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4821 | 2 | null | 4816 | 24 | null | The Sharpest bound I know is that of [Massart and Laurent](https://projecteuclid.org/journals/annals-of-statistics/volume-28/issue-5/Adaptive-estimation-of-a-quadratic-functional-by-model-selection/10.1214/aos/1015957395.full) Lemma 1 p1325.
A corollary of their bound is:
$$P(X-k\geq 2\sqrt{kx}+2x)\leq \exp (-x) $$
$$P... | null | CC BY-SA 4.0 | null | 2010-11-23T12:02:06.880 | 2022-08-12T20:27:22.900 | 2022-08-12T20:27:22.900 | 79696 | 223 | null |
4822 | 2 | null | 4818 | 16 | null | A close-to-intuitive answer:
Take a closer look at the formula for the McNemar test, given the table
```
pos | neg
----|-----|-----
pos | a | b
----|-----|-----
neg | c | d
```
The McNemar statistic `M` is calculated as:
$$ M = {(b-c)^2 \over b+c} $$
The definition of a $\chi^2$ distribution with k degree... | null | CC BY-SA 2.5 | null | 2010-11-23T12:33:32.047 | 2010-11-23T21:09:56.640 | 2010-11-23T21:09:56.640 | 1124 | 1124 | null |
4823 | 1 | 4824 | null | 13 | 43131 | I've got a dataframe like the following:
```
case simulation temp plank oxygen
1 1 1 8 7 11
2 2 1 16 10 15
...
17 17 2 26 12 17
18 18 2 15 8 12
19 19 2 28 11 21
20 20 2 24 6 14
```
I'd lik... | Automatically produce summary by factor variable in R | CC BY-SA 2.5 | null | 2010-11-23T12:44:40.340 | 2010-11-23T14:31:45.363 | null | null | 261 | [
"r"
] |
4824 | 2 | null | 4823 | 19 | null | Check out the `by()` or `tapply()` functions. Basically,
```
tapply(y, g, mean)
```
will give you the mean of `y` by levels of `g`. If you want to get a data.frame from the resulting aggregated measures, use `aggregate()`.
A more elaborated solution is available through the `summary.formula()` function in the [Hmisc]... | null | CC BY-SA 2.5 | null | 2010-11-23T12:49:02.623 | 2010-11-23T12:49:02.623 | null | null | 930 | null |
4825 | 2 | null | 3113 | 7 | null | Some people seem to use a concept of Pitman Asymptotic Relative Efficiency (ARE) to inflate the sample size obtained by using a sample size formula for a parametric test. Ironically, in order to compute it, one has to assume a distribution again... see e.g. [Sample size for the Mann-Whitney U test](http://www.childrens... | null | CC BY-SA 2.5 | null | 2010-11-23T14:04:56.840 | 2010-11-23T19:45:50.180 | 2010-11-23T19:45:50.180 | 1573 | 1573 | null |
4826 | 2 | null | 4823 | 3 | null | package `doBy` has a `summaryBy` function that has a formula based syntax like the one you tried.
Also, i think that question would have been better asked on stackexchange.
| null | CC BY-SA 2.5 | null | 2010-11-23T14:31:45.363 | 2010-11-23T14:31:45.363 | null | null | 1979 | null |
4827 | 2 | null | 4817 | 7 | null | Good (2005) defines the one-sample sign-test for the location parameter $\theta$ for a continuous symmetric variable $X$ as follows:
- Take the difference $D_i$ of each observation to the location parameter $\theta_0$ under the null hypothesis.
- Define an indicator variable $Z_i$ as $0$ when $D_i < 0$, and as $1$ wh... | null | CC BY-SA 2.5 | null | 2010-11-23T14:35:44.533 | 2010-11-23T15:31:39.493 | 2010-11-23T15:31:39.493 | 1909 | 1909 | null |
4828 | 2 | null | 4807 | 2 | null | When I initially wrote the comments below I had assumed that the Heckman estimator could be used for dichotomous outcomes, but the second paper I cite says there is no direct analog. Hopefully someone can point to different and more applicable resources. I still leave my initial comment up as I still feel those papers ... | null | CC BY-SA 3.0 | null | 2010-11-23T15:05:47.490 | 2013-07-25T12:09:30.270 | 2013-07-25T12:09:30.270 | 930 | 1036 | null |
4829 | 2 | null | 4753 | 2 | null | I might suggest non-negative matrix factorization. The iterative algorithm of Lee and Seung is easy to implement and should be amenable to sparse matrices (although it involves Hadamard products, which some sparse matrix packages may not support.).
| null | CC BY-SA 2.5 | null | 2010-11-23T17:00:09.463 | 2010-11-23T17:00:09.463 | null | null | 795 | null |
4830 | 1 | 4873 | null | 13 | 7835 | Full Disclosure: This is homework. I've included a link to the dataset ( [http://www.bertelsen.ca/R/logistic-regression.sav](http://www.bertelsen.ca/R/logistic-regression.sav) )
My goal is to maximize the prediction of loan defaulters in this data set.
Every model that I have come up with so far, predicts >90% of non... | Better Classification of default in logistic regression | CC BY-SA 2.5 | null | 2010-11-23T17:35:51.753 | 2010-12-07T16:30:00.257 | 2010-11-24T11:46:49.317 | 159 | 776 | [
"r",
"logistic",
"spss",
"self-study"
] |
4831 | 1 | 4833 | null | 51 | 42253 | When transforming variables, do you have to use all of the same transformation? For example, can I pick and choose differently transformed variables, as in:
Let, $x_1,x_2,x_3$ be age, length of employment, length of residence, and income.
```
Y = B1*sqrt(x1) + B2*-1/(x2) + B3*log(x3)
```
Or, must you be consistent wi... | Regression: Transforming Variables | CC BY-SA 3.0 | null | 2010-11-23T17:41:19.050 | 2021-05-20T11:40:23.770 | 2017-10-04T13:24:10.083 | 919 | 776 | [
"r",
"regression",
"logistic",
"data-transformation"
] |
4832 | 1 | 4835 | null | 7 | 7205 | In SPSS output there is a pretty little classification table available when you perform a logistic regression, is the same possible with R? If so, how?
| Logistic Regression: Classification Tables a la SPSS in R | CC BY-SA 2.5 | null | 2010-11-23T17:44:28.190 | 2010-11-23T21:33:23.423 | 2010-11-23T17:52:19.663 | 776 | 776 | [
"r",
"logistic",
"spss"
] |
4833 | 2 | null | 4831 | 71 | null | One transforms the dependent variable to achieve approximate symmetry and homoscedasticity of the residuals. Transformations of the independent variables have a different purpose: after all, in this regression all the independent values are taken as fixed, not random, so "normality" is inapplicable. The main objectiv... | null | CC BY-SA 2.5 | null | 2010-11-23T17:55:52.673 | 2010-11-23T18:03:33.650 | 2010-11-23T18:03:33.650 | 919 | 919 | null |
4834 | 1 | 4836 | null | 8 | 2397 | One of the assumptions for using the Wilcoxon sign-rank test is that the underlying distribution is continuous ([see here](http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test#Assumptions).)
However, there are cases (for example, when analyzing Likert scale data) where this assumption might not necessarily hold. In ... | Alternative to the Wilcoxon test when the distribution isn't continuous? | CC BY-SA 2.5 | null | 2010-11-23T17:59:32.200 | 2010-11-23T18:18:57.223 | 2010-11-23T18:18:57.223 | 919 | 253 | [
"r",
"nonparametric",
"sign-test",
"wilcoxon-signed-rank"
] |
4835 | 2 | null | 4832 | 8 | null | I'm not aware of a specific command, but this might be a start:
```
# generate some data
> N <- 100
> X <- rnorm(N, 175, 7)
> Y <- 0.4*X + 10 + rnorm(N, 0, 3)
# dichotomize Y
> Yfac <- cut(Y, breaks=c(-Inf, median(Y), Inf), labels=c("lo", "hi"))
# logistic regression
> glmFit <- glm(Yfac ~ X, family=binomial(link="lo... | null | CC BY-SA 2.5 | null | 2010-11-23T18:00:23.230 | 2010-11-23T18:00:23.230 | null | null | 1909 | null |
4836 | 2 | null | 4834 | 7 | null | I have found that the Wilcoxon statistic is still fine for this purpose and that small simulations do a good job of estimating the size and the power of the test. I suspect this is more powerful than just comparing the two medians. The main concern is lack of power due to extensive numbers of ties, but that concern a... | null | CC BY-SA 2.5 | null | 2010-11-23T18:17:49.580 | 2010-11-23T18:17:49.580 | null | null | 919 | null |
4837 | 2 | null | 4364 | 25 | null | There's an entire book written about this: "Characterizations of the normal probability law", A. M. Mathai & G. Perderzoli. A brief review in [JASA (Dec. 1978)](http://www.jstor.org/pss/2286313) mentions the following:
>
Let $X_1, \ldots, X_n$ be independent random variables. Then $\sum_{i=1}^n{a_i x_i}$ and $\sum_{... | null | CC BY-SA 3.0 | null | 2010-11-23T18:37:54.900 | 2016-11-04T16:46:51.817 | 2016-11-04T16:46:51.817 | 1569 | 919 | null |
4838 | 2 | null | 47 | 2 | null | This is problem with clustering, you can't tell what is considered a cluster. I would see what is the reason behind clustering users, specify my threshold value, and use hierarchical clustering.
In my experience, one has to set either the number of cluster needed, or the threshold value (the distance value that binds t... | null | CC BY-SA 2.5 | null | 2010-11-23T18:59:58.170 | 2010-11-23T18:59:58.170 | null | null | 2108 | null |
4839 | 1 | null | null | 4 | 1870 | I have a survey where I have asked people which type of computer games they enjoy and whether they consider themselves a hardcore gamer. I allowed people to select multiple genres, but now I am unsure what to do with my data.
I initially thought factor analysis, with the idea being if there were genres that belonged to... | Whether to use factor analysis based on binary multiple response data? | CC BY-SA 3.0 | null | 2010-11-23T19:10:52.133 | 2019-07-20T21:48:37.407 | 2019-07-20T21:48:37.407 | 11887 | 2127 | [
"binary-data",
"factor-analysis",
"survey",
"correspondence-analysis"
] |
4840 | 1 | null | null | 1 | 235 | my friend is a chemist and his problem is to predict the level of ozone concentration in a single site. We have the data for the last 12 years.
We want to predict the concentration for the coming years (as much as possible).
I know this data set is small, so my question is is this possible? And how much of data is requ... | How to predict Ozone concentration in few years time? | CC BY-SA 2.5 | null | 2010-11-23T19:11:35.047 | 2011-05-19T21:20:30.913 | 2010-11-27T15:10:15.177 | 2108 | 2108 | [
"machine-learning",
"forecasting",
"predictive-models"
] |
4841 | 2 | null | 4803 | 3 | null | A Gamma distribution with an integer shape parameter is an [Erlang distribution](http://en.wikipedia.org/wiki/Erlang_distribution) which in turn is a generalization of the Exponential distribution often used to model waiting times. Thus, perhaps one approach is to use an Erlang distribution instead of a Gamma to model ... | null | CC BY-SA 2.5 | null | 2010-11-23T20:00:49.273 | 2010-11-23T20:00:49.273 | null | null | null | null |
4842 | 2 | null | 4818 | 8 | null | Won't the two approaches come to the same thing? The relevant chi-square distribution has one degree of freedom so is simply the distribution of the square of a random variable with a standard normal distribution. I'd have to go through the algebra to check, which I haven't got time to do right now, but I'd be surprise... | null | CC BY-SA 2.5 | null | 2010-11-23T20:23:40.877 | 2010-11-23T20:23:40.877 | null | null | 449 | null |
4843 | 1 | 4848 | null | 3 | 2477 | The "standard" way to compute Kappa for a predictive classification model (Witten and Frank page 163) is to construct the random confusion matrix in such a way that the number of predictions for each class is the same as the model predicted.
For a visual, see (right side is the random):
![alt text](https://i.stack.img... | Kappa for Predictive Model | CC BY-SA 2.5 | null | 2010-11-23T20:47:07.533 | 2010-11-23T22:09:40.567 | 2010-11-23T21:30:51.223 | 2040 | 2040 | [
"predictive-models"
] |
4844 | 1 | null | null | 5 | 1780 | I have a timeseries of data which was gathered by driving a car around randomly.
One data point was gathered every minute and each data point is either "Yes" or "No". Yes = temperature is above a threshold, No = it is below the threshold. Assume the car never stops.
I'm interested in the proportion of Yeses and relat... | Confidence interval based on time series | CC BY-SA 2.5 | null | 2010-11-23T20:52:02.350 | 2022-08-18T20:59:10.793 | 2010-11-23T20:58:18.350 | null | 1784 | [
"time-series",
"confidence-interval",
"independence"
] |
4845 | 2 | null | 4830 | 4 | null | I'm not a logistic regression expert, but isn't it just a problem of unbalanced data? Probably you have much more non-defaulters than defaulters what may shift the prediction to deal better with larger class. Try to kick out some non-defaulters and see what happens.
| null | CC BY-SA 2.5 | null | 2010-11-23T20:53:52.237 | 2010-11-23T20:53:52.237 | null | null | null | null |
4846 | 2 | null | 4832 | 5 | null | Thomas D. Fletcher has a function called `ClassLog()` (for "classification analysis for a logistic regression model") in his [QuantPsyc](http://cran.r-project.org/web/packages/QuantPsyc/) package. However, I like @caracal's response because it is self-made and easily customizable.
| null | CC BY-SA 2.5 | null | 2010-11-23T21:33:23.423 | 2010-11-23T21:33:23.423 | null | null | 930 | null |
4847 | 2 | null | 4843 | 2 | null | This is just an outer product of actual and predicted class frequencies times all objects count. In R:
```
A<-c(100,60,40)
P<-c(120,60,20)
numCases<-sum(A) #=sum(P) also
(A/numCases)%o%(P/numCases)*numCases
A%o%P/numCases #same simplified
```
| null | CC BY-SA 2.5 | null | 2010-11-23T21:34:43.747 | 2010-11-23T21:34:43.747 | null | null | null | null |
4848 | 2 | null | 4843 | 6 | null | It might be useful to consider Cohen's $\kappa$ in the context of inter-rater-agreement. Suppose you have two raters individually assigning the same set of objects to the same categories. You can then ask for overall agreement by dividing the sum of the diagonal of the confusion matrix by the total sum. But this does n... | null | CC BY-SA 2.5 | null | 2010-11-23T21:53:33.310 | 2010-11-23T22:09:40.567 | 2010-11-23T22:09:40.567 | 1909 | 1909 | null |
4849 | 2 | null | 4839 | 4 | null | Correspondence analysis might be a good fit. Creating a graph that shows the relationship between Hard-Core gaming (or not) and the different types of games that they are playing. The result would be the ability to say with cautious confidence that Hard Core Gamers play a certain group of game types (or not).
With res... | null | CC BY-SA 2.5 | null | 2010-11-23T22:12:47.053 | 2010-11-23T22:12:47.053 | 2017-04-13T12:44:46.680 | -1 | 776 | null |
4850 | 2 | null | 1525 | 27 | null | I have hesitated to wade into this discussion, but because it seems to have gotten sidetracked over a trivial issue concerning how to express numbers, maybe it's worthwhile refocusing it. A point of departure for your consideration is this:
>
A probability is a hypothetical property. Proportions summarize observatio... | null | CC BY-SA 3.0 | null | 2010-11-23T22:38:38.997 | 2013-12-10T15:01:22.863 | 2020-06-11T14:32:37.003 | -1 | 919 | null |
4851 | 1 | 4853 | null | 4 | 397 | I'm trying to precompute the distributions of several random variables. In particular, these random variables are the results of functions evaluated at locations in a genome, so there will be on the order of 10^8 or 10^9 values for each. The functions are pretty smooth, so I don't think I'll lose much accuracy by only ... | Efficient Empirical CDF Computation / Storage | CC BY-SA 2.5 | null | 2010-11-23T23:02:12.753 | 2010-11-23T23:29:44.537 | null | null | 2111 | [
"distributions",
"python",
"bioinformatics",
"cumulative-distribution-function",
"biostatistics"
] |
4852 | 2 | null | 4851 | 2 | null | Here's a possible computing solution to your problem.
If the pdf is sparse (by sparse I mean that long stretches are zero) as your graph indicates, then you could perhaps exploit that structure.
For example, you could have one vector which gives the pdf for non-zero regions and another vector indicating where the non-z... | null | CC BY-SA 2.5 | null | 2010-11-23T23:28:58.350 | 2010-11-23T23:28:58.350 | null | null | 8 | null |
4853 | 2 | null | 4851 | 6 | null | Why not create bins a priori to cover the expected range of a variable, scan through the data once to record all the bin counts, and then estimate the percentiles via interpolation of the EDF? If the bins are evenly distributed, locating the bin for a value is a $O(1)$ operation; in the worst case it's a binary search... | null | CC BY-SA 2.5 | null | 2010-11-23T23:29:44.537 | 2010-11-23T23:29:44.537 | null | null | 919 | null |
4854 | 1 | 4861 | null | 26 | 17819 | In Logistic Regression, is there a need to be as concerned about multicollinearity as you would be in straight up OLS regression?
For example, with a logistic regression, where multicollinearity exists, would you need to be cautious (as you would in OLS regression) with taking inference from the Beta coefficients?
Fo... | Logistic Regression - Multicollinearity Concerns/Pitfalls | CC BY-SA 2.5 | null | 2010-11-23T23:50:10.213 | 2013-05-31T08:26:39.323 | 2010-11-24T15:01:27.013 | 919 | 776 | [
"regression",
"logistic",
"multicollinearity"
] |
4855 | 2 | null | 4839 | 6 | null | The first step is to define your research question.
A few possible research questions given your data include:
- How can genres of video games be grouped into a smaller set?
- How are genres of video games or groups of genres related to self-identifying as a hard-core gamer?
Then, you could present a table of frequ... | null | CC BY-SA 2.5 | null | 2010-11-24T01:22:36.983 | 2010-11-24T07:56:08.127 | 2010-11-24T07:56:08.127 | 930 | 183 | null |
4856 | 1 | 4857 | null | 3 | 2580 | How can I rewrite an AR(p) model in state-space form?
Max(p)=5 and I want to use Kalman Predictor.
| Rewriting AR model in State-Space form | CC BY-SA 2.5 | null | 2010-11-24T01:36:32.617 | 2010-11-24T12:49:59.810 | null | null | 1637 | [
"time-series",
"kalman-filter"
] |
4857 | 2 | null | 4856 | 4 | null | I suggest you buy the excellent book by G. Petris, S. Petrone and P. Campagnoli [Dynamic Linear Models with R](http://www.springer.com/statistics/statistical+theory+and+methods/book/978-0-387-77237-0).
You will learn that any ARMA model
$Y_t = \sum_{j=1}^{r}\phi_jY_{t-j} + \sum_{j=1}^{r-1}\psi_{j}\epsilon_t$
can be e... | null | CC BY-SA 2.5 | null | 2010-11-24T02:11:51.600 | 2010-11-24T10:57:39.583 | 2010-11-24T10:57:39.583 | 1709 | 1709 | null |
4858 | 1 | null | null | 10 | 8633 | I have been reading a good book called [Applied Longitudinal Data Analysis: Modeling Change and Event Occurrence](http://gseacademic.harvard.edu/alda/) by Judith Singer and John Willet. The book shows that by modeling in 2 levels, we can model the individual change in level 1 and in level 2 model for systematic interin... | How to test random effects in a multilevel model in R | CC BY-SA 3.0 | null | 2010-11-24T04:12:41.550 | 2013-07-01T02:02:33.493 | 2013-07-01T02:02:33.493 | 7290 | 1663 | [
"r",
"hypothesis-testing",
"mixed-model",
"multilevel-analysis",
"panel-data"
] |
4859 | 2 | null | 4830 | 2 | null | You might just try including all of the interaction effects. You can then use L1/L2-regularized logistic regression to minimize over-fitting and take advantage of any helpful features. I really like Hastie/Tibshirani's glmnet package (http://cran.r-project.org/web/packages/glmnet/index.html).
| null | CC BY-SA 2.5 | null | 2010-11-24T04:31:36.440 | 2010-11-24T04:31:36.440 | null | null | 2077 | null |
4860 | 2 | null | 3061 | 4 | null | This statistic measures a [kind of correlation](http://www.jstor.org/pss/2532051) between two sets of data. Its calculation requires no assumptions about what their scatterplot looks like (if that's what you mean by "tendency").
| null | CC BY-SA 2.5 | null | 2010-11-24T05:05:17.127 | 2010-11-24T05:05:17.127 | null | null | 919 | null |
4861 | 2 | null | 4854 | 21 | null | All of the same principles concerning multicollinearity apply to logistic regression as they do to OLS. The same diagnostics assessing multicollinearity can be used (e.g. VIF, condition number, auxiliary regressions.), and the same dimension reduction techniques can be used (such as combining variables via principal co... | null | CC BY-SA 2.5 | null | 2010-11-24T05:13:29.350 | 2010-11-24T13:14:11.800 | 2017-04-13T12:44:21.160 | -1 | 1036 | null |
4862 | 2 | null | 4858 | 3 | null | One of the best resources on multilevel analysis in R is John Fox's web appendix to the text "An R and S-PLUS Companion to Applied Regression". It provides a great overview of the method and means to calculate some of the more familiar measures from the R NLME output. The appendix is available on CRAN.
Here is the li... | null | CC BY-SA 2.5 | null | 2010-11-24T05:32:41.863 | 2010-11-24T05:32:41.863 | null | null | 485 | null |
4863 | 2 | null | 4810 | 2 | null | I guess it depends on what statistics or findings you are going to find out, research, study, or report. I'm assuming you will prob be using these graphs to represent findings for your university topic, right?
Like for example, if you want to present your finding about say, 'How long users stays on a a certain website... | null | CC BY-SA 2.5 | null | 2010-11-24T05:35:57.333 | 2010-11-24T05:35:57.333 | null | null | null | null |
4864 | 2 | null | 4830 | 2 | null | I know your question is about logistic regression and as it is a homework assignment so your approach may be constrained. However, if your interest is in interactions and accuracy of classification, it might be interesting to use something like CART to model this.
Here's some R code to do produce the basic tree. I've... | null | CC BY-SA 2.5 | null | 2010-11-24T05:41:16.503 | 2010-11-24T06:51:58.127 | 2010-11-24T06:51:58.127 | 485 | 485 | null |
4866 | 2 | null | 4858 | 5 | null | The `intervals()` function should provide you with $100(1-\alpha)$ confidence intervals for the random effects in your model, see `help(intervals.lme)` for more information.
You can also test if any of the variance components can be droped from the model by using `anova()` (which amounts to do an LRT between two nested... | null | CC BY-SA 2.5 | null | 2010-11-24T07:54:23.777 | 2010-11-24T07:54:23.777 | null | null | 930 | null |
4867 | 2 | null | 4839 | 3 | null | As an alternative to CA suggested by @Brandon, you could also try [Multiple Correspondence Analysis](http://en.wikipedia.org/wiki/Multiple_correspondence_analysis) which has the advantage of considering all types of games at the same time (unlike CA), which are probably scored as binary variable (in this case, the MCA ... | null | CC BY-SA 2.5 | null | 2010-11-24T08:13:19.087 | 2010-11-24T08:13:19.087 | null | null | 930 | null |
4868 | 1 | 4871 | null | 6 | 2250 | Is it possible to do k-fold cross-validation to test all data, rather than using kfcv to find the optimal hypothesis as is typically done.
Example:
Say I want to use a svms on a dataset of size 1000. Could I use 900 events to train the svm for the testing of the other 100 events. Then use a separate 900 events to train... | Using k-fold cross-validation to test all data | CC BY-SA 2.5 | null | 2010-11-24T08:47:26.607 | 2012-08-19T15:38:39.877 | 2010-11-24T12:47:15.397 | null | 2113 | [
"machine-learning",
"cross-validation"
] |
4869 | 2 | null | 4184 | 2 | null | The paper
[http://www.ricam.oeaw.ac.at/Groebner-Bases-Bibliography/gbbib_files/publication_582.pdf](http://www.ricam.oeaw.ac.at/Groebner-Bases-Bibliography/gbbib_files/publication_582.pdf)
by Pistone et al seems to be the source paper for this.
| null | CC BY-SA 2.5 | null | 2010-11-24T08:52:57.620 | 2010-11-24T08:52:57.620 | null | null | null | null |
4870 | 2 | null | 4858 | 19 | null | First point:
You need to be careful if you want to test whether the variance of a random effect is 0:
The standard $\chi^2$-asymptotics for the LR that are used in `anova()` do not apply for LRTs or restricted likelihood ratio tests on variance components since the null hypothesis is on the edge of the parameter space... | null | CC BY-SA 2.5 | null | 2010-11-24T09:28:58.507 | 2010-11-24T09:28:58.507 | null | null | 1979 | null |
4871 | 2 | null | 4868 | 6 | null | As far as I understand your question, it can be formulated this way:
Instead of calculating a quality measure for each of the k validation-folds and then calculate the average, may I aggregate all folds an then calculate my quality measure, hence getting only one instead of k values?
This question requires two perspect... | null | CC BY-SA 2.5 | null | 2010-11-24T10:43:38.363 | 2010-11-24T12:48:29.310 | 2010-11-24T12:48:29.310 | null | 264 | null |
4872 | 1 | null | null | 2 | 331 | what is the minimum entry degree to be employed as a statistician
| degree to become a statistician | CC BY-SA 2.5 | null | 2010-11-24T10:56:16.390 | 2010-11-24T10:56:16.390 | null | null | null | [
"careers"
] |
4873 | 2 | null | 4830 | 8 | null | In unbalanced datasets such as this, you can usually improve classification performance by moving away from using a fitted probability of .5 as your cutpoint for classifying cases into defaulters and non-defaulters. For example, I get correct classification rates of .88 and .58 with a cutpoint of .4 for a glm with all ... | null | CC BY-SA 2.5 | null | 2010-11-24T11:43:04.703 | 2010-11-24T11:43:04.703 | null | null | 1979 | null |
4874 | 2 | null | 4868 | 1 | null | Yes it is; and while this is a very reliable way of reporting error, I would say it is even encouraged.
| null | CC BY-SA 2.5 | null | 2010-11-24T12:43:13.730 | 2010-11-24T12:43:13.730 | null | null | null | null |
4875 | 2 | null | 4856 | 4 | null | Another good book that covers this is [Time Series Analysis by State Space Methods](http://rads.stackoverflow.com/amzn/click/0198523548) by Durbin and Koopman (pp 46-49.)
| null | CC BY-SA 2.5 | null | 2010-11-24T12:49:59.810 | 2010-11-24T12:49:59.810 | null | null | 439 | null |
4876 | 1 | null | null | 0 | 75 | I am trying to track the performance of a set of homogeneous subsystems vs the system as a whole. The performance metric is measured by a mean value based on a set of tasks performed by a the subsystem, for example, the mean wait time before scheduling task.
Each subsystem may be responsible for varying numbers of task... | Comparing and visualizing rates of subsystems | CC BY-SA 2.5 | null | 2010-11-24T13:20:08.147 | 2010-11-24T16:15:31.593 | null | null | 2122 | [
"data-visualization",
"mean"
] |
4877 | 1 | 4879 | null | 3 | 13448 | I have imported a datafile in R. It has different columns. There is one column that has the name of Operating System belonging to that row information. I wanted to get the percentage share of each unique OS (windows, linux, ios etc) from this column and plot it. I am very new to R and would like to know if there is any... | Creating plots for String type columns in R | CC BY-SA 2.5 | null | 2010-11-24T14:15:28.017 | 2010-11-24T20:49:04.323 | 2010-11-24T14:56:18.557 | 2101 | 2101 | [
"r",
"data-visualization"
] |
4878 | 1 | null | null | 2 | 4190 | We're using Excel Solver to model a DEA problem. We've worked with a few resources and feel confident that the model is returning correct results but I need some help interpreting them.
When solver completes its analysis, it provides an answer report that gives the slack found in some of the constraints -- which are b... | Slack values in Data Envelopment Analysis | CC BY-SA 2.5 | null | 2010-11-24T14:21:34.157 | 2010-11-24T15:44:21.667 | 2010-11-24T15:16:06.487 | null | 1256 | [
"econometrics",
"excel"
] |
4879 | 2 | null | 4877 | 2 | null | It seems the `barplot()` will be your friend in that case, e.g.
```
x <- sample(c("Win","Linux","Mac"), 100, replace=TRUE)
barplot(table(x))
```
This will work for variables of type `character` or `factor`. Another option is to use Cleveland's [dotplot](http://www.b-eye-network.com/view/2468), see `dotchart()` (or `do... | null | CC BY-SA 2.5 | null | 2010-11-24T14:29:50.333 | 2010-11-24T15:04:50.940 | 2010-11-24T15:04:50.940 | 930 | 930 | null |
4880 | 2 | null | 4877 | 4 | null | Simulating some data for a good start
```
x<-sample(c("Win","Linux","Mac"),721,replace=TRUE)
```
The good practice is to hold such categorical variables as factors; one can convert character vector to factor with `factor` function:
```
x<-factor(x)
```
Then, `plot` called on it will make a barplot with counts; to con... | null | CC BY-SA 2.5 | null | 2010-11-24T15:06:04.903 | 2010-11-24T20:49:04.323 | 2010-11-24T20:49:04.323 | 930 | null | null |
4881 | 2 | null | 4878 | 3 | null | As I understand data envelopment analysis, you optimize efficiency of the decision making unit (DMU) subject to a set of constraints on input availability and output requirements.
Since you have identified the maximum efficiency the non-binding input constraints indicate that the amount of slack in inputs is unnecessa... | null | CC BY-SA 2.5 | null | 2010-11-24T15:30:23.087 | 2010-11-24T15:30:23.087 | null | null | null | null |
4882 | 2 | null | 4878 | 3 | null | >
Is it correct to assert that those inputs found to be binding could not be further optimized but those that contained slack are where the inefficiencies are found?
First of all, this is a multivariate problem, so it is really hard to conclude anything about individual inputs/outputs. However I would rather say the... | null | CC BY-SA 2.5 | null | 2010-11-24T15:44:21.667 | 2010-11-24T15:44:21.667 | null | null | 279 | null |
4883 | 2 | null | 4876 | 2 | null | It sounds like you should lookup the topic of [statistical process control](http://en.wikipedia.org/wiki/Statistical_process_control). Its origins lie in manufacturing where deviations from expected process output are used to raise flags to ensure that quality of manufactured items is as per pre-established norms. In p... | null | CC BY-SA 2.5 | null | 2010-11-24T16:15:31.593 | 2010-11-24T16:15:31.593 | null | null | null | null |
4884 | 1 | 9742 | null | 5 | 2212 | Hello fellow number crunchers
I hope this a valid question for this forum. I am a lonesome quarter-statistician and have trouble finding someone to ask.
Introduction:
- Wikipedia Explanation of AB-Tests
- A ton of examples for typical use of AB-Tests
The AB-Test has become really popular since it is so easy to impl... | Aggregation-Level in AB-Tests | CC BY-SA 2.5 | null | 2010-11-24T16:23:43.947 | 2022-02-04T12:00:11.743 | 2011-05-09T17:54:04.670 | 1036 | 264 | [
"hypothesis-testing",
"aggregation",
"ab-test"
] |
4885 | 2 | null | 4810 | 19 | null | It's partly a matter of taste and convention, but theory, attention to your objectives, and a smidgen of cognitive neuroscience [see the references] can provide some guidance.
Because a pdf and a cdf convey the same information, the distinction between them arises from how they do it: a pdf represents probability with... | null | CC BY-SA 3.0 | null | 2010-11-24T16:29:10.620 | 2012-05-01T19:32:12.390 | 2012-05-01T19:32:12.390 | 919 | 919 | null |
4886 | 1 | 4888 | null | 3 | 292 | I have a data set that I need to analyze in R. A simplified version of it would be like this
```
SessionNo. Objects OtherColumns
A 2 .
A 3
B 4
C 1
D 2
D 1
D 2
D 3
E 5
```
here each ... | Getting aggregated share in R | CC BY-SA 2.5 | null | 2010-11-24T17:33:01.623 | 2010-11-25T13:46:52.060 | null | null | 2101 | [
"r",
"aggregation"
] |
4887 | 2 | null | 4886 | 2 | null | If I am reading your question correctly, something like the following should do it:
`aggregate(x$Objects,by=list(x$SessionNo.),sum)`
where `x` is the data frame containing your data. This will give you, for each unique session number, the sum of the object counts.
You can of course substitute other functions (including... | null | CC BY-SA 2.5 | null | 2010-11-24T17:45:13.697 | 2010-11-25T07:42:10.487 | 2010-11-25T07:42:10.487 | 439 | 439 | null |
4888 | 2 | null | 4886 | 3 | null | Perhaps this might help:
```
tapply(df$Objects, df$SessionNo., sum)
```
| null | CC BY-SA 2.5 | null | 2010-11-24T17:48:57.147 | 2010-11-25T10:31:03.683 | 2010-11-25T10:31:03.683 | 1050 | 1050 | null |
4889 | 2 | null | 4886 | 1 | null | I personally like using the plyr and or reshape packages for tasks like this. If you're just starting with R, I would highly recommend getting to know them well. They've solved nearly all of my data manipulation tasks.
```
ddply(df, .(sessionNo.) function(x) data.frame(
obj.count=sum(Objects)
))
```
OR, cast
```
colna... | null | CC BY-SA 2.5 | null | 2010-11-24T17:53:47.613 | 2010-11-25T02:08:57.187 | 2010-11-25T02:08:57.187 | 776 | 776 | null |
4890 | 2 | null | 4884 | 1 | null | The right level of aggregation depends on the time period over which you wish to generalize.
For example, you want to deploy A during nights across several sites but are unsure about its effectiveness relative to the existing option B. Thus, you may deploy A over a small number of sites and see its effects relative to... | null | CC BY-SA 2.5 | null | 2010-11-24T18:09:45.540 | 2010-11-24T18:09:45.540 | null | null | null | null |
4891 | 2 | null | 4810 | 8 | null | I agree with whuber's answer, but have one additional minor point:
The CDF has a simple non-parametric estimator that needs no choices to be made: the [empirical distribution function](http://en.wikipedia.org/wiki/Empirical_distribution_function). It's not quite so simple to estimate a PDF. If you use a histogram you n... | null | CC BY-SA 2.5 | null | 2010-11-24T18:41:20.210 | 2010-11-24T18:41:20.210 | null | null | 449 | null |
4892 | 1 | 5052 | null | 7 | 6672 | What is the [VC dimension](http://en.wikipedia.org/wiki/VC_dimension) of [SVM](http://en.wikipedia.org/wiki/Support_vector_machine) with the polynomial kernel $k(x,x')=(1+<x,x'>_{\mathbb{R^{2}}})^{2}$ for binary classification in $\mathbb{R^{2}}$?
It would be equal or more than v iff there exists a set of v points such... | VC dimension of SVM with polynomial kernel in $\mathbb{R^{2}}$ | CC BY-SA 2.5 | null | 2010-11-24T18:47:26.677 | 2017-05-23T17:29:05.427 | 2015-09-30T03:42:36.400 | 12359 | 1351 | [
"self-study",
"classification",
"svm",
"kernel-trick",
"vc-dimension"
] |
4893 | 2 | null | 4892 | 4 | null | If I'm not mistaken:
$[1 + \langle x,y\rangle]^2$ = $\langle[1, \sqrt{2}y_1, y_1^2, \sqrt{2}y_2, y_2^2, \sqrt{2}y_1y_2], [1, \sqrt{2}x_1, x_1^2, \sqrt{2}x_2, x_2^2, \sqrt{2}x_1x_2]^T\rangle$
So the separating rule is linear in $R^6$. Consequently, VC-dim is 7.
| null | CC BY-SA 3.0 | null | 2010-11-24T20:21:22.247 | 2017-05-23T17:29:05.427 | 2017-05-23T17:29:05.427 | 86522 | 1725 | null |
4894 | 2 | null | 4858 | 11 | null | (I would post this as a comment to the previous answer by 'chi', but don't see that that is possible here.)
Be very careful with `intervals()` and `anova()` from the `nlme` package, these have exactly the flaws that @fabians points out above -- they rely on the standard chi-squared distribution applied to a quadratic a... | null | CC BY-SA 2.5 | null | 2010-11-24T20:48:13.937 | 2010-11-24T20:48:13.937 | null | null | 2126 | null |
4895 | 2 | null | 118 | 30 | null | The answer that best satisfied me is that it falls out naturally from the generalization of a sample to n-dimensional euclidean space. It's certainly debatable whether that's something that should be done, but in any case:
Assume your $n$ measurements $X_i$ are each an axis in $\mathbb R^n$. Then your data $x_i$ define... | null | CC BY-SA 2.5 | null | 2010-11-24T20:49:39.210 | 2010-11-24T20:49:39.210 | null | null | 2456 | null |
4896 | 2 | null | 4886 | 2 | null | The function that is specifically designed for this task is ave(). By default it returned the mean within a group and returns a vector of the same length as the two input arguments. It is designed to fill in columns of either means or deviations from means. If this is in a dataframe with name "tst":
```
> tst$tmn <- wi... | null | CC BY-SA 2.5 | null | 2010-11-25T00:00:21.353 | 2010-11-25T00:00:21.353 | null | null | 2129 | null |
4897 | 2 | null | 4663 | 10 | null | LASSO is not an algorithm per se, but an operator.
There are many different ways to derive efficient algorithms for $\ell_1$ regularized problems. For instance, one can use quadratic programming to them tackle directly. I guess this is what you refer to as LASSO.
Another one is LARS, very popular because of its simpl... | null | CC BY-SA 2.5 | null | 2010-11-25T00:29:06.963 | 2010-11-25T00:29:06.963 | null | null | 358 | null |
4898 | 1 | 4917 | null | 12 | 4150 | If you choose to analyse a pre-post treatment-control design with a continuous dependent variable using a mixed ANOVA, there are various ways of quantifying the effect of being in the treatment group.
The interaction effect is one main option.
In general, I particularly like Cohen's d type measures (i.e., ${\frac{\mu_1... | Effect size for interaction effect in pre-post treatment-control design | CC BY-SA 2.5 | null | 2010-11-25T01:33:26.563 | 2011-04-19T02:15:25.713 | null | null | 183 | [
"anova",
"mixed-model",
"effect-size",
"cohens-d"
] |
4899 | 2 | null | 118 | 11 | null | Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:
$$d_n((X)_{i=1,\ldots,I},\mu)=\left(\sum | X-\mu|^n\right)^{1/n}$$
We usually use the natural euclidean distance ($n=2$), which is the one everybody uses in daily life.
The distance that you pro... | null | CC BY-SA 3.0 | null | 2010-11-25T03:01:40.620 | 2014-07-31T17:00:06.640 | 2014-07-31T17:00:06.640 | 5176 | 1709 | null |
4900 | 2 | null | 4898 | 3 | null | I believe that generalized eta-square ([Olejnik & Algena, 2003](http://www.ncbi.nlm.nih.gov/pubmed/14664681); [Bakeman, 2005](http://brm.psychonomic-journals.org/content/37/3/379)) provides a reasonable solution to the quantification of effect size that generalizes across between-Ss and within-Ss designs. If I read tho... | null | CC BY-SA 2.5 | null | 2010-11-25T03:49:17.473 | 2010-11-25T03:49:17.473 | null | null | 364 | null |
4901 | 1 | 4940 | null | 44 | 36762 | Other than literally testing each possible combination of variable(s) in a model (`x1:x2` or `x1*x2 ... xn-1 * xn`). How do you identify if an interaction SHOULD or COULD exist between your independent (hopefully) variables?
What are best practices in attempting to identify interactions?
Is there a graphical technique... | What are best practices in identifying interaction effects? | CC BY-SA 3.0 | null | 2010-11-25T05:32:53.903 | 2020-08-14T22:49:38.713 | 2011-09-27T05:28:07.813 | 183 | 776 | [
"regression",
"modeling",
"interaction"
] |
4902 | 2 | null | 4901 | 8 | null | How large is $n$ ? how many observations do you have ? this is crucial ...
Sobol indices will tell you the proportion of variance explained by interaction if you have a lot of observations and a few $n$, otherwise you will have to do modelling (linear to start with). You have a nice R package for that called sensitivi... | null | CC BY-SA 3.0 | null | 2010-11-25T07:07:59.057 | 2016-04-21T09:55:28.640 | 2016-04-21T09:55:28.640 | 17230 | 223 | null |
4903 | 2 | null | 4901 | 17 | null | My best practice would be to think about the problem to hand before fitting the model. What is a plausible model given the phenomenon you are studying? Fitting all possible combinations of variables and interactions sounds like data dredging to me.
| null | CC BY-SA 2.5 | null | 2010-11-25T07:50:59.160 | 2010-11-25T07:50:59.160 | null | null | 1390 | null |
4904 | 1 | 4905 | null | 6 | 1636 | I'm searching for a statistical method to determine if a player is cheating in an online game. The game is a Quake3 like game (ego-shooter).
Given a number of positive points and a number of negative points per player (score) and given n players (n<=64).
The score comes together like this (positive/negative seen from ... | Method to reliably determine abnormal statistical values | CC BY-SA 2.5 | null | 2010-11-25T11:16:41.540 | 2010-12-03T04:58:22.370 | 2010-11-25T11:21:15.593 | 2132 | 2132 | [
"outliers"
] |
4905 | 2 | null | 4904 | 11 | null | Your use of the stddev indicates you look at every variable seperately. If you look at them together, you might have more chance. An outlier in one dimension can be coincidence, an outlier in more dimensions is more surely an anomaly. I don't know much about games, but I reckon that you could use find extra variables l... | null | CC BY-SA 2.5 | null | 2010-11-25T12:01:24.780 | 2010-11-25T12:46:04.647 | 2010-11-25T12:46:04.647 | 1124 | 1124 | null |
4906 | 2 | null | 4886 | 1 | null | I often use `by`, which is a wrapper around `tapply`. Result is a list with a weee different print method compared to `tapply`.
```
df <- data.frame(SessionNo. = sample(x = c("A", "B", "C", "D", "E"),
size = 20, replace = TRUE,
prob = c(0.05, 0.1, 0.4, 0.4, 0.05)),
Objects = samp... | null | CC BY-SA 2.5 | null | 2010-11-25T13:46:52.060 | 2010-11-25T13:46:52.060 | null | null | 144 | null |
4907 | 2 | null | 4901 | 17 | null | Fitting a tree model (i.e. using R), will help you identify complex interactions between the explanatory variables. Read the example on page 30 [here](http://www.math.chs.nihon-u.ac.jp/%7Etanaka/files/kenkyuu/MultipleRegression.pdf).
| null | CC BY-SA 4.0 | null | 2010-11-25T13:51:42.120 | 2020-08-14T18:30:10.400 | 2020-08-14T18:30:10.400 | 93702 | 339 | null |
4908 | 1 | null | null | 5 | 272 | Lets imagine that we have some experiments. Each experiment may result in one of the outcomes: A, B, C. So we have probabilities distribution for each experiment $P_A, P_B, P_C$ which is context-dependent.
E.g.:
- $Context_1 \Rightarrow \{P_A^1, P_B^1, P_C^1\},$ experimental outcome is A
- $Context_2 \Rightarrow \{... | Chi-square analog for context-dependent distributions | CC BY-SA 2.5 | null | 2010-11-25T13:55:10.873 | 2010-11-28T23:09:43.057 | 2010-11-26T08:27:35.030 | 1314 | 1314 | [
"distributions",
"probability",
"chi-squared-test",
"predictor"
] |
4909 | 1 | null | null | 6 | 3776 | The empirical cumulative distribution function of a random variable, given observations $x_\left( k \right) > x_\left( k-1 \right)$, $k \in \mathbb N$, $k \le n$, is defined as $F_{emp}(x_\left( k \right) > X \ge x_\left( k-1 \right)) = \frac k {n+1}$ and $F_{emp}(X \ge x_\left(n\right))=1$.
Why? As long as we're inter... | Interpolating the empirical cumulative function | CC BY-SA 2.5 | null | 2010-11-25T14:13:31.967 | 2019-09-27T16:39:20.190 | 2017-09-28T18:27:46.007 | 60613 | 2456 | [
"estimation",
"random-variable",
"interpolation"
] |
4910 | 2 | null | 4909 | 4 | null | The EDF is the CDF of the population constituted by the data themselves. This is exactly what you need to describe and analyze any resampling process from the dataset, including nonparametric bootstrapping, jackknifing, cross-validation, etc. Not only that, it's perfectly general: any kind of interpolation would be i... | null | CC BY-SA 2.5 | null | 2010-11-25T14:34:31.573 | 2010-11-25T14:34:31.573 | null | null | 919 | null |
4911 | 2 | null | 4360 | 2 | null | For the Bayesian explanation, you need a prior probability distribution on the reported results by a lying coin flipper, as well as a prior probability of lying. When you see a value that is much more likely under the lying distribution than the random flips one, that makes your posterior probability of lying much high... | null | CC BY-SA 2.5 | null | 2010-11-25T15:18:15.367 | 2010-11-25T15:18:15.367 | null | null | 2134 | null |
4912 | 1 | 4948 | null | 4 | 11715 | Suppose my sample comprises 350 instances. Each instance has two independent binary (`pos` and `neg`) evaluations, A and B. If my alternative hypothesis is that A is more often positive than B, how would I express whether the difference is statistically significant?
---
To be more precise on the data: This question ... | Calculating significance of difference between two binary values on one dataset | CC BY-SA 2.5 | null | 2010-11-25T16:32:33.287 | 2010-11-28T19:19:27.373 | 2010-11-25T20:28:48.560 | 1785 | 1785 | [
"hypothesis-testing",
"statistical-significance",
"sample",
"binary-data"
] |
4913 | 2 | null | 4912 | 5 | null | Your addendum suggests that A and B are dependent samples since they come from the same "instance", i.e., patient. In that case, I propose the McNemar-Test which tests for a (two-sided) hypothesis of unequal proportions:
```
> N <- 350
> A <- factor(rbinom(N, size=1, p=0.6), labels=c("pos", "neg"))
> B <- fact... | null | CC BY-SA 2.5 | null | 2010-11-25T17:16:36.760 | 2010-11-25T20:10:00.573 | 2010-11-25T20:10:00.573 | 1909 | 1909 | null |
4914 | 1 | null | null | 5 | 528 | In a data set with thousands of data points, I am testing different short-term and longer term data outputs based on 5 rolling data points all the way to 100 rolling data points (which each value being a separate column in excel: 5, 6,..., 100).
The test I developed (with rudimentary knowledge of the whole thing) is t... | Developing a statistical test to ascertain better "fit" | CC BY-SA 2.5 | null | 2010-11-25T17:26:03.753 | 2010-11-27T18:30:58.310 | 2010-11-25T22:44:35.630 | null | 2137 | [
"time-series",
"standard-deviation",
"goodness-of-fit"
] |
4915 | 2 | null | 4914 | 3 | null | Based on the information given, I think that you could consider AIC, a measure of likelihood that is penalized by degrees of freedom.
| null | CC BY-SA 2.5 | null | 2010-11-25T19:49:13.047 | 2010-11-25T21:41:36.517 | 2010-11-25T21:41:36.517 | 930 | 1381 | null |
4916 | 2 | null | 4360 | 6 | null | I think your intuition is flawed. It seems you are implicitly comparing a single, "very special" result (exactly 10000 heads) with a set of many results (all "non-special" numbers of heads close to 10000). However, the definition of "special" is an arbitrary choice based on our psychology. How about binary 100000000000... | null | CC BY-SA 2.5 | null | 2010-11-25T20:33:30.693 | 2010-11-25T23:48:14.830 | 2010-11-25T23:48:14.830 | 1909 | 1909 | null |
4917 | 2 | null | 4898 | 7 | null | Yes, what you are suggesting is exactly what has been suggested in the literature. See, for example: Morris, S. B. (2008). Estimating effect sizes from pretest-posttest-control group designs. Organizational Research Methods, 11(2), 364-386 ([link](http://orm.sagepub.com/cgi/content/abstract/11/2/364), but unfortunately... | null | CC BY-SA 2.5 | null | 2010-11-25T23:37:14.657 | 2010-11-25T23:37:14.657 | null | null | 1934 | null |
4918 | 1 | null | null | -3 | 166 | The input is a bunch of numbers. How can I calculate a threshold that there are 10% of the totally amount of numbers are above this threshold. I think this is equally as to calculate the distribution function, but what I need is the threshold number, not the
| How to get the required data? | CC BY-SA 2.5 | 0 | 2010-11-26T00:46:30.300 | 2010-11-27T11:49:30.547 | 2010-11-27T11:49:30.547 | 930 | 2141 | [
"distributions"
] |
4919 | 2 | null | 4918 | 0 | null | My understanding of the question is that given a series of numbers, which is the number that has 10% of all the given numbers larger than this number and 90% smaller? If this understanding is correct, a software percentile function [(see here)](http://en.wikipedia.org/wiki/Percentile) or a decile function [(see here)](... | null | CC BY-SA 2.5 | null | 2010-11-26T00:51:35.250 | 2010-11-27T11:47:55.777 | 2010-11-27T11:47:55.777 | 930 | 226 | null |
4920 | 1 | 4926 | null | 21 | 33440 | Using Pearson's Correlation Coefficient, I have several variables that are highly correlated ($\rho = 0.978$ and $\rho = 0.989$ for 2 pairs of variables that are in my model).
The reason some of the variables are highly correlated is because one variable is used in the calculation for another variable.
Example:
$B = V ... | Can I simply remove one of two predictor variables that are highly linearly correlated? | CC BY-SA 2.5 | null | 2010-11-26T00:58:39.783 | 2014-04-16T22:07:23.160 | 2010-11-26T01:03:50.200 | 1894 | 1894 | [
"regression",
"correlation",
"modeling"
] |
4921 | 2 | null | 4920 | 0 | null | If D is not a constant, then B and E are effectively two different variables because of the variations in D. The high correlation indicates that D is practically constant throughout the training data. If that is the case, then you can discard either B or E.
| null | CC BY-SA 2.5 | null | 2010-11-26T01:17:49.297 | 2010-11-26T01:17:49.297 | null | null | 2071 | null |
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