idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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52,901 | Faster option than glmnet for elastic net regularized regression [closed] | First, you have a good idea that you can get a feel for things with a smaller sample, to work out the kinks in your approach. And it's definitely true, as you've noticed, that things that take a while to run really break up your focus. It's annoying. But...
Second, you have to define "interactively". Do you mean instan... | Faster option than glmnet for elastic net regularized regression [closed] | First, you have a good idea that you can get a feel for things with a smaller sample, to work out the kinks in your approach. And it's definitely true, as you've noticed, that things that take a while | Faster option than glmnet for elastic net regularized regression [closed]
First, you have a good idea that you can get a feel for things with a smaller sample, to work out the kinks in your approach. And it's definitely true, as you've noticed, that things that take a while to run really break up your focus. It's annoy... | Faster option than glmnet for elastic net regularized regression [closed]
First, you have a good idea that you can get a feel for things with a smaller sample, to work out the kinks in your approach. And it's definitely true, as you've noticed, that things that take a while |
52,902 | Faster option than glmnet for elastic net regularized regression [closed] | Following @Wayne's comments on parallelization, see Zhou et al. (2014), "A Reduction of the Elastic Net to Support Vector Machines with an Application to GPU Computing", arXiv:1409.1976. They say
The state-of-the-art single-core implementation for solving the
Elastic Net problem is the glmnet package developed by Fr... | Faster option than glmnet for elastic net regularized regression [closed] | Following @Wayne's comments on parallelization, see Zhou et al. (2014), "A Reduction of the Elastic Net to Support Vector Machines with an Application to GPU Computing", arXiv:1409.1976. They say
The | Faster option than glmnet for elastic net regularized regression [closed]
Following @Wayne's comments on parallelization, see Zhou et al. (2014), "A Reduction of the Elastic Net to Support Vector Machines with an Application to GPU Computing", arXiv:1409.1976. They say
The state-of-the-art single-core implementation f... | Faster option than glmnet for elastic net regularized regression [closed]
Following @Wayne's comments on parallelization, see Zhou et al. (2014), "A Reduction of the Elastic Net to Support Vector Machines with an Application to GPU Computing", arXiv:1409.1976. They say
The |
52,903 | Post hoc power analysis for a non significant result? | Power analyses exploit an equation with four variables ($\alpha$, power, $N$, and the effect size). When you solve for power by stipulating the others, it is called "post hoc" power analysis. People often use post hoc power analysis to determine the power they had to detect the effect observed in their study after fi... | Post hoc power analysis for a non significant result? | Power analyses exploit an equation with four variables ($\alpha$, power, $N$, and the effect size). When you solve for power by stipulating the others, it is called "post hoc" power analysis. People | Post hoc power analysis for a non significant result?
Power analyses exploit an equation with four variables ($\alpha$, power, $N$, and the effect size). When you solve for power by stipulating the others, it is called "post hoc" power analysis. People often use post hoc power analysis to determine the power they had... | Post hoc power analysis for a non significant result?
Power analyses exploit an equation with four variables ($\alpha$, power, $N$, and the effect size). When you solve for power by stipulating the others, it is called "post hoc" power analysis. People |
52,904 | Post hoc power analysis for a non significant result? | I can't believe people are still asking for post-hoc power analyses!
Please, do not include this in your paper. The post-hoc power analysis is not going to tell you anything, and people reading your paper will think that you do not know what you are doing!
Power analyses can only be performed before you collect your da... | Post hoc power analysis for a non significant result? | I can't believe people are still asking for post-hoc power analyses!
Please, do not include this in your paper. The post-hoc power analysis is not going to tell you anything, and people reading your p | Post hoc power analysis for a non significant result?
I can't believe people are still asking for post-hoc power analyses!
Please, do not include this in your paper. The post-hoc power analysis is not going to tell you anything, and people reading your paper will think that you do not know what you are doing!
Power ana... | Post hoc power analysis for a non significant result?
I can't believe people are still asking for post-hoc power analyses!
Please, do not include this in your paper. The post-hoc power analysis is not going to tell you anything, and people reading your p |
52,905 | How can I determine Gamma distribution parameters from data | There is a number of ways you can estimate the parameters of a gamma distribution. The most popular is maximum likelihood estimation. The resulting estimators are known to have optimal properties for moderately large samples, such as asymptotic normality and minimum variance. The problem with the mle for a gamma distri... | How can I determine Gamma distribution parameters from data | There is a number of ways you can estimate the parameters of a gamma distribution. The most popular is maximum likelihood estimation. The resulting estimators are known to have optimal properties for | How can I determine Gamma distribution parameters from data
There is a number of ways you can estimate the parameters of a gamma distribution. The most popular is maximum likelihood estimation. The resulting estimators are known to have optimal properties for moderately large samples, such as asymptotic normality and m... | How can I determine Gamma distribution parameters from data
There is a number of ways you can estimate the parameters of a gamma distribution. The most popular is maximum likelihood estimation. The resulting estimators are known to have optimal properties for |
52,906 | Student looking to practice | I cannot stress this enough, look for real data.
If you are using a particular statistical method provided in an R stats package for example, often the package will contain a sample dataset which the authors will demonstrate their methods on. Similarly, if you look in R's datasets, a whole bunch of classic datasets ar... | Student looking to practice | I cannot stress this enough, look for real data.
If you are using a particular statistical method provided in an R stats package for example, often the package will contain a sample dataset which the | Student looking to practice
I cannot stress this enough, look for real data.
If you are using a particular statistical method provided in an R stats package for example, often the package will contain a sample dataset which the authors will demonstrate their methods on. Similarly, if you look in R's datasets, a whole ... | Student looking to practice
I cannot stress this enough, look for real data.
If you are using a particular statistical method provided in an R stats package for example, often the package will contain a sample dataset which the |
52,907 | Student looking to practice | If you want to "get your hands dirty" with statistics, there is an infinite number of ways to mine data, set up hypotheses, experiments, analyze data sets using various styles of analysis (i.e., Bayesian vs. frequentist), etc. However, your question indicates that you're having trouble figuring out where to begin.
Find... | Student looking to practice | If you want to "get your hands dirty" with statistics, there is an infinite number of ways to mine data, set up hypotheses, experiments, analyze data sets using various styles of analysis (i.e., Bayes | Student looking to practice
If you want to "get your hands dirty" with statistics, there is an infinite number of ways to mine data, set up hypotheses, experiments, analyze data sets using various styles of analysis (i.e., Bayesian vs. frequentist), etc. However, your question indicates that you're having trouble figur... | Student looking to practice
If you want to "get your hands dirty" with statistics, there is an infinite number of ways to mine data, set up hypotheses, experiments, analyze data sets using various styles of analysis (i.e., Bayes |
52,908 | Student looking to practice | The best method in my opinion to practice the skills you have is getting some data (this is the last of your problem, you can find dataset on the internet or you can create your own dataset) and try to figure out if there are relations or difference in some groups of the data (assuming you want to practice with inferen... | Student looking to practice | The best method in my opinion to practice the skills you have is getting some data (this is the last of your problem, you can find dataset on the internet or you can create your own dataset) and try t | Student looking to practice
The best method in my opinion to practice the skills you have is getting some data (this is the last of your problem, you can find dataset on the internet or you can create your own dataset) and try to figure out if there are relations or difference in some groups of the data (assuming you w... | Student looking to practice
The best method in my opinion to practice the skills you have is getting some data (this is the last of your problem, you can find dataset on the internet or you can create your own dataset) and try t |
52,909 | Student looking to practice | Several of the free (or at least inexpensive) certifications (Coursera, etc.) have small projects that can get you a lot of practice. You can do them quick and dirty, or you can do them in way more detail than is asked for. The advantage over doing it alone is there is a forum where you can discuss your results and oft... | Student looking to practice | Several of the free (or at least inexpensive) certifications (Coursera, etc.) have small projects that can get you a lot of practice. You can do them quick and dirty, or you can do them in way more de | Student looking to practice
Several of the free (or at least inexpensive) certifications (Coursera, etc.) have small projects that can get you a lot of practice. You can do them quick and dirty, or you can do them in way more detail than is asked for. The advantage over doing it alone is there is a forum where you can ... | Student looking to practice
Several of the free (or at least inexpensive) certifications (Coursera, etc.) have small projects that can get you a lot of practice. You can do them quick and dirty, or you can do them in way more de |
52,910 | Portmanteau test results R | Let me use the term "autocorrelation" as a synonym for "serial correlation".
1) How does one interpret the results of the below demonstration?
Most of the interpretation is already in the comments to the code. First, a VAR(1) model is estimated. It is tested for autocorrelation in errors using a portmanteau test. The... | Portmanteau test results R | Let me use the term "autocorrelation" as a synonym for "serial correlation".
1) How does one interpret the results of the below demonstration?
Most of the interpretation is already in the comments t | Portmanteau test results R
Let me use the term "autocorrelation" as a synonym for "serial correlation".
1) How does one interpret the results of the below demonstration?
Most of the interpretation is already in the comments to the code. First, a VAR(1) model is estimated. It is tested for autocorrelation in errors us... | Portmanteau test results R
Let me use the term "autocorrelation" as a synonym for "serial correlation".
1) How does one interpret the results of the below demonstration?
Most of the interpretation is already in the comments t |
52,911 | Are complete statistics always sufficient? | For completeness you need to consider only the distribution of the statistic $T$ (more correctly, the family of distributions indexed by $\theta$), whereas for sufficiency you need to consider the complete likelihood for $\theta$ as a function of the sample $X$; so it's trivial to come up with examples where $T$ isn't ... | Are complete statistics always sufficient? | For completeness you need to consider only the distribution of the statistic $T$ (more correctly, the family of distributions indexed by $\theta$), whereas for sufficiency you need to consider the com | Are complete statistics always sufficient?
For completeness you need to consider only the distribution of the statistic $T$ (more correctly, the family of distributions indexed by $\theta$), whereas for sufficiency you need to consider the complete likelihood for $\theta$ as a function of the sample $X$; so it's trivia... | Are complete statistics always sufficient?
For completeness you need to consider only the distribution of the statistic $T$ (more correctly, the family of distributions indexed by $\theta$), whereas for sufficiency you need to consider the com |
52,912 | Independence of $\min(X,Y)$ and $\max(X,Y)$ for independent $X$, $Y$? | If $X$ and $Y$ are independent continuous random variables, then $\max(X,Y)$ and
$\min(X,Y)$ are independent random variables if and only if one of the
following two conditions holds:
$P(X > Y) = 1$
$P(X < Y) = 1$
Note that the above conditions mean that $P(X=Y) = 0$ but this does
not mean that $(X=Y)$ is the same as... | Independence of $\min(X,Y)$ and $\max(X,Y)$ for independent $X$, $Y$? | If $X$ and $Y$ are independent continuous random variables, then $\max(X,Y)$ and
$\min(X,Y)$ are independent random variables if and only if one of the
following two conditions holds:
$P(X > Y) = 1$
| Independence of $\min(X,Y)$ and $\max(X,Y)$ for independent $X$, $Y$?
If $X$ and $Y$ are independent continuous random variables, then $\max(X,Y)$ and
$\min(X,Y)$ are independent random variables if and only if one of the
following two conditions holds:
$P(X > Y) = 1$
$P(X < Y) = 1$
Note that the above conditions mea... | Independence of $\min(X,Y)$ and $\max(X,Y)$ for independent $X$, $Y$?
If $X$ and $Y$ are independent continuous random variables, then $\max(X,Y)$ and
$\min(X,Y)$ are independent random variables if and only if one of the
following two conditions holds:
$P(X > Y) = 1$
|
52,913 | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices? | Instead of predicting how much the stock gains or loses, the models are predicting the sign of the gain or loss, i.e. a binary outcome. | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices? | Instead of predicting how much the stock gains or loses, the models are predicting the sign of the gain or loss, i.e. a binary outcome. | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices?
Instead of predicting how much the stock gains or loses, the models are predicting the sign of the gain or loss, i.e. a binary outcome. | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices?
Instead of predicting how much the stock gains or loses, the models are predicting the sign of the gain or loss, i.e. a binary outcome. |
52,914 | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices? | The poor phrasing of the abstract* suggests a possible misuse of the term; I've often seen Linear Regression of the logarithm used to predict asset price movements (the idea being that asset prices tend to change by percentages of their current value, rather than by consistent nominal values).
*Full disclosure: I only ... | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices? | The poor phrasing of the abstract* suggests a possible misuse of the term; I've often seen Linear Regression of the logarithm used to predict asset price movements (the idea being that asset prices te | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices?
The poor phrasing of the abstract* suggests a possible misuse of the term; I've often seen Linear Regression of the logarithm used to predict asset price movements (the idea being that asset prices tend to change by percentages o... | Why is Logistic Regression mentioned by many sources as useful in predicting stock prices?
The poor phrasing of the abstract* suggests a possible misuse of the term; I've often seen Linear Regression of the logarithm used to predict asset price movements (the idea being that asset prices te |
52,915 | Reference for xgboost | Source: Tianqi Chen's Quora answer
Both xgboost and gbm follows the principle of gradient boosting.
There are however, the difference in modeling details. Specifically,
xgboost used a more regularized model formalization to control
over-fitting, which gives it better performance.
We have updated a comprehensi... | Reference for xgboost | Source: Tianqi Chen's Quora answer
Both xgboost and gbm follows the principle of gradient boosting.
There are however, the difference in modeling details. Specifically,
xgboost used a more reg | Reference for xgboost
Source: Tianqi Chen's Quora answer
Both xgboost and gbm follows the principle of gradient boosting.
There are however, the difference in modeling details. Specifically,
xgboost used a more regularized model formalization to control
over-fitting, which gives it better performance.
We have... | Reference for xgboost
Source: Tianqi Chen's Quora answer
Both xgboost and gbm follows the principle of gradient boosting.
There are however, the difference in modeling details. Specifically,
xgboost used a more reg |
52,916 | Reference for xgboost | There is now an arXiv article XGBoost: A Scalable Tree Boosting System that describes the algorithm. At the time of the original question and answer this had not yet been posted. | Reference for xgboost | There is now an arXiv article XGBoost: A Scalable Tree Boosting System that describes the algorithm. At the time of the original question and answer this had not yet been posted. | Reference for xgboost
There is now an arXiv article XGBoost: A Scalable Tree Boosting System that describes the algorithm. At the time of the original question and answer this had not yet been posted. | Reference for xgboost
There is now an arXiv article XGBoost: A Scalable Tree Boosting System that describes the algorithm. At the time of the original question and answer this had not yet been posted. |
52,917 | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate] | There is a reason why the 'two-tailed chi-squared' is seldomly used: if you do a $\chi^2$ test for contingency tables, then the test statistic is (without the continuity correction):
$X^2 = \sum_{i,j}\frac{(o_{ij}-e_{ij})^2}{e_{ij}}$
where $o_{ij}$ are the observed counts in cell $i,j$ and $e_{ij}$ are the expected ... | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate] | There is a reason why the 'two-tailed chi-squared' is seldomly used: if you do a $\chi^2$ test for contingency tables, then the test statistic is (without the continuity correction):
$X^2 = \sum_{i, | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate]
There is a reason why the 'two-tailed chi-squared' is seldomly used: if you do a $\chi^2$ test for contingency tables, then the test statistic is (without the continuity correction):
$X^2 = \sum_{i,j}\frac{(o_{ij}-e_{ij})^2}{e_{ij}}$
where $... | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate]
There is a reason why the 'two-tailed chi-squared' is seldomly used: if you do a $\chi^2$ test for contingency tables, then the test statistic is (without the continuity correction):
$X^2 = \sum_{i, |
52,918 | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate] | You are correct that the one-tailed p-value is for any hypothesis that is formulated as one-sided and that two-tailed p-values are for two-sided hypotheses, in which either group (in this case) can prove to be "better" than the other.
Since you're talking about treatments, I assume that your field is in medicine, psych... | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate] | You are correct that the one-tailed p-value is for any hypothesis that is formulated as one-sided and that two-tailed p-values are for two-sided hypotheses, in which either group (in this case) can pr | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate]
You are correct that the one-tailed p-value is for any hypothesis that is formulated as one-sided and that two-tailed p-values are for two-sided hypotheses, in which either group (in this case) can prove to be "better" than the other.
Since you... | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate]
You are correct that the one-tailed p-value is for any hypothesis that is formulated as one-sided and that two-tailed p-values are for two-sided hypotheses, in which either group (in this case) can pr |
52,919 | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate] | As far as I can see, the twosidedness is not referring to the chisquare test at all, but rather to the corresponding two-sided test of two proportions. The chisquare part is indeed a onetailed test, which it should be. The same kind of vocabulary is used in i e openepi.com. Some more details were covered in another com... | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate] | As far as I can see, the twosidedness is not referring to the chisquare test at all, but rather to the corresponding two-sided test of two proportions. The chisquare part is indeed a onetailed test, w | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate]
As far as I can see, the twosidedness is not referring to the chisquare test at all, but rather to the corresponding two-sided test of two proportions. The chisquare part is indeed a onetailed test, which it should be. The same kind of vocabula... | Chi-Square-Test: Why is the chi-squared test a one-tailed test? [duplicate]
As far as I can see, the twosidedness is not referring to the chisquare test at all, but rather to the corresponding two-sided test of two proportions. The chisquare part is indeed a onetailed test, w |
52,920 | Are "covariance function" and "kernel function" synonyms? | Yes, "covariance function" and "positive-definite kernel" refer to the same concept. (Authors in the SVM literature sometimes omit the qualification "positive-definite", since it's typically by far the most relevant type of kernel.)
For example, see page 80 of Rasmussen and Williams, Gaussian Processes for Machine Lear... | Are "covariance function" and "kernel function" synonyms? | Yes, "covariance function" and "positive-definite kernel" refer to the same concept. (Authors in the SVM literature sometimes omit the qualification "positive-definite", since it's typically by far th | Are "covariance function" and "kernel function" synonyms?
Yes, "covariance function" and "positive-definite kernel" refer to the same concept. (Authors in the SVM literature sometimes omit the qualification "positive-definite", since it's typically by far the most relevant type of kernel.)
For example, see page 80 of R... | Are "covariance function" and "kernel function" synonyms?
Yes, "covariance function" and "positive-definite kernel" refer to the same concept. (Authors in the SVM literature sometimes omit the qualification "positive-definite", since it's typically by far th |
52,921 | Are "covariance function" and "kernel function" synonyms? | Danica's answer is correct. Precisely stated, covariance matrices and Mercer kernels are both matrices which are (1) positive definite and (2) symmetric. However, there is some research into matrices otherwise than Mercer kernels, that is, matrices which are not positive-definite but which may be useful in machine lear... | Are "covariance function" and "kernel function" synonyms? | Danica's answer is correct. Precisely stated, covariance matrices and Mercer kernels are both matrices which are (1) positive definite and (2) symmetric. However, there is some research into matrices | Are "covariance function" and "kernel function" synonyms?
Danica's answer is correct. Precisely stated, covariance matrices and Mercer kernels are both matrices which are (1) positive definite and (2) symmetric. However, there is some research into matrices otherwise than Mercer kernels, that is, matrices which are not... | Are "covariance function" and "kernel function" synonyms?
Danica's answer is correct. Precisely stated, covariance matrices and Mercer kernels are both matrices which are (1) positive definite and (2) symmetric. However, there is some research into matrices |
52,922 | A good description of the random forests method | When getting up to speed on a topic, I find it helpful to start at the beginning and work forward chronologically. Breiman's original paper on random forests is where I would recommend starting.
Leo Breiman. "Random Forests." Machine Learning (2001). 45, 5-32. | A good description of the random forests method | When getting up to speed on a topic, I find it helpful to start at the beginning and work forward chronologically. Breiman's original paper on random forests is where I would recommend starting.
Leo B | A good description of the random forests method
When getting up to speed on a topic, I find it helpful to start at the beginning and work forward chronologically. Breiman's original paper on random forests is where I would recommend starting.
Leo Breiman. "Random Forests." Machine Learning (2001). 45, 5-32. | A good description of the random forests method
When getting up to speed on a topic, I find it helpful to start at the beginning and work forward chronologically. Breiman's original paper on random forests is where I would recommend starting.
Leo B |
52,923 | A good description of the random forests method | Did you check out "The Elements of Statistical Learning" http://statweb.stanford.edu/~tibs/ElemStatLearn/ (free online pdf). It is the more advanced version of "An Introduction to Statistical Learning with Applications in R "
If that is not appropriate I would probably just start reading the original journal articles t... | A good description of the random forests method | Did you check out "The Elements of Statistical Learning" http://statweb.stanford.edu/~tibs/ElemStatLearn/ (free online pdf). It is the more advanced version of "An Introduction to Statistical Learning | A good description of the random forests method
Did you check out "The Elements of Statistical Learning" http://statweb.stanford.edu/~tibs/ElemStatLearn/ (free online pdf). It is the more advanced version of "An Introduction to Statistical Learning with Applications in R "
If that is not appropriate I would probably ju... | A good description of the random forests method
Did you check out "The Elements of Statistical Learning" http://statweb.stanford.edu/~tibs/ElemStatLearn/ (free online pdf). It is the more advanced version of "An Introduction to Statistical Learning |
52,924 | A good description of the random forests method | There is a PhD thesis from one of the Kaggle guys about Understanding Random Forests. And that's actually the title of his thesis. This is the link and i think its a pretty new PhD:
http://www.montefiore.ulg.ac.be/~glouppe/pdf/phd-thesis.pdf
Hope this helps, it's more specific and starts from basics as well. | A good description of the random forests method | There is a PhD thesis from one of the Kaggle guys about Understanding Random Forests. And that's actually the title of his thesis. This is the link and i think its a pretty new PhD:
http://www.montefi | A good description of the random forests method
There is a PhD thesis from one of the Kaggle guys about Understanding Random Forests. And that's actually the title of his thesis. This is the link and i think its a pretty new PhD:
http://www.montefiore.ulg.ac.be/~glouppe/pdf/phd-thesis.pdf
Hope this helps, it's more spe... | A good description of the random forests method
There is a PhD thesis from one of the Kaggle guys about Understanding Random Forests. And that's actually the title of his thesis. This is the link and i think its a pretty new PhD:
http://www.montefi |
52,925 | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting is unequal variance? | First, let me assure you that - as mentioned by @amoeba - you are on the right path to land in "research hell", that is the place where researchers (should) go when they let their p-value to decide what to include or not in the analysis.
Reason 1.
You have to decide a priori whether you consider Levene's test a good te... | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting | First, let me assure you that - as mentioned by @amoeba - you are on the right path to land in "research hell", that is the place where researchers (should) go when they let their p-value to decide wh | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting is unequal variance?
First, let me assure you that - as mentioned by @amoeba - you are on the right path to land in "research hell", that is the place where researchers (should) go when they let their p-value to decide w... | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting
First, let me assure you that - as mentioned by @amoeba - you are on the right path to land in "research hell", that is the place where researchers (should) go when they let their p-value to decide wh |
52,926 | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting is unequal variance? | First, ask yourself why IQ is always included in such models. There is probably some reason. It might be that IQ is a mediator (see below)
Second, from what you say, it sounds like IQ is a type of mediator of the relationship between glutamate concentration and whatever your group variable is. Matching will not deal w... | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting | First, ask yourself why IQ is always included in such models. There is probably some reason. It might be that IQ is a mediator (see below)
Second, from what you say, it sounds like IQ is a type of me | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting is unequal variance?
First, ask yourself why IQ is always included in such models. There is probably some reason. It might be that IQ is a mediator (see below)
Second, from what you say, it sounds like IQ is a type of m... | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting
First, ask yourself why IQ is always included in such models. There is probably some reason. It might be that IQ is a mediator (see below)
Second, from what you say, it sounds like IQ is a type of me |
52,927 | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting is unequal variance? | I'm a bit worried about the y-axis label on your plot, "C glutamate SD less than 20 - extremes," which has two potentially important implications.
For one, it might be taken to suggest that there has already been some removal of "outlier" determinations, which is tricky business. This should usually only be done if you... | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting | I'm a bit worried about the y-axis label on your plot, "C glutamate SD less than 20 - extremes," which has two potentially important implications.
For one, it might be taken to suggest that there has | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting is unequal variance?
I'm a bit worried about the y-axis label on your plot, "C glutamate SD less than 20 - extremes," which has two potentially important implications.
For one, it might be taken to suggest that there has... | Do I have a justified reason to exclude a non-significant covariate from my ANCOVA? How interesting
I'm a bit worried about the y-axis label on your plot, "C glutamate SD less than 20 - extremes," which has two potentially important implications.
For one, it might be taken to suggest that there has |
52,928 | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplicate] | You can use the function mvrnorm from the MASS package to sample values from a multivariate normal distrbution.
Your data:
mu <- c(4.23, 3.01, 2.91)
stddev <- c(1.23, 0.92, 1.32)
corMat <- matrix(c(1, 0.78, 0.23,
0.78, 1, 0.27,
0.23, 0.27, 1),
ncol = 3)
corMat
# ... | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplica | You can use the function mvrnorm from the MASS package to sample values from a multivariate normal distrbution.
Your data:
mu <- c(4.23, 3.01, 2.91)
stddev <- c(1.23, 0.92, 1.32)
corMat <- matrix(c(1 | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplicate]
You can use the function mvrnorm from the MASS package to sample values from a multivariate normal distrbution.
Your data:
mu <- c(4.23, 3.01, 2.91)
stddev <- c(1.23, 0.92, 1.32)
corMat <- matrix(c(1, 0.78, 0.23,
... | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplica
You can use the function mvrnorm from the MASS package to sample values from a multivariate normal distrbution.
Your data:
mu <- c(4.23, 3.01, 2.91)
stddev <- c(1.23, 0.92, 1.32)
corMat <- matrix(c(1 |
52,929 | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplicate] | Assuming normality, you could draw samples from Multivariate Normal distribution. What you need for that is a vector of means $\boldsymbol{\mu} = (\mu_1, ..., \mu_k)$ and a covariance matrix $\boldsymbol{\Sigma}$. If you recall that covariance matrix has variances on the diagonal and values of covariance in the rest of... | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplica | Assuming normality, you could draw samples from Multivariate Normal distribution. What you need for that is a vector of means $\boldsymbol{\mu} = (\mu_1, ..., \mu_k)$ and a covariance matrix $\boldsym | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplicate]
Assuming normality, you could draw samples from Multivariate Normal distribution. What you need for that is a vector of means $\boldsymbol{\mu} = (\mu_1, ..., \mu_k)$ and a covariance matrix $\boldsymbol{\Sigma}$. If... | Generating a simulated dataset from a correlation matrix with means and standard deviations [duplica
Assuming normality, you could draw samples from Multivariate Normal distribution. What you need for that is a vector of means $\boldsymbol{\mu} = (\mu_1, ..., \mu_k)$ and a covariance matrix $\boldsym |
52,930 | Why am I getting 100% accuracy for SVM and Decision Tree (scikit) | I was able to reproduce your results:
> clf = svm.SVC()
> scores = cross_validation.cross_val_score(clf, X, Y, cv=10)
I didn't get perfect out of fold classification, but close:
> print(scores)
array([ 1. , 1. , 1. , 0.99152542, 1. ,
1. , 1. , 1. , 1. ... | Why am I getting 100% accuracy for SVM and Decision Tree (scikit) | I was able to reproduce your results:
> clf = svm.SVC()
> scores = cross_validation.cross_val_score(clf, X, Y, cv=10)
I didn't get perfect out of fold classification, but close:
> print(scores)
array | Why am I getting 100% accuracy for SVM and Decision Tree (scikit)
I was able to reproduce your results:
> clf = svm.SVC()
> scores = cross_validation.cross_val_score(clf, X, Y, cv=10)
I didn't get perfect out of fold classification, but close:
> print(scores)
array([ 1. , 1. , 1. , 0.99152542, ... | Why am I getting 100% accuracy for SVM and Decision Tree (scikit)
I was able to reproduce your results:
> clf = svm.SVC()
> scores = cross_validation.cross_val_score(clf, X, Y, cv=10)
I didn't get perfect out of fold classification, but close:
> print(scores)
array |
52,931 | Why am I getting 100% accuracy for SVM and Decision Tree (scikit) | Oh, my God, I came across the same issue. Maybe my answer is not the best answer for you, but it may help other people.
Here is my code with Scikit-Learn
clf = DecisionTreeClassifier(criterion='entropy', max_depth=10)
clf.fit(X, y)
And I got 100% accuracy score.
However, when I got the feature_importances_ of clf, an... | Why am I getting 100% accuracy for SVM and Decision Tree (scikit) | Oh, my God, I came across the same issue. Maybe my answer is not the best answer for you, but it may help other people.
Here is my code with Scikit-Learn
clf = DecisionTreeClassifier(criterion='entro | Why am I getting 100% accuracy for SVM and Decision Tree (scikit)
Oh, my God, I came across the same issue. Maybe my answer is not the best answer for you, but it may help other people.
Here is my code with Scikit-Learn
clf = DecisionTreeClassifier(criterion='entropy', max_depth=10)
clf.fit(X, y)
And I got 100% accur... | Why am I getting 100% accuracy for SVM and Decision Tree (scikit)
Oh, my God, I came across the same issue. Maybe my answer is not the best answer for you, but it may help other people.
Here is my code with Scikit-Learn
clf = DecisionTreeClassifier(criterion='entro |
52,932 | Generating and Working with Random Vectors in R | The mvtnorm package in R has the rmvnorm function (analogous to rnorm) that produces arbitrary-dimensional Gaussian random variables. It also provides the option to use three different algorithms. A quick comparison using your exact setup:
library(mvtnorm)
library(microbenchmark)
sigma <- matrix(c(20, 8, 8, 20), 2)
mu ... | Generating and Working with Random Vectors in R | The mvtnorm package in R has the rmvnorm function (analogous to rnorm) that produces arbitrary-dimensional Gaussian random variables. It also provides the option to use three different algorithms. A q | Generating and Working with Random Vectors in R
The mvtnorm package in R has the rmvnorm function (analogous to rnorm) that produces arbitrary-dimensional Gaussian random variables. It also provides the option to use three different algorithms. A quick comparison using your exact setup:
library(mvtnorm)
library(microbe... | Generating and Working with Random Vectors in R
The mvtnorm package in R has the rmvnorm function (analogous to rnorm) that produces arbitrary-dimensional Gaussian random variables. It also provides the option to use three different algorithms. A q |
52,933 | Generating and Working with Random Vectors in R | In addition to @ssdecontrol's answer, I've been using the MASS package's mvrnorm. Adding to @ssdecontrol's code (with my slower compy):
library(mvtnorm)
library(MASS)
library(microbenchmark)
sigma <- matrix(c(20, 8, 8, 20), 2)
mu <- c(3, -5)
microbenchmark(v1 <- rmvnorm(1e5, mu, sigma, "eigen"),
v2 <-... | Generating and Working with Random Vectors in R | In addition to @ssdecontrol's answer, I've been using the MASS package's mvrnorm. Adding to @ssdecontrol's code (with my slower compy):
library(mvtnorm)
library(MASS)
library(microbenchmark)
sigma | Generating and Working with Random Vectors in R
In addition to @ssdecontrol's answer, I've been using the MASS package's mvrnorm. Adding to @ssdecontrol's code (with my slower compy):
library(mvtnorm)
library(MASS)
library(microbenchmark)
sigma <- matrix(c(20, 8, 8, 20), 2)
mu <- c(3, -5)
microbenchmark(v1 <- rmvnor... | Generating and Working with Random Vectors in R
In addition to @ssdecontrol's answer, I've been using the MASS package's mvrnorm. Adding to @ssdecontrol's code (with my slower compy):
library(mvtnorm)
library(MASS)
library(microbenchmark)
sigma |
52,934 | Generating and Working with Random Vectors in R | For completeness sake, here's a follow-up note on how to generate random vectors regardless of the marginal distribution of the individual components. I'm going to stick with the bivariate case:
Generate a bivariate vector from a standard normal random distribution following a predetermined correlation*. I'll stick wi... | Generating and Working with Random Vectors in R | For completeness sake, here's a follow-up note on how to generate random vectors regardless of the marginal distribution of the individual components. I'm going to stick with the bivariate case:
Gene | Generating and Working with Random Vectors in R
For completeness sake, here's a follow-up note on how to generate random vectors regardless of the marginal distribution of the individual components. I'm going to stick with the bivariate case:
Generate a bivariate vector from a standard normal random distribution follo... | Generating and Working with Random Vectors in R
For completeness sake, here's a follow-up note on how to generate random vectors regardless of the marginal distribution of the individual components. I'm going to stick with the bivariate case:
Gene |
52,935 | Truncating data reduces correlation? | There's a number of ways to look at it, but this is a pretty straightforward one:
Imagine for a moment we're looking at a regression problem. The squared correlation between the two variables ($r^2$) is $R^2$, the coefficient of determination, which is $1-\frac{s^2_\epsilon}{\text{Var}(y)}$. When you restrict the range... | Truncating data reduces correlation? | There's a number of ways to look at it, but this is a pretty straightforward one:
Imagine for a moment we're looking at a regression problem. The squared correlation between the two variables ($r^2$) | Truncating data reduces correlation?
There's a number of ways to look at it, but this is a pretty straightforward one:
Imagine for a moment we're looking at a regression problem. The squared correlation between the two variables ($r^2$) is $R^2$, the coefficient of determination, which is $1-\frac{s^2_\epsilon}{\text{V... | Truncating data reduces correlation?
There's a number of ways to look at it, but this is a pretty straightforward one:
Imagine for a moment we're looking at a regression problem. The squared correlation between the two variables ($r^2$) |
52,936 | Truncating data reduces correlation? | Thinking of a 2D plot of one variable plotted against the other, limiting the range for one variable means only looking at a vertical or horizontal "slice".
So my intuition is that the overall shape of the "cloud" of points will be more vertical or more horizontal, instead of "diagonal". A vertical or horizontal-lookin... | Truncating data reduces correlation? | Thinking of a 2D plot of one variable plotted against the other, limiting the range for one variable means only looking at a vertical or horizontal "slice".
So my intuition is that the overall shape o | Truncating data reduces correlation?
Thinking of a 2D plot of one variable plotted against the other, limiting the range for one variable means only looking at a vertical or horizontal "slice".
So my intuition is that the overall shape of the "cloud" of points will be more vertical or more horizontal, instead of "diago... | Truncating data reduces correlation?
Thinking of a 2D plot of one variable plotted against the other, limiting the range for one variable means only looking at a vertical or horizontal "slice".
So my intuition is that the overall shape o |
52,937 | Why does Pearson's chi-squared test detect differences that the GLM model fails to detect? | I don't see a big difference in the results:
d = read.table(text="Group Black Red
A 296 14
B 292 16
C 301 7
D 289 23", header=T)
chisq.test(d[,2:3])
# Pearson's Chi-squared test
#
# data: d[, 2:3]... | Why does Pearson's chi-squared test detect differences that the GLM model fails to detect? | I don't see a big difference in the results:
d = read.table(text="Group Black Red
A 296 14
B 292 16
C 301 7
| Why does Pearson's chi-squared test detect differences that the GLM model fails to detect?
I don't see a big difference in the results:
d = read.table(text="Group Black Red
A 296 14
B 292 16
C 301 7
D 2... | Why does Pearson's chi-squared test detect differences that the GLM model fails to detect?
I don't see a big difference in the results:
d = read.table(text="Group Black Red
A 296 14
B 292 16
C 301 7
|
52,938 | How to draw a random sample from a Generalized Beta distribution of the second kind | If you consider the density$$f(y;a,b)=\frac{|a|y^{ap-1}}{b^{ap}B(p,q)(1+(y/b)^a)^{p+q}},$$b appears as a scale parameter. This means that, if $Z\sim f(z;a,1)$, then $bZ\sim f(z;a,b)$. So we can assume $b=1$ wlog. Now,$$f(y;a,1)=\frac{|a|y^{ap-1}}{B(p,q)(1+y^a)^{p+q}}=\frac{|a|\{y^{a}\}^{(ap-1)/a}}{B(p,q)(1+y^a)^{p+q}}$... | How to draw a random sample from a Generalized Beta distribution of the second kind | If you consider the density$$f(y;a,b)=\frac{|a|y^{ap-1}}{b^{ap}B(p,q)(1+(y/b)^a)^{p+q}},$$b appears as a scale parameter. This means that, if $Z\sim f(z;a,1)$, then $bZ\sim f(z;a,b)$. So we can assume | How to draw a random sample from a Generalized Beta distribution of the second kind
If you consider the density$$f(y;a,b)=\frac{|a|y^{ap-1}}{b^{ap}B(p,q)(1+(y/b)^a)^{p+q}},$$b appears as a scale parameter. This means that, if $Z\sim f(z;a,1)$, then $bZ\sim f(z;a,b)$. So we can assume $b=1$ wlog. Now,$$f(y;a,1)=\frac{|a... | How to draw a random sample from a Generalized Beta distribution of the second kind
If you consider the density$$f(y;a,b)=\frac{|a|y^{ap-1}}{b^{ap}B(p,q)(1+(y/b)^a)^{p+q}},$$b appears as a scale parameter. This means that, if $Z\sim f(z;a,1)$, then $bZ\sim f(z;a,b)$. So we can assume |
52,939 | How to draw a random sample from a Generalized Beta distribution of the second kind | While Xi'an's answer of course already addresses the underlying theory, you may also find the GB2 package in R to be convenient.
In particular, rgb2(x, shape1, scale, shape2, shape3) will produce random draws. | How to draw a random sample from a Generalized Beta distribution of the second kind | While Xi'an's answer of course already addresses the underlying theory, you may also find the GB2 package in R to be convenient.
In particular, rgb2(x, shape1, scale, shape2, shape3) will produce rand | How to draw a random sample from a Generalized Beta distribution of the second kind
While Xi'an's answer of course already addresses the underlying theory, you may also find the GB2 package in R to be convenient.
In particular, rgb2(x, shape1, scale, shape2, shape3) will produce random draws. | How to draw a random sample from a Generalized Beta distribution of the second kind
While Xi'an's answer of course already addresses the underlying theory, you may also find the GB2 package in R to be convenient.
In particular, rgb2(x, shape1, scale, shape2, shape3) will produce rand |
52,940 | Is "Confidence Level" just 1 minus P-Value? | Is "Confidence Level" just 1 minus P-Value?
No.
If the results of a Chi-Square test give a P-Value of 0.01 then can we say that the confidence level in their being a difference is (1-0.01) = 99% confidence.
This is misusing the terms. Words like "confidence" and "p-value" have a specific meaning in frequentist stat... | Is "Confidence Level" just 1 minus P-Value? | Is "Confidence Level" just 1 minus P-Value?
No.
If the results of a Chi-Square test give a P-Value of 0.01 then can we say that the confidence level in their being a difference is (1-0.01) = 99% co | Is "Confidence Level" just 1 minus P-Value?
Is "Confidence Level" just 1 minus P-Value?
No.
If the results of a Chi-Square test give a P-Value of 0.01 then can we say that the confidence level in their being a difference is (1-0.01) = 99% confidence.
This is misusing the terms. Words like "confidence" and "p-value"... | Is "Confidence Level" just 1 minus P-Value?
Is "Confidence Level" just 1 minus P-Value?
No.
If the results of a Chi-Square test give a P-Value of 0.01 then can we say that the confidence level in their being a difference is (1-0.01) = 99% co |
52,941 | Is "Confidence Level" just 1 minus P-Value? | With Bayesian statistics, you could conceivably turn it into a statement about "99% confident that version B performed better than version A." However, neither the p-value nor the 99% confidence interval will let you conclude that. See Why does a 95% Confidence Interval (CI) not imply a 95% chance of containing the m... | Is "Confidence Level" just 1 minus P-Value? | With Bayesian statistics, you could conceivably turn it into a statement about "99% confident that version B performed better than version A." However, neither the p-value nor the 99% confidence inte | Is "Confidence Level" just 1 minus P-Value?
With Bayesian statistics, you could conceivably turn it into a statement about "99% confident that version B performed better than version A." However, neither the p-value nor the 99% confidence interval will let you conclude that. See Why does a 95% Confidence Interval (CI... | Is "Confidence Level" just 1 minus P-Value?
With Bayesian statistics, you could conceivably turn it into a statement about "99% confident that version B performed better than version A." However, neither the p-value nor the 99% confidence inte |
52,942 | If Manhattan distance always performs better on a dataset...what does it mean? | Also use the search terms l1 norm, l1 distance, absolute deviance etc all of which refer to the same thing as manhattan distance.
The properties of the l1-norm (manhattan distance) can largely be deduced from its shape (ie it is V shaped instead of U shaped like the parabola of the l2-norm (euclidian distance). The l1-... | If Manhattan distance always performs better on a dataset...what does it mean? | Also use the search terms l1 norm, l1 distance, absolute deviance etc all of which refer to the same thing as manhattan distance.
The properties of the l1-norm (manhattan distance) can largely be dedu | If Manhattan distance always performs better on a dataset...what does it mean?
Also use the search terms l1 norm, l1 distance, absolute deviance etc all of which refer to the same thing as manhattan distance.
The properties of the l1-norm (manhattan distance) can largely be deduced from its shape (ie it is V shaped ins... | If Manhattan distance always performs better on a dataset...what does it mean?
Also use the search terms l1 norm, l1 distance, absolute deviance etc all of which refer to the same thing as manhattan distance.
The properties of the l1-norm (manhattan distance) can largely be dedu |
52,943 | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$ | Let's do this the roundabout way (the direct way is what @JohnK's answer remarked).
To consider the expected value, we need to treat the variables involved as random variables. To stress this we write
$$E[F(X_{(n)})-F(X_{(1)})]$$
and we set $F(X_{(n)}) \equiv Z$, $F(X_{(1)}) \equiv Y$,
so we want to calculate
$$E[F(X... | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$ | Let's do this the roundabout way (the direct way is what @JohnK's answer remarked).
To consider the expected value, we need to treat the variables involved as random variables. To stress this we wri | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$
Let's do this the roundabout way (the direct way is what @JohnK's answer remarked).
To consider the expected value, we need to treat the variables involved as random variables. To st... | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$
Let's do this the roundabout way (the direct way is what @JohnK's answer remarked).
To consider the expected value, we need to treat the variables involved as random variables. To stress this we wri |
52,944 | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$ | You need to consider the Probability Integral Transform. For ease of notation denote the order statistics from smallest to largest by $Y_1, \ldots, Y_n $
You can afterwards see that
$$E \left[ F \left(Y_n \right)-F \left( Y_1 \right) \right] $$
is basically the expected value of the difference between the maximum and... | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$ | You need to consider the Probability Integral Transform. For ease of notation denote the order statistics from smallest to largest by $Y_1, \ldots, Y_n $
You can afterwards see that
$$E \left[ F \lef | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$
You need to consider the Probability Integral Transform. For ease of notation denote the order statistics from smallest to largest by $Y_1, \ldots, Y_n $
You can afterwards see that
$... | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$
You need to consider the Probability Integral Transform. For ease of notation denote the order statistics from smallest to largest by $Y_1, \ldots, Y_n $
You can afterwards see that
$$E \left[ F \lef |
52,945 | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$ | You can distribute the expectation function. Then you would have the E[x_n]-E[x_1]. Then you need to find the CDF of the max and min order statistic. From there you can find the expectation of each. To find the CDF of the max and min order statistic use the CDF substitution method and use the fact that x_1,...,x_n are ... | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$ | You can distribute the expectation function. Then you would have the E[x_n]-E[x_1]. Then you need to find the CDF of the max and min order statistic. From there you can find the expectation of each. T | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$
You can distribute the expectation function. Then you would have the E[x_n]-E[x_1]. Then you need to find the CDF of the max and min order statistic. From there you can find the expect... | $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$
You can distribute the expectation function. Then you would have the E[x_n]-E[x_1]. Then you need to find the CDF of the max and min order statistic. From there you can find the expectation of each. T |
52,946 | Comparison of machine learning algorithms | For classification algorithms this would be a good start: Statistical Comparisons of Classifiers over Multiple Data Sets.
To summarize this excellent paper: perform a Friedman test to determine if there is any significant difference between the classifiers and follow-up with an appropriate post-hoc test if there is:
t... | Comparison of machine learning algorithms | For classification algorithms this would be a good start: Statistical Comparisons of Classifiers over Multiple Data Sets.
To summarize this excellent paper: perform a Friedman test to determine if the | Comparison of machine learning algorithms
For classification algorithms this would be a good start: Statistical Comparisons of Classifiers over Multiple Data Sets.
To summarize this excellent paper: perform a Friedman test to determine if there is any significant difference between the classifiers and follow-up with an... | Comparison of machine learning algorithms
For classification algorithms this would be a good start: Statistical Comparisons of Classifiers over Multiple Data Sets.
To summarize this excellent paper: perform a Friedman test to determine if the |
52,947 | Comparison of machine learning algorithms | The notion of which Machine Learning algorithm is best is not universal, rather specific to the problem or the dataset you are dealing with.
In case of a single dataset or a problem, apply all learning algorithms and check the performance on out of sample data. Calculate Root mean Square errors between the predicted a... | Comparison of machine learning algorithms | The notion of which Machine Learning algorithm is best is not universal, rather specific to the problem or the dataset you are dealing with.
In case of a single dataset or a problem, apply all learni | Comparison of machine learning algorithms
The notion of which Machine Learning algorithm is best is not universal, rather specific to the problem or the dataset you are dealing with.
In case of a single dataset or a problem, apply all learning algorithms and check the performance on out of sample data. Calculate Root ... | Comparison of machine learning algorithms
The notion of which Machine Learning algorithm is best is not universal, rather specific to the problem or the dataset you are dealing with.
In case of a single dataset or a problem, apply all learni |
52,948 | Does Stationarity for Time Series extend to Independent Variables? | Stationarity should be sought for both to avoid any spurious correlations (that some or all variables actually just increase/decrease over time, independent of other factors), and there are methods of correction that can be applied to the entire model (i.e., including a trend as a DV) or just to specific variables (i.e... | Does Stationarity for Time Series extend to Independent Variables? | Stationarity should be sought for both to avoid any spurious correlations (that some or all variables actually just increase/decrease over time, independent of other factors), and there are methods of | Does Stationarity for Time Series extend to Independent Variables?
Stationarity should be sought for both to avoid any spurious correlations (that some or all variables actually just increase/decrease over time, independent of other factors), and there are methods of correction that can be applied to the entire model (... | Does Stationarity for Time Series extend to Independent Variables?
Stationarity should be sought for both to avoid any spurious correlations (that some or all variables actually just increase/decrease over time, independent of other factors), and there are methods of |
52,949 | Does Stationarity for Time Series extend to Independent Variables? | Assume you have time series on a variable $Y$ and on a variable $X$, say both non-negative, and based on some idea of yours you believe that there is a reasonable argument that they can be linked by a linear relationship
$$y_t = \beta x_t + u_t \tag{1}$$
Now say that your theoretical argument is basically sound, but i... | Does Stationarity for Time Series extend to Independent Variables? | Assume you have time series on a variable $Y$ and on a variable $X$, say both non-negative, and based on some idea of yours you believe that there is a reasonable argument that they can be linked by a | Does Stationarity for Time Series extend to Independent Variables?
Assume you have time series on a variable $Y$ and on a variable $X$, say both non-negative, and based on some idea of yours you believe that there is a reasonable argument that they can be linked by a linear relationship
$$y_t = \beta x_t + u_t \tag{1}... | Does Stationarity for Time Series extend to Independent Variables?
Assume you have time series on a variable $Y$ and on a variable $X$, say both non-negative, and based on some idea of yours you believe that there is a reasonable argument that they can be linked by a |
52,950 | Does Stationarity for Time Series extend to Independent Variables? | Stationarity means, among other things, that a
All the random variables $X_t$ have the same distribution
All pairs of random variables $(X_t, X_s)$ have a joint distribution that depends only on the difference $t-s$ and not on the
individual values of $t$ and $s$: the joint distribution of $(X_1,X_3)$ is the
same as t... | Does Stationarity for Time Series extend to Independent Variables? | Stationarity means, among other things, that a
All the random variables $X_t$ have the same distribution
All pairs of random variables $(X_t, X_s)$ have a joint distribution that depends only on the | Does Stationarity for Time Series extend to Independent Variables?
Stationarity means, among other things, that a
All the random variables $X_t$ have the same distribution
All pairs of random variables $(X_t, X_s)$ have a joint distribution that depends only on the difference $t-s$ and not on the
individual values of ... | Does Stationarity for Time Series extend to Independent Variables?
Stationarity means, among other things, that a
All the random variables $X_t$ have the same distribution
All pairs of random variables $(X_t, X_s)$ have a joint distribution that depends only on the |
52,951 | Expected value of q given y is weighted average of mean q and and y | All you need to know is that the regression of $q$ on $y$ is determined by standardizing both variables and their correlation coefficient will be the slope.
(In particular this result owes nothing to the assumptions that distributions are Normal; the independence of $q$ and $u$ is sufficient. Thus it will be most reve... | Expected value of q given y is weighted average of mean q and and y | All you need to know is that the regression of $q$ on $y$ is determined by standardizing both variables and their correlation coefficient will be the slope.
(In particular this result owes nothing to | Expected value of q given y is weighted average of mean q and and y
All you need to know is that the regression of $q$ on $y$ is determined by standardizing both variables and their correlation coefficient will be the slope.
(In particular this result owes nothing to the assumptions that distributions are Normal; the i... | Expected value of q given y is weighted average of mean q and and y
All you need to know is that the regression of $q$ on $y$ is determined by standardizing both variables and their correlation coefficient will be the slope.
(In particular this result owes nothing to |
52,952 | Expected value of q given y is weighted average of mean q and and y | The model implies that $y\sim\mathcal{N}(q,\sigma^2_u)$ and $q\sim\mathcal{N}(a,\sigma^2_q)$. By Bayes' rule:
$$p(q\mid y)\propto p(y\mid q,\sigma^2_u)p(q)$$
Ignoring constant factors (see here for a similar development):
$$\begin{align}p(q\mid y) & \propto \exp\left\{-\frac{(y-q)^2}{2\sigma^2_u}-\frac{(q-a)^2}{\sigma^... | Expected value of q given y is weighted average of mean q and and y | The model implies that $y\sim\mathcal{N}(q,\sigma^2_u)$ and $q\sim\mathcal{N}(a,\sigma^2_q)$. By Bayes' rule:
$$p(q\mid y)\propto p(y\mid q,\sigma^2_u)p(q)$$
Ignoring constant factors (see here for a | Expected value of q given y is weighted average of mean q and and y
The model implies that $y\sim\mathcal{N}(q,\sigma^2_u)$ and $q\sim\mathcal{N}(a,\sigma^2_q)$. By Bayes' rule:
$$p(q\mid y)\propto p(y\mid q,\sigma^2_u)p(q)$$
Ignoring constant factors (see here for a similar development):
$$\begin{align}p(q\mid y) & \p... | Expected value of q given y is weighted average of mean q and and y
The model implies that $y\sim\mathcal{N}(q,\sigma^2_u)$ and $q\sim\mathcal{N}(a,\sigma^2_q)$. By Bayes' rule:
$$p(q\mid y)\propto p(y\mid q,\sigma^2_u)p(q)$$
Ignoring constant factors (see here for a |
52,953 | Expected value of q given y is weighted average of mean q and and y | Another way, the shortest one ;-)
In general, if $X$ and $Y$ have a bivariate normal distribution, then (Anderson, Theorem 2.5.1):
$$E[X\mid Y]=E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}(Y-E[X])
=\left(1-\frac{\text{Cov}(X,Y)}{V[Y]}\right)E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}Y$$
i.e. "expected value of X given Y is weighted average... | Expected value of q given y is weighted average of mean q and and y | Another way, the shortest one ;-)
In general, if $X$ and $Y$ have a bivariate normal distribution, then (Anderson, Theorem 2.5.1):
$$E[X\mid Y]=E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}(Y-E[X])
=\left(1-\frac | Expected value of q given y is weighted average of mean q and and y
Another way, the shortest one ;-)
In general, if $X$ and $Y$ have a bivariate normal distribution, then (Anderson, Theorem 2.5.1):
$$E[X\mid Y]=E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}(Y-E[X])
=\left(1-\frac{\text{Cov}(X,Y)}{V[Y]}\right)E[X]+\frac{\text{Cov}(... | Expected value of q given y is weighted average of mean q and and y
Another way, the shortest one ;-)
In general, if $X$ and $Y$ have a bivariate normal distribution, then (Anderson, Theorem 2.5.1):
$$E[X\mid Y]=E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}(Y-E[X])
=\left(1-\frac |
52,954 | Expected value of q given y is weighted average of mean q and and y | I think the following argument shows why, unfortunately it's a messy. Much more elegant derivations are certainly out there somewhere as the linear Gaussian case is the best understood statistical model in existence.
Anyway, we have that:
U~N(0, sigma²)
Q~N(alpha, beta^2)
Y=Q+U.
U and Q are independent.
Because a li... | Expected value of q given y is weighted average of mean q and and y | I think the following argument shows why, unfortunately it's a messy. Much more elegant derivations are certainly out there somewhere as the linear Gaussian case is the best understood statistical mod | Expected value of q given y is weighted average of mean q and and y
I think the following argument shows why, unfortunately it's a messy. Much more elegant derivations are certainly out there somewhere as the linear Gaussian case is the best understood statistical model in existence.
Anyway, we have that:
U~N(0, sigm... | Expected value of q given y is weighted average of mean q and and y
I think the following argument shows why, unfortunately it's a messy. Much more elegant derivations are certainly out there somewhere as the linear Gaussian case is the best understood statistical mod |
52,955 | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant? | There is less to this issue than it seems. The real answer isn't that you cannot interpret the main effects at all, but rather that it is very difficult to interpret them correctly. The reason for the warning not to interpret the main effects is because people will inevitably interpret them incorrectly.
If there is... | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the inter | There is less to this issue than it seems. The real answer isn't that you cannot interpret the main effects at all, but rather that it is very difficult to interpret them correctly. The reason for t | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant?
There is less to this issue than it seems. The real answer isn't that you cannot interpret the main effects at all, but rather that it is very difficult to interpret them correctly. The re... | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the inter
There is less to this issue than it seems. The real answer isn't that you cannot interpret the main effects at all, but rather that it is very difficult to interpret them correctly. The reason for t |
52,956 | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant? | This is an interesting question. Since I don't like gross generalization, I am going to disagree with the suggestion that you should "never" interpret the main effects at all if an interaction is present. Never is just to strong (even if some might argue that there are clear situations where the interaction tells you w... | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the inter | This is an interesting question. Since I don't like gross generalization, I am going to disagree with the suggestion that you should "never" interpret the main effects at all if an interaction is pres | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant?
This is an interesting question. Since I don't like gross generalization, I am going to disagree with the suggestion that you should "never" interpret the main effects at all if an interacti... | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the inter
This is an interesting question. Since I don't like gross generalization, I am going to disagree with the suggestion that you should "never" interpret the main effects at all if an interaction is pres |
52,957 | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant? | When you have an interaction, the real issue is whether the main effects (significant or not) are “descriptive” or “misleading”.
Here is an example of data with a significant interaction and “descriptive” main effect for Task Presentation (Computer does better than Paper overall, and for Easy Tasks and for Hard tasks)
... | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the inter | When you have an interaction, the real issue is whether the main effects (significant or not) are “descriptive” or “misleading”.
Here is an example of data with a significant interaction and “descript | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant?
When you have an interaction, the real issue is whether the main effects (significant or not) are “descriptive” or “misleading”.
Here is an example of data with a significant interaction and... | How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the inter
When you have an interaction, the real issue is whether the main effects (significant or not) are “descriptive” or “misleading”.
Here is an example of data with a significant interaction and “descript |
52,958 | The value of adding the ROC graph if the AUC is given | I usually give the ROC plot but not the AUC: For my applications it is usually clear that either a specific or a sensitive regocognition is needed. The ROC is different for classifiers that are specific but not sensitive vs. sensitive but not specific while the AUC hides this information.
Besides, one can put a whole l... | The value of adding the ROC graph if the AUC is given | I usually give the ROC plot but not the AUC: For my applications it is usually clear that either a specific or a sensitive regocognition is needed. The ROC is different for classifiers that are specif | The value of adding the ROC graph if the AUC is given
I usually give the ROC plot but not the AUC: For my applications it is usually clear that either a specific or a sensitive regocognition is needed. The ROC is different for classifiers that are specific but not sensitive vs. sensitive but not specific while the AUC ... | The value of adding the ROC graph if the AUC is given
I usually give the ROC plot but not the AUC: For my applications it is usually clear that either a specific or a sensitive regocognition is needed. The ROC is different for classifiers that are specif |
52,959 | The value of adding the ROC graph if the AUC is given | I have not seen a single example where the graph changes our actions or the way we think. I think the ink:information ratio in an ROC graph is enormous. But the worst thing about it is that it tempts us to try to select a cutoff for the predicted risk, which is arbitrary, and inconsistent with optimum decision making... | The value of adding the ROC graph if the AUC is given | I have not seen a single example where the graph changes our actions or the way we think. I think the ink:information ratio in an ROC graph is enormous. But the worst thing about it is that it tempt | The value of adding the ROC graph if the AUC is given
I have not seen a single example where the graph changes our actions or the way we think. I think the ink:information ratio in an ROC graph is enormous. But the worst thing about it is that it tempts us to try to select a cutoff for the predicted risk, which is ar... | The value of adding the ROC graph if the AUC is given
I have not seen a single example where the graph changes our actions or the way we think. I think the ink:information ratio in an ROC graph is enormous. But the worst thing about it is that it tempt |
52,960 | The value of adding the ROC graph if the AUC is given | The ROC curve is the specificity/sensitivity plot; the AUC is the Area Under Curve.
To be brief, the ROC curve can be interesting because it allows comparison of the sensitivity/specificity behaviour of the model. More simply:
$ROC = (x,y) \in R^2 \Rightarrow AUC = z$ but $AUC = z \nRightarrow ROC = (x,y) \in R^2$ | The value of adding the ROC graph if the AUC is given | The ROC curve is the specificity/sensitivity plot; the AUC is the Area Under Curve.
To be brief, the ROC curve can be interesting because it allows comparison of the sensitivity/specificity behaviour | The value of adding the ROC graph if the AUC is given
The ROC curve is the specificity/sensitivity plot; the AUC is the Area Under Curve.
To be brief, the ROC curve can be interesting because it allows comparison of the sensitivity/specificity behaviour of the model. More simply:
$ROC = (x,y) \in R^2 \Rightarrow AUC =... | The value of adding the ROC graph if the AUC is given
The ROC curve is the specificity/sensitivity plot; the AUC is the Area Under Curve.
To be brief, the ROC curve can be interesting because it allows comparison of the sensitivity/specificity behaviour |
52,961 | The value of adding the ROC graph if the AUC is given | There is great value in showing the entire ROC esp. when comparing two different classifiers, as it helps us to see whether different curves cross each other. One is not superior to the other, overall, if they cross - see the Figure 3 (screenshot shown below) from
Seong Ho Park, J. M. G. C.-H. J. (2004). Receiver Oper... | The value of adding the ROC graph if the AUC is given | There is great value in showing the entire ROC esp. when comparing two different classifiers, as it helps us to see whether different curves cross each other. One is not superior to the other, overall | The value of adding the ROC graph if the AUC is given
There is great value in showing the entire ROC esp. when comparing two different classifiers, as it helps us to see whether different curves cross each other. One is not superior to the other, overall, if they cross - see the Figure 3 (screenshot shown below) from
... | The value of adding the ROC graph if the AUC is given
There is great value in showing the entire ROC esp. when comparing two different classifiers, as it helps us to see whether different curves cross each other. One is not superior to the other, overall |
52,962 | Is there a better way than side-by-side barplots to compare binned data from different series | I agree with the principle that using more detail, as in looking at the entire distributions or sets of quantiles, would be much better if the data were available. Conversely, converting what you have to quartiles just discards yet more information and is not a good idea here.
You are right that side-by-side or back-t... | Is there a better way than side-by-side barplots to compare binned data from different series | I agree with the principle that using more detail, as in looking at the entire distributions or sets of quantiles, would be much better if the data were available. Conversely, converting what you have | Is there a better way than side-by-side barplots to compare binned data from different series
I agree with the principle that using more detail, as in looking at the entire distributions or sets of quantiles, would be much better if the data were available. Conversely, converting what you have to quartiles just discard... | Is there a better way than side-by-side barplots to compare binned data from different series
I agree with the principle that using more detail, as in looking at the entire distributions or sets of quantiles, would be much better if the data were available. Conversely, converting what you have |
52,963 | Is there a better way than side-by-side barplots to compare binned data from different series | The problem with bars is they don't overlay well. Dots are one alternative and lines are another. If you have the full data there are still others (box plots, violin plots, ...). Nick Cox's answer shows dots, and it's worth highlighting lines in this case since a it's so similar to the frequency polygon use.
I don't k... | Is there a better way than side-by-side barplots to compare binned data from different series | The problem with bars is they don't overlay well. Dots are one alternative and lines are another. If you have the full data there are still others (box plots, violin plots, ...). Nick Cox's answer sho | Is there a better way than side-by-side barplots to compare binned data from different series
The problem with bars is they don't overlay well. Dots are one alternative and lines are another. If you have the full data there are still others (box plots, violin plots, ...). Nick Cox's answer shows dots, and it's worth hi... | Is there a better way than side-by-side barplots to compare binned data from different series
The problem with bars is they don't overlay well. Dots are one alternative and lines are another. If you have the full data there are still others (box plots, violin plots, ...). Nick Cox's answer sho |
52,964 | Understanding the role of the chi-squared distribution in the confidence interval for the variance | Variance is not normally distributed, because variance is the average of the squared deviations of each datum from the mean of the distribution. If all data points in your dataset are identical, then the deviations would each be zero, and so would the squared deviations and their average. Thus, $0$ is the lowest vari... | Understanding the role of the chi-squared distribution in the confidence interval for the variance | Variance is not normally distributed, because variance is the average of the squared deviations of each datum from the mean of the distribution. If all data points in your dataset are identical, then | Understanding the role of the chi-squared distribution in the confidence interval for the variance
Variance is not normally distributed, because variance is the average of the squared deviations of each datum from the mean of the distribution. If all data points in your dataset are identical, then the deviations would... | Understanding the role of the chi-squared distribution in the confidence interval for the variance
Variance is not normally distributed, because variance is the average of the squared deviations of each datum from the mean of the distribution. If all data points in your dataset are identical, then |
52,965 | Understanding the role of the chi-squared distribution in the confidence interval for the variance | The chi squared distribution is indeed related to the normal distribution. Specifically, a chi squared variable with N degrees of freedom is equivalent to the distribution of the sum of N squared independent standard normal random variables, which would be the same as the sum of N squared independent Z scores from a no... | Understanding the role of the chi-squared distribution in the confidence interval for the variance | The chi squared distribution is indeed related to the normal distribution. Specifically, a chi squared variable with N degrees of freedom is equivalent to the distribution of the sum of N squared inde | Understanding the role of the chi-squared distribution in the confidence interval for the variance
The chi squared distribution is indeed related to the normal distribution. Specifically, a chi squared variable with N degrees of freedom is equivalent to the distribution of the sum of N squared independent standard norm... | Understanding the role of the chi-squared distribution in the confidence interval for the variance
The chi squared distribution is indeed related to the normal distribution. Specifically, a chi squared variable with N degrees of freedom is equivalent to the distribution of the sum of N squared inde |
52,966 | How to extract values of the forecasted times series of auto.arima [closed] | You didn't provide your data. So I am guessing what you are looking for: I think you want to get something like this:
> library(forecast)
> fit=Arima(WWWusage,c(3,1,0))
> AA11a<-forecast(fit)
> AA11a$lower
80% 95%
[1,] 215.7393 213.6634
[2,] 209.9265 205.0016
[3,] 203.8380 196.1947
... | How to extract values of the forecasted times series of auto.arima [closed] | You didn't provide your data. So I am guessing what you are looking for: I think you want to get something like this:
> library(forecast)
> fit=Arima(WWWusage,c(3,1,0))
> AA11a<-forecast(fit)
> AA11a$ | How to extract values of the forecasted times series of auto.arima [closed]
You didn't provide your data. So I am guessing what you are looking for: I think you want to get something like this:
> library(forecast)
> fit=Arima(WWWusage,c(3,1,0))
> AA11a<-forecast(fit)
> AA11a$lower
80% 95%
[1,] ... | How to extract values of the forecasted times series of auto.arima [closed]
You didn't provide your data. So I am guessing what you are looking for: I think you want to get something like this:
> library(forecast)
> fit=Arima(WWWusage,c(3,1,0))
> AA11a<-forecast(fit)
> AA11a$ |
52,967 | How to extract values of the forecasted times series of auto.arima [closed] | To extract the mean of the prediction interval as a numeric vector use:
> as.numeric(AA11a$mean)
[1] 219.6608 219.2299 218.2766 217.3484 216.7633 216.3785 216.0062 215.6326 215.3175
[10] 215.0749
The mean itself is a tsclass
> class(AA11a$mean)
[1] "ts" | How to extract values of the forecasted times series of auto.arima [closed] | To extract the mean of the prediction interval as a numeric vector use:
> as.numeric(AA11a$mean)
[1] 219.6608 219.2299 218.2766 217.3484 216.7633 216.3785 216.0062 215.6326 215.3175
[10] 215.0749
Th | How to extract values of the forecasted times series of auto.arima [closed]
To extract the mean of the prediction interval as a numeric vector use:
> as.numeric(AA11a$mean)
[1] 219.6608 219.2299 218.2766 217.3484 216.7633 216.3785 216.0062 215.6326 215.3175
[10] 215.0749
The mean itself is a tsclass
> class(AA11a$mea... | How to extract values of the forecasted times series of auto.arima [closed]
To extract the mean of the prediction interval as a numeric vector use:
> as.numeric(AA11a$mean)
[1] 219.6608 219.2299 218.2766 217.3484 216.7633 216.3785 216.0062 215.6326 215.3175
[10] 215.0749
Th |
52,968 | Present numerical results with variable system parameters | Because contour plots--especially 3D contour plots--are usually difficult to interpret and plots of $I$ against $u$ are familiar to physicists, consider a small multiple of such plots where $A$ and $B$ range through selected values.
Experimentation is in order. You might, for instance, overlay multiple graphs for fixe... | Present numerical results with variable system parameters | Because contour plots--especially 3D contour plots--are usually difficult to interpret and plots of $I$ against $u$ are familiar to physicists, consider a small multiple of such plots where $A$ and $B | Present numerical results with variable system parameters
Because contour plots--especially 3D contour plots--are usually difficult to interpret and plots of $I$ against $u$ are familiar to physicists, consider a small multiple of such plots where $A$ and $B$ range through selected values.
Experimentation is in order. ... | Present numerical results with variable system parameters
Because contour plots--especially 3D contour plots--are usually difficult to interpret and plots of $I$ against $u$ are familiar to physicists, consider a small multiple of such plots where $A$ and $B |
52,969 | Present numerical results with variable system parameters | What would be "typical" values? What would be extreme values?
You can think of I as a function of three variables $I(u,A,B)$. Since $u, A, B \in \mathbb{R}$, it's not that easy to represent. You can either make a 1d plot of $I(u;A,B)$ vs $u$ for fixed values of $A$ and $B$, where you then have to take some representat... | Present numerical results with variable system parameters | What would be "typical" values? What would be extreme values?
You can think of I as a function of three variables $I(u,A,B)$. Since $u, A, B \in \mathbb{R}$, it's not that easy to represent. You can | Present numerical results with variable system parameters
What would be "typical" values? What would be extreme values?
You can think of I as a function of three variables $I(u,A,B)$. Since $u, A, B \in \mathbb{R}$, it's not that easy to represent. You can either make a 1d plot of $I(u;A,B)$ vs $u$ for fixed values of... | Present numerical results with variable system parameters
What would be "typical" values? What would be extreme values?
You can think of I as a function of three variables $I(u,A,B)$. Since $u, A, B \in \mathbb{R}$, it's not that easy to represent. You can |
52,970 | Multicollinearity when adding a confounding variable | First, just because two variables are fairly highly correlated does not mean they are colinear to a problematic degree. Problematic collinearity is best examined with condition indices. See the work of David Belsley or see my dissertation Collinearity Diagnostics in Multiple Regression: A Monte Carlo Study.
Second, i... | Multicollinearity when adding a confounding variable | First, just because two variables are fairly highly correlated does not mean they are colinear to a problematic degree. Problematic collinearity is best examined with condition indices. See the work | Multicollinearity when adding a confounding variable
First, just because two variables are fairly highly correlated does not mean they are colinear to a problematic degree. Problematic collinearity is best examined with condition indices. See the work of David Belsley or see my dissertation Collinearity Diagnostics in... | Multicollinearity when adding a confounding variable
First, just because two variables are fairly highly correlated does not mean they are colinear to a problematic degree. Problematic collinearity is best examined with condition indices. See the work |
52,971 | Multicollinearity when adding a confounding variable | If you cannot use a linear regression with a controlling variable, or related, the partial correlation, another option is to stratify on the temperature variable.
To do so, categorize temperature into multiple, say five, categories. Then you run five conditional regression models, i.e. for the units in each category o... | Multicollinearity when adding a confounding variable | If you cannot use a linear regression with a controlling variable, or related, the partial correlation, another option is to stratify on the temperature variable.
To do so, categorize temperature int | Multicollinearity when adding a confounding variable
If you cannot use a linear regression with a controlling variable, or related, the partial correlation, another option is to stratify on the temperature variable.
To do so, categorize temperature into multiple, say five, categories. Then you run five conditional reg... | Multicollinearity when adding a confounding variable
If you cannot use a linear regression with a controlling variable, or related, the partial correlation, another option is to stratify on the temperature variable.
To do so, categorize temperature int |
52,972 | Why is the average the right way to deal with Gaussian noise? | There are several senses in which the sample mean might be regarded as optimal as an estimator of $\theta$, some of which are optimal more generally than for the normal, and then there's senses in which it's optimal specifically for the normal.
If we take the Gauss-Markov theorem, the sample mean is the best linear unb... | Why is the average the right way to deal with Gaussian noise? | There are several senses in which the sample mean might be regarded as optimal as an estimator of $\theta$, some of which are optimal more generally than for the normal, and then there's senses in whi | Why is the average the right way to deal with Gaussian noise?
There are several senses in which the sample mean might be regarded as optimal as an estimator of $\theta$, some of which are optimal more generally than for the normal, and then there's senses in which it's optimal specifically for the normal.
If we take th... | Why is the average the right way to deal with Gaussian noise?
There are several senses in which the sample mean might be regarded as optimal as an estimator of $\theta$, some of which are optimal more generally than for the normal, and then there's senses in whi |
52,973 | Why is the average the right way to deal with Gaussian noise? | Here's the argument that the average is the maximum-likelihood estimator (MLE) for the parameter $\theta$.
Suppose we are given observations $x_1,\dots,x_n$ of the r.v.s $X_1,\dots,X_n$. The likelihood of $\hat{\theta}$, given $x_1,\dots,x_n$, is
$$L(\hat{\theta}) = \prod_i \Pr[X_i=x_i|\hat{\theta}] = \prod_i \Pr[Y_i=... | Why is the average the right way to deal with Gaussian noise? | Here's the argument that the average is the maximum-likelihood estimator (MLE) for the parameter $\theta$.
Suppose we are given observations $x_1,\dots,x_n$ of the r.v.s $X_1,\dots,X_n$. The likeliho | Why is the average the right way to deal with Gaussian noise?
Here's the argument that the average is the maximum-likelihood estimator (MLE) for the parameter $\theta$.
Suppose we are given observations $x_1,\dots,x_n$ of the r.v.s $X_1,\dots,X_n$. The likelihood of $\hat{\theta}$, given $x_1,\dots,x_n$, is
$$L(\hat{\... | Why is the average the right way to deal with Gaussian noise?
Here's the argument that the average is the maximum-likelihood estimator (MLE) for the parameter $\theta$.
Suppose we are given observations $x_1,\dots,x_n$ of the r.v.s $X_1,\dots,X_n$. The likeliho |
52,974 | Why is the average the right way to deal with Gaussian noise? | A quick remark in addition to the excellent answers above: the type of situation you describe is often discussed in classical test theory. In CTT you regard $X$ as independent repeated measures of the true score $\theta$. The major difference between your question and CTT is that $\theta$ is assumed to have a distribut... | Why is the average the right way to deal with Gaussian noise? | A quick remark in addition to the excellent answers above: the type of situation you describe is often discussed in classical test theory. In CTT you regard $X$ as independent repeated measures of the | Why is the average the right way to deal with Gaussian noise?
A quick remark in addition to the excellent answers above: the type of situation you describe is often discussed in classical test theory. In CTT you regard $X$ as independent repeated measures of the true score $\theta$. The major difference between your qu... | Why is the average the right way to deal with Gaussian noise?
A quick remark in addition to the excellent answers above: the type of situation you describe is often discussed in classical test theory. In CTT you regard $X$ as independent repeated measures of the |
52,975 | Developing a prediction model for bus stops | It seems like simple logistic regression would work well. Hopefully this matches well to the data you have. I've tried to lay off jargon as much as possible.
Let's confine our analysis to a single bus route for simplicity (you can simply repeat this procedure for other routes). The dependent/predicted variable you are ... | Developing a prediction model for bus stops | It seems like simple logistic regression would work well. Hopefully this matches well to the data you have. I've tried to lay off jargon as much as possible.
Let's confine our analysis to a single bus | Developing a prediction model for bus stops
It seems like simple logistic regression would work well. Hopefully this matches well to the data you have. I've tried to lay off jargon as much as possible.
Let's confine our analysis to a single bus route for simplicity (you can simply repeat this procedure for other routes... | Developing a prediction model for bus stops
It seems like simple logistic regression would work well. Hopefully this matches well to the data you have. I've tried to lay off jargon as much as possible.
Let's confine our analysis to a single bus |
52,976 | Developing a prediction model for bus stops | In order to predict any outcome (bus stop in your case), you need some information other than the outcome you want to predict. These variables are often called predictors / covariates / independent variables.
So the answer to your question depends on what information you have.
1. If, the bus GPS signal and door open ... | Developing a prediction model for bus stops | In order to predict any outcome (bus stop in your case), you need some information other than the outcome you want to predict. These variables are often called predictors / covariates / independent va | Developing a prediction model for bus stops
In order to predict any outcome (bus stop in your case), you need some information other than the outcome you want to predict. These variables are often called predictors / covariates / independent variables.
So the answer to your question depends on what information you hav... | Developing a prediction model for bus stops
In order to predict any outcome (bus stop in your case), you need some information other than the outcome you want to predict. These variables are often called predictors / covariates / independent va |
52,977 | Developing a prediction model for bus stops | I'm assuming, from your description of your data, that you effectively have a perfect record of where the bus opens its doors. So, for predicting what a bus will do at a particular stop, your data is what the bus did on previous shifts (entire routes), and what the bus did for previous stops on the current shift.
For t... | Developing a prediction model for bus stops | I'm assuming, from your description of your data, that you effectively have a perfect record of where the bus opens its doors. So, for predicting what a bus will do at a particular stop, your data is | Developing a prediction model for bus stops
I'm assuming, from your description of your data, that you effectively have a perfect record of where the bus opens its doors. So, for predicting what a bus will do at a particular stop, your data is what the bus did on previous shifts (entire routes), and what the bus did fo... | Developing a prediction model for bus stops
I'm assuming, from your description of your data, that you effectively have a perfect record of where the bus opens its doors. So, for predicting what a bus will do at a particular stop, your data is |
52,978 | Developing a prediction model for bus stops | In your question, you need to describe what data you have that are complete, jagged, and missing.
Firstly, how much data do you have for a single bus?
Does a bus always use the same route (for the data you have)? Or does the route change over time?
If you can assemble data that are not jagged, you may be able to deve... | Developing a prediction model for bus stops | In your question, you need to describe what data you have that are complete, jagged, and missing.
Firstly, how much data do you have for a single bus?
Does a bus always use the same route (for the d | Developing a prediction model for bus stops
In your question, you need to describe what data you have that are complete, jagged, and missing.
Firstly, how much data do you have for a single bus?
Does a bus always use the same route (for the data you have)? Or does the route change over time?
If you can assemble data ... | Developing a prediction model for bus stops
In your question, you need to describe what data you have that are complete, jagged, and missing.
Firstly, how much data do you have for a single bus?
Does a bus always use the same route (for the d |
52,979 | Different regularization parameter per parameter | Yes, it had been tried (including by myself - I tried it with neural nets, with rather mixed success). The Relevance Vector Machine (RVM) does pretty much exactly that, and the regularisation parameters are tuned by maximising the marginal likelihood. The advantage of this is that it leads to a sparse model where uni... | Different regularization parameter per parameter | Yes, it had been tried (including by myself - I tried it with neural nets, with rather mixed success). The Relevance Vector Machine (RVM) does pretty much exactly that, and the regularisation paramet | Different regularization parameter per parameter
Yes, it had been tried (including by myself - I tried it with neural nets, with rather mixed success). The Relevance Vector Machine (RVM) does pretty much exactly that, and the regularisation parameters are tuned by maximising the marginal likelihood. The advantage of ... | Different regularization parameter per parameter
Yes, it had been tried (including by myself - I tried it with neural nets, with rather mixed success). The Relevance Vector Machine (RVM) does pretty much exactly that, and the regularisation paramet |
52,980 | Different regularization parameter per parameter | Adaptive Lasso (H.Zou, JASA 2006, Vol. 101, No. 476) achieves consistency in parameter estimates by using individual lambda for each variable. Lambda values are tuned based on OLS solution (which unfortunately is not available in many practical cases where Lasso is used). | Different regularization parameter per parameter | Adaptive Lasso (H.Zou, JASA 2006, Vol. 101, No. 476) achieves consistency in parameter estimates by using individual lambda for each variable. Lambda values are tuned based on OLS solution (which unfo | Different regularization parameter per parameter
Adaptive Lasso (H.Zou, JASA 2006, Vol. 101, No. 476) achieves consistency in parameter estimates by using individual lambda for each variable. Lambda values are tuned based on OLS solution (which unfortunately is not available in many practical cases where Lasso is used)... | Different regularization parameter per parameter
Adaptive Lasso (H.Zou, JASA 2006, Vol. 101, No. 476) achieves consistency in parameter estimates by using individual lambda for each variable. Lambda values are tuned based on OLS solution (which unfo |
52,981 | Different regularization parameter per parameter | If you want to/are able to go nonparametric, this is implemented in the mgcv package, which implements penalized splines. If you use the option select=TRUE, the optimizer that selects smoothness penalties also adds a penalty term to the "main effect" of each smooth term, in addition to the penalty used for smoothness ... | Different regularization parameter per parameter | If you want to/are able to go nonparametric, this is implemented in the mgcv package, which implements penalized splines. If you use the option select=TRUE, the optimizer that selects smoothness pena | Different regularization parameter per parameter
If you want to/are able to go nonparametric, this is implemented in the mgcv package, which implements penalized splines. If you use the option select=TRUE, the optimizer that selects smoothness penalties also adds a penalty term to the "main effect" of each smooth term... | Different regularization parameter per parameter
If you want to/are able to go nonparametric, this is implemented in the mgcv package, which implements penalized splines. If you use the option select=TRUE, the optimizer that selects smoothness pena |
52,982 | Neural network modeling sample size | There are two rules of thumb that I know of:
There should be approximately 30 times more training cases than the number of weights (Neural Network FAQ)
General generalization rule: there should be 10 times more training cases than the VC dimension of the hypothesis set. In NN case the VC dimension is usually assumed t... | Neural network modeling sample size | There are two rules of thumb that I know of:
There should be approximately 30 times more training cases than the number of weights (Neural Network FAQ)
General generalization rule: there should be 10 | Neural network modeling sample size
There are two rules of thumb that I know of:
There should be approximately 30 times more training cases than the number of weights (Neural Network FAQ)
General generalization rule: there should be 10 times more training cases than the VC dimension of the hypothesis set. In NN case t... | Neural network modeling sample size
There are two rules of thumb that I know of:
There should be approximately 30 times more training cases than the number of weights (Neural Network FAQ)
General generalization rule: there should be 10 |
52,983 | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance functions | The proof is very similar to how you prove a bm is a gaussian process.
What i did not say at the end of proof is that, notice, by the stationary independent increment property of a BM, the increments in the brackets i lablelled normal are INDEPENDENT normal distributions, so linear combinations of independent normal di... | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance | The proof is very similar to how you prove a bm is a gaussian process.
What i did not say at the end of proof is that, notice, by the stationary independent increment property of a BM, the increments | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance functions
The proof is very similar to how you prove a bm is a gaussian process.
What i did not say at the end of proof is that, notice, by the stationary independent increment property of a BM, the increments in the bra... | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance
The proof is very similar to how you prove a bm is a gaussian process.
What i did not say at the end of proof is that, notice, by the stationary independent increment property of a BM, the increments |
52,984 | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance functions | Maybe I'm missing something, but this question seems to be easier than it, at first, appears to be. It's not a change of variable problem (which would be messy) but simply a change in labeling with a change of scale. $t$ is an index, not a random variable.
A process is Gaussian if every random subset of variables from ... | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance | Maybe I'm missing something, but this question seems to be easier than it, at first, appears to be. It's not a change of variable problem (which would be messy) but simply a change in labeling with a | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance functions
Maybe I'm missing something, but this question seems to be easier than it, at first, appears to be. It's not a change of variable problem (which would be messy) but simply a change in labeling with a change of ... | I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process and I find mean and covariance
Maybe I'm missing something, but this question seems to be easier than it, at first, appears to be. It's not a change of variable problem (which would be messy) but simply a change in labeling with a |
52,985 | Displaying fine variation across multiple orders of magnitude | Let's step back, and think how to represent data instead of how to visualize data. I love data visualization but I'd say that graph is not a suitable solution here.
Let's evaluate your requests:
I'd like to be able to plot the data in such a way that the viewer will be able to tell how the whole sequence increases
[F... | Displaying fine variation across multiple orders of magnitude | Let's step back, and think how to represent data instead of how to visualize data. I love data visualization but I'd say that graph is not a suitable solution here.
Let's evaluate your requests:
I'd | Displaying fine variation across multiple orders of magnitude
Let's step back, and think how to represent data instead of how to visualize data. I love data visualization but I'd say that graph is not a suitable solution here.
Let's evaluate your requests:
I'd like to be able to plot the data in such a way that the v... | Displaying fine variation across multiple orders of magnitude
Let's step back, and think how to represent data instead of how to visualize data. I love data visualization but I'd say that graph is not a suitable solution here.
Let's evaluate your requests:
I'd |
52,986 | Displaying fine variation across multiple orders of magnitude | Your idea of two plots with different scales is better than a broken axis since it gives you a perceptual view of how disparate the values are. Below is a quick mock-up. For presentation, the two graphs should have more separation and some cue or text that one is a zoomed in view of the other.
Log or some other transf... | Displaying fine variation across multiple orders of magnitude | Your idea of two plots with different scales is better than a broken axis since it gives you a perceptual view of how disparate the values are. Below is a quick mock-up. For presentation, the two grap | Displaying fine variation across multiple orders of magnitude
Your idea of two plots with different scales is better than a broken axis since it gives you a perceptual view of how disparate the values are. Below is a quick mock-up. For presentation, the two graphs should have more separation and some cue or text that o... | Displaying fine variation across multiple orders of magnitude
Your idea of two plots with different scales is better than a broken axis since it gives you a perceptual view of how disparate the values are. Below is a quick mock-up. For presentation, the two grap |
52,987 | When to use residual plots? | They are still useful in assessing whether the relationship between the explanatory variables and the dependent variable is linear (or modeled properly given the equation). For an extreme example, I generated some data with a quadratic relationship and fit a linear regression of the form $Y = \alpha + \beta(X) + e$. (B... | When to use residual plots? | They are still useful in assessing whether the relationship between the explanatory variables and the dependent variable is linear (or modeled properly given the equation). For an extreme example, I g | When to use residual plots?
They are still useful in assessing whether the relationship between the explanatory variables and the dependent variable is linear (or modeled properly given the equation). For an extreme example, I generated some data with a quadratic relationship and fit a linear regression of the form $Y ... | When to use residual plots?
They are still useful in assessing whether the relationship between the explanatory variables and the dependent variable is linear (or modeled properly given the equation). For an extreme example, I g |
52,988 | When to use residual plots? | Assume for simplicity that you have fitted some line $\hat y = b_0 + b_1 x$ given a dependent or response variable $y$ and a predictor or independent variable $x$. This specific assumption can be relaxed, which we will get to in good time.
With one variable on each side, a residual plot (meaning, a plot of residual $y... | When to use residual plots? | Assume for simplicity that you have fitted some line $\hat y = b_0 + b_1 x$ given a dependent or response variable $y$ and a predictor or independent variable $x$. This specific assumption can be rela | When to use residual plots?
Assume for simplicity that you have fitted some line $\hat y = b_0 + b_1 x$ given a dependent or response variable $y$ and a predictor or independent variable $x$. This specific assumption can be relaxed, which we will get to in good time.
With one variable on each side, a residual plot (me... | When to use residual plots?
Assume for simplicity that you have fitted some line $\hat y = b_0 + b_1 x$ given a dependent or response variable $y$ and a predictor or independent variable $x$. This specific assumption can be rela |
52,989 | Can I compare AIC values of a linear function with a non-linear function? | Looks like one of them doesn't really fit the data.
As long as you did not transform the response variable -- for example, replacing $y$ by $\log(y)$ -- you can use very different models, for example you can compare $y = b \cdot x$ with $y = e^{b \cdot x}$ or $y = a\cdot x^b$.
However, you are not allowed to use AIC f... | Can I compare AIC values of a linear function with a non-linear function? | Looks like one of them doesn't really fit the data.
As long as you did not transform the response variable -- for example, replacing $y$ by $\log(y)$ -- you can use very different models, for example | Can I compare AIC values of a linear function with a non-linear function?
Looks like one of them doesn't really fit the data.
As long as you did not transform the response variable -- for example, replacing $y$ by $\log(y)$ -- you can use very different models, for example you can compare $y = b \cdot x$ with $y = e^{... | Can I compare AIC values of a linear function with a non-linear function?
Looks like one of them doesn't really fit the data.
As long as you did not transform the response variable -- for example, replacing $y$ by $\log(y)$ -- you can use very different models, for example |
52,990 | Can I compare AIC values of a linear function with a non-linear function? | Well, they seem to be fitted with different algorithms and the likelihoods (or AIC) are calculated with different methods. Please check about it with your software.
Different software usually uses different methods to calculate the likelihood and often adds a constant to the likelihood or AIC for convenience. That mean... | Can I compare AIC values of a linear function with a non-linear function? | Well, they seem to be fitted with different algorithms and the likelihoods (or AIC) are calculated with different methods. Please check about it with your software.
Different software usually uses dif | Can I compare AIC values of a linear function with a non-linear function?
Well, they seem to be fitted with different algorithms and the likelihoods (or AIC) are calculated with different methods. Please check about it with your software.
Different software usually uses different methods to calculate the likelihood and... | Can I compare AIC values of a linear function with a non-linear function?
Well, they seem to be fitted with different algorithms and the likelihoods (or AIC) are calculated with different methods. Please check about it with your software.
Different software usually uses dif |
52,991 | Can I compare AIC values of a linear function with a non-linear function? | From the description you have given, yes. This is exactly the case where you would want to use AIC and the like, differing models to the same data. The model with the higher value is the worse model. And if the $\Delta \mbox{AIC} >10$ there is a hardly any evidence for the worse model.
However, to make sure there is no... | Can I compare AIC values of a linear function with a non-linear function? | From the description you have given, yes. This is exactly the case where you would want to use AIC and the like, differing models to the same data. The model with the higher value is the worse model. | Can I compare AIC values of a linear function with a non-linear function?
From the description you have given, yes. This is exactly the case where you would want to use AIC and the like, differing models to the same data. The model with the higher value is the worse model. And if the $\Delta \mbox{AIC} >10$ there is a ... | Can I compare AIC values of a linear function with a non-linear function?
From the description you have given, yes. This is exactly the case where you would want to use AIC and the like, differing models to the same data. The model with the higher value is the worse model. |
52,992 | Difference between Sobol indices and total Sobol indices? | The reason why total Sobol' indices are interesting is interactions.
Two inputs $x_1$ and $x_2$ are interacting when their joint effect on the output is different from the sum of their individual effects.
Consider for instance the following model
$$ f(\mathbf{x}) = x_1 . x_2 $$
It is possible to measure interactions by... | Difference between Sobol indices and total Sobol indices? | The reason why total Sobol' indices are interesting is interactions.
Two inputs $x_1$ and $x_2$ are interacting when their joint effect on the output is different from the sum of their individual effe | Difference between Sobol indices and total Sobol indices?
The reason why total Sobol' indices are interesting is interactions.
Two inputs $x_1$ and $x_2$ are interacting when their joint effect on the output is different from the sum of their individual effects.
Consider for instance the following model
$$ f(\mathbf{x}... | Difference between Sobol indices and total Sobol indices?
The reason why total Sobol' indices are interesting is interactions.
Two inputs $x_1$ and $x_2$ are interacting when their joint effect on the output is different from the sum of their individual effe |
52,993 | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$ | $$
(1) \quad M_X(t) = \mathrm{E}\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x)\,dx \quad
$$
$$
(2) \quad M'_X(t) = \frac{d}{dt} \int_{-\infty}^\infty e^{tx} f_X(x)\,dx = \int_{-\infty}^\infty \frac{d}{dt} e^{tx} f_X(x)\,dx = \int_{-\infty}^\infty x\,e^{tx} f_X(x)\,dx
$$
$$
M'_X(0) = \int_{-\infty}^\in... | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$ | $$
(1) \quad M_X(t) = \mathrm{E}\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x)\,dx \quad
$$
$$
(2) \quad M'_X(t) = \frac{d}{dt} \int_{-\infty}^\infty e^{tx} f_X(x)\,dx = \int_{-\infty} | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$
$$
(1) \quad M_X(t) = \mathrm{E}\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x)\,dx \quad
$$
$$
(2) \quad M'_X(t) = \frac{d}{dt} \int_{-\infty}^\infty e^{tx} f_X(x)\,dx = \int_{-\infty}^\infty \frac{d}{dt} e^{tx} f_X(x)\,dx = \int_{-\infty}^\in... | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$
$$
(1) \quad M_X(t) = \mathrm{E}\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x)\,dx \quad
$$
$$
(2) \quad M'_X(t) = \frac{d}{dt} \int_{-\infty}^\infty e^{tx} f_X(x)\,dx = \int_{-\infty} |
52,994 | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$ | You have an error:
$$Ee^{pX}=\int_{-\infty}^{\infty}e^{px}f_X(x)dx$$
Note the limits of integration, you had them wrong. This way you can differentiate under integral. Of course there are conditions, when you can do this exactly, but you can assume that they are met.
The fact that you are integrating over infinite int... | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$ | You have an error:
$$Ee^{pX}=\int_{-\infty}^{\infty}e^{px}f_X(x)dx$$
Note the limits of integration, you had them wrong. This way you can differentiate under integral. Of course there are conditions, | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$
You have an error:
$$Ee^{pX}=\int_{-\infty}^{\infty}e^{px}f_X(x)dx$$
Note the limits of integration, you had them wrong. This way you can differentiate under integral. Of course there are conditions, when you can do this exactly, but you can assume that they ... | Show that $E(x)=M'_X(0)$, where $M'_X(p)=\frac{dM_X(p)}{dp}$
You have an error:
$$Ee^{pX}=\int_{-\infty}^{\infty}e^{px}f_X(x)dx$$
Note the limits of integration, you had them wrong. This way you can differentiate under integral. Of course there are conditions, |
52,995 | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset | You are talking about two distinct problems here
How do I visualise what k-means is doing in N>2 dimensions
How do I calculate k-means in N>2 dimensions
The second one is much easier than the first to answer.
To calculate the Euclidean distance when you have X, Y and Z, you simply sum the squares and square root. Th... | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset | You are talking about two distinct problems here
How do I visualise what k-means is doing in N>2 dimensions
How do I calculate k-means in N>2 dimensions
The second one is much easier than the first | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset
You are talking about two distinct problems here
How do I visualise what k-means is doing in N>2 dimensions
How do I calculate k-means in N>2 dimensions
The second one is much easier than the first to answer.
To calculate the Euclidea... | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset
You are talking about two distinct problems here
How do I visualise what k-means is doing in N>2 dimensions
How do I calculate k-means in N>2 dimensions
The second one is much easier than the first |
52,996 | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset | Don't compute Euclidean distance.
K-means minimizes the within-cluster variance aka: WCSS.
http://en.wikipedia.org/wiki/K-means_clustering
Then your question should be obvious. Sum of squared deviations, sum over all dimensions.
It is equivalent, but misleading, to think of k-means as of "minimizing the squared distanc... | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset | Don't compute Euclidean distance.
K-means minimizes the within-cluster variance aka: WCSS.
http://en.wikipedia.org/wiki/K-means_clustering
Then your question should be obvious. Sum of squared deviatio | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset
Don't compute Euclidean distance.
K-means minimizes the within-cluster variance aka: WCSS.
http://en.wikipedia.org/wiki/K-means_clustering
Then your question should be obvious. Sum of squared deviations, sum over all dimensions.
It is e... | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset
Don't compute Euclidean distance.
K-means minimizes the within-cluster variance aka: WCSS.
http://en.wikipedia.org/wiki/K-means_clustering
Then your question should be obvious. Sum of squared deviatio |
52,997 | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset | Generally speaking Kmeans algorithm can work for any dimensions just make sure that you when calculating the distance you take into account all the N features.
You can still use euclidean distance as a similarity measure have a look at the n dimensional equation http://en.wikipedia.org/wiki/Euclidean_distance.
In ord... | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset | Generally speaking Kmeans algorithm can work for any dimensions just make sure that you when calculating the distance you take into account all the N features.
You can still use euclidean distance as | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset
Generally speaking Kmeans algorithm can work for any dimensions just make sure that you when calculating the distance you take into account all the N features.
You can still use euclidean distance as a similarity measure have a look at... | K-Means Clustering - Calculating Euclidean distances in a multiple variable dataset
Generally speaking Kmeans algorithm can work for any dimensions just make sure that you when calculating the distance you take into account all the N features.
You can still use euclidean distance as |
52,998 | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations between random factors | (Italics represent corrected text)
You are making a 'mistake' in your model specification given what you say you want.
Random effects:
Groups Name Variance Std.Dev. Corr
Item (Intercept) 273.508 16.5381
Subject Gramgram 0.000 0.0000
... | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations betw | (Italics represent corrected text)
You are making a 'mistake' in your model specification given what you say you want.
Random effects:
Groups Name Variance Std.Dev. Corr
Item | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations between random factors
(Italics represent corrected text)
You are making a 'mistake' in your model specification given what you say you want.
Random effects:
Groups Name Variance Std.Dev. Corr
Item ... | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations betw
(Italics represent corrected text)
You are making a 'mistake' in your model specification given what you say you want.
Random effects:
Groups Name Variance Std.Dev. Corr
Item |
52,999 | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations between random factors | Luke (2016) Evaluating significance in linear mixed-effects models in R reports that the most optimal (certainly most conservative) test is p values based on Kenward-Roger approximation for degrees of freedom (in lmer). With large samples, p values based on the likelihood ratio (through anova()) are just as good. | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations betw | Luke (2016) Evaluating significance in linear mixed-effects models in R reports that the most optimal (certainly most conservative) test is p values based on Kenward-Roger approximation for degrees of | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations between random factors
Luke (2016) Evaluating significance in linear mixed-effects models in R reports that the most optimal (certainly most conservative) test is p values based on Kenward-Roger approximation for degrees of ... | I want to use pvals.fnc() to get p-values for a lmer() model but cannot get rid of correlations betw
Luke (2016) Evaluating significance in linear mixed-effects models in R reports that the most optimal (certainly most conservative) test is p values based on Kenward-Roger approximation for degrees of |
53,000 | Statistical meaning of pearsonr() output in Python | The second number is the p value. It can be interpreted as the probability to observe a correlation that extreme in the sample (i.e. that high if it is positive or that low if it is negative) if the true correlation was 0. A low value therefore correspond to stronger evidence that the correlation is different from 0 an... | Statistical meaning of pearsonr() output in Python | The second number is the p value. It can be interpreted as the probability to observe a correlation that extreme in the sample (i.e. that high if it is positive or that low if it is negative) if the t | Statistical meaning of pearsonr() output in Python
The second number is the p value. It can be interpreted as the probability to observe a correlation that extreme in the sample (i.e. that high if it is positive or that low if it is negative) if the true correlation was 0. A low value therefore correspond to stronger e... | Statistical meaning of pearsonr() output in Python
The second number is the p value. It can be interpreted as the probability to observe a correlation that extreme in the sample (i.e. that high if it is positive or that low if it is negative) if the t |
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