?
@mike, first install R (if you have not already), run R and install the TeachingDemos package (exactly how depends on your system), load the package with `library(TeachingDemos)` then type `?loess.demo` to bring up the help page to see how to run it, you can scroll to the bottom where the example are and copy and paste that code to R's command line to see the examples, then run with your own data to further explore.
Another question: In a mixed-model using `rma` is the between study variance by default specified as a random effect? Or is there a way to do that in the model specification?
Thanks, Manuel.  This actually sounds like a very intriguing possibility.  I need to think about whether the hierarchical partitioning would impose certain constraints that would prevent the algorithm from achieving the optimal cluster partitioning directly under the size constraint.   But intuitively, I can see that this might work.
Thanks Neil - But couldn't I say the same about the continuous case? The $x$ don't show up in the $H_C(X)$ equation either, except to find their densities with $f(x)$
Can you post the data online ?
And a good introduction can be found in Morgan (1984), *Elements of Simulation*, Ch 4 & 5.
@ttnphns, I am not too sure your comment has anything to do with my question. is it that you find my question is unclear?
As @jbowman gets at below, I think your theory about how the world is likely to work is important for choosing how you specify your model... do you have a good theoretical reason to believe a variable Z = (a-b)/(a+b) is related to Y?
I'm sorry, I can't follow your question. It sounds like you're trying to assess normality. Why? You make several assertions in your third paragraph but I really can't tell what you're getting at. What parametric tests are you referring to?
It depends on what is important to you. Also if you possibly have an idea how the data could be distributed. In that case see http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf
Only looking at log-likelihood and AIC/BIC is maybe not the best. E.g. if you are interested in the tail of the distribution it would be good to plot the histogram (small bins) and your fitted distribution and start by looking at it. In general consider GOF http://en.wikipedia.org/wiki/Goodness_of_fit
Hi @DrStevenMason, welcome to CV! Could you expand on your answer a little bit? Perhaps by indicating what the "Guttman's split Lambda 4 coefficient" is? Also, nothing in the question indicates that the OP uses SPSS.
As with @Maarten Buis's answer, pedants will want to insert "within the range of the data" in your first sentence.
Give me that parameter and I will define a distribution for it. :-)
@bgbg The total size of the two samples, $N = n_0 + n_1 + n_2 + n_3$ and $M = m_0 + m_1 + m_2 + m_3$. In your case $N = M = 100$.
I agree with the OP, I don't know what it would mean in a model as an IV, I still stick with my statement that some sort of spatially structured model would be better suited for this kind of thing
That is data dredging. What if you identify the best model purely because of some feature of the particular sample of data you collected?
Thanks for your answer. After reading through your comments, I went to Amazon and purchased "Multiple Correspondence Analysis: 163 (Quantitative Applications in the Social Sciences)". Using that text, I was able to visualize the survey data appropriately. Thanks for steering me in the right direction!
@LadislavNado: Table 3 in that article looks at *pairwise* statistics (sequence distances and pairwise Fst values), so they had no choice but to use that method. However, what you have is one aggression value per nest, and determining whether one set of nests has a higher aggression value than another. That's exactly what ANOVA is designed to do
@NeilG maximizing the sum of log marginal predictive data likelihood on cross-validation sets (k is integrated out).
For what it's worth, I'm uncomfortable with the statement "a confidence interval that is 95%... likely to contain the true SD" (or, stated more explicitly in the linked page: "you can be 95% sure that the CI computed from the sample SD contains the true population SD"). I think these statements flirt w/ reinforcing a popular misconception, see [here](http://stats.stackexchange.com/questions/26450/why-does-a-95-ci-not-imply-a-95-chance-of-containing-the-mean), eg, for a related discussion on CV.
Could you explain, what do you mean by semi-analytic model? Is this is a physical variable governed by differential equations? Also what do you mean by "free parameters that are optimised against a set of training data"?
@whuber - Sorry I forgot it! It is a distribution of survival times!
I don't think it's a good idea to standardize the removal of outliers. Rather, you should work on a method for identifying **possible** outliers. Then look at them more closely. 

However, even identifying possible outliers is tricky and not easily automated. For example, there can be masking, where the existence of an outlier masks other outliers. Also, depending on sample size, various deviations will be **expected** to occur.
Thank you for your answer. I'll update my post to give more information. I used the word multinomial because of a word confusion, I'll correct that.
I know how to do it on SPSS, my question is this permissible statistically
Guess I should have worded things better. The paper is on the topic of "How should you fit ANOVA and linear regression models, if the equal variance assumption is violated?"

(that topic, word for word, was assigned to me by my professor)
The reason why I asked the second question was to establish whether normal or Poisson approximations were justified. It would be important to approximately know the number of events and the total number of events for both the nominator and denominator.
A list of a few model selection books and journal papers can be found here [link](http://stats.stackexchange.com/a/33429/12318).
I understand that effect size plays into it as well.  My issue is that I can't find a satisfactory explanation for this calculation that does not at some point simply fall back and refer to software.  It seems to me that this is something that should not be too difficult.  I suppose this begs a further question: why is it so difficult to find a good explanation?  Is it because it is considered too elementary?  Thanks!
One reason why a value can differ from a prediction is that the experimenter messed up. That is like a spelling mistake. But why is your weight different than the average weight of all people of your gender and age? Why is your income different than the mean income? In statistics, this deviation from a mean is an "error" but it isn't a mistake, just variation.
Okay, fantastic. I'll look into this. I had a feeling logistic regression needed to be used, but I was unsure how to implement it in this structure. Thank you.
It's not clear where your difficulties lie. Can you ask more specific questions?
The question does not provide much information but clearly states that the comparison of interest is between two groups. I don't see how a dependent sample T-test is useful in this situation.
You could try a clustering algorithm. It may even be enough to look at the first dimension returned by PCA to get an idea of how the players are differing.