tttjjj/elt_bench huggingface dataset:
I agree--but finding the counterexample is the hard part, @Michael, not using the software!
Have you looked at AIC weights?  http://avesbiodiv.mncn.csic.es/estadistica/ejemploaic.pdf
One other way to consider is that the better of the two models should have the lower residual explained variation with respect to model complexities you hope you don't need.  You might do a pooled many-degree-of-freedom likelihood ratio test of a set of nonlinear and interaction terms and see if one model needs these less than the other.  I would choose the model needing them less.
I just had a think about the linear models that I was doing in first instance. I was using a quasibinomial distribution, but just figured out that I don't necessarily have to use that. My data are proportions but not based on counts from a binary outcome. They are the percentage of a volume. They might as well be noted as litres or qubic meters. I read an interesting [post](http://www.r-bloggers.com/not-all-proportion-data-are-binomial-outcomes/) on r-blogger about it.
Typically with some resampling procedure like bootstrap or cross-validation.
Thank you.  I just modified the question to reflect this advice.
@Engr I probably could have made one like it in `R`, perhaps using `layout`, but in fact it was done with *Mathematica* 9.
@CodeGuy You can't.  You need the raw data.  If you want to do component-wise tests and are willing to assume normality (not sure it is a good assumption) you can do unpaired t tests.  That would give 11 tests for each of the comparisons (c with b and b with a).  That would be a total of 22 hypothesis tests.  The multiplicity correction becomes very important now.  Of course given the magnitude of the difference in means relative to their standard deviations the p-values should be very small.
Sorry for the late delay; anyways, your approach doesn't really change my standard errors and your proposed solution is going again in the direction of the link I provided above.  That is, computing the standard errors yourself. I am just wondering why plm is not doing that automatically?
Thanks, that makes sense. In R I can do `library(pracma);trapz(dens$x[dens$x < 2], dens$y[dens$x < 2])` and I get `.139`. Close enough
Could you try this again with a sample of size 12?  And add a little fuzz to the x-values, so they aren't quite so close to being on a grid... I'm thinking your small sample size might, but only might, be the problem.
That's a great point, Tomas: N[i]==0 corresponds to a certainty that the outcome will be 0.  Consider streamlining your question (in another edit) to focus on this issue: namely, how can one persuade BUGS to accept this degenerate case?
Your answers seems Likert like. Given that, the "no comment' "no opinion"  answers are valid answers and correspond to the middle of the scale. Ordered Logistical Regression as suggested by @Walter is a more correct method but I think OLS is also applicable. Be aware that you may face some problems such a OLS result that is above your maximum value (2?) This is discussed in this CV question http://stats.stackexchange.com/questions/92902/likert-item-regression-analysis?rq=1
I totally agree with the sentiments here.  I agree that my result may be biased, and that the actual answer will depend on the reason the company is displaying the CVs.  I am trying to approach this from both a theoretical and real world perspective.

I am not sure that I have the likelihood function right.  For the reasons you have outlined, $N$ could be much bigger than $m$.  The only way to make this happen is either a) there is a prior $p(N)$ which is some rapidly increasing function of $N$, or b) my likelihood function above is not correct.
@Alexis, amazon has Bradley for under $10 (that's where I got mine, formerly at Community College of Beaver Library - so barely used 8-) ). While it seems his distinctions may not be mainstream anymore, his 300+ pages on nothing but distribution-free (or as most say, non-parametric tests) is pretty thorough, and how much have non-parametric tests changed since 1968? (I honestly don't know, but I doubt the world has shifted).
@shen As to the rest of your question, you actually *do* have two groups - Patients at Baseline and Patients After Treatment. The patients at baseline are essentially serving as their own controls. There are ways to analyze this, but some of them are relatively sophisticated. I **strongly** suggest going and finding a statistician or experienced methodologist who can advise you more directly.
I had noticed that too and checked it to make sure there wasn't some kind of error in calculation. It occurs because in going from the fourth to the fifth attempt, a factor of 20% is removed (chance of success at fourth attempt) to be replaced by 25% (chance of success at fifth attempt) while a factor of 80% is thrown in (chance of failure at fourth attempt). The net change is to replace 20% by 80% of 25%, which is just 20% again. Thinking of a generalization of this problem where 5% is replaced by 100/n% (with n=20 here), that should be considered a coincidence.
You need to be much more specific than how to find "the pattern." Frequently people mistakenly assume that giving a toy characterization of their problem it going to make giving an answer easier. In reality the opposite is normally true, the more details you provide about what you're actually trying to accomplish, the easier it is to give a useful answer.
Hey this helped a lot. I really needed some help with the basic subset commands.
I changed it to be $\\neq 0.5$.
@desperate-about-statistics: no worries! Just a suggestion for improvement. I'll write a brief tutorial on these `IF` statements too since many users seem unaware of them. I'll let you know when it's done if you like. If anything else is not (entirely) clear, let me know, we're obviously still developing the site and feedback is appreciated!
Sorry for the vague question, I naively/unconsciously presumed that you knew about the technical analysis indicators.
@drknexus As far as I remember MSerror=SS/df, thus it is also affected.
I'm recreating two data sets and try your answer with R. I will keep you posted!
If this is homework, please use the tag "self-study" and indicate your initial thoughts on a solution.
You want to read up on *non-response bias.*  That may help you elaborate on this question, because to get a suitable answer you need to disclose the nature of the "invalidity" as well what you're really trying to estimate.
See my edit below, @ric
If your only interest here is how to get Stata to do this, this question is off-topic for CV (it may be on-topic for [Stack Overflow](http://stackoverflow.com/), I'm not sure). You may want to read our [FAQ](http://stats.stackexchange.com/faq). Are you wondering / do you have a question about garch more generally?
@David That's what you stipulated!  The notation you use (and your sample code) imply alpha and beta are constants.  If they are not, then you need to tell us what their distributions are.
When you apply MLE repeatedly to your data then--barring any computational errors--you will get *exactly* the same results each time.  The way to think about this instead is to contemplate the ways in which your data could have turned out differently.  When the data vary, so do the ML estimates based on them and it is this resulting variation in estimates that is of great interest.
One thing one could add is to advise against the danger of spurious regressions. Since this is a school project it will probably not matter, but you need to be careful about your specification. After all if you estimate $$y = a + bx$$, you will always find a solution, but it may not be significant even with decent $R^2$.
There are plenty of non-distance-based algorithms, too. And when you consider minPts to be the same as e.g. `k` for nearest neighbor classification, then you could say the same for the minPts parameter. I guess the main difference is that for distance, there is an "often" sensible default: Euclidean distance; whereas for minPts the value will be data specific.
@drwho I've updated my answer with a few thoughts.
@cardinal You're right.  Actually, I use whatever is handy and seems like it will do a good job quickly.  FWIW, here's the code: `Module[
{y, w, a = 0.1, z = 3.24, f = ChiDistribution[7.6], xmin=0, xmax=5},
y = CDF[f,z];
w = InverseCDF[f, y^(1/a)];
Show[
Plot[{CDF[f, x],CDF[f,x]^a} , {x, xmin, xmax}, Filling->{1->{2}}],
Graphics[{
Dashed,
Arrow[{{z,0}, {z,y}}], 
Arrow[{{z,y}, {w,y}}],
Arrow[{{w,y}, {w,0}}]
}]
]
]`
@jbowman, Not really. The states are evaluated in the order of fittest-first (i.e. an A-star search), so where a state falls in the sequence is highly dependent on its fitness.
+1 Good description and commentary.  In the cases I have seen, the parallel straight lines are merely set one or two standard errors above and below the fitted line: crude and misleading, but better than nothing at all.
@MånsT Yes, it starts off sounding like capture-recapture, but at the end it apparently is not. After all, people are not marked when they are encountered, so it is not possible to determine who has been recaptured. Because we don't know the rates at which people go on cruises, it looks like question (1) is not answerable with these data, and therefore neither is (2).
rmANOVA assumes [sphericity](http://homepages.gold.ac.uk/aphome/spheric.html), which doesn't seem likely to me, given the temporally-ordered nature of your data. You may want to use more general mixed-effects methods.
@chi I was thinking of the users each rating each of a few dozen sports as one of two levels (like or not). Does that answer your question? If not, I should probably review how Cohen's $\\kappa$ works.
That gives the importance for the overall model. My problem is, once you start scoring individual instances, can you get some measure of feature importance by instance? I thought I read somewhere that this might be possible with an alternating decision tree, but can't find the reference anymore.
So pick a few.  Some is better than nothing.
+1 for your comment to the OP's question. I suggest to make this reply a comment too... it does not address one of the OP's (still unclear) questions but may help clarify his/her intentions.
Outstanding answer, @Scortchi!
@whuber thank you so much for this. I've wondered idly about similar questions in the past.
Whether to test them together or apart depends on whether you  want to control for the other one. But probably apart.
@Glen_b Thank you for the clarification.  My comment was based on an apparently incorrect reading of the question, in which all $10+F_{HH}$ tosses had been made.  If only first ten are made then you are completely correct that this asks for a prediction interval.
... and I've only just noticed what you're forecasting. There are relatively widely used models for forecasting mortality curves (and hence, life expectancy) already, including some R packages. You might want to take a look at [this](http://cran.r-project.org/web/packages/lifecontingencies/vignettes/mortality_projection.pdf) as a starting point. There are numerous relevant papers and some books
@Procrastinator, why is there any need to use distribution testing *for normality* in order to justify using the t-test? That is not one of the usual assumptions for applying a t-test.  (Approximate normality of the distribution of the *sample mean* is the usual assumption.)
Are you trying to identify (1) people who will or do use the substance or (2) the level of use of people who are users?  You say you have data on users only.  So it would seem you can only address #2 above.  Also, have you calculated a correlation between the scale score and the level of usage and looked at a scatter plot of the relationship between scale score and level of usage?  If so, what did you find?
http://www.stat.cmu.edu/~cshalizi/uADA/13/lectures/ch02.pdf. I am referring to the notes here concerning prediction error under 2.13. I am not sure how the result was arrived at.
Welcome! Having the ability to reproduce the error will improve your chances of getting advice on how to fix it. This question on SO gives great tips on such a topic: http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
@gung: I read it the opposite way because of the question of a relationship, but the "e.g." section seems to ask for a test of group differences. The difference between these approaches is part of what I meant to refer to as subtle. I wonder if we have any good questions yet about when to prefer one or the other...
Perhaps I am going blind, but are you certain that your conditional densities integrate to unity over their domains and for an arbitrarily fixed value of the conditioning variable?
Sounds like the book uses a definition of Inverse-Gamma different than your one.
+1 to Robert Alberts, & +1 to Peter Flom (Cleveland's books are *definitely* better for statisticians, although Tufte's are beautiful as well, and I have read all of them). @AndyW, *Elements* is introductory, e.g., it has  guidelines for making an informative graphic. *Visualizing* demonstrates how to center your data exploration process around graphics; it starts with preliminary visualization of the data, talks about the issues at hand and walks all the way through to assessing the final model (e.g., residual analysis) via graphics. The latter is much more informative than the former.
This question appears better suited to for a site that discusses legal questions, such as the impact of patents. It doesn't seem to contain any statistical questions. Even though random forests are on topic here, it looks like questions relating to the impact of patents on them aren't --  any more than asking about the likely medical effects of swallowing a small computer really isn't a computing question.
The $p$-value for the hypothesis $H_{a}:\\mu_{\\text{Women}}<\\mu_{\\text{Men}}$ vs. $H_{0}:\\mu_{\\text{Women}}\\geq\\mu_{\\text{Men}}$ is about $0.034/2\\approx0.017$. So you have evidence *against* the null hypothesis. But could you explain what you hope to achieve with this? Normally, you decide *before* the analysis whether you do a two-sided or one-sided test
Models have more than one possible purpose. Is the purpose of the model prediction?
I have now done so!
Thanks for this explanation! I also restructured to question. Regarding question 2: How would I take this into account? (Multiple testing?)
Isn't that (part of) what the theorem asserts? Perhaps you could be a little more precise about what your difficulty is?
@whuber Because I'll do inside a loop, something like: if( uniform_random([0,1]) < $P_{x,c1}$ ) then "create a a new center at x"; else "e.g. assign x to c1"
I would like to know why my question got downvoted, in order to be able to ask a better question later. I have been trying to understand a concept that is difficult for me, and described the issue in my own terms. I was able to receive a sufficient answer for my question. So where is the problem?
Look into [distance correlation](http://cran.r-project.org/web/packages/energy/energy.pdf).
A similar question but rotated by 90 degrees was asked here: http://stackoverflow.com/q/13327489/1700987
How i come to know whether there is unobserved variables in my panel data or not?
btw re terminology. "barriers to entry" is used to refer to companies wanting to release new products and compete in a new area, similar concept applies "on other side of transaction" to customers switching products but maybe there is a different term there? anyway the authors need to tie in their ideas with ***marketing*** which is indeed using more "viral" models. also a key concept in this area [should have mentioned this above] is ***market share***.
Have you read the following paper by Morris: "The BLUPs are not the best when it comes to bootstrapping". It may relate to your work. [link](http://odin.mdacc.tmc.edu/~jmorris/papers_files/Morris%202002%20SP%20Letters.pdf)
Thanks for your reply. I'm comparing the model season with the model species, camera and season. I've noticed that when I run models with for instance season included the p values can change. For instance within my first model season, the p value for summer sr is 0.18 and thus insignificant but within the 2nd model, species, camera and season the p value for summer sr is 0.03 and thus significant. I'm wondering if it is correct to assume that the p value from the 2nd model is due to collinearity and thus not representative of the actual significance of this variable (or level?) summer?
I believe the reference implementation of random forests stop with ~5 observations in the nodes when the response variable is continuous.  That would still introduce a small amount of bias if the splitting variable were continuous as well.  Similar to why LOESS often looks better than a centered moving average at the edges of the support...
The data frame in `df` is for illustration purposes, not your original one, right? For your real data, what is the approximate number of observations that you have for a given month?
+1: Sorry I'd just found Rscript and edited my answer to include this, so we've ended up with a similar answer. Your `read.table` idea is a nice way to get around one-item-per-line.
Thanks! I actually own the Kruschke book already and just went back to review it and realized it's exactly what I need right now. Thanks for the pointer!
thanks for the advice. i'll read on the reference first. thnaks again for the advice.
It is no different, in general, than when $\\mu\\le 0$. The reason is that the critical region corresponds to the worst case among this (compound) null hypothesis and its size is the *supremum* of its probabilities among all elements of the null. In most contexts, leaving or including $\\mu=0$ will not change the supremum at all. Your comment does, though, get to what I think is the critical point: the null hypothesis *must* allow us to construct a critical region of the desired size (and usually that means it corresponds to a definite distribution of the test statistic).
thanks for the illustration. But as mentioned in the comments above, we are aware that without further constraints the solution is not guaranteed to be unique. These are the kinds of constraints the question asks about. Also, the question refers to global optima of the kmeans objective.
"unskwed" = "unskewed"?  Symmetric?  A problem with this question is that it requests a quantitative answer (a distribution or family of them) but supplies no quantitative criteria for selecting among the infinitely many possibilities.  *How* "heavy" should the tails be?  What exactly is the "purpose"?  ("Algorithm" is far too general to be a useful description.)  You might find [this search](http://stats.stackexchange.com/search?q=heavy+%2B+\\[distributions\\]) useful.
Thanks for the reply.

I conduct two series of analyses. In the first, I run four sets of regressions, each with a different relational outcome (e.g., relational closeness). For each DV, there are four regressions, each with a set of control and behavioral IVs as well one of four relationship maintenance strategies. These strategies are correlated, so they cannot be included in the same regression, and I want to see how each individually relates to the DV.
No zero inflation, no one inflation; just that observed zeros and ones are not fatal. Fractional logit would be one name. The reference to binomial in software calls is just a trick but the point of the literature is that the trick is a fair one. The model goes back to Wedderburn at least.
Can you say more about what you are doing? What are your variables? The decision of whether to label a variable nominal, ordinal, continuous, etc. is partially a choice of the analyst.  What is the purpose of your analysis? What statistical procedure are you performing?
@whuber, it would be worth mentioning the software options. I am guessing that you used Stata (if I am trained well enough to Stata graphs and tell them from R and SAS graphs), where this model is fitted with `ologit`. In R, you can do this with `polr` in `MASS` package.
Ah, sorry, didn't fully answer your question.  $cov(\\hat{\\beta}_1,\\hat{\\beta}_2)$ would be the entry in the matrix with column name, in your case, `as.factor(X1)1` row name `as.factor(X2)1`, assuming what you wished to find the covariance for were the coefficients with values -0.12795 and 0.05666, and similarly for the other covariances.
Also, I was thinking that because we tested at different times of day that could have confounded the results. would it be appropriate to suggest this? My other question: do you think a N=67 sample was too small for a 2*2 between subjects anova?
Minor point: I'd argue that neither needs to establish causality unless that's your goal. Right now I'm working on a project where we couldn't care less about causality (yet), we just want to find patterns of association.
Upon re-reading the question I see that it has two versions. One asks about flops that contain "four sequential cards." The other asks about flops that can draw into a straight.  The latter include sets like A-2-4-5 in which drawing a 3 will produce a straight.  I have addressed the latter question, believing it to be more relevant (as well as substantially more difficult to answer). The former version can be addressed using the same techniques shown here, as well as simpler ones akin to the derivations shown on [Wikipedia](http://en.wikipedia.org/wiki/Poker_probability).
Like also the equation in the book says that its an "approximate" confidence interval, what makes it approximate. Forgive my ignorance but I wasnt able to see how you got the scatter plot, are you just plotting you independent normally distributed random variables?
Thank you. I haven't know the notion of "stacked generalization." I have added your idea as the third approach.
Of course that's right @cardinal - lots of theoretically justified things go bad when you're trying to do things numerically with nearly singular matrices. I was (conveniently) imagining the situation where the target correlation matrix was not in the realm where this was an issue. It's good that you pointed this out - thanks (and thanks for the edit to my other answer)
You may want to look up "choice-based modeling".
When you say "variable is different" among species, what exactly do you mean? Different at one time? Different at any time? Different on average? Has a different growth rate? Has a different starting point? Or what?
You certainly should not use the burn-in samples for anything.
Very interesting points - thanks for the links. I'm wondering more if the structure of the problem could influence one's decision (e.g. method X may be a natural fit for inferring network structure based on node attributes)
Where is this in Chung? Did you mean pages 161-165, by chance? Even so, that deals with *characteristic functions*, not *moment-generating functions*, as requested by the OP.
If you watch a lot of game shows you would know that contestants are not rational and often don't even appreciate the odds.  They see a great chance to become a millionaire.
@Momo R comes without warranty, true, but I would like to see a reference that SAS has that strong of one. I could not find the SAS license text on their website, but was able to find something for one component: http://support.sas.com/documentation/onlinedoc/sasc/doc650/common/agreement.htm That has a 90 day warranty on the media that the software is shipped on and an all caps disclaimer of any other warranty. Please give a reference other than "someone once told me".
(+1) Are these regions known and annotated as such in your data or do you have to "identify" them after the fact? Is it reasonable/sensible to estimate via a resampling approach, e.g., redistributing the points at random within the censored region? I can imagine how that might not be satisfactory either. Presumably the nature reserve would have well-travelled trails/paths.
I'm glad you're not downvoting, because the misunderstanding might be on your part: while *multiple regression* deals with multiple independent variables, *multivariate regression* concerns a vector-valued *dependent* (response) variable. It is not the same as a set of independently fitted multiple regressions to the components of the response.