tttjjj/elt_bench huggingface dataset:
It's a fair explanation of the method, but it's a different design: a systematic PPS sample. You cannot have unbiased variance estimates with systematic samples unless you utilize some special tricks (e.g., running two threads with two independent seeds and twice the sampling interval).
internal consistency is a test scale characteristic, not a peson characteristic. It sounds like what Laurens is asking for is a person related characteristic.
(1) Do you have before & after measurements, or at least multiple measurements, for each store?  (2) Significant correlation of wait times within stores is equivalent to finding a significant among-store variance ...
In case of multivariate, at least two of the variables need to be I(1), not all. Yes you can.
@Amin what do you mean by reverse? do you mean getting back to your level-2 data? I think what I would do would be this: `level2data <- combined[duplicated(combined$SchoolID) == FALSE, c("SchoolID", "resources")]`. The bit `duplicated(combined$SchoolID) == FALSE` selects rows for the first occurrence of each SchoolID, then `c("SchoolID", "resources")` only keep the SchoolID and resources columns.
@Glen - thanks! Why don't you post your comment as an answer?
How much data? How much is it skewed?
The second calculation should give you a random variable rather than a number.  Did you notice that the right hand side of the second formula depends on the unknown $x$?
Thanks muchly, especially for the R info!
I agree it possibly isn't the most commonly held one. In the references given (esp. [3]) in the question, however, they seem to use "randomization model" (and randomization test) for when a non-uniform sample of the underlying distribution is taken, and "population model" (and permutation test) for when a uniform sample is taken. I think this is where the answer comes from.
If I decide to carry out a regression, would I still report the Spearman correlation results? Is it appropriate to have both n a study? Furthermore, if the results are statistically significant in the Spearman test but they are not in the regression, what does that mean exactly? Thank you for help, it is greatly appreciated.
@SamLivingstone: Yes, it is. Thanks for informing me. I've fixed it now.
Hey chl - looking at RMSEA, SRMS, and BIC.
@Ted it's hard to say without seeing a reproducible example in your question
Section 3 in [this paper](http://econ.ucsd.edu/~ikomunje/files/papers/hdbk_quantile.pdf) might be useful.
You haven't told us what data you have, what the purpose of the project is, or anything else. However, obviously, you  will get the "quarter ended" information when the rolling quarters happen to include e.g. Jan-March.
And I think your criterion for 'intervality' is valid only for a uniform distribution.
Did you keep all 100 sample averages for each of the two runs? Have you plotted them? Are there outliers? Are the data long-tailed? Do you have access to the internal code that you are trying to time? Is there a case in the code in which it could get lost in a loop for a long time?
Yes, you have a sliding window of time and look at how many events occur in the window. If the usual average is 100, `qpois (0.975, 100)` is 120, and `qpois (0.025, 100)` is 81, which says that more than 120 or fewer than 81 events in the window is "suspicious" (at the 95% level of confidence).
@SheldonCooper: 1. and 2. may be a little over my head. I think you are right about 3. and 4.
Do you have the total population for each pixel?
Hi there, are x1 and x2 both repeated measures, and how do they relate to the timing of the treatment? It would be good to get more detail about your study design.
Possible Duplicate: http://stats.stackexchange.com/q/2002/1390
Please explain to us how this question differs from your previous one at http://stats.stackexchange.com/questions/89476/including-confounders-in-a-model, apart from the change in variable count from $4$ to $10$.
But the Wilcoxon test is not a test of means (or even of medians).  The null for the Wilcoxon is actually the same as for KS, that the 2 distributions are the same.
create dummy for variable, which is 1 for females, then include female specific variables multiplied by this dummy `y ~ x + is.female:nmonths`
If you had the published study's data there are tests you could run, if you had the median there are tests you can run. I'm having a hard time thinking of a nonparametric procedure that would work with what you have from the published study. Is it not possible to ask the original researcher for a copy of their data set? If I think of something I'll post it.
Beware of the recommendation for `cv.lm()` from package `DAAG` for k-fold cross-validation though. It silently uses only the first predictor in the validation, i.e., is only suitable for simple linear regression. As is mentioned in its help file, use `CVlm()`instead.
I notice the question says negative *relation* rather than negative *correlation.*  My guess is that the O.P. has been careful here not to fall into the trap you describe. Once we can see an example we will be much better informed about the situation.  (Welcome to our site, by the way, and thanks for being an answerer as well as an asker!)
These are only tiny local $\\alpha$s. GWAS have still a overall type I error of 5% for claiming a success that isn't there in reality.
@cardinal posted a useful comment to the [earlier version of this question](http://stats.stackexchange.com/questions/21684/high-r2-but-lack-of-fit-in-multiple-regression).  What research have you done in following up on that comment?  BTW, please do not post multiple versions of the same question.  You have editing privileges for questions you post: use them to improve the question as you see fit.
Standard asymptotics for likelihood ratio tests are based on the idea of nested models; practically the smaller model is a special case of the larger model. On the other hand AIC is based on an information theoretical approach that is in a way structure agnostic. The book by Burnham
& Anderson "Model Selection and Multimodel Inference" gives a very good overview on the matter.
Have you ever heard of finite mixture modeling? You may think of the observed distribution as a mixture of two distributions, and you can make inference about where the "split" occurs within this framework.
Do you mean you want to estimate the expected number of visits per month of a user; or do you just want the expected value of the "days since last visit" variable?
As you are looking for applied help, it might be useful to note what software you would prefer using.
"*absolutely no assumptions can be made about the coin*" -- well, you seem to at least assume it has exactly two sides.
@SteveS Actually i want to know why we need hat matrix?
Just out of curiosity, why would you want to use the binomial distribution instead of just using `rnbinom`?
http://www.stats.ox.ac.uk/~meinshau/quantregforests.pdf describes the technique.
@Nicole Ok, now it is clear what you are asking about.
How many different samples?
I can see your point, the mom is not a great estimate to begin with, therefore whether 'n' or 'n-1' is used pretty much depends on the equation, created by equating the sample variance by the appropriate function of parameters being estimated.To investigate further, I assumed that the 'better' estimate (whether with denominator 'n' or 'n-1') was the one closest to the m.l.e 1/$\\bar X$. I attempted to calculate mathematically which mom was closer to the mle but failed to do so >< (it may be that for different $n$ and different $X_i$s the 'better' mom varies).
Questions / misunderstandings about the marginal distribution of $Y$ & the distribution of $X$ are common. In addition to the good answers below, this thread: [what-if-residuals-are-normally-distributed-but-y-is-not](http://stats.stackexchange.com/questions/12262/) might be helpful.
In my case I used AIC, as I don't trust the p-values that much
By "doesn't predict," I mean that when I plot both the predicted win/loss probability for each amount against the actual win/loss for each amount, the line for the model predictions are not even remotely close to following the trend of the actual values.
I would speculate that, in restricting the algorithm to combining clusters you may get a poorer result than if you started from scratch, because you disallow "interactions" between the geo-spatial and temporal effects.
(+1) Yes, there exist some penalized versions of logistic regression (or more generally, GLMs), see some references there: http://stats.stackexchange.com/questions/4272/when-to-use-regularization-methods-for-regression/4274#4274.
That's a good suggestion, @Peter, but I am concerned about subtler yet important violations of OLS assumptions. A student's rank in an exam will depend on the attributes of all the other students taking the exam. This interdependence is not captured by any (routine) application of OLS.
On a completely different track, another difference between the coins is the design on the front/back (although some may have the same design on one side) What about cross correlating the set of design patterns with the coin pixels (or using mutual information) to help determine which coin you are looking at. With a combination of ratios, pixel colour and this design matching you will probably be able to reduce the false discovery rate.
I have update my question with a code. Basically, I am using nearest neighbor and euclidean distance to find the minimum value which is used to update an equation. This search technique is inefficient when the data size is huge+when I need to do monte carlo simulation. Hence, was looking for suitable robust search techniques which will be applicable for my case. Will be greateful for ideas about what algorithm to use instead of nearest neighbor search to improvise the search technique into an iterative manner.
But my problem is why would you want to transform continous data to discrete. It is loosing information essentially. Also when a simple `log` transform would have worked, why discretize your data? Using `glm` works, but every result is asymptotics based (which may or may not hold)
@Corone, I appreciate the work you're doing to find, edit, & revive the dead Q's in CV's backlogs. This edit was rather extensive, & perhaps controversial. I tried to split the difference by using the explication of the Q that you had derived from the comments, but maintaining the example the OP had wondered about. I hope that's OK.
+1 I think this most directly answers the question. Because although multicollinearity affects interpretation. Why it is an issue imho is the stability in estimation.
Hmm...We're always glad of contributions, but you seem so uncertain, perhaps you should do a little research before committing to a reply?
The elephant in the room is the representativeness of the "big data." Many huge datasets collected on the Internet are (at best) samples of convenience; there are hidden but well-known dangers lurking in the effort to generalize from the sample to a larger population or ongoing process.
Laura... go with the two-sided Pr.  http://www.quantitativeskills.com/sisa/statistics/fishrhlp.htm (link "go to procedure").  The online calculator lacks the fidelity to give you a p-value that low.
I have edited my post to address your questions
I've actually run into this exact same situation (with the same package).  Unfortunately, the only solution I could find was to manually loop through the dataset.
IP's are not going to match up to people particularly well because of dynamic addressing and various other issues (such as people using multiple computers). If I wanted to test this hypotheses, I would probably use visitor ID rather than IP address.
@Scortchi, http://en.wikipedia.org/wiki/Ljung%E2%80%93Box_test, see last paragraph under formal defintion.
I also use rlm, and use the bisquare psi function because of its redescending property.  Sometimes there are convergence issues with it, though, especially with smaller samples.
@user1417: You should edit a question rather than posting a new one.
Chi should be closer to the degrees of freedom, which is given by $n-p$ where $n$ is the number of unique $x$ values and $p$ is the number of betas.
@remo: Because quadratic mean (RMS) is the square root of the average squared value; not the other way round. On your lucky day (="nicely behaved data") quadratic mean and arithmetic mean do not make much of a difference.
TPR=True Positive Rate
"Exact opposite" in what sense?
Remember to check if those two vectors are normally distributed first.
all probability measures assign measure one to the entire sample space (by definition). Is that what you're looking for?
Could you clarify how time comes into the experiment - you say you expected differences "to become larger as blocks advanced" - what does this mean?  Also, are you plotting the points using geom_point() or geom_jitter() as well as stat_smooth() - if so it should help for a reality check.  If there's too much data to see try something like + geom_point(alpha=0.1) to add transparency.
+1 This analysis and its results generalize to $k$-sided dice (such as $k$=2, a "fair coin") with only trivial changes. An alternative formula for the answer can be obtained by noting that A's probability of winning is just one-half the chance a tie does not occur and then finding the chance that a tie *does* occur.
But if you simulate approximately collinear and non-collinear data and create the traceplot I mentioned, you will see the traceplot is an effective diagnostic tool. Collinear data shows a large change over the range of the tuning parameter. That change is consistent and bootstrap iterations over various levels will show this. Non-collinear data does not demonstrate that large of a change. It also generalizes to any GLM using the likelihood based L2 penalty.
Thank you for this answer. I'm going to look into these other methods.  For one, I've realized a MANOVA won't work with 3 continuous variables, so I'm starting from the point of a multivariate regression, and will work from there if that is insufficient.
Please explain more about your regression algorithm. How does it produce a classification?
Your biplot above shows cases and variables. But you said you have only casesXcases distance matrix. How do you think a biplot can be drawn? Where you get variables from?
This is a well-written and clear answer. Please note, though, that if the question being posed truly is about Gaussian distributions only, then it is a duplicate of http://stats.stackexchange.com/questions/24257/how-to-simulate-multivariate-outcomes-in-r and eventually will be closed as such (and your answer can be merged with that thread). If it is not just about Gaussians, then the issue becomes much more complicated and needs a different kind of answer.
A very frequently discussed misunderstanding, which might be operating here (search for it), is to suppose that skewness in the responses is a problem.  In general, it is not.  What matters is the distribution of *residuals*.  What can you say about that?
"*The fit does not model the original time series precisely, in my opinion*" -- why would you expect it to??
Thanks! I feel... at home
What exactly are you modelling, and which probability distribution are you using?
It seems to me this problem has nothing to do with a random variable $Z$.  There is an unknown *number* $z$ and two pieces of "evidence" $x$ and $y$: that's it.  This simple description makes it abundantly clear that your question cannot be answered until you explain how $x$ and $y$ were obtained.  Did you consult the oracle of Apollo at Delphi?  Ask two passing strangers on the street?  Look up the results of two double-blinded controlled studies? Pick the two highest values out of a collection of 100 Web articles? :-)
@ Glen_b; Thanks for your great answer.
+1 You will want to re-learn the remaining parameters after removing some of them. Think of your current solution as an extremum point on a manifold. If you reduce the dimensionality of the manifold, why should the same point remain an extremum?
Thanks Blain Waan. You are accurate with my foundational issue. Early on I recognized that grouping objects (people) with similar attributes is not the same as grouping identical objects (flags). You have provided to me some useful techniques, and I am grateful.
The sampling distribution of what statistic?
Can you say more about the process you are asking about, or provide a reference regarding this estimator? I've never heard of it.
As of 2013, the ISU statistics 401 videos are marked *private* on YouTube.
2/2  The other option is to do this in R, which has ready-made packages for imputation (not just the simple one I illustrated above).
It is true I have two sites. I have done the plots it's pretty messy but no real visual differences that I can tell.
In the replies, **please explain _why_ you are recommending a book** as "the best."
Please note that you do not need to perform the Johansen test as indicated in an answer given as you only have two time series. It was designed to test multiple cointegrating relationships and can be sensitive to misspecifications of your model. Also, the Johansen test is more complicated than the above mentioned. models.
Thanks for reading in-between the lines. Yes, x1 and x2 are in fact time-series results of scenarios based on the same input data. Thus the first equation is more descriptive of describing the mean differences between individual days.
But that's the wrong t-test isn't it?  They're using a one-sided t test that will return a low p value only for a high value of r.  So any negative value of r (ie negative correlation of TV watching and IQ, which may well be the question) will return a value of p >0.5.
A regression will work just fine with causal predictors. You will just need to be careful that you don't over-parameterize your model if you only have a short history.
Both of your answers are different. Does dimensionality means multivariate signal where each variable = dimension?So, in linear regression I will have 3 equations if dimension = 3?Can you please explain with AR model?
This is a fill-in-the-wizard application called a "Pivot Table." Just follow the directions.
Before modelling, you should look at correlations between your (assumed) independent variables. I think you should rephrase your question to "how to identify multicolinearity" and ask on crossvalidated.com, instead.
He means there's two different variables, $x$ and $x'$, and he's just taking their dot product.
Michael, I closed a [duplicate thread](http://stats.stackexchange.com/questions/23411/how-can-i-help-ensure-testing-data-does-not-leak-into-training-data) from the ML site and merged the answers here.  Please feel free to edit this question to reflect any changes you wanted to make--I don't want to undertake that for fear of inadvertently changing your meaning.
Another great answer! Thanks whuber.
+1 for the restraint...
Effectively, you're asking if a random forest can be represented equivalently by a matrix or a decision tree. The answer is generally no for a matrix (which would necessarily be a linear classifier, which random forests are not), and yes for a decision tree, though probably not compactly. [This paper](http://www.cs.cornell.edu/~caruana/compression.kdd06.pdf) gives a way to approximate forests with a neural net; I'm not aware of implementations that do this or a similar thing, though.
Assuming equally shaped distributions of ranks between A and B will allow you to conclude such things.
There isn't an analytic expression - as you'll end up with digamma and trigamma functions.  You could approximate these with the stirling approximation though.
There is no such thing as _zero independence_. Random variables are independent or not independent; there are no fine distinctions. This is different from correlation where random variables can be uncorrelated or correlated, but for correlated random variables, we can talk of the _degree_ of correlation (highly correlated, weakly correlated, perfectly correlated etc) and well as positively or negatively correlated. Independence is absolute and random variables cannot be "somewhat" independent or highly independent.
@IrishStat Thanks, but I already have code that does something similar to my satisfaction (I wrote the answer to the thread cited as duplicate to this). I was just curious about any hidden assumptions behind your suggestion.
You got it correctly, it is very easy to sample a variance from the distribution (by taking out the mode value). But I am not sure how to do that in multivariate case. I am using an Inverse Wishart distribution for sampling Covariance matrix.
Thank you so much everyone, I'll send this to my supervisor and we shall have a go at all your suggestions!
@Mauricio_Fuentes, welcome to the site. `ologit` is a valid link for `gllamm`. It must have been something else that was breaking down when you ran it. Can you show the code and the output? Alternatively, you can ask at statalist... although in all likelihood it will be me who will answer it there ;). Another Stata (user-written) command that may be useful is `reoprob`.
I am a little bit confused by your data . For example what is the meaning of 6325.76... and -.247831. What is being observed ? Isthe data chronological and equally spaced in time ? 6325.76666666667 -0.247831949032426 "770" 6326.78333333333 0.0282060159445925 "771" 6327.78333333333 0.162348034962085 "772" 6328.78333333333 -0.206057978534193 "773" 6329.78333333333 -0.427928912063427 "774" 6330.78333333333 NA "775" 6331.78333333333 NA "776" 6332.78333333333 NA "777" 6333.78333333333 NA "778" 6334.78333333333 NA
If you have otherwise identified a model as IrishStat posted, then the corresponding test would be a "Chow test" if you know possible break points and a "Quandt Likelikehood Ratio Test (QLR)" if you do not know the break point. Since this is a financial timeline and most likely discrete, I would recommend checking out econometrics books. Basic would be the Stock&Watson, more advanced Hayashi or Greene, advanced insights using the Hamilton. This should explain what you need to know. Sadly there is little online about the QLR.
I'm wondering whether this extremely simplified approximation would make sense, given that zeros form the absolute majority:
1) find the proportion of zeros in each subset. 2) assume that in the subset with the smallest number of zeros all zeros are true. 3) from each subset, remove the proportion of zeros equal to the proportion of zeros in the most "zero-rich" dataset. 4) run standard non-parametric stats on this modified dataset.
I'd argue from a standpoint of trying to create good guess as to what video game a person likes coding 1s and 0s for like and didn't like isn't a good approach.  A scale of 1-5 or 1-7 is quite easy to elicit and will require fewer datapoints to generate a good model (because each data point provides more information).  With caveats about treating ordinal data as interval data of course applying, but probably not really that important in this context.
So nice explian, thx whulber.@whuber
It depends entirely on how you value the outcomes, which relates to how you value money and risk.
Additionally, it is generally believed that the more hidden neurons you have, the easier it is to escape local minima. That is because the probability of a subset of hidden neurons being near a better minima will increase with the number of units and tend to dominate during training.
is this homework?.. if it is, flag it as such.
You can fit an ARIMA model along with external regressors. If you see that the residuals from the linear regression are not stationary, then this may confirm that you should use an ARIMA model (not an ARMA model) with external regressors. You could split the process in two steps and fit an ARMA model to the differenced residuals, but a joint estimate of the parameters may yield better results.
The t-test works for any sample size. There's nothing magical about n=30; indeed in my book the t-test isn't much like the z near typical significance levels (5% to 1%) until well past n=30. Eventually the tables become very close to the z-tables a fair way into the tail.