tttjjj/elt_bench huggingface dataset:
First you ask us to simplify, then you complain that our answers are simple!  If you want to understand the formula for U (or anything else) LOOK at it. If you want something simple, then don't ask for complexities! The Wikipedia entry is an excellent, detailed entry with all the details. You  don't understand it. So. What do you want?
Your answer seems to contradict leonbloy's answer given earlier @Henry. I've explained this better in the original question.
While I tend to agree with the answer, are you aware of any empirical studies that show this to be the case? RF definitely has issues wrt importance scores and highly correlated variables (e.g., [Strobl et al](http://www.biomedcentral.com/1471-2105/9/307)).
There will still be bias, because only a portion of the contamination distribution will typically be identified as outlying.
@dafeda Great!  Also, you may need to convert back from a `matrix` to a `data.frame`.  I'm not sure if you can use a `matrix` with `rfe`.
Would you care to add a bit of detail to your question? I'm willing to start a bounty, as your question is perhaps related to mine: http://stats.stackexchange.com/questions/29133/effects-of-ignoring-the-sampling-strategy
@Peter Flom I dont think it is broad, if your regressor give you a p-value of 0.6 with 20000 test samples, then I seriously doubt the problem lies in the regression methods he used, lasso or not, much more likely scenarios are either OP miss critical observations in his data set or OP try to "invent" some determinstic knowledge/trend from some largely stochastic process/random data and failed as expected.
Your notation is mystifying because $\\mathrm{E}(Y - \\hat{Y})^2 = \\mathrm{E}[f(X) + \\epsilon - \\hat{f}(X)]^2$ literally means the square of the expectation. Assuming $\\mathrm{E}(\\epsilon)=0$, this immediately reduces to $(f(X)-\\hat{f}(X)+\\mathrm{E}(\\epsilon))^2$ = $(f(X)-\\hat{f}(X))^2$. Evidently, then, what you really want to compute is the expectation of the square, $\\mathrm{E}[(f(X)-\\hat{f}(X)+\\epsilon)^2]$. But if so, the very first step in your derivation makes no sense.  Could you edit the question to clear this up?
n a sample size and Zeff is defined in Eqn (2.41), but where are v, z, and w defined? Are you sure that z is a realisation from a N(1, 1) random variable?
Thanks Glen_b for your guidance. Regards Katherine
Automatic trimming in either tail is a bad idea except where extensive experience with a *specific* data-generation process supports it--and even then it's a bad idea. But think about the regression situation: when there really is a linear relationship, the 1% are there in the tails because they are the responses at extreme (or almost extreme) values of the independent variables. That can be where you want to characterize the regression most accurately. Thus, trimming dooms you inevitably to using the middle values of your data to *extrapolate* into the extreme values: yet another bad idea.
Are you sure that this has nothing to do with the statement (in the example of repeated measurements in SAS I gave you just now) : "*The REPEATED statement contains no effects, taking advantage of the default assumption that the observations are ordered similarly for each subject.*" On second thought, have you read the line in the page you gave : "*We also use the repeated statement to indicate which variable indicates the different subjects (via subject=sub)*". Really get a simple example working first and then go for repeated measurements ANOVA (and read the documentation more carefully).
Because the "confidence" (by almost any definition) depends on $N$, the last parenthetical remark may perhaps be asking for the impossible.  Note, too, that $X$ and $N$ are intimately linked via $X = N P$, so any MoE for $P$ depending on $X$ depends *a fortiori* on $N$.
@whuber Thanks - indeed I saw there would be simpler examples of this new case, but I wanted to show the OP that it was already clear from my previous example that making the covariance positive wouldn't help. I decided that showing a tiny modification of that case might serve to highlight that. I think it's nice that there are so many counterexamples presented; the different approaches to identifying counterexamples in the answers make this more instructive than the question would indicate.
@chl  Thanks!  I studied this stuff in grad school, but that was long ago - my last class was in 1996 or so.
Thank you this has certainly helped me understand.
How would this bootstrap work, exactly? How do you resample from, say, a set of $(t,x)$ values?  (Although there are some well-known ways, one has to make choices and assumptions.  I believe that addressing my question will force you to consider assumptions about the errors in the data, which may have some relevance to the second part of the question about a normal distribution.)
I am confused about how you are evaluating the ANN without cross validation. You must be evaluating on something...For cross validation you are averaging R-sq from each round - without cross validation - how are you obtaining your R sq value?
Have to break my comment in two for length. The *Cauchy* is all the things you said. The distributions here *aren't Cauchy*. The marginals are standard normal. The only thing 'unusual' about it is the bivariate distribution, specifically the manner in which the two normals are associated.
Thank you - I will try this.  I agree that this doesn't seem like the best idea.  The argument I got was that each series does not have enough observations (~28) for a good estimation and that it would be more robust to estimate over all the series.  I'm not sure I agree with that argument.
Slope graphs! Ok that is a great tool! Worth looking at this...much thanks!! The design challenge for me now is that the audience wants to do both between and intra group comparison....Maybe I will do two pages? Or is there another option?
What *specifically* is the problem with just two observations?  Before you answer, check out the [sample-of-one confidence interval technique](http://stats.stackexchange.com/a/1836/): it can be applied to obtain a CI around the slope when only two points are available.
Can you explain *why* this is the case? We want our answers to be a little more substantial than this (please see our [FAQ](http://stats.stackexchange.com/faq), which contains info about answering Q's among other topics). If you're just going to agree w/ @whuber's comment, this itself is best put as a comment instead of an answer.
@user34790 yes, you would have computational issues. Computational challenges are quite big deal for GPs. Rasmussen and Williams have a book on GPs with an entire chapter dedicated to this, and if you google hard enough you can find it online for free. See my updated post for some minimal details.
@Dikran Marsupial, I will soon head to IJCNN 2014 to talk about how and why PSO fails to train high-dimensional NNs. So the answer is yes and know: I took a big detour from the original research vector, and I wonder if I will still come back to the adjustible architectures. Only time and empirical results will tell!
(ctd) second, is that the distribution of the difference of two correlated normal RVs ($X_1, X_2$ ) with means $\\mu_1, \\mu_2$, variance $\\sigma^2_1, \\sigma_2^2$ and covariance $\\mbox{Cov}(X_1, X_2)$, is a normal RV with mean $\\mu_1 - \\mu_2$ and variance $\\sigma^2_1 + \\sigma^2_2 - 2 \\mbox{Cov}(X_1, X_2)$
+1 great question.
Does it report fit statistics well (e.g., p values, variance inflation factors, ...)?
It seems like the random effects model is more practicable here and is probably a more appropriate choice. Whether or not the collinearity presents a problem depends on what your goals are with the analysis.
I don't think so. The relationship between the three are deterministic but nonlinear. There is no way to obtain a variable as a linear combination of the other two
(1) You're right about the low power, but the problem here (it seems to me) is that there is no gradual step from "finite" to "infinite": the problem seems not to have a natural scale to tell us what constitutes a "small" departure from the null compared to a "large" departure. (2) The distributional form is independent of considerations of iid.  I don't mean that, say, 1% of the data will come from a Cauchy and 99% from a Normal.  I mean that 100% of the data come from a distribution that is almost normal but has Cauchy tails. In this sense the data can be iid for a contaminated distribution.
@Grega: You're somewhat doomed in that you need more data.  However, the electrical demand data that you need may be easier to find than you think.  For instance, in Texas here's a link: http://www.ercot.com/gridinfo/load/index     What area of the world are you trying to forecast?
@PeterFlom In fact the signed rank test tests whether a *univariate* (one-sample) median is different from zero, assuming symmetry under the null. When applied to *differences among pairs*, it's a paired test that satisfies the symmetry-in-differences by assuming the pairs have the same distributional shape. This is the same issue as we have with the sign test - it's *actually* a univariate test for a median of zero, but when applied to differences between pairs becomes a paired test. I'm guessing you already knew all that - but it might mislead people with less statistical background.
It was more a general stats question, so I'll edit the question!
(+1) I came across O'Connor's website some years ago and it has a lot of useful resources. Nice that you link it here.
Wooldridge in his [book](http://books.google.com/books?id=cdBPOJUP4VsC&printsec=frontcover&dq=wooldridge+panel+data&hl=fr&ei=6JHCTcW0K4_tOaPW6J0I&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC0Q6AEwAA#v=onepage&q&f=false) gives a very detailed explanation of this. Look in to chapters 10 and 11. In short the key is correlation of lags of independent variables with the error term, not with dependend variable. If the independent variables are strictly exogenous then there is no bias. If they are not the bias is of order 1/T, assuming of course that contemporaneous model holds.
No, there's no contradiction: there's not a whole lot of useful information, either.
@juliohm, Have you encountered cases like mine problem? I mean have you check the trace plots if they are the same in several different runs but with same seed ?
yes.. thank you very much
Instead of dropping one of the items, you could replace one of the two by their difference. The resulting variables are usually only weakly correlated and (in contrast e.g. to principal components) easy to interprete.
Let's see now: there are 22,451,004,309,013,280 subsets of 10 columns you can pick from 200 columns.  I wonder how long it would take Excel to process all those to find the subset with the best correlation to the target? :-)
Andy W
Dr Wolfgang Langer. The assessment of fit in the class of logistic regression models: A pathway out of the jungle of Pseudo R's. Just google that document, you will get the information and citations there
+1 This is very nice. I don't think I could have done it better! I would just add that I have some papers on the performance of the bootstrap in error rate estimation and for confidence intervals when estimating variances. These results can be found in the second edition of my bootstrap methods book and my book on bootstrapping with applications to R.
Please explain how you are conducting a chi-squared test. Ordinarily (in this goodness-of-fit context) it is conducted on counts, but since your "un-normalized graph" evidently contains no counts you must be doing something different. Your description of that process will help people provide good answers to this question.
+1, I've added the code to calculate `suma2` etc. for easy reproduction of the code. Side note: the formula for the combined variance in the first paragraph is only valid, if $X$ and $Y$ are *independent*.
In R, the `var1*var2` terms translate into both main effects plus the interaction term, namely: `var1 + var2 + var1:var2` (that is, in R, the interactions are denoted with `:`, while `*` includes all possible main effects and interactions). This is not reflected in your equations.
@aginensky. Hi, i have posted my code in the answer now. the main result i havent posted it before becoz .....hmm.....my code look too bulky and ugly. dont want anyone to see. wahhaha.
Use average instaed of the sum. You will have to calculate an average manually, but most programming tools do this easily. Edited my answer to correspond with this comment.
Dear chl - any of them.  My real motivation is to find bloggers in other languages, but I would feel fine with other websites from which to find bloggers (see more about my motivation here:  http://www.r-bloggers.com/r-bloggers-now-in-your-language/)
If the interactions are only significant when the main effects are in the model, it may be that the main effects are significant and the interactions not. Consider one highly significant main effect with variance on the order of 100 and another insignificant main effect for which all values are approximately one with very low variance. Their interaction is not significant, but the interaction effect will appear to be significant if the main effects are removed from the model.
I would not trust Excel for anything statistical.
Thank you, but now there is an another problem...multiple definition of node tsn[1,1]! Why? I don't understand this type of error!
+1 You've beat me by few seconds.
Thank you Adam for your response! I do base most of my analisis on odds ratios interpretation and predicted probabilities. But since Im not too confortable yet with logistic regression, I fear my analysis could be dismissed because Im missing some general model fit test. But as you said, I also belive I should focus on the more substancial interpretation of the model. I will consider your recomentdations for forest plots and Cronbach´s Alpha as well. Thanks again!
Depending on established norms in your specific field, you often don't need to do any post-hoc analyses on significant interactions unless there are unanswered questions; you can just report the interaction as significant (usually after stating it specifically like in my answer).  Also, if my answer was helpful, please accept it so your question shows up as answered.
+1 Thank you so much for your answer. I made corrections to my question. By now, I understand you are a pioneer in time series. It would be great if you have some suggestions on my specific case which I put in one of the comments: The time series data that I have is of an internal twitter-like network and the series itself represents the number of messages generated on a particular topic. I want to find other topics that have a similar timeline as the given one. Thank you once again for your time.
A continuous proportion can take any value on the interval $(0,1)$, whereas a discrete proportion has 'gaps'--values w/i that interval that can never be observed. Eg, on 4 coin flips, you can get 0, .25, .5, .75, or 1, but none of the values in between. You have Y discrete items, of which X are tagged correctly, that means the only proportions possible are from 0 to 1 in increments of 1/Y. Note also, the binomial can have 0 & 1, but the beta can't.
see also [here](http://stats.stackexchange.com/questions/206/what-is-the-difference-between-discrete-data-and-continuous-data)
No problem. The above is pretty much all I know though. We used (borrowing your terminology) $f(x)=x^k$ ("power gravity model") once, but unfortunately the paper lies currently on ice.
Please include a *reproducible* example. That is, one that when we run your code, we get what you get.
If you made some assumptions about how the data were distributed within the bins, perhaps.
Yes, that's what I did in the example.
@Adesh Coding a binary 1/2 or 0/1 will get you not affect your correlation coefficient.  0/1 also has the advantage that the mean of the variable would be the percent male or female, depending on which is which.  Other coding schemes may be useful for interpreting different types of analysis.
That's interesting - I never would have guessed that result.
Could you explain how you got to the "this state" displayed equation in your question?  It seems to be equivalent to $$P(B|AC) = P(B)P(A|B)P(C|B)$$ which doesn't seem quite right to me.
@whuber Yes, it should be, but I must be missing something. I understand why there are $n-1$ bars but why are there $2n-1$ possible positions?
+1.  (There's actually a tenth one represented as a single point near the upper right corner.)  The values, of course, are $0, 1, \\ldots, 9$.
If you include a continuous predictor in your logistic regression, the exponentiated coefficient represents the odds ratio for one unit change in the predictor. Often, one unit isn't meaningful and you want the odds ratio for, say, 10 units. To calculate this, just exponentiate the coefficient multiplied by 10: $OR_{10}=\\exp(\\beta\\cdot 10)$.
The polar transformation corresponds to $(x,y) \\rightarrow \\left(\\sqrt{x^{2}+y^{2}}, \\arctan(x,y)\\right)$. The closest related polynomial kernel might be $(x,y) \\rightarrow \\left( x^{2},y^{2},\\sqrt{2}xy\\right)$. In any case, this will work for that toy case.
$r^2$ and r and so on relate to the relationship between two (sometimes more, but here 2) variables; you seem to be after something else - but it isn't completely clear what; I am confused by your title and by "in favor of TV watching".
Can you add links to "some of the papers" you mentioned in your 2nd point? Thanks :)
@robin You said you have a forecast, which implies a model for a random variable. That model comprises a family of distributions for the variable and a function to identify the distribution you predict from the data. The parameter is the processed data that identifies the predictive distribution - without one, there would be no variability in your predictions.
ANCOVA but others are recommending Multiple Regression... The concern is the use of so many non-parametric variables for a parametric test
Thank you! it was very informative!
+1 This answer contains some interesting and clever suggestions.  The last one looks particularly flexible and powerful.
For all the Kappas (and all the contingency table measures discussed so far) they can have the cells divided through by N, and be expressed in terms of probabilities, viz. Kappa = [Acc - E(Acc)] / [1 - E(Acc)].  Confidence/Significance brings back the N - ChiSquared being closely related to MutualInformation*N and Correlation*N (cf. Kramer's V). Sample size is thus irrelevant except for precision of the probability estimates and smoothing away of 0 cells - you could wait till you at least 5 in each cell, or at least an expectation of 5 in each cell.
Do you have the original counts from before the differences were taken? One problem - of several here - is that counts generally don't have constant variance, but a variance which is related to the mean (e.g. consider binomial or Poisson or negative binomial distributions... or even just the fact that counts are necessarily non-negative). This will then impact the variances of the differences.
Hmmm... re-reading your answer, I see that you focus on the question of how to determine which model is more complex. I am thinking in terms of how to incorporate model complexity in the process of using goodness of fit to select a model. This would be the "data informed" version.
Is this self-study or homework?  If so, please add the appropriate tag - we'll guide you to an answer, rather than just providing one outright.
@dassouki: sorry, I saw the "statewide" on the graphic's title and presumed (incorrectly) that it referred to a US state.  The birth/death data for the last 10 years would be excellent, because it will be a reasonably good predictor of birth/death rates into the near future.
This is discussed in detail in the book by Davison and Hinkley, *Bootstrap Methods and Their Application*, along with an explicit algorithm (Algorithm 6.4). They explain concepts, pitfalls and details at greater length than is possible in a reasonable answer here.
If you have a specific thing you're reading in papers you'd like comment on, you could post a question asking about any contradiction you see in answers here and what you can quote from a paper. Incidentally, I'm about to update my answer with further information.
That depends on the context. A nice feature of this approach is that it's embarrassingly parallel, so it scales very easily on many computing architectures: multicore PCs, computing clusters, etc. Assuming that fitting one model is doable, you can fit an ensemble of dozens/hundreds of models without taking much more time.
There's also an answer of me to a similar question: http://stats.stackexchange.com/a/65564/3293
Jerad... That makes a whole lot of sense now. You should write a book. Many thanks.
Based on "*the decision has to be a box*" you pretty much answered your own question: you need a decision tree.
@jbowman The sample size can be very high depending on the code being benchmarked. Each benchmark will repeat a test as many times as possible in `5` seconds. So the sample size in Chrome may be something like `86` to `93`.
@whuber, yes copy-paste. The same error is in the OP, will fix it.
I wouldn't call a single time series like this two-dimensional. You need two dimensions to plot these data, but I doubt that there is any other sense in which statistical people would call them two-dimensional. In statistics, you count dimensions according to how many variables there are of the same kind. Time here is an index rather than a separate measurement or count. I am not sure that much hinges on the answer, however.
I've had luck in the past with time-series regression. It is sensitive to preprocessing via differencing, detrending and removing seasonality. For most problems I've encountered it works better than linear models. For a primer see: http://www.ulb.ac.be/di/map/gbonte/ftp/time_ser.pdf
A typical very simple Bayesian approach would be to form an interval for the difference in means, based off the posterior for it. One might make all the same kinds of assumptions as in the t-test or one might make a variety of changes toward more robustness/wider applicability (e.g. considering possible contamination of the data or heavier tailed distributions, putting hyperiors in and so on). Of course there's also a more formal Bayesian decision-theoretic approach and a variety of more nearly hypothesis-testing style things one might do.
Unless my eyes are deceiving me, you are considering a sum of products, not a product of sums. :-)
If this is done in $R$, you should try replotting it with `plot(...,pch=".")`. That is a quick-and-dirty way to see the density of the points better.
@probabilityislogic Thanks for that advice
Thanks for your calculations. I just wanted to add that $\\rho_{max}=1$ could have been found immediately, noticing that $X_1$ and $X_2$ are of the same type: $\\frac{\\lambda_1}{\\lambda_2}X_1$ has distribution $Exp(\\lambda_2)$, i.e. the same distribution of $X_2$.
After ensuring stationarity (for example by means of identifying suitable ARIMA-models), is it possible to simply conduct a t-test for two groups? Does it make sense/is it allowed if there is dependence within one group? Isn't it possible to treat one person as one population?
This is a cool example where a test for equality of variances is really needed.
WOW! Thanks so much.
I am not sure I understand you correctly but it seems that you *already* have multilevel data. I am just suggesting another way to look at it that would readily accommodate the factors you mentioned yourself (e.g. instead of a correlation you can fit a linear model of the relationship between group membership and response time, add age or gender as predictors etc.)
okay. Thanks. But it has to be like bins of 50 to get it clear in the beginning and since it goes to 25.000 they are really just lines. I reckon i'll stick to the cutting strategy then.
With *decompression rule* the author means *decompression program* or *algorithm*. Otherwise this seems to be written perfectly clear.