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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6,001 | Time series 'clustering' in R | In data streaming and mining of time series databases, a common approach is to transform the series to a symbolic representation, then use a similarity metric, such as Euclidean distance, to cluster the series. The most popular representations are SAX (Keogh & Lin) or the newer iSAX (Shieh & Keogh):
Symbolic Aggregat... | Time series 'clustering' in R | In data streaming and mining of time series databases, a common approach is to transform the series to a symbolic representation, then use a similarity metric, such as Euclidean distance, to cluster t | Time series 'clustering' in R
In data streaming and mining of time series databases, a common approach is to transform the series to a symbolic representation, then use a similarity metric, such as Euclidean distance, to cluster the series. The most popular representations are SAX (Keogh & Lin) or the newer iSAX (Shie... | Time series 'clustering' in R
In data streaming and mining of time series databases, a common approach is to transform the series to a symbolic representation, then use a similarity metric, such as Euclidean distance, to cluster t |
6,002 | Time series 'clustering' in R | Another way of saying "tend to move in sympathy" is "cointegrated".
There are two standard ways of calculating cointegration: Engle-Granger method and the Johansen procedure. These are covered in "Analysis of Integrated and Cointegrated Time Series with R" (Pfaff 2008) and the related R urca package. I highly recom... | Time series 'clustering' in R | Another way of saying "tend to move in sympathy" is "cointegrated".
There are two standard ways of calculating cointegration: Engle-Granger method and the Johansen procedure. These are covered in " | Time series 'clustering' in R
Another way of saying "tend to move in sympathy" is "cointegrated".
There are two standard ways of calculating cointegration: Engle-Granger method and the Johansen procedure. These are covered in "Analysis of Integrated and Cointegrated Time Series with R" (Pfaff 2008) and the related R... | Time series 'clustering' in R
Another way of saying "tend to move in sympathy" is "cointegrated".
There are two standard ways of calculating cointegration: Engle-Granger method and the Johansen procedure. These are covered in " |
6,003 | Time series 'clustering' in R | Clustering time series is done fairly commonly by population dynamacists, particularily those that study insects to understand trends in outbreak and collapse. Look for work on Gypsy moth, Spruce budoworm, mountain pine beetle and larch budmoth.
For the actual clustering you can choose whatever distance metric you lik... | Time series 'clustering' in R | Clustering time series is done fairly commonly by population dynamacists, particularily those that study insects to understand trends in outbreak and collapse. Look for work on Gypsy moth, Spruce budo | Time series 'clustering' in R
Clustering time series is done fairly commonly by population dynamacists, particularily those that study insects to understand trends in outbreak and collapse. Look for work on Gypsy moth, Spruce budoworm, mountain pine beetle and larch budmoth.
For the actual clustering you can choose wh... | Time series 'clustering' in R
Clustering time series is done fairly commonly by population dynamacists, particularily those that study insects to understand trends in outbreak and collapse. Look for work on Gypsy moth, Spruce budo |
6,004 | Time series 'clustering' in R | See my answer to a similar question here. Long story short, do a fast Fourier transform of the data, discard redundant frequencies if your input data was real valued, separate the real and imaginary parts for each element of the fast Fourier transform, and use the Mclust package in R to do model-based clustering on the... | Time series 'clustering' in R | See my answer to a similar question here. Long story short, do a fast Fourier transform of the data, discard redundant frequencies if your input data was real valued, separate the real and imaginary p | Time series 'clustering' in R
See my answer to a similar question here. Long story short, do a fast Fourier transform of the data, discard redundant frequencies if your input data was real valued, separate the real and imaginary parts for each element of the fast Fourier transform, and use the Mclust package in R to do... | Time series 'clustering' in R
See my answer to a similar question here. Long story short, do a fast Fourier transform of the data, discard redundant frequencies if your input data was real valued, separate the real and imaginary p |
6,005 | Time series 'clustering' in R | You could also use the package clustDDist, which performs the leaders method and the hierarchical clustering method with different error measures:
http://r-forge.r-project.org/projects/clustddist/
The squared Euclidean distance favors patterns of distributions that have one steep high peak and therefore measure
$$
d_... | Time series 'clustering' in R | You could also use the package clustDDist, which performs the leaders method and the hierarchical clustering method with different error measures:
http://r-forge.r-project.org/projects/clustddist/
The | Time series 'clustering' in R
You could also use the package clustDDist, which performs the leaders method and the hierarchical clustering method with different error measures:
http://r-forge.r-project.org/projects/clustddist/
The squared Euclidean distance favors patterns of distributions that have one steep high peak... | Time series 'clustering' in R
You could also use the package clustDDist, which performs the leaders method and the hierarchical clustering method with different error measures:
http://r-forge.r-project.org/projects/clustddist/
The |
6,006 | How to interpret glmnet? | Here's an unintuitive fact - you're not actually supposed to give glmnet a single value of lambda. From the documentation here:
Do not supply a single value for lambda (for predictions after CV use predict() instead).
Supply instead a decreasing sequence of lambda values. glmnet relies on its
warms starts for spe... | How to interpret glmnet? | Here's an unintuitive fact - you're not actually supposed to give glmnet a single value of lambda. From the documentation here:
Do not supply a single value for lambda (for predictions after CV use | How to interpret glmnet?
Here's an unintuitive fact - you're not actually supposed to give glmnet a single value of lambda. From the documentation here:
Do not supply a single value for lambda (for predictions after CV use predict() instead).
Supply instead a decreasing sequence of lambda values. glmnet relies on i... | How to interpret glmnet?
Here's an unintuitive fact - you're not actually supposed to give glmnet a single value of lambda. From the documentation here:
Do not supply a single value for lambda (for predictions after CV use |
6,007 | How to interpret glmnet? | Q1) I am not sure about how to choose lambda.
Q2) Should I use the non (.) variables to fit another model? In my case I would like to keep as much variables as possible.
As per @BenOgorek's great answer, typically you let fitting use an entire lambda sequence, then when extracting the optimal coefficients use the lambd... | How to interpret glmnet? | Q1) I am not sure about how to choose lambda.
Q2) Should I use the non (.) variables to fit another model? In my case I would like to keep as much variables as possible.
As per @BenOgorek's great answ | How to interpret glmnet?
Q1) I am not sure about how to choose lambda.
Q2) Should I use the non (.) variables to fit another model? In my case I would like to keep as much variables as possible.
As per @BenOgorek's great answer, typically you let fitting use an entire lambda sequence, then when extracting the optimal c... | How to interpret glmnet?
Q1) I am not sure about how to choose lambda.
Q2) Should I use the non (.) variables to fit another model? In my case I would like to keep as much variables as possible.
As per @BenOgorek's great answ |
6,008 | How to quasi match two vectors of strings (in R)? | I've had similar problems. (seen here: https://stackoverflow.com/questions/2231993/merging-two-data-frames-using-fuzzy-approximate-string-matching-in-r)
Most of the recommendations that I received fell around:
pmatch(), and agrep(), grep(), grepl() are three functions that if you take the time to look through will pro... | How to quasi match two vectors of strings (in R)? | I've had similar problems. (seen here: https://stackoverflow.com/questions/2231993/merging-two-data-frames-using-fuzzy-approximate-string-matching-in-r)
Most of the recommendations that I received fel | How to quasi match two vectors of strings (in R)?
I've had similar problems. (seen here: https://stackoverflow.com/questions/2231993/merging-two-data-frames-using-fuzzy-approximate-string-matching-in-r)
Most of the recommendations that I received fell around:
pmatch(), and agrep(), grep(), grepl() are three functions ... | How to quasi match two vectors of strings (in R)?
I've had similar problems. (seen here: https://stackoverflow.com/questions/2231993/merging-two-data-frames-using-fuzzy-approximate-string-matching-in-r)
Most of the recommendations that I received fel |
6,009 | How to quasi match two vectors of strings (in R)? | There are many ways to measure distances between two strings. Two important (standard) approaches widely implemented in R are the Levenshtein and the Hamming distance. The former is avalaible in package 'MiscPsycho' and the latter in 'e1071'. Using these, i would simply compute a 92 by 55 matrix of pairwise distances, ... | How to quasi match two vectors of strings (in R)? | There are many ways to measure distances between two strings. Two important (standard) approaches widely implemented in R are the Levenshtein and the Hamming distance. The former is avalaible in packa | How to quasi match two vectors of strings (in R)?
There are many ways to measure distances between two strings. Two important (standard) approaches widely implemented in R are the Levenshtein and the Hamming distance. The former is avalaible in package 'MiscPsycho' and the latter in 'e1071'. Using these, i would simply... | How to quasi match two vectors of strings (in R)?
There are many ways to measure distances between two strings. Two important (standard) approaches widely implemented in R are the Levenshtein and the Hamming distance. The former is avalaible in packa |
6,010 | How to quasi match two vectors of strings (in R)? | To supplement Kwak's useful answer, allow me to add some simple principles and ideas. A good way to determine the metric is by considering how the strings might vary from their target. "Edit distance" is useful when the variation is a combination of typographic errors like transposing neighbors or mis-typing a single... | How to quasi match two vectors of strings (in R)? | To supplement Kwak's useful answer, allow me to add some simple principles and ideas. A good way to determine the metric is by considering how the strings might vary from their target. "Edit distanc | How to quasi match two vectors of strings (in R)?
To supplement Kwak's useful answer, allow me to add some simple principles and ideas. A good way to determine the metric is by considering how the strings might vary from their target. "Edit distance" is useful when the variation is a combination of typographic errors... | How to quasi match two vectors of strings (in R)?
To supplement Kwak's useful answer, allow me to add some simple principles and ideas. A good way to determine the metric is by considering how the strings might vary from their target. "Edit distanc |
6,011 | How to quasi match two vectors of strings (in R)? | I would also suggest you check out N-grams and the Damerau–Levenshtein distance besides the other suggestions of Kwak.
This paper compares the accuracy of a few different edit distances mentioned here (and is highly cited according to google scholar).
As you can see there are many different ways to approach this, and ... | How to quasi match two vectors of strings (in R)? | I would also suggest you check out N-grams and the Damerau–Levenshtein distance besides the other suggestions of Kwak.
This paper compares the accuracy of a few different edit distances mentioned her | How to quasi match two vectors of strings (in R)?
I would also suggest you check out N-grams and the Damerau–Levenshtein distance besides the other suggestions of Kwak.
This paper compares the accuracy of a few different edit distances mentioned here (and is highly cited according to google scholar).
As you can see th... | How to quasi match two vectors of strings (in R)?
I would also suggest you check out N-grams and the Damerau–Levenshtein distance besides the other suggestions of Kwak.
This paper compares the accuracy of a few different edit distances mentioned her |
6,012 | How to quasi match two vectors of strings (in R)? | I researched some packages and ways how to solve this problem and I think the best candidate is the fuzzywuzzyR package.
The fuzzywuzzyR package is a fuzzy string matching implemenation of the fuzzywuzzy python package. It uses the Levenshtein Distance to calculate the differences between sequences. More details on th... | How to quasi match two vectors of strings (in R)? | I researched some packages and ways how to solve this problem and I think the best candidate is the fuzzywuzzyR package.
The fuzzywuzzyR package is a fuzzy string matching implemenation of the fuzzyw | How to quasi match two vectors of strings (in R)?
I researched some packages and ways how to solve this problem and I think the best candidate is the fuzzywuzzyR package.
The fuzzywuzzyR package is a fuzzy string matching implemenation of the fuzzywuzzy python package. It uses the Levenshtein Distance to calculate the... | How to quasi match two vectors of strings (in R)?
I researched some packages and ways how to solve this problem and I think the best candidate is the fuzzywuzzyR package.
The fuzzywuzzyR package is a fuzzy string matching implemenation of the fuzzyw |
6,013 | How to quasi match two vectors of strings (in R)? | Based on function adist
Compute the approximate string distance between character vectors.
The distance is a generalized Levenshtein (edit) distance, giving
the minimal possibly weighted number of insertions, deletions and
substitutions needed to transform one string into another
Function stringdist from a package of... | How to quasi match two vectors of strings (in R)? | Based on function adist
Compute the approximate string distance between character vectors.
The distance is a generalized Levenshtein (edit) distance, giving
the minimal possibly weighted number of in | How to quasi match two vectors of strings (in R)?
Based on function adist
Compute the approximate string distance between character vectors.
The distance is a generalized Levenshtein (edit) distance, giving
the minimal possibly weighted number of insertions, deletions and
substitutions needed to transform one string i... | How to quasi match two vectors of strings (in R)?
Based on function adist
Compute the approximate string distance between character vectors.
The distance is a generalized Levenshtein (edit) distance, giving
the minimal possibly weighted number of in |
6,014 | Improve classification with many categorical variables | Random forests should be able to handle categorical values natively so look for a different implementation so you don't have to encode all of those features and use up all your memory.
The problem with high cardinality categorical features is that it is easy to over fit with them. You may have enough data that this is... | Improve classification with many categorical variables | Random forests should be able to handle categorical values natively so look for a different implementation so you don't have to encode all of those features and use up all your memory.
The problem wi | Improve classification with many categorical variables
Random forests should be able to handle categorical values natively so look for a different implementation so you don't have to encode all of those features and use up all your memory.
The problem with high cardinality categorical features is that it is easy to ov... | Improve classification with many categorical variables
Random forests should be able to handle categorical values natively so look for a different implementation so you don't have to encode all of those features and use up all your memory.
The problem wi |
6,015 | Improve classification with many categorical variables | Instead of dummifying your categories, why wouldn't you simply use a single numerical variable for each? In the context of random forests, I've often been wondering about the consequence of doing that (because I agree that it sounds suspcious to introduce ordinality in categorical data with which if often doesn't make ... | Improve classification with many categorical variables | Instead of dummifying your categories, why wouldn't you simply use a single numerical variable for each? In the context of random forests, I've often been wondering about the consequence of doing that | Improve classification with many categorical variables
Instead of dummifying your categories, why wouldn't you simply use a single numerical variable for each? In the context of random forests, I've often been wondering about the consequence of doing that (because I agree that it sounds suspcious to introduce ordinalit... | Improve classification with many categorical variables
Instead of dummifying your categories, why wouldn't you simply use a single numerical variable for each? In the context of random forests, I've often been wondering about the consequence of doing that |
6,016 | Improve classification with many categorical variables | I think you should consider a/more variable reduction technique(s). It gets rid of the not so influent predictors.
I have been reading a lot about data pre-processing and it is a great solution to reduce the n° of your variables.
My suggestions are as follows:
for qualitative variables, replace missing values with cat... | Improve classification with many categorical variables | I think you should consider a/more variable reduction technique(s). It gets rid of the not so influent predictors.
I have been reading a lot about data pre-processing and it is a great solution to red | Improve classification with many categorical variables
I think you should consider a/more variable reduction technique(s). It gets rid of the not so influent predictors.
I have been reading a lot about data pre-processing and it is a great solution to reduce the n° of your variables.
My suggestions are as follows:
for... | Improve classification with many categorical variables
I think you should consider a/more variable reduction technique(s). It gets rid of the not so influent predictors.
I have been reading a lot about data pre-processing and it is a great solution to red |
6,017 | Improve classification with many categorical variables | You may want to consider mixed-effects models. They are popular in social science due to their performance on high-cardinality categorical data, and I have used them to make great predictive models outperforming popular machine learning approaches like gradient boosted trees, random forests, and elastic-net regularized... | Improve classification with many categorical variables | You may want to consider mixed-effects models. They are popular in social science due to their performance on high-cardinality categorical data, and I have used them to make great predictive models ou | Improve classification with many categorical variables
You may want to consider mixed-effects models. They are popular in social science due to their performance on high-cardinality categorical data, and I have used them to make great predictive models outperforming popular machine learning approaches like gradient boo... | Improve classification with many categorical variables
You may want to consider mixed-effects models. They are popular in social science due to their performance on high-cardinality categorical data, and I have used them to make great predictive models ou |
6,018 | Improve classification with many categorical variables | When you say "build dummy variable for each categorical one", sounds like you're using Python not R? R randomforest can natively handle categoricals, also consequential memory reduction. Try R.
Next, you don't need to manually prune/merge the categorical levels, that sounds like a major pain. And even if you did, you'r... | Improve classification with many categorical variables | When you say "build dummy variable for each categorical one", sounds like you're using Python not R? R randomforest can natively handle categoricals, also consequential memory reduction. Try R.
Next, | Improve classification with many categorical variables
When you say "build dummy variable for each categorical one", sounds like you're using Python not R? R randomforest can natively handle categoricals, also consequential memory reduction. Try R.
Next, you don't need to manually prune/merge the categorical levels, th... | Improve classification with many categorical variables
When you say "build dummy variable for each categorical one", sounds like you're using Python not R? R randomforest can natively handle categoricals, also consequential memory reduction. Try R.
Next, |
6,019 | Improve classification with many categorical variables | You should look at the H2O.ai package. It handles categorical variables out of the box without having to do any encoding (make sure the variables are factors).
I particularly like their implementation of Gradient Boosted Machine (GBM) because you can then look at the variable importance after building the model. GBM'... | Improve classification with many categorical variables | You should look at the H2O.ai package. It handles categorical variables out of the box without having to do any encoding (make sure the variables are factors).
I particularly like their implementatio | Improve classification with many categorical variables
You should look at the H2O.ai package. It handles categorical variables out of the box without having to do any encoding (make sure the variables are factors).
I particularly like their implementation of Gradient Boosted Machine (GBM) because you can then look at ... | Improve classification with many categorical variables
You should look at the H2O.ai package. It handles categorical variables out of the box without having to do any encoding (make sure the variables are factors).
I particularly like their implementatio |
6,020 | How to interpret the output of the summary method for an lm object in R? [duplicate] | It sounds like you need a decent basic statistics text that covers at least basic location tests, simple regression and multiple regression.
Std. Error,t value and Pr.
Std. Error is the standard deviation of the sampling distribution
of the estimate of the coefficient under the standard regression
assumptions. Such... | How to interpret the output of the summary method for an lm object in R? [duplicate] | It sounds like you need a decent basic statistics text that covers at least basic location tests, simple regression and multiple regression.
Std. Error,t value and Pr.
Std. Error is the standard d | How to interpret the output of the summary method for an lm object in R? [duplicate]
It sounds like you need a decent basic statistics text that covers at least basic location tests, simple regression and multiple regression.
Std. Error,t value and Pr.
Std. Error is the standard deviation of the sampling distributi... | How to interpret the output of the summary method for an lm object in R? [duplicate]
It sounds like you need a decent basic statistics text that covers at least basic location tests, simple regression and multiple regression.
Std. Error,t value and Pr.
Std. Error is the standard d |
6,021 | How to interpret the output of the summary method for an lm object in R? [duplicate] | The Standard error is an estimate of the variance of the strength of the effect, or the strength of the relationship between each causal variable and the predicted variable. If it's high, then the effect size will have to be stronger for us to be able to be sure that it's a real effect, and not just an artefact of rand... | How to interpret the output of the summary method for an lm object in R? [duplicate] | The Standard error is an estimate of the variance of the strength of the effect, or the strength of the relationship between each causal variable and the predicted variable. If it's high, then the eff | How to interpret the output of the summary method for an lm object in R? [duplicate]
The Standard error is an estimate of the variance of the strength of the effect, or the strength of the relationship between each causal variable and the predicted variable. If it's high, then the effect size will have to be stronger f... | How to interpret the output of the summary method for an lm object in R? [duplicate]
The Standard error is an estimate of the variance of the strength of the effect, or the strength of the relationship between each causal variable and the predicted variable. If it's high, then the eff |
6,022 | How does R handle missing values in lm? | Edit: I misunderstood your question. There are two aspects:
a) na.omit and na.exclude both do casewise deletion with respect to both predictors and criterions. They only differ in that extractor functions like residuals() or fitted() will pad their output with NAs for the omitted cases with na.exclude, thus having an o... | How does R handle missing values in lm? | Edit: I misunderstood your question. There are two aspects:
a) na.omit and na.exclude both do casewise deletion with respect to both predictors and criterions. They only differ in that extractor funct | How does R handle missing values in lm?
Edit: I misunderstood your question. There are two aspects:
a) na.omit and na.exclude both do casewise deletion with respect to both predictors and criterions. They only differ in that extractor functions like residuals() or fitted() will pad their output with NAs for the omitted... | How does R handle missing values in lm?
Edit: I misunderstood your question. There are two aspects:
a) na.omit and na.exclude both do casewise deletion with respect to both predictors and criterions. They only differ in that extractor funct |
6,023 | How does R handle missing values in lm? | I can think of two ways. One is combine the data use the na.exclude and then separate data again:
A = matrix(1:20, nrow=10, ncol=2)
colnames(A) <- paste("A",1:ncol(A),sep="")
B = matrix(1:10, nrow=10, ncol=1)
colnames(B) <- paste("B",1:ncol(B),sep="")
C <- cbind(A,B)
C[1,1] <- NA
C.ex <- na.exclude(C)
A.ex <- C[,co... | How does R handle missing values in lm? | I can think of two ways. One is combine the data use the na.exclude and then separate data again:
A = matrix(1:20, nrow=10, ncol=2)
colnames(A) <- paste("A",1:ncol(A),sep="")
B = matrix(1:10, nrow=10 | How does R handle missing values in lm?
I can think of two ways. One is combine the data use the na.exclude and then separate data again:
A = matrix(1:20, nrow=10, ncol=2)
colnames(A) <- paste("A",1:ncol(A),sep="")
B = matrix(1:10, nrow=10, ncol=1)
colnames(B) <- paste("B",1:ncol(B),sep="")
C <- cbind(A,B)
C[1,1] <-... | How does R handle missing values in lm?
I can think of two ways. One is combine the data use the na.exclude and then separate data again:
A = matrix(1:20, nrow=10, ncol=2)
colnames(A) <- paste("A",1:ncol(A),sep="")
B = matrix(1:10, nrow=10 |
6,024 | How does R handle missing values in lm? | The following example shows how to make predictions and residuals that conform to the original dataframe (using the "na.action=na.exclude" option in lm() to specify that NA's should be placed in the residual and prediction vectors where the original dataframe had missing values. It also shows how to specify whether pr... | How does R handle missing values in lm? | The following example shows how to make predictions and residuals that conform to the original dataframe (using the "na.action=na.exclude" option in lm() to specify that NA's should be placed in the r | How does R handle missing values in lm?
The following example shows how to make predictions and residuals that conform to the original dataframe (using the "na.action=na.exclude" option in lm() to specify that NA's should be placed in the residual and prediction vectors where the original dataframe had missing values. ... | How does R handle missing values in lm?
The following example shows how to make predictions and residuals that conform to the original dataframe (using the "na.action=na.exclude" option in lm() to specify that NA's should be placed in the r |
6,025 | Training loss goes down and up again. What is happening? | Your learning rate could be to big after the 25th epoch. This problem is easy to identify. You just need to set up a smaller value for your learning rate. If the problem related to your learning rate than NN should reach a lower error despite that it will go up again after a while. The main point is that the error rate... | Training loss goes down and up again. What is happening? | Your learning rate could be to big after the 25th epoch. This problem is easy to identify. You just need to set up a smaller value for your learning rate. If the problem related to your learning rate | Training loss goes down and up again. What is happening?
Your learning rate could be to big after the 25th epoch. This problem is easy to identify. You just need to set up a smaller value for your learning rate. If the problem related to your learning rate than NN should reach a lower error despite that it will go up a... | Training loss goes down and up again. What is happening?
Your learning rate could be to big after the 25th epoch. This problem is easy to identify. You just need to set up a smaller value for your learning rate. If the problem related to your learning rate |
6,026 | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | This answer follows up on my answer in Bias and variance in leave-one-out vs K-fold cross validation that discusses why LOOCV does not always lead to higher variance. Following a similar approach, I will attempt to highlight a case where LOOCV does lead to higher variance in the presence of outliers and an "unstable mo... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | This answer follows up on my answer in Bias and variance in leave-one-out vs K-fold cross validation that discusses why LOOCV does not always lead to higher variance. Following a similar approach, I w | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
This answer follows up on my answer in Bias and variance in leave-one-out vs K-fold cross validation that discusses why LOOCV does not always lead to higher variance. Following a similar approach, I will attempt to highlight a c... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
This answer follows up on my answer in Bias and variance in leave-one-out vs K-fold cross validation that discusses why LOOCV does not always lead to higher variance. Following a similar approach, I w |
6,027 | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | I will give my answer in context of the paragraph you cite:
With K=N, the cross-validation estimator is approximately unbiased for
the true (expected) prediction error, but can have high variance
because the N "training sets" are so similar to one another.
The CV estimator of the true (expected) prediction error ... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | I will give my answer in context of the paragraph you cite:
With K=N, the cross-validation estimator is approximately unbiased for
the true (expected) prediction error, but can have high variance
| Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
I will give my answer in context of the paragraph you cite:
With K=N, the cross-validation estimator is approximately unbiased for
the true (expected) prediction error, but can have high variance
because the N "training set... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
I will give my answer in context of the paragraph you cite:
With K=N, the cross-validation estimator is approximately unbiased for
the true (expected) prediction error, but can have high variance
|
6,028 | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | We've been through this before -- you're getting too mathematical about a dead horse. See Ron Kohavi's (Stanford-Univ) classic paper on CV and the bias-variance dilemma here. When you're done reading this, you won't want to perform LOOCV, and will likely be attracted to 10-fold CV and/or bootstrap-bias CV.
You al... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | We've been through this before -- you're getting too mathematical about a dead horse. See Ron Kohavi's (Stanford-Univ) classic paper on CV and the bias-variance dilemma here. When you're done readin | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
We've been through this before -- you're getting too mathematical about a dead horse. See Ron Kohavi's (Stanford-Univ) classic paper on CV and the bias-variance dilemma here. When you're done reading this, you won't want to pe... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
We've been through this before -- you're getting too mathematical about a dead horse. See Ron Kohavi's (Stanford-Univ) classic paper on CV and the bias-variance dilemma here. When you're done readin |
6,029 | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | This answer focuses not on stability, but on a different related issue that I have not seen addressed in the answers/comments above.
There is "conventional wisdom" about LOOCV having higher variance, e.g. quoted from ESL in the question above. This conventional wisdom makes sense to me only if CV with different fold si... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"? | This answer focuses not on stability, but on a different related issue that I have not seen addressed in the answers/comments above.
There is "conventional wisdom" about LOOCV having higher variance, | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
This answer focuses not on stability, but on a different related issue that I have not seen addressed in the answers/comments above.
There is "conventional wisdom" about LOOCV having higher variance, e.g. quoted from ESL in the ... | Variance of $K$-fold cross-validation estimates as $f(K)$: what is the role of "stability"?
This answer focuses not on stability, but on a different related issue that I have not seen addressed in the answers/comments above.
There is "conventional wisdom" about LOOCV having higher variance, |
6,030 | When teaching statistics, use "normal" or "Gaussian"? | Even though I tend to say 'normal' more often (since that's what I was taught when first learning), I think "Gaussian" is a better choice, as long as students/readers are quite familiar with both terms:
The normal isn't particularly typical, so the name is itself misleading. It certainly plays an important role (not l... | When teaching statistics, use "normal" or "Gaussian"? | Even though I tend to say 'normal' more often (since that's what I was taught when first learning), I think "Gaussian" is a better choice, as long as students/readers are quite familiar with both term | When teaching statistics, use "normal" or "Gaussian"?
Even though I tend to say 'normal' more often (since that's what I was taught when first learning), I think "Gaussian" is a better choice, as long as students/readers are quite familiar with both terms:
The normal isn't particularly typical, so the name is itself m... | When teaching statistics, use "normal" or "Gaussian"?
Even though I tend to say 'normal' more often (since that's what I was taught when first learning), I think "Gaussian" is a better choice, as long as students/readers are quite familiar with both term |
6,031 | When teaching statistics, use "normal" or "Gaussian"? | I would use Gaussian.
One problem that faces people learning statistics is that we use everyday English words to mean different things (power, significant, distribution etc). To the extent that we can minimize this, we should. "Normal" already has a bunch of meanings. | When teaching statistics, use "normal" or "Gaussian"? | I would use Gaussian.
One problem that faces people learning statistics is that we use everyday English words to mean different things (power, significant, distribution etc). To the extent that we can | When teaching statistics, use "normal" or "Gaussian"?
I would use Gaussian.
One problem that faces people learning statistics is that we use everyday English words to mean different things (power, significant, distribution etc). To the extent that we can minimize this, we should. "Normal" already has a bunch of meaning... | When teaching statistics, use "normal" or "Gaussian"?
I would use Gaussian.
One problem that faces people learning statistics is that we use everyday English words to mean different things (power, significant, distribution etc). To the extent that we can |
6,032 | When teaching statistics, use "normal" or "Gaussian"? | One argument in favor of normal is the entrenched $N(\mu, \sigma^2)$ notation for the distribution, in which $N$ stands for "normal". I haven't seen anyone propose changing this to $G(\mu, \sigma^2)$. | When teaching statistics, use "normal" or "Gaussian"? | One argument in favor of normal is the entrenched $N(\mu, \sigma^2)$ notation for the distribution, in which $N$ stands for "normal". I haven't seen anyone propose changing this to $G(\mu, \sigma^2)$ | When teaching statistics, use "normal" or "Gaussian"?
One argument in favor of normal is the entrenched $N(\mu, \sigma^2)$ notation for the distribution, in which $N$ stands for "normal". I haven't seen anyone propose changing this to $G(\mu, \sigma^2)$. | When teaching statistics, use "normal" or "Gaussian"?
One argument in favor of normal is the entrenched $N(\mu, \sigma^2)$ notation for the distribution, in which $N$ stands for "normal". I haven't seen anyone propose changing this to $G(\mu, \sigma^2)$ |
6,033 | When teaching statistics, use "normal" or "Gaussian"? | In German it is often called Gaußsche Normalverteilung so it is nearly impossible to conflict easily.
Would it be appropriate for you to combine gaussian and normal? | When teaching statistics, use "normal" or "Gaussian"? | In German it is often called Gaußsche Normalverteilung so it is nearly impossible to conflict easily.
Would it be appropriate for you to combine gaussian and normal? | When teaching statistics, use "normal" or "Gaussian"?
In German it is often called Gaußsche Normalverteilung so it is nearly impossible to conflict easily.
Would it be appropriate for you to combine gaussian and normal? | When teaching statistics, use "normal" or "Gaussian"?
In German it is often called Gaußsche Normalverteilung so it is nearly impossible to conflict easily.
Would it be appropriate for you to combine gaussian and normal? |
6,034 | When teaching statistics, use "normal" or "Gaussian"? | According to the Wolfram encyclopedia:
While statisticians and mathematicians uniformly use the term "normal distribution" for this
distribution, physicists sometimes call it a Gaussian distribution and, because of its
curved flaring shape, social scientists refer to it as the "bell curve."
I agree that "... | When teaching statistics, use "normal" or "Gaussian"? | According to the Wolfram encyclopedia:
While statisticians and mathematicians uniformly use the term "normal distribution" for this
distribution, physicists sometimes call it a Gaussian distrib | When teaching statistics, use "normal" or "Gaussian"?
According to the Wolfram encyclopedia:
While statisticians and mathematicians uniformly use the term "normal distribution" for this
distribution, physicists sometimes call it a Gaussian distribution and, because of its
curved flaring shape, social scien... | When teaching statistics, use "normal" or "Gaussian"?
According to the Wolfram encyclopedia:
While statisticians and mathematicians uniformly use the term "normal distribution" for this
distribution, physicists sometimes call it a Gaussian distrib |
6,035 | When teaching statistics, use "normal" or "Gaussian"? | I'd like to point out that S. Stigler uses Normal / Gauss / Laplace-Gauss distribution to prove 'Stigler's law of eponymy' published in Statistics on the Table (some pages are available on books.google).
Particularly interesting and relevant to this questions is that on pg 287-288 there are tables of the historical usa... | When teaching statistics, use "normal" or "Gaussian"? | I'd like to point out that S. Stigler uses Normal / Gauss / Laplace-Gauss distribution to prove 'Stigler's law of eponymy' published in Statistics on the Table (some pages are available on books.googl | When teaching statistics, use "normal" or "Gaussian"?
I'd like to point out that S. Stigler uses Normal / Gauss / Laplace-Gauss distribution to prove 'Stigler's law of eponymy' published in Statistics on the Table (some pages are available on books.google).
Particularly interesting and relevant to this questions is tha... | When teaching statistics, use "normal" or "Gaussian"?
I'd like to point out that S. Stigler uses Normal / Gauss / Laplace-Gauss distribution to prove 'Stigler's law of eponymy' published in Statistics on the Table (some pages are available on books.googl |
6,036 | When teaching statistics, use "normal" or "Gaussian"? | An answer I haven't seen yet among all the good answers:
I mostly use "normal" for reasons of previous familiarity, but I like to capitalize it to emphasize its technical meaning: "... if the data are Normally distributed ..." (I don't know whether I copied this practice from somewhere else or (re-)invented it myself) | When teaching statistics, use "normal" or "Gaussian"? | An answer I haven't seen yet among all the good answers:
I mostly use "normal" for reasons of previous familiarity, but I like to capitalize it to emphasize its technical meaning: "... if the data are | When teaching statistics, use "normal" or "Gaussian"?
An answer I haven't seen yet among all the good answers:
I mostly use "normal" for reasons of previous familiarity, but I like to capitalize it to emphasize its technical meaning: "... if the data are Normally distributed ..." (I don't know whether I copied this pra... | When teaching statistics, use "normal" or "Gaussian"?
An answer I haven't seen yet among all the good answers:
I mostly use "normal" for reasons of previous familiarity, but I like to capitalize it to emphasize its technical meaning: "... if the data are |
6,037 | When teaching statistics, use "normal" or "Gaussian"? | Which to use depends on the level of statistics being taught. Unfortunately, my teaching experience indicates that the majority of undergraduate students never fully grasp the concept of a probability distribution. However, they all must somehow come to grips with the CLT and ways to think about uncertainty. For an u... | When teaching statistics, use "normal" or "Gaussian"? | Which to use depends on the level of statistics being taught. Unfortunately, my teaching experience indicates that the majority of undergraduate students never fully grasp the concept of a probability | When teaching statistics, use "normal" or "Gaussian"?
Which to use depends on the level of statistics being taught. Unfortunately, my teaching experience indicates that the majority of undergraduate students never fully grasp the concept of a probability distribution. However, they all must somehow come to grips with ... | When teaching statistics, use "normal" or "Gaussian"?
Which to use depends on the level of statistics being taught. Unfortunately, my teaching experience indicates that the majority of undergraduate students never fully grasp the concept of a probability |
6,038 | When teaching statistics, use "normal" or "Gaussian"? | The name normal came from some of the observations that errors behave normally. You will find more details here. If that is the reason to call this distribution a normal distribution, it may create new confusion as normal distribution for counts of accidents is poisson. I believe we should move forward and start callin... | When teaching statistics, use "normal" or "Gaussian"? | The name normal came from some of the observations that errors behave normally. You will find more details here. If that is the reason to call this distribution a normal distribution, it may create ne | When teaching statistics, use "normal" or "Gaussian"?
The name normal came from some of the observations that errors behave normally. You will find more details here. If that is the reason to call this distribution a normal distribution, it may create new confusion as normal distribution for counts of accidents is pois... | When teaching statistics, use "normal" or "Gaussian"?
The name normal came from some of the observations that errors behave normally. You will find more details here. If that is the reason to call this distribution a normal distribution, it may create ne |
6,039 | What is the minimum recommended number of groups for a random effects factor? | Short answer: Yes, you can use ID as random effect with 6 levels.
Slightly longer answer: The @BenBolker's GLMM FAQ says (among other things) the following under the headline "Should I treat factor xxx as fixed or random?":
One point of particular relevance to 'modern' mixed model estimation
(rather than 'classical'... | What is the minimum recommended number of groups for a random effects factor? | Short answer: Yes, you can use ID as random effect with 6 levels.
Slightly longer answer: The @BenBolker's GLMM FAQ says (among other things) the following under the headline "Should I treat factor xx | What is the minimum recommended number of groups for a random effects factor?
Short answer: Yes, you can use ID as random effect with 6 levels.
Slightly longer answer: The @BenBolker's GLMM FAQ says (among other things) the following under the headline "Should I treat factor xxx as fixed or random?":
One point of part... | What is the minimum recommended number of groups for a random effects factor?
Short answer: Yes, you can use ID as random effect with 6 levels.
Slightly longer answer: The @BenBolker's GLMM FAQ says (among other things) the following under the headline "Should I treat factor xx |
6,040 | What is the minimum recommended number of groups for a random effects factor? | In an attempt to figure out the minimum number of groups for a multilevel model I looked at the book Data Analysis Using Regression and Mulitilevel/Hierarchical models by Gelman and Hill (2007).
They appear to address this topic in Chapter 11, Section 5 (page 247) where they write that when there are < 5 groups then mu... | What is the minimum recommended number of groups for a random effects factor? | In an attempt to figure out the minimum number of groups for a multilevel model I looked at the book Data Analysis Using Regression and Mulitilevel/Hierarchical models by Gelman and Hill (2007).
They | What is the minimum recommended number of groups for a random effects factor?
In an attempt to figure out the minimum number of groups for a multilevel model I looked at the book Data Analysis Using Regression and Mulitilevel/Hierarchical models by Gelman and Hill (2007).
They appear to address this topic in Chapter 11... | What is the minimum recommended number of groups for a random effects factor?
In an attempt to figure out the minimum number of groups for a multilevel model I looked at the book Data Analysis Using Regression and Mulitilevel/Hierarchical models by Gelman and Hill (2007).
They |
6,041 | What is the minimum recommended number of groups for a random effects factor? | For what it's worth, I did a bit of a simulation study to look at the stability of the variance estimate for a relatively simple LMM (using the sleepstudy dataset available through lme4). The first way generates all possible subject combinations for ngroups number of subjects, and refits the model for each possible com... | What is the minimum recommended number of groups for a random effects factor? | For what it's worth, I did a bit of a simulation study to look at the stability of the variance estimate for a relatively simple LMM (using the sleepstudy dataset available through lme4). The first wa | What is the minimum recommended number of groups for a random effects factor?
For what it's worth, I did a bit of a simulation study to look at the stability of the variance estimate for a relatively simple LMM (using the sleepstudy dataset available through lme4). The first way generates all possible subject combinati... | What is the minimum recommended number of groups for a random effects factor?
For what it's worth, I did a bit of a simulation study to look at the stability of the variance estimate for a relatively simple LMM (using the sleepstudy dataset available through lme4). The first wa |
6,042 | What is the minimum recommended number of groups for a random effects factor? | You could also use a Bayesian mixed model - in that case the uncertainty in the estimation of the random effects is fully taken care of in the calculation of the 95% prediction credible intervals. The new R package brms and function brm, for example, allow for a very easy transition from an lme4 frequentist mixed model... | What is the minimum recommended number of groups for a random effects factor? | You could also use a Bayesian mixed model - in that case the uncertainty in the estimation of the random effects is fully taken care of in the calculation of the 95% prediction credible intervals. The | What is the minimum recommended number of groups for a random effects factor?
You could also use a Bayesian mixed model - in that case the uncertainty in the estimation of the random effects is fully taken care of in the calculation of the 95% prediction credible intervals. The new R package brms and function brm, for ... | What is the minimum recommended number of groups for a random effects factor?
You could also use a Bayesian mixed model - in that case the uncertainty in the estimation of the random effects is fully taken care of in the calculation of the 95% prediction credible intervals. The |
6,043 | What is the minimum recommended number of groups for a random effects factor? | Angrist and Pischke's "Mostly Harmless Econometrics" has a section titled, "Fewer than 42 clusters", in which they semi-jokingly say,
Therefore, following the... dictum that the answer to life, the universe and everything is 42, we believe the question is: How many clusters are enough for reliable inference using the s... | What is the minimum recommended number of groups for a random effects factor? | Angrist and Pischke's "Mostly Harmless Econometrics" has a section titled, "Fewer than 42 clusters", in which they semi-jokingly say,
Therefore, following the... dictum that the answer to life, the un | What is the minimum recommended number of groups for a random effects factor?
Angrist and Pischke's "Mostly Harmless Econometrics" has a section titled, "Fewer than 42 clusters", in which they semi-jokingly say,
Therefore, following the... dictum that the answer to life, the universe and everything is 42, we believe th... | What is the minimum recommended number of groups for a random effects factor?
Angrist and Pischke's "Mostly Harmless Econometrics" has a section titled, "Fewer than 42 clusters", in which they semi-jokingly say,
Therefore, following the... dictum that the answer to life, the un |
6,044 | What is the minimum recommended number of groups for a random effects factor? | I would not use a random effects model with only 6 levels. Models using a 6-level random effect can sometime be run using many statistical programs and sometimes give unbiased estimates, but:
I think there is an arbitrary consensus in the statistical community that 10-20 is the minimum number. If you want to have yo... | What is the minimum recommended number of groups for a random effects factor? | I would not use a random effects model with only 6 levels. Models using a 6-level random effect can sometime be run using many statistical programs and sometimes give unbiased estimates, but:
I thi | What is the minimum recommended number of groups for a random effects factor?
I would not use a random effects model with only 6 levels. Models using a 6-level random effect can sometime be run using many statistical programs and sometimes give unbiased estimates, but:
I think there is an arbitrary consensus in the ... | What is the minimum recommended number of groups for a random effects factor?
I would not use a random effects model with only 6 levels. Models using a 6-level random effect can sometime be run using many statistical programs and sometimes give unbiased estimates, but:
I thi |
6,045 | What is the minimum recommended number of groups for a random effects factor? | It has been a long time since the original question but I thought I might add a few points pertinent to model selection.
1 - As long as the model is identified (i.e. you have degrees of freedom in the parameter space) you should be able to TRY to fit the model. Depending on the optimization method model may or may not ... | What is the minimum recommended number of groups for a random effects factor? | It has been a long time since the original question but I thought I might add a few points pertinent to model selection.
1 - As long as the model is identified (i.e. you have degrees of freedom in the | What is the minimum recommended number of groups for a random effects factor?
It has been a long time since the original question but I thought I might add a few points pertinent to model selection.
1 - As long as the model is identified (i.e. you have degrees of freedom in the parameter space) you should be able to TR... | What is the minimum recommended number of groups for a random effects factor?
It has been a long time since the original question but I thought I might add a few points pertinent to model selection.
1 - As long as the model is identified (i.e. you have degrees of freedom in the |
6,046 | What does negative R-squared mean? | $R^2$ can be negative, it just means that:
The model fits your data very badly
You did not set an intercept
To the people saying that $R^2$ is between 0 and 1, this is not the case. While a negative value for something with the word 'squared' in it might sound like it breaks the rules of maths, it can happen in an $R... | What does negative R-squared mean? | $R^2$ can be negative, it just means that:
The model fits your data very badly
You did not set an intercept
To the people saying that $R^2$ is between 0 and 1, this is not the case. While a negative | What does negative R-squared mean?
$R^2$ can be negative, it just means that:
The model fits your data very badly
You did not set an intercept
To the people saying that $R^2$ is between 0 and 1, this is not the case. While a negative value for something with the word 'squared' in it might sound like it breaks the rul... | What does negative R-squared mean?
$R^2$ can be negative, it just means that:
The model fits your data very badly
You did not set an intercept
To the people saying that $R^2$ is between 0 and 1, this is not the case. While a negative |
6,047 | What does negative R-squared mean? | Neither answer so far is entirely correct, so I will try to give my understanding of R-Squared. I have given a more detailed explanation of this on my blog post here "What is R-Squared"
Sum Squared Error
The objective of ordinary least squared regression is to get a line which minimized the sum squared error. The def... | What does negative R-squared mean? | Neither answer so far is entirely correct, so I will try to give my understanding of R-Squared. I have given a more detailed explanation of this on my blog post here "What is R-Squared"
Sum Squared E | What does negative R-squared mean?
Neither answer so far is entirely correct, so I will try to give my understanding of R-Squared. I have given a more detailed explanation of this on my blog post here "What is R-Squared"
Sum Squared Error
The objective of ordinary least squared regression is to get a line which minimi... | What does negative R-squared mean?
Neither answer so far is entirely correct, so I will try to give my understanding of R-Squared. I have given a more detailed explanation of this on my blog post here "What is R-Squared"
Sum Squared E |
6,048 | What does negative R-squared mean? | The question is asking about "a model (a non-linear regression)". In this case there is no bound of how negative R-squared can be.
R-squared = 1 - SSE / TSS
As long as your SSE term is significantly large, you will get an a negative R-squared. It can be caused by overall bad fit or one extreme bad prediction.
For exam... | What does negative R-squared mean? | The question is asking about "a model (a non-linear regression)". In this case there is no bound of how negative R-squared can be.
R-squared = 1 - SSE / TSS
As long as your SSE term is significantly | What does negative R-squared mean?
The question is asking about "a model (a non-linear regression)". In this case there is no bound of how negative R-squared can be.
R-squared = 1 - SSE / TSS
As long as your SSE term is significantly large, you will get an a negative R-squared. It can be caused by overall bad fit or o... | What does negative R-squared mean?
The question is asking about "a model (a non-linear regression)". In this case there is no bound of how negative R-squared can be.
R-squared = 1 - SSE / TSS
As long as your SSE term is significantly |
6,049 | What does negative R-squared mean? | Negative $R^2$ may indicate poor relative calibration of the model. This strictly applies to Efron's $R^2$ in the context of logistic regression. I think that this can be extrapolated (at least as an intuition) to standard $R^2$ for linear probability model.
This follows from the equivalence of Efron's $R^2$ and Brier ... | What does negative R-squared mean? | Negative $R^2$ may indicate poor relative calibration of the model. This strictly applies to Efron's $R^2$ in the context of logistic regression. I think that this can be extrapolated (at least as an | What does negative R-squared mean?
Negative $R^2$ may indicate poor relative calibration of the model. This strictly applies to Efron's $R^2$ in the context of logistic regression. I think that this can be extrapolated (at least as an intuition) to standard $R^2$ for linear probability model.
This follows from the equi... | What does negative R-squared mean?
Negative $R^2$ may indicate poor relative calibration of the model. This strictly applies to Efron's $R^2$ in the context of logistic regression. I think that this can be extrapolated (at least as an |
6,050 | What does negative R-squared mean? | Means the model is worse the horizontal line (mean).
See this explanation pag. 31: https://scholarsmine.mst.edu/masters_theses/7913/ | What does negative R-squared mean? | Means the model is worse the horizontal line (mean).
See this explanation pag. 31: https://scholarsmine.mst.edu/masters_theses/7913/ | What does negative R-squared mean?
Means the model is worse the horizontal line (mean).
See this explanation pag. 31: https://scholarsmine.mst.edu/masters_theses/7913/ | What does negative R-squared mean?
Means the model is worse the horizontal line (mean).
See this explanation pag. 31: https://scholarsmine.mst.edu/masters_theses/7913/ |
6,051 | What does negative R-squared mean? | As the previous commenter notes, r^2 is between [0,1], not [-1,+1], so it is impossible to be negative. You cannot square a value and get a negative number. Perhaps you are looking at r, the correlation? It can be between [-1,+1], where zero means there is no relationship between the variables, -1 means there is a ... | What does negative R-squared mean? | As the previous commenter notes, r^2 is between [0,1], not [-1,+1], so it is impossible to be negative. You cannot square a value and get a negative number. Perhaps you are looking at r, the correl | What does negative R-squared mean?
As the previous commenter notes, r^2 is between [0,1], not [-1,+1], so it is impossible to be negative. You cannot square a value and get a negative number. Perhaps you are looking at r, the correlation? It can be between [-1,+1], where zero means there is no relationship between ... | What does negative R-squared mean?
As the previous commenter notes, r^2 is between [0,1], not [-1,+1], so it is impossible to be negative. You cannot square a value and get a negative number. Perhaps you are looking at r, the correl |
6,052 | How to efficiently generate random positive-semidefinite correlation matrices? | You can do it backward: every matrix $C \in \mathbb{R}_{++}^p$ (the set of all symmetric $p \times p$ PSD matrices) can be decomposed as
$C=O^{T}DO$ where $O$ is an orthonormal matrix
To get $O$, first generate a random basis $(v_1,...,v_p)$ (where $v_i$ are random vectors, typically in $(-1,1)$). From there, use th... | How to efficiently generate random positive-semidefinite correlation matrices? | You can do it backward: every matrix $C \in \mathbb{R}_{++}^p$ (the set of all symmetric $p \times p$ PSD matrices) can be decomposed as
$C=O^{T}DO$ where $O$ is an orthonormal matrix
To get $O$, f | How to efficiently generate random positive-semidefinite correlation matrices?
You can do it backward: every matrix $C \in \mathbb{R}_{++}^p$ (the set of all symmetric $p \times p$ PSD matrices) can be decomposed as
$C=O^{T}DO$ where $O$ is an orthonormal matrix
To get $O$, first generate a random basis $(v_1,...,v_... | How to efficiently generate random positive-semidefinite correlation matrices?
You can do it backward: every matrix $C \in \mathbb{R}_{++}^p$ (the set of all symmetric $p \times p$ PSD matrices) can be decomposed as
$C=O^{T}DO$ where $O$ is an orthonormal matrix
To get $O$, f |
6,053 | How to efficiently generate random positive-semidefinite correlation matrices? | A paper Generating random correlation matrices based on vines and extended
onion method by Lewandowski, Kurowicka, and Joe (LKJ), 2009, provides a unified treatment and exposition of the two efficient methods of generating random correlation matrices. Both methods allow to generate matrices from a uniform distribution... | How to efficiently generate random positive-semidefinite correlation matrices? | A paper Generating random correlation matrices based on vines and extended
onion method by Lewandowski, Kurowicka, and Joe (LKJ), 2009, provides a unified treatment and exposition of the two efficien | How to efficiently generate random positive-semidefinite correlation matrices?
A paper Generating random correlation matrices based on vines and extended
onion method by Lewandowski, Kurowicka, and Joe (LKJ), 2009, provides a unified treatment and exposition of the two efficient methods of generating random correlatio... | How to efficiently generate random positive-semidefinite correlation matrices?
A paper Generating random correlation matrices based on vines and extended
onion method by Lewandowski, Kurowicka, and Joe (LKJ), 2009, provides a unified treatment and exposition of the two efficien |
6,054 | How to efficiently generate random positive-semidefinite correlation matrices? | An even simpler characterization is that for real matrix $A$, $A^TA$ is positive semidefinite. To see why this is the case, one only has to prove that $y^T (A^TA) y \ge 0$ for all vectors $y$ (of the right size, of course). This is trivial: $y^T (A^TA) y = (Ay)^T Ay = ||Ay||$ which is nonnegative. So in Matlab, simply ... | How to efficiently generate random positive-semidefinite correlation matrices? | An even simpler characterization is that for real matrix $A$, $A^TA$ is positive semidefinite. To see why this is the case, one only has to prove that $y^T (A^TA) y \ge 0$ for all vectors $y$ (of the | How to efficiently generate random positive-semidefinite correlation matrices?
An even simpler characterization is that for real matrix $A$, $A^TA$ is positive semidefinite. To see why this is the case, one only has to prove that $y^T (A^TA) y \ge 0$ for all vectors $y$ (of the right size, of course). This is trivial: ... | How to efficiently generate random positive-semidefinite correlation matrices?
An even simpler characterization is that for real matrix $A$, $A^TA$ is positive semidefinite. To see why this is the case, one only has to prove that $y^T (A^TA) y \ge 0$ for all vectors $y$ (of the |
6,055 | How to efficiently generate random positive-semidefinite correlation matrices? | As a variation on kwak's answer: generate a diagonal matrix $\mathbf{D}$ with random nonnegative eigenvalues from a distribution of your choice, and then perform a similarity transformation $\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^T$ with $\mathbf{Q}$ a Haar-distributed pseudorandom orthogonal matrix. | How to efficiently generate random positive-semidefinite correlation matrices? | As a variation on kwak's answer: generate a diagonal matrix $\mathbf{D}$ with random nonnegative eigenvalues from a distribution of your choice, and then perform a similarity transformation $\mathbf{A | How to efficiently generate random positive-semidefinite correlation matrices?
As a variation on kwak's answer: generate a diagonal matrix $\mathbf{D}$ with random nonnegative eigenvalues from a distribution of your choice, and then perform a similarity transformation $\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^T$ with ... | How to efficiently generate random positive-semidefinite correlation matrices?
As a variation on kwak's answer: generate a diagonal matrix $\mathbf{D}$ with random nonnegative eigenvalues from a distribution of your choice, and then perform a similarity transformation $\mathbf{A |
6,056 | How to efficiently generate random positive-semidefinite correlation matrices? | You haven't specified a distribution for the matrices. Two common ones are the Wishart and inverse Wishart distributions. The Bartlett decomposition gives a Cholesky factorisation of a random Wishart matrix (which can also be efficiently solved to obtain a random inverse Wishart matrix).
In fact, the Cholesky space is... | How to efficiently generate random positive-semidefinite correlation matrices? | You haven't specified a distribution for the matrices. Two common ones are the Wishart and inverse Wishart distributions. The Bartlett decomposition gives a Cholesky factorisation of a random Wishart | How to efficiently generate random positive-semidefinite correlation matrices?
You haven't specified a distribution for the matrices. Two common ones are the Wishart and inverse Wishart distributions. The Bartlett decomposition gives a Cholesky factorisation of a random Wishart matrix (which can also be efficiently sol... | How to efficiently generate random positive-semidefinite correlation matrices?
You haven't specified a distribution for the matrices. Two common ones are the Wishart and inverse Wishart distributions. The Bartlett decomposition gives a Cholesky factorisation of a random Wishart |
6,057 | How to efficiently generate random positive-semidefinite correlation matrices? | The simplest method is the one above, which is a simulation of a random dataset and the computation of the Gramian. A word of caution: The resulting matrix will not be uniformly random, in that its decomposition, say $U^TSU$ will have rotations not distributed according to the Haar Measure. If you want to have "uniform... | How to efficiently generate random positive-semidefinite correlation matrices? | The simplest method is the one above, which is a simulation of a random dataset and the computation of the Gramian. A word of caution: The resulting matrix will not be uniformly random, in that its de | How to efficiently generate random positive-semidefinite correlation matrices?
The simplest method is the one above, which is a simulation of a random dataset and the computation of the Gramian. A word of caution: The resulting matrix will not be uniformly random, in that its decomposition, say $U^TSU$ will have rotati... | How to efficiently generate random positive-semidefinite correlation matrices?
The simplest method is the one above, which is a simulation of a random dataset and the computation of the Gramian. A word of caution: The resulting matrix will not be uniformly random, in that its de |
6,058 | How to efficiently generate random positive-semidefinite correlation matrices? | If you'd like to have more control over your generated symmetric PSD matrix, e.g. generate a synthetic validation dataset, you have a number of parameters available.
A symmetric PSD matrix corresponds to a hyper-ellipse in the N-dimensional space, with all the related degrees of freedom:
Rotations.
Lengths of axes.
S... | How to efficiently generate random positive-semidefinite correlation matrices? | If you'd like to have more control over your generated symmetric PSD matrix, e.g. generate a synthetic validation dataset, you have a number of parameters available.
A symmetric PSD matrix corresponds | How to efficiently generate random positive-semidefinite correlation matrices?
If you'd like to have more control over your generated symmetric PSD matrix, e.g. generate a synthetic validation dataset, you have a number of parameters available.
A symmetric PSD matrix corresponds to a hyper-ellipse in the N-dimensional ... | How to efficiently generate random positive-semidefinite correlation matrices?
If you'd like to have more control over your generated symmetric PSD matrix, e.g. generate a synthetic validation dataset, you have a number of parameters available.
A symmetric PSD matrix corresponds |
6,059 | How to efficiently generate random positive-semidefinite correlation matrices? | If you want to sample random correlation matrices from an empirical distribution, you can try to use generative models from machine learning to do that.
One such example is CorrGAN: Sampling Realistic Financial Correlation Matrices Using Generative Adversarial Networks
The basic idea is to fit generative adversarial ne... | How to efficiently generate random positive-semidefinite correlation matrices? | If you want to sample random correlation matrices from an empirical distribution, you can try to use generative models from machine learning to do that.
One such example is CorrGAN: Sampling Realistic | How to efficiently generate random positive-semidefinite correlation matrices?
If you want to sample random correlation matrices from an empirical distribution, you can try to use generative models from machine learning to do that.
One such example is CorrGAN: Sampling Realistic Financial Correlation Matrices Using Gen... | How to efficiently generate random positive-semidefinite correlation matrices?
If you want to sample random correlation matrices from an empirical distribution, you can try to use generative models from machine learning to do that.
One such example is CorrGAN: Sampling Realistic |
6,060 | How to efficiently generate random positive-semidefinite correlation matrices? | you can create an arbitrary covariance matrix from the Wishart distribution using the stats function rWishart (included in base R) and then transform it to a correlation matrix
p <- 4
S <- drop(rWishart(1,p,diag(p)))
S * (diag(S)^(-1/2) %o% diag(S)^(-1/2))
The Wishart distribution takes two parameters: the degrees of ... | How to efficiently generate random positive-semidefinite correlation matrices? | you can create an arbitrary covariance matrix from the Wishart distribution using the stats function rWishart (included in base R) and then transform it to a correlation matrix
p <- 4
S <- drop(rWisha | How to efficiently generate random positive-semidefinite correlation matrices?
you can create an arbitrary covariance matrix from the Wishart distribution using the stats function rWishart (included in base R) and then transform it to a correlation matrix
p <- 4
S <- drop(rWishart(1,p,diag(p)))
S * (diag(S)^(-1/2) %o% ... | How to efficiently generate random positive-semidefinite correlation matrices?
you can create an arbitrary covariance matrix from the Wishart distribution using the stats function rWishart (included in base R) and then transform it to a correlation matrix
p <- 4
S <- drop(rWisha |
6,061 | How to efficiently generate random positive-semidefinite correlation matrices? | A cheap and cheerful approach I've used for testing is to generate m N(0,1) n-vectors V[k] and then use P = d*I + Sum{ V[k]*V[k]'} as an nxn psd matrix. With m < n this will be singular for d=0, and for small d will have high condition number. | How to efficiently generate random positive-semidefinite correlation matrices? | A cheap and cheerful approach I've used for testing is to generate m N(0,1) n-vectors V[k] and then use P = d*I + Sum{ V[k]*V[k]'} as an nxn psd matrix. With m < n this will be singular for d=0, and f | How to efficiently generate random positive-semidefinite correlation matrices?
A cheap and cheerful approach I've used for testing is to generate m N(0,1) n-vectors V[k] and then use P = d*I + Sum{ V[k]*V[k]'} as an nxn psd matrix. With m < n this will be singular for d=0, and for small d will have high condition numbe... | How to efficiently generate random positive-semidefinite correlation matrices?
A cheap and cheerful approach I've used for testing is to generate m N(0,1) n-vectors V[k] and then use P = d*I + Sum{ V[k]*V[k]'} as an nxn psd matrix. With m < n this will be singular for d=0, and f |
6,062 | "Model failed to converge" warning in lmer() | "Solving" the issue you experience in the sense of not receiving warnings about failed convergence is rather straightforward: you do not use the default BOBYQA optimiser but instead you opt to use the Nelder-Mead optimisation routine used by default in earlier 1.0.x previous versions. Or you install the package optimx ... | "Model failed to converge" warning in lmer() | "Solving" the issue you experience in the sense of not receiving warnings about failed convergence is rather straightforward: you do not use the default BOBYQA optimiser but instead you opt to use the | "Model failed to converge" warning in lmer()
"Solving" the issue you experience in the sense of not receiving warnings about failed convergence is rather straightforward: you do not use the default BOBYQA optimiser but instead you opt to use the Nelder-Mead optimisation routine used by default in earlier 1.0.x previous... | "Model failed to converge" warning in lmer()
"Solving" the issue you experience in the sense of not receiving warnings about failed convergence is rather straightforward: you do not use the default BOBYQA optimiser but instead you opt to use the |
6,063 | "Model failed to converge" warning in lmer() | The question is statistical rather than technical. Actually, I used a random effect model instead of a fixed effect one.None of the factors, I think, should be treated as the random factor as we need at least 5 or 6 levels or replicates to treat a factor as random effect (see here What is the minimum recommended number... | "Model failed to converge" warning in lmer() | The question is statistical rather than technical. Actually, I used a random effect model instead of a fixed effect one.None of the factors, I think, should be treated as the random factor as we need | "Model failed to converge" warning in lmer()
The question is statistical rather than technical. Actually, I used a random effect model instead of a fixed effect one.None of the factors, I think, should be treated as the random factor as we need at least 5 or 6 levels or replicates to treat a factor as random effect (se... | "Model failed to converge" warning in lmer()
The question is statistical rather than technical. Actually, I used a random effect model instead of a fixed effect one.None of the factors, I think, should be treated as the random factor as we need |
6,064 | "Model failed to converge" warning in lmer() | I've had success with the change of optimizer suggested here in larger samples, but this fix does not always work in relatively small samples in my experience. Matuschek et al., 2017 (Journal of Memory and Language) note that failed model convergence can also result from over specification of the random effects structu... | "Model failed to converge" warning in lmer() | I've had success with the change of optimizer suggested here in larger samples, but this fix does not always work in relatively small samples in my experience. Matuschek et al., 2017 (Journal of Memor | "Model failed to converge" warning in lmer()
I've had success with the change of optimizer suggested here in larger samples, but this fix does not always work in relatively small samples in my experience. Matuschek et al., 2017 (Journal of Memory and Language) note that failed model convergence can also result from ove... | "Model failed to converge" warning in lmer()
I've had success with the change of optimizer suggested here in larger samples, but this fix does not always work in relatively small samples in my experience. Matuschek et al., 2017 (Journal of Memor |
6,065 | L1 regression estimates median whereas L2 regression estimates mean? | There is a simple geometric explanation for why the L1 loss function yields the median.
Recall that we are working in one dimension, so imagine a number line spreading horizontally. Plot each of the data points on the number line. Put your finger somewhere on the line; your finger will be your current candidate estim... | L1 regression estimates median whereas L2 regression estimates mean? | There is a simple geometric explanation for why the L1 loss function yields the median.
Recall that we are working in one dimension, so imagine a number line spreading horizontally. Plot each of the | L1 regression estimates median whereas L2 regression estimates mean?
There is a simple geometric explanation for why the L1 loss function yields the median.
Recall that we are working in one dimension, so imagine a number line spreading horizontally. Plot each of the data points on the number line. Put your finger so... | L1 regression estimates median whereas L2 regression estimates mean?
There is a simple geometric explanation for why the L1 loss function yields the median.
Recall that we are working in one dimension, so imagine a number line spreading horizontally. Plot each of the |
6,066 | L1 regression estimates median whereas L2 regression estimates mean? | This explanation is a summation of muratoa and Yves's comments on D.W.'s answer. Though it is based on calculus, I found it straightforward and easy to understand.
Assuming we have $y_1, y_2, ... y_k$ and to want get a new estimate $\beta$ based on them. The smallest loss is obtained when we find $\beta$ which makes th... | L1 regression estimates median whereas L2 regression estimates mean? | This explanation is a summation of muratoa and Yves's comments on D.W.'s answer. Though it is based on calculus, I found it straightforward and easy to understand.
Assuming we have $y_1, y_2, ... y_k$ | L1 regression estimates median whereas L2 regression estimates mean?
This explanation is a summation of muratoa and Yves's comments on D.W.'s answer. Though it is based on calculus, I found it straightforward and easy to understand.
Assuming we have $y_1, y_2, ... y_k$ and to want get a new estimate $\beta$ based on th... | L1 regression estimates median whereas L2 regression estimates mean?
This explanation is a summation of muratoa and Yves's comments on D.W.'s answer. Though it is based on calculus, I found it straightforward and easy to understand.
Assuming we have $y_1, y_2, ... y_k$ |
6,067 | L1 regression estimates median whereas L2 regression estimates mean? | In addition to the already-posted answers (which have been very helpful to me!), there is a geometric explanation for the connection between the L2 norm and the mean.
To use the same notation as chefwen, the formula for L2 loss is:
$$ L2 = \frac{1}{k} \sum^{k}_{i=1} (y_i - \beta)^2$$
We wish to find the value of $\beta... | L1 regression estimates median whereas L2 regression estimates mean? | In addition to the already-posted answers (which have been very helpful to me!), there is a geometric explanation for the connection between the L2 norm and the mean.
To use the same notation as chefw | L1 regression estimates median whereas L2 regression estimates mean?
In addition to the already-posted answers (which have been very helpful to me!), there is a geometric explanation for the connection between the L2 norm and the mean.
To use the same notation as chefwen, the formula for L2 loss is:
$$ L2 = \frac{1}{k}... | L1 regression estimates median whereas L2 regression estimates mean?
In addition to the already-posted answers (which have been very helpful to me!), there is a geometric explanation for the connection between the L2 norm and the mean.
To use the same notation as chefw |
6,068 | L1 regression estimates median whereas L2 regression estimates mean? | Adding to D.W.'s answer with an even more practical example (for L2 loss function as well):
Imagine a small village made of 4 houses close to each other (e.g. 10 meters). At 1 kilometer from those, you have another very isolated house. Now, you arrive in that town and want to build your own house somewhere. You want to... | L1 regression estimates median whereas L2 regression estimates mean? | Adding to D.W.'s answer with an even more practical example (for L2 loss function as well):
Imagine a small village made of 4 houses close to each other (e.g. 10 meters). At 1 kilometer from those, yo | L1 regression estimates median whereas L2 regression estimates mean?
Adding to D.W.'s answer with an even more practical example (for L2 loss function as well):
Imagine a small village made of 4 houses close to each other (e.g. 10 meters). At 1 kilometer from those, you have another very isolated house. Now, you arrive... | L1 regression estimates median whereas L2 regression estimates mean?
Adding to D.W.'s answer with an even more practical example (for L2 loss function as well):
Imagine a small village made of 4 houses close to each other (e.g. 10 meters). At 1 kilometer from those, yo |
6,069 | Why law of large numbers does not apply in the case of Apple share price? | Here is the rub: Apple is so big, it’s running up against the law of
large numbers.
Also known as the golden theorem, with a proof attributed to the
17th-century Swiss mathematician Jacob Bernoulli, the law states that
a variable will revert to a mean over a large sample of results. In
the case of the largest c... | Why law of large numbers does not apply in the case of Apple share price? | Here is the rub: Apple is so big, it’s running up against the law of
large numbers.
Also known as the golden theorem, with a proof attributed to the
17th-century Swiss mathematician Jacob Bernoull | Why law of large numbers does not apply in the case of Apple share price?
Here is the rub: Apple is so big, it’s running up against the law of
large numbers.
Also known as the golden theorem, with a proof attributed to the
17th-century Swiss mathematician Jacob Bernoulli, the law states that
a variable will rever... | Why law of large numbers does not apply in the case of Apple share price?
Here is the rub: Apple is so big, it’s running up against the law of
large numbers.
Also known as the golden theorem, with a proof attributed to the
17th-century Swiss mathematician Jacob Bernoull |
6,070 | Why law of large numbers does not apply in the case of Apple share price? | Humorously enough, I just wrote a blog post on this very subject:
http://confounding.net/2012/03/12/thats-not-how-the-law-of-large-numbers-works/
Essentially, the Law of Large Numbers is that as the number of trials of a random process increases, the mean of those trials will approach the actual mean (or expectation, f... | Why law of large numbers does not apply in the case of Apple share price? | Humorously enough, I just wrote a blog post on this very subject:
http://confounding.net/2012/03/12/thats-not-how-the-law-of-large-numbers-works/
Essentially, the Law of Large Numbers is that as the n | Why law of large numbers does not apply in the case of Apple share price?
Humorously enough, I just wrote a blog post on this very subject:
http://confounding.net/2012/03/12/thats-not-how-the-law-of-large-numbers-works/
Essentially, the Law of Large Numbers is that as the number of trials of a random process increases,... | Why law of large numbers does not apply in the case of Apple share price?
Humorously enough, I just wrote a blog post on this very subject:
http://confounding.net/2012/03/12/thats-not-how-the-law-of-large-numbers-works/
Essentially, the Law of Large Numbers is that as the n |
6,071 | Why law of large numbers does not apply in the case of Apple share price? | There is no reason to think that stock price draws over time for a particular company represent independent, identically distributed random variables. | Why law of large numbers does not apply in the case of Apple share price? | There is no reason to think that stock price draws over time for a particular company represent independent, identically distributed random variables. | Why law of large numbers does not apply in the case of Apple share price?
There is no reason to think that stock price draws over time for a particular company represent independent, identically distributed random variables. | Why law of large numbers does not apply in the case of Apple share price?
There is no reason to think that stock price draws over time for a particular company represent independent, identically distributed random variables. |
6,072 | When should one include a variable in a regression despite it not being statistically significant? | Yes!
That a coefficient is statistically indistinguishable from zero does not imply that the coefficient actually is zero, that the coefficient is irrelevant. That an effect does not pass some arbitrary cutoff for statistical significance does not imply one should not attempt to control for it.
Generally speaking, the ... | When should one include a variable in a regression despite it not being statistically significant? | Yes!
That a coefficient is statistically indistinguishable from zero does not imply that the coefficient actually is zero, that the coefficient is irrelevant. That an effect does not pass some arbitra | When should one include a variable in a regression despite it not being statistically significant?
Yes!
That a coefficient is statistically indistinguishable from zero does not imply that the coefficient actually is zero, that the coefficient is irrelevant. That an effect does not pass some arbitrary cutoff for statist... | When should one include a variable in a regression despite it not being statistically significant?
Yes!
That a coefficient is statistically indistinguishable from zero does not imply that the coefficient actually is zero, that the coefficient is irrelevant. That an effect does not pass some arbitra |
6,073 | When should one include a variable in a regression despite it not being statistically significant? | Yes, there are. Any variable that could correlate with your response variable in a meaningful way, even at a statistically insignificant level, could confound your regression if it is not included. This is known as underspecification, and leads to parameter estimates that are not as accurate as they could otherwise be.... | When should one include a variable in a regression despite it not being statistically significant? | Yes, there are. Any variable that could correlate with your response variable in a meaningful way, even at a statistically insignificant level, could confound your regression if it is not included. Th | When should one include a variable in a regression despite it not being statistically significant?
Yes, there are. Any variable that could correlate with your response variable in a meaningful way, even at a statistically insignificant level, could confound your regression if it is not included. This is known as unders... | When should one include a variable in a regression despite it not being statistically significant?
Yes, there are. Any variable that could correlate with your response variable in a meaningful way, even at a statistically insignificant level, could confound your regression if it is not included. Th |
6,074 | When should one include a variable in a regression despite it not being statistically significant? | Usually you do not include or exclude variables for linear regression because of their significance. You include them because you assume that the selected variables are (good) predictors of the regression criteria. In other words, the predictor selection is based on theory.
Statistical insignificance in linear regressi... | When should one include a variable in a regression despite it not being statistically significant? | Usually you do not include or exclude variables for linear regression because of their significance. You include them because you assume that the selected variables are (good) predictors of the regres | When should one include a variable in a regression despite it not being statistically significant?
Usually you do not include or exclude variables for linear regression because of their significance. You include them because you assume that the selected variables are (good) predictors of the regression criteria. In oth... | When should one include a variable in a regression despite it not being statistically significant?
Usually you do not include or exclude variables for linear regression because of their significance. You include them because you assume that the selected variables are (good) predictors of the regres |
6,075 | When should one include a variable in a regression despite it not being statistically significant? | In econometrics this happens left and right. For instance, if you are using quarterly seasonality dummies Q2,Q3, and Q4, it happens often that as a group they're significant, but some of them are not significant individually. In this case you usually keep them all.
Another typical case is interactions. Consider a mode... | When should one include a variable in a regression despite it not being statistically significant? | In econometrics this happens left and right. For instance, if you are using quarterly seasonality dummies Q2,Q3, and Q4, it happens often that as a group they're significant, but some of them are not | When should one include a variable in a regression despite it not being statistically significant?
In econometrics this happens left and right. For instance, if you are using quarterly seasonality dummies Q2,Q3, and Q4, it happens often that as a group they're significant, but some of them are not significant individua... | When should one include a variable in a regression despite it not being statistically significant?
In econometrics this happens left and right. For instance, if you are using quarterly seasonality dummies Q2,Q3, and Q4, it happens often that as a group they're significant, but some of them are not |
6,076 | When should one include a variable in a regression despite it not being statistically significant? | You are asking two different questions:
When does statistical significance not matter?
When should we include a variable in a regression in spite of it not being statistically significant?
Edit: this was true about the original post, but might no longer be true after the edits.
Regarding Q1, I think it is on the bor... | When should one include a variable in a regression despite it not being statistically significant? | You are asking two different questions:
When does statistical significance not matter?
When should we include a variable in a regression in spite of it not being statistically significant?
Edit: thi | When should one include a variable in a regression despite it not being statistically significant?
You are asking two different questions:
When does statistical significance not matter?
When should we include a variable in a regression in spite of it not being statistically significant?
Edit: this was true about the ... | When should one include a variable in a regression despite it not being statistically significant?
You are asking two different questions:
When does statistical significance not matter?
When should we include a variable in a regression in spite of it not being statistically significant?
Edit: thi |
6,077 | When should one include a variable in a regression despite it not being statistically significant? | I'll add another "yes". I've always been taught -- and I've tried to pass it along -- that the primary consideration in covariate choice is domain knowledge, not statistics. In biostatistics, for instance, if I'm modelling some health outcome on individuals, then no matter what the regression says, you'll need some dar... | When should one include a variable in a regression despite it not being statistically significant? | I'll add another "yes". I've always been taught -- and I've tried to pass it along -- that the primary consideration in covariate choice is domain knowledge, not statistics. In biostatistics, for inst | When should one include a variable in a regression despite it not being statistically significant?
I'll add another "yes". I've always been taught -- and I've tried to pass it along -- that the primary consideration in covariate choice is domain knowledge, not statistics. In biostatistics, for instance, if I'm modellin... | When should one include a variable in a regression despite it not being statistically significant?
I'll add another "yes". I've always been taught -- and I've tried to pass it along -- that the primary consideration in covariate choice is domain knowledge, not statistics. In biostatistics, for inst |
6,078 | When should one include a variable in a regression despite it not being statistically significant? | The only thing that the result of "statistical insignificance" truly says is that, at the selected level of Type I error, we cannot even tell whether the effect of the regressor on the dependent variable is positive or negative (see this post).
So, if we keep this regressor, any discussion about its own effect on the d... | When should one include a variable in a regression despite it not being statistically significant? | The only thing that the result of "statistical insignificance" truly says is that, at the selected level of Type I error, we cannot even tell whether the effect of the regressor on the dependent varia | When should one include a variable in a regression despite it not being statistically significant?
The only thing that the result of "statistical insignificance" truly says is that, at the selected level of Type I error, we cannot even tell whether the effect of the regressor on the dependent variable is positive or ne... | When should one include a variable in a regression despite it not being statistically significant?
The only thing that the result of "statistical insignificance" truly says is that, at the selected level of Type I error, we cannot even tell whether the effect of the regressor on the dependent varia |
6,079 | When should one include a variable in a regression despite it not being statistically significant? | You may include a variable of particular interest if it is the focus of research, even if not statistically significant. Also, in biostatistics, clinical significance is often different than statistical significance. | When should one include a variable in a regression despite it not being statistically significant? | You may include a variable of particular interest if it is the focus of research, even if not statistically significant. Also, in biostatistics, clinical significance is often different than statistic | When should one include a variable in a regression despite it not being statistically significant?
You may include a variable of particular interest if it is the focus of research, even if not statistically significant. Also, in biostatistics, clinical significance is often different than statistical significance. | When should one include a variable in a regression despite it not being statistically significant?
You may include a variable of particular interest if it is the focus of research, even if not statistically significant. Also, in biostatistics, clinical significance is often different than statistic |
6,080 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | Covariances are variances.
Since by the polarization identity
$$\newcommand{\c}{\text{Cov}}\newcommand{\v}{\text{Var}}
\c(X,Y) = \v\left(\frac{X+Y}{2}\right) - \v\left(\frac{X-Y}{2}\right),$$
the denominators must be the same. | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | Covariances are variances.
Since by the polarization identity
$$\newcommand{\c}{\text{Cov}}\newcommand{\v}{\text{Var}}
\c(X,Y) = \v\left(\frac{X+Y}{2}\right) - \v\left(\frac{X-Y}{2}\right),$$
the den | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
Covariances are variances.
Since by the polarization identity
$$\newcommand{\c}{\text{Cov}}\newcommand{\v}{\text{Var}}
\c(X,Y) = \v\left(\frac{X+Y}{2}\right) - \v\left(\frac{X-Y}{2}\right),$$
the denominators must be the same. | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
Covariances are variances.
Since by the polarization identity
$$\newcommand{\c}{\text{Cov}}\newcommand{\v}{\text{Var}}
\c(X,Y) = \v\left(\frac{X+Y}{2}\right) - \v\left(\frac{X-Y}{2}\right),$$
the den |
6,081 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | A special case ought to give you an intuition; think about the following:
$$\hat{\mathbb{Cov}}\left(X, X\right)=
\hat{\mathbb{V}}\left(X\right)$$
You are happy that the latter is $\frac{\sum_{i=1}^{n}\left(X_{i}-\overline{X}\right)^{2}}{n-1}$ due to the Bessel correction.
But replacing $Y$ by $X$ in $\hat{\mathbb{Cov}}... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | A special case ought to give you an intuition; think about the following:
$$\hat{\mathbb{Cov}}\left(X, X\right)=
\hat{\mathbb{V}}\left(X\right)$$
You are happy that the latter is $\frac{\sum_{i=1}^{n} | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
A special case ought to give you an intuition; think about the following:
$$\hat{\mathbb{Cov}}\left(X, X\right)=
\hat{\mathbb{V}}\left(X\right)$$
You are happy that the latter is $\frac{\sum_{i=1}^{n}\left(X_{i}-\overline{X}\right)^{2}}{n... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
A special case ought to give you an intuition; think about the following:
$$\hat{\mathbb{Cov}}\left(X, X\right)=
\hat{\mathbb{V}}\left(X\right)$$
You are happy that the latter is $\frac{\sum_{i=1}^{n} |
6,082 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | A quick and dirty answer... Let’s consider first $\text{var}(X)$; if you had $n$ observations with known expected value $E(X) = 0$ you would use ${1\over n}\sum_{i=1}^n X_i^2$ to estimate the variance.
The expected value being unknown, you can transform your $n$ observations into $n-1$ observations with known expected... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | A quick and dirty answer... Let’s consider first $\text{var}(X)$; if you had $n$ observations with known expected value $E(X) = 0$ you would use ${1\over n}\sum_{i=1}^n X_i^2$ to estimate the variance | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
A quick and dirty answer... Let’s consider first $\text{var}(X)$; if you had $n$ observations with known expected value $E(X) = 0$ you would use ${1\over n}\sum_{i=1}^n X_i^2$ to estimate the variance.
The expected value being unknown, y... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
A quick and dirty answer... Let’s consider first $\text{var}(X)$; if you had $n$ observations with known expected value $E(X) = 0$ you would use ${1\over n}\sum_{i=1}^n X_i^2$ to estimate the variance |
6,083 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | Here is a proof that the p-variate sample covariance estimator with denominator $\frac{1}{n-1}$ is an unbiased estimator of the covariance matrix:
$ x' = (x_1,...,x_p) $.
$\Sigma= E((x-\mu)(x-\mu)') $
$S = \frac{1}{n} \sum (x_i - \bar{x})(x_i - \bar{x})'$
To show: $E(S) = \frac{n-1}{n}\Sigma$
Proof: $S= \frac{1}{n}\su... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | Here is a proof that the p-variate sample covariance estimator with denominator $\frac{1}{n-1}$ is an unbiased estimator of the covariance matrix:
$ x' = (x_1,...,x_p) $.
$\Sigma= E((x-\mu)(x-\mu)') | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
Here is a proof that the p-variate sample covariance estimator with denominator $\frac{1}{n-1}$ is an unbiased estimator of the covariance matrix:
$ x' = (x_1,...,x_p) $.
$\Sigma= E((x-\mu)(x-\mu)') $
$S = \frac{1}{n} \sum (x_i - \bar{x}... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
Here is a proof that the p-variate sample covariance estimator with denominator $\frac{1}{n-1}$ is an unbiased estimator of the covariance matrix:
$ x' = (x_1,...,x_p) $.
$\Sigma= E((x-\mu)(x-\mu)') |
6,084 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | I guess one way to build intuition behind using 'n-1' and not 'n-2' is - that for calculating co-variance we do not need to de-mean both X and Y, but either of the two, i.e. | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | I guess one way to build intuition behind using 'n-1' and not 'n-2' is - that for calculating co-variance we do not need to de-mean both X and Y, but either of the two, i.e. | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
I guess one way to build intuition behind using 'n-1' and not 'n-2' is - that for calculating co-variance we do not need to de-mean both X and Y, but either of the two, i.e. | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
I guess one way to build intuition behind using 'n-1' and not 'n-2' is - that for calculating co-variance we do not need to de-mean both X and Y, but either of the two, i.e. |
6,085 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | 1) Start $df=2n$.
2) Sample covariance is proportional to $\Sigma_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y})$. Lose two $df$; one from $\bar{X}$, one from $\bar{Y}$ resulting in $df=2(n-1)$.
3) However, $\Sigma_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y})$ only contains $n$ separate terms, one from each product. When two numbers are mul... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | 1) Start $df=2n$.
2) Sample covariance is proportional to $\Sigma_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y})$. Lose two $df$; one from $\bar{X}$, one from $\bar{Y}$ resulting in $df=2(n-1)$.
3) However, $\Si | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
1) Start $df=2n$.
2) Sample covariance is proportional to $\Sigma_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y})$. Lose two $df$; one from $\bar{X}$, one from $\bar{Y}$ resulting in $df=2(n-1)$.
3) However, $\Sigma_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y})$... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
1) Start $df=2n$.
2) Sample covariance is proportional to $\Sigma_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y})$. Lose two $df$; one from $\bar{X}$, one from $\bar{Y}$ resulting in $df=2(n-1)$.
3) However, $\Si |
6,086 | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | My simple way of thinking about the intuition behind the 'n-1' degree of freedom is as follow. An observation is a pair $\{x_i, y_i\}$. Therefore, we can always infer the last observation $\{x_n, y_n\}$ if we know 'n-1' previous observations and the sample mean $\{\overline{x}, \overline{y}\}$. The last observation can... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1? | My simple way of thinking about the intuition behind the 'n-1' degree of freedom is as follow. An observation is a pair $\{x_i, y_i\}$. Therefore, we can always infer the last observation $\{x_n, y_n\ | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
My simple way of thinking about the intuition behind the 'n-1' degree of freedom is as follow. An observation is a pair $\{x_i, y_i\}$. Therefore, we can always infer the last observation $\{x_n, y_n\}$ if we know 'n-1' previous observati... | Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
My simple way of thinking about the intuition behind the 'n-1' degree of freedom is as follow. An observation is a pair $\{x_i, y_i\}$. Therefore, we can always infer the last observation $\{x_n, y_n\ |
6,087 | Correlation does not imply causation; but what about when one of the variables is time? | I'll provide another answer, since I think the ones currently provided miss an important point of the statement the physicist made. The quoted statement is:
"correlation does not imply causation, UNLESS one of the variables is time. So, if there is a strong correlation between some independent variable and time, then ... | Correlation does not imply causation; but what about when one of the variables is time? | I'll provide another answer, since I think the ones currently provided miss an important point of the statement the physicist made. The quoted statement is:
"correlation does not imply causation, UNL | Correlation does not imply causation; but what about when one of the variables is time?
I'll provide another answer, since I think the ones currently provided miss an important point of the statement the physicist made. The quoted statement is:
"correlation does not imply causation, UNLESS one of the variables is time... | Correlation does not imply causation; but what about when one of the variables is time?
I'll provide another answer, since I think the ones currently provided miss an important point of the statement the physicist made. The quoted statement is:
"correlation does not imply causation, UNL |
6,088 | Correlation does not imply causation; but what about when one of the variables is time? | We don't know what the physicist meant. Two different interpretations follow.
The claim that $X$ preceding $Y$ and being correlated with $Y$ implies that $X$ causes $Y$ is wrong. It's not enough for $X$ and $Y$ to be dependent even if $X$ precedes $Y$. For example, $X$ and $Y$ can both be caused by some other varia... | Correlation does not imply causation; but what about when one of the variables is time? | We don't know what the physicist meant. Two different interpretations follow.
The claim that $X$ preceding $Y$ and being correlated with $Y$ implies that $X$ causes $Y$ is wrong. It's not enough fo | Correlation does not imply causation; but what about when one of the variables is time?
We don't know what the physicist meant. Two different interpretations follow.
The claim that $X$ preceding $Y$ and being correlated with $Y$ implies that $X$ causes $Y$ is wrong. It's not enough for $X$ and $Y$ to be dependent ev... | Correlation does not imply causation; but what about when one of the variables is time?
We don't know what the physicist meant. Two different interpretations follow.
The claim that $X$ preceding $Y$ and being correlated with $Y$ implies that $X$ causes $Y$ is wrong. It's not enough fo |
6,089 | Correlation does not imply causation; but what about when one of the variables is time? | I speculate that your guest lecturer meant that in physics the only correlations that survive replication are the ones where there is an underlying causal relationship. Time variable is an exception because it is the only variable that is not controlled by the physicist. Here's why.
In physics we usually deal with repe... | Correlation does not imply causation; but what about when one of the variables is time? | I speculate that your guest lecturer meant that in physics the only correlations that survive replication are the ones where there is an underlying causal relationship. Time variable is an exception b | Correlation does not imply causation; but what about when one of the variables is time?
I speculate that your guest lecturer meant that in physics the only correlations that survive replication are the ones where there is an underlying causal relationship. Time variable is an exception because it is the only variable t... | Correlation does not imply causation; but what about when one of the variables is time?
I speculate that your guest lecturer meant that in physics the only correlations that survive replication are the ones where there is an underlying causal relationship. Time variable is an exception b |
6,090 | Correlation does not imply causation; but what about when one of the variables is time? | I have not heard this before, and it would not be true according to the conceptions of causality that I am familiar with (though I am not a physicist).
Typically, for $X$ to cause $Y$ it is necessary that $X$ precedes $Y$ in time. So if $Y$ precedes $X$ then it cannot be "caused" by $X$, regardless of any correlation. ... | Correlation does not imply causation; but what about when one of the variables is time? | I have not heard this before, and it would not be true according to the conceptions of causality that I am familiar with (though I am not a physicist).
Typically, for $X$ to cause $Y$ it is necessary | Correlation does not imply causation; but what about when one of the variables is time?
I have not heard this before, and it would not be true according to the conceptions of causality that I am familiar with (though I am not a physicist).
Typically, for $X$ to cause $Y$ it is necessary that $X$ precedes $Y$ in time. S... | Correlation does not imply causation; but what about when one of the variables is time?
I have not heard this before, and it would not be true according to the conceptions of causality that I am familiar with (though I am not a physicist).
Typically, for $X$ to cause $Y$ it is necessary |
6,091 | Correlation does not imply causation; but what about when one of the variables is time? | I don't think time is necessarily unique in this, but it's certainly a good example. The point is that typically if A & B are correlated, you can surmise that there is some common causality, but you don't know whether A causes B or B causes A, or perhaps a third variable C causes both A & B. However, in certain cases, ... | Correlation does not imply causation; but what about when one of the variables is time? | I don't think time is necessarily unique in this, but it's certainly a good example. The point is that typically if A & B are correlated, you can surmise that there is some common causality, but you d | Correlation does not imply causation; but what about when one of the variables is time?
I don't think time is necessarily unique in this, but it's certainly a good example. The point is that typically if A & B are correlated, you can surmise that there is some common causality, but you don't know whether A causes B or ... | Correlation does not imply causation; but what about when one of the variables is time?
I don't think time is necessarily unique in this, but it's certainly a good example. The point is that typically if A & B are correlated, you can surmise that there is some common causality, but you d |
6,092 | Correlation does not imply causation; but what about when one of the variables is time? | Actually, correlation does imply a causal relationship.
Perhaps A caused B, or C caused A and B.
However, correlation does not prove causation.
This is self-evident. | Correlation does not imply causation; but what about when one of the variables is time? | Actually, correlation does imply a causal relationship.
Perhaps A caused B, or C caused A and B.
However, correlation does not prove causation.
This is self-evident. | Correlation does not imply causation; but what about when one of the variables is time?
Actually, correlation does imply a causal relationship.
Perhaps A caused B, or C caused A and B.
However, correlation does not prove causation.
This is self-evident. | Correlation does not imply causation; but what about when one of the variables is time?
Actually, correlation does imply a causal relationship.
Perhaps A caused B, or C caused A and B.
However, correlation does not prove causation.
This is self-evident. |
6,093 | Correlation does not imply causation; but what about when one of the variables is time? | I would interpret this as a semantic rather than mathematical/statistical argument. I would also take it as a rather severe generalization.
The Bradford Hill Criteria, often used in epidemiology, provide a good framework for thinking about causation. Nothing can definitively prove causation, whether or not time is a fa... | Correlation does not imply causation; but what about when one of the variables is time? | I would interpret this as a semantic rather than mathematical/statistical argument. I would also take it as a rather severe generalization.
The Bradford Hill Criteria, often used in epidemiology, prov | Correlation does not imply causation; but what about when one of the variables is time?
I would interpret this as a semantic rather than mathematical/statistical argument. I would also take it as a rather severe generalization.
The Bradford Hill Criteria, often used in epidemiology, provide a good framework for thinkin... | Correlation does not imply causation; but what about when one of the variables is time?
I would interpret this as a semantic rather than mathematical/statistical argument. I would also take it as a rather severe generalization.
The Bradford Hill Criteria, often used in epidemiology, prov |
6,094 | Correlation does not imply causation; but what about when one of the variables is time? | The sentence is quite simple and not worth overthinking (and has nothing to do with precedence).
If there is an established correlation between a variable and time (i.e. we know that an increase in time is accompanied by an increase in the variable, and this is a given), then we know the "causal" direction: i.e. time ... | Correlation does not imply causation; but what about when one of the variables is time? | The sentence is quite simple and not worth overthinking (and has nothing to do with precedence).
If there is an established correlation between a variable and time (i.e. we know that an increase in t | Correlation does not imply causation; but what about when one of the variables is time?
The sentence is quite simple and not worth overthinking (and has nothing to do with precedence).
If there is an established correlation between a variable and time (i.e. we know that an increase in time is accompanied by an increas... | Correlation does not imply causation; but what about when one of the variables is time?
The sentence is quite simple and not worth overthinking (and has nothing to do with precedence).
If there is an established correlation between a variable and time (i.e. we know that an increase in t |
6,095 | Correlation does not imply causation; but what about when one of the variables is time? | This is really a question of how to establish causality, because events which are related but not causative WILL likely be correlated in time or space. So looking at some correlated data, how can we determine if the relationship is dependent? A wise research adviser once told me, "correlation does not imply causation, ... | Correlation does not imply causation; but what about when one of the variables is time? | This is really a question of how to establish causality, because events which are related but not causative WILL likely be correlated in time or space. So looking at some correlated data, how can we d | Correlation does not imply causation; but what about when one of the variables is time?
This is really a question of how to establish causality, because events which are related but not causative WILL likely be correlated in time or space. So looking at some correlated data, how can we determine if the relationship is ... | Correlation does not imply causation; but what about when one of the variables is time?
This is really a question of how to establish causality, because events which are related but not causative WILL likely be correlated in time or space. So looking at some correlated data, how can we d |
6,096 | Is LSTM (Long Short-Term Memory) dead? | Maybe. But RNNs aren't.
Transformers learn "pseudo-temporal" relationships; they lack the true recurrent gradient that RNNs have, and thus extract fundamentally different features. This paper, for example, shows that the standard transformers are difficult to optimize in reinforcement learning settings, especially in m... | Is LSTM (Long Short-Term Memory) dead? | Maybe. But RNNs aren't.
Transformers learn "pseudo-temporal" relationships; they lack the true recurrent gradient that RNNs have, and thus extract fundamentally different features. This paper, for exa | Is LSTM (Long Short-Term Memory) dead?
Maybe. But RNNs aren't.
Transformers learn "pseudo-temporal" relationships; they lack the true recurrent gradient that RNNs have, and thus extract fundamentally different features. This paper, for example, shows that the standard transformers are difficult to optimize in reinforce... | Is LSTM (Long Short-Term Memory) dead?
Maybe. But RNNs aren't.
Transformers learn "pseudo-temporal" relationships; they lack the true recurrent gradient that RNNs have, and thus extract fundamentally different features. This paper, for exa |
6,097 | Is LSTM (Long Short-Term Memory) dead? | It is funny that you ask now, since just today I came across a paper by Wang, Khabsa, and Ma (2020) To Pretrain or Not to Pretrain who show that if you have large enough training set, the difference in performance between huge, "SOTA" model (RoBERTa), and LSTMs is small for NLP task. There was another recent paper by M... | Is LSTM (Long Short-Term Memory) dead? | It is funny that you ask now, since just today I came across a paper by Wang, Khabsa, and Ma (2020) To Pretrain or Not to Pretrain who show that if you have large enough training set, the difference i | Is LSTM (Long Short-Term Memory) dead?
It is funny that you ask now, since just today I came across a paper by Wang, Khabsa, and Ma (2020) To Pretrain or Not to Pretrain who show that if you have large enough training set, the difference in performance between huge, "SOTA" model (RoBERTa), and LSTMs is small for NLP ta... | Is LSTM (Long Short-Term Memory) dead?
It is funny that you ask now, since just today I came across a paper by Wang, Khabsa, and Ma (2020) To Pretrain or Not to Pretrain who show that if you have large enough training set, the difference i |
6,098 | Is LSTM (Long Short-Term Memory) dead? | Our group recently built an LSTM model in a real world application. At first we had used other approaches, but then we decided to include features that were measurements taken over time, but of variable length - so for one person, we would have 15 measurements (of the same parameter) taken over a 3-month period, for an... | Is LSTM (Long Short-Term Memory) dead? | Our group recently built an LSTM model in a real world application. At first we had used other approaches, but then we decided to include features that were measurements taken over time, but of variab | Is LSTM (Long Short-Term Memory) dead?
Our group recently built an LSTM model in a real world application. At first we had used other approaches, but then we decided to include features that were measurements taken over time, but of variable length - so for one person, we would have 15 measurements (of the same paramet... | Is LSTM (Long Short-Term Memory) dead?
Our group recently built an LSTM model in a real world application. At first we had used other approaches, but then we decided to include features that were measurements taken over time, but of variab |
6,099 | Is LSTM (Long Short-Term Memory) dead? | LSTM is a statistical method. It is not alive so it cannot be dead. It can be useful though. Any statistical method is another tool in a box. If one does not work it is good to have an alternative.
LSTM is good for language recognition tasks where context is important. It is also good for forecasting time series. The M... | Is LSTM (Long Short-Term Memory) dead? | LSTM is a statistical method. It is not alive so it cannot be dead. It can be useful though. Any statistical method is another tool in a box. If one does not work it is good to have an alternative.
LS | Is LSTM (Long Short-Term Memory) dead?
LSTM is a statistical method. It is not alive so it cannot be dead. It can be useful though. Any statistical method is another tool in a box. If one does not work it is good to have an alternative.
LSTM is good for language recognition tasks where context is important. It is also ... | Is LSTM (Long Short-Term Memory) dead?
LSTM is a statistical method. It is not alive so it cannot be dead. It can be useful though. Any statistical method is another tool in a box. If one does not work it is good to have an alternative.
LS |
6,100 | How trustworthy are the confidence intervals for lmer objects through effects package? | All of the results are essentially the same (for this particular example). Some theoretical differences are:
as @rvl points out, your reconstruction of CIs without taking account of covariance among parameters is just wrong (sorry)
confidence intervals for parameters can be based on Wald confidence intervals (assumin... | How trustworthy are the confidence intervals for lmer objects through effects package? | All of the results are essentially the same (for this particular example). Some theoretical differences are:
as @rvl points out, your reconstruction of CIs without taking account of covariance among | How trustworthy are the confidence intervals for lmer objects through effects package?
All of the results are essentially the same (for this particular example). Some theoretical differences are:
as @rvl points out, your reconstruction of CIs without taking account of covariance among parameters is just wrong (sorry)... | How trustworthy are the confidence intervals for lmer objects through effects package?
All of the results are essentially the same (for this particular example). Some theoretical differences are:
as @rvl points out, your reconstruction of CIs without taking account of covariance among |
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