tttjjj/elt_bench huggingface dataset:
@Scortchi: I think your comment highlights the ambiguity inherent in "weakly informative prior." Your interpretation makes sense and is in a similar vein to user777. The relationship between "informativeness" and probability is a tricky thing, with only partially satifactory solutions (e.g., Shannon entropy). I get your point though...they are not necessarily synonymous, as weakly informative priors may use only some of the information, while uninformative priors explicity ignore all available information.
I want to predict how the crowdsourced data can be used to predict the future wait-times of the coffee shop. I also have a ground truth wait-times for the coffee shop, so I can make comparisons.
Also, I don't see an email address on your profile.  I'm quite new to the site and ergo, am not sure how to send you my paper?  Could you let me know how to send you a direct or private message attachment?
As the `main` title asserts, this is a probability *density* function.  As you might imagine, this basic tool [appears in thousands of posts here](http://stats.stackexchange.com/search?q=probability+density+function), many of which will illuminate the question by showing how such plots are to be understood and read. The basics are explained in the apparent duplicate.
This is an example of what I meant by it depending on what you mean when you say "parameter" and "model".  Non-parametric statistics is more based on flexible forms of model functions rather than having no "parameters".  In fact many non-parametric statistical method are heavily parameterised.  And resampling is a parametric method based on the empirical cdf and taylor series expansion - at least that what I got from Efrons paper on bootstrap and jacknife.
The [mlogit](http://cran.r-project.org/web/packages/mlogit/index.html) package fits multinomial logit models. Just to be sure, are you confident that your dependent variable is nominal, as opposed to ordinal?
Thank you for your answer and driving my attention to regressors. It would definetely help me to explain diferrent results. Anyway I decided to go with Lag=12, as it is used in Analysis_of_Financial_Time_Series by TSAY and Modeling_financial_time_series_with_s-plus by Zivot, Wang where stationarity was made by using logarithmic differention.  Thank you once more for finding time and answer my question in great detail. Good luck
@AdamOmidpanah: I believe nlme can't do GLMs, and lme4 can't do different variance structures.
I am aware it is designed as a Q&A, but with questions like "What is your favorite statistics cartoon?" being voted up highly, I figured this wouldn't be terribly inappropriate. Especially pedagogically, if someone is trying to learn about data analysis and exploratory techniques, it might be useful to get some feedback on public datasets which offer a rich structure and have a large amount of history and research behind them.
It's practically impossible to have more power between S when you're measuring the same kind of thing either way.  If you do then THAT's a meaningful effect.  It means there's a negative correlation between the conditions. A positive correlation across subjects is expected, and how with within gets more power.  But a negative would be surprising.  Either way, you seem to be discussing the difference between significant and not significant, which is not itself significant (see Gelman).
Read this: http://people.psych.cornell.edu/~jec7/pcd%20pubs/simmonsetal11.pdf
I still don't know what you mean by the variance of the sample given that the 100 Xs can all have different variances. Your notation is wrong.  It is the true mean that you are testing for 0 not the sample mean.
As per my (non-automatically-generated) comment on the OP, I agree with the decision to decline my flag on this question. The other question was put on hold as too broad (not my opinion), and this question is somewhat less broad, so I hope it won't be put on hold at all.
It would be preferable to incorporate this into your previous question. Have you searched the site for duplicate and related questions?
@James the identity matrix implies independent *identically* distributed features. If you are only certain of independence, that implies a diagonal matrix, where features may have different variances. Is it possible the procedure you have used has also normalized your features? This should be easy to check, their variances should be 1.
I spent a lot of time searching in internet how to do that but I failed. Has anyone any idea how to proceed ?
@John Christie : Could you show me how your last bootstrap code should be modified in order to draw resamples of size 5 with replacement from the first sample and separate resamples of size 6 from the second sample (instead of resampling from the total number of 11 values). As I read on page 18 of the linked pdf, this should be the correct way to draw the bootstrap samples.
How is Mean Density of Numbers and Quartiles different? I mean, statistically, dont you think quartiles convey this information in a better way?
@Scortchi I believe $H_0:\\, \\mu=\\mu_0$, and the OP is talking about two one-sided and one two-sided alternatives.
(The Wiki quote does not speak about the purpose of the validation at all which may be different from selection.)
As you say, much discussed e.g. http://stats.stackexchange.com/questions/33780/r-vs-sas-why-is-sas-prefered-by-private-companies See also the site http://r4stats.com/ (which I find a little partisan, but everyone has biases of some kind).
This is hard to answer without some context about what these variables are, etc., but I can say that the values of the coefficients are entirely scale dependent so it's not clear that these coefficients are large, or small, or what. For example, if typical values of `LITRATE` were on the order of $10^6$, then the coefficient values actually seem pretty large.
And just in case anyone stumbles upon this and is working in Python (as I am): http://docs.scipy.org/doc/scipy/reference/generated/scipy.misc.logsumexp.html
Yes, for the first data type you need more and strong assumptions. For the standard errors, the Newey West correction should work. Actually it's analogous to one of the correction methods proposed by Bertrand et al. (they use clustered standard errors). A more recent method uses the bootstrap which works pretty well (see http://www.rbnz.govt.nz/research_and_publications/seminars_and_workshops/feb2012/4682249.pdf). Hope this helps!
(+1), especially for the examples.
Thanks for the answer! I will look into this and also update my question with an example.
possible duplicate of [Cox Baseline Hazard](http://stats.stackexchange.com/questions/46532/cox-baseline-hazard)
Assuming that $X$ and $Y$ are independent discrete random variable equally likely to take on any of the $256$ numerical values $\\{0, 1, 2, \\ldots, 255\\}$ (which is **not** the same as your $\\sim U(0,255)$ but matches your comment), then $$P\\{X-Y = 17\\} = \\sum_{i=17}^{255}P\\{X=i,Y=i-17\\}=\\sum_{i=17}^{255}P\\{X=i\\}\\{Y=i-17\\}\\\\=\\sum_{i=17}^{255}\\left(\\frac{1}{256}\\right)\\times \\left(\\frac{1}{256}\\right)$$
Also the `contrast` function in the `rms` package does quite general comparisons.
I see no figure for the person you are marrying. That is the most important quantity.
$$M= \\left[ \\begin{matrix}
1 & 0\\\\
0 & 1\\\\
\\end{matrix}\\right] + \\alpha\\left[ \\begin{matrix}
\\exp{\\{-D_{11}/\\lambda\\}} & \\exp{\\{-D_{12}/\\lambda\\}}\\\\
\\exp{\\{-D_{21}/\\lambda\\}} & \\exp{\\{-D_{22}/\\lambda\\}}\\\\
\\end{matrix}\\right]$$
The problem is that confidence intervals aren't produced using a Bayesian methodology. You don't start with a prior. I'll edit the post to add something which might help.
One thing I can see is that the estimated intraclass correlation for EmpID is $\\hat{\\rho} = 680.236/(680.236+13.562+774.256)$. That is, the estimated correlation between two individuals in the same level of EmpID is $\\hat{\\rho}$. I agree with @guest that more output (and some context) would be helpful.
SE=(SD/√n) (√[(N-n)/N]
Using MLE under what distributional assumption? You can't even write down a likelihood if there's no assumed distribution for $y$ at a given $x$ value.
@histelheim Well, you could tell us what you know in your question. Then we could give better answers. We can't read your mind or divine what your background is.
Not really.  You seem set on your model with the covariates.  It is ok to explore how the correlation between BMI and weight affect the results.  This is something you should mention in the write up.
Nope, you've got it in one.  I won't bother updating my answer as it is already in your comment. :-)
I did try linear in log-log space and then took those results as the initial conditions to the non-linear power fit. So even if I used $a_0=exp(-24.203)=3.0815e-11$, $a$ would converge to something but $b$ wouldn't change. And then when I used $a_0=1$, it would converge to the same value of $a$ but $b$ still didn't change. That's why I said $a_0$ doesn't seem to make a difference. It turned out that my tolerance (1e-6) was too large. When I lower it to (1e-16) I get values identical to yours even with $a_0=1$.
The study is the experiment set out in the top box, i.e 50 people throwing a dice 36 times each and I am interested in what the power is of this set up to detect the difference observed between the groups
Beautiful, exactly what I was looking for.

The two lines:

w <- colSums(coef(svp)[[1]] * x[unlist(alphaindex(svp)),])
b <- b(svp)

were a godsend.  Thank you!
... and don't forget to validate the optimal model with new independent test data (or an outer cross validation loop)
^^^ The program should more or less correctly predict the known past.
@Heisenberg The answer treated the likelihood function as the joint probability function of the sample (for which the ML provides the max w.r.t the parameters, and so maximizes probability for any sample). And this is correct when the r.v's are discrete, but not when they are continuous, since the joint density, _by construction_ is not a joint probability. I wouldn't characterize this as a "nuance", it is a fundamental difference between the discrete and the continuous worlds.
Maybe your fitted model is not what you think it is. Repeated observations on the same unit does not constitute replication of evidence. What are the experimental units in your model?
What I have done so far is I have removed this species from my dataset and run the nbglm without it to see how results are affected...
+1, however, "30 points are enough to assume that the data is normally distributed" is not true. It may well be that people believe this, but how much data are required for the CLT to make the sampling distribution converge adequately to a normal depends on the nature of the data distribution (see [here](http://stats.stackexchange.com/questions/29731/regression-when-the-ols-residuals-are-not-normally-distributed/29748#29748)). Instead, 30 (may be) approximately enough if the data are already normal, but the SD is estimated from the same data set (cf, the t-distribution).
If you are asking whether it is valid to perform k-means clustering on non-normal data, the answer is yes if the data are assumed to be continuous. Binary data isn't continuous. Some people do k-means on such data, which thing is heuristically permissible, but theoretically invalid.
@StasK Indeed, that was the purpose behind my question "*What were the specified population mean and variance of the residuals for your KS test and how did you get such population values?*"
+1 this is neat, I'd forgotten about the vcdExtra package.
Hint: did the professor use a Normal distribution function or a Student t distribution function to compute the p-value?
Karl, Thanks for your excellent reply. I'll see whether I can get your first suggestion to work. As I mentioned in my reply to Henry (above), I'm reluctant to collapse across participants as in this case (unlike with dice rolls), the outcomes could well vary from person to person.
@chl, thanks for the tip, that book looks interesting.
I edited references to time-points, hopefully it's a little easier to read now.  If you are after a fun and readable chapter on repeated-measures ANOVA, Discovering Statistics Using SPSS by Andy Field is a good start.  With multiple time points, you might also consider how you would deal with missing data, as repeated measures ANOVA does list-wise deletion.
This question appears to be off-topic because it is about how to do something with the mice package in R, not about statistics.
Your item 1 is incorrect: the P value is the probability of obtaining data as extreme, or more extreme, given the null hypothesis. As far as I know that means that P _is_ a probability--conditional but a probability nonetheless. Your statement is correct in the circumstances that one is working within the Neyman-Pearson paradigm of errors, but not is one is working within the Fisherian paradigm where P values are idices of evidence against the null hypothesis. It is true that the paradigms are regularly mixed into an incoherent mish-mash, but both are 'correct' when used alone and intact.
"*For finite moments, I mean all the moments are finite. Otherwise, student-t distribution with 2 degree of freedom works*" - but note that your thereby exclude, for example, a student t distribution with $10^{100}$ degrees of freedom. That's a pretty strong condition.
(+1) Good answer. One can restrict attention to orthogonal, instead of unitary, $V$ since $Q$ is real and symmetric. Also, what is $\\mathrm{SCR}$? I do not see it defined. By slightly rejiggering the argument, one can also avoid the use of a degenerate normal, in case that causes some consternation to those not familiar with it.
Can you explain this a little more clearly?  Are you calculating correlation estimates between two variable that are paired?  What is your null hypothesis?  What bootstrap distribution are you using under the null hypothesis?  It is definitely true that the more variables you test the more likely you are to commit a type 1 error unless you adjust for multiplicity.
That's a good start! Here are some thoughts on how to continue: For each $x \\in [a,b]$, is $x$ in every set on the RHS? Why or why not? Suppose $x > b$. Write $x = b + \\delta$ for some $\\delta > 0$. Recall the Archimedean property of the reals. Now, do something analogous for $x < a$. Conclusion?
Thanks a lot Bill, a very clear exposition.
I'm not sure I'd say this was `basic-concepts` - even though the manipulations aren't particularly difficult, there's quite a few ideas to put together before this stuff starts to look easy/basic. Indeed I'm not sure it applies to even one of the many questions you have posted under that tag.
@forecaster In order to identify an ARIMA model via an AIC criteria or to formally test the significance of estimated parameters via a T test or an F test the residuals should not exhibit non-randomness. THe Gaussian assumptions have all to do with the model's error process and nothing to do with the distribution of the observed series. For sure the originally observed series is functionally related to the error process via the model BUT there are no parametric distributional requirements for the observed series. My comments are all supportive of Glen B's reflections.
"*we always calculate the probability that we would obtain the measured dataset*" -- not so. See wikipedia's definition of what a [p-value](http://en.wikipedia.org/wiki/P-value) is. The test in question simply *conditions* on the margins. Not everyone agrees that conditioning is always appropriate in all the situations the test is used for -- that's a debate that's been running for many decades.
In practice, you're probably better off making use of a tool that can handle the problem formulation for you.  For example, CVX would do this directly for you.
The prior distribution is a specification for the parameter(s) of the model for the data. To specify a prior one needs to know what model you use for the data. So, what is it?
@MånsT yes this i understand, but I'm confused about which of the two i need to report.
...however, it might be better to determine if there is an interaction between task and visualisation. This might show that some visualisations work particularly well with some kinds of task. For example, in a one way ANOVA you might find Map is best. But in an ANOVA with visualisation and task as factors (e.g a 3x3 ANOVA - if you have 3 tasks) you might see that Map is the best technique with tasks 1 & 2, but worst for task 3. This would mean that results from the oneway ANOVA are misleading. Map is not best overall - only for task 1 and 2. It is, in fact, worst for task 3.
I think you need extra cautions given that the response is measured, not counted. That doesn't rule out these kinds of models, but at least some of the software available may expect integers only. (@Peter Flom's answer is pertinent too.)
I'm sorry if that wasn't clear, by the notation I mean, you have 4 variables, $(y_1,y_2,y_3,y_4)$ and they are all of lenght $n=1000$, is this correct?
A continuous distribution can approach a discrete one (in cdf terms) arbitrarily closely (e.g. construct a continuous density from a given discrete one by placing a little Beta(4,4)-shaped kernel centered at each mass point - of the appropriate area - and let the standard deviation of each such kernel shrink toward zero while keeping its area constant). Such discrete bounds as discussed here will thereby also act as bounds on continuous distributions. I expect you're thinking about continuous *unimodal* distributions... which indeed have different upper bounds.
ARIMA is a model structure.  It stands for autoregressive integrated moving average process.  This modle include differening the series "the I part"  A difference would be done by constructing Y(t)=X(t)- X(t-1).  First differencing removes linear trends and drifting that comes about in a random walk.  The AR terms simply provide a regression of the time seires on itself at earlier times.  The moving average component creates correaltion by creating a linear combination of random noise terms.  Each AR and MA term has a coeffocoent associated with it.
I guess that this question could be answered by technical understanding of [Wozniak, T. (2012)](http://cadmus.eui.eu/bitstream/handle/1814/25136/2012_Wozniak.pdf?sequence=2) with the findings of [Hamilton, J. D. (2008)](http://dss.ucsd.edu/~jhamilto/JHamilton_Engle.pdf) regarding inference in the presence of ARCH effects.
Classification is the same thing as prediction on "old" data.
Matt do you have a bit of time free to go into this a little more? My stats knowledge isn't strong here.
It sounds to me like you might be interested in geographically weighted regression. http://ncg.nuim.ie/ncg/GWR/whatis.htm
@whuber - Without some reasonable guesses at the various distributions involved, it's not clear to me that we *can* cope properly with the noise, but we can still try something.  If the signal-to-noise ratio in the $f_i$ is fairly high (whatever that means), I wouldn't worry, since we don't care about the coefficient estimates as such.  If we have low predictive power, then time to think of something else.  If it's high, then the noise likely hasn't had an overwhelming effect.  I should emphasize that I'd try this first and see what happens, rather than commit to it as the only approach.
histelheim - (i) the point of the answer is to illustrate that the question seemingly relies on faulty assumptions (that transformation in any form is a good approach for this problem); (ii) Correct, I took the square root of C (as you should be able see from the axis labels), but there's no simple transformation of both variables either, unless you know the very thing you're trying to estimate; (iii) I believe it should be possible to include $b$ and $b^2$ as predictors of $C$ in your path analysis, but it depends on what exactly you're doing and what you're using to do it.
(Good!) Yes! and Yes!
I don't yet see "two" Maximum Likelihood Estimators. I see one, that comes from a sample of i.i.d Paretos, and I see another function, the $\\tilde L$, which will become a "log-likelihood" only if it can be shown to represent the logarithm of a joint density function. If it doesn't, then its maximizer is not a "maximum likelihood estimator", it is just the maximizer of some function.
@whuber: I started from Andrews and went forward. Perhaps I should work backwards.
Adding to mpiktas comment, that link says that OLS is a lower variance estimator of y than any other linear combination of variables.  It does not discuss the estimation of a linear combination of parameters.
Thank you for your interesting answer. In fact, I am trying to improve the first algorithm because, when interpreting the data, we realized it was possible to refine it. Since the whole dataset is 'big' and it takes days to run the algorithms on it, I took one cluster generated by the first algorithm and applied the second to it. It generated 3 clusters out of it.
a t-test is used to compare means of two distributions. The observations in each distribution have to be independent of each other thus time series presents an opportunity/compication. Your question is very vague and it would help if you actually posted your data.
@B_Miner: don't worry, no problem. PS: depending on the type/amount of data you have, you may want to have a look at LOESS regression.
We look for replies that provide support or at least references.  What are yours?
Thanks mrbcuda. Your biplot looks identical to Lindsay Smith's so I presume he/she used the same method 12 years ago!  I'm also aware of [some other higher level methods](http://gastonsanchez.com/blog/how-to/2012/06/17/PCA-in-R.html), but as you rightly point out this is an exercise to make the underlying PCA maths explicit.
@myu.ng, if Glen's answer has helped you, you might consider *accepting* it by clicking the check mark below the vote total. You might also want to upvote it, if you haven't already.
If $X \\sim N(\\mu, \\sigma^2)$, then $X + \\textrm{cst} \\sim N(\\mu + \\textrm{cst}, \\sigma^2)$. Here "constant" means "not a random variable". Regarding your example, $X=\\epsilon_i$, $\\mu = 0$ and $\\textrm{cst} = ax_i^3$.
Interesting package, I'll look into it in the future. The graphs seem a little complicated though for someone new to the field. My audience is MD's without any mathematical or statistical background knowledge. Thanks!
In a sense, a discriminant accounts for a variability as a p. component does, the eigenvalue being the amount of it. However, the "variability" in LDA is of special sort - it is the _ratio_ of between-class variabilty to the within-class variability. Each discriminant tries to account for as much as possible of that ratio. [Read further](http://stats.stackexchange.com/a/48859/3277)
Thanks you,I thought that model selection is correct,but the thing is selected variables have some problem.I removed some variables and I found that there are some outliers in the variables,after remove that variable I could get significant p values.
@Matt's suggestion works well to distinguish weekends (or initial weekly segments) from other days. A subtler issue lurking here concerns the autoregressive assumption: why can we assume that the serial correlation (in the *random* components of the process) is the same between Fridays and Mondays as it is between any other two successive pairs of days? That seems to call for some exploratory data analysis and possibly some very special treatment of this problem to handle such subtle heteroscedasticity.
Jeromy, allow me to take all of these with one blow: I'm a psychology student. And I reckon you're familiar with required statistical background... So there... =)
(I'm good with R and SPSS... but R has a greater priority)
"The last factor" is $\\prod_{i=1}^n I(0\\le x_i \\le \\theta)$ which clearly is the same as $\\prod_{i=1}^n I(0\\le x_{(i)} \\le \\theta)$ so depends on the order statistics. The point is that the product is one only if all the $n$ factors are one, only if $x_{(n)} \\le \\theta$.
(+1) Nice answer. As hinted to in my comments to the OP, this is the answer I was hoping would be posted (even the counterexample was carefully chosen with this in mind). :)