tttjjj/elt_bench huggingface dataset:
If the errors were correlated over time you could model the noise component as a mving average with the coefficients of the moving average determining the autocorrelation function for the noise. Model parameters for time series which include the coefficients of the moving average terms can then be fit using maximum likelihood.  If the data do not form a time series there still may be a way to model the noise term so that correlation is part of the structure.
I have made some minor edits to my answer.
I am struggling at this moment and I guess I got the idea but I cannot make it to work
If you have cross-over interactions, this might et difficult, indeed. But I think in your case the main question would be: did you expect a positive main effect of 'year'? Then add it to all slopes.
@Andy W: thanks for you edit :)
@FirhatNawfanH. Yes. Usually ARCH/GARCH mainly used in financial time series.
So you have a fitted curve ($\\hat{g}(x)$), and you want to draw a tangent line at some known point, $x^*$?
you'll scare the poor man away forever!
@vonjd The author seems to have posted it on both sites now.
I am sorry if I didn't make my self clear. I want to make a classification 3x3 table similar to the link I posted at the end (classification table a la SPSS in R). I posted the probability table just to show the result I take when I run the olr and in case it might help anyone
"subtract -1" would be the same as "add 1". Did you really mean that? It would seem to make more sense to subtract 1
Do you really have 21 data points in total / s?  It's common in a repeated measures design to get multiple measures of each datapoint for each subject.  In an ANOVA those are aggregate out but in a mixed model you're best off keeping each individual data point.
If initially  $run/pcp = pcp + ant$ etc then it is logically _necessary_ to include also $(pcp)^2$ when the dependent variable is $run$ only -you essentially bring the denominator in the left-hand side to the right-hand side.
(+1) This is a nice answer, though I have a couple of comments. First, I think this may be selling the probabilist a little bit short. It is not at all uncommon to consider a set of probability spaces in a probabilistic model, and indeed, the possible measures can even be *random* (constructed on a suitably larger space). Second, a Bayesian (in particular) might find this answer slightly disconcerting in that a Bayesian statistical model can often be viewed as a single probability model on a suitable product space $\\Omega \\times \\Theta$.
Hi everybody,

I've been searching for a way to combine two hazard ratios from the same study for a meta-analysis, but found nothing. Does anyone know how to do so ?

Any thoughts would be grate.

Good Day,

Simon
...of having a randomly selected point falling inside this ellipse (your acceptance rate) will become ever smaller. You would have to make exponentially more tries to have a fixed number of PSD matrices. THat's why it's better to have an explicit formulation of the boundary of the ellipse.
Suppose the dimension of $\\beta$ is 3 x 1, how would you approach this problem ? Thanks. Because I am completely clueless right now.
(+1) Nice answer.
After doing that which could be the best next move from your perspective in terms of calculating distance?
oh...shame on me :(
@ThePawn-i saw you are good in probability.Can you help me?
Since you are dealing with a 2-way ANOVA, you need to pay particular attention to Gung's last paragraph.  Your question of whether you can trust the ANOVA results may be obliquely asking about the power of your test, that is, the probability that you will reject a false null hypothesis.  In some research areas, research with N's such as your will have good power. In other areas, you need much larger N's.  Power depends on the effect size as well as N.
@scottyaz As mentioned in my answer, the usual reason I've found is an *extremely* expensive lab assay, exposure measurement, etc. that would be impractical to do on the full cohort.
@Jeromy [Silhouette measure](http://en.wikipedia.org/wiki/Silhouette_(clustering)) is quite usable, at least for start; it often offers clear minimum, so it is good for automatic optimization.
@aboabo, Updated my answer.
Very broad question, unfortunately. Some research shows that item order does not matter very much (e.g., Likert scale items assessing anxiety).
Shouldn't q1 <- pnorm((mu*t + S)/sigma*sqrt(t)) be pnorm((mu*t+S)/(sigma*sqrt(t)))? As sqrt(t) is also in the denominator
Not sure to understand why you can't. Is that a question on how to do a pivot table in excel ? The problem of creating a rank is interesting though, have a look at Voting theory and Arrow's impossibility theorem
@jona I'm all for it! As I said in my OP, I'm not quite sure of the questions to ask, hence my correlation-centric focus on a solution. What would your answer be to this using regression techniques?
Just as an example, http://static.googleusercontent.com/media/research.google.com/en//pubs/archive/35179.pdf?
You are not precise terminologically. Eigenvalues/vectors that you imply here are _not_ a characteristic of non-square matrix X, they are property of square matrix X'X.
There is no valid demonstration of this equality, because the expectation is $\\sigma\\sqrt{\\frac{2}{\\pi}}.$  The indefinite integral is simple and can be expressed in terms of elementary functions.
How would you account for the within-person correlation between item responses within such models? (I'm assuming that items are ordered by "difficulty")
If we talk about fat tailed distributions, then the existence of expected value is a very strong assumption.
It is surprising to hear that somebody using SPSS knows only Type III or II SS. And you sound like that.
Thank you for the information. I will follow up your suggestions. If there are no more suggestions I will accept your answer.
@NickCox you're right, I did mean 'fewer bins' there. Thanks. Fixed.
In my work, the easy parallelization was the single most important factor in using a genetic algorithm rather than something like simulated annealing.
I can't consider $X_1$ and $X_2$ to be independent soley because of the fact that they can both be greater than $\\frac{1}{2}$ and $X_3$ needs to be positive, which means that $X_1$ and $X_2$ can't be greater than $\\frac{1}{2}$ at the same time, which means they're not independent.
There have been implementations of LASSO for nonlinear data.
Hey I used tsDyn to fit a threshold model :) It does the same thing as nonlinearTest :)
As far as I can tell, encog is a lot more accurate. I need to do a proper test, but trying the 'Abalone' dataset resulted in 82% acc for `nnet` in `R` and 96.8 for encog in 4k iterations. Might be some settings but still...
your answer is a good one.  The OP should have remarked that going from n=51 to n=33 is a strong signal than an inappropriate analysis has been chosen.  There is no justification for matching, especially when it discards valuable data.  Covariate adjustment is much less arbitrary and is more efficient.
thanks @jona - very insightful answer
Nice vulgarisation (+1)
Thank you, @Glen_b!
I think the sentence *But this does not connect to relative frequencies of measurable events in a probability space* is (potentially) confusing, especially since you pretty much demonstrate the opposite immediately following. :)
John, please change the sampleSize to 12 and give me your thought.
You should only have two dummy variables for the three shops.
Thank you very much. I can now screw up the many pages of scribbles that I have made. (Some bits are similar to yours, but certainly not all of them!)
How about a paired t-test on the square-root transformed measurements?  This will account for the non-stationarity of the distribution.
As Singer and willet point on in their text, the interesting value is the hazard, which is interpreted as in the discrete time case as the probability of event, in the interval of time, given that it has not happened prior.
and the question is?
This is a nice try but unfortunately Part 2 is incorrect (even after interpreting \\Phi as the Gaussian probability in the *upper* tail): the right answer has to involve an *integral* of the Gaussian CDF, not the product of two of them (as the iterated expectation indicates).  Alternatively, consider what happens when both variances are 1, the correlation coefficient approaches 1, and u1=u2: the joint probability that X1>u1 and X2>u2 should converge to Pr[X1>u1] (due to near-perfect correlation), but on the contrary the first factor in the right hand side goes to zero.
I have problems understanding why you perform a Tukey test but then use the non-adjusted p-values and correct them manually. I usually prefer Tukey over Bonferroni for this.
I didn't knew that the estimator that I should show that is sufficient is $(x_1+x_2)/2$.
Whether you can trust the outcome depends on a lot more than what you've written here, but what you've done here should not cause problems in and of itself. Most statistical programs will create dummy codes for you.
You have implicitly specified nine categories--all 3 * 3 combinations of the two variables--along with their probabilities.  Why not use the answer to your [previous question](http://stats.stackexchange.com/questions/14158/how-to-generate-random-categorical-data)?
Could people choose more than one activity?
Why not using PLS discriminant analysis in this case?
@gung thanks for the improvement!
Gung, many thanks for the suggestion. I were dealing with star distances and parallaxes, but will read the paper and tell my Lab teacher about it (I asked him this same question)
similar question: http://stats.stackexchange.com/q/33315/5509
Thanks.This is exactly what I want @PGreen
Perhaps you could expand on why it's not an answer. He's looking for a java library and I pointed him to one that has the majority of features he's looking for.
I'm a bit confused by what you mean by "the standard deviation is known to have bias that depends on the number of samples observed". Do you mean that a specific estimate of the standard deviation has bias? In this case, there exist unbiased estimates. Perhaps you mean the MLE? 

Also, are you assuming that the true distribution of the two population are identical?
@ Tamas 
I can't thank you enough for your great help
"but how did you get to the lower bound?" From the same equation. I added a explanation.
Or written slightly differently, the odds of 1 in L1=4 is $\\frac{1328}{44} = 30.182$. The odds of 1 in L1=0 is $\\frac{1284}{17537} = .073$. The ratio of these odds is $\\frac{30.182}{0.073} = 413.45$. Result differences are due to rounding.
Is this a statistical question or a software (SmartPLS) question?
I'm sorry, you're right! It was E-5.
But if you're going to make a draw from that distribution, don't you need to first re-scale it so that it integrates to 1 (or, once discretized, sums to 1)?  That converts it from a joint distribution to a conditional distribution, at which point you shouldn't have the underflow problem.
"Near the borderline" is unknown to a classifier that is not provided the continuous $Y$ values.
@IrishStat: I meant that I just happened to retrieve the records in a certain order (think `ORDER BY`). The rows have no inherent special order. So you can safely re-arrange them without losing any information. Sorry if I confused you :-)
How likely are your Poisson counts to be small?
Hi again, more information on your study design, such as what your data is, how you collected it, what your groups are, how the observations were selected. All I know is that you did an experiment with 9 observations (people? rats? neurons? blocks of cheese? radiation frequencies?) that are from two groups.
Can you tell us what aspect(s) of your data you are interested in. Do you want to make a statement about attempts alone, hits alone, or only percent of attempts that are hits?  From the little data you present, it seems the siblings are making fewer attempts by a factor of 10. That might be of interest to you.  If so, you could do a paired t-test on attempts.
check out this older post: http://stats.stackexchange.com/questions/3113/parametric-sample-size-calculation-and-non-parametric-analysis
2 thoughts and a question. **1)** You can certainly do this type of forecasting in Excel (see duke.edu/~rnau/411outbd.htm for a good primer), but R is "better". **2)** Know that you'll probably have a number of people recommend R for this type of work-and if you're pretty comfortable on a computer, it may work for you (and its free)-but Excel will do it too, and you don't have to learn something new. **Question:** How comfortable are you in Excel, and which version do you use? While this type of forecasting is possible, it is somewhat complex and takes a fair bit of work to accomplish.
What would it mean for AB to cause AC to cause Y? That is the causal pattern you seem to be asking about. It may well be possible to set up such a thing in some software, but it isn't necessarily worth running an analysis it you can't say what you think it means for your phenomenon &/or theory.
Thanks for your answer, but could you please be a bit more specific about it? In particular, what you mean by expertise? If it's just probability that the person will answer correctly, then we already have its estimate as an average of previous answers, so it's not latent. If you mean than expertise incorporates both average and confidence about our estimate, then how can we propagate probabilities to get expertise and result?
I mean, I'm sure a graduate student exists out there in the world who doesn't know the equation for a line, but presumably an audience to which you would want to explain generalized linear models would at least have half a clue about high school level algebra! :-o
Greg: $\\sqrt{\\text{Var}(\\hat\\beta+\\hat\\delta)}=\\sqrt{\\text{Var}(\\hat\\beta)+\\text{Var}(\\hat\\delta)+2 \\text{Cov}(\\hat\\beta,\\hat\\delta)}$
Initially I understood you exactly this way. So, my answer remains: alpha, formally, should not be used for your A+B+C _composite score_ (that's the word for it, not "unidimensional index"). Reliability (in the sense of internal homogeneity herein) applies separately for A, B, C. Despite that your interest might be with A+B+C only.
A simple geometric approach works for your previous question, too, and shows that the two questions are essentially the same one. Graph $F_X$. Cut out the vertical band over $(-c,c)$. Sew the two halves of the graph back together. Relabel the axes: you have to subtract $c$ from numbers at the right and add $c$ to numbers at the left of the origin. That's the graph of $F_Y$. You can check that this is not what you have written, so stare at it and write a correct formula. (In the previous question the entire negative half of the graph over the interval $(-\\infty,0)$ would be excised.)
A t-test does not measure correlation. A t-test tests whether the means of two samples are equal.
What types of data have you got?
Note that multicolinearity is not a problem for predictive purposes as it just means there is more than one optimal model.  you can write the least squares estimate of $\\beta$ as $\\hat{\\beta}=X^{+}Y$ where $X^{+}$ is the moore penrose pseudo inverse of $X$.  this reduces to the usual least squares estimate when $X$ has full column rank but still exists uniquely in case of multicolinearity.  it resolves the non uniqueness of the original least squares estimates by taking the beta with the smallest sum of squares and corresponds to limiting ridge regression as penalty goes to zero.
Oh yes, that's it. I should blame myself.
This is a bit confusing.  Perhaps you could take some of your comments to the answer provided and have another go at expanding (editing) the question so your problem is clearer.
Can you say more about how the groups were formed?
Is your only problem that the data aren't normal?  The Gauss-Markov theorem tells us that normality isn't necessary--you'd just have to bootstrap your standard errors.  (Of course, I have no idea how you'd do that in SPSS.)  On the other hand, if your samples are large enough, SE's based on asymptotics would probably be OK, but your samples might need to be awfully large depending on how non-normal your data is.
If you find the duplicate does not have enough detail, check out other similar questions by searching on [regression prediction limit](http://stats.stackexchange.com/search?tab=relevance&q=prediction%20limit%20regression).
@gung These are not missing values in the usual sense: they are a particular form of interdependent response.  All you have to do is introduce a dummy variable to distinguish the two cases and interact it with $X_{\\cdot 2}$ as described at http://stats.stackexchange.com/a/1795 and http://stats.stackexchange.com/a/1795.  That is effectively what is proposed in this question.
Hmm, so this question cannot be answers for 1 week alone? E.g. "In _this_ week the number of people present depends more on the day than on the hour?"
Factor scores, dear @Cyrus, is another name for values of latent variables. The scores are indeed computed with the help of "Component Score Coefficient Matrix", but the program does the computation for you and saves the scores as new variables in the working data window. Check what is under "Scores" button in Factor analysis dialog box.
Maybe I missed something, but didn't you mean $log \\prod_i f(x_i)$ rather than $log \\sum_i f(x_i)$ ?
This question has been asked at [math](http://math.stackexchange.com/questions/457866/a-question-on-raining-probability-using-conditional-probability) with an answer.
I'd say that's an advantage, not a disadvantage.  It stops you from doing something you probably shouldn't be doing.
More specifically, the asker may be interested in [JFreeChart](http://www.jfree.org/jfreechart/) which powers a lot of Incanter graphics.
@RasmusBååth I agree, I've edited it to hopefully make it clearer what I would like. Thanks.
(i) Likert scale variable are discrete  --  transformation usually won't do much on individual scale items; ...  (ii) Why are you transforming these variables?
It seems that you have a mathematical generalization. Therefore if you want to name it a generalized R square how could we argue against it. Now r square has the interpretation of percentage variance explained by the model. What does you generalized R square represent analogous to this interpretation? Is it something like a percentage loss explained?? I am having a hard time making such an analogy.
Good idea, output added for clarity. Thanks for the suggestion Andy.
The counts will be more stable, but they will also depend on that assumption of "closeness" and on how it is quantified. If you have external reasons to suppose the values are serially correlated, exploiting that correlation potentially can improve the accuracy of the estimates. The windowing approach also is an approximation (to the true ML estimator), but if the window is short compared to the full range of data it should remain a good approximation.
@brianjd Nice link, thanks; when I've needed expectations I've just been typing out the integral but I think I'd be better off using the probability distribution facilities (via ProbabilityDistribution[]) which takes a pdf + support and gives you not just expectations but also moments, chf, etc. Needless to say I haven't properly learned the software yet :)
Bingo! Do I need to use a statistical model (logit, probit, etc) or can I simply run a few functions in R and get the desired results?
I agree with most of what Tufte says, but I have to say, his [low data:ink boxplots](http://indiemaps.com/blog/2008/02/props-to-box-plots/) are just plain idiotic. I think they take me 3-4 times longer to figure out than standard boxplots. The R defaults are much better (although the lines on the ends of the tails are unnecessary). Traditional boxplots have the added advantage that they can represent sample size (with width), and standard deviations (with notches).
@KarlOskar It will be a covariance matrix of $(X_{1,1},X_{2,1}, \\ldots, X_{1,T}, X_{2,T})^{\\prime}$. In the notation I used, the "element"  $(m,n)$ in this covariance matrix really denotes the 2x2 covariance matrix of two vector variables $X_{m}$ and $X_{n}$.
Thanks. I majored in applied math as an undergrad (and have extensive programming experience), so I have all of that, except for a rigorous course in abstract algebra, which I'm teaching myself...I ended up going with Neural Networks for Pattern Recognition, by Bishop.  For anyone else who's interested, I highly recommend it...
-------thanks !
Please tell us what you are trying to find out. What is your research question(s)? What hypotheses?
I'm comparing samples drawn from two days and each data point can be considered independent within each day and between days.  My thinking was to first see if the two samples were drawn from the same distribution and the two-sample Kolmogorov-Smirnov test seemed to indicate that they were.  Then I tested to see if both samples could be considered drawn from a Normal distribution.  Testing both samples seemed to support this.  So my thinking was, if I combined the two datasets, I could get more precise estimates.  Is this even valid?
Gotcha, I've edited the post as people sometimes don't look in comments.
I would add that some researchers consider that the way we treat data (e.g., 'ordinal'--that is, a set of ordered response categories--, or 'interval'--leaving alone the controversy that data are truly continuous in very rare cases--) depends on the underlying *measurement model*. Beside Borsboom, De Boeck and coworkers have a nice paper on [distinguishing between categories and dimensions](https://ppw.kuleuven.be/okp/_pdf/DeBoeck2005ACAPF.pdf). A 'practical' aspect of that distinction is the way we conceptualize several mental constructs, e.g. [mental disorders](http://bit.ly/17c4J1N).
Could you update your question text and question with what you are trying to do? The question suggests you're interested in income, but you are looking at something different given your comment below.
Do you take a fixed sample of size $n$ from the underlying distribution and then throw away the bad observations or do you continue sampling until you get a fixed sample size of "good" observations?
Examples where this cannot be done would be where the data are given without any description of the data generating mechanism or the model is not presented in a parametric form such as when you are given two iid data sets and are asked to test whether they come from the same distribution where you only specify that the distributions are absolutely continuous.
Perhaps it wasn't clear in my OP that all I wanted to determine in the end was whether one model was closer to the "accepted" model than the other (and I do not care so much about whether each was "close enough" in absolute terms). As such, your above comment essentially allows for your original answer to be applied as a "mechanism to compare the two models against each other in a relative way," which is exactly what I want. So as a technical note: I mark this answer plus the above comment as being the accepted answer for my question. P.S. I can't upvote due to my small rep
The solution quoted at http://stats.stackexchange.com/questions/10407/probability-for-finding-a-double-as-likely-event/10830#10830 is $p(100, 6, 50) - p(100, 6, 49)$ = $\\frac{8473424541566438848424241044494401095413860748334784411104375}{21004328173227588287573632560379945361919486576419588205474278754163359744}$ which is approximately $8.22963054197003118712621974351591 31460490899570690189896043772762691815344035140089 \\times 10^{-14}$. I found this by searching our site on [multinomial maximum](http://stats.stackexchange.com/search?q=multinomial+maximum).
I am not totally clear from your question however it is likely that SPSS automatically decides which partitions to use for each run of Cross-validation and so you do not need to partition beforehand.
A little.  At least, I did not know the i's were different.  (Really they are?)  Ok, I can imagine little normal distributions all lined up along the regression line . . .  but then what is going in the error covariance matrix?