tttjjj/elt_bench huggingface dataset:
Closely related (and highly voted): http://stats.stackexchange.com/questions/29682/how-to-rigorously-define-the-likelihood, http://stats.stackexchange.com/questions/31238/what-is-the-reason-that-a-likelihood-function-is-not-a-pdf/31241#31241, http://stats.stackexchange.com/questions/2641/what-is-the-difference-between-likelihood-and-probability/2647#2647.
For integral values of $\\alpha_1,\\beta_1,\\alpha_2,\\beta_2$ this probability is a linear function of $\\log(\\text{val})$ with coefficients that are polynomials in $\\text{val}$. Approximation is not necessary, though, because numerical integration can obtain the values as accurately as you like.
$\\sum d_i^2/\\sigma_i^2$ isn't quite a variance; that $\\sigma_i$ makes it something else, and a variance would be scaled for sample size. No, that's not normal. Simpler question: what would be the assumed distribution of $d_i/\\sigma_i$? On the other thing - presumably the $\\sigma_i$ come from the observational error calculations.
Thank you for your answers. I was thinking of Box M, but I wasn't sure if it applies into repeated measures. Had to get that Rencher's book. I'm quite certain that nested model comparison can be done used for example proc mixed of SAS as well. Nevertheless, thank you! I'm new here and hopefully I'm someday able to provide some answers as well:o)
If you mean evidence *for a point null against the alternative of anything else* ... then, no. The uncountably infinite number of alternatives between the observed very small value and the null are still going to be more likely than the null. If you mean something else, then perhaps in some circumstances.
FairMiles, whuber: Let me express it a different way.  Let's say we take a picture of 1000 cars that pass by our road and measure the diameter of the wheels.  Now somebody tells us a diameter X, and we want to know whether or not X is likely to be the diameter of a car wheel.  In this case, we don't exactly know the distribution of the data, but we know that there are multiple peaks in the data.
It doesn't really change much since the x's are observable, you can collect them on the right hand side (so instead of > 0 it is > x_{i1} - x_{i2})
The thread from the alternative forum looks good Nick. Appreciate it
Thanks for your answer. I recognized those packages but I hoped that I can easily specify nonlinear kernels with them. But I guess I have to make my own.
I believe you made a typographical error and wish to ask (in shorthand notation) whether $c_1 \\ge c_2 \\ge c_3$ implies $c_1 \\ge c_3$. The answer depends on what you mean by "statistically better." Be careful how you answer that!--a natural meaning of "better" in this context is closely related to [Efron's nontransitive dice](http://en.wikipedia.org/wiki/Nontransitive_dice).
@Placidia: just came accross this link http://andrewgelman.com/2012/11/16554/#comment-108310 which provides a better elaboration (recall with 30 features a sampel size of hundreds is still very small due to curse of dimensionality)
Hi, thanks for the link, do you know how it compares against Apache Commons Math.
2. See [here](http://stats.stackexchange.com/questions/67903/does-down-sampling-change-logistic-regression-coefficients/68726). The only reason to do that would be constraints on computational resources.
As for the forest plot, heterogeneity is not only present when the confidence intervals don't overlap, but can be present when they partially overlap, but the studies have small standard errors. Without any bias in the individual trials, they should all be measuring a similar (if not exactly the same) 'true' point estimate. Therefore, if each trial has a small standard error and they are different from each other, the amount of heterogeneity can be large. We see this often in reviews of observational studies were the sample size can often be very large in each study. Hope this helps. abousetta
The blue curve on the left corresponds to a hypothetical line of "best fit" on the right. The line on the right is *not* that best fit line: it is one that best fits the *middle* values in the distribution. If you were to fit a "bell curve" to the middle two-thirds or so of the histogram, instead of to the whole thing, that curve would come close to following the peak and the steeply sloped sides, but then it would be far too low at the shoulders and tails. That's exactly what the line in the qq plot is showing, too: those plots are in perfect agreement; it is the *fits* that differ.
This looks like the right *graphical* idea, but it seems unlikely that a smooth or density of chemical concentrations would be meaningful.  It sounds more like some form of *interpolation* of the data is desired. Methods could range from regression (to fit a surface) through various forms of kriging to model-free approaches with splines, natural neighbors, etc.
#3: As pointed, be careful about using the data twice. On thin is a hierarchical model, for instance, and another one is to pick a prior that is in agreement with the likelihood. In the later, I'd be concerned with such analysis. I see the choice of a prior more as a regularization tool.
@Martin. Yes, that's right. The imputed value is calculated as the predicted value + a random draw from the residual distribution.
+1 Great link, I missed this reference -- so I guess now it can be handled within R without any problem :) I put the PDF at this address in case someone want to read it: http://j.mp/bnnwsu
thankyou @David Marx
Thanks for the references. I was able to find a "pre-print" version of the Campbell paper [here](http://www.iancampbell.co.uk/twobytwo/files/campbell_twobytwo_preprint.pdf) for those who can't access Pub Med.
There's no moral hazard.  The calculation of the main effects with the interaction included is quite different from the calculation without it.  You have to do the additive model to report the main effects and then include the interaction in a separate model anyway.  You ignore the main effects in the model that includes the interaction because they're not really main effects, they're effects at specific levels of the other predictors (including the interaction).
@llorgge I changed the example to use the iris dataset.  It would appear there is a very specific bug in caret when using the formula interface for classification models that contain factors.
@DilipSarwate So if this isn't really a measure theory question, what area do you think I should be looking in? I have seen papers that deal with judging metrics to say that measure x is more sensitive than y, but beyond the common sense reasoning for the simple ones, I don't know how they arrive at their interpretations.
I think most statistical packages will output sth like $\\log\\big(\\Pr(Y=2)/\\Pr(Y=1)\\big)=b_{10}+\\sum_ib_{1i}x_i$ and $\\log\\big(\\Pr(Y=3)/\\Pr(Y=1)\\big)=b_{20}+\\sum_ib_{2i}x_i$; and the interpretation of a regression coefficient is straightforward: for a one-unit increase in a (continuous) $x$, the log of the ratio of either of these probabilities will be $b$.
By "strength of schedule", I mean the strength of the opponents that a player would have to face in the first round. For instance, if I had a matchup of the 1-13-36 seeds in the first round, and each player had strength (37-seed), the 1st seed would have schedule strength (37-13)+(37-36)=25.
Look up *prior elicitation*, e.g. [here](http://stats.stackexchange.com/questions/1/eliciting-priors-from-experts)
Thanks @MarkHall. I tried LMT and it seems to do the job of providing class probabilities as a function of input variables. I will plot the class probabilities on a test data and see how it behaves.
@Glen_b:  Z is essentially a 2D array of log-likelihood values, and should not have zeros. The values of Z are all negative in the example shown in the figure.  I have included additional plots, including log(-Z), in the edited question.
Thanks for your help in clarification - this is my first post.
This answer is incorrect.  See my answer.
@Jake Westfall I should not spoil a little joke, but sodium would always be Na.
@GeekOnAcid, [here](http://cross-validated-blog.googlecode.com/files/ggplot2_dynamite.r) I have posted some code to generate a similar plot in `ggplot2` in `R`, I will I will let you know in an update when I figure out how to generate a legend.
Many thanks for your detailed & helpful reply! About the 1 group = 1 subject comment: I concurrently recorded 3 other groups, each were diff species and two groups were housed another building. I didn't plan on analyzing them all together because of the species/location differences (hence only mentioning 1 group in my initial question), but now I'm wondering if using RM ANOVA could control for these differences? So rather than doing a chi-squared for each group of birds, do a repeated measures with all groups (4 subjects) and post-hoc for species/location/time investigations?
Measuring IQ isn't the most important characteristic per say, just that the items selected have the same correlation with the latent trait and hence can reasonably be fit by a Rasch-type model. However, I tend to think of the Rasch/1PL model as a kind of mythical unicorn that is hard to obtain in practice without a lot of discipline and patience. Formally, it's a nonlinear equivalent to having loading exactly equal in standard factor analysis (tau equiv). 2PL likely is what you should strive for, and if you can reduce to 1PL then do it via item diagnostics and nested model comparisons tests
This question is just another version of one you have already asked at http://stats.stackexchange.com/posts/13213/edit (concerning how to interpret the p-value of another test).
Some comments: (a) Looks good, although, strictly speaking, the null hypothesis would be $H_{0}: U_{d}\\leq 0$, (c) the $p$-value is wrong. Because you have a one-sided hypothesis, the $p$-value is here $\\mathrm{P}(t_{\\mathrm{crit}}>t_{\\mathrm{obs}})$ which is $\\approx 1$ in this case. So you have no evidence against the null hypothesis. You can calculate a $t$-test in `R` with the command `t.test(AM,PM, paired=TRUE, alternative="greater")`.
@ttnphns this has been done, and it's called k-medoids. Indeed it is not k-means anymore, as you can't rely on the mean to minimize your objective function if it is an arbitrary distance function.
No it is not sparse
Can you be more precise about what you mean by "the cointegration technique"? As far as I know, cointegration is when multiple time series, very roughly, are influenced by a common stochastic process. The Johansen test is a way of testing whether there is cointegration.
Sophia-ml seems relevant here as well.
No problem, @s94, & you're welcome. If you [register](http://stats.stackexchange.com/help/why-register) your [account](http://stats.stackexchange.com/help/creating-accounts), you'll be able to come back & ask more questions when you need help, & you'll accrue reputation from your various questions.
You probably want `library(plyr); d <- data.frame(x, y, year); ddply(d, .(year), summarize, r=cor(x, y))` :-)
Neither the distribution nor its moments are likely to have any simple closed form. It is definitely the case that the first two moments will not adequately describe the distribution in many circumstances (it can be rather skewed). Given these significant limitations, perhaps you could explain why you want to do this?
@steffen I think you're right: complete unpredictability can be worse than random.  There is a difference!  But although there is no probability model available, we can analyze this situation with game theory.  It's you *vs* the classifier.  Assume the classifier is out to get you: it will do its worst.  What is your optimal strategy?  The answer, in the absence of any other information, is to flip the coin: that guarantees you won't lose more than half the time in the long run. What ruins this analysis is the eternal hope that *our* classifier is actually good, so we rarely use this strategy.
Could you explain the data a bit more or maybe provide a shortened sample?  When you say the "variables contain longitudinal data" do you mean they are all repeated measurements on the same person or thing over 7 days (and thus likely to be correlated).
Please change your title to reflect the actual nature of your question.
All of the above sounds right. I would throw my weight behind the Klein and Moeschberger recommendation for survival analysis. I seem to remember that Thomas Lumley used it as the notation basis when he ported Therneau's survival package to R.
Thanks for ur help, +1 :)
Thanks for pointing out about stack-overflow community rules, I'll keep them in mind in future.
Do you mean "the prior *mean* for $\\alpha_{p,w} $ is..."?  This phrase doesn't imply that you can estimate $\\alpha_{p,w} \\forall $ combinations..., that's implied by "Because this is a Bayesian model", so you don't need the "the prior for..." clause.
Your question contains the seeds of an answer. Calculate the fraction of values that are impossibly negative with a normal distribution with stated mean and SD. I'd recommend biting the bullet and using logit as a link function (not transformation) to avoid impossible predictions http://www.stata-journal.com/sjpdf.html?articlenum=st0147 gives a concise review, but you would need to identify R equivalents. That way you have scope for an innovative paper to show the way to your field. Note that dividing by 0.4 would make the problem that models might predict impossible values worse, not better.
To elaborate on the second paragraph, my understanding is when comparing models, we normally cross validate to make sure we're not overestimating the predictive ability of model 1 relative to model 2 based on model 1 having been overfit. But in the situation I've described above, overfitting doesn't seem like much of a concern, since the models are of the same complexity and they were trained on the same data set. I'm new to this stuff, so perhaps I'm thinking about this wrong.
Do you mean "latitude" instead of "tilt" (which is, for these purposes, essentially constant)?  I suppose the x-axes of the upper row of graphs (temperature) could represent latitude in degrees, but the units on the x-axes on the lower row (rainfall) are unrecognizable; they range from 0 to 8000.  What do these graphs actually represent?  These graphs are not scatterplots of rainfall or temperature against latitude, which is what one would expect from the question, so it's difficult or impossible to tell what you are really asking here.
Given the context, I'm pretty sure $E(Y - f(X))^2$ is supposed to mean $E((Y - f(X))^2)$ in the book. We're trying to minimize the distance $(Y - f(X))^2$ by finding f. The full context is linked from the question. Let me know if I'm wrong.
@user2238: Your last statement is true, but it would be extrordinarily rare. In fact, seeing a run of 5 heads in 5 tosses would happen just 3% of the time if the coin is fair. It is always possible that the null is true and we have witnessed a rare event.
Ah, from looking at the results of a model to find out which p and q produced it, right? Sorry, I'm not aware of any techniques to do this. You can try asking at http://quant.stackexchange.com
@StéphaneLaurent Have a look at the comments of the OP in the referred question. The code passed through a couple of modifications. The idea is to produce a modified Inverse Gamma with support on $(par[3],\\infty)$.
Formally yes, there is 15,462 PCs. But only 1,188 of them exist really - as variables with nonzero variance, so the rest are not displayed. Also, you might want to read http://stats.stackexchange.com/q/20101/3277. I'm not R user and don't know function `prcomp`.
Hello, thanks for your answer. Could you please be so kind to make an example in R with the glm function you mentioned? So I will understand better.
@chl Thanks for your attention. Now that I got the basics, I’d like to see explicit writing of the likelihood, and conditions for identifiability. Something about binary and ordinal indicators would be much welcome also: Fox uses polychoric correlations, which seems simple and efficient, but obviously more complex solutions are possible... I found some interesting papers but I lack time to go into an extensive bibliographic search; a textbook or a "paper of reference" would be welcome.
If you have ordinal data (as in your 4 point rating scale), you lose power by using a statistical test that is designed for categorical data.
Thanks for your comment. It's a great start for me. However, I just want to point out that, when developing the question, I wasn't thinking of creating a scale at all. The options are distinct responses and seek a "best fit" response. For example, if I have the multiple choices, "< 25%", "< 50%", "< 75%", "< 100%", one can see that "< 25%" is also "< 100%", but "< 25%" is more precise than "< 100%". I am seeking a "chose the most correct" response than a scale. Does it make sense? Do I still read about scales if this is the case?
@whuber Okay, then please correct me If I'm wrong, $E(Y|\\bar{X})=E(\\bar{X}|\\bar{X})+E(Y-\\bar{X}|\\bar{X})$ And now the second term is zero because the median is symmetric around $\\bar{x}$. Therefore, the expectation reduces to $\\bar{x}$
Thanks Nick. Indeed, today that's what I've done. I feel now more confortable once you've suggested it. As I said to whuber, results were not what I've expected. I mean, i saw no strong evidence of downtime decrease with age increase. I will tomorow try to shape data differently.
Dear Adam, thank you for your reply! I made some changes in my post. It is a shame I can't upload an image. Do you think for stimulus -1 will be used as the reference you are talking about? Or you think both 'one' and 'minus one' are tested against zero? Maybe my model is incorrect in that case. -1 refers to low threatening faces, 0 to neutral, 1 to highly threatening face stimuli. I have 88 participants equally divided over the two groups.
I don't mean to sound negative, but in your paragraph you've progressed from what sound like interesting and useful questions (do the groups show differences in this or that respect) to a much less interesting one (can they be grouped together--yes or no).  What makes you dissatisfied with the nuanced approach and steers you toward that seemingly sterile, either-or question?
up please help me
Do you know the exact stochastic process, or do you only have observations? Can you describe your signal in details? Do you want the entropy rate as a function of time? Do you have independent trials?
What you are calling stable here is closer under sums which seems to me more like the term infinitely divisible. In what fields is the term stable used for this? Is it becoming used in probability and statistics?
When you tag this as self-study, are you answering a question like this (e.g. from a textbook), or just interested in this question because of something that came up in the course of study?  The first would be self-study, the second wouldn't be.
@whuber thank you for the comment.  I tried to use Mathematica to invert the cf.  However, it had a problem in integration under some parameter combinations.  I then switched to FFT-based approximation, which did not suffer the same problem.  However, the latter approach did not give densities at dense enough points.  Probably, can I try interpolation to fill the gaps?
Firstly notice that, in time series context you actually are dealing with *spurious relationships* (if the latter are linear than *spurious regressions*) rather than *spurious correlations*. What you will typically see in time-series analysis are auto-correlations, that are obviously not spurious. Well all your questions are actually covered by 3 responses below. By the way even more sound Granger (non)causality concept does not imply theory based causation.
I like how you've done it. I have the same loop but with one exception at the end : I look at how frequently different features pop up as opposed to the lowest MSE from all iterations. I pick an arbitrary cut point (i.e. show up 50/100 iterations) and use those features. Curious contrast the two approaches.
@StephanKolassa Done. If you prefer a different document type, let me know.
@subhashc.davar, yes, that's another way to say it.
Thank you, done.
I used kruskal-wallis test to know if there are gender differences, income differences and so on. Now i have a trouble on how to interpret the results.if the there are no significant differences for each group, is it possible to use the calculated mean rank to see if there is a increasing or decreasing trend?
I'm sorry, but I can't understand what exactly you are looking for. What is your independent variable (the function), the dependent variables (the input values) and the parameters are?
@Matt: that's correct
Another possibility would perhaps be to consider a finite mixture.
Thanks @erik, I had a look at the Gelman book and will explore the hierarchical approach.
@MartinZotter, one thing is to look at what others do, e.g. that paper on German GDP. However the best approach is to come up with a good theory explaining the underlying phenomenon. For example, we have a theory for radioactive decay, and when we process its data we know which model to fit (exponential growth).
Yes, your comment on the other question directed me back to here, where I saw this answer, and have just now successfully (I hope) used it to derive what I wanted.  Thanks.
And thank you, too!
(2) isn't a necessary feature of a meaningful sample that you can associate with every element of the population a probability of showing up in the sample? I wouldn't know how to calculate that for a population of hypothetical bookings. But your point is valid - but I assume that on average dentists are considerably consistent.
I don't know why this question got a downvote without a reason. Please leave a reason before a simple click.
Yep, a direct application of the law of total probability using the conditional probability $P( \\cdot | C)$ gives you $P(A|C) = P(A|B_1,C) P(B_1|C) + P(A|B_2,C)P(B_2|C)$. So you need to know $P(C)$ to answer.
You're welcome.  It doesn't matter how you assign the indices as long as you do it consistently.
I did see, while trying to research an answer to this question, a method in which the difference between risk ratios was divided by the standard error, but I wasn't sure what the standard method was.
One clarifying question, is the same data for the single patient being used in each comparison, or is it 10 different data from the same patient?
Then all your censored data is all missing, and your results will be biased, potentially by a lot (after all you don't know how bad it might be, so you can't even bound the effect).
This would seem to have a connection with repeated measures designs and with growth curves.
Is this homework? If so, it should have the self-study tag
@GrantBlue You need to write down the full model if you want more specific pointers. I am affraid it is very likely you will not have full conditionals. Please, take a look at the Metropolis-within-Gibbs algorithm as an alternative solution. An adaptive version of this is implemented in the 'spBayes' R package.
I don't understand what will be $O_i$ and $E_i$ in terms of the observed cell frequencies a, b, c and d.Please help.
You have a fair chance of creating such a chart in R.  But this would require learning R, and that is not an something you can do in a day or two.
I'm not sure I understand exactly. Is the problem simply that the two variables have many levels? Also, is person_id unique for each row? If so, it won't do much good as a predictor.
I didn't have time for more than a pithy answer this morning. I was planning to expand on it this evening but you've done the the job very nicely, whuber.
In the Paper on page 25 they start creating the new variables and show he graphs in the appendix (p.42+)
@Michael: Almost undoubtedly it (i.e., IT) stands for "information theoretic".
Type II sum of squares is a good approach.