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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
7240 | 1 | null | null | 21 | 23580 | I do not know if this has been asked before, but I do not found anything about it. My question is if anyone can provide a good reference to learn how to obtain the proportion of variance explained by each one of the fixed and random factors in a mixed-effects model.
| Proportion of explained variance in a mixed-effects model | CC BY-SA 2.5 | null | 2011-02-15T09:11:10.857 | 2022-04-03T03:14:27.767 | null | null | 221 | [
"mixed-model",
"variance"
] |
7241 | 2 | null | 7240 | 13 | null | I can provide some references:
Xu, R. (2003). Measuring explained variation in linear mixed effects models. Statistics in Medicine, 22, 3527-3541. [DOI:10.1002/sim.1572](http://dx.doi.org/10.1002/sim.1572)
Edwards, L. J., Muller, K. E., Wolfinger, R. D., Qaqish, B. F., & Schabenberger, O. (2008). An $R^2$ statistic for... | null | CC BY-SA 3.0 | null | 2011-02-15T10:27:16.157 | 2017-03-28T21:01:05.427 | 2017-03-28T21:01:05.427 | 1934 | 1934 | null |
7243 | 2 | null | 7233 | 5 | null | As @suncooolsu has already pointed out, we need more information. Residuals are often used when it comes to identifying outliers and to gain a better understanding of the appropriateness of a certain model (here: FEM vs REM). Given your "Std Residual", I would say that a REM seems more appropriate. However, you definit... | null | CC BY-SA 2.5 | null | 2011-02-15T13:01:56.817 | 2011-02-15T14:29:12.967 | 2011-02-15T14:29:12.967 | 449 | 307 | null |
7244 | 1 | 7245 | null | 6 | 5620 | Lets say I have a website which gets 100 hits per day (mu = 100). Yesterday my website got 130 hits (x = 130). If I assume a Poisson distribution, then the probability of getting 130 hits is:
```
> dpois(130, 100)
[1] 0.0005752527 # about 0.06%
```
So this tells me that getting 130 hits is quite unusual for my website... | Poisson distribution and statistical significance | CC BY-SA 2.5 | null | 2011-02-15T13:48:07.757 | 2011-02-15T15:35:20.957 | null | null | 3253 | [
"r",
"distributions",
"statistical-significance",
"poisson-distribution"
] |
7245 | 2 | null | 7244 | 8 | null | There are two points to make:
- It is not the specific value of 130 that is unusual, but that it is much larger than 100. If you got more than 130 hits, that would have been even more surprising. So we usually look at the P(X>=130), not just P(X=130). By your logic even 100 hits would be unusual, because dpois(100,100... | null | CC BY-SA 2.5 | null | 2011-02-15T14:15:43.147 | 2011-02-15T14:15:43.147 | null | null | 279 | null |
7246 | 1 | null | null | 7 | 884 | It seems non-convexity of loss function is not such a problem for [boosting with a normalized sigmoid loss function](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.19.2603&rep=rep1&type=pdf).
Do you know any further work showing better results with this kind of boosting than [Adaboost](http://en.wikipedia.org... | Normalized sigmoid loss function for boosting? | CC BY-SA 4.0 | null | 2011-02-15T14:22:27.763 | 2021-01-11T15:18:55.843 | 2021-01-11T15:18:55.843 | 22452 | 1351 | [
"boosting",
"loss-functions"
] |
7247 | 2 | null | 7244 | 3 | null | First, note that `dpois(130, 100)` will give you the probability of exactly 130 hits if you are assuming that the true rate is 100. That probability is indeed very low. However, in the usual hypothesis testing framework, what we calculate is the probability of the observed outcome or an even more extreme outcome. You c... | null | CC BY-SA 2.5 | null | 2011-02-15T14:24:58.077 | 2011-02-15T15:35:20.957 | 2011-02-15T15:35:20.957 | 1934 | 1934 | null |
7249 | 1 | 11361 | null | 7 | 1918 | I was wondering if there is a free tool to build a decision tree in interactive fashion like in SAS Enterprise Mining. I'm used to work with Weka. But nothing fits to my needs. I would like that before splitting every node, the program asks to user which attribute (maybe from a list of the "best" attributes) to choose.... | Interactive decision trees | CC BY-SA 2.5 | null | 2011-02-15T15:52:09.717 | 2013-08-11T21:16:09.673 | 2011-02-15T17:21:45.207 | null | 2719 | [
"sas",
"cart",
"weka"
] |
7250 | 1 | 7273 | null | 10 | 9158 | I'm having difficulty understanding one or two aspects of the cluster package. I'm following the example from [Quick-R](http://www.statmethods.net/advstats/cluster.html) closely, but don't understand one or two aspects of the analysis. I've included the code that I am using for this particular example.
```
## Libraries... | Using the stats package in R for kmeans clustering | CC BY-SA 2.5 | null | 2011-02-15T16:02:30.703 | 2016-02-16T08:46:41.250 | 2011-02-22T19:05:52.250 | 2635 | 2635 | [
"r",
"clustering"
] |
7251 | 1 | null | null | 6 | 1081 | Framework. Fix $\alpha\in ]0,1[$. Imagine you have $n$ $\alpha$-quantile forecast methodologies that give you, at time $t$ for look ahead time $t+h$, an estimation of the quantile of wind power. Formally, for $i=1,\dots,n$, you know how to produce $\hat{q}_{t+h|t}^{(i)}$ at time $t$ for look ahead time $t+h$ an estimat... | How to make a combination (aggregation) of quantile forecast? | CC BY-SA 2.5 | null | 2011-02-15T16:09:10.773 | 2017-05-11T17:39:13.173 | 2017-05-11T17:39:13.173 | 53690 | 223 | [
"time-series",
"forecasting",
"quantiles",
"aggregation",
"forecast-combination"
] |
7252 | 2 | null | 7236 | 0 | null | I think no extra package is needed for the task, just use the basic `sample` function, e.g.:
Get sample from the first group:
```
sample <- sample(data[data$"Care Type" == "Acute Care",], size = 25)
```
Get the choosen IDs out of the orig. dataset (making a backup could be a good idea before that):
```
data <- data[se... | null | CC BY-SA 2.5 | null | 2011-02-15T16:12:24.807 | 2011-02-15T16:12:24.807 | null | null | 2714 | null |
7253 | 2 | null | 7249 | 6 | null | Try the examples under dendrogram. You can make it as interactive as you want.
```
require(graphics); require(utils)
hc <- hclust(dist(USArrests), "ave")
(dend1 <- as.dendrogram(hc)) # "print()" method
str(dend1) # "str()" method
str(dend1, max = 2) # only the first two sub-levels
op <- par(mfrow= c(2,2), ... | null | CC BY-SA 3.0 | null | 2011-02-15T16:41:24.487 | 2011-05-31T19:42:30.857 | 2011-05-31T19:42:30.857 | 2775 | 2775 | null |
7255 | 2 | null | 4364 | 10 | null | This is not a characterization but a conjecture, which dates back from 1917 and is due to Cantelli:
>
If $f$ is a positive function on $\mathbb{R}$ and $X$ and $Y$ are $N(0,1)$ independent random variables such that $X+f(X)Y$ is normal, then $f$ is a constant almost everywhere.
Mentioned by Gérard Letac [here](http... | null | CC BY-SA 2.5 | null | 2011-02-15T17:43:50.737 | 2011-02-15T18:07:59.233 | 2017-04-13T12:58:32.177 | -1 | 2592 | null |
7256 | 1 | 7283 | null | 8 | 2746 | I am working on binary classification problem. Data set is very large and highly imbalanced.
Data dimensionality is also very high.
Now I want to balance data by under-sampling the majority class, and I also want to reduce data dimensionality by applying PCA, etc...
So my question is that which one should be applied f... | Which one should be applied first: data sampling or dimensionality reduction? | CC BY-SA 2.5 | null | 2011-02-15T18:20:56.520 | 2011-02-16T20:14:40.750 | 2011-02-15T19:03:27.827 | null | 2534 | [
"classification",
"sampling",
"dataset"
] |
7257 | 1 | null | null | 14 | 2049 | I'm trying to understand how Boltzmann machines work, but I'm not quite sure how weights are learned, and haven't been able to find a clear description. Is the following correct? (Also, pointers to any good Boltzmann machine explanations would also be great.)
We have a set of visible units (e.g., corresponding to black... | Learning weights in a Boltzmann machine | CC BY-SA 2.5 | null | 2011-02-15T18:39:29.390 | 2015-01-07T14:19:29.467 | 2011-02-15T20:35:20.667 | 1106 | 1106 | [
"neural-networks"
] |
7258 | 1 | null | null | 5 | 160 | As I have written in my question "[How much undersampling should be done?](https://stats.stackexchange.com/questions/7209/how-much-undersampling-should-be-done)", I want to predict defaults, where a default is per se really unlikely (average ~ 0.3 percent). My models are not affected by the unequal distribution: It's a... | Information content of examples and undersampling | CC BY-SA 2.5 | null | 2011-02-15T18:50:27.707 | 2011-03-10T09:26:59.513 | 2017-04-13T12:44:33.310 | -1 | 2549 | [
"sampling"
] |
7259 | 1 | 7309 | null | 4 | 3845 | I have a data-table that has about 26000 rows and about 35 columns. The columns are paired, so the values in columns 6 and 7 (for example) are related to each other, so are 8 and 9 and so on. There are 23 different types of annotations in the table, which I have read in as "factor". The ratio of these pairs of colum... | Combine multiple boxplots in a lattice | CC BY-SA 2.5 | null | 2011-02-15T18:54:10.333 | 2011-02-22T07:57:34.973 | 2011-02-22T07:57:34.973 | 2116 | 3263 | [
"r",
"data-visualization",
"boxplot"
] |
7260 | 2 | null | 7256 | 5 | null | Do the dimensionality reduction first: Your error in estimating the principal components will be smaller due to the larger sample (your Corr/Cov-matrix used in PCA has to be estimated!).
The other way around only makes sense for computational reasons.
| null | CC BY-SA 2.5 | null | 2011-02-15T18:55:09.617 | 2011-02-15T18:55:09.617 | null | null | 2549 | null |
7261 | 1 | 7262 | null | 26 | 29480 | My question relates mostly around the practical differences between
General Linear Modeling (GLM) and Generalized Linear Modelling (GZLM).
In my
case it would be a few continuous variables as covariates and a few
factors in an ANCOVA, versus GZLM. I want to examine the main effects
of each variable, as well as one thre... | General Linear Model vs. Generalized Linear Model (with an identity link function?) | CC BY-SA 4.0 | null | 2011-02-15T19:28:22.333 | 2020-06-28T15:49:34.997 | 2020-06-28T15:49:34.997 | 154402 | 3262 | [
"generalized-linear-model",
"modeling",
"linear-model"
] |
7262 | 2 | null | 7261 | 24 | null | A generalized linear model specifying an identity link function and a normal family distribution is exactly equivalent to a (general) linear model. If you're getting noticeably different results from each, you're doing something wrong.
Note that specifying an identity link is not the same thing as specifying a normal d... | null | CC BY-SA 2.5 | null | 2011-02-15T20:10:32.780 | 2011-02-18T10:23:28.347 | 2011-02-18T10:23:28.347 | 449 | 449 | null |
7263 | 1 | 7269 | null | 16 | 7284 | I am writing about using a 'joint probability distribution' for an audience that would be more likely to understand 'multivariate distribution' so I am considering using the later. However, I do not want to loose meaning while doing this.
[Wikipedia](http://en.wikipedia.org/wiki/Joint_probability_distribution) seems to... | Difference between the terms 'joint distribution' and 'multivariate distribution'? | CC BY-SA 2.5 | null | 2011-02-15T20:21:36.447 | 2017-12-01T13:01:50.277 | 2013-07-12T13:03:29.237 | 22468 | 1381 | [
"probability",
"terminology",
"joint-distribution",
"definition"
] |
7264 | 2 | null | 7239 | 4 | null | disclaimer I still don't fully understand your model; but without at least a reproducible example, this is the best I can offer. It is not clear exactly what you are doing here. For example, how are `pvr` and `pir` calculated? Would it make sense to calculate them inside the same model?
Answer
I am assuming that your d... | null | CC BY-SA 2.5 | null | 2011-02-15T20:30:35.940 | 2011-02-18T21:58:49.337 | 2011-02-18T21:58:49.337 | 1381 | 1381 | null |
7265 | 2 | null | 7263 | 2 | null | I'd be inclined to say that "multivariate" describes the random variable, i.e., it is a vector, and that the components of a multivariate random variable have a joint distribution. "Multivariate random variable" sounds a bit strange, though; I'd call it a random vector.
| null | CC BY-SA 2.5 | null | 2011-02-15T20:39:58.100 | 2011-02-15T20:39:58.100 | null | null | 401 | null |
7266 | 2 | null | 7263 | 0 | null | I think they are mostly synonyms, and that if there is any difference, it lies in details that are likely irrelevant to your audience.
| null | CC BY-SA 2.5 | null | 2011-02-15T20:45:55.713 | 2011-02-15T20:45:55.713 | null | null | 2044 | null |
7267 | 2 | null | 7263 | 1 | null | The [canonical textbooks describing properties of the various probability distributions by Johnson & Kotz](http://www.google.com/search?hl=en&tbs=bks%3A1&q=inauthor%3Ajohnson+inauthor%3Akotz+intitle%3Adistributions) and later co-authors are entitled Univariate Discrete Distributions, Continuous Univariate Distributions... | null | CC BY-SA 2.5 | null | 2011-02-15T20:56:41.460 | 2011-02-15T20:56:41.460 | null | null | 449 | null |
7268 | 1 | 7306 | null | 15 | 42334 | How would you get hourly means for multiple data columns, for a daily period, and show results for twelve "Hosts" in the same graph? That is, I'd like to graph what a 24 hour period looks like, for a weeks worth of data. The eventual goal would be to compare two sets of this data, before and after samplings.
```
... | How to aggregate by minute data for a week into hourly means? | CC BY-SA 4.0 | null | 2011-02-15T21:00:59.150 | 2020-05-01T09:42:41.837 | 2020-05-01T09:42:41.837 | 18417 | 2770 | [
"r",
"time-series",
"aggregation"
] |
7269 | 2 | null | 7263 | 14 | null | The terms are basically synonyms, but the usages are slightly different. Think about the univariate case: you may talk about "distributions" in general, you might more specifically refer to "univariate distributions", and you refer to "the distribution of $X$". You don't normally say "the univariate distribution of $X... | null | CC BY-SA 2.5 | null | 2011-02-15T21:05:16.413 | 2011-02-15T21:43:16.040 | 2011-02-15T21:43:16.040 | 89 | 89 | null |
7270 | 1 | 7274 | null | 9 | 3230 | I have a CSV file with 4 million edges of a directed network representing people communicating with each other (e.g. John sends a message to Mary, Mary sends a message to Ann, John sends another message to Mary, etc.). I would like to do two things:
- Find degree, betweeness and (maybe) eigenvector centrality measures... | How to calculate centrality measures in a 4 million edge network using R? | CC BY-SA 2.5 | null | 2011-02-15T22:11:35.660 | 2011-10-08T21:14:42.427 | 2011-02-16T16:25:07.737 | 1762 | 1762 | [
"r",
"data-visualization",
"networks"
] |
7271 | 1 | 7284 | null | 6 | 2840 | Hellwig's method is a method of selection of variables in linear model. It is widely used in Poland, probably only in Poland because it is really hard to find it in any scientific paper written in English.
Description of method:
$m_{k}$ - set of variables in k'th combination (there are $2^{p}-1$ combinations, where p... | Hellwig's method of selection of variables | CC BY-SA 2.5 | null | 2011-02-15T22:30:47.803 | 2011-02-16T08:49:24.693 | null | null | 1643 | [
"model-selection"
] |
7272 | 2 | null | 7256 | 1 | null | Devil's advocate: I could imagine the principal components differing depending on who's sampled. I'd think this validity issue would take precedence over the precision issue Richard points out.
| null | CC BY-SA 2.5 | null | 2011-02-15T23:43:27.807 | 2011-02-15T23:43:27.807 | null | null | 2669 | null |
7273 | 2 | null | 7250 | 8 | null | I did not grasp question 1 completely, but I'll attempt an answer. The plot of Q1 shows how the within sum of squares (wss) changes as cluster number changes. In this kind of plots you must look for the kinks in the graph, a kink at 5 indicates that it is a good idea to use 5 clusters.
WSS has a relationship with your... | null | CC BY-SA 2.5 | null | 2011-02-15T23:48:05.417 | 2011-02-15T23:48:05.417 | null | null | 2902 | null |
7274 | 2 | null | 7270 | 7 | null | What you have is an edge list, which can be converted to a network object using the network library. Here is an example using fictitious data.
```
library(network)
src <- c("A", "B", "C", "D", "E", "B", "A", "F")
dst <- c("B", "E", "A", "B", "B", "A", "F", "A")
edges <- cbind(src, dst)
Net <- as.network(edges, matrix... | null | CC BY-SA 2.5 | null | 2011-02-16T00:40:01.187 | 2011-02-16T00:40:01.187 | null | null | 3265 | null |
7276 | 2 | null | 4999 | 2 | null | I believe that in the specific case of L2 loss (ordinary linear regression), the convergence rate of coordinate descent will depend on the correlation structure of the predictors ($X_i$’s). Consider the case where they are uncorrelated. Then cyclic coordinate descent converges after one cycle.
Another heuristic that ... | null | CC BY-SA 2.5 | null | 2011-02-16T02:07:31.860 | 2011-02-16T02:07:31.860 | null | null | 1670 | null |
7277 | 2 | null | 7261 | 5 | null | I would like to include my experience in this discussion. I have seen that a generalized linear model (specifying an identity link function and a normal family distribution) is identical to a general linear model only when you use the maximum likelihood estimate as scale parameter method. Otherwise if "fixed value = 1"... | null | CC BY-SA 2.5 | null | 2011-02-16T03:22:22.480 | 2011-02-16T03:22:22.480 | null | null | null | null |
7278 | 1 | 7280 | null | 8 | 35909 | I'm tasked with deriving the MGF of a $\chi^2$ random variable.
I think the way to do is is by using the fact that $\Sigma_{j=1}^{m} Z^2_j$ is a $\chi^2$ R.V. and that MGF of a sum is the product of the MGFs of the individual terms. Although that may not be right and it may be $E(e^{tX})$ way.
I don't need it solved... | Finding the Moment Generating Function of chi-squared distribution | CC BY-SA 4.0 | null | 2011-02-16T04:02:25.177 | 2021-02-05T12:52:40.250 | 2021-02-05T12:52:40.250 | 11887 | 2387 | [
"mathematical-statistics",
"moments",
"moment-generating-function",
"chi-squared-distribution"
] |
7279 | 1 | null | null | 18 | 21889 | What is paired t-test, and under which circumstances should I use paired t-test?
Is there any difference between paired t-test and pairwise t-test?
| Is there any difference between the terms "paired t-test" and "pairwise t-test"? | CC BY-SA 3.0 | null | 2011-02-16T04:34:46.397 | 2017-06-29T04:48:37.377 | 2011-09-23T05:43:12.140 | 183 | 3269 | [
"hypothesis-testing",
"anova",
"t-test"
] |
7280 | 2 | null | 7278 | 9 | null | Yes, since $\chi^2$ is a sum of $Z_i^2$ the MGF is a product of individual summands. But then you need the MGF of $Z_i^2$ which is $\chi^2$ with 1 degree of freedom. The obvious way of calculating the MGF of $\chi^2$ is by integrating. It is not that hard:
$$Ee^{tX}=\frac{1}{2^{k/2}\Gamma(k/2)}\int_0^\infty x^{k/2-1}e^... | null | CC BY-SA 2.5 | null | 2011-02-16T05:46:50.920 | 2011-02-16T16:58:31.657 | 2011-02-16T16:58:31.657 | 279 | 2116 | null |
7281 | 2 | null | 7279 | 15 | null | Roughly, [paired t-test](http://en.wikipedia.org/wiki/Paired_difference_test) is a t-test in which each subject is compared with itself or, in [other words](http://mathworld.wolfram.com/Pairedt-Test.html), determines whether they differ from each other in a significant way under the assumptions that the paired differen... | null | CC BY-SA 2.5 | null | 2011-02-16T05:57:30.550 | 2011-02-16T05:57:30.550 | null | null | 1496 | null |
7282 | 2 | null | 7270 | 3 | null | I don't think that R is a first choice here (maybe I'm wrong). You will need huge arrays here to index and prepare your networks files in the appropriate data format. First of all, I will try to use Jure's (Rob mention him in the post above) [SNAP](http://goo.gl/L4jSO) library; it's written in C++ and works very well o... | null | CC BY-SA 2.5 | null | 2011-02-16T06:11:07.573 | 2011-02-16T06:11:07.573 | null | null | 609 | null |
7283 | 2 | null | 7256 | 4 | null | Generally, you want your training and validation data sets be separate as much as possible. Ideally, the validation set data would have been obtained only after the model has been trained. If you perform dimensionality reduction before splitting your data to separate sets, you break this isolation between the training ... | null | CC BY-SA 2.5 | null | 2011-02-16T06:21:29.233 | 2011-02-16T06:21:29.233 | null | null | 1496 | null |
7284 | 2 | null | 7271 | 5 | null | After spending too long on web research, I'm pretty sure the source of 'Hellwig's method' is:
Hellwig, Zdzisław. [On the optimal choice of predictors.](http://www.worldcat.org/title/on-the-optimal-choice-of-predictors/oclc/217223232) Study VI in Z. Gostkowski (ed.): Toward a system of quantitative indicators of compone... | null | CC BY-SA 2.5 | null | 2011-02-16T08:49:24.693 | 2011-02-16T08:49:24.693 | null | null | 449 | null |
7285 | 1 | 7303 | null | 10 | 447 |
### My Aim:
I'd like to a have a function that takes an email address and outputs a quasi-random number of 1, 2, 3, or 4.
### A little detail:
By quasi-random number I mean that given a typical population of email addresses, the probabilities of getting a value of 1, 2, 3, or 4 are roughly equal, and that obvious... | From an email address to a quasi-random number | CC BY-SA 2.5 | null | 2011-02-16T08:54:55.157 | 2018-02-15T19:26:23.593 | 2018-02-14T22:53:43.290 | 11887 | 183 | [
"algorithms",
"random-generation"
] |
7286 | 1 | 7294 | null | 15 | 3290 | Random walk Metropolis-Hasitings with symmetric proposal
$q(x|y)= g(|y-x|)$ has the property that the acceptance probability
$$P(accept\ y) = \min\{1, f(y)/f(x)\}$$
does not depend on proposal $g(\cdot)$.
Does that mean that I can change the $g(\cdot)$ as a function of previous performance of the chain, without affect... | Can I change the proposal distribution in random-walk MH MCMC without affecting Markovianity? | CC BY-SA 4.0 | null | 2011-02-16T10:20:43.147 | 2022-08-22T04:22:20.033 | 2022-06-30T23:49:55.560 | 79696 | 1542 | [
"markov-chain-montecarlo",
"metropolis-hastings"
] |
7287 | 2 | null | 7285 | 3 | null | Why not just have a look-up table of numbers for each possible character in an email. Then concatenate the numbers to form a seed. For example,
```
A 1
B 2
C 3
....
@ 27
....
```
So abc@ccc, would be converted to 12327333. This would give you a unique seed for each person. You would then use this to generate the 1, 2... | null | CC BY-SA 2.5 | null | 2011-02-16T10:59:12.323 | 2011-02-16T10:59:12.323 | null | null | 8 | null |
7288 | 2 | null | 7141 | 1 | null | Ok, I've thought of two possible ways to answer to this problem using Bayesian analysis. I will assume $\sigma$ to be known throughout this answer. First start with a "baby" case, where $n=2$ (or alternatively, only using the last two observations as a first approximation). You would usually start this by assuming a... | null | CC BY-SA 2.5 | null | 2011-02-16T12:31:45.973 | 2011-03-15T07:16:37.997 | 2011-03-15T07:16:37.997 | 2392 | 2392 | null |
7289 | 2 | null | 7223 | 2 | null | Regarding 1) Yes, you do lose this. See e.g. Harrell Regression Modeling Strategies, a book published by Wiley or a paper I presented with David Cassell called "Stopping Stepwise" available e.g. www.nesug.org/proceedings/nesug07/sa/sa07.pdf
| null | CC BY-SA 2.5 | null | 2011-02-16T13:18:20.643 | 2011-02-16T13:18:20.643 | null | null | 686 | null |
7290 | 2 | null | 7225 | 4 | null | My first reaction to Jelle's comments given is "bias-schmias". You have to be careful about what you mean by "large amount of predictors". This could be "large" with respect to:
- The number of data points ("big p small n")
- The amount of time you have to investigate the variables
- The computational cost of inve... | null | CC BY-SA 2.5 | null | 2011-02-16T14:10:31.810 | 2011-02-16T14:10:31.810 | null | null | 2392 | null |
7291 | 2 | null | 7285 | 0 | null | You could try converting each character to an ascii number, multiplying them all together to force overflow, and then performing a modulus operation on the least significant digits. If this is not pseudo-random enough, you can perform a bit-shift the numbers a bit...
-Ralph Winters
| null | CC BY-SA 2.5 | null | 2011-02-16T14:25:46.130 | 2011-02-16T14:25:46.130 | null | null | 3489 | null |
7292 | 1 | 7327 | null | 14 | 4297 | If you can measure a time series of observations at any level of precision in time, and your goal of the study is to identify a relationship between X and Y, is there any empirical justification for choosing a specific level of aggregation over another, or should the choice be simply taken based on theory and/or practi... | How do you choose a unit of analysis (level of aggregation) in a time series? | CC BY-SA 2.5 | null | 2011-02-16T14:47:53.653 | 2019-02-07T02:10:00.637 | 2019-02-07T02:10:00.637 | 11887 | 1036 | [
"time-series",
"aggregation",
"disaggregation"
] |
7293 | 1 | 7296 | null | 5 | 247 | The following problem comes from a max likelihood calculation for gaussian families, but is of independent interest.
Is it possible to find a closed-form approximation for small values of $x$ for
$\text{det}(B + xI)$
where I is the identity matrix and B is hermitian rank-deficient positive semidefinite?
| Determinant perturbation approximation | CC BY-SA 2.5 | null | 2011-02-16T14:57:58.397 | 2011-02-16T18:07:22.083 | 2011-02-16T18:07:22.083 | null | 30 | [
"maximum-likelihood",
"matrix"
] |
7294 | 2 | null | 7286 | 7 | null | I think that this [paper](https://projecteuclid.org/journals/bernoulli/volume-7/issue-2/An-adaptive-Metropolis-algorithm/bj/1080222083.full) from Heikki Haario et al. will give you the answer you need. The markovianity of the chain is affected by the adaptation of the proposal density, because then a new proposed value... | null | CC BY-SA 4.0 | null | 2011-02-16T15:15:42.040 | 2022-08-22T04:22:20.033 | 2022-08-22T04:22:20.033 | 79696 | 3108 | null |
7295 | 1 | 7299 | null | 15 | 20512 | Currently i am using RF toolbox on MATLAB for a binary classification Problem
Data Set: 50000 samples and more than 250 features
So what should be the number of trees and randomly selected feature on each split to grow the trees?
can any other parameter greatly affect the results?
| What should be the optimal parameters for Random Forest classifier? | CC BY-SA 2.5 | null | 2011-02-16T15:20:21.510 | 2017-05-31T08:44:39.347 | 2011-02-16T17:39:26.287 | null | 2534 | [
"machine-learning",
"classification",
"random-forest"
] |
7296 | 2 | null | 7293 | 7 | null | I'll assume that you already know the eigenvalues of $B$. Since $B$ is symmetric positive semidefinite, it can be decomposed as
$$
B = U D U^T
$$
where $U$ is an orthogonal matrix and $D$ is the diagonal of nonnegative eigenvalues (some of which may be exactly zero).
Now
$$
B+xI = U D U^T + x U U^T = U (D + x I) U^T... | null | CC BY-SA 2.5 | null | 2011-02-16T15:23:49.660 | 2011-02-16T15:23:49.660 | null | null | 2970 | null |
7297 | 1 | null | null | 4 | 1654 | I have a smooth but rather complex curve, sampled with a good frequency. I apply the discrete Fourier transform to it using the fast Fourier transform (FFT) algorithm and get its Fourier image. I need to find peaks on the resulting Fourier transform curve, but the image I get contains a substantial amount of noise, whi... | Noise in the Fourier transform | CC BY-SA 2.5 | null | 2011-02-16T15:58:10.347 | 2011-02-17T09:11:52.597 | 2011-02-16T22:57:59.120 | 3272 | 3272 | [
"data-transformation",
"fourier-transform"
] |
7298 | 2 | null | 7286 | 3 | null | You can improve the acceptance rate using delayed rejection as described in [Tierney, Mira (1999)](http://scholar.google.fr/scholar?cluster=8248856686379090018). It is based on a second proposal function and a second acceptance probability, which guarantees the Markov chain is still reversible with the same invariant d... | null | CC BY-SA 2.5 | null | 2011-02-16T16:18:35.263 | 2011-02-16T16:18:35.263 | null | null | 1351 | null |
7299 | 2 | null | 7295 | 8 | null | Pick a large number of trees, say 100. From what I have read on the Internet, pick $\sqrt{250}$ randomly selected features. However, in [the original paper](http://www.stat.berkeley.edu/~breiman/randomforest2001.pdf), Breiman used about the closest integer to $\frac{\log{M}}{\log{2}}$.
I would say cross-validation is u... | null | CC BY-SA 3.0 | null | 2011-02-16T16:26:18.673 | 2017-05-31T08:44:39.347 | 2017-05-31T08:44:39.347 | 1351 | 1351 | null |
7300 | 2 | null | 7270 | 3 | null | Gephi ( [http://gephi.org/](http://gephi.org/) ) might be an easy way to explore the data. You can almost certainly visualize it, and perform some calculations (though I have not used it for some time so I can't remember all the functions).
| null | CC BY-SA 2.5 | null | 2011-02-16T16:52:02.890 | 2011-02-16T16:52:02.890 | null | null | 2635 | null |
7301 | 2 | null | 7259 | 0 | null | Are you familiar with [ggplot2](http://had.co.nz/ggplot2)? I'm not certain I understand the question, but you can look at histograms colored by a certain paramater ([http://had.co.nz/ggplot2/geom_histogram.html](http://had.co.nz/ggplot2/geom_histogram.html)), and also it has a useful facetting function ([http://had.co.... | null | CC BY-SA 2.5 | null | 2011-02-16T16:55:04.833 | 2011-02-16T19:50:51.380 | 2011-02-16T19:50:51.380 | 2635 | 2635 | null |
7302 | 1 | null | null | 5 | 257 | suppose you might buy some software to be used in a ML/NLP/DM laboratory. What software package would you ask for? Let's say: MATLAB (with some toolboxes), SPSS, and what else?
I know that there is a lot of free software one can use like R, Weka, Rapidminer, python packages, and so on. However, the above question targe... | Software for ML/NLP/DM laboratory | CC BY-SA 2.5 | null | 2011-02-16T17:29:10.427 | 2012-06-05T17:17:26.633 | 2011-02-16T18:00:10.717 | null | 976 | [
"machine-learning",
"data-mining",
"software"
] |
7303 | 2 | null | 7285 | 10 | null | Look up hash functions, for example at [http://en.wikipedia.org/wiki/Hash_function](http://en.wikipedia.org/wiki/Hash_function)
| null | CC BY-SA 2.5 | null | 2011-02-16T17:31:19.580 | 2011-02-16T17:31:19.580 | null | null | 247 | null |
7304 | 2 | null | 7295 | 12 | null | Number of trees the bigger, the better. You almost can't overshoot with this parameter, but of course the upper limit depends on the computational time you want to spend on RF.
The good idea is to make a long forest first and then see (I hope it is available in MATLAB implementation) when the OOB accuracy converges.
Nu... | null | CC BY-SA 2.5 | null | 2011-02-16T17:39:07.300 | 2011-02-16T17:39:07.300 | null | null | null | null |
7305 | 2 | null | 7293 | 4 | null | I second @cardinal's answer, but provide a simple trick: If $p(z)$ is a polynomial (with integer powers), and $\mathbf{v}, \lambda$ are eigenvector and corresponding eigenvalue of matrix $M$, then $\mathbf{v}, p(\lambda)$ are eigenvector and corresponding eigenvalue of $p(M)$. The proof is a simple exercise. The polyno... | null | CC BY-SA 2.5 | null | 2011-02-16T18:01:53.177 | 2011-02-16T18:01:53.177 | 2017-04-13T12:44:39.283 | -1 | 795 | null |
7306 | 2 | null | 7268 | 14 | null | Here is one approach using cut() to create the appropriate hourly factors and ddply() from the plyr library for calculating the means.
```
library(lattice)
library(plyr)
## Create a record and some random data for every 5 seconds
## over two days for two hosts.
dates <- seq(as.POSIXct("2011-01-01 00:00:00", tz = "GMT... | null | CC BY-SA 2.5 | null | 2011-02-16T19:21:35.107 | 2011-02-16T19:21:35.107 | null | null | 3265 | null |
7307 | 1 | 114363 | null | 21 | 27201 | Can somebody explain me clear the mathematical logic that would link two statements (a) and (b) together? Let us have a set of values (some distribution). Now,
a) Median does not depend on every value [it just depends on one or two middle values];
b) Median is the locus of minimal sum-of-absolute-deviations from it.
An... | Mean and Median properties | CC BY-SA 3.0 | null | 2011-02-16T19:33:34.640 | 2022-01-25T15:20:16.673 | 2020-07-25T12:12:51.043 | 7290 | 3277 | [
"mean",
"median",
"robust",
"sensitivity-analysis",
"types-of-averages"
] |
7308 | 1 | 7629 | null | 18 | 3930 | Jeffrey Wooldridge in his Econometric Analysis of Cross Section and Panel Data (page 357) says that the empirical Hessian "is not guaranteed to be positive definite, or even positive semidefinite, for the particular sample we are working with.".
This seems wrong to me as (numerical problems apart) the Hessian must be p... | Can the empirical Hessian of an M-estimator be indefinite? | CC BY-SA 2.5 | null | 2011-02-16T19:52:33.287 | 2018-07-07T19:08:49.507 | 2011-02-28T04:09:22.650 | 1393 | 1393 | [
"estimation",
"maximum-likelihood",
"econometrics",
"asymptotics"
] |
7309 | 2 | null | 7259 | 4 | null | Sam,
I think I understood what you are after, so let me know if I've misinterpreted anything:
- You want a separate box_plot for the ratio of each pairs of columns. There are 15 ratios we are interested in...(column 6 / column 7, column 8 / column 9, etc.)
- This plot should have a separate "window" or facet for each... | null | CC BY-SA 2.5 | null | 2011-02-16T20:01:30.047 | 2011-02-16T22:11:00.050 | 2011-02-16T22:11:00.050 | 696 | 696 | null |
7311 | 2 | null | 7256 | 0 | null | You should perform sampling and dimensionality reduction in combination.
The best way to do this is undersample the majority class, and run a decision tree. It is the best variable selector you can imagine.
Perform this a number of times (each time another sample). The result will be a number of list of candidate pre... | null | CC BY-SA 2.5 | null | 2011-02-16T20:14:40.750 | 2011-02-16T20:14:40.750 | null | null | null | null |
7312 | 2 | null | 7307 | 3 | null |
- Roughly speaking, the median is the "middle value". Now, if you change the highest value (which is supposed to be positive here) from $x_{(n)}$ to $2 * x_{(n)}$, say, it does not change the median. But it does change the arithmetic mean. This shows, in simple terms, that the median does not depend on every value wh... | null | CC BY-SA 2.5 | null | 2011-02-16T20:19:38.270 | 2011-02-16T20:26:15.590 | 2011-02-16T20:26:15.590 | 3019 | 3019 | null |
7313 | 1 | 31835 | null | 4 | 1063 | Per my [earlier question](https://stats.stackexchange.com/q/7115/1026) I'm trying to find a reasonable metric for the semantic distance between two short text strings. One metric mentioned in the answers of that question was to use shortest hypernym path to create a metric for phrases. So for instance, if I was to fi... | Closest distance in hypernym tree as measure of semantic distance between phrases | CC BY-SA 2.5 | null | 2011-02-16T21:18:23.097 | 2012-07-07T10:51:14.693 | 2017-04-13T12:44:33.977 | -1 | 1026 | [
"text-mining",
"distance-functions"
] |
7314 | 1 | null | null | 2 | 1305 | Winbugs seem to support either stochastic or deterministic relationship between variables.
However, many Bayesian Networks represent relationships between variables using conditional probability tables. The "visit to Asia", "burglar alarm", "smoking & cancer" examples are classic introductory material.
However, condi... | How can I represent Conditional Probability Table based Bayesian Networks in Winbugs? | CC BY-SA 3.0 | null | 2011-02-16T22:30:17.503 | 2014-08-30T20:13:48.720 | 2014-08-30T20:13:48.720 | 3280 | 3280 | [
"bayesian",
"causality",
"bugs"
] |
7315 | 2 | null | 7307 | 12 | null | For the computation of the median, let $x_1,x_2,\ldots,x_n$ be the data. Assume, for simplicity, that $n$ is even, and the points are distinct! Let $y$ be some number. Let $f(y)$ be the 'sum-of-absolute deviations' of $y$ to the points $x_i$. This means that $f(y) = |x_1 - y| + |x_2 - y| + \ldots + |x_n - y|$. Your goa... | null | CC BY-SA 2.5 | null | 2011-02-16T23:13:35.653 | 2011-02-17T00:59:26.753 | 2011-02-17T00:59:26.753 | 795 | 795 | null |
7316 | 1 | 7317 | null | 27 | 36779 | I have data with many correlated features, and I want to start by reducing the features with a smooth basis function, before running an LDA. I'm trying to use natural cubic splines in the `splines` package with the `ns` function. How do I go about assigning the knots?
Here's the basic R code:
```
library(splines)
lda.... | Setting knots in natural cubic splines in R | CC BY-SA 3.0 | null | 2011-02-17T03:01:41.173 | 2018-05-28T01:54:47.840 | 2017-08-18T16:54:06.147 | 7290 | 988 | [
"r",
"splines"
] |
7317 | 2 | null | 7316 | 45 | null | How to specify the knots in R
The `ns` function generates a natural regression spline basis given an input vector. The knots can be specified either via a degrees-of-freedom argument `df` which takes an integer or via a knots argument `knots` which takes a vector giving the desired placement of the knots. Note that in ... | null | CC BY-SA 3.0 | null | 2011-02-17T04:00:30.850 | 2011-09-27T00:25:53.663 | 2011-09-27T00:25:53.663 | 2970 | 2970 | null |
7318 | 1 | null | null | 6 | 4606 | I have an experiment producing results (dependent variables) that don't pass tests of normality, thus I am testing hypotheses using non-parametric tests.
My DVs are continuous, while my factors (independent variables) are ordinal or nominal.
I've been using the Kruskal-Wallis test and Friedman test (using Matlab).
Most... | Which non-parametric test can I use to identify significant interactions of independent variables? | CC BY-SA 2.5 | null | 2011-02-17T04:34:41.540 | 2017-05-28T16:55:43.337 | null | null | 3285 | [
"hypothesis-testing",
"anova",
"statistical-significance",
"nonparametric",
"interaction"
] |
7319 | 1 | 7331 | null | 5 | 4049 | I am new to statistics, so pardon any mistakes in my question.
I have two time series $X_i$ and $Y_i$. Assuming that they're stationary AR(1) processes with possibly different means, how do I test for difference of means?
I found this link (but tests only one sample) which discusses using gls: [https://stat.ethz.ch/pip... | Two sample t-test for data (maybe time series) with autocorrelation? | CC BY-SA 2.5 | null | 2011-02-17T04:50:17.777 | 2011-08-16T12:18:34.573 | 2011-02-17T08:49:27.607 | null | null | [
"hypothesis-testing",
"mean",
"autocorrelation"
] |
7320 | 2 | null | 5430 | 1 | null | You could simply smooth the data and find the peaks.
Since there are presumably several pertinent, distinct (larger) objects amongst many irrelevant, indistinct (smaller) objects providing the noisy distance environment, you could probably assume that the distance distribution of pertinent objects is likely to be unif... | null | CC BY-SA 2.5 | null | 2011-02-17T04:51:40.427 | 2011-02-17T04:51:40.427 | null | null | 3285 | null |
7321 | 1 | null | null | 5 | 313 | Consider a single x axis representing points along a line in space. I've got a set of data (in this case the receptive fields of hippocampal place cells recorded as a rat runs along a linear track), which are themselves spatially extensive, ie. if displayed graphically against the axis they would look like a series of ... | Determining whether a group of spatially extensive data is centered around one point in space | CC BY-SA 3.0 | null | 2011-02-17T05:29:19.383 | 2011-07-25T10:29:10.680 | 2011-06-25T09:39:29.480 | null | null | [
"data-visualization",
"spatial",
"continuous-data"
] |
7322 | 1 | 7329 | null | 5 | 1369 | I am working with the dataset of some heights and weights at different ages. My professor wants me to plot the residuals from regression of `soma.WT9` against the residuals from regression of `HT9.WT9` (don't mind the notation it's just two columns where soma is being regressed on `WT9` and `HT9` being regressed on `W... | What does plotting residuals from one regression against the residuals from another regression give us? | CC BY-SA 2.5 | null | 2011-02-17T06:46:02.780 | 2011-04-29T00:51:09.930 | 2011-04-29T00:51:09.930 | 3911 | 3008 | [
"regression",
"residuals"
] |
7323 | 2 | null | 7322 | 6 | null | Judging by the details and variable names, soma.WT9 and HT9.WT9, you are obtaining the residuals by first regressing, soma on WT9 and HT9 on WT9 (right?). If I understood you correctly, the scatter plot between soma.WT9 and HT9.WT9 will tell you -- if after removing the effects of WT9 (possibly linear effects in your c... | null | CC BY-SA 2.5 | null | 2011-02-17T07:00:28.667 | 2011-02-17T07:00:28.667 | null | null | 1307 | null |
7324 | 2 | null | 7297 | 5 | null | This seems like $e^{-ax}\sin(bx)$ function -- FT of such are two Dirac deltas, so it is not surprising at all that they appear as a noisy peaks after DFT (this is a variation of ultraviolet crisis). So, well, don't worry -- you can do nothing wise about it, at least smooth the transform (for instance with moving mean) ... | null | CC BY-SA 2.5 | null | 2011-02-17T09:11:52.597 | 2011-02-17T09:11:52.597 | null | null | null | null |
7325 | 2 | null | 4451 | 1 | null | [This paper](http://arxiv.org/abs/1102.2166) uses a [facebook dataset](http://people.maths.ox.ac.uk/~porterm/data/facebook100.zip) that is available here. Here is the description from the authors:
>
The data includes the complete set of nodes and links (and some
demographic information) from 100 US colleges and univ... | null | CC BY-SA 2.5 | null | 2011-02-17T10:55:37.620 | 2011-02-17T10:55:37.620 | null | null | 3291 | null |
7326 | 1 | 7363 | null | 4 | 454 | I am new here, so I might have missed a similar question.
I am trying to do things good in my research and using proper statistical approaches, but in computer science, we had quite a weak training on statistical methods applied to our research.
So, what I am asking is probably trivial. Here is the problem:
We have a ... | Checking the distribution of classes in a sample of a dataset | CC BY-SA 2.5 | null | 2011-02-17T11:12:27.553 | 2011-02-18T06:07:14.057 | null | null | 3291 | [
"distributions",
"sampling",
"dataset"
] |
7327 | 2 | null | 7292 | 11 | null | Introduction
My interest in the topic is now about 7 years and resulted in PhD thesis [Time series: aggregation, disaggregation and long memory](http://uosis.mif.vu.lt/~celov/DC/dcelov%20thesis%2009_05.pdf), where attention was paid to a specific question of cross-sectional disaggregation problem for AR(1) scheme.
Da... | null | CC BY-SA 2.5 | null | 2011-02-17T11:23:36.643 | 2011-02-17T11:23:36.643 | null | null | 2645 | null |
7329 | 2 | null | 7322 | 4 | null | This sounds like what I call an "added variable" plot. The idea behind these is to provide a visual way of whether adding a variable to a model (ht9 in your case) is likely to add anything to the model (soma on wt9 in your case).
It was explained to me like this. When you fit a linear regression, the order of the var... | null | CC BY-SA 2.5 | null | 2011-02-17T13:03:10.287 | 2011-02-17T13:03:10.287 | null | null | 2392 | null |
7330 | 1 | 7332 | null | 4 | 389 | I have a question about a rotation matrix, which can be represented in 2 dimensions as:
$$R_{2}(\theta)=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$
For some arbitrary angle $\theta$. This can be extended to an arbitrary number of dimensions by adding an identity matrix:
$$R_{n}... | Rotation matrices and prior invariance for arbitrary dimensions | CC BY-SA 2.5 | null | 2011-02-17T13:55:51.813 | 2011-02-17T22:00:09.717 | 2011-02-17T19:33:00.870 | 2970 | 2392 | [
"prior",
"rotation"
] |
7331 | 2 | null | 7319 | 4 | null | Since $X_i$ and $Y_i$ are AR(1) processes, we can treat $(X_i,Y_i)$ as a VAR(1) process. Then testing the difference of means means testing the restrictions on VAR coefficients. For this simple case there is a [formula](http://books.google.com/books?id=B8_1UBmqVUoC&lpg=PP1&dq=hamilton%20time%20series%20analysis&hl=fr&p... | null | CC BY-SA 2.5 | null | 2011-02-17T14:33:45.927 | 2011-02-20T12:25:06.973 | 2011-02-20T12:25:06.973 | 2116 | 2116 | null |
7332 | 2 | null | 7330 | 7 | null | The answer, I believe, to your first question is "yes". This can be accomplished with [Givens rotations](http://en.wikipedia.org/wiki/Givens_rotation), which allow for the annihilation of arbitrary elements of a matrix via a $2\times 2$ rotation matrix. The implication is that if you start with a rotation matrix, then ... | null | CC BY-SA 2.5 | null | 2011-02-17T14:41:10.700 | 2011-02-17T19:44:10.390 | 2011-02-17T19:44:10.390 | 449 | 2970 | null |
7333 | 2 | null | 7084 | 3 | null | The best argument I have heard about using the normal distribution is one of "sufficient statistics" given by Larry Bretthorst in his PhD thesis on spectral estimation.
Basically because the normal distribution has a set of sufficient statistics (mean and covariance matrix), then this is the only thing that matters whe... | null | CC BY-SA 2.5 | null | 2011-02-17T15:47:56.670 | 2011-02-17T15:47:56.670 | null | null | 2392 | null |
7334 | 2 | null | 4978 | 3 | null | It is true that in the past, MaxEnt and Bayes have dealt with different types or forms of information. I would say that Bayes uses "hard" constraints as well though, the likelihood.
In any case, it is not an issue anymore as Bayes Rule (not the product rule) can be obtained from Maximum relative Entropy (MrE), and not ... | null | CC BY-SA 2.5 | null | 2011-02-17T16:03:19.943 | 2011-02-17T16:03:19.943 | null | null | null | null |
7335 | 2 | null | 7270 | 3 | null | From past experience with a network of 7 million nodes, I think visualizing your complete network will give you an uninterpretable image. I might suggest different visualizations using subsets of your data such as just using the top 10 nodes with the most inbound or outbound links. I second celenius's suggestion on usi... | null | CC BY-SA 2.5 | null | 2011-02-17T16:06:24.620 | 2011-02-17T16:06:24.620 | null | null | 3298 | null |
7336 | 1 | 7341 | null | 10 | 754 | I'm running a simulation on R and a cluster of computers and have the following problem. On each of X computers I run:
```
fxT2 <- function(i) runif(10)
nessay <- 100
c(mclapply(1:nessay, fxT2), recursive=TRUE)
```
There are 32 computers, each with 16 cores. However, around 2% of the random numbers are identical. What... | RNG, R, mclapply and cluster of computers | CC BY-SA 2.5 | null | 2011-02-17T16:18:29.617 | 2011-02-17T18:27:08.447 | 2011-02-17T17:52:48.100 | 8 | 603 | [
"r",
"random-generation",
"parallel-computing",
"multicore"
] |
7337 | 2 | null | 7336 | 3 | null | You need to use a RNG specifically designed for parallel computing. See the "Parallel computing: Random numbers" section of the [High Performance Computing Task View](http://cran.r-project.org/web/views/HighPerformanceComputing.html).
| null | CC BY-SA 2.5 | null | 2011-02-17T16:39:34.187 | 2011-02-17T16:39:34.187 | null | null | 1657 | null |
7338 | 2 | null | 7208 | 4 | null | The "chance correction" in Cohen's $\kappa$ estimates probabilities with which each rater chooses the existing categories. The estimation comes from the marginal frequencies of the categories. When you only have 1 judgement for each rater, this means that $\kappa$ assumes the category chosen for this single judgement i... | null | CC BY-SA 2.5 | null | 2011-02-17T17:14:48.070 | 2011-02-17T17:14:48.070 | null | null | 1909 | null |
7340 | 2 | null | 7326 | 0 | null | What you might want to try is to take many bootstrap samples of 500 pairs and then build a confidence interval of this distribution to see if the population mean is included.
However you really first need to go through your tags and spell correct, replace synonyms etc.
www.kamalnigam.com/papers/bootstrap-ijcaiws99.pdf... | null | CC BY-SA 2.5 | null | 2011-02-17T18:25:00.827 | 2011-02-17T18:25:00.827 | null | null | 3489 | null |
7341 | 2 | null | 7336 | 6 | null | The [snow](http://cran.r-project.org/package=snow) has explicit support to initialise the given number of RNG streams in a cluster computation.
It can employ one of two RNG implementations:
- rsprng and
- rlecuyer
Otherwise you have to do the coordination by hand.
| null | CC BY-SA 2.5 | null | 2011-02-17T18:27:08.447 | 2011-02-17T18:27:08.447 | null | null | 334 | null |
7342 | 2 | null | 7268 | 2 | null | You might checkout the `aggregate.zoo` function from the package `zoo`:
[http://cran.r-project.org/web/packages/zoo/zoo.pdf](http://cran.r-project.org/web/packages/zoo/zoo.pdf)
Charlie
| null | CC BY-SA 2.5 | null | 2011-02-17T20:19:02.183 | 2011-02-17T20:19:02.183 | null | null | 401 | null |
7343 | 1 | 7345 | null | 5 | 1519 | Suppose I have a function `f`, and I want to sample it at 100 points in the interval `[0, 100]`. For some reason (that seemed smart to me at the time), I decided to not sample at equidistant intervals, but rather use the following function to determine the sample points:
```
log2(x)*(100/log2(100))
```
This gives me a... | How to correct uneven sampling distribution when calculating the mean? | CC BY-SA 2.5 | null | 2011-02-17T20:44:05.473 | 2011-02-17T21:41:38.403 | null | null | 977 | [
"sampling",
"error",
"mean"
] |
7344 | 1 | 7369 | null | 6 | 16190 | I'm trying to write a function to graphically display predicted vs. actual relationships in a linear regression. What I have so far works well for linear models, but I'd like to extend it in a few ways.
- Handle glm models
- Deal with NAs in the predicted values
Does what I have so far seem like a good solution, o... | How to graphically compare predicted and actual values from multivariate regression in R? | CC BY-SA 2.5 | null | 2011-02-17T20:44:25.817 | 2011-02-18T10:04:53.950 | 2011-02-17T23:03:57.857 | 2817 | 2817 | [
"r",
"regression",
"multivariate-analysis",
"multiple-regression"
] |
7345 | 2 | null | 7343 | 7 | null | The answer depends on the characteristics of $f$. Regardless, its average (by definition) is $\frac{1}{100}\int_0^{100}{f(x)dx}$, so your problem is one of estimating that integral from values at a discrete set of points. For a highly discontinuous function you can use any [Riemann sum](http://en.wikipedia.org/wiki/R... | null | CC BY-SA 2.5 | null | 2011-02-17T21:41:38.403 | 2011-02-17T21:41:38.403 | null | null | 919 | null |
7346 | 6 | null | null | 0 | null | I did not participate to the beta, but I have been very happy to contribute (Q&A, edits, votes, etc.) during the last six months. Despite being also very happy with our actual moderators, I would like to propose myself for fulfilling this task if one of them feel the need to get some time off.
Needless to say, I will v... | null | CC BY-SA 2.5 | null | 2011-02-17T21:56:24.370 | 2011-02-17T21:56:24.370 | 2011-02-17T21:56:24.370 | 930 | 930 | null |
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