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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
8013 | 1 | null | null | 9 | 1911 | I have data that is equivalent to:
```
shopper_1 = ['beer', 'eggs', 'water',...]
shopper_2 = ['diapers', 'beer',...]
...
```
I would like to do some analysis on this data set to get a correlation matrix that would have an implication similar to: if you bought x, you are likely to buy y.
Using python (or perhaps anythi... | How to do a 'beer and diapers' correlation analysis | CC BY-SA 2.5 | null | 2011-03-08T12:51:04.077 | 2011-04-11T16:13:02.430 | 2011-03-09T05:54:19.903 | 3618 | 3618 | [
"correlation",
"econometrics",
"python",
"cross-correlation"
] |
8014 | 2 | null | 8000 | 51 | null | What you describe is in fact a "sliding time window" approach and is different to recurrent networks. You can use this technique with any regression algorithm. There is a huge limitation to this approach: events in the inputs can only be correlatd with other inputs/outputs which lie at most t timesteps apart, where t i... | null | CC BY-SA 2.5 | null | 2011-03-08T12:51:29.233 | 2011-03-08T15:43:55.373 | 2011-03-08T15:43:55.373 | 2860 | 2860 | null |
8015 | 1 | null | null | 12 | 3115 | I was wondering if anyone could help me with information about kurtosis (i.e. is there any way to transform your data to reduce it?)
I have a questionnaire dataset with a large number of cases and variables. For a few of my variables, the data shows pretty high kurtosis values (i.e. a leptokurtic distribution) which is... | Treatment of outliers produced by kurtosis | CC BY-SA 4.0 | null | 2011-03-08T13:16:00.873 | 2020-09-27T09:03:05.620 | 2020-09-27T09:03:05.620 | 22047 | 3619 | [
"distributions",
"assumptions",
"discriminant-analysis",
"kurtosis"
] |
8016 | 2 | null | 8010 | 4 | null | The initial problem is:
$$\min_{f\in H}\frac{1}{n}\sum_{i=1}^{n}\phi\left(y_{i}f\left(x_{i}\right)\right)+\lambda\left\Vert f\right\Vert _{H}^{2}$$
$$\lambda\geq0$$
$$\phi\left(u\right)=\max\left(1-u,\,0\right)^{2}$$
---
Since, $f=\sum_{i=1}^{n}\alpha_i K_{x_i}$ and we are considering a RKHS, the primal problem is:
... | null | CC BY-SA 2.5 | null | 2011-03-08T13:19:37.123 | 2011-03-09T18:19:41.447 | 2011-03-09T18:19:41.447 | 1351 | 1351 | null |
8017 | 2 | null | 7989 | 5 | null | One thing that you must recognise is that the term "noise" is relative not absolute. Thinking about an obvious example (not anything to do with statistics per se), imagine you are at a night club. What is "noise" in here?
If you trying to have a conversation with someone, then the music is "noise", and so are the o... | null | CC BY-SA 2.5 | null | 2011-03-08T13:35:37.283 | 2011-03-08T13:35:37.283 | null | null | 2392 | null |
8018 | 2 | null | 8015 | 10 | null | The obvious "common sense" way to resolving your problem is to
- Get the conclusion using the full data set. i.e. what results will you declare ignoring intermediate calculations?
- Get the conclusion using the data set with said "outliers" removed. i.e. what results will you declare ignoring intermediate calculati... | null | CC BY-SA 2.5 | null | 2011-03-08T14:11:17.500 | 2011-03-08T14:11:17.500 | null | null | 2392 | null |
8019 | 1 | 8023 | null | 26 | 34009 | Let's say I test how variable `Y` depends on variable `X` under different experimental conditions and obtain the following graph:
The dash lines in the graph above represent linear regression for each data series (experimental setup) and the numbers i... | Averaging correlation values | CC BY-SA 2.5 | null | 2011-03-08T15:06:44.510 | 2019-01-11T05:47:12.990 | 2011-03-08T15:57:47.940 | 919 | 1496 | [
"regression",
"correlation",
"mean"
] |
8021 | 1 | null | null | 12 | 13977 | I have posted a [previous question](https://stats.stackexchange.com/questions/7977/how-to-generate-uniformly-distributed-points-on-the-surface-of-the-3-d-unit-spher), this is related but I think it is better to start another thread. This time, I am wondering how to generate uniformly distributed points inside the 3-d u... | How to generate uniformly distributed points in the 3-d unit ball? | CC BY-SA 3.0 | null | 2011-03-08T15:34:34.147 | 2018-06-26T19:35:56.567 | 2017-04-13T12:44:23.203 | -1 | 3552 | [
"random-generation"
] |
8022 | 2 | null | 8021 | 16 | null | The easiest way is to sample points uniformly in the corresponding hypercube and discard those that do not lie within the sphere. In 3D, this should not happen that often, about 50% of the time. (Volume of the hypercube is 1, volume of the sphere is $\frac{4}{3}\pi r^3 = 0.523...$.)
| null | CC BY-SA 2.5 | null | 2011-03-08T15:52:11.650 | 2011-03-08T15:52:11.650 | null | null | 2860 | null |
8023 | 2 | null | 8019 | 17 | null | The simple way is to add a categorical variable $z$ to identify the different experimental conditions and include it in your model along with an "interaction" with $x$; that is, $y \sim z + x\#z$. This conducts all five regressions at once. Its $R^2$ is what you want.
To see why averaging individual $R$ values may be... | null | CC BY-SA 3.0 | null | 2011-03-08T15:57:22.620 | 2018-03-13T17:55:04.707 | 2018-03-13T17:55:04.707 | 919 | 919 | null |
8025 | 1 | 8026 | null | 45 | 63590 | Precision is defined as:
>
p = true positives / (true positives + false positives)
What is the value of precision if (true positives + false positives) = 0? Is it just undefined?
Same question for recall:
>
r = true positives / (true positives + false negatives)
In this case, what is the value of recall if (true ... | What are correct values for precision and recall when the denominators equal 0? | CC BY-SA 2.5 | null | 2011-03-08T16:31:51.660 | 2017-10-02T11:17:57.327 | 2017-04-13T12:44:40.807 | -1 | 3604 | [
"precision-recall"
] |
8026 | 2 | null | 8025 | 21 | null | The answers to the linked earlier question apply here too.
If (true positives + false negatives) = 0 then no positive cases in the input data, so any analysis of this case has no information, and so no conclusion about how positive cases are handled. You want N/A or something similar as the ratio result, avoiding a di... | null | CC BY-SA 2.5 | null | 2011-03-08T17:02:35.940 | 2011-03-08T17:02:35.940 | null | null | 2958 | null |
8028 | 1 | null | null | 0 | 128 | My second question is about Test Statistics.
The questions and answers are on this PDF: [http://www.mediafire.com/?b74e633lxdb49rb](http://www.mediafire.com/?b74e633lxdb49rb)
I understand that they work out the join density and I know how to calculate this.
What I really don’t understand is how they figure out the test... | Test Statistics | CC BY-SA 2.5 | null | 2011-03-08T18:35:12.617 | 2011-03-08T19:20:35.300 | null | null | null | [
"descriptive-statistics"
] |
8029 | 1 | 8030 | null | 3 | 2370 | I am writing a program in C# that requires me to use the Ttest formula. I have to effectively interpret the Excel formula:
```
=TTEST(range1,range2,1,3)
```
I am using the formula given [here](http://www.monarchlab.umn.edu/lab/research/stats/2SampleT.aspx)
and have interpreted into code as such:
```
double TStatisti... | How to implement formula for a independent groups t-test in C#? | CC BY-SA 3.0 | null | 2011-03-08T18:17:59.787 | 2013-12-18T11:17:36.437 | 2011-09-23T05:42:13.263 | 183 | 3624 | [
"t-test"
] |
8030 | 2 | null | 8029 | 5 | null | There are at least two problems with what you have done.
- You have misinterpreted the formula
$$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{s_1^2 / n_1 + s_2^2 / n_2}}$$ since $s^2$ is already a variance (square of standard deviation) and does not need to be squared again.
- You are comparing eggs and omelettes: you ne... | null | CC BY-SA 2.5 | null | 2011-03-08T19:04:14.603 | 2011-03-08T19:04:14.603 | null | null | 2958 | null |
8032 | 2 | null | 8028 | 2 | null | They don't randomly multiply by 1. What they do is split the joint density into the product of two functions: $g(\text{sufficient statistic},\text{parameter})$ and $h(\text{data})$.
The advantage of $h()$ is that it removes sometimes complicated parts of the density function which provide no useful information about... | null | CC BY-SA 2.5 | null | 2011-03-08T19:20:35.300 | 2011-03-08T19:20:35.300 | null | null | 2958 | null |
8033 | 1 | null | null | 4 | 4607 | A lecturer wishes to "grade on the curve". The students' marks seem to be normally distributed with mean 70 and standard deviation 8. If the lecturer wants to give 20% A's, what should be the threshold between an A grade and a B grade?
| How to find percentiles of a Normal distribution? | CC BY-SA 3.0 | null | 2011-03-08T20:00:17.407 | 2011-11-16T20:37:44.703 | 2011-11-16T20:37:44.703 | 919 | null | [
"self-study",
"normal-distribution"
] |
8034 | 2 | null | 8013 | 7 | null | In addition to the links that were given in comments, here are some further pointers:
- Association rules and frequent itemsets
- Survey on Frequent Pattern Mining -- look around Table 1, p. 4
About Python, I guess now you have an idea of what you should be looking for, but the [Orange](http://orange.biolab.si/) da... | null | CC BY-SA 3.0 | null | 2011-03-08T20:59:01.840 | 2011-04-11T16:13:02.430 | 2011-04-11T16:13:02.430 | 930 | 930 | null |
8035 | 2 | null | 7975 | 4 | null | When you have supporting/causal/helping/right-hand side/exogenous/predictor series, the approach that is preferred is to construct a single equation, multiple-input Transfer Function. One needs to examine possible model residuals for both unspecified/omitted deterministic inputs i.e. do Intervention Detection ala Ruey ... | null | CC BY-SA 2.5 | null | 2011-03-08T21:06:10.590 | 2011-03-08T21:06:10.590 | null | null | 3382 | null |
8036 | 1 | null | null | 3 | 170 | I have gazillion documents (your tax return) that I need to check for correctness, but I don't the manpower nor the will power to read through all of it. Even if I do, I can not guarantee the quality and consistency of the proof reading process.
The only thing I can do is to pick a sample collection of document to p... | Proofreading lots of documents based on small sample | CC BY-SA 2.5 | null | 2011-03-08T21:08:10.473 | 2011-03-31T12:47:51.767 | 2011-03-31T07:36:08.540 | null | 3625 | [
"confidence-interval",
"quality-control"
] |
8037 | 2 | null | 7975 | 22 | null | Based upon the comments that you've offered to the responses, you need to be aware of spurious causation. Any variable with a time trend is going to be correlated with another variable that also has a time trend. For example, my weight from birth to age 27 is going to be highly correlated with your weight from birth to... | null | CC BY-SA 2.5 | null | 2011-03-08T22:02:19.000 | 2011-03-08T22:02:19.000 | null | null | 401 | null |
8040 | 1 | 8048 | null | 11 | 108357 | Is there an easy way to apply the trend line formula from a chart to any given X value in Excel?
For example, I want to get the Y value for a given X = $2,006.00. I've already taken the formula and retyped it out be:
```
=-0.000000000008*X^3 - 0.00000001*X^2 + 0.0003*X - 0.0029
```
I am continually making adjustments ... | Use a trendline formula to get values for any given X with Excel | CC BY-SA 2.5 | null | 2011-03-08T22:32:45.597 | 2011-03-11T00:04:11.817 | 2011-03-11T00:04:11.817 | null | 3626 | [
"regression",
"excel"
] |
8041 | 1 | null | null | 3 | 1918 | I have a bunch of experiments in which I am calculating precision and recall. I want to present a mean precision and recall for these experiments. Should these values be weighted by anything?
| Should the mean precision and recall be weighted? | CC BY-SA 2.5 | null | 2011-03-08T22:40:06.737 | 2011-09-08T19:46:24.037 | null | null | 3604 | [
"precision-recall"
] |
8042 | 2 | null | 7996 | 3 | null | EDIT: After some reflection, I modified my answer substantially.
The best thing to do would be to try to find a reasonable model for your data (for example, by using multiple linear regression). If you cannot get enough data to do this, I would try the following "non-parametric" approach. Suppose that in your data se... | null | CC BY-SA 2.5 | null | 2011-03-08T22:49:23.450 | 2011-03-08T23:40:46.820 | 2011-03-08T23:40:46.820 | 3567 | 3567 | null |
8043 | 2 | null | 7996 | 5 | null | A few comments:
- Why did you go with your particular experimental design set-up? For example, fix A+B and vary C. What would you fix A + B at? If you are interesting in determining the effect of A and B, it seems a bit strange that you can fix them at "optimal values". There are standard statistical techniques for sa... | null | CC BY-SA 2.5 | null | 2011-03-08T23:15:32.683 | 2011-03-09T12:46:36.880 | 2020-06-11T14:32:37.003 | -1 | 8 | null |
8044 | 1 | 8050 | null | 18 | 2425 | I use R. Every day. I think in terms of data.frames, the apply() family of functions, object-oriented programming, vectorization, and ggplot2 geoms/aesthetics. I just started working for an organization that primarily uses SAS. I know there's a book about [learning R for SAS users](http://rads.stackoverflow.com/amzn/cl... | Resources for an R user who must learn SAS | CC BY-SA 2.5 | null | 2011-03-08T23:27:13.440 | 2011-03-09T14:29:35.183 | null | null | 36 | [
"r",
"sas"
] |
8045 | 5 | null | null | 0 | null | Overview
A [distribution](http://en.wikipedia.org/wiki/Distribution) is a mathematical description of probabilities or frequencies. It can be applied to observed frequencies, estimated probabilities or frequencies, and theoretically hypothesized probabilities or frequencies. Distributions can be univariate, describin... | null | CC BY-SA 3.0 | null | 2011-03-08T23:41:16.543 | 2013-09-01T23:15:41.880 | 2013-09-01T23:15:41.880 | 27581 | 919 | null |
8046 | 4 | null | null | 0 | null | A distribution is a mathematical description of probabilities or frequencies. | null | CC BY-SA 3.0 | null | 2011-03-08T23:41:16.543 | 2014-05-22T03:02:41.767 | 2014-05-22T03:02:41.767 | 7290 | 919 | null |
8048 | 2 | null | 8040 | 16 | null | Use `LINEST`, as shown:
The method is to create new columns (C:E here) containing the variables in the fit. Perversely, `LINEST` returns the coefficients in the reverse order, as you can see from the calculation in the "Fit" column. An example predi... | null | CC BY-SA 2.5 | null | 2011-03-09T00:12:52.290 | 2011-03-09T00:12:52.290 | null | null | 919 | null |
8049 | 2 | null | 8040 | 8 | null | Try trend(known_y's, known_x's, new_x's, const).
Column A below is X. Column B is X^2 (the cell to the left squared). Column C is X^3 (two cells to the left cubed). The trend() formula is in Cell E24 where the cell references are shown in red.
The "known_y's" are in E3:E22
The "known_x's" are in A3:C22
The "new_x's... | null | CC BY-SA 2.5 | null | 2011-03-09T00:46:15.910 | 2011-03-09T01:42:27.743 | 2011-03-09T01:42:27.743 | 2775 | 2775 | null |
8050 | 2 | null | 8044 | 20 | null | 15 months ago, I started my current job as someone who had been using R exclusively for about 3 years; I had used SAS in my first-ever stats class, loathed it, and never touched it again until I started here. Here's what has been helpful for me, and what hasn't:
Helpful:
---
- Colleagues' code. This is the single... | null | CC BY-SA 2.5 | null | 2011-03-09T00:47:43.870 | 2011-03-09T01:07:51.007 | 2011-03-09T01:07:51.007 | 71 | 71 | null |
8051 | 2 | null | 7990 | 3 | null | EDIT 2: We have to simplify the problem by removing the restriction that $x \ne y$ always in case two. Otherwise correlation issues vastly complicate the answer.
Let $|v|$, the 1-norm, denote the sum of the coordinates of the vector.
Let $v * w$ denote the coordinate-wise product of the vectors $v$ and $w$.
(So the do... | null | CC BY-SA 2.5 | null | 2011-03-09T00:50:01.100 | 2011-03-09T10:28:37.683 | 2011-03-09T10:28:37.683 | 3567 | 3567 | null |
8052 | 1 | 8053 | null | 5 | 5722 | Suppose I'm doing binary classification, and I want to test whether using feature X is significant or not. (For example, I could be building a decision tree, and I want to see whether I should prune feature X or not.)
I believe the standard method is to use a chi-square test on the 2x2 table
```
X = 0 X... | 2x2 chi-square test vs. binomial proportion statistic | CC BY-SA 2.5 | null | 2011-03-09T00:54:14.237 | 2011-06-26T01:43:59.280 | null | null | 1106 | [
"statistical-significance",
"chi-squared-test"
] |
8053 | 2 | null | 8052 | 3 | null | Suppose you have
Case 1:
>
A=200, B=100
C=100, D=200
versus
Case 2:
>
A=200, B=0
C=200, D=200
The B=0 in case 2 means that case 2 provides much stronger evidence than case 1 of a relationship between X and Outcome; but in your test, both cases would be scored the same.
The Chi-Square test, informally speaking, no... | null | CC BY-SA 2.5 | null | 2011-03-09T01:14:39.337 | 2011-03-09T02:04:47.480 | 2020-06-11T14:32:37.003 | -1 | 3567 | null |
8054 | 2 | null | 8005 | 2 | null | I'm not sure I understand your question. But if you are working with nontraditional distributions you might want to look at nonparametric methods--in particular, you can use the nonparametric bootstrap to get confidence intervals for "4yrMean - 1yrMean" or whatever statistic you need.
| null | CC BY-SA 2.5 | null | 2011-03-09T02:52:53.240 | 2011-03-09T02:52:53.240 | null | null | 3567 | null |
8055 | 1 | null | null | 21 | 24903 | Could someone walk me through an example on how to use DLM Kalman filtering in R on a time series. Say I have a these values (quarterly values with yearly seasonality); how would you use DLM to predict the next values? And BTW, do I have enough historical data (what is the minimum)?
```
89 2009Q1
82 2009Q2
89 20... | How to use DLM with Kalman filtering for forecasting | CC BY-SA 4.0 | null | 2011-03-08T18:50:07.840 | 2018-08-28T16:14:47.997 | 2018-08-28T16:14:47.997 | 128677 | 3645 | [
"r",
"time-series",
"forecasting"
] |
8056 | 2 | null | 8055 | 17 | null | The paper at [JSS 39-02](http://www.jstatsoft.org/v39/i02) compares 5 different Kalman filtering R packages and gives sample code.
| null | CC BY-SA 2.5 | null | 2011-03-08T21:47:06.360 | 2011-03-08T21:47:06.360 | null | null | 4704 | null |
8057 | 2 | null | 8055 | 8 | null | I suggest you read the dlm vignette [http://cran.r-project.org/web/packages/dlm/vignettes/dlm.pdf](http://cran.r-project.org/web/packages/dlm/vignettes/dlm.pdf) especially the chapter 3.3
| null | CC BY-SA 2.5 | null | 2011-03-09T05:56:56.280 | 2011-03-09T05:56:56.280 | null | null | 1709 | null |
8058 | 1 | 8060 | null | 6 | 11039 | I'm building a logistic regression, and two of my variables are categorical with three levels each. (Say one variable is male, female, or unknown, and the other is single, married, or unknown.)
How many dummy variables am I supposed to make? Do I make 4 in total (2 for each of the categorical variables, e.g., a male va... | How to choose number of dummy variables when encoding several categorical variables? | CC BY-SA 2.5 | null | 2011-03-09T06:04:31.097 | 2017-07-12T11:28:18.607 | 2017-07-12T11:28:18.607 | 3277 | null | [
"logistic",
"categorical-data",
"categorical-encoding"
] |
8059 | 1 | 8076 | null | 6 | 950 | I'm doing some work on modelling transition matrices, and for this I need a measure of discrepancy or lack of fit: that is, if I have a matrix $T$ and a target matrix $T_0$, I want to be able to calculate how far $T$ is from $T_0$. Would anyone be able to provide pointers on what measure I should be using?
I've seen so... | Discrepancy measures for transition matrices | CC BY-SA 2.5 | null | 2011-03-09T06:14:27.973 | 2011-03-09T16:06:28.627 | 2011-03-09T16:06:28.627 | 223 | 1569 | [
"estimation",
"markov-process",
"distance-functions"
] |
8060 | 2 | null | 8058 | 9 | null | You would make k-1 dummy variables for each of your categorical variables. The textbook argument holds; if you were to make k dummies for any of your variables, you would have a collinearity. You can think of the k-1 dummies as being contrasts between the effects of their corresponding levels, and the level whose dummy... | null | CC BY-SA 2.5 | null | 2011-03-09T06:43:14.660 | 2011-03-09T06:43:14.660 | null | null | 1569 | null |
8063 | 1 | null | null | 3 | 3213 | How can I measure linear correlation of non-normally distributed variables?
Pearson coefficient is not valid for non-normally distributed data, and Spearman's rho does not capture linear correlation.
Thank you
| Measuring linear correlation of non-normally distributed variables | CC BY-SA 2.5 | null | 2011-03-09T09:43:14.043 | 2012-11-05T17:50:45.337 | 2011-03-09T09:59:04.997 | 930 | 3465 | [
"correlation",
"measurement"
] |
8064 | 2 | null | 8063 | 3 | null | Why do you require normality for computing a correlation? How about a simple scatterplot?
As long as the data are continuous, ordinary (Pearson) correlation should be fine. All that it is measuring is the strength of the linear relationship between two variables (if indeed there is such a relationship).
| null | CC BY-SA 2.5 | null | 2011-03-09T09:49:24.237 | 2011-03-09T09:49:24.237 | null | null | 1945 | null |
8065 | 1 | 8069 | null | 1 | 137 | I have the following data of 2 diseases from 5 areas. I want to see if there is any relationship between the 2 diseases.
Incidence Rates of 2 diseases (cases per million per year)
```
Areas Disease 1 Disease 2
1 4.653 0.751
2 6.910 1.121
3 4.957 0.... | Establishing relationship between 2 diseases | CC BY-SA 2.5 | null | 2011-03-09T10:11:34.210 | 2011-03-09T12:36:20.820 | 2011-03-09T12:36:20.820 | 2956 | 2956 | [
"correlation"
] |
8066 | 2 | null | 8063 | 2 | null | What @galit said is absolutely right. You can find the linear correlation between any two continuously distributed variables.
But perhaps you are thinking of the meaning of such a correlation? Indeed, [Anscombe's quartet](http://en.wikipedia.org/wiki/Anscombe%27s_quartet) shows that while the correlation is defined for... | null | CC BY-SA 2.5 | null | 2011-03-09T10:11:52.887 | 2011-03-09T10:11:52.887 | null | null | 686 | null |
8067 | 2 | null | 8044 | 6 | null | A couple things to add to what @matt said:
In addition to SUGI (which is now renamed SAS Global Forum, and will be held this year in Las Vegas) there are numerous local and regional SAS user groups. These are smaller, more intimate, and (usually) a lot cheaper. Some local groups are even free. See [here](http://suppo... | null | CC BY-SA 2.5 | null | 2011-03-09T10:18:53.657 | 2011-03-09T10:18:53.657 | null | null | 686 | null |
8068 | 2 | null | 8059 | 5 | null | Why does one want the measure of discrepancy to be a true metric? There is a huge literature on axiomatic characterizations of I-divergence as measure of distance. It is neither symmetric nor satisfies triangle inequality.
I hope by 'transition matrix' you mean 'probability transition matrix'. Never mind, as long as th... | null | CC BY-SA 2.5 | null | 2011-03-09T10:34:57.833 | 2011-03-09T10:34:57.833 | null | null | 3485 | null |
8069 | 2 | null | 8065 | 3 | null | There is no simple answer for how to "establish a relationship between two variables;" indeed, your question is one of the central issues in statistics and research is still going on on how to do this. But some basics: first you will want to plot your data, and then you will want to carry out a linear regression to t... | null | CC BY-SA 2.5 | null | 2011-03-09T10:51:36.343 | 2011-03-09T10:56:46.197 | 2011-03-09T10:56:46.197 | 3567 | 3567 | null |
8070 | 2 | null | 7994 | 2 | null | See the papers by Sebastian Mika and co-authors (I think this was the subject of Mika's PhD thesis). The original paper is [here](http://dx.doi.org/10.1109/NNSP.1999.788121) and for free [here](http://ml.cs.tu-berlin.de/publications/MikRaeWesSchMue99.pdf).
| null | CC BY-SA 2.5 | null | 2011-03-09T11:16:18.363 | 2011-03-09T11:16:18.363 | null | null | 887 | null |
8071 | 1 | null | null | 156 | 591713 | How do I know when to choose between Spearman's $\rho$ and Pearson's $r$? My variable includes satisfaction and the scores were interpreted using the sum of the scores. However, these scores could also be ranked.
| How to choose between Pearson and Spearman correlation? | CC BY-SA 2.5 | null | 2011-03-09T11:28:52.587 | 2021-01-30T01:07:10.827 | 2017-03-02T22:43:24.067 | 28666 | null | [
"correlation",
"pearson-r",
"spearman-rho"
] |
8072 | 1 | 8102 | null | 4 | 1731 | If I want to achieve a margin-of-error of <= 5 % for a representative population sample, how large a sample do I need when: The interviewees are picked from X regions and from Y age groups? That is, how many samples from each age group & region?
I know how many people live in each region and how they are distributed in... | Choosing sample size to achieve pre-specified margin-of-error | CC BY-SA 2.5 | null | 2011-03-09T11:34:14.167 | 2011-03-09T22:35:21.137 | 2011-03-09T11:52:29.903 | 3401 | 3401 | [
"statistical-significance",
"sampling"
] |
8073 | 2 | null | 8071 | 45 | null | This happens often in statistics: there are a variety of methods which could be applied in your situation, and you don't know which one to choose. You should base your decision the pros and cons of the methods under consideration and the specifics of your problem, but even then the decision is usually subjective with ... | null | CC BY-SA 4.0 | null | 2011-03-09T11:34:52.310 | 2021-01-30T01:04:45.773 | 2021-01-30T01:04:45.773 | 22047 | 3567 | null |
8074 | 2 | null | 8071 | 6 | null | While agreeing with Charles' answer, I would suggest (on a strictly practical level) that you compute both of the coefficients and look at the differences. In many cases, they will be exactly the same, so you don't need to worry.
If however, they are different then you need to look at whether or not you met the assumpt... | null | CC BY-SA 4.0 | null | 2011-03-09T11:54:45.967 | 2021-01-30T01:07:10.827 | 2021-01-30T01:07:10.827 | 22047 | 656 | null |
8075 | 2 | null | 8072 | 3 | null | If you weight your measurements (proportion of subpopulation/proportion of subpopulation in sample), your estimates will be unbiased. I assume this is what you meant by "poll results being skewed".
If I interpret your question correctly, your goal is the simultaneous estimation of multiple population proportions, wher... | null | CC BY-SA 2.5 | null | 2011-03-09T12:13:57.863 | 2011-03-09T12:28:18.687 | 2020-06-11T14:32:37.003 | -1 | 3567 | null |
8076 | 2 | null | 8059 | 4 | null | As long as your matrix represent conditional probability I think that using a general matrix norm is a bit artificial. Using some sort of geodesic distance on the set of transition matrix might be more relevant but I clearly prefer to come back to probabilities.
I assume you want to compare $Q=(Q_{ij})$ and $P=(P_{ij}... | null | CC BY-SA 2.5 | null | 2011-03-09T13:09:16.753 | 2011-03-09T13:09:16.753 | null | null | 223 | null |
8077 | 2 | null | 8071 | 59 | null | Shortest and mostly correct answer is:
Pearson benchmarks linear relationship, Spearman benchmarks monotonic relationship (few infinities more general case, but for some power tradeoff).
So if you assume/think that the relation is linear (or, as a special case, that those are a two measures of the same thing, so the re... | null | CC BY-SA 2.5 | null | 2011-03-09T13:16:03.530 | 2011-03-09T13:16:03.530 | null | null | null | null |
8078 | 2 | null | 8000 | 10 | null | You may also consider simply using a number of transforms of time series for the input data. Just for one example, the inputs could be:
- the most recent interval value
(7)
- the next most recent interval
value (6)
- the delta between most
recent and next most recent (7-6=1)
- the third most recent interv... | null | CC BY-SA 2.5 | null | 2011-03-09T13:57:54.917 | 2011-03-09T13:57:54.917 | null | null | 2917 | null |
8079 | 1 | null | null | 0 | 146 | Let say we have only one sample in which we are interested only on one statistics c of this sample. This is usually true in practice when we usually have only one data set. So, most probably the usual estimating method like MLE cant be applied.
And let say we want to estimate 2 parameters (i.e. a & b) of a certain dist... | Bootstrap for estimating parameters with only one sample | CC BY-SA 2.5 | 0 | 2011-03-09T14:15:07.337 | 2011-03-09T14:15:07.337 | null | null | null | [
"estimation",
"bootstrap"
] |
8080 | 2 | null | 8044 | 4 | null | In addition to Matt Parkers excellent advice (particularly about reading colleagues code), the actual SAS documentation can be surprisingly helpful (once you've figured out the name of what you want):
[http://support.sas.com/documentation/](http://support.sas.com/documentation/)
And the Global Forum/SUGI proceedings ar... | null | CC BY-SA 2.5 | null | 2011-03-09T14:29:35.183 | 2011-03-09T14:29:35.183 | null | null | 495 | null |
8082 | 1 | 8084 | null | 3 | 1496 | I am running `ivprobit` in Stata to look at the determinants of enrolment in health insurance (cbhi). I have several exogenous regressors and one endogenous regressor (consumption). I am using wealthindex as an intrumental variable for consumption. However, when I run the `ivprobit` model all my exogenous regressors ap... | Why is Stata automatically converting regressors to instrumental variables in ivprobit model? | CC BY-SA 2.5 | null | 2011-03-09T15:10:25.447 | 2011-03-09T16:32:43.060 | 2011-03-09T16:32:43.060 | 930 | 834 | [
"stata",
"instrumental-variables"
] |
8083 | 1 | null | null | 1 | 1723 | Revised Version
Situation:
We have a sample of size $r$. Then, we count how many out of these $r$ to be "success".
Let $X=$ the no. of the "successes" and say $X$ follow Poisson dist.
Goal: To test whether $X$ is really Poisson using Dispersion Test with test statistics given by
$D=\frac{\sum_{i=1}^{n}(X_{i}-\overline... | One-sample-dispersion-test for Poisson parameter | CC BY-SA 2.5 | null | 2011-03-09T15:12:20.300 | 2011-03-10T13:31:19.273 | 2011-03-10T03:59:32.407 | null | null | [
"hypothesis-testing",
"poisson-distribution",
"bootstrap"
] |
8084 | 2 | null | 8082 | 6 | null | In general all exogenous variables are always included as instruments. Usually instruments are picked for variables which are endogenous, but we can think (it follows from the mathematical derivation of instrumental variable estimation) that we need to choose the instruments for all the variables. Instruments for exoge... | null | CC BY-SA 2.5 | null | 2011-03-09T15:27:12.967 | 2011-03-09T15:27:12.967 | null | null | 2116 | null |
8086 | 2 | null | 8083 | 4 | null | First of all, I don't think you can estimate (over)dispersion from a sample of size 1. Overdispersion means that the sample variance is more than predicted based on other properties of the sample (mean) and properties of the assumed model (for Poisson models, variance = mean). But for a sample of size one, the popula... | null | CC BY-SA 2.5 | null | 2011-03-09T16:40:43.410 | 2011-03-09T16:40:43.410 | null | null | 2975 | null |
8087 | 1 | 44407 | null | 7 | 527 | I am trying to study hidden conditional random fields but I still have some fundamental questions about those methods. I would be immensely grateful if someone could provide some clarification over the notation used on most papers about the topic.
In several papers the most common form of the the HCRF model is given as... | Hidden states in hidden conditional random fields | CC BY-SA 2.5 | null | 2011-03-09T16:46:25.777 | 2014-09-02T13:56:03.697 | 2011-03-09T16:53:24.623 | null | 1538 | [
"machine-learning",
"classification",
"image-processing"
] |
8088 | 1 | null | null | 8 | 1128 | I'm working on a project where we observe behaviour on a task (eg. response time) and model this behaviour as a function of several experimentally manipulated variables as well as several observed variable (participant sex, participant IQ, responses on a follow-up questionnaire). I don't have concerns about multicollin... | Assessing multicollinearity of dichotomous predictor variables | CC BY-SA 2.5 | null | 2011-03-09T17:14:51.320 | 2011-03-11T17:47:46.867 | 2011-03-09T18:54:53.200 | 364 | 364 | [
"regression",
"binomial-distribution",
"independence",
"multicollinearity",
"likelihood-ratio"
] |
8089 | 1 | 8129 | null | 11 | 1246 | Why doesn't recall take into account true negatives? In experiments where true negatives are just as important as true positives, is their a comparable metric that does take it into account?
| Why doesn't recall take into account true negatives? | CC BY-SA 2.5 | null | 2011-03-09T17:28:58.713 | 2011-03-10T14:51:01.483 | null | null | 3604 | [
"precision-recall"
] |
8090 | 1 | 8092 | null | 8 | 1158 | I have a dataset that's nominally 16-dimensional. I have about 100 samples in one case and about 20,000 in another. Based on various exploratory analyses I've conducted using PCA and heat maps, I'm convinced that the true dimensionality (i.e. the number of dimensions needed to capture most of the "signal") is around ... | How to visualize the true dimensionality of the data? | CC BY-SA 2.5 | null | 2011-03-09T17:49:53.070 | 2011-03-10T04:47:44.177 | 2011-03-09T17:57:03.650 | null | 1347 | [
"data-visualization",
"pca",
"dimensionality-reduction"
] |
8091 | 2 | null | 8090 | 4 | null | One way to visualize this would be as follows:
- Perform a PCA on the data.
- Let $V$ be the vector space spanned by the first two principal component vectors, and let $V^\top$ be the complement.
- Decompose each vector $x_i$ in your data set as the sum of an element in $V$ plus a remainder term (which is in $V^\to... | null | CC BY-SA 2.5 | null | 2011-03-09T18:03:12.767 | 2011-03-09T18:19:03.830 | 2011-03-09T18:19:03.830 | 795 | 795 | null |
8092 | 2 | null | 8090 | 6 | null | A standard approach would be to do PCA and then show a scree plot, which you ought to be able to get that out of any software you might choose. A little tinkering and you could make it more interpretable for your particular audience if necessary. Sometimes they can be convincing, but often they're ambiguous and there'a... | null | CC BY-SA 2.5 | null | 2011-03-09T18:11:31.550 | 2011-03-10T04:47:44.177 | 2011-03-10T04:47:44.177 | 26 | 26 | null |
8094 | 2 | null | 7973 | 2 | null | You just asked for a mathematical shortcut. But I want to take a step back and point out that your approach could give you misleading results.
Are you planning this approach? Add additional observations only if you aren't happy with the initial results. Once the results make you happy, presumably because the P value i... | null | CC BY-SA 2.5 | null | 2011-03-09T18:50:07.717 | 2011-03-09T18:50:07.717 | null | null | 25 | null |
8096 | 1 | null | null | 1 | 6174 | How do I establish the +/- in terms of a set of mean ages. My mean age is 27.2. The ages are 20 23 24 43 22 26 18 32 18 41 22 20 26 46 21 27 19 19 39 40 19 39 18 38 24 24 23 30.
Thank you.
| How to calculate confidence interval of the mean age of a sample? | CC BY-SA 2.5 | null | 2011-03-09T19:54:06.577 | 2014-08-25T17:05:57.230 | 2011-03-10T03:29:13.523 | 183 | null | [
"mean"
] |
8097 | 2 | null | 8090 | 0 | null | I've done similar using PROC Varclus in SAS. The basic idea is to generate a 4 cluster solution, pick the highest correlated variable with each cluster, and then to demonstrate that this 4 cluster solution explains more of the variation than the two cluster solution. For the 2 cluster solution you could use either Va... | null | CC BY-SA 2.5 | null | 2011-03-09T20:11:09.467 | 2011-03-09T20:18:42.687 | 2011-03-09T20:18:42.687 | 3489 | 3489 | null |
8099 | 1 | 8101 | null | 4 | 181 | I have a linear string of unit length, and I randomly sample two locations a and b from Uniform(0, 1). Then I cut the string at these two locations to get a sub-string. What is the distribution for the size of the sub-string (i.e. |a-b|)? Otherwise, I want to at least know the mean and variance.
| Distribution of a random segment on a string | CC BY-SA 2.5 | null | 2011-03-09T20:39:48.563 | 2011-03-10T01:29:29.180 | null | null | 578 | [
"distributions",
"variance",
"mean"
] |
8100 | 2 | null | 8099 | 7 | null | Here's a snippet of Maple code that shows that the PDF of $\lvert a - b \rvert$ is $2 - 2 t$:
```
with(Statistics):
a := RandomVariable(Uniform(0,1));
b := RandomVariable(Uniform(0,1));
PDF(abs(X - Y), t) assuming 0 < t, t < 1;
```
returns $2 - 2 t$. You can see that it's true by computing the probability that $\lvert... | null | CC BY-SA 2.5 | null | 2011-03-09T21:12:21.203 | 2011-03-09T21:12:21.203 | null | null | 2898 | null |
8101 | 2 | null | 8099 | 8 | null | $(a,b)$ constitute one draw from the uniform distribution on the unit square. The region where $|a - b| \gt x$ is a pair of equilateral right triangles, one at $(0,1)$ and the other at $(1,0)$ with side lengths $1-x$; they fit together into a square with area $(1-x)^2$. Thus the CDF of $x = |a-b|$ equals $1 - (1-x)^2$... | null | CC BY-SA 2.5 | null | 2011-03-09T21:13:30.397 | 2011-03-10T01:29:29.180 | 2011-03-10T01:29:29.180 | 919 | 919 | null |
8102 | 2 | null | 8072 | 6 | null | Your problem can be solved using the info in the [Wikipedia article for 'Margin of error'](http://en.wikipedia.org/wiki/Margin_of_error#Different_confidence_levels). As it says, the margin of error is largest when the proportion is 0.5. The maximum margin of error at 95% confidence is $m = 0.98 / \sqrt{n}.$ Rearranging... | null | CC BY-SA 2.5 | null | 2011-03-09T22:35:21.137 | 2011-03-09T22:35:21.137 | null | null | 449 | null |
8103 | 1 | 15249 | null | 2 | 1473 | Asking this for a colleague:
They have two groups of people with various characteristics and to address the issue of comparability of outcomes they would like to reference a paper or book that discusses the concept and problems of common support.
| What is a good reference that discusses the problem of common support? | CC BY-SA 2.5 | 0 | 2011-03-10T01:12:57.133 | 2011-09-06T13:55:24.433 | null | null | 1144 | [
"distributions",
"propensity-scores"
] |
8104 | 1 | 8108 | null | 23 | 13186 |
## Background
One of the most commonly used weak prior on variance is the inverse-gamma with parameters $\alpha =0.001, \beta=0.001$ [(Gelman 2006)](http://www.stat.columbia.edu/~gelman/research/published/taumain.pdf).
However, this distribution has a 90%CI of approximately $[3\times10^{19},\infty]$.
```
library(psc... | Why is a $p(\sigma^2)\sim\text{IG(0.001, 0.001)}$ prior on variance considered weak? | CC BY-SA 2.5 | null | 2011-03-10T03:00:48.210 | 2013-10-23T16:17:08.657 | 2011-04-29T00:40:19.050 | 3911 | 1381 | [
"bayesian",
"multilevel-analysis",
"prior"
] |
8105 | 2 | null | 8104 | 10 | null | It's pretty close to flat. Its median is 1.9 E298, almost the largest number one can represent in double precision floating arithmetic. As you point out, the probability it assigns to any interval that isn't really huge is really small. It's hard to get less informative than that!
| null | CC BY-SA 2.5 | null | 2011-03-10T03:15:51.573 | 2011-03-10T03:15:51.573 | null | null | 919 | null |
8106 | 1 | 8131 | null | 44 | 57218 | I am currently reading a [paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6VG2-4Y648C8-1&_user=10&_coverDate=11/30/2009&_rdoc=1&_fmt=high&_orig=gateway&_origin=gateway&_sort=d&_docanchor=&view=c&_searchStrId=1672491428&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=da95476... | What is the significance of logistic regression coefficients? | CC BY-SA 2.5 | null | 2011-03-10T03:45:41.767 | 2018-10-30T16:01:00.207 | 2020-06-11T14:32:37.003 | -1 | 3652 | [
"regression",
"logistic",
"interpretation"
] |
8107 | 1 | 8162 | null | 11 | 23526 | In `R` I can perform a Breusch–Pagan test for heteroscedasticity using the `ncvTest` function of the `car` package. A Breusch–Pagan test is a type of chi squared test.
How do I interpret these results:
```
> require(car)
> set.seed(100)
> x1 = runif(100, -1, 1)
> x2 = runif(100, -1, 1)
> ncvTest(lm(x1 ~ x2))
Non-const... | How do I interpret the results of a Breusch–Pagan test? | CC BY-SA 2.5 | null | 2011-03-10T04:09:10.763 | 2017-11-12T17:21:11.933 | 2017-11-12T17:21:11.933 | 11887 | 179 | [
"interpretation",
"heteroscedasticity"
] |
8108 | 2 | null | 8104 | 42 | null | Using the inverse gamma distribution, we get:
$$p(\sigma^2|\alpha,\beta) \propto (\sigma^2)^{-\alpha-1} \exp(-\frac{\beta}{\sigma^2})$$
You can see easily that if $\beta \rightarrow 0$ and $\alpha \rightarrow 0$ then the inverse gamma will approach the Jeffreys prior. This distribution is called "uninformative" becaus... | null | CC BY-SA 3.0 | null | 2011-03-10T04:27:49.310 | 2013-10-23T16:17:08.657 | 2013-10-23T16:17:08.657 | 17230 | 2392 | null |
8109 | 2 | null | 8106 | 22 | null | The idea here is that in logistic regression, we predict not the actual probability that, say, a southerner votes Republican, but a transformed version of it, the "log odds". Instead of the probability $p$, we deal with $\log p/(1-p)$ and find linear regression coefficients for the log odds.
So for example, let's assum... | null | CC BY-SA 2.5 | null | 2011-03-10T04:30:00.697 | 2011-03-10T05:05:51.337 | 2011-03-10T05:05:51.337 | 98 | 98 | null |
8110 | 2 | null | 8106 | 6 | null | The coefficients in the logistic regression represent the tendency for a given region/demographic to vote Republican, compared to a reference category. A positive coefficent means that region is more likely to vote Republican, and vice-versa for a negative coefficient; a larger absolute value means a stronger tendency ... | null | CC BY-SA 2.5 | null | 2011-03-10T05:42:36.100 | 2011-03-10T05:42:36.100 | null | null | 1569 | null |
8111 | 2 | null | 8088 | 3 | null | I think you are trying to interpret P(A|B) and P(B|A) as if they should be the same thing. There is no reason for them to be equal, because of the product rule:
$$P(AB)=P(A|B)P(B)=P(B|A)P(A)$$
unless $P(B)=P(A)$ then $P(A|B) \neq P(B|A)$ in general. This explains the difference in the "yn" case. Unless you have a "b... | null | CC BY-SA 2.5 | null | 2011-03-10T05:44:26.280 | 2011-03-10T05:51:25.883 | 2011-03-10T05:51:25.883 | 2392 | 2392 | null |
8112 | 2 | null | 8063 | 3 | null | In addition to Anscombe's quartet as mentioned by Peter Flom, [here](http://ferrari.dmat.fct.unl.pt/personal/mle/GestaoRisco/Embrechts/risk.pdf) is a very nice paper in the risk-management context illustrating the problems of using linear correlation with non-normally distributed variables. In a nutshell, much of our i... | null | CC BY-SA 2.5 | null | 2011-03-10T06:01:42.227 | 2011-03-10T06:07:40.783 | 2011-03-10T06:07:40.783 | 1569 | 1569 | null |
8113 | 2 | null | 8107 | 5 | null | First application of `ncvTest` reports that there is no heteroscedasticity, as it should. The second is not meaningful, since your dependent random variable is random walk. [Breusch-Pagan test](http://en.wikipedia.org/wiki/Breusch%E2%80%93Pagan_test) is assymptotic, so I suspect that it cannot be readily applied for ra... | null | CC BY-SA 2.5 | null | 2011-03-10T07:53:02.863 | 2011-03-10T07:53:02.863 | null | null | 2116 | null |
8114 | 1 | 8120 | null | 9 | 8154 | According to [this document](http://www.stat.psu.edu/online/courses/stat509/09_interim/09_interim_print.htm): The estimated logarithm of the hazard ratio is approximately normally distributed with variance (1/d1) + (1/d2), where d1 and d2 are the numbers of events in the two treatment groups.
Do you have a reference fo... | The estimated logarithm of the hazard ratio is approximately normally distributed | CC BY-SA 4.0 | null | 2011-03-10T08:07:51.637 | 2022-07-12T00:52:14.443 | 2022-07-12T00:52:14.443 | 11887 | 3019 | [
"survival",
"logarithm",
"hazard"
] |
8115 | 2 | null | 7258 | 1 | null | If I understand, you have a classification problem where you observe $(X_1,Y_1),\dots,(X_n,Y_n)$ ($Y_i=1$ correspond to default and $Y_i=0$ no default), and you have much more $Y_i=0$ than $Y_i=1$. You want to take a subsample $S$ of the $(X_i,Y_i)$ that have $Y_i=0$.
I am not sure I understand all the points in your ... | null | CC BY-SA 2.5 | null | 2011-03-10T08:41:57.213 | 2011-03-10T08:41:57.213 | null | null | 223 | null |
8117 | 1 | 8118 | null | 37 | 12598 | Our small team was having a discussion and got stuck. Does anyone know whether Cox regression has an underlying Poisson distribution. We had a debate that maybe Cox regression with constant time at risk will have similarities with Poisson regression with a robust variance. Any ideas?
| Does Cox Regression have an underlying Poisson distribution? | CC BY-SA 2.5 | null | 2011-03-10T09:17:00.893 | 2013-10-25T22:07:58.210 | 2013-10-25T22:07:58.210 | 7515 | 2961 | [
"regression",
"poisson-distribution",
"cox-model"
] |
8118 | 2 | null | 8117 | 37 | null | Yes, there is a link between these two regression models. Here is an illustration:
Suppose the baseline hazard is constant over time: $h_{0}(t) = \lambda$. In that case, the survival function is
$S(t) = \exp\left(-\int_{0}^{t} \lambda du\right) = \exp(-\lambda t)$
and the density function is
$f(t) = h(t) S(t) = \lambda... | null | CC BY-SA 2.5 | null | 2011-03-10T09:34:42.957 | 2011-03-10T09:34:42.957 | null | null | 3019 | null |
8119 | 1 | 13846 | null | 5 | 1756 | I am trying to sample from a Gamma distribution in JAGS
```
gd[i] ~ dgamma(k,r)
```
where k and r are have priors such that they are positive. However I get the following error when I compile the model:
```
Error in node gd[9]
Unobserved node inconsistent with unobserved parents at initialization
```
Any ideas?
EDI... | Jags error with dgamma | CC BY-SA 2.5 | null | 2011-03-10T10:09:33.283 | 2011-11-02T14:22:11.550 | 2011-03-11T08:03:16.257 | 1443 | 1443 | [
"bayesian",
"markov-chain-montecarlo",
"jags"
] |
8120 | 2 | null | 8114 | 10 | null | The fact that this is approximately normally distributed relies on the central limit theorem (CLT), so will be a better approximation in large samples. The CLT works better for the log of any ratio (risk ratio, odds ratio, hazard ratio..) than for the ratio itself.
In suitably large samples, I think this is a good app... | null | CC BY-SA 2.5 | null | 2011-03-10T10:38:25.433 | 2011-03-10T10:44:15.927 | 2011-03-10T10:44:15.927 | 449 | 449 | null |
8121 | 2 | null | 8103 | 1 | null | If I understand, you have a classification problem where you observe $(X_1,Y_1),\dots,(X_n,Y_n)$. Denoting by $\mathcal{X}$ the space where $X$ takes values, you assume the existance of $\mathcal{X}_{10}$ the smallest closed subset of $\mathcal{X}$ such that:
$\mathcal{X}\setminus \mathcal{X}_{10}$ can be partinione... | null | CC BY-SA 2.5 | null | 2011-03-10T10:39:31.160 | 2011-03-10T10:39:31.160 | 2017-04-13T12:44:24.677 | -1 | 223 | null |
8123 | 2 | null | 8106 | 6 | null | You also asked "how do I know what is significant and what is not." (I assume you mean statistically significant, since practical or substantive significance is another matter.) The asterisks in the table refer to the footnote: some effects are noted as having small p-values. These are obtained using a Wald test of... | null | CC BY-SA 2.5 | null | 2011-03-10T13:08:08.867 | 2011-03-10T13:08:08.867 | null | null | 2669 | null |
8124 | 2 | null | 8083 | 3 | null | Your question contains a "contradiction". The way you have described it, the variable defined as $X$ cannot have a Poisson distribution. I'll explain why.
So, you choose a cut-off point, $y_{0}$, and define the original variables $Y_{i}\:(i=1,\dotsc,r)$. You then take a new variable $X_{i}$ and define it as:
$$
X... | null | CC BY-SA 2.5 | null | 2011-03-10T13:31:19.273 | 2011-03-10T13:31:19.273 | null | null | 2392 | null |
8125 | 2 | null | 8012 | 1 | null | First let us introduce notation $y_{ki}=t_{k,i,2}-t_{k,i,1}$. Then $\bar\tau=\frac{1}{K}\sum_k\frac{1}{n_k}\sum_iy_{ki}$. The variance then is
\begin{align}
Var(\bar\tau)&=E(\bar\tau-E\bar\tau)^2=E\left(\frac{1}{K}\sum_k\frac{1}{n_k}\sum_i(y_{ki}-Ey_{ki})\right)^2\\
&=E\frac{1}{K^2}\sum_k\sum_l\frac{1}{n_kn_l}\sum_i\su... | null | CC BY-SA 2.5 | null | 2011-03-10T13:55:34.027 | 2011-03-10T13:55:34.027 | null | null | 2116 | null |
8126 | 1 | 8128 | null | 9 | 5693 | I want to calculate the hat matrix directly in R for a logit model. According to Long (1997) the hat matrix for logit models is defined as:
$$H = VX(X'VX)^{-1} X'V$$
X is the vector of independent variables, and V is a diagonal matrix with $\sqrt{\pi(1-\pi)}$ on the diagonal.
I use the `optim` function to maximize the... | How to calculate the hat matrix for logistic regression in R? | CC BY-SA 2.5 | null | 2011-03-10T14:04:22.007 | 2011-03-10T14:59:47.350 | 2011-03-10T14:55:03.557 | 2704 | 2704 | [
"r",
"logistic",
"deviance"
] |
8127 | 1 | null | null | 1 | 835 | I have two variables and 1000 cases. How can I statistically find representative cases from total of 1000, based on statistical properties of both variables and correlation between them. Perhaps something based on T-test and 95% (or 99%) interval but for both variables? I would like to know which statistical method can... | How to make representative sample in dataset with two variables? | CC BY-SA 3.0 | null | 2011-03-10T14:12:47.033 | 2013-12-22T08:58:57.147 | 2013-12-22T08:58:57.147 | 930 | null | [
"distributions",
"sample"
] |
8128 | 2 | null | 8126 | 14 | null | For logistic regression $\pi$ is calculated using formula
$$\pi=\frac{1}{1+\exp(-X\beta)}$$
So diagonal values of $V$ can be calculated in the following manner:
```
pi <- 1/(1+exp(-X%*%beta))
v <- sqrt(pi*(1-pi))
```
Now multiplying by diagonal matrix from left means that each row is multiplied by corresponding elemen... | null | CC BY-SA 2.5 | null | 2011-03-10T14:20:46.650 | 2011-03-10T14:59:47.350 | 2011-03-10T14:59:47.350 | 2116 | 2116 | null |
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