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2599 | 2 | null | 2597 | 19 | null | The following Wikipedia entry actually does a pretty good job of explaining the most popular and relatively simple methods:
- Determining the number of clusters in a data set
The [Elbow Method](http://en.wikipedia.org/wiki/Determining_the_number_of_clusters_in_a_data_set#The_Elbow_Method) heuristic described there i... | null | CC BY-SA 2.5 | null | 2010-09-12T20:26:40.637 | 2010-09-12T20:26:40.637 | null | null | 251 | null |
2600 | 5 | null | null | 0 | null | Psychometrics has evolved as a subfield of psychology to become the science of measurement of unobservable individual characteristics.
Unlike psychophysics which mostly focus on perceptual attributes, psychometrics aims at providing reproducible methods for assessing high-level or cognitive skills through the scientifi... | null | CC BY-SA 3.0 | null | 2010-09-12T20:54:55.157 | 2015-08-29T00:06:49.490 | 2015-08-29T00:06:49.490 | 7290 | 930 | null |
2601 | 4 | null | null | 0 | null | Psychometrics has evolved as a subfield of psychology to become the science of measurement of unobservable individual characteristics. | null | CC BY-SA 2.5 | null | 2010-09-12T20:54:55.157 | 2010-09-12T20:54:55.157 | 2010-09-12T20:54:55.157 | 930 | 930 | null |
2602 | 1 | 2607 | null | 4 | 1477 | I often find parts of financial correlation matrices not statistically significantly different from zero. Sometimes, these correlations have a tangible effect on results - low correlations lead to high diversification benefit. Also, the sign of correlations can skew the results.
Would you consider it appropriate to mak... | How to deal with correlations not statistically significantly different from zero? | CC BY-SA 2.5 | null | 2010-09-12T21:38:14.563 | 2010-09-17T21:30:52.620 | null | null | 1250 | [
"correlation",
"statistical-significance"
] |
2603 | 2 | null | 2598 | 4 | null | Is the location of the treasure changed after each attempt? Wouldn't it be $(10\%)^5$?
| null | CC BY-SA 2.5 | null | 2010-09-12T21:54:02.223 | 2010-09-13T12:38:17.643 | 2010-09-13T12:38:17.643 | null | 1250 | null |
2604 | 2 | null | 2580 | 8 | null | You can't compare them directly since the Negative Binomial has more parameters. Indeed the Poisson is "nested" within the Negative Binomial in the sense that it's a limiting case, so the NegBin will always fit better than the Poisson. However, that makes it possible to consider something like a likelihood ratio test b... | null | CC BY-SA 3.0 | null | 2010-09-12T23:31:55.703 | 2013-03-22T00:29:55.287 | 2013-03-22T00:29:55.287 | 805 | 805 | null |
2605 | 2 | null | 2572 | 4 | null | Of course, there are limits to stress testing due to the requirement of the correlation matrix to remain positive semidefinite.
finding these limits is, imho, the critical part of what you are trying to do.
The answer to this problem depends on how many of the correlation coefficient you want to stress test simultaneou... | null | CC BY-SA 2.5 | null | 2010-09-13T00:13:30.363 | 2010-09-13T10:36:05.453 | 2010-09-13T10:36:05.453 | 603 | 603 | null |
2606 | 2 | null | 2580 | 4 | null | You can use a metric such as [Mean Squared Error](http://en.wikipedia.org/wiki/Mean_squared_error) between actual vs predicted values to compare the two models.
| null | CC BY-SA 2.5 | null | 2010-09-13T00:45:03.360 | 2010-09-13T00:45:03.360 | null | null | null | null |
2607 | 2 | null | 2602 | 2 | null | I often find parts of financial correlation matrices not statistically significantly different from zero
You might want to be careful with that. First of all, statistically significant and large enough to be meaningful are not the same [thing](https://stats.stackexchange.com/questions/2516/are-large-data-sets-inappropr... | null | CC BY-SA 2.5 | null | 2010-09-13T00:56:45.050 | 2010-09-13T16:19:52.300 | 2017-04-13T12:44:39.283 | -1 | 603 | null |
2608 | 2 | null | 2602 | 1 | null | I am not sure if this is feasible in your context but one thing you can do to avoid these issues is to use bayesian estimation and to compute the expected values of different investing decisions based on the posterior distribution of the parameters.
| null | CC BY-SA 2.5 | null | 2010-09-13T01:48:17.720 | 2010-09-13T01:48:17.720 | null | null | null | null |
2609 | 2 | null | 2597 | 5 | null | I recently became fund of the [clustergram visualization](http://www.r-statistics.com/2010/06/clustergram-visualization-and-diagnostics-for-cluster-analysis-r-code/) method (implemented in R).
I use it for an extra method to assess a "good" number of clusters. Extending it to other clustering methods is not so hard (I... | null | CC BY-SA 2.5 | null | 2010-09-13T05:22:06.293 | 2010-09-13T05:22:06.293 | null | null | 253 | null |
2610 | 2 | null | 2597 | 17 | null | It is rather difficult to provide a clear-cut solution about how to choose the "best" number of clusters in your data, whatever the clustering method you use, because Cluster Analysis seeks to isolate groups of statistical units (whether it be individuals or variables) for exploratory or descriptive purpose, essentiall... | null | CC BY-SA 2.5 | null | 2010-09-13T08:26:14.417 | 2010-09-13T08:26:14.417 | 2017-04-13T12:44:48.803 | -1 | 930 | null |
2611 | 1 | null | null | 21 | 3506 | The logic of multiple imputation (MI) is to impute the missing values not once but several (typically M=5) times, resulting in M completed datasets. The M completed datasets are then analyzed with complete-data methods upon which the M estimates and their standard errors are combined using Rubin's formulas to obtain th... | How to combine confidence intervals for a variance component of a mixed-effects model when using multiple imputation | CC BY-SA 2.5 | null | 2010-09-13T11:47:27.230 | 2010-10-06T14:31:47.217 | 2010-09-13T12:20:07.773 | 930 | 1266 | [
"modeling",
"confidence-interval",
"mixed-model",
"data-imputation"
] |
2612 | 2 | null | 2611 | 8 | null | This is a great question! Not sure this is a full answer, however, I drop these few lines in case it helps.
It seems that Yucel and Demirtas (2010) refer to an older paper published in the JCGS, [Computational strategies for multivariate linear mixed-effects models with missing values](http://sci2s.ugr.es/keel/pdf/spec... | null | CC BY-SA 2.5 | null | 2010-09-13T12:57:32.943 | 2010-09-13T14:14:58.480 | 2010-09-13T14:14:58.480 | 930 | 930 | null |
2613 | 1 | 2618 | null | 7 | 864 | I am advising a small medical study (two groups, treatment is a dummy variable), i.e. a 2x2 contingency table. I'm comparing the value of the [Pearson's $\chi^2$ test](http://en.wikipedia.org/wiki/Pearson%27s_chi-square_test) and a non parametric competitor [McNemar's $\chi^2$ test](http://en.wikipedia.org/wiki/McNemar... | How to explain such a big difference between parametric and non parametric test (and other questions)? | CC BY-SA 2.5 | null | 2010-09-13T13:14:51.620 | 2010-09-13T16:28:22.580 | 2010-09-13T16:28:22.580 | 603 | 603 | [
"hypothesis-testing",
"nonparametric"
] |
2614 | 2 | null | 372 | 1 | null | Understanding the causes of over-fitting, and how it can be avoided, is absoluetly vital in datamining. If you mine for statistical associations in data, you will always be able to find them (even if the data are purely random) but that doesn't mean they are of any predictive value. The more you mine, the more likel... | null | CC BY-SA 2.5 | null | 2010-09-13T13:16:33.200 | 2010-09-13T13:16:33.200 | null | null | 887 | null |
2615 | 1 | 2664 | null | 7 | 613 | Suppose I have some initial correlation matrix. I want to stress each correlation in the matrix by the same constant simultaneously (except the diagonal; lets call this global parallel stress since it affects the entire matrix by the same constant at the same time). I start with adding 0.01% to each correlation and che... | Analytical solutions to limits of correlation stress testing | CC BY-SA 2.5 | null | 2010-09-13T13:35:11.777 | 2010-09-20T01:50:20.380 | 2017-04-13T12:44:33.977 | -1 | 1250 | [
"correlation"
] |
2616 | 2 | null | 485 | 1 | null | I just came across this website, [CensusAtSchool -- Informal inference](http://www.censusatschool.org.nz/2009/informal-inference/). Maybe worth looking at the videos and handouts...
| null | CC BY-SA 2.5 | null | 2010-09-13T14:07:11.150 | 2010-09-13T14:07:11.150 | null | null | 930 | null |
2617 | 1 | null | null | 4 | 4144 | I'm trying to take a normal distribution of points, and force them to become a uniform distribution. I've had little success on [S.O.](https://stackoverflow.com/questions/3696747/generation-of-uniformly-distributed-random-noise), so I thought I'd ask here.
Basically, I have a hash function which takes an `X`, `Y`, and ... | How can I determine the best fit normal distribution from this information? | CC BY-SA 2.5 | null | 2010-09-13T14:07:17.403 | 2010-09-16T05:33:58.557 | 2017-05-23T12:39:26.593 | -1 | 1267 | [
"distributions",
"algorithms",
"data-transformation"
] |
2618 | 2 | null | 2613 | 11 | null | Because the null hypothesis your McNemar test tests, is not the same as the one tested by the $\chi^2$ test. The McNemar test actually tests whether the probability of 1-2 equals that of 2-1 (say first number is the row and second the column). If you switch columns, then you get a completely different outcome. The $\ch... | null | CC BY-SA 2.5 | null | 2010-09-13T14:58:53.460 | 2010-09-13T14:58:53.460 | null | null | 1124 | null |
2619 | 1 | 2622 | null | 10 | 21601 | I have three variables:
- distance (continuous, variable range negative infinity to positive infinity)
- isLand (discrete categorical/ Boolean, variable range 1 or 0)
- occupants (discrete categorical, variable range 0-7)
I want to answer the following statistical questions:
- How to I compare distributions tha... | Continuous and Categorical variable data analysis | CC BY-SA 2.5 | null | 2010-09-13T14:59:18.623 | 2019-07-01T08:39:16.593 | 2010-09-16T06:31:27.430 | null | 559 | [
"categorical-data",
"continuous-data"
] |
2620 | 2 | null | 2617 | 3 | null | The [inverse probability transform](http://en.wikipedia.org/wiki/Probability_integral_transform) seems to be what you want.
To get a better uniform fit, the CDF $F(x)$ in the inverse probabiliy transform needs to be the actual distribution generating the data. If this isn't actually normal, then that could account f... | null | CC BY-SA 2.5 | null | 2010-09-13T15:12:28.100 | 2010-09-13T19:54:36.780 | 2017-04-13T12:44:54.643 | -1 | 251 | null |
2621 | 2 | null | 2573 | 25 | null | To clarify Dirks point :
Say your data is a sample of a normal distribution. You could construct the following plot:
The red line is the empirical density estimate, the blue line is the theoretical pdf of the underlying normal distribution. Note that the histogram is exp... | null | CC BY-SA 2.5 | null | 2010-09-13T15:14:47.923 | 2010-09-13T19:52:04.797 | 2010-09-13T19:52:04.797 | 1124 | 1124 | null |
2622 | 2 | null | 2619 | 5 | null | I would recommend reading about logistic or log-linear models in particular, and methods of categorical data analysis in general. The notes on the following course are pretty good for a start: [Analysis of Discrete Data](http://www.stat.psu.edu/online/courses/stat504/). The [textbook](http://rads.stackoverflow.com/am... | null | CC BY-SA 2.5 | null | 2010-09-13T16:18:06.563 | 2010-09-13T16:18:06.563 | null | null | 251 | null |
2623 | 1 | 2627 | null | 16 | 13322 | Does the autocorrelation function have any meaning with a non-stationary time series?
The time series is generally assumed to be stationary before autocorrelation is used for Box and Jenkins modeling purposes.
| Autocorrelation in the presence of non-stationarity? | CC BY-SA 2.5 | null | 2010-09-13T16:56:42.960 | 2016-09-21T13:32:55.927 | 2016-05-24T09:43:57.127 | 1352 | 284 | [
"time-series",
"autocorrelation",
"box-jenkins"
] |
2624 | 2 | null | 2615 | 3 | null | This is not a complete answer but it was too long to fit in as a comment and hopefully gives you some ideas.
What I am about to say is not really a proof but more of an intuitive sketch as to why the minimum eigenvalue may matter as far as the upper stress boundary is concerned.
Every correlation matrix can be decompos... | null | CC BY-SA 2.5 | null | 2010-09-13T17:27:13.523 | 2010-09-13T18:23:25.273 | 2010-09-13T18:23:25.273 | null | null | null |
2625 | 2 | null | 2617 | 4 | null | Interpreting your question as asking for the "inverse probability transform," as ars has indicated and you have confirmed in comments to the question, gives a simple solution: perform a sort to rank all the $z$ values in ascending order from $1$ to $N$ (the amount of data), then convert each $z$ into $2 Rank(z)/(N+1) -... | null | CC BY-SA 2.5 | null | 2010-09-13T17:37:30.807 | 2010-09-13T17:37:30.807 | null | null | 919 | null |
2626 | 2 | null | 2623 | 8 | null | In its alternative form as a variogram, the rate at which the function grows with large lags is roughly the square of the average trend. This can sometimes be a useful way to decide whether you have adequately removed any trends.
You can think of the variogram as the squared correlation multiplied by an appropriate va... | null | CC BY-SA 3.0 | null | 2010-09-13T17:44:20.733 | 2016-05-24T11:43:50.740 | 2017-04-13T12:44:44.767 | -1 | 919 | null |
2627 | 2 | null | 2623 | 15 | null | @whuber gave a nice answer. I would just add, that you can simulate this very easily in R:
```
op <- par(mfrow = c(2,2), mar = .5 + c(0,0,0,0))
N <- 500
# Simulate a Gaussian noise process
y1 <- rnorm(N)
# Turn it into integrated noise (a random walk)
y2 <- cumsum(y1)
plot(ts(y1), xlab="", ylab="", main="", axes=F);... | null | CC BY-SA 2.5 | null | 2010-09-13T17:55:10.243 | 2010-09-13T20:00:10.513 | 2010-09-13T20:00:10.513 | 5 | 5 | null |
2628 | 1 | 2629 | null | 57 | 30797 | Imagine you have to do reporting on the numbers of candidates who yearly take a given test. It seems rather difficult to infer the observed % of success, for instance, on a wider population due to the specifity of the target population. So you may consider that these data represent the whole population.
Are results of... | Statistical inference when the sample "is" the population | CC BY-SA 2.5 | null | 2010-09-13T18:35:23.810 | 2021-08-19T20:40:06.823 | 2010-09-14T07:59:54.997 | 183 | 1154 | [
"hypothesis-testing",
"population",
"sampling"
] |
2629 | 2 | null | 2628 | 39 | null | There may be varying opinions on this, but I would treat the population data as a sample and assume a hypothetical population, then make inferences in the usual way. One way to think about this is that there is an underlying data generating process responsible for the collected data, the "population" distribution.
I... | null | CC BY-SA 2.5 | null | 2010-09-13T19:30:09.810 | 2010-09-13T19:30:09.810 | null | null | 251 | null |
2630 | 2 | null | 2628 | 31 | null | Actually, if you're really positive you have the whole population, there's even no need to go into statistics. Then you know exactly how big the difference is, and there is no reason whatsoever to test it any more. A classical mistake is using statistical significance as "relevant" significance. If you sampled the popu... | null | CC BY-SA 2.5 | null | 2010-09-13T20:12:57.343 | 2010-09-13T21:52:22.893 | 2010-09-13T21:52:22.893 | 1124 | 1124 | null |
2631 | 1 | 2633 | null | 9 | 3902 | This is a followup to [this](https://stats.stackexchange.com/q/2597/977) question. I am currently trying to implement the C-Index in order to find a near-optimal number of clusters from a hierarchy of clusters. I do this by calculating the C-Index for every step of the (agglomerative) hierarchical clustering. The probl... | Can someone explain the C-Index in the context of hierarchical clustering? | CC BY-SA 2.5 | null | 2010-09-13T20:20:30.583 | 2010-09-13T21:04:03.930 | 2017-04-13T12:44:24.947 | -1 | 977 | [
"clustering"
] |
2632 | 2 | null | 2617 | 3 | null | OK, your hash function is generating numbers according to some distribution. You're thinking of it as normal, but it's really not.
There's a simple way to convert between any univariate distribution and a uniform distribution. That is to use the cumulative distribution function CDF, which is a simple monotonic function... | null | CC BY-SA 2.5 | null | 2010-09-13T20:53:17.000 | 2010-09-13T21:01:50.457 | 2010-09-13T21:01:50.457 | 1270 | 1270 | null |
2633 | 2 | null | 2631 | 2 | null | This may be one of the cases where there's more art than science to clustering. I would suggest that you let your clustering algorithm run for a short time before letting the C-Index calculations kick in. "Short time" may be after processing a few pairs, just when it starts to exceed 0, or some other heuristic. (Aft... | null | CC BY-SA 2.5 | null | 2010-09-13T21:04:03.930 | 2010-09-13T21:04:03.930 | null | null | 251 | null |
2634 | 2 | null | 2619 | 2 | null |
- To examine the relationship between a continuous and categorical factor, a good start is to use side-by-side box plots, continuous on the left, categorical on the bottom. Are the means different? Use ANOVA to check.
- To examine the relationship between categorical factors, a good start is to use a mosaic plot, as... | null | CC BY-SA 2.5 | null | 2010-09-13T23:32:13.713 | 2010-09-13T23:32:13.713 | null | null | 74 | null |
2635 | 1 | null | null | 12 | 27355 | I need to update the failure rate (given as deterministic) based on new rate of failure about the same system (it is a deterministic one too). I read about conjugate priors and Gamma distribution as a conjugate for the Poisson process.
Also, I can equate the mean value of Gamma dist. ($\beta/\alpha$) to the new rate (... | Conjugate prior for a Gamma distribution | CC BY-SA 3.0 | null | 2010-09-14T00:20:44.673 | 2022-02-23T01:10:27.043 | 2013-08-02T05:12:13.260 | 805 | 1272 | [
"bayesian"
] |
2636 | 2 | null | 2547 | 2 | null | You're getting good answers here, but let me just add my 2 cents. I work in pharmacometrics, which deals in things like blood volume, elimination rate, base level of drug effect, maximum drug effect, and parameters like that.
We make a distinction between variables that can take on any value plus or minus, versus value... | null | CC BY-SA 2.5 | null | 2010-09-14T00:23:26.010 | 2010-09-14T00:23:26.010 | null | null | 1270 | null |
2637 | 2 | null | 166 | 2 | null | As a rough generalization, any time you sample a fraction of the people in a population, you're going to get a different answer than if you sample the same number again (but possibly different people).
So if you want to find out how many people in Australia are >= 30 years old, and if the true fraction (God told us) ju... | null | CC BY-SA 3.0 | null | 2010-09-14T00:55:16.510 | 2016-04-09T12:49:45.397 | 2016-04-09T12:49:45.397 | 5739 | 1270 | null |
2638 | 2 | null | 155 | 3 | null | Fun question.
Someone found out I work in biostatistics, and they asked me (basically) "Isn't statistics just a way of lying?"
(Which brings back the Mark Twain quote about Lies, Damn Lies, and Statistics.)
I tried to explain that statistics allows us to say with 100 percent precision that, given assumptions, and given... | null | CC BY-SA 2.5 | null | 2010-09-14T01:02:43.727 | 2010-09-14T18:11:45.067 | 2010-09-14T18:11:45.067 | 1270 | 1270 | null |
2639 | 1 | 2766 | null | 2 | 1377 | I have values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains values from a different method as:
```
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
```
The number of columns indicate the number of... | How to compare different distributions with reference truth value in Matlab? | CC BY-SA 3.0 | 0 | 2010-09-14T01:11:32.053 | 2016-07-11T23:24:31.623 | 2016-07-11T23:24:31.623 | null | null | [
"distributions",
"confidence-interval",
"multiple-comparisons",
"matlab"
] |
2641 | 1 | 2647 | null | 648 | 471077 | The [wikipedia page](http://en.wikipedia.org/wiki/Likelihood_function) claims that likelihood and probability are distinct concepts.
>
In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability... | What is the difference between "likelihood" and "probability"? | CC BY-SA 2.5 | null | 2010-09-14T03:24:01.207 | 2021-11-28T21:59:44.893 | 2021-08-22T20:05:49.100 | 5679 | 386 | [
"probability",
"terminology",
"likelihood",
"intuition"
] |
2642 | 2 | null | 2641 | 76 | null | If I have a fair coin (parameter value) then the probability that it will come up heads is 0.5. If I flip a coin 100 times and it comes up heads 52 times then it has a high likelihood of being fair (the numeric value of likelihood potentially taking a number of forms).
| null | CC BY-SA 2.5 | null | 2010-09-14T03:44:11.397 | 2010-09-14T03:44:11.397 | null | null | 601 | null |
2644 | 2 | null | 2617 | 1 | null | Why does your transformation need to be a smooth deformation? Just take the binary representation of each number in 64-bit two's complement fixed point, or 64-bit IEEE floating point, and throw that into SHA-1. Boom, instant hash with uniform distribution of the outputs (assuming no duplicates).
| null | CC BY-SA 2.5 | null | 2010-09-14T04:40:27.327 | 2010-09-14T04:40:27.327 | null | null | 1122 | null |
2645 | 2 | null | 2641 | 79 | null | I will give you the perspective from the view of Likelihood Theory which originated with [Fisher](http://en.wikipedia.org/wiki/Ronald_Fisher) -- and is the basis for the statistical definition in the cited Wikipedia article.
Suppose you have random variates $X$ which arise from a parameterized distribution $F(X; \the... | null | CC BY-SA 3.0 | null | 2010-09-14T05:16:17.883 | 2015-10-23T10:09:59.643 | 2015-10-23T10:09:59.643 | 48595 | 251 | null |
2646 | 2 | null | 2641 | 61 | null | Suppose you have a coin with probability $p$ to land heads and $(1-p)$ to land tails. Let $x=1$ indicate heads and $x=0$ indicate tails. Define $f$ as follows
$$f(x,p)=p^x (1-p)^{1-x}$$
$f(x,2/3)$ is probability of x given $p=2/3$, $f(1,p)$ is likelihood of $p$ given $x=1$. Basically likelihood vs. probability tells yo... | null | CC BY-SA 4.0 | null | 2010-09-14T06:04:27.093 | 2019-01-10T12:08:06.180 | 2019-01-10T12:08:06.180 | 79696 | 511 | null |
2647 | 2 | null | 2641 | 470 | null | The answer depends on whether you are dealing with discrete or continuous random variables. So, I will split my answer accordingly. I will assume that you want some technical details and not necessarily an explanation in plain English.
Discrete Random Variables
Suppose that you have a stochastic process that takes disc... | null | CC BY-SA 4.0 | null | 2010-09-14T06:08:02.933 | 2019-03-24T14:27:29.807 | 2019-03-24T14:27:29.807 | null | null | null |
2648 | 1 | null | null | 7 | 420 | Imagine
- having a dependent variable $Y$ that is a proportion (i.e., the proportion of observations made at the given time point that satisfy a condition, where each time point involves 50 to 250 observations)
- $Y$ is measured at a series of time time points $X$, where $X = 1, 2, 3, ...$, typically to around 400.
... | Predicting proportions from time with a discontinuity | CC BY-SA 2.5 | null | 2010-09-14T08:43:15.420 | 2017-11-01T11:31:57.867 | 2017-11-01T11:31:57.867 | 28666 | 183 | [
"time-series",
"proportion",
"change-point",
"sigmoid-curve"
] |
2649 | 2 | null | 2641 | 155 | null | I'll try and minimise the mathematics in my explanation as there are some good mathematical explanations already.
As Robin Girard comments, the difference between probability and likelihood is closely related to the [difference between probability and statistics](https://stats.stackexchange.com/questions/665/whats-the-... | null | CC BY-SA 4.0 | null | 2010-09-14T08:45:32.590 | 2020-07-28T09:27:25.707 | 2020-07-28T09:27:25.707 | 125268 | 521 | null |
2650 | 1 | 2651 | null | 2 | 1826 | I'm working with a panel dataset, I've used many models,
homogeneous (fixed effect, pooled ols and Driscoll and Kraay)
heterogeneous (swamy random coefficients) and would like to do a
post-estimation to select the model that best fit my regression.
Is there any method, command that may allow me to do this?
Any hint ... | Panel data and selection models issue | CC BY-SA 4.0 | null | 2010-09-14T09:05:22.790 | 2019-07-14T08:29:01.283 | 2019-07-14T08:29:01.283 | 11887 | 1251 | [
"model-selection",
"econometrics",
"panel-data"
] |
2651 | 2 | null | 2650 | 3 | null | According to my experience, model selection is not so much about seeking the model that best fits the regression. The first question to ask is : which model reflects my experimental design the best? The second question would then be : which model reflects the covariance structure in my data the best?
Only then you can ... | null | CC BY-SA 2.5 | null | 2010-09-14T10:02:54.613 | 2010-09-14T10:02:54.613 | null | null | 1124 | null |
2652 | 2 | null | 2648 | 3 | null | Ignoring the "change point" your description suggests to me a (non-linear) mixed effects model of the following form:
```
g(E(Yi)) = Xi*beta + Zi*U
```
Where The betas are fixed effects, the U's are random effects, g(E(yi)) is the (logit) link to a binomial mean.
This will deal with logitudinal correlation of data and... | null | CC BY-SA 2.5 | null | 2010-09-14T11:59:07.703 | 2010-09-14T12:07:16.107 | 2010-09-14T12:07:16.107 | 521 | 521 | null |
2653 | 2 | null | 2579 | 4 | null | I think you use Stata, given your other post about [Panel data and selection models issue](https://stats.stackexchange.com/questions/2650/panel-data-and-selection-models-issue). Did you look at the following paper, [From the help desk: Swamy’s random-coefficients model](http://www.stata-journal.com/sjpdf.html?articlenu... | null | CC BY-SA 2.5 | null | 2010-09-14T13:02:36.380 | 2010-09-14T13:28:24.557 | 2017-04-13T12:44:45.640 | -1 | 930 | null |
2654 | 2 | null | 2648 | 4 | null | Sounds to me that Y(X) is a [sigmiodal](http://en.wikipedia.org/wiki/Sigmoid_function) process. Thus logistic regression should be suitable for this data. If you model this in R with:
```
glm(Y~X,family=binomial)
```
you will find that the "sharpness" of the transition is determined by the magnitude of the X coefficie... | null | CC BY-SA 2.5 | null | 2010-09-14T13:05:45.937 | 2010-09-14T13:05:45.937 | null | null | 229 | null |
2657 | 2 | null | 2628 | 2 | null | If you consider whatever it is that you are measuring to be a random process, then yes statistical tests are relevant. Take for example, flipping a coin 10 times to see if it is fair. You get 6 heads and 4 tails - what do you conclude?
| null | CC BY-SA 2.5 | null | 2010-09-14T14:58:42.770 | 2010-09-14T14:58:42.770 | null | null | 229 | null |
2658 | 2 | null | 2648 | 2 | null | James' approach looks good: each observation, according to your description, might have a Binomial(n[i], p[i]) distribution where n[i] is known and--to be fully general--p[i] is a completely unknown function that rises from near 0 to near 1 as i increases. A logistic regression (GLM with binomial response and logistic... | null | CC BY-SA 2.5 | null | 2010-09-14T15:28:33.867 | 2010-09-14T15:28:33.867 | null | null | 919 | null |
2659 | 2 | null | 2641 | 203 | null | This is the kind of question that just about everybody is going to answer and I would expect all the answers to be good. But you're a mathematician, Douglas, so let me offer a mathematical reply.
A statistical model has to connect two distinct conceptual entities: data, which are elements $x$ of some set (such as a ve... | null | CC BY-SA 3.0 | null | 2010-09-14T15:45:09.347 | 2019-07-30T12:19:12.380 | 2019-07-30T12:19:12.380 | 919 | 919 | null |
2660 | 2 | null | 2639 | 2 | null | This is two questions; (1) has good standard explanations available anywhere, so I'll address (2). You can compare distributions using a Kolmogorov-Smirnov statistic: it is the maximum difference between their empirical distribution functions. (The value of the edf at a number $x$ is the proportion of data less than ... | null | CC BY-SA 2.5 | null | 2010-09-14T16:43:26.823 | 2010-09-28T19:19:41.043 | 2010-09-28T19:19:41.043 | 919 | 919 | null |
2661 | 1 | 2665 | null | 7 | 18082 | I am doing some basic fitting of data and exploring different fits. I understand that the residual is the difference between the sample and the estimated function value. The norm of the residuals is a measure of the deviation between the correlation and the data. A lower norm signifies a better fit.
Suppose a cubic fi... | Difference between Norm of Residuals and what is a "good" Norm of Residual | CC BY-SA 2.5 | null | 2010-09-14T17:18:10.680 | 2010-10-14T17:12:45.643 | 2010-10-14T17:12:45.643 | 930 | 559 | [
"regression",
"residuals"
] |
2663 | 2 | null | 2628 | 19 | null | Traditionally, statistical inference is taught in the context of probability samples and the nature of sampling error. This model is the basis for the test of significance. However, there are other ways to model systematic departures from chance and it turns out that our parametric (sampling based) tests tend to be g... | null | CC BY-SA 2.5 | null | 2010-09-14T18:15:20.813 | 2010-09-14T18:15:20.813 | null | null | 485 | null |
2664 | 2 | null | 2615 | 2 | null | As kwak has pointed out, my question was answered on another forum:
[http://www.or-exchange.com/questions/695/analytical-solutions-to-limits-of-correlation-stress-testing](http://www.or-exchange.com/questions/695/analytical-solutions-to-limits-of-correlation-stress-testing)
I did several quick calculations and the sugg... | null | CC BY-SA 2.5 | null | 2010-09-14T19:18:26.480 | 2010-09-14T19:18:26.480 | null | null | 1250 | null |
2665 | 2 | null | 2661 | 11 | null | So, I would recommend using standard method for comparing nested models. In your case, you consider two alternative models, the cubic fit being the more "complex" one. An F- or $\chi^2$-test tells you whether the residual sum of squares or deviance significantly decrease when you add further terms. It is very like comp... | null | CC BY-SA 2.5 | null | 2010-09-14T20:21:34.290 | 2010-10-08T07:22:07.983 | 2010-10-08T07:22:07.983 | 930 | 930 | null |
2667 | 1 | 2674 | null | 7 | 786 | When given samples of a discrete random variable, the entropy of the distribution may be estimated by $- \sum \hat{P_i} \log{\hat{P_i}}$, where $\hat{P_i}$ is the sample estimate of the frequency of the $i$th value. (this is up to a constant determined by the base of the log.) This estimate should not be applied to obs... | All-Purpose Sample Entropy | CC BY-SA 2.5 | null | 2010-09-14T21:43:41.883 | 2010-09-15T09:34:46.873 | null | null | 795 | [
"discrete-data",
"continuous-data",
"entropy"
] |
2668 | 2 | null | 665 | 8 | null | Probability is the embrace of uncertainty, while statistics is an empirical, ravenous pursuit of the truth (damned liars excluded, of course).
| null | CC BY-SA 2.5 | null | 2010-09-14T21:56:16.450 | 2010-09-14T21:56:16.450 | null | null | null | null |
2670 | 1 | null | null | 4 | 5839 | There are 100 people in a room.
What are the odds at least 1 person is born in January?
What is the best way to calculate this?
I used Binomial Distribution.
$E(x) = np = 100 \frac{1}{12} = 8.3$
$std.dev. = \sqrt{npq} = 2.76$
$Z(\text{1 person}) = (1 - 8.3) / 2.76 = -2.64$
$p = P(Z<-2.64) = .004$ or $.4%$ odds that les... | Odds at least 1 person is born in January? | CC BY-SA 2.5 | null | 2010-09-15T00:45:11.290 | 2022-12-06T01:00:14.010 | 2010-09-16T12:29:04.900 | null | 1279 | [
"probability",
"confidence-interval",
"binomial-distribution"
] |
2671 | 2 | null | 2670 | 12 | null | Do you mean [odds or probability](http://en.wikipedia.org/wiki/Odds#Gambling_odds_versus_probabilities)?
$$p(\text{at least 1 person born in January}) = 1 - p(\text{no one born in January})$$
Assuming Independence:
\begin{align}
p(\text{no one born in January}) &= P(\text{person 1 not born in Jan}) \\
... | null | CC BY-SA 4.0 | null | 2010-09-15T01:07:21.113 | 2022-12-06T01:00:14.010 | 2022-12-06T01:00:14.010 | 11887 | 521 | null |
2672 | 2 | null | 2670 | 6 | null | I will shamelessly propose using simulation (in R) to answer your questions.
`rbinom(n = 100000, size = 100, prob = 1/12)` draws 100 trials from the binomial distribution 100,000 times and returns a vector of the number of 'successes' (i.e., January births, using the naive 1/12 probability).
You should be able to appro... | null | CC BY-SA 2.5 | null | 2010-09-15T06:30:41.257 | 2010-09-15T06:30:41.257 | null | null | 71 | null |
2673 | 2 | null | 2670 | 8 | null | Just to add few comments to Thylacoleo excellent answer. From the CDC’s National Vital Statistics System [recent report](http://www.cdc.gov/nchs/data/nvsr/nvsr54/nvsr54_02.pdf), we can estimate the probability of being born on January 1st with a bit more precision.
Month
From page 50 of the report, we the probability ... | null | CC BY-SA 2.5 | null | 2010-09-15T06:49:48.500 | 2010-09-15T06:49:48.500 | null | null | 8 | null |
2674 | 2 | null | 2667 | 6 | null | Entropy is entropy with respect to a measure
As noticed in the answer to this question [https://mathoverflow.net/questions/33088/entropy-of-a-general-prob-measure/33090#33090](https://mathoverflow.net/questions/33088/entropy-of-a-general-prob-measure/33090#33090) , entropy is only defined with respect to a given measu... | null | CC BY-SA 2.5 | null | 2010-09-15T09:34:46.873 | 2010-09-15T09:34:46.873 | 2017-04-13T12:58:32.177 | -1 | 223 | null |
2675 | 1 | 2701 | null | 3 | 1801 | We want to work out a value for "Average time in service" for our users.
However, as we have many subscribers still active, how can we do that? If we just look at average time of deactivated subscribers, that won't show the picture.
Is there some "correct" way to do this? We know our churn rate by month, so maybe we n... | How to calculate average time in service | CC BY-SA 3.0 | null | 2010-09-15T10:49:45.213 | 2011-05-24T15:30:25.013 | 2011-05-24T15:30:25.013 | 919 | 12517 | [
"survival"
] |
2678 | 2 | null | 2675 | 5 | null | Sounds like [Survival analysis](http://en.wikipedia.org/wiki/Time_to_event_analysis) to me. The still active subscribers have censored times.
| null | CC BY-SA 2.5 | null | 2010-09-15T11:50:29.130 | 2010-09-15T11:50:29.130 | null | null | 449 | null |
2679 | 1 | 2680 | null | 4 | 3549 | Car analogy:
Assume the traffic (number of cars per hour) on a road has a Poisson distribution, and the time between cars has the matching exponential distribution. If the chance of each car being red is independent from both time and the color of other cars, will the number of red cars per hour also have a Poisson dis... | Is a random sample of a Poisson distribution also Poisson distributed? | CC BY-SA 2.5 | null | 2010-09-15T14:18:53.813 | 2010-09-18T05:07:02.720 | null | null | 1287 | [
"poisson-distribution"
] |
2680 | 2 | null | 2679 | 5 | null | The answer to your first question is yes. If the sum of two independent variables is poisson then the individual variables are also poisson. See [Raikov's Theorem](http://en.wikipedia.org/wiki/Raikov%27s_theorem)
| null | CC BY-SA 2.5 | null | 2010-09-15T14:28:02.933 | 2010-09-15T18:15:34.653 | 2010-09-15T18:15:34.653 | 919 | null | null |
2681 | 2 | null | 2679 | 4 | null | What you have described is a standard and important result from the theory of stochastic processes. Pick up any good book in the library on those, flip to the "Poisson Processes" chapter, and look for "decomposition" or sometimes "thinning".
Two of my favorites are Introduction to Stochastic Processes by Cinlar and Ess... | null | CC BY-SA 2.5 | null | 2010-09-15T14:34:29.747 | 2010-09-15T14:34:29.747 | null | null | null | null |
2682 | 2 | null | 2679 | 1 | null | I think you need additional assumptions to draw this conclusion.
Imagine this: the cars on the road are originally unpainted, but just before you observe them a demon decides what color they will become: red or some other color (of which there could be many). The demon does not have a clock (to ensure that whatever he... | null | CC BY-SA 2.5 | null | 2010-09-15T14:39:11.517 | 2010-09-15T14:39:11.517 | null | null | 919 | null |
2684 | 1 | 2740 | null | 5 | 17763 | I am using g*power to calculate the a priori power for a 2 group repeated measures anova with a continuous dependent variable with 3 time points. I think that higher values for the "correlation among rep measures" should indicate a better test (i.e., high test-retest correltaion) and thus yield better power and a lowe... | How does "correlation among repeated measures" work for repeated measures power analysis in g*power? | CC BY-SA 2.5 | null | 2010-09-15T16:40:39.463 | 2016-03-04T05:59:37.200 | 2010-09-16T12:27:41.330 | null | 1290 | [
"hypothesis-testing",
"anova"
] |
2685 | 2 | null | 1799 | 2 | null | I don't think that repeated measures is appropriate here - as I understand your protocol, the five temperatures (either hot or cold) are not repeated measures as they occur at different temperatures.
However, it does feel like you need some data reduction across the five levels in each temperature group. You might be ... | null | CC BY-SA 2.5 | null | 2010-09-15T16:53:28.813 | 2010-09-15T16:53:28.813 | null | null | 1290 | null |
2686 | 1 | null | null | 9 | 335 | Help me here, please. Perhaps before even giving me an answer you may need to help me ask the question. I have never learned about time series analysis and do not know if that is indeed what I need. I have never learned about time smoothed averages and do not know if that is indeed what I need. My statistics background... | Understanding productivity or expenses over time without falling victim to stochastic interruptions | CC BY-SA 2.5 | null | 2010-09-15T18:42:29.020 | 2010-09-24T17:57:25.143 | null | null | 104 | [
"time-series",
"finance",
"count-data"
] |
2687 | 2 | null | 2684 | 1 | null | Higher correlation among repeated measures means that each repeat is adding less information to the first measurement. If the correlation were 1, the repeats are no extra information to the original measurement, and the sample size could be calculated ignoring the repeats. If the correlation were 0, each repeated measu... | null | CC BY-SA 2.5 | null | 2010-09-15T19:03:14.480 | 2010-09-15T19:03:14.480 | null | null | 449 | null |
2688 | 1 | null | null | 11 | 2399 | I have a dataset where I need to do linear regression. Unfortunately there is a problem with heteroscedasticity. I´ve rerun the analysis using robust regression with the HC3 estimator for the variance and also done bootstrapping with the bootcov function in Hmisc for R. The results are quite close. What is generally re... | Whether to use robust linear regression or bootstrapping when there is heteroscedasticity? | CC BY-SA 3.0 | null | 2010-09-15T19:20:45.563 | 2011-09-28T00:02:30.217 | 2011-09-28T00:02:30.217 | 183 | 1291 | [
"regression",
"bootstrap",
"heteroscedasticity"
] |
2689 | 1 | 2707 | null | 7 | 3183 | I'm doing this only for learning purposes. I've no intentions of reversing the methods of IMDB.
I asked myself I owned IMDB or similar website. How would I compute the movie rating?
All I can think of is Arithmetic Mean
For a movie data provided below computation would be
>
(38591*10 + 27994*9 + 32732*8 + 17864*7 + 7... | How would YOU compute IMDB movie rating? | CC BY-SA 2.5 | null | 2010-09-15T19:54:28.173 | 2010-09-17T02:08:35.420 | 2010-09-15T22:39:07.443 | null | 851 | [
"rating"
] |
2690 | 2 | null | 2684 | 1 | null | I would guess the explanation is something like this: The higher the correlation between the time points, the lower the proportion of unshared variance. That is, the lower the proportion of variance that can contain the effect (i.e., the unshared variance). Given that in this unshared variance the ratio of effect to no... | null | CC BY-SA 2.5 | null | 2010-09-15T20:02:18.693 | 2010-09-15T20:02:18.693 | null | null | 442 | null |
2691 | 1 | 140579 | null | 1313 | 894129 | In today's pattern recognition class my professor talked about PCA, eigenvectors and eigenvalues.
I understood the mathematics of it. If I'm asked to find eigenvalues etc. I'll do it correctly like a machine. But I didn't understand it. I didn't get the purpose of it. I didn't get the feel of it.
I strongly beli... | Making sense of principal component analysis, eigenvectors & eigenvalues | CC BY-SA 4.0 | null | 2010-09-15T20:05:55.993 | 2022-12-04T10:39:59.730 | 2019-09-27T13:25:06.037 | 919 | 851 | [
"pca",
"intuition",
"eigenvalues",
"faq"
] |
2692 | 2 | null | 2689 | 8 | null | Partial answer. See the help page entitled: [The vote average for film "X" should be Y! Why are you displaying another rating?](http://www.imdb.com/help/show_leaf?voteaverage)
In short, IMDb uses:
>
a complex voter weighting system to make sure
that the final rating is
representative of the general voting
popul... | null | CC BY-SA 2.5 | null | 2010-09-15T20:42:28.683 | 2010-09-15T20:42:28.683 | null | null | 938 | null |
2693 | 2 | null | 2688 | 5 | null | In economics, the Eicker-White or "robust" standard errors are typically reported. Bootstrapping (unfortunately, I'd say) is less common. I'd say that the robust estimates are the standard version.
| null | CC BY-SA 2.5 | null | 2010-09-15T20:45:01.947 | 2010-09-15T20:45:01.947 | null | null | 401 | null |
2695 | 2 | null | 2691 | 56 | null | Alright, I'll give this a try. A few months back I dug through a good amount of literature to find an intuitive explanation I could explain to a non-statistician. I found the derivations that use Lagrange multipliers the most intuitive.
Let's say we have high dimension data - say 30 measurements made on an insect. The ... | null | CC BY-SA 4.0 | null | 2010-09-15T21:07:58.990 | 2022-12-04T10:39:59.730 | 2022-12-04T10:39:59.730 | 362671 | 7 | null |
2696 | 2 | null | 1 | 10 | null | I'd recommend letting the subject expert specify the mean or mode of the prior but I'd feel free to adjust what they give as a scale. Most people are not very good at quantifying variance.
And I would definitely not let the expert determine the distribution family, in particular the tail thickness. For example, supp... | null | CC BY-SA 2.5 | null | 2010-09-15T21:08:26.077 | 2010-09-15T21:08:26.077 | null | null | 319 | null |
2697 | 2 | null | 2691 | 127 | null | Hmm, here goes for a completely non-mathematical take on PCA...
Imagine you have just opened a cider shop. You have 50 varieties of cider and you want to work out how to allocate them onto shelves, so that similar-tasting ciders are put on the same shelf. There are lots of different tastes and textures in cider - swe... | null | CC BY-SA 2.5 | null | 2010-09-15T21:14:37.067 | 2010-09-15T21:14:37.067 | null | null | 266 | null |
2698 | 2 | null | 2691 | 161 | null | Let's do (2) first. PCA fits an ellipsoid to the data. An ellipsoid is a multidimensional generalization of distorted spherical shapes like cigars, pancakes, and eggs. These are all neatly described by the directions and lengths of their principal (semi-)axes, such as the axis of the cigar or egg or the plane of the... | null | CC BY-SA 3.0 | null | 2010-09-15T21:33:12.360 | 2014-11-11T16:53:10.420 | 2014-11-11T16:53:10.420 | 919 | 919 | null |
2699 | 2 | null | 2689 | 5 | null | >
Whats wrong with my score? Why is it not ideal (because IMDB didn't
use it)?
If the score was only for your use, then nothing is wrong with your calculation.
However, IMDB try to make it difficult for people to obviously influence the final score.
>
If you had to compute. How would you have done it? What facto... | null | CC BY-SA 2.5 | null | 2010-09-15T21:39:44.023 | 2010-09-15T21:39:44.023 | null | null | 8 | null |
2700 | 2 | null | 2691 | 439 | null | The manuscript ["A tutorial on Principal Components Analysis" by Lindsay I Smith](https://ourarchive.otago.ac.nz/bitstream/handle/10523/7534/OUCS-2002-12.pdf?sequence=1&isAllowed=y) really helped me grok PCA. I think it's still too complex for explaining to your grandmother, but it's not bad. You should skip first few ... | null | CC BY-SA 4.0 | null | 2010-09-15T21:42:36.200 | 2019-01-06T11:54:14.150 | 2019-01-06T11:54:14.150 | -1 | 29 | null |
2701 | 2 | null | 2675 | 2 | null | onestop is right, you're looking to do survival analysis. In general, you can use a nonparametric [Kaplan-Meier estimator](http://en.wikipedia.org/wiki/Kaplan-Meier_estimator) to plot the survival curve and then derive the "average time in service". The area underneath the survival curve works if you don't have any ce... | null | CC BY-SA 2.5 | null | 2010-09-16T00:20:19.813 | 2010-09-16T16:23:10.463 | 2010-09-16T16:23:10.463 | 251 | 251 | null |
2702 | 2 | null | 665 | 12 | null | Probability answers questions about what will happen, statistics answers questions about what did happen.
| null | CC BY-SA 2.5 | null | 2010-09-16T00:54:49.907 | 2010-09-16T00:54:49.907 | null | null | 9426 | null |
2703 | 2 | null | 2691 | 18 | null | I might be a bad person to answer this because I'm the proverbial grandmother who has had the concept explained to me and not much more, but here goes:
Suppose you have a population. A large portion of the population is dropping dead of heart attacks. You are trying to figure out what causes the heart attacks.
You have... | null | CC BY-SA 3.0 | null | 2010-09-16T02:04:43.360 | 2013-06-20T10:05:32.863 | 2013-06-20T10:05:32.863 | 22047 | 239 | null |
2705 | 2 | null | 2684 | 2 | null |
- All else being equal, bigger correlations in a repeated measures design means more power
- A larger correlation means that more of the variance in the dependent variable is explained by systematic effects of cases/individuals
- Prior research or previous datasets are useful for estimating the correlation. Estimate... | null | CC BY-SA 3.0 | null | 2010-09-16T03:26:35.713 | 2011-06-07T04:24:49.023 | 2011-06-07T04:24:49.023 | 183 | 183 | null |
2706 | 2 | null | 2691 | 8 | null | Basically PCA finds new variables which are linear combinations of the original variables such that in the new space, the data has fewer dimensions. Think of a data set consisting of the points in 3 dimensions on the surface of a flat plate held up at an angle. In the original x, y, z axes you need 3 dimensions to re... | null | CC BY-SA 2.5 | null | 2010-09-16T04:10:44.910 | 2010-09-16T04:10:44.910 | null | null | 1303 | null |
2707 | 2 | null | 2689 | 7 | null | First, define the theoretical construct of interest. There are many ways that a rating can be defined:
- What is the theoretical target population? The entire world, English speakers, people who visit IMDB, people who have seen the movie in question?
- What is the target time frame? It is the rating of the movie now ... | null | CC BY-SA 2.5 | null | 2010-09-16T05:11:05.977 | 2010-09-17T02:08:35.420 | 2010-09-17T02:08:35.420 | 183 | 183 | null |
2708 | 2 | null | 2691 | 9 | null | I view PCA as a geometric tool. If you are given a bunch of points in 3-space which are pretty much all on a straight line, and you want to figure out the equation of that line, you get it via PCA (take the first component). If you have a bunch of points in 3-space which are mostly planar, and want to discover the equa... | null | CC BY-SA 2.5 | null | 2010-09-16T05:15:42.383 | 2010-09-16T05:15:42.383 | null | null | 795 | null |
2709 | 2 | null | 2617 | 1 | null | To recap, you have a function $f(x,y)$ which is continuous, and pseudo-random uniform at integer $(x,y)$, but nearly normal over the plane. You wish to make $f$ pseudo-random uniform over the whole plane. Suppose you could somehow make $f$ parametrized by some parameter $\lambda$, say by using $\theta$ to twiddle the p... | null | CC BY-SA 2.5 | null | 2010-09-16T05:33:58.557 | 2010-09-16T05:33:58.557 | null | null | 795 | null |
2710 | 2 | null | 2688 | 3 | null | You could use generalized least squares, such as the gls() function from the nlme package, which allows you to specify a variance function using the weight argument.
| null | CC BY-SA 2.5 | null | 2010-09-16T06:36:14.457 | 2010-09-16T06:36:14.457 | null | null | 187 | null |
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