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12591 | 2 | null | 12588 | 5 | null | A few quick thoughts:
- The most important thing is that you state clearly what the error bar represents. Too often, it is left up to the reader to guess what the error bars might represent (e.g., confidence intervals or standard errors? between subjects or within subjects standard errors?).
- I think Estes' paper provides some good advice.
| null | CC BY-SA 3.0 | null | 2011-07-03T11:14:19.037 | 2011-07-03T11:14:19.037 | null | null | 183 | null |
12592 | 1 | 12593 | null | 4 | 8423 | If i have the relative proportion for two sample.
For example :
group 1:
- Male: 50%
- Female: 25%
- children: 25%
group 2:
- Male: 50%
- Female: 20%
- children: 30%
How can I measure the % of similarity between the two group?
| Comparing relative frequencies between two groups | CC BY-SA 3.0 | null | 2011-07-03T15:00:09.053 | 2011-07-03T16:03:44.080 | 2011-07-03T16:03:44.080 | 930 | 5244 | [
"independence",
"proportion"
] |
12593 | 2 | null | 12592 | 3 | null | You can use a Pearson's chi-squared test. Calculate the average number of people in each demographic across both groups (e.g., how many women were there overall)? Suppose that it's 23%. See how many women that would imply would be in group 1, 11 say. But you actually observed 13 (I'm assuming that there are 50 people in group 1 for concreteness). Take 13, subtract off 11, square this number (2 squared is 4), then divide by the number of women that you expected on average (11). Do this for group 2's women and add these fractions together. Then do the same procedure for men and children and add all these number together. Your result has a chi-squared distribution with 3 degrees of freedom. See [the Wikipedia article here](http://en.wikipedia.org/wiki/Pearson%27s_chi-square_test) for more information
| null | CC BY-SA 3.0 | null | 2011-07-03T15:11:01.873 | 2011-07-03T15:11:01.873 | null | null | 401 | null |
12594 | 1 | 12595 | null | 1 | 193 | I am not a statistics expert so please bear with me. It is about making predictions about how long something should take based on measurements of how long things took in the past. (Things like fixing software bugs)
Up until now I have been using a normal standard deviation to be able to state things like "the average bug will be fixed within 40 days, and 98% of bugs will be fixed within 115 days" since 115 is 3 standard deviations away from the mean. Then we can do things like establish SLAs (service level agreements) and promise that bugs will be fixed within 115 days or we pay a fee.
However I have been told that it is not appropriate to use standard deviation as lead times are not normally distributed, which makes sense, the normal distribution would enable us to say a small percentage of bugs will be fixed in -10 days for example. Lead times cannot be less than 0. I was told they might be Poisson distributed, which I am not sure about either, and that I could use Shewhart's method for calculating "the mean plus 2 sigma" and that would be a better value to use for lead times.
I haven't been able to find any information about this Shewhart's method. I did find something that said the standard deviation of a Poisson distribution was simply the square root of the sample size, but this doesn't make sense either, as I can imagine two different teams for example that both have 300 samples, and an average of say 40 days, but one team has 95% of lead times between 30 and 50, and none over 60, whereas the other team has 95% of lead times between 30 and 100, and some over 110. Simply using the square root of 300 will give the same standard deviation for both which does not seem to make sense.
Anyway I guess my questions are:
- What is Shewhart's method for working out something like "the mean plus 2 sigma", and
- If you don't know Shewhart's method, what would be a good formula to use to make statements like "x% of bugs are fixed within n days" based on measured lead time data. (Knowing that lead times cannot be less than 0, and are probably not normally distributed etc)
Thanks a lot, I am really out of my depth here.
| Shewhart's method for working out "sigma" of a sample of lead times | CC BY-SA 3.0 | null | 2011-07-03T15:24:49.777 | 2011-07-03T20:01:01.123 | 2011-07-03T19:44:28.940 | null | 4591 | [
"poisson-distribution"
] |
12595 | 2 | null | 12594 | 2 | null | Assuming you have an iid sample, you have a estimate of the distribution of lead times. Instead of using a normal approximation (what you were originally doing) or something similar to it, you could just used the empirical percentiles, which are consistent estimators of the true percentiles (by the law of large numbers), to make your claims. Specifically, if $X$ is the $p$-th percentile of your sample, then the claim that "$p$ proportion of bugs will be fixed within $X$ time" will be true with probability 1 as the sample size increases.
If you insist on using the poisson distribution, which specifies that if $X$ has a poisson distribution (say $X$ is your lead time in this case), then
$$
P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}
$$
where $\lambda$ is a parameter estimated from the data (as the sample mean happens to be the maximum likelihood estimator). So your guess as the mass function would be
$$
P(X=k) = \frac{40^k e^{-40}}{k!},
$$
since you said the sample mean was 40. You'd then sum over this mass function to calculate quantities like the probability of a lead time being greater or less than a particular value. I don't see the need for some fancy method.
Edit: As noted in the comments below, if your lead times are discrete then something like the geometric distribution would be a much more natural parametric choice for their distribution than poisson.
| null | CC BY-SA 3.0 | null | 2011-07-03T16:24:13.417 | 2011-07-03T20:01:01.123 | 2011-07-03T20:01:01.123 | 4856 | 4856 | null |
12596 | 2 | null | 6896 | 7 | null | you can calculate the covariance matrix for each set and then calculate the Hausdorff distance between the two set using the Mahalanobis distance.
The Mahalanobis distance is a useful way of determining similarity of an unknown sample set to a known one. It differs from Euclidean distance in that it takes into account the correlations of the data set and is scale-invariant.
| null | CC BY-SA 3.0 | null | 2011-07-03T17:33:12.993 | 2011-07-03T17:33:12.993 | null | null | 5244 | null |
12597 | 1 | 17833 | null | 4 | 3403 | This question is in some way similar to [this one](https://stats.stackexchange.com/questions/3589/correlation-between-two-variables-of-unequal-size), but about another nuance.
I have two time series (update: stationary - with both mean and variance equal over time) with missing values in one of them, such as
- 1,2,3
- 1,Absent,3
(in reality I have many more observations). I want to compute their correlation, so I will not use the 2nd time-point in these data. But should I use it for computing the mean of the first series?
I think that including this point in computing the mean will lead to a more precise estimate of the population mean for the first variable, but don't know whether it is correct though.
| Mean when computing correlation between samples of unequal size | CC BY-SA 3.0 | null | 2011-07-03T17:55:49.413 | 2011-11-02T16:46:10.450 | 2017-04-13T12:44:44.530 | -1 | 5271 | [
"time-series",
"correlation"
] |
12599 | 1 | 12603 | null | 6 | 1403 | What are some applications of Chinese restaurant processes?
I'm trying to learn a bit about non-parametric Bayesian methods, starting with Dirichlet processes and CRPs, but all the tutorials I've found are about theory, without describing any applications in depth.
Names of papers would be great. I'm not really looking for state-of-the-art applications, but just some "canonical" examples (say, in natural language processing) of why Dirichlet processes and Chinese restaurant processes are useful and why I should care.
| What are some applications of Chinese restaurant processes? | CC BY-SA 3.0 | null | 2011-07-03T20:42:33.997 | 2011-07-04T17:08:40.403 | 2011-07-04T06:41:58.703 | null | 1106 | [
"nonparametric-bayes"
] |
12600 | 2 | null | 11085 | 2 | null | What I know that people do (and I do it myself in some sort of way in GWAS studies) is that you combine all your permuted p values into a null distribution and then just see how many p values are above a threshold in your real experiment and your permuted null.
So your FDR would then be something like: number of null p values < x / number of real p values < x
| null | CC BY-SA 3.0 | null | 2011-07-03T22:46:20.757 | 2011-07-03T22:46:20.757 | null | null | 5275 | null |
12601 | 2 | null | 12588 | 1 | null | I must admit I'm unfamiliar with Field 2000 but I concur with Jeromy Anglim about Estes and that you must clearly label.
My recommendation is that you plot the effect and within S confidence interval around the effect. In your text include a standard error for the overall mean for meta-analysis purposes but downplay it.
Conveying between S calculated error bars of any kind with repeated measures is generally unwise because that's not what you tried to measure and often the estimates of the means across subjects are very variable with repeated measures because the N is low (you may have a high N but generally repeated measures experiments have low Ns).
| null | CC BY-SA 3.0 | null | 2011-07-04T02:41:08.897 | 2011-07-04T02:41:08.897 | null | null | 601 | null |
12602 | 1 | 27713 | null | 11 | 1249 | I typically deal with data where multiple individuals are each measured multiple times in each of 2 or more conditions. I have recently been playing with mixed effects modelling to evaluate evidence for differences between conditions, modelling `individual` as a random effect. To visualize uncertainty regarding the predictions from such modelling, I have been using bootstrapping, where on each iteration of the bootstrap both individuals and observations-within-conditions-within-individuals are sampled with replacement and a new mixed effect model is computed from which predictions are obtained. This works fine for data that assumes gaussian error, but when the data are binomial, the bootstrapping can take a very long time because each iteration must compute a relatively compute-intensive binomial mixed effects model.
A thought I had was that I could possibly use the residuals from the original model then use these residuals instead of the raw data in the bootstrapping, which would permit me to compute a gaussian mixed effect model on each iteration of the bootstrap. Adding the original predictions from the binomial model of the raw data to the bootstrapped predictions from residuals yields a 95% CI for the original predictions.
However, I recently [coded](http://gist.github.com/06b1daffa33e613f7597) a simple evaluation of this approach, modelling no difference between two conditions and computing the proportion of times a 95% confidence interval failed to include zero, and I found that the above residuals-based bootstrapping procedure yields rather strongly anti-conservative intervals (they exclude zero more than 5% of the time). Furthermore, I then coded (same link as previous) a similar evaluation of this approach as applied to data that is originally gaussian, and it obtained similarly (though not as extreme) anti-conservative CIs. Any idea why this might be?
| Why does bootstrapping the residuals from a mixed effects model yield anti-conservative confidence intervals? | CC BY-SA 3.0 | null | 2011-07-04T03:20:06.327 | 2012-08-14T10:51:39.140 | null | null | 364 | [
"confidence-interval",
"mixed-model",
"bootstrap",
"monte-carlo",
"simulation"
] |
12603 | 2 | null | 12599 | 6 | null | Here are some important papers that have applied nonparametric Bayes to topic modeling for example:
- Blei et al., Hierarchical Topic Models and the Nested Chinese Restaurant Process, NIPS 2003.
- Teh et al., Hierarchical Dirichlet Processes, JASA 2006.
You can find [tons](http://www.google.com/search?sourceid=chrome&ie=UTF-8&q=site%3aaclweb.org%20chinese%20restaurant%20process) of [other papers](http://www.google.com/search?sourceid=chrome&ie=UTF-8&q=site%3aaclweb.org%20dirichlet%20process) at aclweb.org.
| null | CC BY-SA 3.0 | null | 2011-07-04T05:15:57.587 | 2011-07-04T05:15:57.587 | null | null | 5179 | null |
12605 | 1 | 12609 | null | 44 | 62949 | I've been playing around with random forests for regression and am having difficulty working out exactly what the two measures of importance mean, and how they should be interpreted.
The `importance()` function gives two values for each variable: `%IncMSE` and `IncNodePurity`.
Is there simple interpretations for these 2 values?
For `IncNodePurity` in particular, is this simply the amount the RSS increase following the removal of that variable?
| Measures of variable importance in random forests | CC BY-SA 3.0 | null | 2011-07-04T08:25:44.600 | 2018-01-04T17:42:13.930 | 2018-01-04T17:42:13.930 | 128677 | 845 | [
"r",
"machine-learning",
"random-forest",
"importance"
] |
12606 | 1 | 12607 | null | 3 | 226 | As stated in the question, what sort of modeling technique would be most appropriate?
| What if I know for sure that my target variable is not normally but Beta distributed? | CC BY-SA 4.0 | null | 2011-07-04T08:46:38.130 | 2019-03-05T00:13:24.373 | 2019-03-05T00:13:24.373 | 11887 | 333 | [
"modeling",
"beta-distribution"
] |
12607 | 2 | null | 12606 | 5 | null | [beta regression](http://www.jstatsoft.org/v34/i02/paper): see Fransisco Cribari-Neto and Achim Zeileis, "Beta Regression in R", Journal of Statistical Software, volume 34, issue 2, April 2010. If you have linear regression problem with a Beta noise process, then this is the neat solution.
| null | CC BY-SA 3.0 | null | 2011-07-04T08:57:01.777 | 2011-07-04T09:43:34.383 | 2011-07-04T09:43:34.383 | 887 | 887 | null |
12608 | 2 | null | 12606 | 4 | null | In addition to the suggestions, you could also try the logit transformation:
$$\log \frac{p}{1-p}$$
where $p$ is a Beta random variable.
You can now fit a linear regression model. The estimates would be ok, the only thing that might be a problem are inferences regarding the regression coefficients.
In order to avoid this problem, I would choose Bayesian approaches that impose a diffuse t-prior on the regression coefficients. If you don't have a huge data set, you could try either the [bayesglm](http://rss.acs.unt.edu/Rdoc/library/arm/html/bayesglm.html) function in the arm package or fit a hierarchical model using JAGS and rjags package in R. Otherwise if you have a huge data set, normal Gaussian regression should work fine.
However, I must remark that the choice depends on the specific data that you have and the interpretation of the parameter estimates.
| null | CC BY-SA 3.0 | null | 2011-07-04T09:17:02.310 | 2011-07-04T09:43:00.073 | 2011-07-04T09:43:00.073 | 1307 | 1307 | null |
12609 | 2 | null | 12605 | 47 | null | The first one can be 'interpreted' as follows: if a predictor is important in your current model, then assigning other values for that predictor randomly but 'realistically' (i.e.: permuting this predictor's values over your dataset), should have a negative influence on prediction, i.e.: using the same model to predict from data that is the same except for the one variable, should give worse predictions.
So, you take a predictive measure (MSE) with the original dataset and then with the 'permuted' dataset, and you compare them somehow. One way, particularly since we expect the original MSE to always be smaller, the difference can be taken. Finally, for making the values comparable over variables, these are scaled.
For the second one: at each split, you can calculate how much this split reduces node impurity (for regression trees, indeed, the difference between RSS before and after the split). This is summed over all splits for that variable, over all trees.
Note: a good read is [Elements of Statistical Learning](http://www-stat.stanford.edu/~tibs/ElemStatLearn/) by Hastie, Tibshirani and Friedman...
| null | CC BY-SA 3.0 | null | 2011-07-04T09:17:37.780 | 2011-07-04T12:19:55.710 | 2011-07-04T12:19:55.710 | 930 | 4257 | null |
12612 | 1 | null | null | 9 | 5202 | I have almost the same questions like this:
[How can I efficiently model the sum of Bernoulli random variables?](https://stats.stackexchange.com/questions/5347/how-can-i-efficiently-model-the-sum-of-bernoulli-random-variables)
But the setting is quite different:
- $S=\sum_{i=1,N}{X_i}$, $P(X_{i}=1)=p_i$, $N$~20, $p_i$~0.1
- We have the data for the outcomes of Bernoulli random variables: $X_{i,j}$ , $S_j=\sum_{i=1,N}{X_{i,j}}$
- If we estimate the $p_i$ with maximum likelihood estimation (and get $\hat p^{MLE}_i$), it turns out that $\hat P\{S=3\} (\hat p^{MLE}_i)$ is much larger then expected by the other criteria: $\hat P\{S=3\} (\hat p^{MLE}_i) - \hat P^{expected} \{S=3\}\approx 0.05$
- So, $X_{i}$ and $X_{j}$ $(j>k)$ cannot be treated as independent (they have
small dependence).
- There are some constrains like these: $p_{i+1} \ge p_i$ and $\sum_{s \le 2}\hat P\{S=s\}=A$ (known), which should help with the estimation of $P\{S\}$.
How could we try to model the sum of Bernoulli random variables in this case?
What literature could be useful to solve the task?
UPDATED
There are some further ideas:
(1) It's possible to assume that the unknown dependence between ${X_i}$ begins after 1 or more successes in series. So when $\sum_{i=1,K}{X_i} > 0$, $p_{K+1} \to p'_{K+1}$ and $p'_{K+1} < p_{K+1}$.
(2) In order to use MLE we need the least questionable model. Here is an variant:
$P\{X_1,...,X_k\}= (1-p_1) ... (1-p_k)$ if $\sum_{i=1,k}{X_i} = 0$ for any k
$P\{X_1,...,X_k,X_{k+1},...,X_N\}= (1-p_1) ... p_k P'\{X_{k+1},...,X_N\}$ if $\sum_{i=1,k-1}{X_i} = 0$ and $X_k = 1$, and $P'\{X_{k+1}=1,X_{k+2}=1,...,X_N=1\} \le p_{k+1} p_{k+2} ... p_N$ for any k.
(3) Since we interested only in $P\{S\}$ we can set $P'\{X_{k+1},...,X_N\} \approx P''\{\sum_{i=1,k}{X_i}=s' ; N-(k+1)+1=l\}$ (the probability of $\sum_{i=k+1,N}{X_i}$ successes for N-(k+1)+1 summands from the tail). And use parametrization $P''\{\sum_{i=k,N}{X_i}=s' ; N-k+1=l\}= p_{s',l}$
(4) Use MLE for model based on parameters $p_1,...,p_N$ and $p_{0,1}, p_{1,1}; p_{0,2}, p_{1,2}, p_{2,2};...$ with $p_{s',l}=0$ for $s' \ge 6$ (and any $l$) and some other native constrains.
Is everything ok with this plan?
UPDATED 2
Some examples of empirical distribution $P\{S\}$ (red) compared with Poisson distribution (blue) (the poisson means are 2.22 and 2.45, sample sizes are 332 and 259):
 
For samples (A1, A2) with the poisson means 2.28 and 2.51 (sample sizes are 303 and 249):
 
For joined samlpe A1 + A2 (the sample size is 552):
Looks like some correction to Poisson should be the best model :).
| How to model the sum of Bernoulli random variables for dependent data? | CC BY-SA 3.0 | null | 2011-07-04T10:30:39.660 | 2011-12-05T12:40:59.293 | 2017-04-13T12:44:25.243 | -1 | 3670 | [
"distributions",
"modeling",
"binomial-distribution",
"random-variable",
"non-independent"
] |
12613 | 2 | null | 12580 | 2 | null | No idea what you mean by bullet 1. Perhaps an explanation of what the code is doing will help?
You have a data frame (`mtcars`), which you standardise (using the `scale()` function) so that the variables have zero mean and unit variance. This has been performed because the data are measured in different units and you ideally want each variable to contribute equally to the dissimilarity measurements used to form the cluster analysis. Finally, this scaled data frame is converted to a matrix and transposed.
Because you transposed the matrix, the first clustering groups the variables in the original (`mpg`, `cyl` etc) into groups that have similar profiles over the different cars. The ordering implied by this clustering is recorded.
The procedure is repeated on the transpose of `x` so now the clustering is in the more familiar way, finding groups of cars that are similar in terms of the design and performance. Again, the ordering of the data implied by this second clustering is recorded.
Next comes the `levelplot` code. This produces a heatmap of the standardised data `x`, but we reorder the matrix using the two orderings generated by the cluster analyses earlier. The colours on the levelplot represent the actual data in each cell of matrix `x`; you have in fact just plotted the matrix but instead of showing the actual data, the individual data points are shown by colours.
That should now explain what the colour scale is; it is showing the mapping from the actual data (the standardised data, zero mean, unit variance) to colours used on the plot. Pink colours represent lower than average values and blues the opposite. By reordering the data plotted, we emphasise the group structure in the data. The clear pattern is one of two groups of cars, one that is fuel efficient (above average `mpg` etc and lower `cyl`) shown in the lower rows of the levelplot. The upper rows are the sportier, less efficient cars with larger engines and higher fuel consumption.
To the best of my knowledge, such a plot is not possible with ggplot2.
| null | CC BY-SA 3.0 | null | 2011-07-04T10:32:48.560 | 2011-07-04T10:32:48.560 | null | null | 1390 | null |
12615 | 1 | 12628 | null | 3 | 2215 | I have to plot a graph with log distribution on y axis. The values are: 10^-3..10^3.
What software do you suggest me to use.
My OS is Ubuntu, so I prefer software for Linux.
Thanks.
| Software to plot a log graph | CC BY-SA 3.0 | null | 2011-07-04T11:38:12.637 | 2015-10-30T20:32:35.183 | null | null | 5279 | [
"data-visualization",
"logarithm"
] |
12616 | 2 | null | 12615 | 9 | null | R is good and can be freely downloaded from [http://www.r-project.org/](http://www.r-project.org/)
R takes some time to get used to, but here's a simple example ("#" indicates that a comment follows):
```
x <- rnorm(20) # generate a sample of size 20 from N(0,1)
y <- 10^x # define y_i = 10^(x_i) for each i=1,...,20
plot(x, y) # plot y vs x
plot(x, y, log="y") # plot y vs x with log scale for y
```
Edited after comment from @chl
| null | CC BY-SA 3.0 | null | 2011-07-04T11:55:23.640 | 2011-07-04T12:26:11.963 | 2011-07-04T12:26:11.963 | 3835 | 3835 | null |
12617 | 2 | null | 10464 | 1 | null | You can incorporate this in the Bayesian framework by specifying a prior distribution for the age variable. And for the posterior, you have:
$$p(\theta|DI)\propto p(\theta|I)p(D|\theta I)$$
Now you simply take $D\equiv (18+)$ for example. This is no more difficult "in-principle" compared to when you actually do know the ages. The difference is that your likelihood function must be a cumulative distribution function instead of a density. As an example, suppose age is the only regressor you have (denoted $x_i$), and you are fitting a OLS model. This is for my benefit - but the generalisation is just details, rather than conceptual. If you have observed the ages exactly the likelihood function is:
$$p(y_1\dots y_N|x_1\dots x_N\alpha\beta\sigma I)=(2\pi\sigma^2)^{-\frac{N}{2}}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^{N}(y_i-\alpha-\beta x_i)^2\right)$$
But now suppose that the $(N+1)$th observation, you only observe that $L<x_{N+1}<U$. Lets call this piece of information $Z$. Now we can use the brilliant trick of introducing a nuisance parameter and then integrating it out again (via the sum rule). The nuisance parameter we introduce is $x_{N+1}$ (the actual unobserved age), and we have:
$$p(y_1\dots y_N y_{N+1}|x_1\dots x_N Z\alpha\beta\sigma I)=\int_{L}^{U} p(y_1\dots y_N y_{N+1}x_{N+1}|x_1\dots x_N Z\alpha\beta\sigma I)dx_{N+1}$$
Now we can split the integrand by using the product rule $P(AB|C)=P(A|C)P(B|AC)$ and we get:
$$p(x_{N+1}|x_1\dots x_N Z\alpha\beta\sigma I)p(y_1\dots y_N y_{N+1}|x_1\dots x_N x_{N+1}Z\alpha\beta\sigma I)$$
Note that in the second density, the information $Z\equiv L<x_{N+1}<U$ is redundant because we are already conditioning on the true value $x_{N+1}$. So we can remove it. Note that this second term could be called the "clean" data. The first term is basically a statement of how likely the unobserved age is given $L<x_{N+1}<U$, in addition to the position of the "true line" $(\alpha,\beta)$, the noise level $\sigma$, and the values of all other ages $(x_1\dots x_N)$. And so you have an integrated likelihood (sometimes called quasi-likelihood):
$$p(Y|XZ\alpha\beta\sigma I)=(2\pi\sigma^2)^{-\frac{N+1}{2}}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^{N}(y_i-\alpha-\beta x_i)^2\right)$$
$$\times\int_{L}^{U}p(x_{N+1}|X\alpha\beta\sigma I)\exp\left(-\frac{(y_{N+1}-\alpha-\beta x_{N+1})^2}{2\sigma^2}\right)dx_{N+1}$$
Now for every "messy" data, you will have a similar integral. You can take the above integral as multi-dimensional (with appropriate matrix sum of squares in the exponential).
I have heard something like this called the "Missing information Principle". You basically create a "nice" dataset from your "messy" one (i.e. the data set you wish you had), and then average out the "nice" inferences. You give more weight to certain nice data sets according to what your "messy" information is.
| null | CC BY-SA 3.0 | null | 2011-07-04T12:45:42.313 | 2011-07-04T12:45:42.313 | null | null | 2392 | null |
12619 | 2 | null | 12615 | 9 | null | You could also use the open source plotting package [Gnuplot](http://www.gnuplot.info/) for this task. The first section of a very readable tutorial [here](http://t16web.lanl.gov/Kawano/gnuplot/plot6-e.html) shows how to plot with a log scale.
| null | CC BY-SA 3.0 | null | 2011-07-04T13:20:03.970 | 2011-07-04T13:20:03.970 | null | null | 226 | null |
12620 | 2 | null | 12612 | 1 | null | If the dependence is due to clumping, a compound Poisson model could be the solution as a model of $S_j$. A somewhat random reference is [this one](http://www.google.dk/url?sa=t&source=web&cd=4&ved=0CD0QFjAD&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.90.5908%26rep%3Drep1%26type%3Dpdf&rct=j&q=compound%20poisson%20approximations&ei=ArsRTu_3E43oOZOGjbwL&usg=AFQjCNEAARi2Knn9RHMLtsbBq73saa91OQ&sig2=NnyzJhj3ZS9o0GZAdQyDHA&cad=rja) by Barbour and Chryssaphinou.
In a completely different direction, since you indicate that $N$ is 20, and thus relatively small, could be to build a graphical model of the $X_{ij}$'s, but I don't know if your setup and data make it possible. As @chl comments, it will be useful if you describe what the $X_{i,j}$'s are.
If the $X_{i,j}$'s represent sequential measurements, e.g. over time, and the dependence is related to this, a third possibility - and to some extend a compromise between the two suggestions above - is to use a hidden Markov model of the $X_{i,j}$'s.
| null | CC BY-SA 3.0 | null | 2011-07-04T13:23:25.110 | 2011-07-04T13:23:25.110 | null | null | 4376 | null |
12621 | 2 | null | 10464 | 0 | null | You can treat age as an interval censored variable. Some survival routines do this in a straight forward way for the response variable, if age is a predictor then I don't know if there are ready made tools available. But you could still do it using maximum liklihood.
| null | CC BY-SA 3.0 | null | 2011-07-04T14:06:11.700 | 2011-07-04T14:06:11.700 | null | null | 4505 | null |
12622 | 1 | 12624 | null | 5 | 238 | I am currently generating data by simulating a model of chemical system under different conditions (temperature) over time. In each simulation, the starting structure being modeled is exactly the same - only the temperature is different. The system is allowed to propagate over time and the length and number of observations in each simulation is identical. I would like to compare mean values e.g. distances between two atom under the different conditions. I have two questions:
- Should I regard the two simulations I have (high and low temperature) as paired data. How would an analogous human study be treated (e.g. comparing the behaviour of a single human participant under during a 1 hour period under condition 1 and another 1 hour period after an extensive washout period - under condition 2)?
- Since I effectively have two time series, what are the implications of the distance I want to measure being autocorrelated in some way?
| Multiple simulations of a system under different conditions - paired data? | CC BY-SA 3.0 | 0 | 2011-07-04T14:17:55.443 | 2011-07-08T09:37:47.333 | 2011-07-04T14:24:16.090 | 8 | 4054 | [
"time-series",
"hypothesis-testing",
"simulation",
"computational-statistics"
] |
12623 | 1 | 12626 | null | 11 | 28358 | I used my training dataset to fit cluster using kmenas function
```
fit <- kmeans(ca.data, 2);
```
How can I use fit object to predict cluster membership in a new dataset?
Thanks
| Predicting cluster of a new object with kmeans in R | CC BY-SA 3.0 | null | 2011-07-04T14:32:46.033 | 2021-03-24T10:08:23.050 | 2011-07-04T15:16:53.970 | null | 333 | [
"clustering"
] |
12624 | 2 | null | 12622 | 2 | null | Stochastic model
Consider the following scenario:
- You measure the blood pressure of the same man every day for two weeks;
- You measure the blood pressure of the same woman every day for two weeks;
What can you say about gender and blood pressure. Answer: not very much. This example is very similar to what you have. Until your sample size is bigger than one, you can't say very much.
As I commented above, you know that your two simulations are different. If you simulated the process enough times, you would get a "small p-value". What I suspect you want is to estimate the magnitude of these differences. So,
- You need more simulations to estimate the variance. There's no way around this.
- At each time-point store $x$ and $x^2$. This will allow you estimate the mean and variance of $x$.
- Once you have the mean and variance you could plot:
$x$ with a 95% confidence interval over time;
$x_H - x_L$ with a 95% confidence interval over time.
Deterministic model
Consider the following deterministic model for a simple death process:
\begin{equation}
\frac{dX(t)}{dt} = -\mu X(t)
\end{equation}
We can solve this equation to get $X(0) e^{-\mu t}$. Now suppose:
- For a high temperature, X(0) = 50
- For a low temperature, X(0) = 30.
Then for all $t$, the population of the high temperature is always greater than the low temperature. As we have a deterministic model, we have no uncertainty. If you did a t-test, your variance would be zero.
This is your scenario. Your model is deterministic. So the simulations are either the same or they are different. You have no uncertainty in the result.
| null | CC BY-SA 3.0 | null | 2011-07-04T14:36:49.990 | 2011-07-08T09:37:47.333 | 2011-07-08T09:37:47.333 | 8 | 8 | null |
12626 | 2 | null | 12623 | 17 | null | One of your options is to use [cl_predict](http://rss.acs.unt.edu/Rdoc/library/clue/html/predict.html) from the `clue`package (note: I found this through googling "kmeans R predict").
| null | CC BY-SA 3.0 | null | 2011-07-04T14:54:22.763 | 2011-07-04T14:54:22.763 | null | null | 4257 | null |
12627 | 1 | 12632 | null | 2 | 285 | This may be a silly question, but I'm not seeing a clear answer in any of the usual sources. I'm preparing to build a Bayesian model to fit with BUGS/JAGS, currently working through the model logic in plate notation.
I have a few kinds of observed variables, and several latent variables. I know that a Bayesian model has to be a DAG. What other conceptual constraints are there on the network? In particular, can an observed variable be the parent of a latent variable in a Bayesian model?
Thanks, and happy fourth of July to all the Americans out there.
| Can an observed variable be the parent of a latent variable in a Bayesian network? | CC BY-SA 3.0 | null | 2011-07-04T15:52:47.440 | 2011-07-04T16:51:51.153 | null | null | 4110 | [
"bayesian",
"networks",
"jags"
] |
12628 | 2 | null | 12615 | 4 | null | In case you are using LaTeX for your report writing, the package pgfplots can read in data files and plot single or double logarithmic axis. In case you need to do calculations you can escape to gnuplot.
It just looks this tiny bit better if your text font matches your axis labels font.
| null | CC BY-SA 3.0 | null | 2011-07-04T16:07:14.403 | 2011-07-04T16:07:14.403 | null | null | 4342 | null |
12629 | 1 | null | null | 6 | 994 | I have a problem that I don't think I've met before.
I have N observations of the variables v1 and v2 and I assume that there is a function f such as v2 = f(v1).
I want to know if f has a particular 'statistical' monotony (increasing or decreasing) and if it is 'statistically' convex or concave.
By 'statistical' I mean that my observations may include error terms so you may encounter pairs of variable that show a monotony that is the opposite of the global monotony, if that's English.
Should I simply compute a f' and a f'' ? (derivatives of f).
If you have some thoughts on this I'd be glad to read it.
Thanks,
Arthur
(btw I use Stata)
| Is there a test of monotony, convexity or concavity? | CC BY-SA 3.0 | null | 2011-07-04T16:44:07.433 | 2022-09-27T17:27:51.753 | 2011-07-04T18:23:28.340 | null | 4398 | [
"hypothesis-testing"
] |
12631 | 2 | null | 12615 | 1 | null | For a quick and easy graph, you should give [GraphCalc](http://graphcalc.com) a try.
| null | CC BY-SA 3.0 | null | 2011-07-04T16:51:36.597 | 2011-07-04T16:51:36.597 | null | null | 4754 | null |
12632 | 2 | null | 12627 | 5 | null | Sure, parents of latent variables can be observed.
The graph just encodes conditional independence relationships among variables, and is totally separate from the question of which variables are observed or latent. For example, if the graph is $X \to Y \to Z$, this tells you that the density $f(x,y,z)$ factors as $f(x) f(y|x) f(z|y)$, but doesn't tell you anything about which variables are observed.
(It can be easier or harder to do inference depending on which variables are observed, though.)
| null | CC BY-SA 3.0 | null | 2011-07-04T16:51:51.153 | 2011-07-04T16:51:51.153 | null | null | 5179 | null |
12634 | 2 | null | 12599 | 4 | null | Kevin Knight's [Bayesian Inference With Tears](http://www.isi.edu/natural-language/people/bayes-with-tears.pdf) describes applications of the Chinese Restaurant Process (which he calls a "cache model") to tree substitution grammars, Chinese word segmentation, and part-of-speech tagging.
(If anyone knows the original sources for these applications, that'd be great as well.)
| null | CC BY-SA 3.0 | null | 2011-07-04T17:08:40.403 | 2011-07-04T17:08:40.403 | null | null | 1106 | null |
12635 | 2 | null | 6896 | 2 | null | This sounds like it is similar to a certain application of Information Retrieval (IR). A few years ago I attended a talk about gait recognition that sounds similar to what you are doing. In Information Retrieval, "documents" (in your case: a person's angle data) are compared to some query (which in your case could be "is there a person with angle data (.., ..)"). Then the documents are listed in the order of the one that matches the closest down to the one that matches the least. That, in turn, means that one central component of IR is putting a document in some kind of vector space (in your case: angle space) and comparing it to one specific query or example document or measuring their distance. (See below.) If you have a sound definition of the distance between two individual vectors, all you have to do is coming up with a measure for the distance of two data sets. (Traditionally in IR the distance in vector space model is calculated either by the cosine measure or Euclidean distance but I don't remember how they did it in that case.)
In IR there is also a mechanism called "relevance feedback" that, conceptually, works with the distance of two sets of documents. That mechanism normally uses a measure of distance that sums up all individual distances between all pairs of documents (or in your case: person vectors). Maybe that is of use to you.
The following page has some papers that seem relevant to your issue: [http://www.mpi-inf.mpg.de/~mmueller/index_publications.html](http://www.mpi-inf.mpg.de/~mmueller/index_publications.html)
Especially this one [http://www.mpi-inf.mpg.de/~mmueller/publications/2006_DemuthRoederMuellerEberhardt_MocapRetrievalSystem_ECIR.pdf](http://www.mpi-inf.mpg.de/~mmueller/publications/2006_DemuthRoederMuellerEberhardt_MocapRetrievalSystem_ECIR.pdf) seems interesting.
The talk of Müller that I attended mentions similarity measures from Kovar and Gleicher called "point cloud" (see [http://portal.acm.org/citation.cfm?id=1186562.1015760&coll=DL&dl=ACM](http://portal.acm.org/citation.cfm?id=1186562.1015760&coll=DL&dl=ACM)) and one called "quaternions".
Hope, it helps.
| null | CC BY-SA 3.0 | null | 2011-07-04T19:16:51.750 | 2011-07-07T12:22:58.180 | 2011-07-07T12:22:58.180 | 1048 | 1048 | null |
12637 | 1 | null | null | 4 | 564 | I'm trying to estimate the unknown 8x8 covariance matrix X in R using the maximum likehood, but I have problems of figuring out the efficient way of parametrization of X when some of the covariances in X are constrained to zero. When there's no constraints, I have used a cholesky factorization, ie. I have parameters which correspond to the lower triangular matrix L, and I get X by X<-t(L)%*%L which is a proper positive definite matrix, but how to do it now that I want some of the cells of X to be constrained to zero? In my case, X is constrained in a way that the upper left and lower right 4x4 matrices can be anything (but of course having the properties of proper covariance matrices), and the off-diagonal 4x4 matrix is a diagonal matrix, ie. the the first variable x1 correlates with variables x2,x3,x4 and x5, variable x2 correlates with x1,x3,x4 and x6 etc.
Thanks.
| Efficient parametrization of the covariance matrix with some covariances constrained to zero | CC BY-SA 3.0 | null | 2011-07-04T19:58:13.217 | 2011-07-04T21:54:39.327 | null | null | 5286 | [
"r",
"optimization",
"matrix",
"matrix-decomposition",
"covariance"
] |
12638 | 2 | null | 11085 | 0 | null | 1) I agree with suncoolsu that FDR is not estimated (to the best of my knowledge, which is slim), but controlled for.
2) Once you have p values you can use something like - p.adjust(my_p_values, method = "BH"). And you will get the adjusted p values using the Benjamini Hochberg procedure. Any p value that is bellow (let's say) 0.05, can be rejected for a Q=0.05.
| null | CC BY-SA 3.0 | null | 2011-07-04T20:00:40.123 | 2011-07-04T20:00:40.123 | null | null | 253 | null |
12640 | 2 | null | 12637 | 3 | null | You could start from your original approach and impose the equations that the specified coefficients are 0. This leads to a fairly large system of polynomial equations - 12 equations (corresponding to the zeroes in the upper triangle) in 25 unknowns. I can solve this in Maple and obtain a union of 87 parametric solutions of dimensions from 11 to 16, each of which parametrizes a subset of the full solution and the union of which forms that full solution.
For example, one solution (one of the two 16-dimensional ones) is given by:
$$
\begin{gather}
A_{{1,1}}=0,A_{{2,1}}=0,A_{{2,2}}=0,A_{{3,1}}=0,A_{{3,2}}=0,A
_{{3,3}}=0, \\ A_{{5,1}}=-{\frac {A_{{4,2}}A_{{5,2}}+A_{{4,3}}A_{{5,3}}+A_
{{4,4}}A_{{5,4}}}{A_{{4,1}}}}, \\ A_{{6,1}}=-{\frac {A_{{4,2}}A_{{6,2}}+A_
{{4,3}}A_{{6,3}}+A_{{4,4}}A_{{6,4}}}{A_{{4,1}}}}, \\ A_{{7,1}}=-{\frac {A_
{{4,2}}A_{{7,2}}+A_{{4,3}}A_{{7,3}}+A_{{4,4}}A_{{7,4}}}{A_{{4,1}}}}.
\end{gather}
$$
However, even though all of these solutions are needed to describe all lower-triangular matrices for which $L L^T$ is of the form you require, it may well be that to just obtain all such matrices, we can make do with fewer solutions. In fact, I have a sneaking suspicion that the solution above might cover all cases. This is just a hunch that would require a more thorough investigation, though.
In order to reproduce the computation if you have a copy of Maple (15, in my case) available, you can run the following:
```
# Construct a generic lower triangular 8x8 matrix.
lt := Matrix(8, symbol = A, shape = triangular[lower]):
mm := lt . lt^%T:
# Select the positions that should be zero.
positions := [seq(seq([i, j], j = 5 .. 8), i = 1 .. 4)]:
positions := remove(pair -> pair[2] = pair[1] + 4, positions):
# Construct and solve the system of equations.
sys := map(pair -> mm[op(pair)], positions):
solutions := [solve(sys)]:
nops(solutions); # returns 87, so 87 different solutions.
# Split each solution into trivial equations (like A[2,3] = A[2,3]) and
# nontrivial ones. This separates free variables from those determined
# by other ones.
split := map2([selectremove], evalb, solutions):
numbers := map2(map, nops, split):
convert(numbers, set); # returns {[11, 14], [12, 13], ..., [16, 9]},
# showing that the dimension of the components runs
# from 11 to 16.
# Select the solutions of dimension 16.
dim16 := select(pair -> nops(pair[1]) = 16, split):
dim16 := map(pair -> pair[2], dim16): # We're only interested in the
# nontrivial equations.
nops(dim16); # returns 2, showing there are 2. One of these is the one
# printed above.
```
| null | CC BY-SA 3.0 | null | 2011-07-04T21:54:39.327 | 2011-07-04T21:54:39.327 | null | null | 2898 | null |
12641 | 1 | 12649 | null | 25 | 808 | I was wondering how the Bayesians in the CrossValidated community view the problem of model uncertainty and how they prefer to deal with it? I will try to pose my question in two parts:
- How important (in your experience / opinion) is dealing with model uncertainty? I haven't found any papers dealing with this issue in the machine learning community, so I'm just wondering why.
- What are the common approaches for handling model uncertainty (bonus points if you provide references)? I've heard of Bayesian model averaging, though I am not familiar with the specific techniques / limitations of this approach. What are some others and why do you prefer one over another?
| Addressing model uncertainty | CC BY-SA 3.0 | null | 2011-07-04T22:15:12.780 | 2014-06-22T13:16:06.840 | 2013-08-27T01:48:57.613 | 7290 | 1913 | [
"machine-learning",
"bayesian",
"model-selection"
] |
12642 | 1 | 12667 | null | 5 | 433 | I am having trouble understanding the following. Let $\mu$ and $\sigma^2$ be the true mean and variance, $\bar{x}$ and $s^2$ the measured mean and variance for a random variable $x$, where $$\displaystyle s^2 = \frac{1}{N+k}\sum_i (x_i-\bar{x})^2.$$
- If $s^2$ is an unbiased estimate
of variance then $k=-1$.
- If $s^2$ has the smallest
mean square spread from true variance then $k=1$.
| Understanding variance estimators | CC BY-SA 3.0 | null | 2011-07-04T22:20:08.360 | 2011-07-05T13:42:33.260 | 2011-07-05T08:35:56.303 | null | 4714 | [
"estimation",
"variance"
] |
12643 | 1 | null | null | 4 | 3555 | Here is an ANOVA model with one between-subject factor (condition; 4 levels) and one within-subject factor (trial_seq; 20 levels).
```
amod = aov(decision_quality ~ condition*trial_seq +
Error(user_id/trial_seq), data = d.task1)
summary(amod)
```
I want to do a pairwise analysis for both the between-subject factor and the within-subject factor. I found that it is not straightforward to use a function like TukeyHSD() for a within-subject factor. I researched this problem and found some suggestions in [another thread](https://stats.stackexchange.com/questions/575/post-hocs-for-within-subjects-tests).
However, my lack of statistical background prevented me from fully understanding vignettes coming with the multcomp package. Anyway, I tried some random statements like:
```
summary(glht(amod, linfct = mcp(trial_seq = "Tukey")))
```
However, it generates the following error:
```
Error in model.matrix.aovlist(model) :
‘glht’ does not support objects of class ‘aovlist’
Error in summary(glht(amod, linfct = mcp(trial_seq = "Tukey"))) :
error in evaluating the argument 'object' in selecting a method for
function 'summary': Error in factor_contrasts(model) :
no ‘model.matrix’ method for ‘model’ found!
```
- What did I do wrong?
| Can I have an example of pairwise comparisons of data against a within-subject factor? | CC BY-SA 3.0 | null | 2011-07-04T22:52:04.963 | 2011-09-29T09:46:43.493 | 2017-04-13T12:44:41.980 | -1 | 5281 | [
"r",
"anova",
"multiple-comparisons",
"repeated-measures"
] |
12644 | 2 | null | 11085 | 1 | null | The adjusted analogue to the p-values you are probably looking for is the q-value, which is described by Storey as:
>
(...) [giving] the scientist a hypothesis testing error measure for each observed statistic with respect to pFDR.
The p-value accomplishes the same goal with respect to the type I error, and the adjusted p-value with respect to FWER.
So, q-value is to FDR as adjusted p-value is to FWER.
In R, there is a [qvalue](http://cran.r-project.org/web/packages/qvalue/index.html) package which can produce these estimates given p-values.
| null | CC BY-SA 3.0 | null | 2011-07-04T23:49:36.427 | 2011-07-04T23:49:36.427 | null | null | 5240 | null |
12645 | 1 | null | null | 5 | 270 | The title says it all. Specifically: given strictly positive real numbers $a_1,\dots,a_T$ and $b_1,\dots,b_T$, I want to sample from $$\mu:=\text{Uniform}(\{p\in[0,\infty)^T : \sum_{t=1}^T a_t p_t = 1\}\cap\{p\in[0,\infty)^T : \sum_{i=1}^T b_t p_t = 1\}),$$ assuming that the intersection is nonempty.
My best attempt so far is the following rejection sampling algorithm:
- Set $u_t = \min\{\frac{1}{a_t},\frac{1}{b_t}\}$ for $t=1,\dots,T-1$.
- Sample $p_t \sim \text{Uniform}(0,u_t)$ for $t=1,\dots,T-1$.
(Justification: if $p_t>u_t$ then $\max\{a_t,b_t\} p_t > \max\{a_t,b_t\}\min\{\frac{1}{a_t},\frac{1}{b_t}\} = 1$, so $p_t$ cannot be the $t$th coordinate of a sample from $\mu$.)
- Set $s_a = \sum_{t=1}^{T-1} a_t p_t$ and $s_b = \sum_{t=1}^{T-1} b_t p_t$.
- If $s_a>1$ or $s_b>1$, reject.
- If $\frac{1}{a_T}(1-s_a) \neq \frac{1}{b_T}(1-s_b)$, reject.
- Else, set $p_T = \frac{1}{a_T}(1-s_a)$ and accept $p=(p_1,\dots,p_T)$.
The trouble is that in step 5, we will reject with probability 1.
Does anyone know how to do this properly, either using something like the above or via a totally different approach? Thanks in advance for any ideas!
| How to sample uniformly from an intersection of simplices? | CC BY-SA 3.0 | null | 2011-07-05T01:53:56.483 | 2013-02-26T09:26:55.453 | 2011-07-05T04:49:11.280 | 5179 | 5179 | [
"sampling"
] |
12646 | 2 | null | 10361 | 1 | null | part 1:
1:19 (prior odds = 5%) x 10 (likelihood ratio for positive test results = for every 100 true positives there are 10 false positives, so true positive is 10x more likely) = 10:19 (posterior odds) = 34% chance that contaminating agent is present.
part 2:
test a: 1:9 (prior odds = 10%, baserate for presence of agent in poorly maintained plants) x 10 (likelihood ratio for positive test results) = 10:9 (posterior odds) = 53% chance that contaminating agent is present.
test b: 10:9 (prior odds based on test a result & baserate for poorly maintained plants) x 1 (likelihood ratio for negative result-- just as likely to be false negative as true negative) = 10:9 = 53% chance that agent is present.
| null | CC BY-SA 3.0 | null | 2011-07-05T02:12:38.797 | 2011-07-05T14:26:24.717 | 2011-07-05T14:26:24.717 | 11954 | 11954 | null |
12647 | 1 | 12660 | null | 8 | 1518 | Here's the thread I got the idea from: [http://www.quora.com/Do-men-have-a-wider-variance-of-intelligence-than-women/answer/Ed-Yong](http://www.quora.com/Do-men-have-a-wider-variance-of-intelligence-than-women/answer/Ed-Yong)
Basically, this is a model that might be able to explain why there aren't more females in prestigious math/science competitions - it might be a statistical artifact arising from the simple fact that there are far more males than females in math/science. If this model applies, then we may not need to assume that male intelligence has higher variance than female intelligence.
The question I'd like to see addressed: If we assume equal means and equal variances (but different sample sizes), then is the model in the paper still the best model when used for predicting, say, the gender composition of the team of the 5-10 best players? Rather than just the gender composition of the grandmaster?
[http://rspb.royalsocietypublishing.org/content/276/1659/1161.full#sec-3](http://rspb.royalsocietypublishing.org/content/276/1659/1161.full#sec-3) has the diagram and use of the model
They basically used pairing between the top 100 males and top 100 females. Is that a valid assumption to make though? It works for grandmasters - that's true - but would it work if we're trying to select the top 10 people in any field? It's entirely possible, after all, that the expected distributions would be different if we're trying to select from a random distribution of the top 5 players of each gender, rather than the n-th ranked player of each gender.
As you increase the number of players you select for a "winning" team, for example, then maybe the distributions play out in a different way. I would expect the smaller group to have higher variance in mean than the larger group. We know that to be true when averaging over the entire population distribution (as a consequence of the central limit theorem). But what if we just want 10 people from each population instead? The fact is that a lot of "potentially" top people will end up dropping out because they would do something other than spend hours a day to practice for a "winning team"
High variability of the extreme value though - that makes sense if we're talking about the very top. In a large population, the extreme value is going to be very consistent. Whereas in a small population, the extreme value is going to have A LOT of variability - but that extreme value spends far more time in the left part of the (mean of extreme values) as compared to the right part of it. So if you had a head-to-head match up most years, the population with the larger sample size will win.
The thing is, what about a head-to-head matchup of the top 10 members of each distribution? It would be some sort of average between the model the paper used (1 to 1 matchups) and the model where we simply had matchups of the two entire populations with each other.
| Assuming two Gaussian distributions of equal mean and variance, then how different can we expect the top X members of each group to be? | CC BY-SA 3.0 | null | 2011-07-05T02:15:16.063 | 2012-11-08T14:54:29.453 | 2011-07-06T00:01:05.500 | 5288 | 5288 | [
"normal-distribution"
] |
12648 | 1 | null | null | 1 | 1970 | I have carried out PCA and then clustered the 6 resultant components using K-means clustering technique using SPSS. Normally SPSS adds a class variable for each case indicating its assigned group.
Is there any other method I can "calculate" the class variable (i.e. using component scores and REGR factor scores for the K means analysis given in the "Final Cluster centers" table????
| Assigning class to the cases after K means cluster analysis (SPSS) | CC BY-SA 3.0 | null | 2011-07-05T02:38:20.980 | 2015-04-01T23:49:13.037 | 2011-07-05T06:44:49.363 | 3277 | 5187 | [
"spss",
"k-means"
] |
12649 | 2 | null | 12641 | 17 | null | There are two cases which arise in dealing with model-selection:
- When the true model belongs in the model space.
This is very simple to deal with using BIC. There are results which show that BIC will select the true model with high probability.
However, in practice it is very rare that we know the true model. I must remark BIC tends to be misused because of this (probable reason is its similar looks as [AIC](http://en.wikipedia.org/wiki/Akaike_information_criterion)). These issues have been addressed on this forum before in various forms. A good discussion is [here](https://stats.stackexchange.com/questions/577/is-there-any-reason-to-prefer-the-aic-or-bic-over-the-other).
- When the true model is not in the model space.
This is an active area of research in the Bayesian community. However, it is confirmed that people know that using BIC as a model selection criteria in this case is dangerous. Recent literature in high dimension data analysis shows this. One such example is this. Bayes factor definitely performs surprisingly well in high dimensions. Several modifications of BIC have been proposed, such as mBIC, but there is no consensus. Green's RJMCMC is another popular way of doing Bayesian model selection, but it has its own short-comings. You can follow-up more on this.
There is another camp in Bayesian world which recommends model averaging. Notable being, Dr. Raftery.
- Bayesian model averaging.
This website of Chris Volinksy is a comprehensive source of Bayesian model averging. Some other works are here.
Again, Bayesian model-selection is still an active area of research and you may get very different answers depending on who you ask.
| null | CC BY-SA 3.0 | null | 2011-07-05T03:06:08.320 | 2014-06-22T13:16:06.840 | 2017-04-13T12:44:37.583 | -1 | 1307 | null |
12650 | 2 | null | 12546 | 1 | null | @cardinal's answer is well-stated and has been accepted, but, for the sake of closing this thread completely I'll offer the following: The [IMSL Numerical Libraries](http://www.roguewave.com/products/imsl-numerical-libraries.aspx) contain a routine for performing L-infinity norm regression. The routine is available in Fortran, C, Java, C# and Python. I have used the C and Python versions for which the method is call lnorm_regression, which also supports general $L_p$-norm regression, $p >= 1$.
Note that these are commercial libraries but the Python versions are free (as in beer) for non-commercial use.
| null | CC BY-SA 3.0 | null | 2011-07-05T03:44:13.463 | 2011-10-19T03:01:46.837 | 2011-10-19T03:01:46.837 | 1080 | 1080 | null |
12651 | 1 | 12737 | null | 15 | 1733 | The Box-Jenkins model selection procedure in time series analysis begins by looking at the autocorrelation and partial autocorrelation functions of the series. These plots can suggest the appropriate $p$ and $q$ in an ARMA$(p,q)$ model. The procedure continues by asking the user to apply the AIC/BIC criteria to select the most parsimonious model among those that produce a model with a white noise error term.
I was wondering how these steps of visual inspection and criterion-based model selection impact the estimated standard errors of the final model. I know that many search procedures in a cross-sectional domain can bias standard errors downward, for example.
On the first step, how does selecting the appropriate number of lags by looking at the data (ACF/PACF) impact the standard errors for time series models?
I would guess that selecting the model based upon AIC/BIC scores would have an impact analogous to that for cross-sectional methods. I actually don't know much about this area either, so any comments would be appreciated on this point as well.
Lastly, if you wrote down the precise criterion used for each step, could you bootstrap the entire process to estimate the standard errors and eliminate these concerns?
| Box-Jenkins model selection | CC BY-SA 3.0 | null | 2011-07-05T05:02:10.367 | 2016-05-23T10:01:03.543 | 2016-05-23T10:01:03.543 | 1352 | 401 | [
"regression",
"time-series",
"arima",
"model-selection",
"box-jenkins"
] |
12652 | 2 | null | 12587 | 0 | null | In Information Retrieval research, an experiment is often repeated several times (in system-oriented IR e.g. about 50 times with slightly varying data). In that case the solution is normally to establish some measure of central tendency (e.g. arithmetic mean) and do the significance test on that. Your population is human beings so you would want to come up with a measure for a single human being, which, in your case, could be the mean answer.
| null | CC BY-SA 3.0 | null | 2011-07-05T05:12:05.167 | 2011-07-05T05:12:05.167 | null | null | 1048 | null |
12653 | 2 | null | 12587 | 4 | null | You find yourself in a common case where you have measured a certain outcome in response to various parameters.
This can be analysed using a generalized linear model, a fairly wide category of statistical models (that, for instance, also include ANOVA). This types of model will create a linear relationship between your output and the variables (regressors).
In this case, as the outcome is discrete, I would use a logistic regression; as you measured the same subject twice we will have to use a mixed effect (repeated measures) logistic model, including the subject as a random effect
I'm sorry, but I don't know anything about SPSS, so I will give you an example in R.
Let's make some mock data:
```
# Make data reproducible
set.seed(51280)
# number of the experimental subject
subject <- rep(1:20, each=2)
# 10 men, 10 women (twice, as you did two trials)
gender <- rep(c("M", "F"), each=40)
# 1: attractive dialog, 0 neutral dialog
trial <- rep(1:0, 40)
# outcome: red or green, I'm sampling it randomly here
outcome <- sample(c("red", "green"), 40, replace=TRUE)
data <- data.frame(subject, gender, trial, outcome)
```
`data` will look something like (first 10 rows only):
```
subject gender trial outcome
1 1 M 0 green
2 1 M 1 green
3 2 M 0 green
4 2 M 1 red
5 3 M 0 green
6 3 M 1 green
7 4 M 0 red
8 4 M 1 red
9 5 M 0 red
10 5 M 1 red
```
If we inspect the data using `table` we get
```
> table(data$outcome,data$gender)
F M
green 19 19
red 21 21
> table(data$outcome,data$trial)
0 1
green 24 14
red 16 26
```
Now we can use the `lmer` (linear mixed model) function of the `lme4` package to generate a model which linearly connects our outcome (red/green) with the regressors (sex and trial), considering the nuisance factor (subject). We use the formula `outcome ~ gender * trial` which tells `glm` to use `outcome` as the independent variable and gender and trial as regressors, counting their interactions. We don't just want to know if `outcome` is different between men and women and between the two trials (case in which we would use `+` instead of `*`), we want to know if it is different when the two factors are considered together.
```
require(lme4)
logit.model <- glmer(outcome ~ gender * trial + (trial | subject), data,
family = binomial(link = "logit"))
```
Finally, `summary(logit.model)` will tell us that in this case there are no differences in the outcome (as expected from random data).
```
Generalized linear mixed model fit by the Laplace approximation
Formula: outcome ~ gender * trial + (trial | subject)
Data: data
AIC BIC logLik deviance
60.5 77.17 -23.25 46.5
Random effects:
Groups Name Variance Std.Dev. Corr
subject (Intercept) 2301.2 47.971
trial 4464.4 66.816 -0.675
Number of obs: 80, groups: subject, 20
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.108e+01 1.733e+01 -0.640 0.523
genderM -5.001e-06 8.240e+00 0.000 1.000
trial 2.244e+01 2.557e+01 0.878 0.380
genderM:trial -2.156e-04 1.218e+01 0.000 1.000
Correlation of Fixed Effects:
(Intr) gendrM trial
genderM -0.238
trial -0.667 0.161
genderM:trl 0.161 -0.677 -0.238
```
| null | CC BY-SA 3.0 | null | 2011-07-05T06:18:48.537 | 2011-07-06T05:42:20.180 | 2011-07-06T05:42:20.180 | 582 | 582 | null |
12654 | 1 | null | null | 3 | 1098 | This is a simple question. I am dealing with a "clipped" normal distribution -- say, $N(0,0.5)$ clipped between $[-1,1]$. I would like to calculate the "probability" of a sample, but I know that in $(-1,1)$, the probability of a single event is 0. However, the probability of 1 is $1-\text{cdf}(1,0,0.5)$ and -1 is $\text{cdf}(-1,0,0.5)$. How does one typically reconcile these sorts of differences?
My intuition would be to use epsilon balls within $(-1,1)$, but it's not quite the RIGHT thing to do...
| Calculating event probabilities in mixed, discrete/continuous distributions | CC BY-SA 3.0 | null | 2011-07-05T07:19:29.237 | 2011-07-05T23:31:44.727 | 2011-07-05T16:45:27.733 | 919 | 4742 | [
"distributions",
"density-function",
"cumulative-distribution-function"
] |
12655 | 2 | null | 12648 | 3 | null | Rerun your clusterization to save the final cluster centers as .SAV data file (check "Write final"). Then you open the data file with new objects to classify (this dataset may contain only new objects or a mix of new and old objects - it will make no difference). Check "Read initial" and choose here that saved file with cluster centers. Check "Classify only" instead of "Iterate and classify". Order to save cluster memberships under "Save" button. Run.
| null | CC BY-SA 3.0 | null | 2011-07-05T07:24:35.493 | 2011-07-05T07:24:35.493 | null | null | 3277 | null |
12656 | 1 | null | null | 12 | 284 | I'm writing a bit of code that makes pretty heavy use of sampling (eg, MCMC, Particle Filters, etc), and I would really like to test it to make sure that it's doing what I think it is before claiming any results. Is there a typical way unit test these methods to ensure correctness?
| Unit testing sampling methods | CC BY-SA 3.0 | null | 2011-07-05T07:33:50.277 | 2023-03-31T19:05:30.720 | 2022-06-09T02:21:15.443 | 11887 | 4742 | [
"hypothesis-testing",
"markov-chain-montecarlo",
"software",
"particle-filter"
] |
12657 | 2 | null | 165 | 122 | null | I think there's a nice and simple intuition to be gained from the (independence-chain) Metropolis-Hastings algorithm.
First, what's the goal? The goal of MCMC is to draw samples from some probability distribution without having to know its exact height at any point. The way MCMC achieves this is to "wander around" on that distribution in such a way that the amount of time spent in each location is proportional to the height of the distribution. If the "wandering around" process is set up correctly, you can make sure that this proportionality (between time spent and height of the distribution) is achieved.
Intuitively, what we want to do is to to walk around on some (lumpy) surface in such a way that the amount of time we spend (or # samples drawn) in each location is proportional to the height of the surface at that location. So, e.g., we'd like to spend twice as much time on a hilltop that's at an altitude of 100m as we do on a nearby hill that's at an altitude of 50m. The nice thing is that we can do this even if we don't know the absolute heights of points on the surface: all we have to know are the relative heights. e.g., if one hilltop A is twice as high as hilltop B, then we'd like to spend twice as much time at A as we spend at B.
The simplest variant of the Metropolis-Hastings algorithm (independence chain sampling) achieves this as follows: assume that in every (discrete) time-step, we pick a random new "proposed" location (selected uniformly across the entire surface). If the proposed location is higher than where we're standing now, move to it. If the proposed location is lower, then move to the new location with probability p, where p is the ratio of the height of that point to the height of the current location. (i.e., flip a coin with a probability p of getting heads; if it comes up heads, move to the new location; if it comes up tails, stay where we are). Keep a list of the locations you've been at on every time step, and that list will(asymptotically) have the right proportion of time spent in each part of the surface. (And for the A and B hills described above, you'll end up with twice the probability of moving from B to A as you have of moving from A to B).
There are more complicated schemes for proposing new locations and the rules for accepting them, but the basic idea is still: (1) pick a new "proposed" location; (2) figure out how much higher or lower that location is compared to your current location; (3) probabilistically stay put or move to that location in a way that respects the overall goal of spending time proportional to the height of the location.
What is this useful for? Suppose we have a probabilistic model of the weather that allows us to evaluate A*P(weather), where A is an unknown constant. (This often happens--many models are convenient to formulate in a way such that you can't determine what A is). So we can't exactly evaluate P("rain tomorrow"). However, we can run the MCMC sampler for a while and then ask: what fraction of the samples (or "locations") ended up in the "rain tomorrow" state. That fraction will be the (model-based) probabilistic weather forecast.
| null | CC BY-SA 4.0 | null | 2011-07-05T08:57:48.513 | 2021-06-04T18:21:47.593 | 2021-06-04T18:21:47.593 | 165987 | 5289 | null |
12659 | 2 | null | 12654 | 4 | null | There are two different object you might be interested in:
- the truncated normal distribution, which is a Gaussian whose range is restricted to lie within $[a,b]$. This is a density, so it's continuous (no need for epsilon balls).
- a random variable $Z$ that results from drawing a Gaussian random variable $X$ and then "thresholding" it, setting $Z=a$ if $X<a$ and to $Z=b$ if $X>b$, to $Z=X$ if $a\leq X \leq b$.
I'm guessing it's the latter object you're interested in. As you've noted, the probability of getting a sample on the border is non-zero: $P(Z=a) = \Phi((a-\mu)/\sigma)$ and $P(Z=b)=\Phi((\mu-b)/\sigma)$, where $\Phi$ is the standard Gaussian cdf, $\mu$ is the mean, and $\sigma$ the standard deviation.
However, there is zero probability of getting any particular sample within the range $(a,b)$, but the probability density within this range is simply that of the original Gaussian. (The probability of getting a sample from somewhere within this range is simply $1-P(Z=a)-P(Z=b)$.
The resulting object is therefore defined by "point masses" at $a$ and $b$ and a continuous density between $a$ and $b$.
| null | CC BY-SA 3.0 | null | 2011-07-05T09:56:37.230 | 2011-07-05T23:31:44.727 | 2011-07-05T23:31:44.727 | 5289 | 5289 | null |
12660 | 2 | null | 12647 | 2 | null | Let's look at
the top 3 of 100 Gaussians vs. the top 3 of 1000.
Real statisticians will give formulas for this and more;
for the rest of us, here's a little Monte Carlo.
The intent of the code is to give a rough
idea of the distributions of $X_{(N-2)} X_{(N-1)} X_{(N)}$;
running it gives
```
# top 3 of 100 Gaussians, medians: [[ 2. 2.1 2.4]]
# top 3 of 1000 Gaussians, medians: [[ 2.8 2.9 3.2]]
```
If someone could do this in R with rug plots, that would certainly
be clearer.
```
#!/usr/bin/env python
# Monte Carlo the top 3 of 100 / of 1000 Gaussians
# top 3 of 100 Gaussians, medians: [[ 2. 2.1 2.4]]
# top 3 of 1000 Gaussians, medians: [[ 2.8 2.9 3.2]]
# http://stats.stackexchange.com/questions/12647/assuming-two-gaussian-distributions-of-equal-mean-and-variance-then-how-differen
# cf. Wikipedia World_record_progression_100_metres_men / women
import sys
import numpy as np
top = 3
Nx = 100
Ny = 1000
nmonte = 100
percentiles = [50]
seed = 1
exec "\n".join( sys.argv[1:] ) # run this.py top= ...
np.set_printoptions( 1) # .1f
np.random.seed(seed)
print "Monte Carlo the top %d of many Gaussians:" % top
# sample Nx / Ny Gaussians, nmonte times --
X = np.random.normal( size=(nmonte,Nx) )
Y = np.random.normal( size=(nmonte,Ny) )
# top 3 or so --
Xtop = np.sort( X, axis=1 )[:,-top:]
Ytop = np.sort( Y, axis=1 )[:,-top:]
# medians (any percentiles, but how display ?) --
Xp = np.array( np.percentile( Xtop, percentiles, axis=0 ))
Yp = np.array( np.percentile( Ytop, percentiles, axis=0 ))
print "top %d of %4d Gaussians, medians: %s" % (top, Nx, Xp)
print "top %d of %4d Gaussians, medians: %s" % (top, Ny, Yp)
```
| null | CC BY-SA 3.0 | null | 2011-07-05T11:05:36.957 | 2011-07-07T13:35:47.030 | 2011-07-07T13:35:47.030 | 557 | 557 | null |
12661 | 1 | null | null | 0 | 3678 | How do you do normalization so that when I get the mean/variance, the flat values wont affect the results? For example, in the figure below, Graph 1 & 2 are both considered to be noisy as compared to Graph 3.
However, as you can see, each of their flat intensity values vary from each other, so it could be possible that a 'clean' graph that contains very high intensity values can be considered 'noisy' as well as a 'noisy graph that contains low intensity values can be considered as 'clean'. I am using mean/variance and a threshold for determining if it is noisy or not, hence the need to 'normalize' the values.
Ive tried to do, getting the total and dividing each intensity to the total (e.g. arr[x] = arr[x] / total) but it does not work properly that way.
Any tips? Thanks!
| How to do 'normalization'? | CC BY-SA 3.0 | null | 2011-07-05T11:10:28.843 | 2011-07-05T14:18:20.093 | null | null | 5290 | [
"variance",
"mean",
"normalization"
] |
12662 | 2 | null | 12651 | 7 | null | In my opinion selecting the appropriate number of lags is no different than selecting the number of input series in a stepwise forward regression procedure. The incremental importance of lags or a specific input series is the basis for the tentative model specification.
Since you have asserted that the acf/pacf is the only basis for Box-Jenkins model selection, let me tell you what some experience has taught me. If a series exhibits an acf that doesn't decay, the Box-Jenkins approach (circa 1965) suggests differencing the data. But if a series has a level shift, like the [Nile data](http://support.sas.com/rnd/app/da/new/802ce/ets/chap3/sect8.htm), then the "visually apparent" non-stationarity is a symptom of needed structure but differencing is not the remedy. This Nile dataset can be modeled without differencing by simply identifying the need for a level shift first. In a similar vein we are taught using 1960 concepts that if the acf exhibits a seasonal structure (i.e. significant values at lags of s,2s,3s,...) then we should incorporate a seasonal ARIMA component. For discussion purposes, consider a series that is stationary around a mean and at fixed intervals, say every June there is a "high value". This series is properly treated by incorporating an "old-fashioned" dummy series of 0 and 1's (at June) in order to treat the seasonal structure. A seasonal ARIMA model would incorrectly use memory instead of an unspecified but waiting-to-be-found X variable. These two concepts of identifying/incorporating unspecified deterministic structure are direct applications of the work of I. Chang, William Bell, George Tiao, [R.Tsay](http://www.unc.edu/~jbhill/tsay.pdf), Chen et al (starting in 1978) under the general concept of Intervention Detection.
Even today some analysts are mindlessly performing memory maximization strategies, calling them Automatic ARIMA, without recognizing that "mindless memory modeling" assumes that deterministic structure such as pulses, level shifts, seasonal pulses and local time trends are non-existent or worse yet play no role in model identification. This is akin to putting one's head in the sand, IMHO.
| null | CC BY-SA 3.0 | null | 2011-07-05T11:11:19.893 | 2011-07-05T20:09:25.697 | 2011-07-05T20:09:25.697 | 919 | 3382 | null |
12663 | 1 | 12668 | null | 5 | 254 | I was recently asked:
- Is there an R implementation of a significance test for testing whether three or more correlations drawn from independent samples are equal?
I found [this formula](http://luna.cas.usf.edu/~mbrannic/files/regression/corr1.html#More%20than%20two) which wouldn't be too difficult to implement, but I was curious whether there was an existing implementation either based on the linked formula or on some other method.
| Test in R of whether three or more correlations from independent samples are equal | CC BY-SA 3.0 | null | 2011-07-05T11:11:47.167 | 2011-07-05T14:20:32.647 | null | null | 183 | [
"r",
"correlation",
"statistical-significance"
] |
12665 | 1 | null | null | 5 | 2061 | Does anybody know how to compare two positive predictive value (PPV) from two different predictive models? I found
some papers reporting $\chi^2$-test or Fisher exact test but I can´t figure out how the contingency tables would look like: They cannot be the same as for sensitivity/specifity tests, can they?
To clarify:
Imagine you want to compare one method for detecting cancer with another, newer one. I know how to estimate sensitivity, specificity etc., and to compare the two sensitivities from the two different models (sensitivity_method1 vs. sensitivity_method2; using McNemar) but regarding the comparison between PPV_method1 vs. PPV_method2 you can not use McNemar, because margins differ. There are possibilities to calculate the comparison in SAS `catmod` but it was too difficult for me to understand. So I thought there may be an easier way.
I would be really thankful for clarifications.
| Comparison of positive predictive value between two models | CC BY-SA 3.0 | 0 | 2011-07-05T12:27:37.283 | 2013-06-20T09:22:50.070 | 2013-06-20T09:22:50.070 | 805 | 5291 | [
"statistical-significance",
"contingency-tables",
"model-comparison"
] |
12666 | 2 | null | 12642 | 4 | null | What's the problem with just dividing by $N$? You don't take into account that you aren't subtracting the true population mean off of each $x_i$, but rather than estimate of it.
One way that I like to think about it is, suppose that I gave you the sample mean $\bar{x}$. How many data points $N-k$ would I have to give you so that you could tell me the exact values of the remaining $k$? Well, a mean gives one equation involving every observation and, if there was only one unknown observation, we could solve the equation. In summary, knowing the mean and $N - 1$ data points is the same as knowing every data point.
In calculating the sample variance, I know the mean, so it's like I only have $N - 1$ effective data points giving me information; the last one I could guess using the others plus the mean. We divide by the number of effective data points that we have, $N-1$. This is known as a degrees-of-freedom correction (we only have $N-1$ degrees of freedom, parameters that we don't know, given that we know the mean).
Now, here's the math: Let
$$\begin{equation*}
s^2 = \frac{1}{N}\sum_{i=1}^N{(y_i - \bar{y})^2}.
\end{equation*}$$
Then,
$$\begin{align*}
E[s^2] &= E\left[ \frac{1}{N}\sum_{i=1}^N{(y_i - \bar{y})^2} \right] \\
&= E\left[\frac{1}{N}\sum_{i=1}^N{\left((y_i - \mu) - (\bar{y} - \mu)\right)^2} \right] \\
&= \frac{1}{N}\sum_{i=1}^NE\left[(y_i - \mu)^2\right] - 2\frac{1}{N}E\left[(\bar{y} - \mu)\sum_{i=1}^N{(y_i - \mu)}\right] \\
&\qquad + E\left[(\bar{y} - \mu)^2\right] \\
&= \frac{1}{N}\sum_{i=1}^NE\left[(y_i - \mu)^2\right] - E\left[(\bar{y} - \mu)^2\right] \\
&= \text{Var}(y_i) - \frac{\text{Var}(y_i)}{N} = \frac{N-1}{N}\text{Var}(y_i).
\end{align*}$$
This uses the fact that the variance of the sample mean is the variance of $y_i$ divided by $N$.
Hence, an unbiased estimator requires multiplying $s^2$ by $N/(N-1)$, giving the equation that you sought.
As you mention, dividing by $N-1$ is close to dividing by $N$; the two get close as $N$ gets big. Hence, $s^2$ is a consistent estimator---its bias goes to 0 as $N$ gets big.
| null | CC BY-SA 3.0 | null | 2011-07-05T12:44:44.943 | 2011-07-05T12:44:44.943 | null | null | 401 | null |
12667 | 2 | null | 12642 | 5 | null | The idea of the unbaised variance estimate, is to have $E(s^{2})=\sigma^{2}$ where the expectation is with respect to the sampling distribution of $s^{2}$ or equivalently, with respect to the sampling distribution of $x_1,\dots,x_N$. So if we knew the true mean and the true variance, but not the value of $s^{2}$, then $s^2$ would have expected value of $\sigma^{2}$.
Now we have:
$$E(s^{2})=E\left[ \frac{1}{N+k}\sum_i (x_i-\bar{x})^2\right]
=\frac{1}{N+k}E\left[\sum_i x_i^2-N\bar{x}^{2}\right]$$
$$=\frac{1}{N+k}\left[\sum_i E(x_i^2)-NE(\bar{x}^{2})\right]$$
We can use the identity $E(Y^{2})=V(Y)+[E(Y)]^{2}$, and the fact that we know $E(x_i)=\mu$ and $V(x_i)=\sigma^{2}$, and the summation becomes:
$$\sum_i E(x_i^2)=\sum_i V(x_i)+[E(x_i)]^2=\sum_i (\sigma^2+\mu^2)=N(\sigma^2+\mu^2)$$
Now to calculate the second expectation, we re-write $NE(\bar{x}^{2})$ as follows:
$$NE(\bar{x}^{2})=NE\left(\left[\frac{1}{N}\sum_i x_i\right]^{2}\right)=\frac{1}{N}E\left(\sum_i \sum_j x_jx_i\right)=\frac{1}{N}\sum_i \sum_j E(x_jx_i)$$
$$=\frac{1}{N}\left(\sum_i E(x_i^2)+\sum_{i\neq j} E(x_jx_i)\right)=(\sigma^2+\mu^2)+\frac{1}{N}\sum_{i\neq j} E(x_jx_i)$$
Now we have another identity we can use $E(YZ)=Cov(Y,Z)+E(Y)E(Z)$. Your problem statement does not specify whether or not the sample is independent, but it does say that they have the same distribution. So we can take $Cov(x_ix_j)=\rho\sigma^2$ for some correlation $-\frac{1}{N-1}\leq\rho\leq 1$ (lower bound required for positive variance). We then get:
$$NE(\bar{x}^{2})=(\sigma^2+\mu^2)+\frac{1}{N}\sum_{i\neq j} (\rho\sigma^2+\mu^2)=(\sigma^2+\mu^2)+\frac{N(N-1)}{N}(\rho\sigma^2+\mu^2)$$
$$=\sigma^2(1+(N-1)\rho)+N\mu^{2}$$
Putting this together we have:
$$E(s^2)=\frac{1}{N+k}\left[N(\sigma^2+\mu^2)-\left(\sigma^2(1+(N-1)\rho)+N\mu^{2}\right)\right]=\frac{N-1}{N+k}\sigma^2(1-\rho)$$
So if we choose $k=-1$ and we assume that $\rho=0$ (i.e. independence), then we have $E(s^2)=\sigma^2$, and we get an unbiased estimate for $s^2$. However, if we assume that $\rho=-\frac{1}{N-1}$ (i.e. the sum is fixed), then we get $E(s^2)=\frac{N-1}{N+k}\sigma^2(1+\frac{1}{N-1})=\frac{N}{N+k}\sigma^2$ and we should set $k=0$ for an unbiased estimate. So one can interpret the intuition behind $N-1$ as accounting for the fact that the true mean has been estimated by the sample mean (and hence is not "fixed").
For minimum mean square error, we require $MSE(s^2)=E\left[(s^2-\sigma^2)^2\right]$ to be a minimum for some choice of $k$. Expanding the mean square error we get:
$$MSE(s^2)=E[s^4]-2\sigma^2E[s^2]+\sigma^4$$
$E(s^2)$ has been calculated already, now to calculate $E(s^4)$. squaring $s^2$ gives us:
$$s^4=\frac{1}{(N+k)^2}\left[\sum_i x_i^2-N\bar{x}^{2}\right]^2$$
$$=\frac{1}{(N+k)^2}\left(\left[\sum_i x_i^2\right]^2-2\left[\sum_i x_i^2\right]\left[N\bar{x}^{2}\right]+\left[N\bar{x}^{2}\right]^2\right)$$
$$=\frac{\left[\sum_i x_i^4+\sum_{i\neq j} x_j^2x_i^2\right]-2\frac{1}{N}\left[\sum_i x_i^2\right]\left[\sum_i x_i^2+\sum_{i\neq j} x_jx_i\right]+\frac{1}{N^2}\left[\sum_i x_i^2+\sum_{i\neq j} x_jx_i\right]^2}{(N+k)^2}$$
$$=\frac{(\frac{1}{N}-1)^2\left[\sum_i x_i^4+\sum_{i\neq j} x_j^2x_i^2\right]+2\frac{1}{N}(\frac{1}{N}-1)\left[\sum_i x_i^2\right]\left[\sum_{i\neq j} x_jx_i\right]+\frac{1}{N^2}\left[\sum_{i\neq j} x_jx_i\right]^2}{(N+k)^2}$$
$$=\frac{f(x_1,\dots,x_N)}{(N+k)^2}$$
And you can see that without some assumptions that the expectation will in general be a function of the fourth order moments $E(x_ix_jx_kx_l),E(x_i^2x_jx_k),E(x_i^3x_j),E(x_i^2x_j^2),E(x_i^4)$ (which are not given in the question). However, its dependence on $k$ is quite simple, so we can still solve the variation problem algebraically with $F=E[f(x_1,\dots,x_N)]$. So we have:
$$MSE(s^2)=\frac{F}{(N+k)^2}-2\sigma^2\left[\frac{N-1}{N+k}\sigma^2(1-\rho)\right]+\sigma^4$$
Taking derivative with respect to $k$, set to zero, and solve for $k$:
$$-2\frac{F}{(N+k)^3}+2\frac{N-1}{(N+k)^2}\sigma^4(1-\rho)=0$$
$$\implies k=\frac{F}{(N-1)\sigma^4(1-\rho)}-N$$
This shows that unless $F=c\sigma^4(1-\rho)$, where $c$ depends only on the sample size, the optimum value of $k$ will be a function of the parameters, and hence you have no "solution" per se because it depends on things you don't know. You can show that if you assume an independent normal distribution for $x_i$ (so $\rho=0$), then $F=(N^2-1)\sigma^4$ and you get $k=+1$ as the optimum value.
| null | CC BY-SA 3.0 | null | 2011-07-05T13:42:33.260 | 2011-07-05T13:42:33.260 | null | null | 2392 | null |
12668 | 2 | null | 12663 | 2 | null | This is a meta-analytical approach. The formula you provided looks like the standard [Q-test](http://digitalcommons.uconn.edu/cgi/viewcontent.cgi?article=1019&context=chip_docs&sei-redir=1#search=%22q-test%20meta-analysis%22) (test of heterogeneity; see p. 11f) which tests for $H_0: \theta_1=\theta_2=\ldots=\theta_k$ (with $k$ being the number of studies or independent correlation coefficients).
Most meta-analysis packages in R can do this test, for instance, [meta](http://cran.r-project.org/web/packages/meta/) or [metafor](http://cran.r-project.org/web/packages/metafor/index.html). I will provide an example for the `meta` package (differences are due to rounding errors):
```
library(meta)
library(psychometric)
dfr <- data.frame(r=c(0.2, 0.5, 0.6), n=c(200, 150, 75))
dfr$z <- r2z(dfr$r) ## Fisher's z transformation
dfr$z.se <- SEz(dfr$n) ## SEs for Fisher's z
## recommended approach
metacor(cor=r, n=n, sm="ZCOR", data=dfr)
## replicating the results from your "Correlation" webpage
metagen(TE=z, seTE=z.se, data=dfr)
> metagen(TE=z, seTE=z.se, data=dfr)
95%-CI %W(fixed) %W(random)
1 0.2027 [0.0631; 0.3424] 47.36 35.07
2 0.5493 [0.3877; 0.7110] 35.34 34.14
3 0.6931 [0.4622; 0.9241] 17.31 30.79
Number of trials combined: 3
95%-CI z p.value
Fixed effect model 0.4101 [0.3140; 0.5062] 8.3640 < 0.0001
Random effects model 0.4721 [0.1796; 0.7645] 3.1637 0.0016
Quantifying heterogeneity:
tau^2 = 0.0584; H = 2.92 [1.75; 4.87]; I^2 = 88.3% [67.5%; 95.8%]
Test of heterogeneity:
Q d.f. p.value
17.09 2 0.0002
Method: Inverse variance method
```
| null | CC BY-SA 3.0 | null | 2011-07-05T14:10:39.067 | 2011-07-05T14:20:32.647 | 2011-07-05T14:20:32.647 | 307 | 307 | null |
12669 | 2 | null | 12661 | 3 | null | So if I understand correctly, you want to detect reliably whether a data sample has few high peaks as opposed to many low peaks? What I would do is sort your data by intensity, then determine (say) the 98th intensity percentile and the 90th intensity percentile and find the ratio between the two. For graph 3, the 98th intensity percentile should still be part of the peak and the 90th percentile deep in a valley so you get a big ratio. For 1 and 2, they should be much closer together, so you get a small ratio. Then play with the three numbers involved (the high percentile, the low percentile, and the threshold for the ratio) till it does something close to what you want on a training set.
| null | CC BY-SA 3.0 | null | 2011-07-05T14:18:20.093 | 2011-07-05T14:18:20.093 | null | null | 2898 | null |
12670 | 1 | 12692 | null | 53 | 21022 | EDIT: The [Web Technologies and Services](http://cran.r-project.org/web/views/WebTechnologies.html) CRAN [task view](http://cran.r-project.org/web/views/) contains a much more comprehensive list of data sources and APIs available in R. You can [submit a pull request on github](https://github.com/ropensci/webservices) if you wish to add a package to the task view.
---
I'm making a list of the various data feeds that are already hooked into R or that are easy to setup. Here's my initial list of packages, and I was wondering what else I'm missing.
I'm trying to limit this list to "real time" or "close to real time" data feeds/APIs, where the underlying data might change between downloads. There's plenty of [lists out there](http://archive.ics.uci.edu/ml/) for static datasets, which only require one download.
This list is currently biased towards financial/time series data, and I could use some help expanding it into other domains.
Free Data:
Data Source - Package
[Google Finance historical data](http://www.google.com/finance/historical?q=NASDAQ:MSFT) - [quantmod](http://www.quantmod.com/)
[Google Finance balance sheets](http://www.google.com/finance?q=NASDAQ:MSFT&fstype=ii) - quantmod
[Yahoo Finance historical data](http://finance.yahoo.com/q/hp?s=MSFT%20Historical%20Prices) - quantmod
Yahoo Finance historical data - [tseries](http://rss.acs.unt.edu/Rdoc/library/tseries/html/get.hist.quote.html)
[Yahoo Finance current options chain](http://finance.yahoo.com/q/op?s=MSFT) - quantmod
[Yahoo Finance historical analyst estimates](http://finance.yahoo.com/q/ud?s=MSFT) - [fImport](http://cran.r-project.org/web/packages/fImport/index.html)
[Yahoo Finance current key stats](http://finance.yahoo.com/q/ks?s=MSFT) - fImport - seems to be broken
[OANDA historic exchange rates/metal prices](http://www.oanda.com/) - quantmod
[FRED historic macroeconomic indicators](http://research.stlouisfed.org/fred2/) - quantmod
[World Bank historic macroeconomic indicators](http://data.worldbank.org/indicator) - [WDI](http://cran.r-project.org/web/packages/WDI/index.html)
[Google Trends historic search volume data](http://www.google.com/trends?q=microsoft) - [RGoogleTrends](http://www.omegahat.org/RGoogleTrends/)
Google Docs - [RGoogleDocs](http://www.omegahat.org/RGoogleDocs/)
Google Storage - [RGoogleStorage](http://www.omegahat.org/RGoogleStorage/)
[Twitter](http://twitter.com/#!/search/microsoft) - [twitteR](http://cran.r-project.org/web/packages/twitteR/index.html)
[Zillow](http://www.zillow.com/) - [Zillow](http://www.omegahat.org/Zillow/)
[New York Times](http://www.nytimes.com/) - [RNYTimes](http://www.omegahat.org/RNYTimes/)
[US Census 2000](http://www.census.gov/main/www/cen2000.html) - [UScensus2000](http://cran.r-project.org/web/packages/UScensus2000/index.html)
[infochimps](http://www.infochimps.com/) - [infochimps](http://cran.r-project.org/web/packages/infochimps/index.html)
[datamarket](http://datamarket.com/) - [rdatamarket](http://cran.r-project.org/web/packages/rdatamarket/index.html) - requires free account
[Factual.com](http://www.factual.com/) - [factualR](http://cran.r-project.org/web/packages/factualR/index.html)
Geocode addresses - [RDSTK](http://thelogcabin.wordpress.com/2011/05/02/r-and-the-data-science-toolkit/)
Map coordinates to political boundaries - RDSTK
[Weather Underground](http://www.wunderground.com/) - [Roll your own](http://casoilresource.lawr.ucdavis.edu/drupal/node/991)
[Google News](http://www.google.com/finance/company_news?q=NASDAQ:MSFT) - [Roll your own](https://stackoverflow.com/questions/5761576/improving-a-function-to-get-stock-news-data-from-google-in-r)
[Earth Sciences](http://opendap.deltares.nl/thredds/catalog/opendap/catalog.html) netCDF [Data](http://www.unidata.ucar.edu/software/netcdf/examples/files.html) - [Roll your own](http://public.deltares.nl/display/OET/KML+overview+of+OPeNDAP+data#KMLoverviewofOPeNDAPdata-AccessingnetCDF/OPeNDAPdatawithR)
[Climate Data](http://processtrends.com/RClimate.htm) - [Roll your own](http://chartsgraphs.wordpress.com/2011/01/24/using-rclimate-to-retrieve-climate-series-data/)
Public health data - [Roll your own](http://www.medepi.net/msamuel/R.Visual.Display.Samuel.10-2008.pdf)
[OAI Harvester](http://cran.r-project.org/web/packages/OAIHarvester/index.html) - Open Archives Initiative harvester
[RAmazonS3](http://www.omegahat.org/RAmazonS3/) - S3 Amazon storage server
[Rflikr](http://www.omegahat.org/Rflickr/) - Flikr api
Requires a subscription:
[Bloomberg](http://www.bloomberg.com/professional/software_support/) - [RBloomberg](http://cran.r-project.org/web/packages/RBloomberg/index.html)
[LIM](http://www.lim.com/) - [LIM](http://www.lim.com/sites/default/files/R_package_LIMWS.pdf)
[Trades and Quotes from NYSE](http://www.nyxdata.com/Data-Products/Daily-TAQ) - [RTAQ](http://cran.r-project.org/web/packages/RTAQ/index.html)
[Interactive Brokers](http://www.interactivebrokers.com/en/p.php?f=marketData) - [IBrokers](http://cran.r-project.org/web/packages/IBrokers/index.html)
| Data APIs/feeds available as packages in R | CC BY-SA 3.0 | null | 2011-07-05T14:31:00.900 | 2017-11-23T10:42:18.043 | 2017-11-23T10:42:18.043 | 17230 | 2817 | [
"r",
"references",
"dataset"
] |
12671 | 2 | null | 12641 | 5 | null | I know people use DIC and Bayes factor, as suncoolsu said. And I was interested when he said "There are results which show that BIC will select the true model with high probability" (references?). But I use the only thing I know, which is posterior predictive check, championed by Andrew Gelman. If you google Andrew Gelman and posterior predictive checks you will find a lot of things. And I'd take a look at what Christian Robert [is writting](http://xianblog.wordpress.com/2011/02/24/lack-of-confidence-in-abc-model-choice/) on ABC [about model choice](http://arxiv.org/abs/1106.5919). In any case, here are some references I like, and some recent posts in Gelman's blog:
Blog
[DIC and AIC](http://www.stat.columbia.edu/~cook/movabletype/archives/2011/06/deviance_dic_ai.html);
[More on DIC](http://www.stat.columbia.edu/~cook/movabletype/archives/2011/06/plummer_on_dic.html).
[Model checking and external validation](http://www.stat.columbia.edu/~cook/movabletype/archives/2010/08/useful_models_m.html)
Papers on posterior predictive checks:
GELMAN, Andrew. (2003a). “A Bayesian Formulation of Exploratory Data Analysis and Goodness-of-fit Testing”. International Statistical Review, vol. 71, n.2, pp. 389-382.
GELMAN, Andrew. (2003b). “Exploratory Data Analysis for Complex Models”. Journal of Computational and Graphic Statistics, vol. 13, n. 4, pp. 755/779.
GELMAN, Andrew; MECHELEN, Iven Van; VERBEKE, Geert; HEITJAN, Daniel F.; MEULDERS, Michel. (2005). “Multiple Imputation for Model Checking: Completed-Data Plots with Missing and Latent Data.” Biometrics 61, 74–85, March
GELMAN, Andrew; MENG, Xiao-Li; STERN, Hal. (1996). “Posterior Predictive Assessment of Model Fitness via Realized Discrepancies”. Statistica Sinica, 6, pp. 733-807.
| null | CC BY-SA 3.0 | null | 2011-07-05T14:59:14.047 | 2011-07-05T14:59:14.047 | null | null | 3058 | null |
12672 | 2 | null | 12641 | 11 | null | A "true" Bayesian would deal with model uncertainty by marginalising (integrating) over all plausble models. So for example in a linear ridge regression problem you would marginalise over the regression parameters (which would have a Gaussian posterior, so it could be done analytically), but then marginalise over the hyper-paremeters (noise level and regularisation parameter) via e.g. MCMC methods.
A "lesser" Bayesian solution would be to marginalise over the model parameters, but to optimise the hyper-parameters by maximising the marginal likelihood (also known as the "Bayesian evidence") for the model. However, this can lead to more over-fitting than might be expected (see e.g. [Cawley and Talbot](http://jmlr.csail.mit.edu/papers/volume11/cawley10a/cawley10a.pdf)). See the work of [David MacKay](http://www.inference.phy.cam.ac.uk/mackay/) for information on evidence maximisation in machine learning. For comparison, see the work of [Radford Neal](http://www.cs.toronto.edu/~radford/) on the "integrate everything out" approach to similar problems. Note that the evidence framework is very handy for situations where integrating out is too computationally expensive, so there is scope for both approaches.
Effectively Bayesians integrate rather than optimise. Ideally, we would state our prior belief regarding the characteristics of the solution (e.g. smoothness) and make predictions notoionally without actually making a model. The Gaussian process "models" used in machine learning are an example of this idea, where the covariance function encodes our prior belief regarding the solution. See the excellent book by [Rasmussen and Williams](http://www.gaussianprocess.org/gpml/).
For practical Bayesians, there is always cross-validation, it is hard to beat for most things!
| null | CC BY-SA 3.0 | null | 2011-07-05T15:24:28.383 | 2011-07-05T15:24:28.383 | null | null | 887 | null |
12673 | 1 | 12674 | null | 17 | 17556 | This wiki page [Simple linear regression](http://en.wikipedia.org/wiki/Simple_linear_regression) has formulas to calculate $\alpha$ and $\beta$. Could anyone tell me how to derive the formulas in weighted case?
| Formula for weighted simple linear regression | CC BY-SA 3.0 | null | 2011-07-05T16:16:54.060 | 2022-09-15T03:44:45.890 | null | null | 5212 | [
"regression"
] |
12674 | 2 | null | 12673 | 21 | null | Think of ordinary least squares (OLS) as a "black box" to minimize
$$\sum_{i=1}^n (y_i - (\alpha 1 + \beta x_i))^2$$
for a data table whose $i^\text{th}$ row is the tuple $(1, x_i, y_i)$.
When there are weights, necessarily positive, we can write them as $w_i^2$. By definition, weighted least squares minimizes
$$\sum_{i=1}^n w_i^2(y_i - (\alpha 1 + \beta x_i))^2$$
$$=\sum_{i=1}^n (w_i y_i - (\alpha w_i + \beta w_i x_i))^2 .$$
But that's exactly what the OLS black box is minimizing when given the data table consisting of the "weighted" tuples $(w_i, w_i x_i, w_i y_i)$. So, applying the OLS formulas to these weighted tuples gives the formulas you seek.
| null | CC BY-SA 3.0 | null | 2011-07-05T16:40:00.000 | 2011-07-05T16:40:00.000 | null | null | 919 | null |
12675 | 1 | null | null | 4 | 165 |
### Context
Let $\mathbf{x}_i \in \mathbb{R}^{100}$ and $\mathbf{z}_i \in \mathbb{R}^{20}$ be input vectors with the same corresponding target $\mathbf{y}_i \in \mathbb{R}^{25}$.
Using ridge regression we obtain two separate mappings, $\mathbf{A}_\mathbf{x}$ and $\mathbf{A}_\mathbf{z}$, such that
$$\mathbf{A}_\mathbf{x}\mathbf{x}_i = \mathbf{y}_i + \mathbf{\epsilon}_\mathbf{x},$$
$$\mathbf{A}_\mathbf{z}\mathbf{z}_i = \mathbf{y}_i + \mathbf{\epsilon}_\mathbf{z},$$
with $\mathbf{\epsilon}$ denoting noise.
I wish to minimise the noise (error) by considering both observations $\mathbf{x}_i$ and $\mathbf{z}_i$.
### Questions
How can I combine the observations ($\mathbf{x}$ and $\mathbf{z}$) to obtain a lower error on average?
I have two approaches, neither of which give satisfactory results on average:
- Append the two vectors [$\mathbf{x}$ ; $\mathbf{z}$] and learn a new mapping.
- Use a weighted average of the two estimates.
For (1), do I need to do any preprocessing before appending the vectors, such as spatial alignment (PCA?) or some form of value normalisation?
For (2), is there a way to automatically determine a suitable weighting?
What other ways are likely to yield better results?
How can I analyse the complementary nature of the two observations in relation to the state being estimated?
| How to combine 2 different observations to improve state estimates? | CC BY-SA 3.0 | null | 2011-07-05T16:49:18.447 | 2011-07-06T01:18:59.177 | 2020-06-11T14:32:37.003 | -1 | 3052 | [
"regression",
"machine-learning",
"ridge-regression"
] |
12676 | 1 | null | null | 0 | 201 | I'm reading in a paper "the goal is to cluster data in a way that is meaningful and at the same time as orthogonal as possible to the given classification"
Could someone have any idea how the notion of orthogonality is used in this case?
| Notion of orthogonality in clustering | CC BY-SA 3.0 | null | 2011-07-05T19:03:58.233 | 2018-09-04T16:26:15.430 | null | null | 4549 | [
"clustering",
"data-mining"
] |
12678 | 1 | null | null | 1 | 355 | I would like to calculate the following conditional probability:
I know that the probability of a BLUE ball being drawn is 0.3.
I receive a message from A or B who saw the ball that has been drawn.
This person tells me that the ball drawn is BLUE.
I know the probabilities:
A observing BLUE and sending BLUE- 0.16
B observing RED and sending BLUE - 0.09
A observing BLUE and sending BLUE- 0.05
B observing RED and sending BLUE- 0.17
I also know the probabilities:
A observing BLUE and sending RED - 0.019
B observing RED and sending RED- 0.323
A observing BLUE and sending RED-0.073
B observing RED and sending RED=0.106
I would like to calculate the conditional probability of the ball drawn actually being BLUE given that I received the message BLUE and the information I know about each player's probability of sending a given message.
| probability question | CC BY-SA 3.0 | 0 | 2011-07-05T19:30:40.490 | 2011-07-08T19:26:21.190 | 2011-07-08T19:26:21.190 | 919 | null | [
"probability",
"bayesian",
"self-study"
] |
12679 | 2 | null | 12678 | 0 | null | If the honest person said "blue", and we are not dealing with the possibility of a misperception (honestly getting confused about colors), then there's a 100% chance it's blue.
But I suspect this isn't the question you intended to ask.
| null | CC BY-SA 3.0 | null | 2011-07-05T20:15:12.217 | 2011-07-05T20:15:12.217 | null | null | 3919 | null |
12680 | 2 | null | 165 | 38 | null | Imagine you want to find a better strategy to beat your friends at the board game Monopoly. Simplify the stuff that matters in the game to the question: which properties do people land on most? The answer depends on the structure of the board, the rules of the game and the throws of two dice.
One way to answer the question is this. Just follow a single piece around the board as you throw the dice and follow the rules. Count how many times you land on each property (or program a computer to do the job for you). Eventually, if you have enough patience or you have programmed the rules well enough in you computer, you will build up a good picture of which properties get the most business. This should help you win more often.
What you have done is a Markov Chain Monte Carlo (MCMC) analysis. The board defines the rules. Where you land next only depends on where you are now, not where you have been before and the specific probabilities are determined by the distribution of throws of two dice. MCMC is the application of this idea to mathematical or physical systems like what tomorrow's weather will be or where a pollen grain being randomly buffeted by gas molecules will end up.
| null | CC BY-SA 3.0 | null | 2011-07-05T20:15:25.317 | 2011-07-05T20:15:25.317 | null | null | 5294 | null |
12681 | 1 | 12693 | null | 3 | 3903 | Our lab used to have a program called GS+ that let us make semi variograms on our data and analyse them. Unfortunately, the licence has expired.
Is there another piece of software that lets you do this? If so, how specifically? We prefer a free and open source licence, but all suggestions are welcome.
| Software for making semi variograms and analyses? | CC BY-SA 3.0 | null | 2011-07-05T20:25:43.703 | 2015-11-19T19:23:56.333 | 2015-11-19T19:23:56.333 | 22468 | 2830 | [
"references",
"spatial",
"software",
"interpolation",
"variogram"
] |
12682 | 2 | null | 12566 | 0 | null | You might try comparing...
-the corr. of x with z while controlling for y
-the corr. of y with z while controlling for x
It won't give you a definitive answer, since as @Macro said statistics are limited in that way, but it could give you clues.
| null | CC BY-SA 3.0 | null | 2011-07-05T20:57:06.470 | 2011-07-05T20:57:06.470 | null | null | 2669 | null |
12683 | 1 | null | null | 2 | 421 |
- I have a population of 300 cases. It's split in three sub-populations 50, 50 and 200 in size.
- I have developed three (different) models resulting in a score variable which rank orders each of the sub-populations separately. The score-ranges for all three segments are different ... 40 to 240, 20-410 and 4 to 600. Within each sub-population, higher score means higher level of toxicity i.e. level of risk. Score are not comparable at the moment (that's what I'm trying to achieve).
- I know (a priori) that the absolute level of toxicity for first sub population is highest, average for middle sub-population and lowest for the third sub-population.
What I'm trying to achieve is single rating scale. Ideally, I should be rescale each score variables to reflect my a priori knowledge ... first population should have highest scores, then second and then third ... something like this
```
range A from 40 to 240 to 600 to 1000
range B from 20 to 410 to 400 to 700
range C from 4 to 600 to 0 to 500
```
NOTE: I intentionally allowed for overlaps
I'm currently stuck ... I know its more art than science, and I will appreciate any ideas you may have.
| How to combine three different scores | CC BY-SA 4.0 | null | 2011-07-05T20:59:24.767 | 2022-06-22T20:53:49.037 | 2022-06-22T20:53:49.037 | 11887 | 333 | [
"distributions",
"data-transformation",
"biostatistics"
] |
12684 | 2 | null | 12678 | 1 | null | If Truth/Lie and color are independent and everything is random, then you would have an 18% chance of an honest person seeing blue and a 28% chance of a lier seeing yellow (and therefore claiming to see blue), you also have a 12% chance of lier seeing blue and 42% chance of honest seeing yellow. So the probability of an honest person seeing blue given that the person claimed to see blue is $\frac{0.18}{0.18+0.28} = 0.39$.
| null | CC BY-SA 3.0 | null | 2011-07-05T20:59:57.837 | 2011-07-05T20:59:57.837 | null | null | 4505 | null |
12685 | 2 | null | 12681 | 1 | null | The [R](http://www.r-project.org/) program is free and open source, there are functions in the nlme package for R that do semi-variograms (and probably other options as well, but nlme came up in a quick search).
| null | CC BY-SA 3.0 | null | 2011-07-05T21:08:27.707 | 2011-07-05T21:08:27.707 | null | null | 4505 | null |
12686 | 2 | null | 12681 | 4 | null | package `geoR` for `R` could be what you're looking for. From the documentation:
"function `variogram` Computes sample (empirical) variograms with options for the classical or robust estimators. Output can be returned as a binned variogram, a variogram cloud or a smoothed variogram. Data transformation (Box-Cox) is allowed. “Trends” can be specified and are fitted by ordinary least squares in which case the variograms are computed using the residuals."
Also `variofit`: "Estimate covariance parameters by fitting a parametric model to a empirical variogram. Variograms models can be fitted by using weighted or ordinary least squares."
There are more geostatistical packages in `R`.
| null | CC BY-SA 3.0 | null | 2011-07-05T21:08:59.170 | 2011-07-05T21:08:59.170 | null | null | 5020 | null |
12687 | 1 | 12732 | null | 18 | 5415 | I am looking for papers or texts that compare and discuss (either empirically or theoretically):
- Boosting and Decision trees algorithms such as Random Forests or AdaBoost, and GentleBoost applied to decision trees.
with
- Deep learning methods such as Restricted Boltzmann Machines, Hierarchical Temporal Memory, Convolutional Neural Networks, etc.
More specifically, does anybody know of a text that discusses or compares these two blocks of ML methods in terms of speed, accuracy or convergence? Also, I am looking for texts that explain or summarize the differences (e.g. pros and cons) between the models or methods in the second block.
Any pointers or answers addressing such comparisons directly would be greatly appreciated.
| Deep learning vs. Decision trees and boosting methods | CC BY-SA 3.0 | null | 2011-07-05T21:11:53.197 | 2017-10-28T16:52:19.093 | 2017-10-28T16:52:19.093 | 128677 | 2798 | [
"machine-learning",
"deep-learning",
"cart",
"restricted-boltzmann-machine",
"adaboost"
] |
12690 | 2 | null | 9872 | 1 | null | You could specify your binding algorithm in a function, define utility function and optimize input parameters...
The ideas for utility function can be:
- Predictive power (weight of evidence and information value)
- Monotonnicly decreasing average default rate from one bin to another (as you increase the age of history...)
You can also constrain your optimization to look only for three to 5 bins for example...
| null | CC BY-SA 3.0 | null | 2011-07-05T21:19:46.747 | 2011-07-05T21:19:46.747 | null | null | 333 | null |
12691 | 2 | null | 12681 | 5 | null | The most comprehensive list has long been maintained at the [AI Geostats pages](http://www.ai-geostats.org/bin/view/AI_GEOSTATS/SoftwareGeostatistical): it includes freeware and commercial software (almost 30 free ones). That page has a [separate link](http://www.ai-geostats.org/bin/view/AI_GEOSTATS/SoftwareCodes) to add-ons, such as any R packages.
AFAIK, there is no R package for handling 3D variography or change of support problems. The R packages are also weak at modeling rich datasets, which often need nested models with varying anisotropy and variogram shapes. For good integration of variography, kriging, simulation, EDA, and statistical modeling, though, the R packages are excellent.
| null | CC BY-SA 3.0 | null | 2011-07-05T21:22:31.813 | 2011-07-05T21:22:31.813 | null | null | 919 | null |
12692 | 2 | null | 12670 | 8 | null |
- Instructions for using R to download netCDF data can be found here, a common format used for storing Earth science data, e.g. as in marine geospatial data from OpenEarth or climate model driver and forecasts from UCAR
- rnpn (under development) enables you to get data from the National Phenology Network - a citizen science project to track the timing of plant green-up, flowering, and senescence. See the developer's blog post.
- -obsolete- RClimate provides tools to download and manipulate flat-file climate data (with tutorials, including here-
- Download historical finance data with tseries::get.hist.quote
- Michael Samuel's documents downloading public health data
- raster::getData provides access to climate variables via worldclim
| null | CC BY-SA 3.0 | null | 2011-07-05T22:15:55.230 | 2014-03-26T03:42:36.527 | 2014-03-26T03:42:36.527 | 1381 | 1381 | null |
12693 | 2 | null | 12681 | 4 | null | Besides R, here are a few links that may be of interest that do not appear to be on the AI Geostats page:
- SpaceStat (not free)
- Passage software (haven't tried it, looks like it has functionality you stated in your question plus some more)
- High performance Geostat library (looks like GUI is in the works, but only beta versions of it are available)
- SAM (Spatial Analysis in Macroecology). It has a kriging module so I assume it must allow you to estimate the semi-variogram.
- Pierre Goovaerts has a book chapter review on Geostatistical software publicly listed on his website. (A note, he works for the makers of the above SpaceStat software, so unsurprisingly he gives it glowing reviews.)
The book, [A Practical Guide to Geostatistical Mapping](http://spatial-analyst.net/book/), has relevent examples using R and Google Earth (of course R is used to estimate the variograms).
| null | CC BY-SA 3.0 | null | 2011-07-06T00:37:17.370 | 2011-07-06T00:37:17.370 | null | null | 1036 | null |
12694 | 2 | null | 1781 | 2 | null | How many compounds are in your sample? (Or, how big is the covariance matrix in question?).
Alan Genz has some very nice code in a variety of languages (R, Matlab, Fortran; see [here](http://www.math.wsu.edu/faculty/genz/software/software.html)) for computing integrals of multivariate normal densities over hyper-rectangles (i.e., the kinds of integrals you need to evaluate the likelihood, as noted by user28).
I've used these functions ("ADAPT" and "QSIMVN") for integrals up to around 10-12 dimensions, and several functions on that page advertise integrals (and associated derivatives you might need) for problems up to dimension 100. I don't know if that's enough dimensions for your purposes, but if so it could presumably allow you to find maximum likelihood estimates by gradient ascent.
| null | CC BY-SA 3.0 | null | 2011-07-06T00:38:28.180 | 2011-07-06T00:38:28.180 | null | null | 5289 | null |
12695 | 2 | null | 10361 | 3 | null | It is simpler to start from scratch in your part 2, rather than proceed sequentially. I will show the answer this way, and then show where the step was "wrong" in your approach.
The independence means:
$$P(A_+B_-|c)=P(A_+|c)P(B_-|c)=P(B_-|c)=50\text{%}$$
$$P(A_+B_-|\neg c)=P(A_+|\neg c)P(B_-|\neg c)=P(A_+|\neg c)=10\text{%}$$
Where we have inserted the information given in the question (A never has false negative but 10% false positive, B never has false positive, but 50% false negative). You can use this to calculate the marginal probabilities
$$P(A_+B_-)=P(A_+B_-|c)P(c)+P(A_+B_-|\neg c)P(\neg c)=50\text{%}P(c)+10\text{%}P(\neg c)=14\text{%}$$
Where we have used $P(c)=10\text{%}$, we then have:
$$P(c|A_+B_-)=P(c)\frac{P(A_+B_-|c)}{P(A_+B_-)}=10\text{%}\frac{50\text{%}}{14\text{%}}=36\text{%}$$
Note that this is not derivable from $P(c|A_+)$ and $P(c|B_-)$ alone. The reason is that while $A_+$ and $B_-$ are conditionally independent, they are not unconditionally independent.
Now if we start from $P(c|A_+)=52\text{%}$, we need to multiply by the ratio $$\frac{P(B_-|A_+c)}{P(B_-|A_+)}=\frac{P(B_-|c)}{P(B_-|A_+)}$$
The likelihood when $c$ is true is unchanged because we have the logical relation $c\implies A_+$ which is the same thing as $c\land A_+=c$.
However, to calculate $P(B_-|A_+)$ we have:
$$P(B_-|A_+)=P(B_-|A_+c)P(c|A_+)+P(B_-|A_+\neg c)P(\neg c|A_+)$$
$$=P(B_-|c)P(c|A_+)+\frac{P(B_-A_+|\neg c)}{P(A_+|\neg c)}P(\neg c|A_+)$$
$$=50\text{%}\times 0.52+\frac{10\text{%}}{10\text{%}}\times 0.48=74\text{%}$$
And this is not the same as $P(B_-)=95\text{%}$. Plugging this in we get:
$$p(c|A_+B_-)=52\text{%}\times\frac{50\text{%}}{74\text{%}}=36\text{%}$$
Which is the same the earlier calculation I gave before.
UPDATE
@dmk has asked for some additional explanation about why this answer is intuitively the correct one not the answer given by the more straight-forward:
$$\frac{P(c|A_+B_-)}{P(\neg c|A_+B_-)}=\frac{P(c)}{P(\neg c)}\frac{P(A_+|c)}{P(A_+|\neg c)}\frac{P(B_-|c)}{P(B_-|\neg c)}$$
The answer is: they give the same answer!
$$\frac{P(c|A_+B_-)}{P(\neg c|A_+B_-)}=\frac{0.1}{0.9}\frac{1}{0.1}\frac{0.5}{1}=0.555\dots$$
Converting this to probabilities and we get $\frac{0.555}{1+0.555}=36\text{%}$
So I must retract my comments to @dmk's answer, noting that his procedure was correct, but his calculation was not. $B_-$ is twice as likely to occur if $c$ is not present compared to if it is present. So in his calculation, we modify from $10:9$ odds (after observing $A_+$) to $5:9$ odds (after observing $B_-$)
| null | CC BY-SA 3.0 | null | 2011-07-06T00:50:09.513 | 2011-07-06T16:44:37.127 | 2011-07-06T16:44:37.127 | 2392 | 2392 | null |
12696 | 2 | null | 1781 | 4 | null | Another more computationally efficient option would be to fit the covariance matrix by moment matching using a model that has been called the "dichomized Gaussian", really just a Gaussian copula model.
A recent paper from [Macke et al 2010](http://www.mitpressjournals.org/doi/abs/10.1162/neco.2008.02-08-713) describes a closed form procedure for fitting this model which involves only the (censored) empirical covariance matrix and the calculation of some bivariate normal probabilities. The same group (Bethge lab at MPI Tuebingen) has also described hybrid discrete/continuous Gaussian models which are probably what you want here (i.e., since the Gaussian RVs aren't fully "dichotomized" -- only those below threshold).
Critically, this is not an ML estimator, and I'm afraid I don't know what its bias properties are.
| null | CC BY-SA 3.0 | null | 2011-07-06T00:54:48.573 | 2011-07-06T00:54:48.573 | null | null | 5289 | null |
12697 | 2 | null | 12675 | 1 | null | First of all, the training error obtained by solution #1 (concatenating $x$ and $z$) can't be higher than the error obtained by regression against $x$ or $z$ alone. But of course, the test error (i.e., prediction performance on new data) may well be worse.
If I were confronted with this problem, I would normalize $x$ and $z$ to have the same marginal variance, then concatenate them and then attempt to vary the ridge parameter $\lambda$ -- I would bet that for some setting of $\lambda$ you can find performance that's better than you get from regressing against $x$ or $z$ separately.
However, if $x$ and $z$ are highly correlated, then you might wish to try [Principal Components Regression](http://en.wikipedia.org/wiki/Principal_component_regression) instead. The basic idea is to compute the principal components of $[x;z]$, project onto the subspace spanned by the first $n$ principal components, then perform standard regression using the projected data as your regressors. You can now vary $n$ and the ridge parameter (if you like) to explore the space of possible solutions using cross-validation.
You can of course also try other regularization methods: Lasso (L1 regularization) is more appropriate if you expect your regression coefficients to be sparse (several large values and many zeros); "elastic net" trades off Ridge and Lasso (L1 and L2) forms of regularization.
| null | CC BY-SA 3.0 | null | 2011-07-06T01:12:16.670 | 2011-07-06T01:18:59.177 | 2011-07-06T01:18:59.177 | 5289 | 5289 | null |
12698 | 2 | null | 10598 | 2 | null | I think there's a slight conceptual problem here: the likelihood is the quantity $P(x,y|\theta)$ as a function of $\theta$. But the correlation coefficient is a function of the data $x$ and $y$. The correlation coefficient is therefore by definition a "statistic" (a function of the data) and not a likelihood (a function of the parameters).
To put it another way, there are values of $\theta$ for which the probability of the observed correlation would be high, and other values of $\theta$ for which the observed correlation would be low. If you have a correlated Gaussian in mind, then the observed correlation might be unlikely due to the fact that the covariance matrix was either more dependent or less dependent than would be expected from the data. (So the relationship between likelihood and $r^2$ will not even be monotonic unless $r^2=1$). Of course, the expected correlation coefficient is monotonically related to the off-diagonal term in a bivariate Gaussian, but this doesn't make $r^2$ itself a proxy for likelihood.
It sounds like you're interested in a test of whether a particular correlation coefficient is likely to have been observed given that $x$ and $y$ were generated independently. For this, you might want to perform a significance test on the value of the observed $r$ or $r^2$, which will tell you how probable it is that you observed that value given a particular model of the data. The Wikipedia article on correlation coefficient [inference](http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient#Inference) discusses a few common tests.
| null | CC BY-SA 3.0 | null | 2011-07-06T01:43:28.030 | 2011-07-06T01:43:28.030 | null | null | 5289 | null |
12699 | 2 | null | 12566 | 1 | null | I think you're looking for an [instrumental variable](http://en.wikipedia.org/wiki/Instrumental_variable) test, which will allow you to assess whether $y$ mediates the correlation between $x$ and $z$. The section on "interpretation as two-stage least squares" provides a nice intuition for this procedure.
| null | CC BY-SA 3.0 | null | 2011-07-06T03:54:13.137 | 2011-07-06T03:54:13.137 | null | null | 5289 | null |
12700 | 1 | null | null | 1 | 708 | I want to calculate Kullback–Leibler divergence on two documents. It requires probability distribution of each document.
I do not know how to calculate probability for each document. Any simple answer with layman example would be much appreciated.
Let's say we have follow two documents:
```
1 - cross validated answers are good
2 - simply validated answers are nice
```
(wording of the documents is just bla bla to give you an example)
How do we calculate probabilities for these documents?
Let's say we add one more document:
```
3 - simply cross is not good answer
```
If we add another document then how would it impact probability distribution?
Thanks
| How to to calculate the topic distribution of a document | CC BY-SA 4.0 | null | 2011-07-06T04:16:14.407 | 2019-12-16T00:10:29.153 | 2019-12-16T00:10:29.153 | 92235 | 3900 | [
"probability",
"mathematical-statistics",
"kullback-leibler",
"information-retrieval"
] |
12701 | 2 | null | 1257 | 1 | null | This is all correct, except you don't need the last term: $P(X_3=X_{31})$. You observe the data $Y_1$ and $Y_2$, so you know whether $X_3=X_{31}$ or $X_3=X_{32}$.
Another way to see this is that the likelihood function for this system is is:
$L(\theta) = P(Y_1,Y_2,Y_3|\theta) = P(Y_1|\theta)P(Y_2|\theta)P(Y_3|Y_1,Y_2,\theta)$.
The right most term is:
$P(Y_3|Y1,Y2,\theta) =
\begin{cases}
f(Y_3|X_{31},-) & \mbox{if }Y_1 Y_2 \gt 0 \\ f(Y_3|X_{32},-) & \mbox{if }Y_1 Y_2 \le 0
\end{cases}$
You could write that out using an indicator function of $Y_1Y_2$ if you wanted to,
$f(Y_3|X_{31},-)\cdot \mathbb{I}(Y_1Y_2>0) + f(Y_3|X_{32},-)\cdot \mathbb{I}(Y_1Y_2\leq 0)$, where $\mathbb{I}(\cdot)$ is 1 if its argument is true and 0 otherwise.
| null | CC BY-SA 3.0 | null | 2011-07-06T04:28:54.450 | 2011-07-06T04:28:54.450 | null | null | 5289 | null |
12702 | 1 | null | null | 43 | 49545 | Consider a scenario where you are provided with KnownLabel Matrix and PredictedLabel matrix. I would like to measure the goodness of the PredictedLabel matrix against the KnownLabel Matrix.
But the challenge here is that KnownLabel Matrix have few rows only one 1 and other few rows have many 1's (those instance are multi labeled). An example of KnownLabel Matrix is given below.
```
A =[1 0 0 0
0 1 0 0
0 1 1 0
0 0 1 1
0 1 1 1]
```
In the above matrix, data instance 1 and 2 are single label data, data instance 3 and 4 are two label data and data instance 5 is the three label data.
Now I have PredictedLabel Matrix of data instance using an algorithm.
I would like to know various measure which can be used to measure the goodness of the PredictedLabel Matrix against KnownLabel Matrix.
I can think of frobeinus norm difference between them as one of the measure.
But im looking for the measure such as accuracy $(= \frac{\text{Correctly_predicted_instance}}{\text{total_instance}})$
Here how do can we define the $\rm Correctly\_predicted$ for multiple data instance?
| What are the measure for accuracy of multilabel data? | CC BY-SA 3.0 | null | 2011-07-06T05:05:12.360 | 2015-08-27T03:36:16.367 | 2015-08-26T23:24:50.793 | 12359 | 4290 | [
"machine-learning",
"data-mining",
"multilabel"
] |
12703 | 1 | null | null | 1 | 792 | I am performing some analysis on understanding various characteristics of a certain population. To narrow down my selection (as I will be going through them manually), I am using different thresholds on certain measurable properties (which tell me how relevant someone is) to draw the following frequency graph. For instance, the line marked `Threshold 1` tells me that when I fix my threshold at `t1`, there are approximately 360 people with at least 1 friend, 290 people with at least 2 friends etc.
I am aware that I will be having a hard time explaining what I am doing but I will try my best. I am trying to fix two thresholds: one that is used to obtain the line and the other that I use to filter noise. The '0' on the graph marks the thresholds. In this particular case, I am using `Threshold 3` and filtering my population to contain only those people that have `>= 2 friends`. While this is working well for me, I am trying to determine if there is systematic procedure to explain why these thresholds are better than others. For instance, `Threshold 2` is not such a good filter because it behaves like `Threshold 1`. Utilizing `Threshold 4` is not so useful because the returns in going from `Threshold 2` to `Threshold 3` is much higher than that going from `Threshold 3` to `Threshold 4`.
Any suggestions on where I should start? Basically, I want to do something similar to cost-benefit analysis (that's the closest I could think off).
| Systematic way of choosing a threshold? | CC BY-SA 3.0 | null | 2011-07-06T07:06:21.887 | 2011-07-06T07:06:21.887 | null | null | 2164 | [
"r",
"statistical-significance",
"dataset",
"data-mining",
"threshold"
] |
12704 | 1 | null | null | 2 | 269 | I have made several experiments, the result graphs are below, I need to figure out what is the distribution function of the following random variables... Where do I start with statistical analysis of my experiment?
How to perform these analysis with spss?
EDIT
Explanations: I'm testing algorithm efficiency for 2 algorithms, the graphs describe efficiency distribution.
- I made 500000 tests for 2 algorithms, which produced
2 sets of 500000 floating point values.
- I've computed MIN and MAX value of
the set (The smaller the values the better the result is, the bigger the values the worser the result is)
- I wanted to see the distribution of values according to 500 groups, so I've computed step = (MAX-MIN) / 500
- I've computed the amount of values that fall into each of 500 segments
[MIN + 0 * Step..MIN + 1 * Step],
[MIN + 1 * Step, MIN + 2 * Step], ...
[MIN + 498 * Step..MIN + 499 * Step]
The graph you see below represents on x-axis an index of the segment, on y-axis the amount values fallen into the segment divided by the total number of values, and also divided by the `Step = (Max - Min) / 500`. As a result this should represent the distribution function of my algorithm efficiency...
EDIT [Here](https://docs.google.com/uc?id=0ByVHzYDa6azxNzEwNDc0N2EtNjFhNy00M2ZiLWJmNjYtZjVkYTQwMGU1NDUz&export=download&hl=uk) is DOC containing SPSS > Analyze > Descriptive Statistics > Explore output. Can anybody help reading that?
| What distribution function might describe the following statistical experiments? | CC BY-SA 3.0 | null | 2011-07-06T07:26:32.410 | 2011-07-07T08:18:39.090 | 2011-07-06T20:52:10.953 | 930 | 5300 | [
"distributions",
"spss"
] |
12706 | 1 | null | null | 1 | 1892 | I have 5 single factor regression models. Each has intuitive (positive) coefficients which is good. If I run multiple logistic regression one of the coefficients turns negative.
If I run stepwise selection, the negative coefficient stays in ... it's not redundant ...
How should I deal with this? Is there a way to enforce positive coefficients?
| How to deal with negative coefficients in logistic regression | CC BY-SA 4.0 | null | 2011-07-06T10:23:10.323 | 2022-06-16T19:19:51.230 | 2022-06-16T19:19:51.230 | 11887 | 333 | [
"logistic",
"multicollinearity",
"regression-coefficients"
] |
12709 | 1 | 12718 | null | 3 | 7128 | I have a large dataset and have performed a multilevel regression in Stata, the model is the following:
`xtmixed dependent independen1 independent2 independent3 independent4 || independendt5:`
So there is one grouping factor: `independent5`
In R I did the following:
```
lmer(dependent ~ independent1 + indepdendent2 + independent3 + independent4 + 1 | independent5, REML=TRUE)
```
A few questions: is this identical?
Stata output gives me number of groups, 100, while R gives number of groups as 99.
Furthermore, the variances and standard deviations are not the same.
Also I would like to know how to obtain p values and coefficients from the R ouput.
I have done `fixef(model)` and `ranef(model)` but when I do `coef(model)` it says: `Error in coef(model) : unable to align random and fixed effects`
Also Stata only gives me 1 coefficient for each predictor while the R `fixef(model)` gives one for each group.
So someone familiar with both Stata and R could help me with this.
| Multilevel regression using lmer function in R and Stata | CC BY-SA 3.0 | null | 2011-07-06T12:38:53.277 | 2011-07-06T16:01:52.827 | 2011-07-06T12:53:53.063 | 1036 | null | [
"r",
"stata",
"multilevel-analysis",
"lme4-nlme"
] |
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