tttjjj/elt_bench huggingface dataset:
@BabakP In this case, perhaps, one shouldn't over-rely on Wikipedia. While it's suitable to use the K-W in that situation, and what's more, one can easily produce estimates and even interval-estimates for location shifts, the K-W is a suitable test for a wider array of circumstances than a pure location-shift-alternative.
well, the problem is that I want a 1-sample test, not a 2-sample test! I am testing the null $E[X] = \\mu$, and not $E[X_1] = E[X_2]$. I will look up the Kubinger et. al., paper (Ich kann schlecht Deutsche).
+1 Thanks for editing; your question is better for it. I made some minor changes myself to reflect your comment, please check that your intended question is still correctly conveyed.
@Ahmet Yılmaz: The point I am making is: algorithm A may "prioritize" precision, while algorithm B may prioritize recall. It is not at all surprising to get a result like yours.  Try comparing the 2 algorithm on a non-threshold dependent metric, like area under the ROC curve.
@Ismail: Fixed.
I see my flaw in my understanding now. I usually think of probability in terms of $accepted/possible$ cases. But there are many more than just 8 ways of having 8 heads in 21 flips.
I'm confused by your edit. Did you mean to put the $\\max$ function inside the sum? Also, in the original equation, are you wanting to allow for an intercept or not? Otherwise, by forcing the line through the origin, the answer is trivial.
So that this can be made into an answerable question, please tell us what *you* mean by "best."
Does that help? http://stackoverflow.com/a/9667122/420055
See my elaboration below.
Note that I am using the predict function from the raster package (i.e., predicting directly from my raster layers) not from a dataframe, because my layers were too big to convert to dataframe. My rasters are floating point data and I think one problem is that there is no attribute table associated with them.
One might add that fitting the regressions separateley is indeed equivalent to the multivariate formulation with a matrix of dependent variables. In R with package mvtnorm installed (1st: multivariate model, 2nd: separate univariate models):
library(mvtnorm);
X <- rmvnorm(100, c(1, 2), matrix(c(4, 2, 2, 3), ncol=2));
Y <- X %*% matrix(1:4, ncol=2) + rmvnorm(100, c(0, 0), diag(c(20, 30)));
lm(Y ~ X[ , 1] + X[ , 2]);
lm(Y[ , 1] ~ X[ , 1] + X[ , 2]);
lm(Y[ , 2] ~ X[ , 1] + X[ , 2])
I'm not sure I quite understand your question yet. Is $\\mu$ a vector, as in $\\mu = (\\mu_1,\\ldots,\\mu_n)$? And, is the notation $Y_i \\to N(\\mu_i,\\sigma^2)$ meant to indicate that $Y_i$ has a normal distribution with mean $\\mu_i$ and $\\sigma^2$. (I ask because this is usually denoted by $\\sim$, i.e., `\\sim` and so I want to make sure your question is being interpreted properly.) The description after the display equation could also be improved with a little more elaboration, I think.
@user1471980 I updated my answer, I am removing my comments to avoid confusion.
2) $P[y_i=0]=0$, you may need to reformulate your 2nd question. perhaps, in intervals or maybe in discrete distributions.
The reason ratio distributions might fail to have moments is due to the possibility that the denominator is zero (and has a nonzero chance). Because computations take finite amounts of time, we may assume *a priori* that the denominator is nonzero, whence it is likely the ratio distribution for this problem is nicer than you think.
@Clarinetist X is a N dim. variable corresponding to MFCCs. MFCCs are computed for each frame of audio, wherein you take a sequence of amplitude values from audio frame, apply a mathematical transformation, and get the MFCC value at the output. Is $\\zeta$ then a set of amplitude values i.e. $\\zeta$ is a vector?
Ahh.  I see.  Then I'd also be (more than a little) reluctant to impute existing data and would recommend it wasn't done at all, despite this sort of argument: http://gking.harvard.edu/gking/files/measure.pdf
I confess to being confused by the reference to typical sets in this context. How does that explain the distinction between 60 sixes and any other 60-element sequence, given that all such sequences have the same probability?
+1 - these comments are *not* meant for a discussion (however well-intentioned) designed to help the OP find the answer for themselves. CV isn't a site for teaching - if a simple hint isn't sufficient either leave it at that or write up an Answer.
@Peter Flom: with the difference that in my case the test sets would be independent, while in CV they are not.
http://pic.dhe.ibm.com/infocenter/spssstat/v20r0m0/index.jsp?topic=%2Fcom.ibm.spss.statistics.help%2Falg_mixed_ci-covariance.htm
It seems that we should make use of the right distance, 20 km.
@whuber Could you demonstrate what you said and provide a complete answer? That can be useful for learning to me and future readers. Furthermore, it can be possible to me to narrow my question according to your answer.
(+1) Welcome to our site, Drew!
@whuber: Thank you for clarifying. So here is what I have done as an initial ad-hoc approach: 1. Model the expected values of the conditional means using a local linear estimator on the logs and optimal BW selection based on cross validation of least squares. 2. Calculate squared deviation from the conditional means so obtained. 3. Model the variance again as a second step using another local linear estimator to estimate conditional expected values of variance. 4. Calculate deviation/expected conditional variance for each point to obtain the final rank score. I wonder if something is missing.
Read this question and the answers http://stats.stackexchange.com/questions/23197/type-iii-sum-of-squares-from-sas-and-r
@Glen_b, with so few data & only 2 means, it's not clear it will make much difference in practice.  An OLS regression just fits the 2 means, as would a LR w/ the right numbers of trials to make sense, & a beta regression does the same as well.
I respectfully disagree regarding your second recommendation. Though certainly an interesting book, it is almost surely *not* the place to start for a non-statistician. Especially someone concerned about clear delivery.
This is a great comment--but it doesn't seem to address the question asking for "statistical reasons" underlying the standard BMI formula.
(+1) "Some sort of regression algorithm" is perhaps broad enough for it to be worth mentioning that techniques that shrink all coefficients together according to their magnitude, such as ridge regression or LASSO, will give different answers when you standardize predictors & when you don't.
Which function in TSA are you using to fit transfer function models?
Thanks a lot!!! for your very great answer!
sometimes its good to just provide a simple concrete example: What is $\\int_0^5 x^2 dx$? What is $\\int_0^\\pi \\sin{x}dx$? Are they functions of $x$?
+1. It may be useful to know that this difference of two betas (with integral parameters) has a piecewise polynomial PDF (and CDF): there are just two formulas, one for the negative values and another for the positive. A simple example: if $X\\sim B(1,2)$ and $Y\\sim B(2,3)$, then the CDF of $z=X-Y$ is $\\frac{1}{5} (1+z)^4 \\left(3-6 z+z^2\\right)$ for $-1\\le z\\le 0$ and $\\frac{1}{5} \\left(3+6 z-5 z^2+2 z^5-z^6\\right)$ for $0\\le z\\le 1$. For typical click-through data, though, approximating this as a Skellam distribution should be accurate and tractable.
+1, this is a really excellent answer.
@ttnphns That suggestion is not very helpful.  You should consider the fact that Pr0no is having trouble understanding time series model fitting and is a bit panicked.  A simple explanation like what I give below seems to me to be more appropriate.
In such large samples it makes almost no difference which you use. However, your chi-square statistic looks wrong.
Even if the constant is included, the AIC (AICc) can be negative.
While the sentence is indeed vague it should be clarified by numbers. If it isn't insist they report numbers.
Thank you for your reply. Whoops, I've fixed the link to Lenth's paper. However, I am very confused by your location estimate. When I plot the bearings in my example, they go northwest, northeast, and north. The estimate location you listed (approx. 553650, 2813504) is quite a bit south of all the points where the bearings were taken. It is extremely off. Did you perhaps input a bearing wrong? Here's everything plotted on a map: http://img.photobucket.com/albums/0903/abovedust/mle_map.png

You're correct about the eigenvalues. I added a PDF to my post which helped me understand the ellipses.
Hi Jon, could you explain how to generate ranks within groups?
Also, is there a way to compare which theta value is closer to the population theta . I mean which theta value is more accurate?  Thanks.
An example why precision is not enough is that you can classify everything as positive and you have a nice fine 1 value for precision and an useless classifier.
"*Plotting the prices into a histogram, it is clear that it is not a normal distribution, so it is not a stationary stochastic variable I am dealing with.*" -- how does this follow (or how would the converse follow, in the other situation?)
@whuber: ...Let $X_i \\sim F$ and $Y_i \\sim G$ be independent iid sequences. Then $n \\hat{F}_n(x) = |\\{i: X_i \\leq x\\}| = |\\{i: F(X_i) \\leq F(x)\\}|$ and $n \\hat{G}_n(x) = |\\{i: Y_i \\leq x \\}| = |\\{i: G(Y_i) \\leq G(x)\\}|$. So, with the aforementioned assumption that $F$ and $G$ are continuous distributions, under the null hypothesis $F=G$, we see that $\\sup |\\hat{F}_n(x) - \\hat{G}_n(x)|$ is equal in distribution to the same statistic obtained from two independent $\\mathcal U(0,1)$ samples of the same size.
Do you think you could provide a little more insight into what your data looks like.  I am a little confused.
@chl it is only full - reduced if you take the -2 into the equation.
I think the idea may be that the height is the sum of the distances between the old means and the new means weighted for the number of elements, less an adjustment for the heights of the earlier subsets to take account of new elements being introduced.  But this is my pattern recognition rather than knowledge.
@Scortchi, I guess I do not get that interpretation when I read the question since they state that they have parameter estimates $\\hat \\beta$ and $x$.  Never the less, if you want to change it to what you are saying then just make the simple adjustment of re-estimating the model parameters each time inside of the for-loop with each bootstrap sample. Not sure why we need to fuss with this anyway though.
Thanks -- that is hugely helpful. I think I will go with the long format.
This question would almost certainly be a better fit for stackoverflow.com, as there is little specifically statistical here.
really need a help on that, someone please?
EMS, I don't think it's an embarrassing question; judging from how you posed it, you've clearly put a significant amount of effort into thinking about your problem. Since this problem is about finding an analytical expression, it is out of scope here, and probably best asked on Stats.SE. I'm going to talk to their mods and see if it's a good migration candidate and get back to you; for the time being, please don't delete your question.
@soufanom I wrote that you need an extra cross validation data set to select best performing classifier, otherwise your results will not be reliable. How can you judge performance of a model by running it on a data set that was used to either train the model or to select parameters. IMHO the only reliable way to judge performance of any classifier is to run it on previously unseen data. I hope this answers your question, if not please refine it and add more information.
I note that the fits are not identical. The standard errors have doubled in the NB model.
+1, this is a good question. Unfortunately, I think the necessary picture would need to be in a high-dimensional space. In other words, I'm not sure how well it can be pictured.
Unfortunately, this is not correct, as written.
@Caroline You can check everything easily in R: `"1"==1` returns true (T) but `" 1"==1` returns false (F). `c(T,F,T) & c(F,F,T)` returns `F,F,T` as expected but `c(T,F,T) && c(F,F,T)` returns `F` because it only compares the first element in each vector.
Ah, okay, I think I get it now. Need to be extra careful with these *variables* :). Thank you for your explanation.
So, looking at @SamLivingstone's answer below, you need to add the analogous second term to your sum.
Please post your code. Something is wrong.
This is the area of (nonlinear) design of experiments. The optimality criteria usually have to do with the joint covariance of the parameters to be estimated (e.g., $\\det {\\mathbf{V}}[ \\hat a, {\\rm vec}\\hat B ]$, aka $D$-optimality criterion).
"Standardization is simple and involves rescaling the density to have unit standard deviation" So it does NOT mean scale 1!
+1 Good points and good guidance.  We could (and should) go even further: if interaction makes a substantial difference, then what else could be done to further improve the model?  Is the model even appropriate for these data?  (One way that interactions can become significant is when the independent variable should really be re-expressed in a nonlinear fashion.) The significant interaction becomes a flag that alerts us to the need for further investigation.
+1 The illustration says it all.
Since you applied chi-squared test, I assume your data is categorical. However, person's correlation coefficient is intended to quantify the association between variables measured on an [interval or ratio scale](http://en.wikipedia.org/wiki/Level_of_measurement).
**An observation**: Let $Z_1, Z_2,\\ldots,Z_n$ be iid $\\mathcal N(0,1)$. Note that $\\mathbb P(|Z| \\leq z) = 2 \\Phi(z) - 1$, so the median $m$ of the distribution of $|Z|$ solves $\\mathbb P(|Z| \\leq m) = 1/2$, hence $m = \\Phi^{-1}(3/4)$. Since the normal distribution is continuous, the sample quantiles converge everywhere, implying $\\hat{m}_n = \\mathrm{median}\\{Z_1,\\ldots,Z_n\\} \\to \\Phi^{-1}(3/4)$ in probability. In particular, this means that if $X_1,\\ldots,X_n$ are iid $\\mathcal N(0,\\sigma^2)$, then $\\mathrm{median}\\{X_1,\\ldots,X_n\\} / \\Phi^{-1}(3/4)$ is consistent for $\\sigma$.
I don't think you can prove this. @Procrastinator offers some tools to give evidence, but really this is a question that revolves around how the variables were drawn, because even independent samples will be correlated sometimes (indeed, if you randomly generate two variables many times and check correlation, e.g., they will be significantly correlated 5% of the time).
I am comparing group1 (my null case, i.e no treatment) vs group2 (my group exposed to treatment). So comparing two contingency tables as opposed to columns within one.
In general I do agree, but what is PPV ?
Where does your lrt() function come from?  Is it: https://stat.ethz.ch/pipermail/r-sig-mixed-models/2008q3/001175.html?
I don't see how. Can you explain? Note in particular, that in the top schedule I give, the 25th seed has only a fractionally harder task than the 24th seed, so if it is unfair on 25, it presumably must also be unfair on 24.
@Erik, thanks, I misread. Anyway, regarding the first part, I don't think $R^2$, in isolation, can be used to "check if my regression model is good", to use the OP's words. Your model could fail miserably to predict effectively on vast majority of the data while still having a high $R^2$. See [here](http://stats.stackexchange.com/questions/32781/how-can-i-use-the-value-of-r2-to-test-the-linearity-assumption-in-multiple-re/32785#32785) for an example - in example (1), there's almost no predictive power but $R^2$ is still high.
@B_Miner Have you tried AUTOBOX as it is much more powerful than SAS or any other time series software to deal with this issue.
@Gavin
You are right. The reason for the differences between the regression coefficients is due to different effects between the instances of `m`.
This behaviour is a result of including at least one random factor in the model formula. Hence, it generally is a basic part of mixed models.
I noticed there are votes to close this question as too broad, but there *is* a general theorem about the asymptotic behaviour of MLEs that can be succinctly stated. I put in a terse answer that I'll expand a bit later.
In the example I gave, consider the variance of $X_{(n)}$ in the limit (it's 0). In other cases, it's the mean you need to worry about.
@MichaelChernick I think there is a missinterpretation of what a Bayesian credibility region means. Suppose that the true value of the parameter is $\\theta_0=1$ and you choose a uniform prior on $(1,100)$. Therefore no credibility interval would contain the true value of the parameter, contradicting your reasoning.
Adressing the question of categorical treatment of age and scores: Did you find out how to let your software treat them as numerical?
Hmmm, that's an unusual scenario. I gather the height of a tree isn't the actual variable you want to predict; that could be measured. Can you state what the actual variable *is*? The more details we have, the more likely someone will be able to provide helpful advice.
For that, the OP needs to switch to probability output for the SVM. IIRC at least with libSVM this is calculated by fitting a logistic, and I'm not sure how closely related this is to the original SVM.
@dcarlson Thanks for your questions. I tried to answer them above, but if you have any other questions before you answer please let me know.
welcome to our site :). While rapidminer questions are in general _on topic_, you might be better off asking the question on forum.rapid-i.com (the official forum), since the rm expertise is not extraordinarily high here ;). If you are going to ask the question there, please tell it by dropping a comment.
Gnedenko's original paper, which is republished in Kotz & Johnson *Breakthroughs in Statistics Volume I*, uses elementary techniques and provides ample insight.
ah, didn't see robin's post!
Can you add a bit more detail to the question itself rather than putting a hyperlink with the question at the end of it? Cheers!
Thanks @Placidia
If there is a physical/biological/experimental way of improving the precision and reducing this natural variability, then it is possible. If not, then you have to deal with this. This is, depending on your goals, you have to include this characteristic in your statistical model. What are you planning to do with these data?
Welcome to the site, @user2747481. Would you mind fleshing this answer out a bit? We would like our answers to be mostly self-contained. Since you are new here, you may want to read our [about page](http://stats.stackexchange.com/about), which contains information for new users.
+1 Beautiful. :)
A sum of $k$ exponentials with parameter $\\lambda$ gives the erlang with parameters $k$ and $\\lambda$. Thus, it is not a asymptotic result but a finitary one.
$P(y_i|x)$ considered as a function of $x$ varies continuously with $x$, but $P$ is the pmf for $y$, and as such is defined over the discrete values $y$ takes.
Beautiful plots
@Jonathan: Thanks. Yes, I could have made the proof quite a bit more terse. Due to the wide range of backgrounds of the participants here, I often tend to indulge in extra (or, perhaps, excessive) detail. :-)
It won't help if just the numerical algorithm is diverging for certain parameters.
Actually if you look at figure 1, the Strome 2002 data shows odds ratio of: 10.2 (2.8, 37.0), so that 18.2 looks like a typo to me. Also in figure 2 they report the OR as 10.2 from Strome data.
From my experience, this message is displayed when there is a very big (infinity for BUGS) number obtained, e.g. 10e100 or something. In addition, the diffuse gamma prior distribution usually is the cause of this. I would recommend to use uniforma distribution instead of gamma. Also, from my experience, gamma prior is is quite informative, i.e. influences posterior more that uniform.
If you did it correctly, centering should have no effect on interaction term estimates and their standard errors.  Centering should only affect main effects.
Thanks, but I'm aware of that, and that's specifically *not* what I was asking for (since a straight line is more or less the simplest model possible, that's implied in my question). I've clarified the question.
It looks to me like there might be an error in your calculation of $\\log f(x)$. I think it should be $\\mbox{const} -\\frac{n}{2} \\log(\\theta) - \\frac{s}{2\\theta} + t - \\frac{n \\theta}{2}$. As is, there is a positive probability chance that $1 - 4 \\frac s n < 0$ which is a problem.
Suppose alpha were .01, the *p* for A were .001, and the *p* for B were .00000000001.  In the binary sense you describe, both results would be the same (on the same side of alpha). Would there be any amount of money you'd be willing to bet that B discriminates the groups better than A does?
I agree, log-normal often seems to be a surprisingly good approximation for all sorts of economic variables.  Definitely in this case it would be my first choice.  Exponential istribution woudl rarely be a good fit.
Regarding this conservative weighted test for *partially correlated data*, the authors seem to distinguish several cases (large sample approximation, unequal ratio of matched/independent samples, small samples) in order to get the distribution of $T_0$ under the null. It would be helpful if you could summarize them in your answer, since the same article is referenced in another (partial) reply.
definitions of [panel](http://en.wikipedia.org/wiki/Panel_data), x-section and TS are standard in stat textbooks.
+1, but I wonder if your equation for $\\beta_0$ could be clarified...
The two copies of this question reference each other and, appropriately, the stats site (here) has a statistical answer and the math site has a mathematical answer.  It seems like a good division: let it stand!
Re: causal connections:  This is too important to skip over on the way to calculating effects.  Please see http://stats.stackexchange.com/questions/11327/dependent-variable-is-a-function-of-independent-variables-can-i-sensibly-includ  and  http://stats.stackexchange.com/questions/47314/variable-entered-in-logistic-regression-model-is-part-of-another-variable-entere
Can you provide a bit more context to your question? What type of categorical missing values are you interested in?
Yes, I was trying to use R and this solved the problem. Thanks
start by looking at the answers to [this question](http://stats.stackexchange.com/questions/125/what-is-the-best-introductory-bayesian-statistics-textbook)
Is this really just an indirect way of asking http://stats.stackexchange.com/questions/33078/data-has-two-trends-how-to-extract-independent-trendlines?
That's it, thank you.
Other responses have answered your question, but I just want to remind you that prevalence is about the phenomenon at a time or during a time period, so if you asked for a "suitable way to present" it, I'd suggest don't forget attaching the time reference.
Blain, your initial intention is correct. If you strongly believe there are 4 factors, specify 4 dimensions on CATPCA. This will quantify your categorical variables optimally to meet the expectation that there is 4 factors. Then (if you want FA model rather than PCA model and/or you want rotations) use standard FA with those quantified variables to extract the 4 factors and to rotate the loadings.
now i can understand it...thanks for the explanation @Michael Chernick ...:)
I would add that if you have high frequency oscillations it could be that 2 months should be high and one month low or vice versa.  So linear interpolation could impute poorly to all three months.  You need to knwo something about what you don't know to do this right.  You are probably making unverifiable assumptions.  If yuo are lucky and the assumptions are approximately right you win.  If not yuo can lose badly!
Regarding bootstrapping for small samples, I found the following PDF these days that might help you: http://www.stat.washington.edu/fritz/Reports/bootstrap-report.pdf