idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k β | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 β | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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49,301 | Anomaly detection based on clustering | ELKI includes a class called KMeansOutlierDetection (and many more).
But of all the methods that I have tried, this one worked worst:
Even on easy, artificial data it doesn't work too well, except for the trivial objects (that literally any method will detect).
The problems with cluster-based outlier detection is that... | Anomaly detection based on clustering | ELKI includes a class called KMeansOutlierDetection (and many more).
But of all the methods that I have tried, this one worked worst:
Even on easy, artificial data it doesn't work too well, except fo | Anomaly detection based on clustering
ELKI includes a class called KMeansOutlierDetection (and many more).
But of all the methods that I have tried, this one worked worst:
Even on easy, artificial data it doesn't work too well, except for the trivial objects (that literally any method will detect).
The problems with c... | Anomaly detection based on clustering
ELKI includes a class called KMeansOutlierDetection (and many more).
But of all the methods that I have tried, this one worked worst:
Even on easy, artificial data it doesn't work too well, except fo |
49,302 | How to calculate probability of observing a value given a permutation distribution? | This is quite straightforward, there is no need to infer a distribution under the Null Hypothesis. Your p-value is just the number of times $x_{permuted}$ get superior or equal to $0.5$, divided by the number of permutations made.
This fits the definition of the p-value: "If H0 is true and a new sample is drawn, what i... | How to calculate probability of observing a value given a permutation distribution? | This is quite straightforward, there is no need to infer a distribution under the Null Hypothesis. Your p-value is just the number of times $x_{permuted}$ get superior or equal to $0.5$, divided by th | How to calculate probability of observing a value given a permutation distribution?
This is quite straightforward, there is no need to infer a distribution under the Null Hypothesis. Your p-value is just the number of times $x_{permuted}$ get superior or equal to $0.5$, divided by the number of permutations made.
This ... | How to calculate probability of observing a value given a permutation distribution?
This is quite straightforward, there is no need to infer a distribution under the Null Hypothesis. Your p-value is just the number of times $x_{permuted}$ get superior or equal to $0.5$, divided by th |
49,303 | Comparison of Bernstain and Chebyshev inequalities applied to Bernoulli distribution - simulation in R gives unexpected results | I ran the code and didn't see any cases for which the theoretical bounds were breached. But even so, it could happen, as the bound is on the probability that the empirical mean deviates from the true mean. Run the experiment $N$ times, and due to bad luck many of the empirical means could deviate from $\mu$, in fact, ... | Comparison of Bernstain and Chebyshev inequalities applied to Bernoulli distribution - simulation in | I ran the code and didn't see any cases for which the theoretical bounds were breached. But even so, it could happen, as the bound is on the probability that the empirical mean deviates from the true | Comparison of Bernstain and Chebyshev inequalities applied to Bernoulli distribution - simulation in R gives unexpected results
I ran the code and didn't see any cases for which the theoretical bounds were breached. But even so, it could happen, as the bound is on the probability that the empirical mean deviates from t... | Comparison of Bernstain and Chebyshev inequalities applied to Bernoulli distribution - simulation in
I ran the code and didn't see any cases for which the theoretical bounds were breached. But even so, it could happen, as the bound is on the probability that the empirical mean deviates from the true |
49,304 | How to identify variables with significant loadings in PCA? | This is not (yet) and answer, only a comment but too long for the box
I do not really know how to determine such significance; but out of couriosity I did a bootstrap-procedure: from a replication of the original data to a pseudo-population of $N=19200$ I draw $t=1000$ randomsamples of $n=150$ (each row of the datase... | How to identify variables with significant loadings in PCA? | This is not (yet) and answer, only a comment but too long for the box
I do not really know how to determine such significance; but out of couriosity I did a bootstrap-procedure: from a replication o | How to identify variables with significant loadings in PCA?
This is not (yet) and answer, only a comment but too long for the box
I do not really know how to determine such significance; but out of couriosity I did a bootstrap-procedure: from a replication of the original data to a pseudo-population of $N=19200$ I dr... | How to identify variables with significant loadings in PCA?
This is not (yet) and answer, only a comment but too long for the box
I do not really know how to determine such significance; but out of couriosity I did a bootstrap-procedure: from a replication o |
49,305 | Bonferroni Adjustment and Assumptions? | When one analyses the proof that Bonferroni controls the type I error ''family-wise'' then you see that no assumptions are needed; it basically uses only the inequality of Boole. So Bonferroni does not need e.g. an independence assumption.
However, the analysis of the proof learns that the probability of a type I erro... | Bonferroni Adjustment and Assumptions? | When one analyses the proof that Bonferroni controls the type I error ''family-wise'' then you see that no assumptions are needed; it basically uses only the inequality of Boole. So Bonferroni does n | Bonferroni Adjustment and Assumptions?
When one analyses the proof that Bonferroni controls the type I error ''family-wise'' then you see that no assumptions are needed; it basically uses only the inequality of Boole. So Bonferroni does not need e.g. an independence assumption.
However, the analysis of the proof learn... | Bonferroni Adjustment and Assumptions?
When one analyses the proof that Bonferroni controls the type I error ''family-wise'' then you see that no assumptions are needed; it basically uses only the inequality of Boole. So Bonferroni does n |
49,306 | Bonferroni Adjustment and Assumptions? | The Bonferroni (and similar corrections like Bonferroni-Holm etc.) assumes that the p-value you provide it follows a uniform distribution under the null hypothesis and would under the null hypothesis only be below 0.05 in 5% of the time, if you repeated your experiment again and again. Pre-tests such as for normality o... | Bonferroni Adjustment and Assumptions? | The Bonferroni (and similar corrections like Bonferroni-Holm etc.) assumes that the p-value you provide it follows a uniform distribution under the null hypothesis and would under the null hypothesis | Bonferroni Adjustment and Assumptions?
The Bonferroni (and similar corrections like Bonferroni-Holm etc.) assumes that the p-value you provide it follows a uniform distribution under the null hypothesis and would under the null hypothesis only be below 0.05 in 5% of the time, if you repeated your experiment again and a... | Bonferroni Adjustment and Assumptions?
The Bonferroni (and similar corrections like Bonferroni-Holm etc.) assumes that the p-value you provide it follows a uniform distribution under the null hypothesis and would under the null hypothesis |
49,307 | Bonferroni Adjustment and Assumptions? | Adjusting for multiple comparisons generally applies to your actual hypotheses, not to tests of assumptions. | Bonferroni Adjustment and Assumptions? | Adjusting for multiple comparisons generally applies to your actual hypotheses, not to tests of assumptions. | Bonferroni Adjustment and Assumptions?
Adjusting for multiple comparisons generally applies to your actual hypotheses, not to tests of assumptions. | Bonferroni Adjustment and Assumptions?
Adjusting for multiple comparisons generally applies to your actual hypotheses, not to tests of assumptions. |
49,308 | Interaction effect in a multiple regression vs split sample | What you have to realize it that a split sample is different from a interaction effect.
Interaction effect with x concerns only a change of the slope of that particular independent variable x, leaving all other slopes constant.
Splitting the sample is equivalent to having an interaction dummy for every independent vari... | Interaction effect in a multiple regression vs split sample | What you have to realize it that a split sample is different from a interaction effect.
Interaction effect with x concerns only a change of the slope of that particular independent variable x, leaving | Interaction effect in a multiple regression vs split sample
What you have to realize it that a split sample is different from a interaction effect.
Interaction effect with x concerns only a change of the slope of that particular independent variable x, leaving all other slopes constant.
Splitting the sample is equivale... | Interaction effect in a multiple regression vs split sample
What you have to realize it that a split sample is different from a interaction effect.
Interaction effect with x concerns only a change of the slope of that particular independent variable x, leaving |
49,309 | Confidence interval for exponential distribution | The asymptotic confidence interval may be based on the (asymptotic) distribution of the mle. The Fisher information for this problem is given by $\frac{1}{\theta^2}$. Hence an asymptotic CI for $\theta$ is given by
$$\bar{X} \pm 1.96 \sqrt{\frac{\bar{X}^2}{n}}$$
where we have replaced $\theta^2$ by its mle, since we do... | Confidence interval for exponential distribution | The asymptotic confidence interval may be based on the (asymptotic) distribution of the mle. The Fisher information for this problem is given by $\frac{1}{\theta^2}$. Hence an asymptotic CI for $\thet | Confidence interval for exponential distribution
The asymptotic confidence interval may be based on the (asymptotic) distribution of the mle. The Fisher information for this problem is given by $\frac{1}{\theta^2}$. Hence an asymptotic CI for $\theta$ is given by
$$\bar{X} \pm 1.96 \sqrt{\frac{\bar{X}^2}{n}}$$
where we... | Confidence interval for exponential distribution
The asymptotic confidence interval may be based on the (asymptotic) distribution of the mle. The Fisher information for this problem is given by $\frac{1}{\theta^2}$. Hence an asymptotic CI for $\thet |
49,310 | Confidence interval for exponential distribution | For ii) you have several options but two of them are particularly appealing. One option is to go for a Wald-type confidence interval the other is to go with an interval based on the log-likelihood ratio statistic.
For the Wald-type you first get the limiting distribution of the MLE of $\theta$, thus something along the... | Confidence interval for exponential distribution | For ii) you have several options but two of them are particularly appealing. One option is to go for a Wald-type confidence interval the other is to go with an interval based on the log-likelihood rat | Confidence interval for exponential distribution
For ii) you have several options but two of them are particularly appealing. One option is to go for a Wald-type confidence interval the other is to go with an interval based on the log-likelihood ratio statistic.
For the Wald-type you first get the limiting distribution... | Confidence interval for exponential distribution
For ii) you have several options but two of them are particularly appealing. One option is to go for a Wald-type confidence interval the other is to go with an interval based on the log-likelihood rat |
49,311 | Confidence interval for exponential distribution | I will explain a bit more deeply the process @JohnK gone through:
$\hat{\lambda}_{MLE}$=$\frac{1}{\bar{X}}$
$l''$=$\log(L)'' = \frac{-n}{\lambda^2}$
therefore, $-E[l''] = Fisher's \ Information = \frac{1}{\lambda^2} (n=1 \ for \ a \ single \ sample \ variance)$
now : $I({\hat{\lambda}_{MLE}}) = \bar{X}^2$
$The \ var... | Confidence interval for exponential distribution | I will explain a bit more deeply the process @JohnK gone through:
$\hat{\lambda}_{MLE}$=$\frac{1}{\bar{X}}$
$l''$=$\log(L)'' = \frac{-n}{\lambda^2}$
therefore, $-E[l''] = Fisher's \ Information = \fra | Confidence interval for exponential distribution
I will explain a bit more deeply the process @JohnK gone through:
$\hat{\lambda}_{MLE}$=$\frac{1}{\bar{X}}$
$l''$=$\log(L)'' = \frac{-n}{\lambda^2}$
therefore, $-E[l''] = Fisher's \ Information = \frac{1}{\lambda^2} (n=1 \ for \ a \ single \ sample \ variance)$
now : ... | Confidence interval for exponential distribution
I will explain a bit more deeply the process @JohnK gone through:
$\hat{\lambda}_{MLE}$=$\frac{1}{\bar{X}}$
$l''$=$\log(L)'' = \frac{-n}{\lambda^2}$
therefore, $-E[l''] = Fisher's \ Information = \fra |
49,312 | Fit exponential distribution with noise | In the absence of a response to my questions relating to the variation about the signal, I'll explain a little about nonlinear least squares.
You can fit a model of the following form:
$y_i = c + \alpha \exp(-\alpha x_i)+\varepsilon_i$, where $E(\varepsilon_i)=0$.
If the $\varepsilon$ values are independent and of cons... | Fit exponential distribution with noise | In the absence of a response to my questions relating to the variation about the signal, I'll explain a little about nonlinear least squares.
You can fit a model of the following form:
$y_i = c + \alp | Fit exponential distribution with noise
In the absence of a response to my questions relating to the variation about the signal, I'll explain a little about nonlinear least squares.
You can fit a model of the following form:
$y_i = c + \alpha \exp(-\alpha x_i)+\varepsilon_i$, where $E(\varepsilon_i)=0$.
If the $\vareps... | Fit exponential distribution with noise
In the absence of a response to my questions relating to the variation about the signal, I'll explain a little about nonlinear least squares.
You can fit a model of the following form:
$y_i = c + \alp |
49,313 | Fit exponential distribution with noise | First of all, are you sure that the noise is additive? For instance, if your noise was multiplicative, then linearization would have worked.
For an additive noise you can't do much. Use nonlinear regression. | Fit exponential distribution with noise | First of all, are you sure that the noise is additive? For instance, if your noise was multiplicative, then linearization would have worked.
For an additive noise you can't do much. Use nonlinear regr | Fit exponential distribution with noise
First of all, are you sure that the noise is additive? For instance, if your noise was multiplicative, then linearization would have worked.
For an additive noise you can't do much. Use nonlinear regression. | Fit exponential distribution with noise
First of all, are you sure that the noise is additive? For instance, if your noise was multiplicative, then linearization would have worked.
For an additive noise you can't do much. Use nonlinear regr |
49,314 | Does summary.aov() of MANOVA object adjust for multiple comparisons? | In any situation where you already have a set of unadjusted P values, you can use p.adjust to obtain adjusted ones for any of the methods in p.adjust.methods:
> p.adjust(c(.002432, .02292, .7524), "fdr")
[1] 0.007296 0.034380 0.752400
I had a much more elaborate answer using the lsmeans package, but it does not seem n... | Does summary.aov() of MANOVA object adjust for multiple comparisons? | In any situation where you already have a set of unadjusted P values, you can use p.adjust to obtain adjusted ones for any of the methods in p.adjust.methods:
> p.adjust(c(.002432, .02292, .7524), "fd | Does summary.aov() of MANOVA object adjust for multiple comparisons?
In any situation where you already have a set of unadjusted P values, you can use p.adjust to obtain adjusted ones for any of the methods in p.adjust.methods:
> p.adjust(c(.002432, .02292, .7524), "fdr")
[1] 0.007296 0.034380 0.752400
I had a much mo... | Does summary.aov() of MANOVA object adjust for multiple comparisons?
In any situation where you already have a set of unadjusted P values, you can use p.adjust to obtain adjusted ones for any of the methods in p.adjust.methods:
> p.adjust(c(.002432, .02292, .7524), "fd |
49,315 | Differences in correlation for individual and aggregated data | The issue is with the binning. When you order the variable $A$ by size, divide it to 100 equal bins and then sum the data in the bins you introduce the order. The bins at the beginning will have lower sums and higher sums at the end. This is perfectly normal, because that is the way bins were constructed.
Here is a sim... | Differences in correlation for individual and aggregated data | The issue is with the binning. When you order the variable $A$ by size, divide it to 100 equal bins and then sum the data in the bins you introduce the order. The bins at the beginning will have lower | Differences in correlation for individual and aggregated data
The issue is with the binning. When you order the variable $A$ by size, divide it to 100 equal bins and then sum the data in the bins you introduce the order. The bins at the beginning will have lower sums and higher sums at the end. This is perfectly normal... | Differences in correlation for individual and aggregated data
The issue is with the binning. When you order the variable $A$ by size, divide it to 100 equal bins and then sum the data in the bins you introduce the order. The bins at the beginning will have lower |
49,316 | BIC in Item Response Theory Models: Using log(N) vs log(N*I) as a weight | I think it's neither. The "textbook" information criteria formulation that you cite are derived for i.i.d. data, while you have a two-way array with weird cross-dependencies: you have the same questions, and you have the same students.
The issue is always there with mixed models. I am not going to try to reproduce thei... | BIC in Item Response Theory Models: Using log(N) vs log(N*I) as a weight | I think it's neither. The "textbook" information criteria formulation that you cite are derived for i.i.d. data, while you have a two-way array with weird cross-dependencies: you have the same questio | BIC in Item Response Theory Models: Using log(N) vs log(N*I) as a weight
I think it's neither. The "textbook" information criteria formulation that you cite are derived for i.i.d. data, while you have a two-way array with weird cross-dependencies: you have the same questions, and you have the same students.
The issue i... | BIC in Item Response Theory Models: Using log(N) vs log(N*I) as a weight
I think it's neither. The "textbook" information criteria formulation that you cite are derived for i.i.d. data, while you have a two-way array with weird cross-dependencies: you have the same questio |
49,317 | Fitting an ARIMA model with conflicting indicators | The null of a unit root is not rejected in the ADF test and the null of stationarity in the KPSS test is not rejected either. This is an inconvenient situation since none of the hypotheses can be rejected.
In principle, as mentioned here, in this situation it may be more cautions to consider the presence of a unit root... | Fitting an ARIMA model with conflicting indicators | The null of a unit root is not rejected in the ADF test and the null of stationarity in the KPSS test is not rejected either. This is an inconvenient situation since none of the hypotheses can be reje | Fitting an ARIMA model with conflicting indicators
The null of a unit root is not rejected in the ADF test and the null of stationarity in the KPSS test is not rejected either. This is an inconvenient situation since none of the hypotheses can be rejected.
In principle, as mentioned here, in this situation it may be mo... | Fitting an ARIMA model with conflicting indicators
The null of a unit root is not rejected in the ADF test and the null of stationarity in the KPSS test is not rejected either. This is an inconvenient situation since none of the hypotheses can be reje |
49,318 | Recovering original regression coefficients from standardized | The models are not the same. Therefore the coefficients should differ. When you recenter, you are taking linear combinations of the columns of $X$ with the vector $\mathbf{1}=(1,1,\ldots, 1)^\prime$. This is fine, provided that $\mathbf{1}$ lies in the column space. In your example it does not.
What is worse, when ... | Recovering original regression coefficients from standardized | The models are not the same. Therefore the coefficients should differ. When you recenter, you are taking linear combinations of the columns of $X$ with the vector $\mathbf{1}=(1,1,\ldots, 1)^\prime$ | Recovering original regression coefficients from standardized
The models are not the same. Therefore the coefficients should differ. When you recenter, you are taking linear combinations of the columns of $X$ with the vector $\mathbf{1}=(1,1,\ldots, 1)^\prime$. This is fine, provided that $\mathbf{1}$ lies in the co... | Recovering original regression coefficients from standardized
The models are not the same. Therefore the coefficients should differ. When you recenter, you are taking linear combinations of the columns of $X$ with the vector $\mathbf{1}=(1,1,\ldots, 1)^\prime$ |
49,319 | Deep learning: representation learning or classification? | In my opinion: it's both. It's referenced many times in the highly cited article on convolutional neural networks Gradient-Based Learning Applied to Document Recognition by Yann LeCun, Yoshua Bengio, Leon Bottou and Patrick Haffner.
The idea is that it is quite hard to hand-design a rich and complex feature hierarchy. ... | Deep learning: representation learning or classification? | In my opinion: it's both. It's referenced many times in the highly cited article on convolutional neural networks Gradient-Based Learning Applied to Document Recognition by Yann LeCun, Yoshua Bengio, | Deep learning: representation learning or classification?
In my opinion: it's both. It's referenced many times in the highly cited article on convolutional neural networks Gradient-Based Learning Applied to Document Recognition by Yann LeCun, Yoshua Bengio, Leon Bottou and Patrick Haffner.
The idea is that it is quite ... | Deep learning: representation learning or classification?
In my opinion: it's both. It's referenced many times in the highly cited article on convolutional neural networks Gradient-Based Learning Applied to Document Recognition by Yann LeCun, Yoshua Bengio, |
49,320 | Deep learning: representation learning or classification? | I would say it's basically representation learning followed by classification at the end.
Consider an Image Classification Problem.
-> We have to find some way(some characteristics/attributes) to tell if the image is a dog or a cat (which in our terms is referred to as features)
-> Extracting the useful features is the... | Deep learning: representation learning or classification? | I would say it's basically representation learning followed by classification at the end.
Consider an Image Classification Problem.
-> We have to find some way(some characteristics/attributes) to tell | Deep learning: representation learning or classification?
I would say it's basically representation learning followed by classification at the end.
Consider an Image Classification Problem.
-> We have to find some way(some characteristics/attributes) to tell if the image is a dog or a cat (which in our terms is referre... | Deep learning: representation learning or classification?
I would say it's basically representation learning followed by classification at the end.
Consider an Image Classification Problem.
-> We have to find some way(some characteristics/attributes) to tell |
49,321 | How to assess if a model is good in multinomial logistic regression? | Let's take apart your modeling approach to see if we can figure out why a certain model is going to "fit" better.
Multinomial vs ordinal: Multinomial I would bet is almost always going to fit better than an ordinal because it gives you coefficients for every level. It is the most flexible here and has the least restri... | How to assess if a model is good in multinomial logistic regression? | Let's take apart your modeling approach to see if we can figure out why a certain model is going to "fit" better.
Multinomial vs ordinal: Multinomial I would bet is almost always going to fit better | How to assess if a model is good in multinomial logistic regression?
Let's take apart your modeling approach to see if we can figure out why a certain model is going to "fit" better.
Multinomial vs ordinal: Multinomial I would bet is almost always going to fit better than an ordinal because it gives you coefficients f... | How to assess if a model is good in multinomial logistic regression?
Let's take apart your modeling approach to see if we can figure out why a certain model is going to "fit" better.
Multinomial vs ordinal: Multinomial I would bet is almost always going to fit better |
49,322 | Strange results in parallel analysis -- weird output by rstudio but not R-Fiddle | Pay attention to the plot's y-axis label. It says: "Eigen values of original and simulated factors and components" (emphasis mine). Parallel analysis (PA) produces separate sets of eigen values for both factors (factor analysis, FA) and components (principal component analysis, PCA). While FA and PCA seem to be similar... | Strange results in parallel analysis -- weird output by rstudio but not R-Fiddle | Pay attention to the plot's y-axis label. It says: "Eigen values of original and simulated factors and components" (emphasis mine). Parallel analysis (PA) produces separate sets of eigen values for bo | Strange results in parallel analysis -- weird output by rstudio but not R-Fiddle
Pay attention to the plot's y-axis label. It says: "Eigen values of original and simulated factors and components" (emphasis mine). Parallel analysis (PA) produces separate sets of eigen values for both factors (factor analysis, FA) and co... | Strange results in parallel analysis -- weird output by rstudio but not R-Fiddle
Pay attention to the plot's y-axis label. It says: "Eigen values of original and simulated factors and components" (emphasis mine). Parallel analysis (PA) produces separate sets of eigen values for bo |
49,323 | Interpreting standard deviation for PCA | From your input, you should use the "Cumulative Proportion" field as a guide how many principal components to keep. You define the percentage of variance and then you select the column (which is also the number of that principal component) which cumulatively accounts the variance you would like to keep. For 85% and mor... | Interpreting standard deviation for PCA | From your input, you should use the "Cumulative Proportion" field as a guide how many principal components to keep. You define the percentage of variance and then you select the column (which is also | Interpreting standard deviation for PCA
From your input, you should use the "Cumulative Proportion" field as a guide how many principal components to keep. You define the percentage of variance and then you select the column (which is also the number of that principal component) which cumulatively accounts the variance... | Interpreting standard deviation for PCA
From your input, you should use the "Cumulative Proportion" field as a guide how many principal components to keep. You define the percentage of variance and then you select the column (which is also |
49,324 | Interpreting standard deviation for PCA | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
I think we should set up a threshold by multiplying wi... | Interpreting standard deviation for PCA | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| Interpreting standard deviation for PCA
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
I think we sho... | Interpreting standard deviation for PCA
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
|
49,325 | Maximum number of alternatives in a discrete choice model | The main issue with asmprobit is the flexibility it provides which relaxes the independence of irrelevant alternatives (IIA) assumption but it comes at the cost of increased computing power. In this sense you allow the odds of choosing one alternative over some other alternative to depend on the remaining alternative, ... | Maximum number of alternatives in a discrete choice model | The main issue with asmprobit is the flexibility it provides which relaxes the independence of irrelevant alternatives (IIA) assumption but it comes at the cost of increased computing power. In this s | Maximum number of alternatives in a discrete choice model
The main issue with asmprobit is the flexibility it provides which relaxes the independence of irrelevant alternatives (IIA) assumption but it comes at the cost of increased computing power. In this sense you allow the odds of choosing one alternative over some ... | Maximum number of alternatives in a discrete choice model
The main issue with asmprobit is the flexibility it provides which relaxes the independence of irrelevant alternatives (IIA) assumption but it comes at the cost of increased computing power. In this s |
49,326 | How to create a QQ plot of azimuths to test rotational symmetry of a spherical point dataset? | You just want to study the azimuths of a set of spherical points $P_i$ relative to their spherical mean $\bar P$. The most straightforward solution solves the spherical triangles $(N,\bar P, P_i)$ where $N$ is the North Pole.
Let the co-latitudes of the points $P_i$ and $\bar P$ (angles from North) be $a$ and $b$ resp... | How to create a QQ plot of azimuths to test rotational symmetry of a spherical point dataset? | You just want to study the azimuths of a set of spherical points $P_i$ relative to their spherical mean $\bar P$. The most straightforward solution solves the spherical triangles $(N,\bar P, P_i)$ wh | How to create a QQ plot of azimuths to test rotational symmetry of a spherical point dataset?
You just want to study the azimuths of a set of spherical points $P_i$ relative to their spherical mean $\bar P$. The most straightforward solution solves the spherical triangles $(N,\bar P, P_i)$ where $N$ is the North Pole.... | How to create a QQ plot of azimuths to test rotational symmetry of a spherical point dataset?
You just want to study the azimuths of a set of spherical points $P_i$ relative to their spherical mean $\bar P$. The most straightforward solution solves the spherical triangles $(N,\bar P, P_i)$ wh |
49,327 | How to choose a regression tree (base learner) at each iteration of Gradient Tree Boosting? | The regions are not split based only on the data features.
In each iteration of gradient boosting, you fit a regression tree to the residuals of the loss function at the current prediction, $\frac{\partial L(y_i, F(x_i))}{\partial F(x_i)},$ where $F$ is the function you have learned so far. Since these residuals change... | How to choose a regression tree (base learner) at each iteration of Gradient Tree Boosting? | The regions are not split based only on the data features.
In each iteration of gradient boosting, you fit a regression tree to the residuals of the loss function at the current prediction, $\frac{\pa | How to choose a regression tree (base learner) at each iteration of Gradient Tree Boosting?
The regions are not split based only on the data features.
In each iteration of gradient boosting, you fit a regression tree to the residuals of the loss function at the current prediction, $\frac{\partial L(y_i, F(x_i))}{\parti... | How to choose a regression tree (base learner) at each iteration of Gradient Tree Boosting?
The regions are not split based only on the data features.
In each iteration of gradient boosting, you fit a regression tree to the residuals of the loss function at the current prediction, $\frac{\pa |
49,328 | using caret and glmnet for variable selection | If you check the lambdas and your best lambda obtained from caret, you will see that it is not present in the model:
lassoFit1$bestTune$lambda
[1] 0.01545996
lassoFit1$bestTune$lambda %in% lassoFit1$finalModel$lambda
[1] FALSE
If you do:
coef(lassoFit1$finalModel,lassoFit1$bestTune$lambda)
8 x 1 sparse Matrix of class... | using caret and glmnet for variable selection | If you check the lambdas and your best lambda obtained from caret, you will see that it is not present in the model:
lassoFit1$bestTune$lambda
[1] 0.01545996
lassoFit1$bestTune$lambda %in% lassoFit1$f | using caret and glmnet for variable selection
If you check the lambdas and your best lambda obtained from caret, you will see that it is not present in the model:
lassoFit1$bestTune$lambda
[1] 0.01545996
lassoFit1$bestTune$lambda %in% lassoFit1$finalModel$lambda
[1] FALSE
If you do:
coef(lassoFit1$finalModel,lassoFit1... | using caret and glmnet for variable selection
If you check the lambdas and your best lambda obtained from caret, you will see that it is not present in the model:
lassoFit1$bestTune$lambda
[1] 0.01545996
lassoFit1$bestTune$lambda %in% lassoFit1$f |
49,329 | Correcting naΓ―ve Sensitivity and Specificity for classifier tested against imperfect gold standard | Hugues,
This should be relatively straightforward given one very crucial assumption, that we will get to. Let's establish some notation. Let's define $X$ to be the random variable obtained by randomly selecting a data point from your set and classifying it using your classifier. $Y$ as the random variable obtained b... | Correcting naΓ―ve Sensitivity and Specificity for classifier tested against imperfect gold standard | Hugues,
This should be relatively straightforward given one very crucial assumption, that we will get to. Let's establish some notation. Let's define $X$ to be the random variable obtained by random | Correcting naΓ―ve Sensitivity and Specificity for classifier tested against imperfect gold standard
Hugues,
This should be relatively straightforward given one very crucial assumption, that we will get to. Let's establish some notation. Let's define $X$ to be the random variable obtained by randomly selecting a data p... | Correcting naΓ―ve Sensitivity and Specificity for classifier tested against imperfect gold standard
Hugues,
This should be relatively straightforward given one very crucial assumption, that we will get to. Let's establish some notation. Let's define $X$ to be the random variable obtained by random |
49,330 | Correct glmer distribution family and link for a continuous zero-inflated data set | Assuming that you are describing conditional and not marginal distributions (i.e., if your response variable is y then hist(mydata$y) will not typically give you what you want; you should be concerned with the distribution around the expected values):
Changing the link function won't help you; it determines the depend... | Correct glmer distribution family and link for a continuous zero-inflated data set | Assuming that you are describing conditional and not marginal distributions (i.e., if your response variable is y then hist(mydata$y) will not typically give you what you want; you should be concerned | Correct glmer distribution family and link for a continuous zero-inflated data set
Assuming that you are describing conditional and not marginal distributions (i.e., if your response variable is y then hist(mydata$y) will not typically give you what you want; you should be concerned with the distribution around the exp... | Correct glmer distribution family and link for a continuous zero-inflated data set
Assuming that you are describing conditional and not marginal distributions (i.e., if your response variable is y then hist(mydata$y) will not typically give you what you want; you should be concerned |
49,331 | Detect periodic events within data | I would not recommend turning the data into 0/1 . I have had a lot of experience with daily bank payment data , ATM access , deposits etc.. If the payments are systematic/regular to a particular day of the month then its is fairly straightforward to identify these patterns. If however the data is non-systematic then I... | Detect periodic events within data | I would not recommend turning the data into 0/1 . I have had a lot of experience with daily bank payment data , ATM access , deposits etc.. If the payments are systematic/regular to a particular day | Detect periodic events within data
I would not recommend turning the data into 0/1 . I have had a lot of experience with daily bank payment data , ATM access , deposits etc.. If the payments are systematic/regular to a particular day of the month then its is fairly straightforward to identify these patterns. If howeve... | Detect periodic events within data
I would not recommend turning the data into 0/1 . I have had a lot of experience with daily bank payment data , ATM access , deposits etc.. If the payments are systematic/regular to a particular day |
49,332 | Measuring length of intervention effect | Given that you seem to have a panel of individuals who you follow over time of which some are treated and others are not you could run a difference in difference analysis. You could run a regression like
$$y_{it} = \beta_1 (\text{treat}_{i}) + \beta_2 (\text{intervention}_t) + \beta_3 (\text{treat}_{i} \cdot \text{inte... | Measuring length of intervention effect | Given that you seem to have a panel of individuals who you follow over time of which some are treated and others are not you could run a difference in difference analysis. You could run a regression l | Measuring length of intervention effect
Given that you seem to have a panel of individuals who you follow over time of which some are treated and others are not you could run a difference in difference analysis. You could run a regression like
$$y_{it} = \beta_1 (\text{treat}_{i}) + \beta_2 (\text{intervention}_t) + \b... | Measuring length of intervention effect
Given that you seem to have a panel of individuals who you follow over time of which some are treated and others are not you could run a difference in difference analysis. You could run a regression l |
49,333 | Measuring length of intervention effect | Intervention Detection http://www.unc.edu/~jbhill/tsay.pdf and elsewhere can be employed with or without a user-suggested intervention variable. In either case one needs to treat any auto-projective process that might be present i.e. the ARIMA structure. Identifying both the ARIMA structure and the response to any user... | Measuring length of intervention effect | Intervention Detection http://www.unc.edu/~jbhill/tsay.pdf and elsewhere can be employed with or without a user-suggested intervention variable. In either case one needs to treat any auto-projective p | Measuring length of intervention effect
Intervention Detection http://www.unc.edu/~jbhill/tsay.pdf and elsewhere can be employed with or without a user-suggested intervention variable. In either case one needs to treat any auto-projective process that might be present i.e. the ARIMA structure. Identifying both the ARIM... | Measuring length of intervention effect
Intervention Detection http://www.unc.edu/~jbhill/tsay.pdf and elsewhere can be employed with or without a user-suggested intervention variable. In either case one needs to treat any auto-projective p |
49,334 | Comparing many performance curves: request for data visualization tips | As was the recommendation in Color and line thickness recommendations for line plots, small multiples are a common solution for plots that have problems with overplotting. Here is an example with the 5 curves you provided.
It is a lot of information, but it is pretty easy to see that A, C & E are all decreasing. C & E... | Comparing many performance curves: request for data visualization tips | As was the recommendation in Color and line thickness recommendations for line plots, small multiples are a common solution for plots that have problems with overplotting. Here is an example with the | Comparing many performance curves: request for data visualization tips
As was the recommendation in Color and line thickness recommendations for line plots, small multiples are a common solution for plots that have problems with overplotting. Here is an example with the 5 curves you provided.
It is a lot of informatio... | Comparing many performance curves: request for data visualization tips
As was the recommendation in Color and line thickness recommendations for line plots, small multiples are a common solution for plots that have problems with overplotting. Here is an example with the |
49,335 | Clustering in Instrumental Variables Regression? | The relevant reference would be Shore-Sheppard (1996) "The Precision of Instrumental Variables Estimates With Grouped Data". You can directly calculate by how much the standard errors in 2SLS are over-estimated by using the Moulton factor
$$\frac{Var(\widehat{\beta}^c)}{Var(\widehat{\beta}^{ols})} = 1 + \left(\frac{Var... | Clustering in Instrumental Variables Regression? | The relevant reference would be Shore-Sheppard (1996) "The Precision of Instrumental Variables Estimates With Grouped Data". You can directly calculate by how much the standard errors in 2SLS are over | Clustering in Instrumental Variables Regression?
The relevant reference would be Shore-Sheppard (1996) "The Precision of Instrumental Variables Estimates With Grouped Data". You can directly calculate by how much the standard errors in 2SLS are over-estimated by using the Moulton factor
$$\frac{Var(\widehat{\beta}^c)}{... | Clustering in Instrumental Variables Regression?
The relevant reference would be Shore-Sheppard (1996) "The Precision of Instrumental Variables Estimates With Grouped Data". You can directly calculate by how much the standard errors in 2SLS are over |
49,336 | Clustering in Instrumental Variables Regression? | I did some background research and found this here which characterizes the clustering issue in IV regression. Naturally, the clustering of errors will only appear in the covariance matrix of the structural errors. Therefore it is non-sensical to write down clustered first-stage errors. Hence
\begin{eqnarray}
Y_{i,g} = ... | Clustering in Instrumental Variables Regression? | I did some background research and found this here which characterizes the clustering issue in IV regression. Naturally, the clustering of errors will only appear in the covariance matrix of the struc | Clustering in Instrumental Variables Regression?
I did some background research and found this here which characterizes the clustering issue in IV regression. Naturally, the clustering of errors will only appear in the covariance matrix of the structural errors. Therefore it is non-sensical to write down clustered firs... | Clustering in Instrumental Variables Regression?
I did some background research and found this here which characterizes the clustering issue in IV regression. Naturally, the clustering of errors will only appear in the covariance matrix of the struc |
49,337 | Clustering in Instrumental Variables Regression? | In the standard instrumental variable case with 2-SLS, you indeed not do need to take into account the errors in the first stage as you say.
However, if you were confronted with weak instruments, or want some more fancy endogeneity tests etc, then the usual weak instruments asymptotic need to be adjusted for the prese... | Clustering in Instrumental Variables Regression? | In the standard instrumental variable case with 2-SLS, you indeed not do need to take into account the errors in the first stage as you say.
However, if you were confronted with weak instruments, or | Clustering in Instrumental Variables Regression?
In the standard instrumental variable case with 2-SLS, you indeed not do need to take into account the errors in the first stage as you say.
However, if you were confronted with weak instruments, or want some more fancy endogeneity tests etc, then the usual weak instrum... | Clustering in Instrumental Variables Regression?
In the standard instrumental variable case with 2-SLS, you indeed not do need to take into account the errors in the first stage as you say.
However, if you were confronted with weak instruments, or |
49,338 | How to interpret this PCA biplot? | Your interpretation is mostly correct. The first PC accounts for most of the variance, and the first eigenvector (principal axis) has all positive coordinates. It probably means that all variables are positively correlated between each other, and the first PC represents this "common factor". The second PC (looks like i... | How to interpret this PCA biplot? | Your interpretation is mostly correct. The first PC accounts for most of the variance, and the first eigenvector (principal axis) has all positive coordinates. It probably means that all variables are | How to interpret this PCA biplot?
Your interpretation is mostly correct. The first PC accounts for most of the variance, and the first eigenvector (principal axis) has all positive coordinates. It probably means that all variables are positively correlated between each other, and the first PC represents this "common fa... | How to interpret this PCA biplot?
Your interpretation is mostly correct. The first PC accounts for most of the variance, and the first eigenvector (principal axis) has all positive coordinates. It probably means that all variables are |
49,339 | Confidence intervals of fitted Weibull survival function? | I just worked this out as I had the same question myself. My answer is largely thanks to Mara Tableman (p65 of Survival Analysis Using S/R).
Let's say you have the survival function $S(t)$ using the Weibull distribution, the lower and upper bounds of the 95% confidence interval can be calculated for every $t$ of the f... | Confidence intervals of fitted Weibull survival function? | I just worked this out as I had the same question myself. My answer is largely thanks to Mara Tableman (p65 of Survival Analysis Using S/R).
Let's say you have the survival function $S(t)$ using the | Confidence intervals of fitted Weibull survival function?
I just worked this out as I had the same question myself. My answer is largely thanks to Mara Tableman (p65 of Survival Analysis Using S/R).
Let's say you have the survival function $S(t)$ using the Weibull distribution, the lower and upper bounds of the 95% co... | Confidence intervals of fitted Weibull survival function?
I just worked this out as I had the same question myself. My answer is largely thanks to Mara Tableman (p65 of Survival Analysis Using S/R).
Let's say you have the survival function $S(t)$ using the |
49,340 | Confidence intervals of fitted Weibull survival function? | I found that @tystanza's answer (Tableman's equations) give a too conservative confidence interval -- that is, they are too wide. When I did a Monte Carlo simulation to estimate the coverage of the Tableman confidence interval, I found that nominal 95% CI's gave 100% coverage.
I found another option from the package ... | Confidence intervals of fitted Weibull survival function? | I found that @tystanza's answer (Tableman's equations) give a too conservative confidence interval -- that is, they are too wide. When I did a Monte Carlo simulation to estimate the coverage of the Ta | Confidence intervals of fitted Weibull survival function?
I found that @tystanza's answer (Tableman's equations) give a too conservative confidence interval -- that is, they are too wide. When I did a Monte Carlo simulation to estimate the coverage of the Tableman confidence interval, I found that nominal 95% CI's gave... | Confidence intervals of fitted Weibull survival function?
I found that @tystanza's answer (Tableman's equations) give a too conservative confidence interval -- that is, they are too wide. When I did a Monte Carlo simulation to estimate the coverage of the Ta |
49,341 | Tsallis and RΓ©nyi Normalized Entropy | Tsallis and RΓ©nyi entropy is the same thing, up to some rescaling. All of them are functions of $\sum_i p_i^\alpha$, with the special case of $\alpha\to1$ giving Shannon entropy.
Look at Tom Leinster's "Entropy, Diversity and Cardinality (Part 2)", especially at the table comparing these properties.
In short:
RΓ©nyi en... | Tsallis and RΓ©nyi Normalized Entropy | Tsallis and RΓ©nyi entropy is the same thing, up to some rescaling. All of them are functions of $\sum_i p_i^\alpha$, with the special case of $\alpha\to1$ giving Shannon entropy.
Look at Tom Leinster' | Tsallis and RΓ©nyi Normalized Entropy
Tsallis and RΓ©nyi entropy is the same thing, up to some rescaling. All of them are functions of $\sum_i p_i^\alpha$, with the special case of $\alpha\to1$ giving Shannon entropy.
Look at Tom Leinster's "Entropy, Diversity and Cardinality (Part 2)", especially at the table comparing ... | Tsallis and RΓ©nyi Normalized Entropy
Tsallis and RΓ©nyi entropy is the same thing, up to some rescaling. All of them are functions of $\sum_i p_i^\alpha$, with the special case of $\alpha\to1$ giving Shannon entropy.
Look at Tom Leinster' |
49,342 | Tsallis and RΓ©nyi Normalized Entropy | The answer provided by Piotr is mostly correct, but there is a tiny mistake.
The maximum Tsallis entropy value is actually given by the expression: $(1βN^{1β\alpha})/(\alpha-1)$.
The denominator order is reversed.
This is because this expression is obtained when dealing with a uniform probability distribution $ P = \{\... | Tsallis and RΓ©nyi Normalized Entropy | The answer provided by Piotr is mostly correct, but there is a tiny mistake.
The maximum Tsallis entropy value is actually given by the expression: $(1βN^{1β\alpha})/(\alpha-1)$.
The denominator order | Tsallis and RΓ©nyi Normalized Entropy
The answer provided by Piotr is mostly correct, but there is a tiny mistake.
The maximum Tsallis entropy value is actually given by the expression: $(1βN^{1β\alpha})/(\alpha-1)$.
The denominator order is reversed.
This is because this expression is obtained when dealing with a unifo... | Tsallis and RΓ©nyi Normalized Entropy
The answer provided by Piotr is mostly correct, but there is a tiny mistake.
The maximum Tsallis entropy value is actually given by the expression: $(1βN^{1β\alpha})/(\alpha-1)$.
The denominator order |
49,343 | Are cross-validated prediction errors i.i.d? | I think you need to be clear what distribution you need to represent. This differers according to what the cross validation is meant for.
In the case that the cross validation is meant to measure (approximate) the performance of the model obtained from this particular training set, the corresponding distribution wou... | Are cross-validated prediction errors i.i.d? | I think you need to be clear what distribution you need to represent. This differers according to what the cross validation is meant for.
In the case that the cross validation is meant to measure ( | Are cross-validated prediction errors i.i.d?
I think you need to be clear what distribution you need to represent. This differers according to what the cross validation is meant for.
In the case that the cross validation is meant to measure (approximate) the performance of the model obtained from this particular tra... | Are cross-validated prediction errors i.i.d?
I think you need to be clear what distribution you need to represent. This differers according to what the cross validation is meant for.
In the case that the cross validation is meant to measure ( |
49,344 | Are cross-validated prediction errors i.i.d? | They can't be independent. Consider adding one extreme outlier sample, then many of your cross validation folds will be skewed in a correlated way. | Are cross-validated prediction errors i.i.d? | They can't be independent. Consider adding one extreme outlier sample, then many of your cross validation folds will be skewed in a correlated way. | Are cross-validated prediction errors i.i.d?
They can't be independent. Consider adding one extreme outlier sample, then many of your cross validation folds will be skewed in a correlated way. | Are cross-validated prediction errors i.i.d?
They can't be independent. Consider adding one extreme outlier sample, then many of your cross validation folds will be skewed in a correlated way. |
49,345 | Is the objective to beat a random classifier when the data set is skewed using PR curves? | A random classifiers randomly selects a subset of the total data and labels it as positive. The size of said subset is associated with the recall of the random classifier. Since predictions are done entirely at random, the expected precision of such a labeling is equal to the fraction of positives in the total data set... | Is the objective to beat a random classifier when the data set is skewed using PR curves? | A random classifiers randomly selects a subset of the total data and labels it as positive. The size of said subset is associated with the recall of the random classifier. Since predictions are done e | Is the objective to beat a random classifier when the data set is skewed using PR curves?
A random classifiers randomly selects a subset of the total data and labels it as positive. The size of said subset is associated with the recall of the random classifier. Since predictions are done entirely at random, the expecte... | Is the objective to beat a random classifier when the data set is skewed using PR curves?
A random classifiers randomly selects a subset of the total data and labels it as positive. The size of said subset is associated with the recall of the random classifier. Since predictions are done e |
49,346 | Asymmetric measure of non-linear dependence/correlation? | The $R^2$ of a multivariate regression model is such an asymmetric measure. The regression model of $Y$ on $X$ leads to a different $R^2$ than the regression model of $X$ on $Y$. This is because the value is computed using the proportion of vertical distance from the mean accounted for by the line of best fit using the... | Asymmetric measure of non-linear dependence/correlation? | The $R^2$ of a multivariate regression model is such an asymmetric measure. The regression model of $Y$ on $X$ leads to a different $R^2$ than the regression model of $X$ on $Y$. This is because the v | Asymmetric measure of non-linear dependence/correlation?
The $R^2$ of a multivariate regression model is such an asymmetric measure. The regression model of $Y$ on $X$ leads to a different $R^2$ than the regression model of $X$ on $Y$. This is because the value is computed using the proportion of vertical distance from... | Asymmetric measure of non-linear dependence/correlation?
The $R^2$ of a multivariate regression model is such an asymmetric measure. The regression model of $Y$ on $X$ leads to a different $R^2$ than the regression model of $X$ on $Y$. This is because the v |
49,347 | Asymmetric measure of non-linear dependence/correlation? | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
It is possible to define nonsymmetric dependence measu... | Asymmetric measure of non-linear dependence/correlation? | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| Asymmetric measure of non-linear dependence/correlation?
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
... | Asymmetric measure of non-linear dependence/correlation?
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
|
49,348 | How does one create a confidence interval for the ratio of the means of two non-normal bounded distributions? | What it seems like you're really trying to do is:
Prove that there is a linear relationship between $X_i$ and $Y_i$
Determine the constant proportion relating the two with some degree of certainty.
I think the separation is important because if your assumption "$X_i/Y_i$ and $Y_i$ are not independent" was
meant to m... | How does one create a confidence interval for the ratio of the means of two non-normal bounded distr | What it seems like you're really trying to do is:
Prove that there is a linear relationship between $X_i$ and $Y_i$
Determine the constant proportion relating the two with some degree of certainty.
| How does one create a confidence interval for the ratio of the means of two non-normal bounded distributions?
What it seems like you're really trying to do is:
Prove that there is a linear relationship between $X_i$ and $Y_i$
Determine the constant proportion relating the two with some degree of certainty.
I think th... | How does one create a confidence interval for the ratio of the means of two non-normal bounded distr
What it seems like you're really trying to do is:
Prove that there is a linear relationship between $X_i$ and $Y_i$
Determine the constant proportion relating the two with some degree of certainty.
|
49,349 | How does one create a confidence interval for the ratio of the means of two non-normal bounded distributions? | For your application, where what you care about is ultimately the probability that an individual has property P, it seems that one of two simpler types of analysis should suffice. These analyses do not depend on distributions of the $X_i$ or $Y_i$, but simply whether an individual has property P and, possibly, to which... | How does one create a confidence interval for the ratio of the means of two non-normal bounded distr | For your application, where what you care about is ultimately the probability that an individual has property P, it seems that one of two simpler types of analysis should suffice. These analyses do no | How does one create a confidence interval for the ratio of the means of two non-normal bounded distributions?
For your application, where what you care about is ultimately the probability that an individual has property P, it seems that one of two simpler types of analysis should suffice. These analyses do not depend o... | How does one create a confidence interval for the ratio of the means of two non-normal bounded distr
For your application, where what you care about is ultimately the probability that an individual has property P, it seems that one of two simpler types of analysis should suffice. These analyses do no |
49,350 | What to do about very unstable mixed-effects models | Actually, in addition to the random intercepts, you are adding 4 random effects (assuming that none of the Var1, Var2, t_before, and t_after variables are factors) that are all allowed to be correlated. That's $5$ variance components plus $(5\times4)/2 = 10$ correlations, so $15$ var-cov parameters for the random effec... | What to do about very unstable mixed-effects models | Actually, in addition to the random intercepts, you are adding 4 random effects (assuming that none of the Var1, Var2, t_before, and t_after variables are factors) that are all allowed to be correlate | What to do about very unstable mixed-effects models
Actually, in addition to the random intercepts, you are adding 4 random effects (assuming that none of the Var1, Var2, t_before, and t_after variables are factors) that are all allowed to be correlated. That's $5$ variance components plus $(5\times4)/2 = 10$ correlati... | What to do about very unstable mixed-effects models
Actually, in addition to the random intercepts, you are adding 4 random effects (assuming that none of the Var1, Var2, t_before, and t_after variables are factors) that are all allowed to be correlate |
49,351 | Confidence interval for polynomial linear regression | Polynomial regression is in effect multiple linear regression: consider $X_1=X$ and $X_2=X^2$ -- then $E(Y) = \beta_1 X + \beta_2 X^2$ is the same as $E(Y) = \beta_1 X_1 + \beta_2 X_2$.
As such, methods for constructing confidence intervals for parameters (and for the mean in multiple regression) carry over directly to... | Confidence interval for polynomial linear regression | Polynomial regression is in effect multiple linear regression: consider $X_1=X$ and $X_2=X^2$ -- then $E(Y) = \beta_1 X + \beta_2 X^2$ is the same as $E(Y) = \beta_1 X_1 + \beta_2 X_2$.
As such, metho | Confidence interval for polynomial linear regression
Polynomial regression is in effect multiple linear regression: consider $X_1=X$ and $X_2=X^2$ -- then $E(Y) = \beta_1 X + \beta_2 X^2$ is the same as $E(Y) = \beta_1 X_1 + \beta_2 X_2$.
As such, methods for constructing confidence intervals for parameters (and for th... | Confidence interval for polynomial linear regression
Polynomial regression is in effect multiple linear regression: consider $X_1=X$ and $X_2=X^2$ -- then $E(Y) = \beta_1 X + \beta_2 X^2$ is the same as $E(Y) = \beta_1 X_1 + \beta_2 X_2$.
As such, metho |
49,352 | Confidence interval for polynomial linear regression | It is somewhat of a lengthy procedure to verify that the linear model obeys a t-distribution. To do so for the quadratic would be tedious. I do not think the above suggestion that one can simply substitute for the quadratic term is sound. There are methods involving Taylor expansions to make the conversion. See the fol... | Confidence interval for polynomial linear regression | It is somewhat of a lengthy procedure to verify that the linear model obeys a t-distribution. To do so for the quadratic would be tedious. I do not think the above suggestion that one can simply subst | Confidence interval for polynomial linear regression
It is somewhat of a lengthy procedure to verify that the linear model obeys a t-distribution. To do so for the quadratic would be tedious. I do not think the above suggestion that one can simply substitute for the quadratic term is sound. There are methods involving ... | Confidence interval for polynomial linear regression
It is somewhat of a lengthy procedure to verify that the linear model obeys a t-distribution. To do so for the quadratic would be tedious. I do not think the above suggestion that one can simply subst |
49,353 | Does the birth date of professional boxers matter? Prove/disprove what an astrologer might predict | If your hypothesis was formulated a priori, then the data are quite strongly significant.
Your null hypothesis is that astrology does not predict anything. This would mean that the probability of a boxer to be born under an "earth" sign is $0.25$, and the same is true for the "ethereal" signs. I assume that you selecte... | Does the birth date of professional boxers matter? Prove/disprove what an astrologer might predict | If your hypothesis was formulated a priori, then the data are quite strongly significant.
Your null hypothesis is that astrology does not predict anything. This would mean that the probability of a bo | Does the birth date of professional boxers matter? Prove/disprove what an astrologer might predict
If your hypothesis was formulated a priori, then the data are quite strongly significant.
Your null hypothesis is that astrology does not predict anything. This would mean that the probability of a boxer to be born under ... | Does the birth date of professional boxers matter? Prove/disprove what an astrologer might predict
If your hypothesis was formulated a priori, then the data are quite strongly significant.
Your null hypothesis is that astrology does not predict anything. This would mean that the probability of a bo |
49,354 | Expected survival time from log-logistic survival model in R from survreg | Seems that this was more of a coding question and might have gotten a more prompt coding response on StackOverflow, but since no close votes have been offered I put in a belated CV response. Most R regression functions have an associated predict method and survival::survreg is no exception. You need to assign the outpu... | Expected survival time from log-logistic survival model in R from survreg | Seems that this was more of a coding question and might have gotten a more prompt coding response on StackOverflow, but since no close votes have been offered I put in a belated CV response. Most R re | Expected survival time from log-logistic survival model in R from survreg
Seems that this was more of a coding question and might have gotten a more prompt coding response on StackOverflow, but since no close votes have been offered I put in a belated CV response. Most R regression functions have an associated predict ... | Expected survival time from log-logistic survival model in R from survreg
Seems that this was more of a coding question and might have gotten a more prompt coding response on StackOverflow, but since no close votes have been offered I put in a belated CV response. Most R re |
49,355 | How to prove that the permutation of the points are the minimal sufficient statistics for Cauchy distribution? | Hint: Apply the ratio test (e.g. Theorem 6.2.13 in Casella and Berger's "Statistical Inference, Second Edition"), and consider the roots of the polynomial that is the denominator of the joint Cauchy distribution. | How to prove that the permutation of the points are the minimal sufficient statistics for Cauchy dis | Hint: Apply the ratio test (e.g. Theorem 6.2.13 in Casella and Berger's "Statistical Inference, Second Edition"), and consider the roots of the polynomial that is the denominator of the joint Cauchy d | How to prove that the permutation of the points are the minimal sufficient statistics for Cauchy distribution?
Hint: Apply the ratio test (e.g. Theorem 6.2.13 in Casella and Berger's "Statistical Inference, Second Edition"), and consider the roots of the polynomial that is the denominator of the joint Cauchy distributi... | How to prove that the permutation of the points are the minimal sufficient statistics for Cauchy dis
Hint: Apply the ratio test (e.g. Theorem 6.2.13 in Casella and Berger's "Statistical Inference, Second Edition"), and consider the roots of the polynomial that is the denominator of the joint Cauchy d |
49,356 | Error bars on log of big numbers | The problem isn't as profound as it may appear. Because
$$\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)} = e^Y\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)-Y} $$
for
$$Y = \max_i\{\phi(X_i)\},$$
an algebraically equivalent expression is
$$\mu = Y -\log n + \log \sum_{i=1}^{n} e^{\phi(X_i)-Y}. $$
In this one there will be no diff... | Error bars on log of big numbers | The problem isn't as profound as it may appear. Because
$$\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)} = e^Y\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)-Y} $$
for
$$Y = \max_i\{\phi(X_i)\},$$
an algebraically | Error bars on log of big numbers
The problem isn't as profound as it may appear. Because
$$\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)} = e^Y\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)-Y} $$
for
$$Y = \max_i\{\phi(X_i)\},$$
an algebraically equivalent expression is
$$\mu = Y -\log n + \log \sum_{i=1}^{n} e^{\phi(X_i)-Y}. $$
... | Error bars on log of big numbers
The problem isn't as profound as it may appear. Because
$$\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)} = e^Y\frac{1}{n} \sum_{i=1}^{n} e^{\phi(X_i)-Y} $$
for
$$Y = \max_i\{\phi(X_i)\},$$
an algebraically |
49,357 | Propensity Score can be used as a covariate in regression? | This would be the standard propensity score estimator. For a binary treatment the conditional independence assumption (CIA) states that
$$
\newcommand\independent{\protect\mathpalette{\protect\independenT}{\perp}}
\def\independent#1#2{\mathrel{\rlap{$#1#2$}\mkern2mu{#1#2}}}
T_i\perp\hspace{-0.28cm}\perp (Y_{i0}, Y_{i1}... | Propensity Score can be used as a covariate in regression? | This would be the standard propensity score estimator. For a binary treatment the conditional independence assumption (CIA) states that
$$
\newcommand\independent{\protect\mathpalette{\protect\indepen | Propensity Score can be used as a covariate in regression?
This would be the standard propensity score estimator. For a binary treatment the conditional independence assumption (CIA) states that
$$
\newcommand\independent{\protect\mathpalette{\protect\independenT}{\perp}}
\def\independent#1#2{\mathrel{\rlap{$#1#2$}\mke... | Propensity Score can be used as a covariate in regression?
This would be the standard propensity score estimator. For a binary treatment the conditional independence assumption (CIA) states that
$$
\newcommand\independent{\protect\mathpalette{\protect\indepen |
49,358 | How to implement reduced-rank regression in R? | A set of S functions for least-squares reduced-rank can be found in the StatLib archive.
See the file rrr.s and this paper:
Splus function for reduced-rank regression and softly shrunk
reduced-rank regression. Submitted by Magne Aldrin
(magne.aldrin@nr.no). [19/Apr/99][8/Mar/00] (14k) | How to implement reduced-rank regression in R? | A set of S functions for least-squares reduced-rank can be found in the StatLib archive.
See the file rrr.s and this paper:
Splus function for reduced-rank regression and softly shrunk
reduced-rank | How to implement reduced-rank regression in R?
A set of S functions for least-squares reduced-rank can be found in the StatLib archive.
See the file rrr.s and this paper:
Splus function for reduced-rank regression and softly shrunk
reduced-rank regression. Submitted by Magne Aldrin
(magne.aldrin@nr.no). [19/Apr/99... | How to implement reduced-rank regression in R?
A set of S functions for least-squares reduced-rank can be found in the StatLib archive.
See the file rrr.s and this paper:
Splus function for reduced-rank regression and softly shrunk
reduced-rank |
49,359 | How to implement reduced-rank regression in R? | There are now R packages for reduced-rank regression: rrpack, rrr. | How to implement reduced-rank regression in R? | There are now R packages for reduced-rank regression: rrpack, rrr. | How to implement reduced-rank regression in R?
There are now R packages for reduced-rank regression: rrpack, rrr. | How to implement reduced-rank regression in R?
There are now R packages for reduced-rank regression: rrpack, rrr. |
49,360 | Deriving the maximum likelihood for the parameters in linear regression | $$
0 = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T - \textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) \tag 3
$$
$$
\textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T \tag a
$$
Recall the Delta Matrix is defined as: $$ \Phi =\beg... | Deriving the maximum likelihood for the parameters in linear regression | $$
0 = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T - \textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) \tag 3
$$
$$
\textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf | Deriving the maximum likelihood for the parameters in linear regression
$$
0 = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T - \textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) \tag 3
$$
$$
\textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) = \sum_{n=1}^N t_n \phi(\textbf{... | Deriving the maximum likelihood for the parameters in linear regression
$$
0 = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T - \textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) \tag 3
$$
$$
\textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf |
49,361 | Unclear area in Convolutional Neural Net | I've stumbled upon this before, and it is is generally poorly explained. It's best to think of images as three dimensional, with a width, a height and a number of channels $w \times h \times c$. An input image for instance might have three channels, one for each color.
The next layer might have 50 different filters, so... | Unclear area in Convolutional Neural Net | I've stumbled upon this before, and it is is generally poorly explained. It's best to think of images as three dimensional, with a width, a height and a number of channels $w \times h \times c$. An in | Unclear area in Convolutional Neural Net
I've stumbled upon this before, and it is is generally poorly explained. It's best to think of images as three dimensional, with a width, a height and a number of channels $w \times h \times c$. An input image for instance might have three channels, one for each color.
The next ... | Unclear area in Convolutional Neural Net
I've stumbled upon this before, and it is is generally poorly explained. It's best to think of images as three dimensional, with a width, a height and a number of channels $w \times h \times c$. An in |
49,362 | Why is my R density plot a bell curve when all datapoints are 0? | You should explain what the intuition is that you have that the behavior runs counter to - it would make it easier to focus the explanation to address that.
A kernel density estimate is the convolution of the sample probability function ($n$ point masses of size $\frac{1}{n}$) and the kernel function (itself, by defaul... | Why is my R density plot a bell curve when all datapoints are 0? | You should explain what the intuition is that you have that the behavior runs counter to - it would make it easier to focus the explanation to address that.
A kernel density estimate is the convolutio | Why is my R density plot a bell curve when all datapoints are 0?
You should explain what the intuition is that you have that the behavior runs counter to - it would make it easier to focus the explanation to address that.
A kernel density estimate is the convolution of the sample probability function ($n$ point masses ... | Why is my R density plot a bell curve when all datapoints are 0?
You should explain what the intuition is that you have that the behavior runs counter to - it would make it easier to focus the explanation to address that.
A kernel density estimate is the convolutio |
49,363 | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable | It is true that in the original papers on co-integration, all variables involved were assumed to be individually $I(1)$, and this is usually the case presented and used. But this is not restrictive. For example, Lutkepohl (1993), defines co-integration as follows:
A K-dimensional process $\mathbf z_t$ is integrated of... | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable | It is true that in the original papers on co-integration, all variables involved were assumed to be individually $I(1)$, and this is usually the case presented and used. But this is not restrictive. F | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable
It is true that in the original papers on co-integration, all variables involved were assumed to be individually $I(1)$, and this is usually the case presented and used. But this is not restrictive. For example, Lutkepohl (1993), defines co-integra... | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable
It is true that in the original papers on co-integration, all variables involved were assumed to be individually $I(1)$, and this is usually the case presented and used. But this is not restrictive. F |
49,364 | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable | +1 for the question! It can be shown that if $y_{t}$
and $x_{1t}$
are cointegrated then the OLS estimator of that equation will be super consistent with a rate of convergance of $T^{-2}
$ compared to the stationary $T^{-1}
$ case. Notice further, that even if you have misspecified stationary terms terms in your r... | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable | +1 for the question! It can be shown that if $y_{t}$
and $x_{1t}$
are cointegrated then the OLS estimator of that equation will be super consistent with a rate of convergance of $T^{-2}
$ compare | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable
+1 for the question! It can be shown that if $y_{t}$
and $x_{1t}$
are cointegrated then the OLS estimator of that equation will be super consistent with a rate of convergance of $T^{-2}
$ compared to the stationary $T^{-1}
$ case. Notice furt... | Modeling an I(1) process with a cointegrating I(1) and an I(0) variable
+1 for the question! It can be shown that if $y_{t}$
and $x_{1t}$
are cointegrated then the OLS estimator of that equation will be super consistent with a rate of convergance of $T^{-2}
$ compare |
49,365 | Why are there two forms for the Mann-Whitney U test statistic? | There are actually more than two forms of the Mann-Whitney-Wilcoxon test.
Given no ties (which I will assume throughout), the two forms you have there correspond to
(i) the number of times an observation in sample 1 exceeds an observation from sample 2, and
(ii) the number of times an observation in sample 2 exceeds a... | Why are there two forms for the Mann-Whitney U test statistic? | There are actually more than two forms of the Mann-Whitney-Wilcoxon test.
Given no ties (which I will assume throughout), the two forms you have there correspond to
(i) the number of times an observa | Why are there two forms for the Mann-Whitney U test statistic?
There are actually more than two forms of the Mann-Whitney-Wilcoxon test.
Given no ties (which I will assume throughout), the two forms you have there correspond to
(i) the number of times an observation in sample 1 exceeds an observation from sample 2, an... | Why are there two forms for the Mann-Whitney U test statistic?
There are actually more than two forms of the Mann-Whitney-Wilcoxon test.
Given no ties (which I will assume throughout), the two forms you have there correspond to
(i) the number of times an observa |
49,366 | Why AUC-PR increases when the number of positives increase? | Remember that PR curves visualize a model's performance over the entire operating range, not just where its classification threshold happens to be. Your reasoning in the final paragraph seems to be based on the model's classification of test instances, rather than their ranking.
PR curves are not computed based on pred... | Why AUC-PR increases when the number of positives increase? | Remember that PR curves visualize a model's performance over the entire operating range, not just where its classification threshold happens to be. Your reasoning in the final paragraph seems to be ba | Why AUC-PR increases when the number of positives increase?
Remember that PR curves visualize a model's performance over the entire operating range, not just where its classification threshold happens to be. Your reasoning in the final paragraph seems to be based on the model's classification of test instances, rather ... | Why AUC-PR increases when the number of positives increase?
Remember that PR curves visualize a model's performance over the entire operating range, not just where its classification threshold happens to be. Your reasoning in the final paragraph seems to be ba |
49,367 | Why AUC-PR increases when the number of positives increase? | precision
$$ prec = \frac{TP}{TP+FP} $$
and in case when we add positive values prec is not deceasing (so we find at least previous TP number of ones. When we find them more, the prec is increasing, like:
$$ \frac{TP+1}{TP+1+FP} \geqslant \frac{TP}{TP+FP} $$ (is strictly greater if FP > 0)
FP is not changing because we... | Why AUC-PR increases when the number of positives increase? | precision
$$ prec = \frac{TP}{TP+FP} $$
and in case when we add positive values prec is not deceasing (so we find at least previous TP number of ones. When we find them more, the prec is increasing, l | Why AUC-PR increases when the number of positives increase?
precision
$$ prec = \frac{TP}{TP+FP} $$
and in case when we add positive values prec is not deceasing (so we find at least previous TP number of ones. When we find them more, the prec is increasing, like:
$$ \frac{TP+1}{TP+1+FP} \geqslant \frac{TP}{TP+FP} $$ (... | Why AUC-PR increases when the number of positives increase?
precision
$$ prec = \frac{TP}{TP+FP} $$
and in case when we add positive values prec is not deceasing (so we find at least previous TP number of ones. When we find them more, the prec is increasing, l |
49,368 | Weighted cases in a cluster analysis for cases in SPSS | Using K-means after hierarchical clustering or hierarchical clustering after K-means may be sometimes a sound trick on its own - not because of weighting.
Frequency weighting of objects when clustering objects
Now about weighting. To do hierarchical cluster analysis of cases with frequency weights attached to the cases... | Weighted cases in a cluster analysis for cases in SPSS | Using K-means after hierarchical clustering or hierarchical clustering after K-means may be sometimes a sound trick on its own - not because of weighting.
Frequency weighting of objects when clusterin | Weighted cases in a cluster analysis for cases in SPSS
Using K-means after hierarchical clustering or hierarchical clustering after K-means may be sometimes a sound trick on its own - not because of weighting.
Frequency weighting of objects when clustering objects
Now about weighting. To do hierarchical cluster analysi... | Weighted cases in a cluster analysis for cases in SPSS
Using K-means after hierarchical clustering or hierarchical clustering after K-means may be sometimes a sound trick on its own - not because of weighting.
Frequency weighting of objects when clusterin |
49,369 | What does the covariance of a quaternion *mean*? | EDIT: So my actual answer is that it is difficult to visualize quaternions (and you should not feel alone in that), thus the conversion to Euler angles. The thesis is there in case you wanted the covariance laws, and to add to the rationale above.
Here is some information from a thesis (the full thesis PDF link is ... | What does the covariance of a quaternion *mean*? | EDIT: So my actual answer is that it is difficult to visualize quaternions (and you should not feel alone in that), thus the conversion to Euler angles. The thesis is there in case you wanted the cov | What does the covariance of a quaternion *mean*?
EDIT: So my actual answer is that it is difficult to visualize quaternions (and you should not feel alone in that), thus the conversion to Euler angles. The thesis is there in case you wanted the covariance laws, and to add to the rationale above.
Here is some inform... | What does the covariance of a quaternion *mean*?
EDIT: So my actual answer is that it is difficult to visualize quaternions (and you should not feel alone in that), thus the conversion to Euler angles. The thesis is there in case you wanted the cov |
49,370 | Metrics for comparing estimated lists to a 'true' list | Your first issue appears to be that for some ranks, it is considered more important to be predicted correctly (usually the upper ranks), than others. So you should be looking into weighted rank correlation coefficients that can give to the top ranks' similarities/dissimilarities greater weight.
Here is some literatur... | Metrics for comparing estimated lists to a 'true' list | Your first issue appears to be that for some ranks, it is considered more important to be predicted correctly (usually the upper ranks), than others. So you should be looking into weighted rank correl | Metrics for comparing estimated lists to a 'true' list
Your first issue appears to be that for some ranks, it is considered more important to be predicted correctly (usually the upper ranks), than others. So you should be looking into weighted rank correlation coefficients that can give to the top ranks' similarities/d... | Metrics for comparing estimated lists to a 'true' list
Your first issue appears to be that for some ranks, it is considered more important to be predicted correctly (usually the upper ranks), than others. So you should be looking into weighted rank correl |
49,371 | Metrics for comparing estimated lists to a 'true' list | The simple approach is simply to construct a loss function over rankings, like squared errors.
However since you are concerned about ties and would like to use voting data as well, you could try to model the cumulative distribution function (CDF) of the votes, which you could do either parametrically or non-parametri... | Metrics for comparing estimated lists to a 'true' list | The simple approach is simply to construct a loss function over rankings, like squared errors.
However since you are concerned about ties and would like to use voting data as well, you could try to | Metrics for comparing estimated lists to a 'true' list
The simple approach is simply to construct a loss function over rankings, like squared errors.
However since you are concerned about ties and would like to use voting data as well, you could try to model the cumulative distribution function (CDF) of the votes, wh... | Metrics for comparing estimated lists to a 'true' list
The simple approach is simply to construct a loss function over rankings, like squared errors.
However since you are concerned about ties and would like to use voting data as well, you could try to |
49,372 | Prediction error in least squares with a linear model | Your last statement provides an important clue: not only would $D$ be diagonal, it would have to have $p$ units on the diagonal and zeros elsewhere. So there must be something special about $X(X^\prime X)^{-1}X^\prime$. To see what, look at the Singular Value Decomposition of $X$,
$$X = U \Sigma V^\prime$$
where $U$ ... | Prediction error in least squares with a linear model | Your last statement provides an important clue: not only would $D$ be diagonal, it would have to have $p$ units on the diagonal and zeros elsewhere. So there must be something special about $X(X^\pri | Prediction error in least squares with a linear model
Your last statement provides an important clue: not only would $D$ be diagonal, it would have to have $p$ units on the diagonal and zeros elsewhere. So there must be something special about $X(X^\prime X)^{-1}X^\prime$. To see what, look at the Singular Value Deco... | Prediction error in least squares with a linear model
Your last statement provides an important clue: not only would $D$ be diagonal, it would have to have $p$ units on the diagonal and zeros elsewhere. So there must be something special about $X(X^\pri |
49,373 | How to choose between exponential and gamma distributions | Fortunately, you're mistaken.
The shape parameter for a gamma ($\alpha$, say) has to be $\ge 0$.
http://en.wikipedia.org/wiki/Gamma_distribution
The exponential has $\alpha=1$.
http://en.wikipedia.org/wiki/Gamma_distribution#Others
So the exponential is not at the boundary and you should be able to apply a likelihood ... | How to choose between exponential and gamma distributions | Fortunately, you're mistaken.
The shape parameter for a gamma ($\alpha$, say) has to be $\ge 0$.
http://en.wikipedia.org/wiki/Gamma_distribution
The exponential has $\alpha=1$.
http://en.wikipedia.or | How to choose between exponential and gamma distributions
Fortunately, you're mistaken.
The shape parameter for a gamma ($\alpha$, say) has to be $\ge 0$.
http://en.wikipedia.org/wiki/Gamma_distribution
The exponential has $\alpha=1$.
http://en.wikipedia.org/wiki/Gamma_distribution#Others
So the exponential is not at ... | How to choose between exponential and gamma distributions
Fortunately, you're mistaken.
The shape parameter for a gamma ($\alpha$, say) has to be $\ge 0$.
http://en.wikipedia.org/wiki/Gamma_distribution
The exponential has $\alpha=1$.
http://en.wikipedia.or |
49,374 | Can I still interpret a Q-Q plot that uses discrete/rounded data? | As you say, a staircase pattern is an inevitable side-effect of discreteness, but that is the only obvious limitation.
The rule for quantile-quantile plots otherwise remains that departures from sameness of distributions are shown by departures from equality of quantiles.
Here are some dopey examples. I simulated som... | Can I still interpret a Q-Q plot that uses discrete/rounded data? | As you say, a staircase pattern is an inevitable side-effect of discreteness, but that is the only obvious limitation.
The rule for quantile-quantile plots otherwise remains that departures from same | Can I still interpret a Q-Q plot that uses discrete/rounded data?
As you say, a staircase pattern is an inevitable side-effect of discreteness, but that is the only obvious limitation.
The rule for quantile-quantile plots otherwise remains that departures from sameness of distributions are shown by departures from equ... | Can I still interpret a Q-Q plot that uses discrete/rounded data?
As you say, a staircase pattern is an inevitable side-effect of discreteness, but that is the only obvious limitation.
The rule for quantile-quantile plots otherwise remains that departures from same |
49,375 | About the derivation of group Lasso | the derivation of L2 norm is follow:
1. $\frac{\beta_j}{\|\beta_j\|}$ when $\beta_j \ne 0 $.
2. any vector with $ \| \beta_j \| \le 1 $ when $beta_j = 0$.
So when combing these two formula together, you can get the plus sign in the formula. | About the derivation of group Lasso | the derivation of L2 norm is follow:
1. $\frac{\beta_j}{\|\beta_j\|}$ when $\beta_j \ne 0 $.
2. any vector with $ \| \beta_j \| \le 1 $ when $beta_j = 0$.
So when combing these two formula together, y | About the derivation of group Lasso
the derivation of L2 norm is follow:
1. $\frac{\beta_j}{\|\beta_j\|}$ when $\beta_j \ne 0 $.
2. any vector with $ \| \beta_j \| \le 1 $ when $beta_j = 0$.
So when combing these two formula together, you can get the plus sign in the formula. | About the derivation of group Lasso
the derivation of L2 norm is follow:
1. $\frac{\beta_j}{\|\beta_j\|}$ when $\beta_j \ne 0 $.
2. any vector with $ \| \beta_j \| \le 1 $ when $beta_j = 0$.
So when combing these two formula together, y |
49,376 | How to model the effect of time in a balanced repeated measures design with 2 measures each at baseline, during instruction and post instruction? | I'd suggest you use Time as a 6-level factor, and then use appropriate contrasts to compare the phases, e.g., Post vs Instruction would be examined using contrast coefficients $(0,0,-.5,-.5,+.5,+.5)$.
It's potentially important to use Time itself in the model because of the repeated measures. For example, some people ... | How to model the effect of time in a balanced repeated measures design with 2 measures each at basel | I'd suggest you use Time as a 6-level factor, and then use appropriate contrasts to compare the phases, e.g., Post vs Instruction would be examined using contrast coefficients $(0,0,-.5,-.5,+.5,+.5)$. | How to model the effect of time in a balanced repeated measures design with 2 measures each at baseline, during instruction and post instruction?
I'd suggest you use Time as a 6-level factor, and then use appropriate contrasts to compare the phases, e.g., Post vs Instruction would be examined using contrast coefficient... | How to model the effect of time in a balanced repeated measures design with 2 measures each at basel
I'd suggest you use Time as a 6-level factor, and then use appropriate contrasts to compare the phases, e.g., Post vs Instruction would be examined using contrast coefficients $(0,0,-.5,-.5,+.5,+.5)$. |
49,377 | How to model the effect of time in a balanced repeated measures design with 2 measures each at baseline, during instruction and post instruction? | There seems to be many possible ways to do this, depending on what you want exactly.
The idea of ANOVA or repeated measure ANOVA only makes sense (to me) if you have different treatment groups (say, half of the 38 received different instructions etc.). Since all participants belong to 1 group, it seems to me all you ne... | How to model the effect of time in a balanced repeated measures design with 2 measures each at basel | There seems to be many possible ways to do this, depending on what you want exactly.
The idea of ANOVA or repeated measure ANOVA only makes sense (to me) if you have different treatment groups (say, h | How to model the effect of time in a balanced repeated measures design with 2 measures each at baseline, during instruction and post instruction?
There seems to be many possible ways to do this, depending on what you want exactly.
The idea of ANOVA or repeated measure ANOVA only makes sense (to me) if you have differen... | How to model the effect of time in a balanced repeated measures design with 2 measures each at basel
There seems to be many possible ways to do this, depending on what you want exactly.
The idea of ANOVA or repeated measure ANOVA only makes sense (to me) if you have different treatment groups (say, h |
49,378 | Alternative to forecast() and ets() in Python? | There is an open PR to add full ETS functionality to statsmodels here. I ran out of steam trying to code up all the heuristics for optimization that Hyndman suggests to make it less fragile. If someone wants to take up the torch here, I have some uncommitted code and thoughts on how to proceed. | Alternative to forecast() and ets() in Python? | There is an open PR to add full ETS functionality to statsmodels here. I ran out of steam trying to code up all the heuristics for optimization that Hyndman suggests to make it less fragile. If someon | Alternative to forecast() and ets() in Python?
There is an open PR to add full ETS functionality to statsmodels here. I ran out of steam trying to code up all the heuristics for optimization that Hyndman suggests to make it less fragile. If someone wants to take up the torch here, I have some uncommitted code and thoug... | Alternative to forecast() and ets() in Python?
There is an open PR to add full ETS functionality to statsmodels here. I ran out of steam trying to code up all the heuristics for optimization that Hyndman suggests to make it less fragile. If someon |
49,379 | Two simple questions regarding GLM | The equation is a general form for the broad class of densities in the exponential family (i.e. that's the pdf).
If it's the distribution for y corresponding to a fixed $x_i$, is it possible that even if the plot of $y$ against $x$ looks like a straight line, I should still use GLM instead of simple regression?
The ... | Two simple questions regarding GLM | The equation is a general form for the broad class of densities in the exponential family (i.e. that's the pdf).
If it's the distribution for y corresponding to a fixed $x_i$, is it possible that ev | Two simple questions regarding GLM
The equation is a general form for the broad class of densities in the exponential family (i.e. that's the pdf).
If it's the distribution for y corresponding to a fixed $x_i$, is it possible that even if the plot of $y$ against $x$ looks like a straight line, I should still use GLM ... | Two simple questions regarding GLM
The equation is a general form for the broad class of densities in the exponential family (i.e. that's the pdf).
If it's the distribution for y corresponding to a fixed $x_i$, is it possible that ev |
49,380 | Two simple questions regarding GLM | That is the equation for the distribution of $y$. Its mean $E[y]$ is a function thereof, when we set that mean equal to $\beta^TX_i$ we call it $E[y_i|X_i]$
The logarithmic link is the "canonical link" for the Poisson. More info here.
This setup is developed in chapter 4 of Categorical Data Analysis by Agresti. It's t... | Two simple questions regarding GLM | That is the equation for the distribution of $y$. Its mean $E[y]$ is a function thereof, when we set that mean equal to $\beta^TX_i$ we call it $E[y_i|X_i]$
The logarithmic link is the "canonical link | Two simple questions regarding GLM
That is the equation for the distribution of $y$. Its mean $E[y]$ is a function thereof, when we set that mean equal to $\beta^TX_i$ we call it $E[y_i|X_i]$
The logarithmic link is the "canonical link" for the Poisson. More info here.
This setup is developed in chapter 4 of Categoric... | Two simple questions regarding GLM
That is the equation for the distribution of $y$. Its mean $E[y]$ is a function thereof, when we set that mean equal to $\beta^TX_i$ we call it $E[y_i|X_i]$
The logarithmic link is the "canonical link |
49,381 | Does the vanishing gradient in RNNs present a problem? | First let's restate the problem of vanishing gradients. Suppose you have a normal multilayer perceptron with sigmoidal hidden units. This is trained by back-propagation. When there are many hidden layers the error gradient weakens as it moves from the back of the network to the front, because the derivative the sigmoid... | Does the vanishing gradient in RNNs present a problem? | First let's restate the problem of vanishing gradients. Suppose you have a normal multilayer perceptron with sigmoidal hidden units. This is trained by back-propagation. When there are many hidden lay | Does the vanishing gradient in RNNs present a problem?
First let's restate the problem of vanishing gradients. Suppose you have a normal multilayer perceptron with sigmoidal hidden units. This is trained by back-propagation. When there are many hidden layers the error gradient weakens as it moves from the back of the n... | Does the vanishing gradient in RNNs present a problem?
First let's restate the problem of vanishing gradients. Suppose you have a normal multilayer perceptron with sigmoidal hidden units. This is trained by back-propagation. When there are many hidden lay |
49,382 | Statistical Significance or Unambiguous Direction of Influence? | I think this way about the relationship between NHSTs and CIs in general, but don't know of any references that describe everything the same way off the top of my head...Seems there's bound to be some out there though, as resolving directional ambiguity of an effect is the most compelling reason to perform a NHST that ... | Statistical Significance or Unambiguous Direction of Influence? | I think this way about the relationship between NHSTs and CIs in general, but don't know of any references that describe everything the same way off the top of my head...Seems there's bound to be some | Statistical Significance or Unambiguous Direction of Influence?
I think this way about the relationship between NHSTs and CIs in general, but don't know of any references that describe everything the same way off the top of my head...Seems there's bound to be some out there though, as resolving directional ambiguity of... | Statistical Significance or Unambiguous Direction of Influence?
I think this way about the relationship between NHSTs and CIs in general, but don't know of any references that describe everything the same way off the top of my head...Seems there's bound to be some |
49,383 | In inverse theory, how do I transform the averaging kernel matrix to a new grid? | This is considered in Calisesi et al. (2005). They derive that
$$\mathbf{A_{z_i}} = \mathbf{W_i^* A_x W_i} \, ,$$
where $\mathbf{A_{z_i}}$ is the averaging kernel for the new grid, $\mathbf{W_i}$ is the interpolation matrix with $\mathbf{W_i^*}$ its Moore-Penrose pseudo-inverse, and $\mathbf{A_x}$ is the averaging ker... | In inverse theory, how do I transform the averaging kernel matrix to a new grid? | This is considered in Calisesi et al. (2005). They derive that
$$\mathbf{A_{z_i}} = \mathbf{W_i^* A_x W_i} \, ,$$
where $\mathbf{A_{z_i}}$ is the averaging kernel for the new grid, $\mathbf{W_i}$ is | In inverse theory, how do I transform the averaging kernel matrix to a new grid?
This is considered in Calisesi et al. (2005). They derive that
$$\mathbf{A_{z_i}} = \mathbf{W_i^* A_x W_i} \, ,$$
where $\mathbf{A_{z_i}}$ is the averaging kernel for the new grid, $\mathbf{W_i}$ is the interpolation matrix with $\mathbf{... | In inverse theory, how do I transform the averaging kernel matrix to a new grid?
This is considered in Calisesi et al. (2005). They derive that
$$\mathbf{A_{z_i}} = \mathbf{W_i^* A_x W_i} \, ,$$
where $\mathbf{A_{z_i}}$ is the averaging kernel for the new grid, $\mathbf{W_i}$ is |
49,384 | How to cope with missing data in logistic regression? | I am afraid you cannot expect to find some "canned" solution to your problem. Most methods for handling missing data assumes "missing at random" or even "missing completely at random" (you can google those terms!). Your problem seems definitely to be a problem of informative missingness. Then you will need to model... | How to cope with missing data in logistic regression? | I am afraid you cannot expect to find some "canned" solution to your problem. Most methods for handling missing data assumes "missing at random" or even "missing completely at random" (you can google | How to cope with missing data in logistic regression?
I am afraid you cannot expect to find some "canned" solution to your problem. Most methods for handling missing data assumes "missing at random" or even "missing completely at random" (you can google those terms!). Your problem seems definitely to be a problem of ... | How to cope with missing data in logistic regression?
I am afraid you cannot expect to find some "canned" solution to your problem. Most methods for handling missing data assumes "missing at random" or even "missing completely at random" (you can google |
49,385 | How to cope with missing data in logistic regression? | @Kjetil gave a good answer.
One possible simple alternative, if you have enough auctions, is to run two models: One with the data that has both highest and second highest and one that has just the highest.
An advantage of this approach would be that each model will be considerably simpler than a full model with both. ... | How to cope with missing data in logistic regression? | @Kjetil gave a good answer.
One possible simple alternative, if you have enough auctions, is to run two models: One with the data that has both highest and second highest and one that has just the hig | How to cope with missing data in logistic regression?
@Kjetil gave a good answer.
One possible simple alternative, if you have enough auctions, is to run two models: One with the data that has both highest and second highest and one that has just the highest.
An advantage of this approach would be that each model will... | How to cope with missing data in logistic regression?
@Kjetil gave a good answer.
One possible simple alternative, if you have enough auctions, is to run two models: One with the data that has both highest and second highest and one that has just the hig |
49,386 | Confidence interval for poisson distributed data | This answer is based on the clarification offered in comments:
I'd like to make a statement such as ... "I am 68% sure the mean is between $3.1βΟ_β$ and $3.1+Ο_+$", and I want to calculate $Ο_+$ and $Ο_β$.
I think that at least in the physics world this is called a confidence interval.
Let's take it as given that in... | Confidence interval for poisson distributed data | This answer is based on the clarification offered in comments:
I'd like to make a statement such as ... "I am 68% sure the mean is between $3.1βΟ_β$ and $3.1+Ο_+$", and I want to calculate $Ο_+$ and | Confidence interval for poisson distributed data
This answer is based on the clarification offered in comments:
I'd like to make a statement such as ... "I am 68% sure the mean is between $3.1βΟ_β$ and $3.1+Ο_+$", and I want to calculate $Ο_+$ and $Ο_β$.
I think that at least in the physics world this is called a con... | Confidence interval for poisson distributed data
This answer is based on the clarification offered in comments:
I'd like to make a statement such as ... "I am 68% sure the mean is between $3.1βΟ_β$ and $3.1+Ο_+$", and I want to calculate $Ο_+$ and |
49,387 | Confidence interval for poisson distributed data | It was mentioned in the comments on the original post but here it is more explicitly. The bootstrap is simple to use and has a ton of nice asymptotic theory behind it, e.g. Shao and Tu, 1995. Here's some R code which does what I think you want:
the_data = c(1,2,3,5,1,2,2,3,7,2,3,4,1,5,7,6,4,1,2,2,3,9,2,1,2,2,3)
n_res... | Confidence interval for poisson distributed data | It was mentioned in the comments on the original post but here it is more explicitly. The bootstrap is simple to use and has a ton of nice asymptotic theory behind it, e.g. Shao and Tu, 1995. Here's | Confidence interval for poisson distributed data
It was mentioned in the comments on the original post but here it is more explicitly. The bootstrap is simple to use and has a ton of nice asymptotic theory behind it, e.g. Shao and Tu, 1995. Here's some R code which does what I think you want:
the_data = c(1,2,3,5,1,2... | Confidence interval for poisson distributed data
It was mentioned in the comments on the original post but here it is more explicitly. The bootstrap is simple to use and has a ton of nice asymptotic theory behind it, e.g. Shao and Tu, 1995. Here's |
49,388 | Confidence interval for poisson distributed data | For skewed distributions the confidence interval is tricky. One way to proceed is by having equal quantiles from tails. So, for instance, if you wish to have 95% confidence interval, you'd get 2.5% and 97.5% quantiles.
Your comment about $\pm\sigma$ being 68% CI in physics is only true when you assume normal-ish distri... | Confidence interval for poisson distributed data | For skewed distributions the confidence interval is tricky. One way to proceed is by having equal quantiles from tails. So, for instance, if you wish to have 95% confidence interval, you'd get 2.5% an | Confidence interval for poisson distributed data
For skewed distributions the confidence interval is tricky. One way to proceed is by having equal quantiles from tails. So, for instance, if you wish to have 95% confidence interval, you'd get 2.5% and 97.5% quantiles.
Your comment about $\pm\sigma$ being 68% CI in physi... | Confidence interval for poisson distributed data
For skewed distributions the confidence interval is tricky. One way to proceed is by having equal quantiles from tails. So, for instance, if you wish to have 95% confidence interval, you'd get 2.5% an |
49,389 | Is there a statistical measure for how much a variable fluctuates over time? | Variance of the first derivative would mean looking for variations in derivative of your variable. Rather, I would recommend taking derivative of your variable with respect to time and see the results. This result would be rate of change of your variable with respect to time.
rate_of_change = d(variable)/d(time)
Note ... | Is there a statistical measure for how much a variable fluctuates over time? | Variance of the first derivative would mean looking for variations in derivative of your variable. Rather, I would recommend taking derivative of your variable with respect to time and see the results | Is there a statistical measure for how much a variable fluctuates over time?
Variance of the first derivative would mean looking for variations in derivative of your variable. Rather, I would recommend taking derivative of your variable with respect to time and see the results. This result would be rate of change of yo... | Is there a statistical measure for how much a variable fluctuates over time?
Variance of the first derivative would mean looking for variations in derivative of your variable. Rather, I would recommend taking derivative of your variable with respect to time and see the results |
49,390 | Convergence in distribution of sum implies marginal convergence? | It is a particular case of the accompanying law theorem.
Let $f$ be a bounded uniformly continuous function on $\mathbf R$. Since
$$|E[f(X_n)]-E[f(X)]|\leqslant |E[f(X_n)]-E[f(X_n+cY)]|+|E[f(X_n+cY)]-E[f(X+cY)]|+|E[f(X+cY)-E[f(X)]]|$$
and $X_n+cY\to X+cY$ in distribution, we obtain for each positive $c$,
$$\limsup_{... | Convergence in distribution of sum implies marginal convergence? | It is a particular case of the accompanying law theorem.
Let $f$ be a bounded uniformly continuous function on $\mathbf R$. Since
$$|E[f(X_n)]-E[f(X)]|\leqslant |E[f(X_n)]-E[f(X_n+cY)]|+|E[f(X_n+cY) | Convergence in distribution of sum implies marginal convergence?
It is a particular case of the accompanying law theorem.
Let $f$ be a bounded uniformly continuous function on $\mathbf R$. Since
$$|E[f(X_n)]-E[f(X)]|\leqslant |E[f(X_n)]-E[f(X_n+cY)]|+|E[f(X_n+cY)]-E[f(X+cY)]|+|E[f(X+cY)-E[f(X)]]|$$
and $X_n+cY\to X+c... | Convergence in distribution of sum implies marginal convergence?
It is a particular case of the accompanying law theorem.
Let $f$ be a bounded uniformly continuous function on $\mathbf R$. Since
$$|E[f(X_n)]-E[f(X)]|\leqslant |E[f(X_n)]-E[f(X_n+cY)]|+|E[f(X_n+cY) |
49,391 | Convergence in distribution of sum implies marginal convergence? | I may prove it under the assumption that $\mathrm{E}\left|Y\right|<\infty$. In order
to prove $X_{n}\rightarrow_{d}X$, we wish to show that $\mathrm{E}f\left(X_{n}\right)\rightarrow\mathrm{E}f\left(X\right)$
for all bounded, Lipschitz functions $f$ (this is Portmanteau lemma).
Hereafter let $f$ be an arbitrary bounded ... | Convergence in distribution of sum implies marginal convergence? | I may prove it under the assumption that $\mathrm{E}\left|Y\right|<\infty$. In order
to prove $X_{n}\rightarrow_{d}X$, we wish to show that $\mathrm{E}f\left(X_{n}\right)\rightarrow\mathrm{E}f\left(X\ | Convergence in distribution of sum implies marginal convergence?
I may prove it under the assumption that $\mathrm{E}\left|Y\right|<\infty$. In order
to prove $X_{n}\rightarrow_{d}X$, we wish to show that $\mathrm{E}f\left(X_{n}\right)\rightarrow\mathrm{E}f\left(X\right)$
for all bounded, Lipschitz functions $f$ (this ... | Convergence in distribution of sum implies marginal convergence?
I may prove it under the assumption that $\mathrm{E}\left|Y\right|<\infty$. In order
to prove $X_{n}\rightarrow_{d}X$, we wish to show that $\mathrm{E}f\left(X_{n}\right)\rightarrow\mathrm{E}f\left(X\ |
49,392 | How do you create variables reflecting the lead and lag impact of holidays / calendar effects in a time-series analysis? | Create a predictor variable (zeroes except a 1 at the beginning of the exceptional period and then specify a poylynomial of order k where k is the expected length. This will form the long response that you are looking for. Make sure that you also accommodate individually tailored windows of response around each major e... | How do you create variables reflecting the lead and lag impact of holidays / calendar effects in a t | Create a predictor variable (zeroes except a 1 at the beginning of the exceptional period and then specify a poylynomial of order k where k is the expected length. This will form the long response tha | How do you create variables reflecting the lead and lag impact of holidays / calendar effects in a time-series analysis?
Create a predictor variable (zeroes except a 1 at the beginning of the exceptional period and then specify a poylynomial of order k where k is the expected length. This will form the long response th... | How do you create variables reflecting the lead and lag impact of holidays / calendar effects in a t
Create a predictor variable (zeroes except a 1 at the beginning of the exceptional period and then specify a poylynomial of order k where k is the expected length. This will form the long response tha |
49,393 | Training and testing on Unbalanced Data Set | Reampling should be applied to training set only, and then you should test on a subset having the same class distribution as the population.
If you oversample the minority class in the test set, you may get a higher hit rate i.e. better performances than in the real case, where positive instances are rare. | Training and testing on Unbalanced Data Set | Reampling should be applied to training set only, and then you should test on a subset having the same class distribution as the population.
If you oversample the minority class in the test set, you m | Training and testing on Unbalanced Data Set
Reampling should be applied to training set only, and then you should test on a subset having the same class distribution as the population.
If you oversample the minority class in the test set, you may get a higher hit rate i.e. better performances than in the real case, whe... | Training and testing on Unbalanced Data Set
Reampling should be applied to training set only, and then you should test on a subset having the same class distribution as the population.
If you oversample the minority class in the test set, you m |
49,394 | How to speed up Kernel density estimation | One trick which I used when I implemented KDEs is to limit the effect of kernel to some values.
Suppose you have a sample $x = {x_1, x_2, .. , x_n}$, and some points where you want to estimate the kernel $k = {k_1, k_2, .., k_n}$. Now, without loosing of generality, we can consider $k$ values as being sorted. If not, ... | How to speed up Kernel density estimation | One trick which I used when I implemented KDEs is to limit the effect of kernel to some values.
Suppose you have a sample $x = {x_1, x_2, .. , x_n}$, and some points where you want to estimate the ke | How to speed up Kernel density estimation
One trick which I used when I implemented KDEs is to limit the effect of kernel to some values.
Suppose you have a sample $x = {x_1, x_2, .. , x_n}$, and some points where you want to estimate the kernel $k = {k_1, k_2, .., k_n}$. Now, without loosing of generality, we can con... | How to speed up Kernel density estimation
One trick which I used when I implemented KDEs is to limit the effect of kernel to some values.
Suppose you have a sample $x = {x_1, x_2, .. , x_n}$, and some points where you want to estimate the ke |
49,395 | How to speed up Kernel density estimation | Suppose you have $x_1,\dots,x_m$ data points, called source points hereafter, and $z_1,\dots,z_n$ points you want to estimate at, called query points hereafter. A naive implementation of
$$ \hat f(z_j)=\frac 1{mh}\sum_{i=1}^mK\left(\frac{x_i-z_j}{h}\right)$$
is then $\mathcal{O}(mn)$ if you want to evaluate it for all ... | How to speed up Kernel density estimation | Suppose you have $x_1,\dots,x_m$ data points, called source points hereafter, and $z_1,\dots,z_n$ points you want to estimate at, called query points hereafter. A naive implementation of
$$ \hat f(z_j | How to speed up Kernel density estimation
Suppose you have $x_1,\dots,x_m$ data points, called source points hereafter, and $z_1,\dots,z_n$ points you want to estimate at, called query points hereafter. A naive implementation of
$$ \hat f(z_j)=\frac 1{mh}\sum_{i=1}^mK\left(\frac{x_i-z_j}{h}\right)$$
is then $\mathcal{O... | How to speed up Kernel density estimation
Suppose you have $x_1,\dots,x_m$ data points, called source points hereafter, and $z_1,\dots,z_n$ points you want to estimate at, called query points hereafter. A naive implementation of
$$ \hat f(z_j |
49,396 | When is Maximum Likelihood the same as Least Squares | Levenberg-Marquardt is a general (nonlinear) optimization technique. It is not specific to LS, although that is probably its widest use. Looking at your referenced paper, they are (mostly) fitting state-space models with additive Normal errors. Forming the likelihood function yields
$\ln L(\theta|X=x) = K - \ln \si... | When is Maximum Likelihood the same as Least Squares | Levenberg-Marquardt is a general (nonlinear) optimization technique. It is not specific to LS, although that is probably its widest use. Looking at your referenced paper, they are (mostly) fitting s | When is Maximum Likelihood the same as Least Squares
Levenberg-Marquardt is a general (nonlinear) optimization technique. It is not specific to LS, although that is probably its widest use. Looking at your referenced paper, they are (mostly) fitting state-space models with additive Normal errors. Forming the likelih... | When is Maximum Likelihood the same as Least Squares
Levenberg-Marquardt is a general (nonlinear) optimization technique. It is not specific to LS, although that is probably its widest use. Looking at your referenced paper, they are (mostly) fitting s |
49,397 | What is the meaning of a large p-value? | How you should 'use' the p-value depends on how you have designed your study with regard to the analyses you will run. I discuss two different philosophies about p-values in my answer here: When to use Fisher and Neyman-Pearson framework? You may find it helpful to read that. If you have, for example, run a power an... | What is the meaning of a large p-value? | How you should 'use' the p-value depends on how you have designed your study with regard to the analyses you will run. I discuss two different philosophies about p-values in my answer here: When to u | What is the meaning of a large p-value?
How you should 'use' the p-value depends on how you have designed your study with regard to the analyses you will run. I discuss two different philosophies about p-values in my answer here: When to use Fisher and Neyman-Pearson framework? You may find it helpful to read that. ... | What is the meaning of a large p-value?
How you should 'use' the p-value depends on how you have designed your study with regard to the analyses you will run. I discuss two different philosophies about p-values in my answer here: When to u |
49,398 | What is the meaning of a large p-value? | In my view, everything boils down to assumptions, that is, how well the models fits them. If it does agree with all of them, treat the p-value as a probability. Then you can compare p-value of 0.06 with 0.99 by concluding which of the two are more likely. Also, a lot depend on circumstances: in some cases, marginal sig... | What is the meaning of a large p-value? | In my view, everything boils down to assumptions, that is, how well the models fits them. If it does agree with all of them, treat the p-value as a probability. Then you can compare p-value of 0.06 wi | What is the meaning of a large p-value?
In my view, everything boils down to assumptions, that is, how well the models fits them. If it does agree with all of them, treat the p-value as a probability. Then you can compare p-value of 0.06 with 0.99 by concluding which of the two are more likely. Also, a lot depend on ci... | What is the meaning of a large p-value?
In my view, everything boils down to assumptions, that is, how well the models fits them. If it does agree with all of them, treat the p-value as a probability. Then you can compare p-value of 0.06 wi |
49,399 | What is the meaning of a large p-value? | It is true that the acceptance range for the p-value of a hypothesis test is rather arbitrary, but nevertheless a lower p-value means that the test result can be accepted with more certainty, because the p-value essentially defines the confidence interval for the estimate, so a narrower confidence interval should be re... | What is the meaning of a large p-value? | It is true that the acceptance range for the p-value of a hypothesis test is rather arbitrary, but nevertheless a lower p-value means that the test result can be accepted with more certainty, because | What is the meaning of a large p-value?
It is true that the acceptance range for the p-value of a hypothesis test is rather arbitrary, but nevertheless a lower p-value means that the test result can be accepted with more certainty, because the p-value essentially defines the confidence interval for the estimate, so a n... | What is the meaning of a large p-value?
It is true that the acceptance range for the p-value of a hypothesis test is rather arbitrary, but nevertheless a lower p-value means that the test result can be accepted with more certainty, because |
49,400 | Overfitting in Genetic Programming | I can "train" my GP on some time period (let's say 2000-2010) and evaluate it's fitness on another time period (let's say 2010-2014). However, won't this just introduce overfitting on the time period from 2010-2014. The GP will just gravitate towards finding good programs on the testing time period.
Yes that is a fun... | Overfitting in Genetic Programming | I can "train" my GP on some time period (let's say 2000-2010) and evaluate it's fitness on another time period (let's say 2010-2014). However, won't this just introduce overfitting on the time period | Overfitting in Genetic Programming
I can "train" my GP on some time period (let's say 2000-2010) and evaluate it's fitness on another time period (let's say 2010-2014). However, won't this just introduce overfitting on the time period from 2010-2014. The GP will just gravitate towards finding good programs on the testi... | Overfitting in Genetic Programming
I can "train" my GP on some time period (let's say 2000-2010) and evaluate it's fitness on another time period (let's say 2010-2014). However, won't this just introduce overfitting on the time period |
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