idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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40,001 | meaning of 'Monte Carlo' in this sentence | What Ng and Russell seem to be saying is that for each policy $\pi$ they simulate $m$ "possible" outcomes for processes starting at point $s_0$. By "trajectories" they seem to mean the possible developments in time of simulated processes -- different possible scenarios created by simulation. So you were correct, Monte ... | meaning of 'Monte Carlo' in this sentence | What Ng and Russell seem to be saying is that for each policy $\pi$ they simulate $m$ "possible" outcomes for processes starting at point $s_0$. By "trajectories" they seem to mean the possible develo | meaning of 'Monte Carlo' in this sentence
What Ng and Russell seem to be saying is that for each policy $\pi$ they simulate $m$ "possible" outcomes for processes starting at point $s_0$. By "trajectories" they seem to mean the possible developments in time of simulated processes -- different possible scenarios created ... | meaning of 'Monte Carlo' in this sentence
What Ng and Russell seem to be saying is that for each policy $\pi$ they simulate $m$ "possible" outcomes for processes starting at point $s_0$. By "trajectories" they seem to mean the possible develo |
40,002 | meaning of 'Monte Carlo' in this sentence | Monte Carlo here simply means use sampling to estimate the values. Practically this means collecting a sequence of (state, action) pairs, i.e. the trajectory using some arbitrary policy, and from this you can compute relevant quantities like V, etc | meaning of 'Monte Carlo' in this sentence | Monte Carlo here simply means use sampling to estimate the values. Practically this means collecting a sequence of (state, action) pairs, i.e. the trajectory using some arbitrary policy, and from this | meaning of 'Monte Carlo' in this sentence
Monte Carlo here simply means use sampling to estimate the values. Practically this means collecting a sequence of (state, action) pairs, i.e. the trajectory using some arbitrary policy, and from this you can compute relevant quantities like V, etc | meaning of 'Monte Carlo' in this sentence
Monte Carlo here simply means use sampling to estimate the values. Practically this means collecting a sequence of (state, action) pairs, i.e. the trajectory using some arbitrary policy, and from this |
40,003 | How can I assess performance of a semi-supervised learning method? | We have addressed this problem in Assessing binary classifiers using only positive and unlabeled data. Specifically, we show how to compute strict bounds on any metric based on contingency tables (accuracy, precision, ROC/PR curves, ...). Our work was accepted by all reviewers at this year's NIPS conference, but then r... | How can I assess performance of a semi-supervised learning method? | We have addressed this problem in Assessing binary classifiers using only positive and unlabeled data. Specifically, we show how to compute strict bounds on any metric based on contingency tables (acc | How can I assess performance of a semi-supervised learning method?
We have addressed this problem in Assessing binary classifiers using only positive and unlabeled data. Specifically, we show how to compute strict bounds on any metric based on contingency tables (accuracy, precision, ROC/PR curves, ...). Our work was a... | How can I assess performance of a semi-supervised learning method?
We have addressed this problem in Assessing binary classifiers using only positive and unlabeled data. Specifically, we show how to compute strict bounds on any metric based on contingency tables (acc |
40,004 | How can I assess performance of a semi-supervised learning method? | Here's a sideways thinking idea: You have some positive labels and you can estimate the natural grouping of data using unsupervised learning. Try to measure the overlap between the known information and the way the data groups together, use the overlap as a ground truth measure.
So, perform unsupervised learning, see h... | How can I assess performance of a semi-supervised learning method? | Here's a sideways thinking idea: You have some positive labels and you can estimate the natural grouping of data using unsupervised learning. Try to measure the overlap between the known information a | How can I assess performance of a semi-supervised learning method?
Here's a sideways thinking idea: You have some positive labels and you can estimate the natural grouping of data using unsupervised learning. Try to measure the overlap between the known information and the way the data groups together, use the overlap ... | How can I assess performance of a semi-supervised learning method?
Here's a sideways thinking idea: You have some positive labels and you can estimate the natural grouping of data using unsupervised learning. Try to measure the overlap between the known information a |
40,005 | Difference in difference with interaction | A triple difference-in-difference is the correct specification for this problem. I'll present a conceptual explanation and then a mathematical one.
Conceptually, the standard (double) difference-in-difference can also be thought of as estimating heterogeneous treatment effect. In this perspective, time is the "treatmen... | Difference in difference with interaction | A triple difference-in-difference is the correct specification for this problem. I'll present a conceptual explanation and then a mathematical one.
Conceptually, the standard (double) difference-in-di | Difference in difference with interaction
A triple difference-in-difference is the correct specification for this problem. I'll present a conceptual explanation and then a mathematical one.
Conceptually, the standard (double) difference-in-difference can also be thought of as estimating heterogeneous treatment effect. ... | Difference in difference with interaction
A triple difference-in-difference is the correct specification for this problem. I'll present a conceptual explanation and then a mathematical one.
Conceptually, the standard (double) difference-in-di |
40,006 | Difference in difference with interaction | I think this (exploring the heterogeneous treatment effects of DD for different groups) could be easily confused with the DDD method.
However, they share the same specification, I'd just run the following:
$y_{it} = \alpha + \beta_1 d_t + \beta_2 Treat_i + \beta_3 d_t \times Treat_i + \\ \beta_4 big_i + \beta_5 big_i\... | Difference in difference with interaction | I think this (exploring the heterogeneous treatment effects of DD for different groups) could be easily confused with the DDD method.
However, they share the same specification, I'd just run the foll | Difference in difference with interaction
I think this (exploring the heterogeneous treatment effects of DD for different groups) could be easily confused with the DDD method.
However, they share the same specification, I'd just run the following:
$y_{it} = \alpha + \beta_1 d_t + \beta_2 Treat_i + \beta_3 d_t \times T... | Difference in difference with interaction
I think this (exploring the heterogeneous treatment effects of DD for different groups) could be easily confused with the DDD method.
However, they share the same specification, I'd just run the foll |
40,007 | Analyzing repeated measures experiment with multiple treatment groups and multiple measures | I think one could write a whole book dealing exclusively with your question (and I am definitely not qualified to write it). So without any attempt at providing a comprehensive answer, here are some points that can hopefully be helpful.
Confirmatory vs. exploratory approach to analysis
As you note yourself, you have a ... | Analyzing repeated measures experiment with multiple treatment groups and multiple measures | I think one could write a whole book dealing exclusively with your question (and I am definitely not qualified to write it). So without any attempt at providing a comprehensive answer, here are some p | Analyzing repeated measures experiment with multiple treatment groups and multiple measures
I think one could write a whole book dealing exclusively with your question (and I am definitely not qualified to write it). So without any attempt at providing a comprehensive answer, here are some points that can hopefully be ... | Analyzing repeated measures experiment with multiple treatment groups and multiple measures
I think one could write a whole book dealing exclusively with your question (and I am definitely not qualified to write it). So without any attempt at providing a comprehensive answer, here are some p |
40,008 | Analyzing repeated measures experiment with multiple treatment groups and multiple measures | It is a multi layered methodological onion to be peeled. I will only be able to deal with the top layers both because of the lack of time and the lack on knowledge. I will base this answer in the very clear statement of the goals of the analysis: in bold in the OP:
We would like to know if Treatment 2 (Dietary Supplem... | Analyzing repeated measures experiment with multiple treatment groups and multiple measures | It is a multi layered methodological onion to be peeled. I will only be able to deal with the top layers both because of the lack of time and the lack on knowledge. I will base this answer in the very | Analyzing repeated measures experiment with multiple treatment groups and multiple measures
It is a multi layered methodological onion to be peeled. I will only be able to deal with the top layers both because of the lack of time and the lack on knowledge. I will base this answer in the very clear statement of the goal... | Analyzing repeated measures experiment with multiple treatment groups and multiple measures
It is a multi layered methodological onion to be peeled. I will only be able to deal with the top layers both because of the lack of time and the lack on knowledge. I will base this answer in the very |
40,009 | Show that a scale mixtures of normals is a power exponential | The marginal distribution of $\beta$ associated with
$$\beta|\tau\sim\mathcal{N}(0,\tau)\quad\tau\sim\mathcal{E}(\lambda^2/2)$$
[with the convention that $\tau$ is the variance] has density
\begin{align*}\mathfrak{t}(\beta) &=\int_0^\infty \tau^{-1/2}\varphi(\beta/\sqrt{\tau})\frac{\lambda^2}{2}\exp\{-\lambda^2\tau/2\}... | Show that a scale mixtures of normals is a power exponential | The marginal distribution of $\beta$ associated with
$$\beta|\tau\sim\mathcal{N}(0,\tau)\quad\tau\sim\mathcal{E}(\lambda^2/2)$$
[with the convention that $\tau$ is the variance] has density
\begin{ali | Show that a scale mixtures of normals is a power exponential
The marginal distribution of $\beta$ associated with
$$\beta|\tau\sim\mathcal{N}(0,\tau)\quad\tau\sim\mathcal{E}(\lambda^2/2)$$
[with the convention that $\tau$ is the variance] has density
\begin{align*}\mathfrak{t}(\beta) &=\int_0^\infty \tau^{-1/2}\varphi(... | Show that a scale mixtures of normals is a power exponential
The marginal distribution of $\beta$ associated with
$$\beta|\tau\sim\mathcal{N}(0,\tau)\quad\tau\sim\mathcal{E}(\lambda^2/2)$$
[with the convention that $\tau$ is the variance] has density
\begin{ali |
40,010 | Show that a scale mixtures of normals is a power exponential | I've always found the direct integration in this case to be a complicated integral. The Moment Generating Function (MGF) approach works too.
The MGF: $\beta | \tau \sim N(0, \tau)$ and then
$M_{\beta|\tau}(t)=e^{\frac{\tau t^2}{2}}.$ Now to get the MGF of $\beta$ marginally, take the expectation with respect to $\tau... | Show that a scale mixtures of normals is a power exponential | I've always found the direct integration in this case to be a complicated integral. The Moment Generating Function (MGF) approach works too.
The MGF: $\beta | \tau \sim N(0, \tau)$ and then
$M_{\bet | Show that a scale mixtures of normals is a power exponential
I've always found the direct integration in this case to be a complicated integral. The Moment Generating Function (MGF) approach works too.
The MGF: $\beta | \tau \sim N(0, \tau)$ and then
$M_{\beta|\tau}(t)=e^{\frac{\tau t^2}{2}}.$ Now to get the MGF of $... | Show that a scale mixtures of normals is a power exponential
I've always found the direct integration in this case to be a complicated integral. The Moment Generating Function (MGF) approach works too.
The MGF: $\beta | \tau \sim N(0, \tau)$ and then
$M_{\bet |
40,011 | Does a lower pvalue mean that test has higher power? | In general, the answer is NO. Suppose you have two different hypothesis tests $T$ and $T'$ for the same hypothesis test problem $H_0$ versus $H_1$ on the same data. Supposedly, $T$ and $T'$ uses different aspects of the data, for example, original data versus ranks. To make a meaningful comparison we must suppose tha... | Does a lower pvalue mean that test has higher power? | In general, the answer is NO. Suppose you have two different hypothesis tests $T$ and $T'$ for the same hypothesis test problem $H_0$ versus $H_1$ on the same data. Supposedly, $T$ and $T'$ uses diff | Does a lower pvalue mean that test has higher power?
In general, the answer is NO. Suppose you have two different hypothesis tests $T$ and $T'$ for the same hypothesis test problem $H_0$ versus $H_1$ on the same data. Supposedly, $T$ and $T'$ uses different aspects of the data, for example, original data versus ranks.... | Does a lower pvalue mean that test has higher power?
In general, the answer is NO. Suppose you have two different hypothesis tests $T$ and $T'$ for the same hypothesis test problem $H_0$ versus $H_1$ on the same data. Supposedly, $T$ and $T'$ uses diff |
40,012 | Does a lower pvalue mean that test has higher power? | The burden, SKAT, and SKAT-O tests represent 3 ways to pool information from low-frequency genetic variants so that relations of genomic loci to a biologic characteristic (phenotype) can be assessed. Burden tests assume that all low-frequency variants at a locus have the same relation to phenotype (unidirectional), so ... | Does a lower pvalue mean that test has higher power? | The burden, SKAT, and SKAT-O tests represent 3 ways to pool information from low-frequency genetic variants so that relations of genomic loci to a biologic characteristic (phenotype) can be assessed. | Does a lower pvalue mean that test has higher power?
The burden, SKAT, and SKAT-O tests represent 3 ways to pool information from low-frequency genetic variants so that relations of genomic loci to a biologic characteristic (phenotype) can be assessed. Burden tests assume that all low-frequency variants at a locus have... | Does a lower pvalue mean that test has higher power?
The burden, SKAT, and SKAT-O tests represent 3 ways to pool information from low-frequency genetic variants so that relations of genomic loci to a biologic characteristic (phenotype) can be assessed. |
40,013 | Am I understanding differences between Bayesian and frequentist inference correctly? | This questions is too broad, but I thought I would respond to a few points where your statements aren't accurate.
Bayesians (typically) believe there is a fixed value for the parameters, but use a probability distribution to represent their uncertainty about what the true value is.
A Bayesian is typically interested... | Am I understanding differences between Bayesian and frequentist inference correctly? | This questions is too broad, but I thought I would respond to a few points where your statements aren't accurate.
Bayesians (typically) believe there is a fixed value for the parameters, but use a p | Am I understanding differences between Bayesian and frequentist inference correctly?
This questions is too broad, but I thought I would respond to a few points where your statements aren't accurate.
Bayesians (typically) believe there is a fixed value for the parameters, but use a probability distribution to represen... | Am I understanding differences between Bayesian and frequentist inference correctly?
This questions is too broad, but I thought I would respond to a few points where your statements aren't accurate.
Bayesians (typically) believe there is a fixed value for the parameters, but use a p |
40,014 | Am I understanding differences between Bayesian and frequentist inference correctly? | A Bayesian would consider the results of the experiments fixed and consider population parameters as stochasts. This in contrast to frequentist, who see the data as "just another sample in an endless stream of samples" and who see the population parameters as fixed (but unknown).
The logical Bayesian order would be:
1... | Am I understanding differences between Bayesian and frequentist inference correctly? | A Bayesian would consider the results of the experiments fixed and consider population parameters as stochasts. This in contrast to frequentist, who see the data as "just another sample in an endless | Am I understanding differences between Bayesian and frequentist inference correctly?
A Bayesian would consider the results of the experiments fixed and consider population parameters as stochasts. This in contrast to frequentist, who see the data as "just another sample in an endless stream of samples" and who see the ... | Am I understanding differences between Bayesian and frequentist inference correctly?
A Bayesian would consider the results of the experiments fixed and consider population parameters as stochasts. This in contrast to frequentist, who see the data as "just another sample in an endless |
40,015 | Variance and covariance in the context of deterministic variables | All five questions have "yes" answers--but we have to be careful about what they mean.
"Variance of a deterministic variable."
Let's understand a "deterministic variable" to be a univariate dataset. It's just a bunch of values $X=x_1, x_2, \ldots, x_n$, with no probability model. By definition its variance is
$$\tex... | Variance and covariance in the context of deterministic variables | All five questions have "yes" answers--but we have to be careful about what they mean.
"Variance of a deterministic variable."
Let's understand a "deterministic variable" to be a univariate dataset. | Variance and covariance in the context of deterministic variables
All five questions have "yes" answers--but we have to be careful about what they mean.
"Variance of a deterministic variable."
Let's understand a "deterministic variable" to be a univariate dataset. It's just a bunch of values $X=x_1, x_2, \ldots, x_n$... | Variance and covariance in the context of deterministic variables
All five questions have "yes" answers--but we have to be careful about what they mean.
"Variance of a deterministic variable."
Let's understand a "deterministic variable" to be a univariate dataset. |
40,016 | Variance and covariance in the context of deterministic variables | The simple answer to your first three questions is no: it makes no sense in general to talk about variance or covariance involving a deterministic variable.
However, if we begin with a deterministic variable but then use some method of imposing a probability distribution on it, it then becomes a random variable, and th... | Variance and covariance in the context of deterministic variables | The simple answer to your first three questions is no: it makes no sense in general to talk about variance or covariance involving a deterministic variable.
However, if we begin with a deterministic v | Variance and covariance in the context of deterministic variables
The simple answer to your first three questions is no: it makes no sense in general to talk about variance or covariance involving a deterministic variable.
However, if we begin with a deterministic variable but then use some method of imposing a probabi... | Variance and covariance in the context of deterministic variables
The simple answer to your first three questions is no: it makes no sense in general to talk about variance or covariance involving a deterministic variable.
However, if we begin with a deterministic v |
40,017 | Understanding 'average slope' regression | Assuming:
$x_i$ and $x_j$ must always be distinct (for simplicity; otherwise the calculations require us to keep conditioning on the distinct pairs and I'd rather leave that out for now)
$y=\beta_0+\beta_1 x+\varepsilon$, with $E(\varepsilon)=0$
$x$'s fixed not random
then the pairwise slopes have expectation $\beta... | Understanding 'average slope' regression | Assuming:
$x_i$ and $x_j$ must always be distinct (for simplicity; otherwise the calculations require us to keep conditioning on the distinct pairs and I'd rather leave that out for now)
$y=\beta_0+ | Understanding 'average slope' regression
Assuming:
$x_i$ and $x_j$ must always be distinct (for simplicity; otherwise the calculations require us to keep conditioning on the distinct pairs and I'd rather leave that out for now)
$y=\beta_0+\beta_1 x+\varepsilon$, with $E(\varepsilon)=0$
$x$'s fixed not random
then th... | Understanding 'average slope' regression
Assuming:
$x_i$ and $x_j$ must always be distinct (for simplicity; otherwise the calculations require us to keep conditioning on the distinct pairs and I'd rather leave that out for now)
$y=\beta_0+ |
40,018 | Linear regression with sine/cosine elements | You simply compute $x_c=\cos(2\pi x)$ and $x_s=\sin(2\pi x)$ and perform a plain multiple linear regression of $y$ on $x, x_c,$ and $x_s$.
That is you supply the original $x$ and the two calculated predictors as if you had three independent variables for your regression, so your now-linear model is:
$$Y = \alpha + \bet... | Linear regression with sine/cosine elements | You simply compute $x_c=\cos(2\pi x)$ and $x_s=\sin(2\pi x)$ and perform a plain multiple linear regression of $y$ on $x, x_c,$ and $x_s$.
That is you supply the original $x$ and the two calculated pr | Linear regression with sine/cosine elements
You simply compute $x_c=\cos(2\pi x)$ and $x_s=\sin(2\pi x)$ and perform a plain multiple linear regression of $y$ on $x, x_c,$ and $x_s$.
That is you supply the original $x$ and the two calculated predictors as if you had three independent variables for your regression, so y... | Linear regression with sine/cosine elements
You simply compute $x_c=\cos(2\pi x)$ and $x_s=\sin(2\pi x)$ and perform a plain multiple linear regression of $y$ on $x, x_c,$ and $x_s$.
That is you supply the original $x$ and the two calculated pr |
40,019 | Linear regression with sine/cosine elements | You can find list of methods used for solving of linear regression problems in this article from Do Q Lee:
Numerically efficient methods for solving Least-Squares problems
Most commonly used methods for these kind of problems are:
Normal equations method using Cholesky factorization. It is the fastest method but numer... | Linear regression with sine/cosine elements | You can find list of methods used for solving of linear regression problems in this article from Do Q Lee:
Numerically efficient methods for solving Least-Squares problems
Most commonly used methods f | Linear regression with sine/cosine elements
You can find list of methods used for solving of linear regression problems in this article from Do Q Lee:
Numerically efficient methods for solving Least-Squares problems
Most commonly used methods for these kind of problems are:
Normal equations method using Cholesky facto... | Linear regression with sine/cosine elements
You can find list of methods used for solving of linear regression problems in this article from Do Q Lee:
Numerically efficient methods for solving Least-Squares problems
Most commonly used methods f |
40,020 | Interpreting percent variance explained in Random Forest output | Yes %explained variance is a measure of how well out-of-bag predictions explain the target variance of the training set. Unexplained variance would be to due true random behaviour or lack of fit.
%explained variance is retrieved by randomForest:::print.randomForest as last element in rf.fit$rsq and multiplied with 100.... | Interpreting percent variance explained in Random Forest output | Yes %explained variance is a measure of how well out-of-bag predictions explain the target variance of the training set. Unexplained variance would be to due true random behaviour or lack of fit.
%exp | Interpreting percent variance explained in Random Forest output
Yes %explained variance is a measure of how well out-of-bag predictions explain the target variance of the training set. Unexplained variance would be to due true random behaviour or lack of fit.
%explained variance is retrieved by randomForest:::print.ran... | Interpreting percent variance explained in Random Forest output
Yes %explained variance is a measure of how well out-of-bag predictions explain the target variance of the training set. Unexplained variance would be to due true random behaviour or lack of fit.
%exp |
40,021 | Interpreting percent variance explained in Random Forest output | To add some details to the content of the other answer, the formula to get the explained variance displayed in the summary is:
#fit.rf <- randomForest(...)
round(100 * fit.rf$rsq[length(fit.rf$rsq)], digits = 2)
You can check this by looking at what randomForest is printing with the command getAnywhere(print.random... | Interpreting percent variance explained in Random Forest output | To add some details to the content of the other answer, the formula to get the explained variance displayed in the summary is:
#fit.rf <- randomForest(...)
round(100 * fit.rf$rsq[length(fit.rf$rsq)], | Interpreting percent variance explained in Random Forest output
To add some details to the content of the other answer, the formula to get the explained variance displayed in the summary is:
#fit.rf <- randomForest(...)
round(100 * fit.rf$rsq[length(fit.rf$rsq)], digits = 2)
You can check this by looking at what rand... | Interpreting percent variance explained in Random Forest output
To add some details to the content of the other answer, the formula to get the explained variance displayed in the summary is:
#fit.rf <- randomForest(...)
round(100 * fit.rf$rsq[length(fit.rf$rsq)], |
40,022 | Interpreting percent variance explained in Random Forest output | This seems to be a misinterpretation of extending $R^2$ to more complicated situations than the usual in-sample OLS linear regression. In particular, the "propotion of variance explained" interpretation of $R^2$ is the exception, not the rule. As is derived in the link, that definition only applies when $\overset{N}{\u... | Interpreting percent variance explained in Random Forest output | This seems to be a misinterpretation of extending $R^2$ to more complicated situations than the usual in-sample OLS linear regression. In particular, the "propotion of variance explained" interpretati | Interpreting percent variance explained in Random Forest output
This seems to be a misinterpretation of extending $R^2$ to more complicated situations than the usual in-sample OLS linear regression. In particular, the "propotion of variance explained" interpretation of $R^2$ is the exception, not the rule. As is derive... | Interpreting percent variance explained in Random Forest output
This seems to be a misinterpretation of extending $R^2$ to more complicated situations than the usual in-sample OLS linear regression. In particular, the "propotion of variance explained" interpretati |
40,023 | Testing the variance component in a mixed effects model | This is usually done with maximum likelihood ratio between original model and a model omitting the variance coefficient to be estimate (random intercept/random slope/random co-variance between slope and intercept).
A good example is in these tutorials:
When model has more than one random coefficient: http://www.bodowi... | Testing the variance component in a mixed effects model | This is usually done with maximum likelihood ratio between original model and a model omitting the variance coefficient to be estimate (random intercept/random slope/random co-variance between slope a | Testing the variance component in a mixed effects model
This is usually done with maximum likelihood ratio between original model and a model omitting the variance coefficient to be estimate (random intercept/random slope/random co-variance between slope and intercept).
A good example is in these tutorials:
When model... | Testing the variance component in a mixed effects model
This is usually done with maximum likelihood ratio between original model and a model omitting the variance coefficient to be estimate (random intercept/random slope/random co-variance between slope a |
40,024 | Testing the variance component in a mixed effects model | Asymptotic test are problematic for variance parameters, because parameter space is bounded by zero. Moreover, the hypothesis you are trying to test, can't be true, as the parameter is continuous. Probability of $\sigma^2 = 0$ is exactly 0.
What you can do to make inference on the variance parameters is to switch to a ... | Testing the variance component in a mixed effects model | Asymptotic test are problematic for variance parameters, because parameter space is bounded by zero. Moreover, the hypothesis you are trying to test, can't be true, as the parameter is continuous. Pro | Testing the variance component in a mixed effects model
Asymptotic test are problematic for variance parameters, because parameter space is bounded by zero. Moreover, the hypothesis you are trying to test, can't be true, as the parameter is continuous. Probability of $\sigma^2 = 0$ is exactly 0.
What you can do to make... | Testing the variance component in a mixed effects model
Asymptotic test are problematic for variance parameters, because parameter space is bounded by zero. Moreover, the hypothesis you are trying to test, can't be true, as the parameter is continuous. Pro |
40,025 | Do logistic population growth models relate to binary logistic regressions? | The name logistic originally comes from the logistic growth equation:
$$ \frac{d N}{d t} = r N (1 - N) $$
which is a simple differential equation model for the growth of a population. The logistic function is its solution:
$$ N(t) = \frac{e^{rt}}{1 + e^{rt}} $$
Which has the attractive property that it is increasing, ... | Do logistic population growth models relate to binary logistic regressions? | The name logistic originally comes from the logistic growth equation:
$$ \frac{d N}{d t} = r N (1 - N) $$
which is a simple differential equation model for the growth of a population. The logistic fu | Do logistic population growth models relate to binary logistic regressions?
The name logistic originally comes from the logistic growth equation:
$$ \frac{d N}{d t} = r N (1 - N) $$
which is a simple differential equation model for the growth of a population. The logistic function is its solution:
$$ N(t) = \frac{e^{r... | Do logistic population growth models relate to binary logistic regressions?
The name logistic originally comes from the logistic growth equation:
$$ \frac{d N}{d t} = r N (1 - N) $$
which is a simple differential equation model for the growth of a population. The logistic fu |
40,026 | $\chi^2 $ of multidimensional data | To analyze a multi-way contingency table, you use log-linear models. In truth, log-linear models are a special case of the Poisson generalized linear model, so you could do that, but log-linear models are more user-friendly. In Python, you may need to use the Poisson GLM, as I gather log-linear models may not be impl... | $\chi^2 $ of multidimensional data | To analyze a multi-way contingency table, you use log-linear models. In truth, log-linear models are a special case of the Poisson generalized linear model, so you could do that, but log-linear model | $\chi^2 $ of multidimensional data
To analyze a multi-way contingency table, you use log-linear models. In truth, log-linear models are a special case of the Poisson generalized linear model, so you could do that, but log-linear models are more user-friendly. In Python, you may need to use the Poisson GLM, as I gathe... | $\chi^2 $ of multidimensional data
To analyze a multi-way contingency table, you use log-linear models. In truth, log-linear models are a special case of the Poisson generalized linear model, so you could do that, but log-linear model |
40,027 | $\chi^2 $ of multidimensional data | I found the answer here under 5: "Three-Way Tables". Obviously, the term I was missing is three-way contingency tables. Determining the expected values in a three dimensional contingency table is actually pretty much analogue to the standard variant.
In normal contingency tables you get the expected values by multiplyi... | $\chi^2 $ of multidimensional data | I found the answer here under 5: "Three-Way Tables". Obviously, the term I was missing is three-way contingency tables. Determining the expected values in a three dimensional contingency table is actu | $\chi^2 $ of multidimensional data
I found the answer here under 5: "Three-Way Tables". Obviously, the term I was missing is three-way contingency tables. Determining the expected values in a three dimensional contingency table is actually pretty much analogue to the standard variant.
In normal contingency tables you g... | $\chi^2 $ of multidimensional data
I found the answer here under 5: "Three-Way Tables". Obviously, the term I was missing is three-way contingency tables. Determining the expected values in a three dimensional contingency table is actu |
40,028 | Stochastic Differential Equations - A Few General Questions | The single best introduction to SDE from numerical angle is this Higham's paper. It will probably give you an approximation to answers to your three questions.
a) In finance the assets are not priced on the basis of knowing exactly their cash flows. Moreover, they're not priced on the basis of knowing their expected ca... | Stochastic Differential Equations - A Few General Questions | The single best introduction to SDE from numerical angle is this Higham's paper. It will probably give you an approximation to answers to your three questions.
a) In finance the assets are not priced | Stochastic Differential Equations - A Few General Questions
The single best introduction to SDE from numerical angle is this Higham's paper. It will probably give you an approximation to answers to your three questions.
a) In finance the assets are not priced on the basis of knowing exactly their cash flows. Moreover, ... | Stochastic Differential Equations - A Few General Questions
The single best introduction to SDE from numerical angle is this Higham's paper. It will probably give you an approximation to answers to your three questions.
a) In finance the assets are not priced |
40,029 | Stochastic Differential Equations - A Few General Questions | a) What is the point of simulating SDEs if the solution is always
different due to the randomness of the Wiener process? I have been
simulating Geometric Brownian Motion and read that it is used in the
Black-Scholes model in finance, so how do they actually price stocks
based on the SDE?
The Black-Scholes mode... | Stochastic Differential Equations - A Few General Questions | a) What is the point of simulating SDEs if the solution is always
different due to the randomness of the Wiener process? I have been
simulating Geometric Brownian Motion and read that it is used i | Stochastic Differential Equations - A Few General Questions
a) What is the point of simulating SDEs if the solution is always
different due to the randomness of the Wiener process? I have been
simulating Geometric Brownian Motion and read that it is used in the
Black-Scholes model in finance, so how do they actua... | Stochastic Differential Equations - A Few General Questions
a) What is the point of simulating SDEs if the solution is always
different due to the randomness of the Wiener process? I have been
simulating Geometric Brownian Motion and read that it is used i |
40,030 | Stochastic Differential Equations - A Few General Questions | "What is the point of sampling random variables at all if the solution is always different due to the randomness..."? When you're simulating an SDE (or any stochastic process in general), you're sampling from a certain distribution of sample paths. (More precisely, a probability measure defined on an infinite dimension... | Stochastic Differential Equations - A Few General Questions | "What is the point of sampling random variables at all if the solution is always different due to the randomness..."? When you're simulating an SDE (or any stochastic process in general), you're sampl | Stochastic Differential Equations - A Few General Questions
"What is the point of sampling random variables at all if the solution is always different due to the randomness..."? When you're simulating an SDE (or any stochastic process in general), you're sampling from a certain distribution of sample paths. (More preci... | Stochastic Differential Equations - A Few General Questions
"What is the point of sampling random variables at all if the solution is always different due to the randomness..."? When you're simulating an SDE (or any stochastic process in general), you're sampl |
40,031 | 2SLS - logit/probit in the second stage? | The reference for this should be Newey (1987) "Efficient estimation of limited dependent variable models with endogenous explanatory variables", Journal of Econometrics, Vol. 36(3), pp. 231–250 (link). This is the estimator that is implemented with the probitiv command in Stata, for instance, where you can have an OLS ... | 2SLS - logit/probit in the second stage? | The reference for this should be Newey (1987) "Efficient estimation of limited dependent variable models with endogenous explanatory variables", Journal of Econometrics, Vol. 36(3), pp. 231–250 (link) | 2SLS - logit/probit in the second stage?
The reference for this should be Newey (1987) "Efficient estimation of limited dependent variable models with endogenous explanatory variables", Journal of Econometrics, Vol. 36(3), pp. 231–250 (link). This is the estimator that is implemented with the probitiv command in Stata,... | 2SLS - logit/probit in the second stage?
The reference for this should be Newey (1987) "Efficient estimation of limited dependent variable models with endogenous explanatory variables", Journal of Econometrics, Vol. 36(3), pp. 231–250 (link) |
40,032 | 2SLS - logit/probit in the second stage? | When googling this problem myself, I found the highly-cited article
Terza, J.V., Basu, A. and Rathouz, P.J., 2008. Two-stage residual inclusion estimation: addressing endogeneity in health econometric modeling. Journal of health economics, 27(3), pp.531-543.
which proposes to use a method called 2-stage residual inc... | 2SLS - logit/probit in the second stage? | When googling this problem myself, I found the highly-cited article
Terza, J.V., Basu, A. and Rathouz, P.J., 2008. Two-stage residual inclusion estimation: addressing endogeneity in health econometr | 2SLS - logit/probit in the second stage?
When googling this problem myself, I found the highly-cited article
Terza, J.V., Basu, A. and Rathouz, P.J., 2008. Two-stage residual inclusion estimation: addressing endogeneity in health econometric modeling. Journal of health economics, 27(3), pp.531-543.
which proposes to... | 2SLS - logit/probit in the second stage?
When googling this problem myself, I found the highly-cited article
Terza, J.V., Basu, A. and Rathouz, P.J., 2008. Two-stage residual inclusion estimation: addressing endogeneity in health econometr |
40,033 | R gives me the error "contrasts can be applied only to factors with 2 or more levels" running an mlogit model, but all my factors have 2 levels [closed] | See Answer Here - https://stackoverflow.com/questions/18171246/error-in-contrasts-when-defining-a-linear-model-in-r
There are factors that you are using that either have only 1 distinct value or are NA | R gives me the error "contrasts can be applied only to factors with 2 or more levels" running an mlo | See Answer Here - https://stackoverflow.com/questions/18171246/error-in-contrasts-when-defining-a-linear-model-in-r
There are factors that you are using that either have only 1 distinct value or are N | R gives me the error "contrasts can be applied only to factors with 2 or more levels" running an mlogit model, but all my factors have 2 levels [closed]
See Answer Here - https://stackoverflow.com/questions/18171246/error-in-contrasts-when-defining-a-linear-model-in-r
There are factors that you are using that either ha... | R gives me the error "contrasts can be applied only to factors with 2 or more levels" running an mlo
See Answer Here - https://stackoverflow.com/questions/18171246/error-in-contrasts-when-defining-a-linear-model-in-r
There are factors that you are using that either have only 1 distinct value or are N |
40,034 | Transformation of Random Variable - Normal Distribution | Since this is self-study I will give you a hint.
$F_{|X|} = P(|X|<x) = P(-x<X<x) = F_X(x) - F_X(-x)$ where $F$ indicates the CDF of the appropriate random variable. Since we know the CDF of a normal random variable, $X$, we now know the CDF the absolute value of $X$. Differentiating will give us the PDF. | Transformation of Random Variable - Normal Distribution | Since this is self-study I will give you a hint.
$F_{|X|} = P(|X|<x) = P(-x<X<x) = F_X(x) - F_X(-x)$ where $F$ indicates the CDF of the appropriate random variable. Since we know the CDF of a normal | Transformation of Random Variable - Normal Distribution
Since this is self-study I will give you a hint.
$F_{|X|} = P(|X|<x) = P(-x<X<x) = F_X(x) - F_X(-x)$ where $F$ indicates the CDF of the appropriate random variable. Since we know the CDF of a normal random variable, $X$, we now know the CDF the absolute value of ... | Transformation of Random Variable - Normal Distribution
Since this is self-study I will give you a hint.
$F_{|X|} = P(|X|<x) = P(-x<X<x) = F_X(x) - F_X(-x)$ where $F$ indicates the CDF of the appropriate random variable. Since we know the CDF of a normal |
40,035 | Transformation of Random Variable - Normal Distribution | I will try to give you some good hints towards the simple solution:
The distribution for $Y=|X|$ must be 0 for $Y<0$.
As $X$ is symmetric about 0 we know that $f_X(-x)=f_X(x)$.
$Y$ can take the value $y\ge 0$ if $x=y$ or $x=-y$.
Hopefully that will get you on your way. | Transformation of Random Variable - Normal Distribution | I will try to give you some good hints towards the simple solution:
The distribution for $Y=|X|$ must be 0 for $Y<0$.
As $X$ is symmetric about 0 we know that $f_X(-x)=f_X(x)$.
$Y$ can take the value | Transformation of Random Variable - Normal Distribution
I will try to give you some good hints towards the simple solution:
The distribution for $Y=|X|$ must be 0 for $Y<0$.
As $X$ is symmetric about 0 we know that $f_X(-x)=f_X(x)$.
$Y$ can take the value $y\ge 0$ if $x=y$ or $x=-y$.
Hopefully that will get you on yo... | Transformation of Random Variable - Normal Distribution
I will try to give you some good hints towards the simple solution:
The distribution for $Y=|X|$ must be 0 for $Y<0$.
As $X$ is symmetric about 0 we know that $f_X(-x)=f_X(x)$.
$Y$ can take the value |
40,036 | Transformation of Random Variable - Normal Distribution | 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.
You just have to double the Right branch and nullify t... | Transformation of Random Variable - Normal Distribution | 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.
| Transformation of Random Variable - Normal Distribution
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.
... | Transformation of Random Variable - Normal Distribution
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.
|
40,037 | Is it possible to use SD instead of entropy? | Why can't we just compute the standard deviation?
Here's why. Let's compare the formulas for entropy and variance:
$H(X) = - \sum\limits_x p(x) \, \log p(x) = - \mathbb E \, [ \log p(X) ]$
$\text{var} (X) = \mathbb E \, \Big[(X - \mathbb E[X])^2 \Big]$
So note that entropy does not care about values that $X$ may tak... | Is it possible to use SD instead of entropy? | Why can't we just compute the standard deviation?
Here's why. Let's compare the formulas for entropy and variance:
$H(X) = - \sum\limits_x p(x) \, \log p(x) = - \mathbb E \, [ \log p(X) ]$
$\text{va | Is it possible to use SD instead of entropy?
Why can't we just compute the standard deviation?
Here's why. Let's compare the formulas for entropy and variance:
$H(X) = - \sum\limits_x p(x) \, \log p(x) = - \mathbb E \, [ \log p(X) ]$
$\text{var} (X) = \mathbb E \, \Big[(X - \mathbb E[X])^2 \Big]$
So note that entrop... | Is it possible to use SD instead of entropy?
Why can't we just compute the standard deviation?
Here's why. Let's compare the formulas for entropy and variance:
$H(X) = - \sum\limits_x p(x) \, \log p(x) = - \mathbb E \, [ \log p(X) ]$
$\text{va |
40,038 | Is it possible to use SD instead of entropy? | When comparing standard deviation and entropy in context of statistics and related areas, I think that it is important to realize the difference between two notions of the entropy concept: entropy as a measure of variability, volatility, chaos (this meaning is usually implied in physics and similar domains) and entropy... | Is it possible to use SD instead of entropy? | When comparing standard deviation and entropy in context of statistics and related areas, I think that it is important to realize the difference between two notions of the entropy concept: entropy as | Is it possible to use SD instead of entropy?
When comparing standard deviation and entropy in context of statistics and related areas, I think that it is important to realize the difference between two notions of the entropy concept: entropy as a measure of variability, volatility, chaos (this meaning is usually implie... | Is it possible to use SD instead of entropy?
When comparing standard deviation and entropy in context of statistics and related areas, I think that it is important to realize the difference between two notions of the entropy concept: entropy as |
40,039 | I need both quadratic and linear coefficients in a GLM with binary response. What's the best option? [closed] | You can add a quadratic term with logistic regression just as you can with regular old linear regression. That is a simple way to include a 'curve' in your model. Be sure you understand what that means. I suspect you want an R tutorial, which is off-topic on CV. The basic approach to adding a quadratic in R is to i... | I need both quadratic and linear coefficients in a GLM with binary response. What's the best option? | You can add a quadratic term with logistic regression just as you can with regular old linear regression. That is a simple way to include a 'curve' in your model. Be sure you understand what that me | I need both quadratic and linear coefficients in a GLM with binary response. What's the best option? [closed]
You can add a quadratic term with logistic regression just as you can with regular old linear regression. That is a simple way to include a 'curve' in your model. Be sure you understand what that means. I su... | I need both quadratic and linear coefficients in a GLM with binary response. What's the best option?
You can add a quadratic term with logistic regression just as you can with regular old linear regression. That is a simple way to include a 'curve' in your model. Be sure you understand what that me |
40,040 | Bootstrap two-sample t test | As @Tim notes, your bootsamples should have the same $n_j$s as your original data.
Next, recognize that there are several ways to bootstrap: e.g., you can bootstrap your data directly or bootstrap a test statistic, you can bootstrap your sampling distribution or a null distribution, etc. You need to make sure you un... | Bootstrap two-sample t test | As @Tim notes, your bootsamples should have the same $n_j$s as your original data.
Next, recognize that there are several ways to bootstrap: e.g., you can bootstrap your data directly or bootstrap a | Bootstrap two-sample t test
As @Tim notes, your bootsamples should have the same $n_j$s as your original data.
Next, recognize that there are several ways to bootstrap: e.g., you can bootstrap your data directly or bootstrap a test statistic, you can bootstrap your sampling distribution or a null distribution, etc. ... | Bootstrap two-sample t test
As @Tim notes, your bootsamples should have the same $n_j$s as your original data.
Next, recognize that there are several ways to bootstrap: e.g., you can bootstrap your data directly or bootstrap a |
40,041 | Bootstrap two-sample t test | First of all, with bootstrap you sample $N$ cases out of $N$ cases in your data. So the number of observations to choose is simple.
And, yes, you can do:
f <- function() {
A <- sample(group_A, 377, replace=T)
B <- sample(group_B, 377, replace=T)
mean(A)-mean(B)
}
replicate(1000, f())
but this is a different ... | Bootstrap two-sample t test | First of all, with bootstrap you sample $N$ cases out of $N$ cases in your data. So the number of observations to choose is simple.
And, yes, you can do:
f <- function() {
A <- sample(group_A, 377, | Bootstrap two-sample t test
First of all, with bootstrap you sample $N$ cases out of $N$ cases in your data. So the number of observations to choose is simple.
And, yes, you can do:
f <- function() {
A <- sample(group_A, 377, replace=T)
B <- sample(group_B, 377, replace=T)
mean(A)-mean(B)
}
replicate(1000, f(... | Bootstrap two-sample t test
First of all, with bootstrap you sample $N$ cases out of $N$ cases in your data. So the number of observations to choose is simple.
And, yes, you can do:
f <- function() {
A <- sample(group_A, 377, |
40,042 | Bootstrap two-sample t test | One more thing: if you're not assuming equal variance, you might want to consider Welch's t-test (http://beheco.oxfordjournals.org/content/17/4/688.full) | Bootstrap two-sample t test | One more thing: if you're not assuming equal variance, you might want to consider Welch's t-test (http://beheco.oxfordjournals.org/content/17/4/688.full) | Bootstrap two-sample t test
One more thing: if you're not assuming equal variance, you might want to consider Welch's t-test (http://beheco.oxfordjournals.org/content/17/4/688.full) | Bootstrap two-sample t test
One more thing: if you're not assuming equal variance, you might want to consider Welch's t-test (http://beheco.oxfordjournals.org/content/17/4/688.full) |
40,043 | Bootstrap two-sample t test | I would condition on the total sample size, not fix the group sizes. Of course, if your group sizes are fixed in advance then do condition on those, my answer is for the case where they are not.
I construct a two column matrix with the first column being an indication of group membership and the second column the obser... | Bootstrap two-sample t test | I would condition on the total sample size, not fix the group sizes. Of course, if your group sizes are fixed in advance then do condition on those, my answer is for the case where they are not.
I con | Bootstrap two-sample t test
I would condition on the total sample size, not fix the group sizes. Of course, if your group sizes are fixed in advance then do condition on those, my answer is for the case where they are not.
I construct a two column matrix with the first column being an indication of group membership and... | Bootstrap two-sample t test
I would condition on the total sample size, not fix the group sizes. Of course, if your group sizes are fixed in advance then do condition on those, my answer is for the case where they are not.
I con |
40,044 | Why do p values for test of likelihood ratio vs Fisher's Exact Test not agree | You have a few issues here. First, understanding what each test is doing, and second interpreting the p-values.
First, each test has different underlying assumptions. The likelihood ratio test statistic is formed by taking the log of the ratio of the likelihood under the null model, divided by the alternative model. ... | Why do p values for test of likelihood ratio vs Fisher's Exact Test not agree | You have a few issues here. First, understanding what each test is doing, and second interpreting the p-values.
First, each test has different underlying assumptions. The likelihood ratio test stati | Why do p values for test of likelihood ratio vs Fisher's Exact Test not agree
You have a few issues here. First, understanding what each test is doing, and second interpreting the p-values.
First, each test has different underlying assumptions. The likelihood ratio test statistic is formed by taking the log of the ra... | Why do p values for test of likelihood ratio vs Fisher's Exact Test not agree
You have a few issues here. First, understanding what each test is doing, and second interpreting the p-values.
First, each test has different underlying assumptions. The likelihood ratio test stati |
40,045 | How to bound a probability with Chernoff's inequality? | Your class is using needlessly complicated expressions for the Chernoff bound
and apparently giving them to you as magical formulas to be applied without
any understanding of how they came about.
Suppose that $X$ is a random variable for which we wish to compute $P\{X \geq t\}$. One way of doing this is to define a re... | How to bound a probability with Chernoff's inequality? | Your class is using needlessly complicated expressions for the Chernoff bound
and apparently giving them to you as magical formulas to be applied without
any understanding of how they came about.
Supp | How to bound a probability with Chernoff's inequality?
Your class is using needlessly complicated expressions for the Chernoff bound
and apparently giving them to you as magical formulas to be applied without
any understanding of how they came about.
Suppose that $X$ is a random variable for which we wish to compute $P... | How to bound a probability with Chernoff's inequality?
Your class is using needlessly complicated expressions for the Chernoff bound
and apparently giving them to you as magical formulas to be applied without
any understanding of how they came about.
Supp |
40,046 | cluster-robust standard errors are smaller than unclustered ones in fgls with cluster fixed effects | Robust clustered standard errors can change your standard errors in both directions. That is, clustered standard errors can be larger or smaller than conventional standard errors. The direction in which standard errors will change depends on the sign of the intra-class correlation. This post explains robust standard er... | cluster-robust standard errors are smaller than unclustered ones in fgls with cluster fixed effects | Robust clustered standard errors can change your standard errors in both directions. That is, clustered standard errors can be larger or smaller than conventional standard errors. The direction in whi | cluster-robust standard errors are smaller than unclustered ones in fgls with cluster fixed effects
Robust clustered standard errors can change your standard errors in both directions. That is, clustered standard errors can be larger or smaller than conventional standard errors. The direction in which standard errors w... | cluster-robust standard errors are smaller than unclustered ones in fgls with cluster fixed effects
Robust clustered standard errors can change your standard errors in both directions. That is, clustered standard errors can be larger or smaller than conventional standard errors. The direction in whi |
40,047 | Power-law fitting and testing | As one of the authors of the methods you're using, I can say with some certainty that the answer to your Question 1 (can you apply the fitting and hypothesis-test methods to a dataset that contains all recorded events in a system) is "yes". In fact, in the 24 datasets that we analyzed in Clauset, Shalizi and Newman, "P... | Power-law fitting and testing | As one of the authors of the methods you're using, I can say with some certainty that the answer to your Question 1 (can you apply the fitting and hypothesis-test methods to a dataset that contains al | Power-law fitting and testing
As one of the authors of the methods you're using, I can say with some certainty that the answer to your Question 1 (can you apply the fitting and hypothesis-test methods to a dataset that contains all recorded events in a system) is "yes". In fact, in the 24 datasets that we analyzed in C... | Power-law fitting and testing
As one of the authors of the methods you're using, I can say with some certainty that the answer to your Question 1 (can you apply the fitting and hypothesis-test methods to a dataset that contains al |
40,048 | Can I model standard deviations in a linear model? | It sounds like you are proposing essentially a two-stage least squares, where stage one reduces each cluster to its standard deviation about a cluster-specific mean. This seems fine, although note that you could actually model on the observational level, ie, let the variance for each observation be a linear function o... | Can I model standard deviations in a linear model? | It sounds like you are proposing essentially a two-stage least squares, where stage one reduces each cluster to its standard deviation about a cluster-specific mean. This seems fine, although note th | Can I model standard deviations in a linear model?
It sounds like you are proposing essentially a two-stage least squares, where stage one reduces each cluster to its standard deviation about a cluster-specific mean. This seems fine, although note that you could actually model on the observational level, ie, let the v... | Can I model standard deviations in a linear model?
It sounds like you are proposing essentially a two-stage least squares, where stage one reduces each cluster to its standard deviation about a cluster-specific mean. This seems fine, although note th |
40,049 | Can I model standard deviations in a linear model? | Yes, you can do this. A GLM of the SDs with a log link and a gamma family is one way to do it, if you think the populations are normal.
It is also not uncommon for people to regress log SD on a bunch of predictors. It is approximate, but all models are. One text where you can see this being done is Box, Hunter, and Hu... | Can I model standard deviations in a linear model? | Yes, you can do this. A GLM of the SDs with a log link and a gamma family is one way to do it, if you think the populations are normal.
It is also not uncommon for people to regress log SD on a bunch | Can I model standard deviations in a linear model?
Yes, you can do this. A GLM of the SDs with a log link and a gamma family is one way to do it, if you think the populations are normal.
It is also not uncommon for people to regress log SD on a bunch of predictors. It is approximate, but all models are. One text where... | Can I model standard deviations in a linear model?
Yes, you can do this. A GLM of the SDs with a log link and a gamma family is one way to do it, if you think the populations are normal.
It is also not uncommon for people to regress log SD on a bunch |
40,050 | Problem with Mann-Whitney U test in scipy | There might be a bug in the package, but if you store the u and the prob output separately, you'll see the u value, although the prob is missing for some reason.
u, prob=scipy.stats.mannwhitneyu(x,y)
u
Out[18]: 193405.5
prob
Out[19]: nan
You could then use the normal approximation of $U$ to get a p-value, though. F... | Problem with Mann-Whitney U test in scipy | There might be a bug in the package, but if you store the u and the prob output separately, you'll see the u value, although the prob is missing for some reason.
u, prob=scipy.stats.mannwhitneyu(x,y) | Problem with Mann-Whitney U test in scipy
There might be a bug in the package, but if you store the u and the prob output separately, you'll see the u value, although the prob is missing for some reason.
u, prob=scipy.stats.mannwhitneyu(x,y)
u
Out[18]: 193405.5
prob
Out[19]: nan
You could then use the normal approx... | Problem with Mann-Whitney U test in scipy
There might be a bug in the package, but if you store the u and the prob output separately, you'll see the u value, although the prob is missing for some reason.
u, prob=scipy.stats.mannwhitneyu(x,y) |
40,051 | Problem with Mann-Whitney U test in scipy | This thread is old, but for those like me who encounter this scipy bug and find themselves here -
The issue is indeed the tiecorrect function in scipy.stats.mannwhitneyu(x,y)
If you do not need tie correcting, the scipy.stats.ranksums(x,y) test works fine.
For your case:
import scipy.stats
x = [1.] * 163 + [2.] * 8... | Problem with Mann-Whitney U test in scipy | This thread is old, but for those like me who encounter this scipy bug and find themselves here -
The issue is indeed the tiecorrect function in scipy.stats.mannwhitneyu(x,y)
If you do not need tie c | Problem with Mann-Whitney U test in scipy
This thread is old, but for those like me who encounter this scipy bug and find themselves here -
The issue is indeed the tiecorrect function in scipy.stats.mannwhitneyu(x,y)
If you do not need tie correcting, the scipy.stats.ranksums(x,y) test works fine.
For your case:
imp... | Problem with Mann-Whitney U test in scipy
This thread is old, but for those like me who encounter this scipy bug and find themselves here -
The issue is indeed the tiecorrect function in scipy.stats.mannwhitneyu(x,y)
If you do not need tie c |
40,052 | Using a chi square test instead of a F test in a linear regression | They are closely related. If you divide the Wald statistic by its degrees of freedom, you in essence have an $F$ statistic with that many numerator df, and infinite denominator df. The Wald statistic is seen in cases where the error variance is known, or where asymptotic (large-sample) approximations are used. Seems su... | Using a chi square test instead of a F test in a linear regression | They are closely related. If you divide the Wald statistic by its degrees of freedom, you in essence have an $F$ statistic with that many numerator df, and infinite denominator df. The Wald statistic | Using a chi square test instead of a F test in a linear regression
They are closely related. If you divide the Wald statistic by its degrees of freedom, you in essence have an $F$ statistic with that many numerator df, and infinite denominator df. The Wald statistic is seen in cases where the error variance is known, o... | Using a chi square test instead of a F test in a linear regression
They are closely related. If you divide the Wald statistic by its degrees of freedom, you in essence have an $F$ statistic with that many numerator df, and infinite denominator df. The Wald statistic |
40,053 | Using a chi square test instead of a F test in a linear regression | This is analogous to the $z$-test vs the $t$-test in the univariate case, where if the variance is known, the distribution of the test statistic is normal ($z$-test), and if it is estimated, the distribution of the test statistic is $t$ ($t$-test), with the $t$-test converging to the $z$-test with large n.
Same thing i... | Using a chi square test instead of a F test in a linear regression | This is analogous to the $z$-test vs the $t$-test in the univariate case, where if the variance is known, the distribution of the test statistic is normal ($z$-test), and if it is estimated, the distr | Using a chi square test instead of a F test in a linear regression
This is analogous to the $z$-test vs the $t$-test in the univariate case, where if the variance is known, the distribution of the test statistic is normal ($z$-test), and if it is estimated, the distribution of the test statistic is $t$ ($t$-test), with... | Using a chi square test instead of a F test in a linear regression
This is analogous to the $z$-test vs the $t$-test in the univariate case, where if the variance is known, the distribution of the test statistic is normal ($z$-test), and if it is estimated, the distr |
40,054 | Name for the special estimate of the mean | It's called the midrange.
It's a good way to estimate the population mean of a $\text{Unif}(\mu-\theta,\mu+\theta)$.
It may be quite good in a variety of other circumstances; they'll generally be ones where the density is both symmetric and 'cuts off' relatively quickly at the bounds, rather than ones that very smoothl... | Name for the special estimate of the mean | It's called the midrange.
It's a good way to estimate the population mean of a $\text{Unif}(\mu-\theta,\mu+\theta)$.
It may be quite good in a variety of other circumstances; they'll generally be ones | Name for the special estimate of the mean
It's called the midrange.
It's a good way to estimate the population mean of a $\text{Unif}(\mu-\theta,\mu+\theta)$.
It may be quite good in a variety of other circumstances; they'll generally be ones where the density is both symmetric and 'cuts off' relatively quickly at the ... | Name for the special estimate of the mean
It's called the midrange.
It's a good way to estimate the population mean of a $\text{Unif}(\mu-\theta,\mu+\theta)$.
It may be quite good in a variety of other circumstances; they'll generally be ones |
40,055 | Standard error of the sampling distribution of the mean | The quoted formula is not quite right. Let's derive the correct one.
Since the population mean (or any other constant) may be subtracted from every value in a population $S$ without changing the variance of the population or of any sample thereof, we might as well assume the population mean is zero. Letting the value... | Standard error of the sampling distribution of the mean | The quoted formula is not quite right. Let's derive the correct one.
Since the population mean (or any other constant) may be subtracted from every value in a population $S$ without changing the vari | Standard error of the sampling distribution of the mean
The quoted formula is not quite right. Let's derive the correct one.
Since the population mean (or any other constant) may be subtracted from every value in a population $S$ without changing the variance of the population or of any sample thereof, we might as wel... | Standard error of the sampling distribution of the mean
The quoted formula is not quite right. Let's derive the correct one.
Since the population mean (or any other constant) may be subtracted from every value in a population $S$ without changing the vari |
40,056 | Standard error of the sampling distribution of the mean | This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write
$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac... | Standard error of the sampling distribution of the mean | This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write
$$\sigma_{\bar{X}} = \sigma \sqrt{\frac | Standard error of the sampling distribution of the mean
This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write
$$\sigma_{\bar{X}} = \sigma \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\fra... | Standard error of the sampling distribution of the mean
This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write
$$\sigma_{\bar{X}} = \sigma \sqrt{\frac |
40,057 | How are standard errors affected in a multivariate regression? | Yes, adding controls can increase the power of your statistical tests and make standard errors smaller. To see this, consider the following two regressions for comparison:
$$
\begin{align}
Y_i &= \alpha + \beta D_i + X'_i\gamma +e_i \newline
Y_i &= \mu + \pi D_i + u_i
\end{align}
$$
Assume that $X,D$ are uncorrelated w... | How are standard errors affected in a multivariate regression? | Yes, adding controls can increase the power of your statistical tests and make standard errors smaller. To see this, consider the following two regressions for comparison:
$$
\begin{align}
Y_i &= \alp | How are standard errors affected in a multivariate regression?
Yes, adding controls can increase the power of your statistical tests and make standard errors smaller. To see this, consider the following two regressions for comparison:
$$
\begin{align}
Y_i &= \alpha + \beta D_i + X'_i\gamma +e_i \newline
Y_i &= \mu + \p... | How are standard errors affected in a multivariate regression?
Yes, adding controls can increase the power of your statistical tests and make standard errors smaller. To see this, consider the following two regressions for comparison:
$$
\begin{align}
Y_i &= \alp |
40,058 | Linear regression with upper and/or lower limits in R? | Yes, you can do this in Lavaan.
Here's an example. We fit a regression model, and find an estimate of 0.10. Then fit a model using lavaan, and get the same parameter estimate. Then fit the model with a constraint that b1 has to be greater than 0.
library(lavaan)
set.seed(1234)
df <- as.data.frame(matrix(rnorm(500), nco... | Linear regression with upper and/or lower limits in R? | Yes, you can do this in Lavaan.
Here's an example. We fit a regression model, and find an estimate of 0.10. Then fit a model using lavaan, and get the same parameter estimate. Then fit the model with | Linear regression with upper and/or lower limits in R?
Yes, you can do this in Lavaan.
Here's an example. We fit a regression model, and find an estimate of 0.10. Then fit a model using lavaan, and get the same parameter estimate. Then fit the model with a constraint that b1 has to be greater than 0.
library(lavaan)
se... | Linear regression with upper and/or lower limits in R?
Yes, you can do this in Lavaan.
Here's an example. We fit a regression model, and find an estimate of 0.10. Then fit a model using lavaan, and get the same parameter estimate. Then fit the model with |
40,059 | Linear regression with upper and/or lower limits in R? | Yes, you can do it by re-defining linear regression as an optimization problem (a residual sum of squares cost function for example) and solving it using the constraints you want. So to use the numbers as the example from @Jeremy:
(Note that in this example we are fitting a model without an intercept)
set.seed(1234)
df... | Linear regression with upper and/or lower limits in R? | Yes, you can do it by re-defining linear regression as an optimization problem (a residual sum of squares cost function for example) and solving it using the constraints you want. So to use the number | Linear regression with upper and/or lower limits in R?
Yes, you can do it by re-defining linear regression as an optimization problem (a residual sum of squares cost function for example) and solving it using the constraints you want. So to use the numbers as the example from @Jeremy:
(Note that in this example we are ... | Linear regression with upper and/or lower limits in R?
Yes, you can do it by re-defining linear regression as an optimization problem (a residual sum of squares cost function for example) and solving it using the constraints you want. So to use the number |
40,060 | Can averaging all the variables be seen as a crude form of PCA? | PCA forms linear combinations of the variables, and averaging all the variables is also taking a linear combination -- namely one where all the weights are equal to $1/d$, where $d$ is the number of variables. So one can view these approaches as conceptually related.
Moreover, under certain conditions averaging can ind... | Can averaging all the variables be seen as a crude form of PCA? | PCA forms linear combinations of the variables, and averaging all the variables is also taking a linear combination -- namely one where all the weights are equal to $1/d$, where $d$ is the number of v | Can averaging all the variables be seen as a crude form of PCA?
PCA forms linear combinations of the variables, and averaging all the variables is also taking a linear combination -- namely one where all the weights are equal to $1/d$, where $d$ is the number of variables. So one can view these approaches as conceptual... | Can averaging all the variables be seen as a crude form of PCA?
PCA forms linear combinations of the variables, and averaging all the variables is also taking a linear combination -- namely one where all the weights are equal to $1/d$, where $d$ is the number of v |
40,061 | Warning message in auto.arima | It looks fine, auto.arima() tries many candidate models. One of them may have been dodgy.
The auto.arima() algorithm follows Hyndman & Khandakar (2008) Automatic time series forecasting (pdf), although the OCSB test is a new development. The algorithm tries different versions of p, q, P and Q and chooses the one with t... | Warning message in auto.arima | It looks fine, auto.arima() tries many candidate models. One of them may have been dodgy.
The auto.arima() algorithm follows Hyndman & Khandakar (2008) Automatic time series forecasting (pdf), althoug | Warning message in auto.arima
It looks fine, auto.arima() tries many candidate models. One of them may have been dodgy.
The auto.arima() algorithm follows Hyndman & Khandakar (2008) Automatic time series forecasting (pdf), although the OCSB test is a new development. The algorithm tries different versions of p, q, P an... | Warning message in auto.arima
It looks fine, auto.arima() tries many candidate models. One of them may have been dodgy.
The auto.arima() algorithm follows Hyndman & Khandakar (2008) Automatic time series forecasting (pdf), althoug |
40,062 | Warning message in auto.arima | [I think this question should be placed on stackexchange]
You should try:
auto.arima(forecast_data_ts, approximation=FALSE,trace=FALSE) | Warning message in auto.arima | [I think this question should be placed on stackexchange]
You should try:
auto.arima(forecast_data_ts, approximation=FALSE,trace=FALSE) | Warning message in auto.arima
[I think this question should be placed on stackexchange]
You should try:
auto.arima(forecast_data_ts, approximation=FALSE,trace=FALSE) | Warning message in auto.arima
[I think this question should be placed on stackexchange]
You should try:
auto.arima(forecast_data_ts, approximation=FALSE,trace=FALSE) |
40,063 | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS) | Both answers posted so far are useful (+1) but let me present this in a slightly different way using the Minimum description length principal. The basic idea behind MDL is related to Kolmogorov Complexity and the concept of the minimum sized program required to reproduce a sequence. The MDL principle states that one sh... | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS) | Both answers posted so far are useful (+1) but let me present this in a slightly different way using the Minimum description length principal. The basic idea behind MDL is related to Kolmogorov Comple | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS)
Both answers posted so far are useful (+1) but let me present this in a slightly different way using the Minimum description length principal. The basic idea behind MDL is related to Kolmogorov Complexity and the concept of the... | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS)
Both answers posted so far are useful (+1) but let me present this in a slightly different way using the Minimum description length principal. The basic idea behind MDL is related to Kolmogorov Comple |
40,064 | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS) | The references you want is The Analysis of Market Demand (JSTOR) by Richard Stone, Journal of the Royal Statistical Society, Vol. 108, No. 3/4 (1945), pp. 286-391. I can't find an ungated link, so here's the gist of it.
He gives a formula for the estimated variance of OLS regressor $\beta_k$ in a regression of $y$ on $... | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS) | The references you want is The Analysis of Market Demand (JSTOR) by Richard Stone, Journal of the Royal Statistical Society, Vol. 108, No. 3/4 (1945), pp. 286-391. I can't find an ungated link, so her | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS)
The references you want is The Analysis of Market Demand (JSTOR) by Richard Stone, Journal of the Royal Statistical Society, Vol. 108, No. 3/4 (1945), pp. 286-391. I can't find an ungated link, so here's the gist of it.
He give... | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS)
The references you want is The Analysis of Market Demand (JSTOR) by Richard Stone, Journal of the Royal Statistical Society, Vol. 108, No. 3/4 (1945), pp. 286-391. I can't find an ungated link, so her |
40,065 | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS) | Basically, you are asking for an interpretation of Occam's razor in therms of probability; quoting from wikipedia, Occam's razor:
is a principle of parsimony, economy, or succinctness used in
problem-solving. It states that among competing hypotheses, the one
with the fewest assumptions should be selected.
I can ... | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS) | Basically, you are asking for an interpretation of Occam's razor in therms of probability; quoting from wikipedia, Occam's razor:
is a principle of parsimony, economy, or succinctness used in
probl | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS)
Basically, you are asking for an interpretation of Occam's razor in therms of probability; quoting from wikipedia, Occam's razor:
is a principle of parsimony, economy, or succinctness used in
problem-solving. It states that ... | Looking for a proof that overfitting a model leads to greater variance estimates (under OLS)
Basically, you are asking for an interpretation of Occam's razor in therms of probability; quoting from wikipedia, Occam's razor:
is a principle of parsimony, economy, or succinctness used in
probl |
40,066 | Is a lower training accuracy possible in overfitting (one class SVM) | Proportion classified correctly is a discontinuous improper scoring rule that is optimized by a bogus model. I would not believe anything that you learn from it. | Is a lower training accuracy possible in overfitting (one class SVM) | Proportion classified correctly is a discontinuous improper scoring rule that is optimized by a bogus model. I would not believe anything that you learn from it. | Is a lower training accuracy possible in overfitting (one class SVM)
Proportion classified correctly is a discontinuous improper scoring rule that is optimized by a bogus model. I would not believe anything that you learn from it. | Is a lower training accuracy possible in overfitting (one class SVM)
Proportion classified correctly is a discontinuous improper scoring rule that is optimized by a bogus model. I would not believe anything that you learn from it. |
40,067 | Is a lower training accuracy possible in overfitting (one class SVM) | UPDATE
There is probably an numerical error with one class nu-svm in LibSVM. At optimum, some training instances should satisfy w'*x - rho = 0. However, numerically they may be slightly smaller than zero Then they are wrongly counted as training errors. Since nu is an upper bound on the ratio of training points on the ... | Is a lower training accuracy possible in overfitting (one class SVM) | UPDATE
There is probably an numerical error with one class nu-svm in LibSVM. At optimum, some training instances should satisfy w'*x - rho = 0. However, numerically they may be slightly smaller than z | Is a lower training accuracy possible in overfitting (one class SVM)
UPDATE
There is probably an numerical error with one class nu-svm in LibSVM. At optimum, some training instances should satisfy w'*x - rho = 0. However, numerically they may be slightly smaller than zero Then they are wrongly counted as training error... | Is a lower training accuracy possible in overfitting (one class SVM)
UPDATE
There is probably an numerical error with one class nu-svm in LibSVM. At optimum, some training instances should satisfy w'*x - rho = 0. However, numerically they may be slightly smaller than z |
40,068 | Is a lower training accuracy possible in overfitting (one class SVM) | Based on Dr. Harrell's suggestion, I tried Logarithmic and Brier scoring rule. Since libsvm does not support probability estimation on nu-svm, I had to do it with binary class SVM.
Some notes on the result image:
The proportion classified correctly is different with the '-b 1' option in training and testing. Since t... | Is a lower training accuracy possible in overfitting (one class SVM) | Based on Dr. Harrell's suggestion, I tried Logarithmic and Brier scoring rule. Since libsvm does not support probability estimation on nu-svm, I had to do it with binary class SVM.
Some notes on the | Is a lower training accuracy possible in overfitting (one class SVM)
Based on Dr. Harrell's suggestion, I tried Logarithmic and Brier scoring rule. Since libsvm does not support probability estimation on nu-svm, I had to do it with binary class SVM.
Some notes on the result image:
The proportion classified correctly... | Is a lower training accuracy possible in overfitting (one class SVM)
Based on Dr. Harrell's suggestion, I tried Logarithmic and Brier scoring rule. Since libsvm does not support probability estimation on nu-svm, I had to do it with binary class SVM.
Some notes on the |
40,069 | Optimal lag length in VECM using vars R package | For VEC models you should select number of lags based on information criteria on VAR model on levels of your time series. For that you can use function VARselect from the same package vars.
Function cajorls does not have the argument $K$. However it has the argument $r$, which denotes cointegration rank. Argument $K$ i... | Optimal lag length in VECM using vars R package | For VEC models you should select number of lags based on information criteria on VAR model on levels of your time series. For that you can use function VARselect from the same package vars.
Function c | Optimal lag length in VECM using vars R package
For VEC models you should select number of lags based on information criteria on VAR model on levels of your time series. For that you can use function VARselect from the same package vars.
Function cajorls does not have the argument $K$. However it has the argument $r$, ... | Optimal lag length in VECM using vars R package
For VEC models you should select number of lags based on information criteria on VAR model on levels of your time series. For that you can use function VARselect from the same package vars.
Function c |
40,070 | Implementing Neural Network for time series | Some Googling for specifically neural networks and seasonality leads to this paper, Neural network forecasting for seasonal and trend time series, Zhang and Qi, European Journal of Operational Research, V.160, 2, 16 January 2005, 501–514. In this paper the authors sought to compare the Box-Jenkins approach with a neur... | Implementing Neural Network for time series | Some Googling for specifically neural networks and seasonality leads to this paper, Neural network forecasting for seasonal and trend time series, Zhang and Qi, European Journal of Operational Researc | Implementing Neural Network for time series
Some Googling for specifically neural networks and seasonality leads to this paper, Neural network forecasting for seasonal and trend time series, Zhang and Qi, European Journal of Operational Research, V.160, 2, 16 January 2005, 501–514. In this paper the authors sought to ... | Implementing Neural Network for time series
Some Googling for specifically neural networks and seasonality leads to this paper, Neural network forecasting for seasonal and trend time series, Zhang and Qi, European Journal of Operational Researc |
40,071 | Implementing Neural Network for time series | There is an editorial by Chris Chatfield in the International Journal of Forecasting 9 (1993) 1-3 entitled "Neural networks: Forecasting breakthrough or passing fad?" It is focused on comparing ARIMA models and neural networks. He warns that sometimes apparently successful applications of neural networks in business an... | Implementing Neural Network for time series | There is an editorial by Chris Chatfield in the International Journal of Forecasting 9 (1993) 1-3 entitled "Neural networks: Forecasting breakthrough or passing fad?" It is focused on comparing ARIMA | Implementing Neural Network for time series
There is an editorial by Chris Chatfield in the International Journal of Forecasting 9 (1993) 1-3 entitled "Neural networks: Forecasting breakthrough or passing fad?" It is focused on comparing ARIMA models and neural networks. He warns that sometimes apparently successful ap... | Implementing Neural Network for time series
There is an editorial by Chris Chatfield in the International Journal of Forecasting 9 (1993) 1-3 entitled "Neural networks: Forecasting breakthrough or passing fad?" It is focused on comparing ARIMA |
40,072 | Interpretation of log-level difference-in-differences specification | You can should treat the interaction variable as a dummy and follow this advice from David Giles:
If $Treat\cdot Post$ switches from 0 to 1, the % impact on $Y$ is $100 \cdot (\exp(\beta_4 - \frac{1}{2} \hat \sigma_{\beta_4}^2)-1).$ | Interpretation of log-level difference-in-differences specification | You can should treat the interaction variable as a dummy and follow this advice from David Giles:
If $Treat\cdot Post$ switches from 0 to 1, the % impact on $Y$ is $100 \cdot (\exp(\beta_4 - \frac{1}{ | Interpretation of log-level difference-in-differences specification
You can should treat the interaction variable as a dummy and follow this advice from David Giles:
If $Treat\cdot Post$ switches from 0 to 1, the % impact on $Y$ is $100 \cdot (\exp(\beta_4 - \frac{1}{2} \hat \sigma_{\beta_4}^2)-1).$ | Interpretation of log-level difference-in-differences specification
You can should treat the interaction variable as a dummy and follow this advice from David Giles:
If $Treat\cdot Post$ switches from 0 to 1, the % impact on $Y$ is $100 \cdot (\exp(\beta_4 - \frac{1}{ |
40,073 | Interpretation of log-level difference-in-differences specification | What's required is that $\beta \cdot \Delta x$ be small. If you know that $\Delta x$ is 1, then that means that $\beta$ has to be small. How small? The true proportionate change in $Outcome$ when the dummy rises by $1$ is $\exp(\beta)-1$. The approximate change is $\beta$. The error from the approximation is:
\begi... | Interpretation of log-level difference-in-differences specification | What's required is that $\beta \cdot \Delta x$ be small. If you know that $\Delta x$ is 1, then that means that $\beta$ has to be small. How small? The true proportionate change in $Outcome$ when t | Interpretation of log-level difference-in-differences specification
What's required is that $\beta \cdot \Delta x$ be small. If you know that $\Delta x$ is 1, then that means that $\beta$ has to be small. How small? The true proportionate change in $Outcome$ when the dummy rises by $1$ is $\exp(\beta)-1$. The appro... | Interpretation of log-level difference-in-differences specification
What's required is that $\beta \cdot \Delta x$ be small. If you know that $\Delta x$ is 1, then that means that $\beta$ has to be small. How small? The true proportionate change in $Outcome$ when t |
40,074 | Flaw in a conditional probability argument | The flaw in the argument is that the conditioning random variable is not well-defined.
The ambiguity lies in how our friend peeking at the dice decides to report the information back to us. If we let $X_1$ and $X_2$ denote the random variables associated with the values of each of the dice, then it is certainly true t... | Flaw in a conditional probability argument | The flaw in the argument is that the conditioning random variable is not well-defined.
The ambiguity lies in how our friend peeking at the dice decides to report the information back to us. If we let | Flaw in a conditional probability argument
The flaw in the argument is that the conditioning random variable is not well-defined.
The ambiguity lies in how our friend peeking at the dice decides to report the information back to us. If we let $X_1$ and $X_2$ denote the random variables associated with the values of ea... | Flaw in a conditional probability argument
The flaw in the argument is that the conditioning random variable is not well-defined.
The ambiguity lies in how our friend peeking at the dice decides to report the information back to us. If we let |
40,075 | Flaw in a conditional probability argument | If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events
$\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$,
(that is, $D_i \cap D_j = \emptyset$ for all $i \neq j$ and
$\bigcup_i D_i = \Omega$), then the
law of total probability tells us that
$$P(B) = \sum_i P(B\mid D_i)P... | Flaw in a conditional probability argument | If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events
$\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$,
(that is, $D_i \cap D_j = \emptyset$ for all | Flaw in a conditional probability argument
If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events
$\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$,
(that is, $D_i \cap D_j = \emptyset$ for all $i \neq j$ and
$\bigcup_i D_i = \Omega$), then the
law of total probability ... | Flaw in a conditional probability argument
If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events
$\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$,
(that is, $D_i \cap D_j = \emptyset$ for all |
40,076 | Variance of sum of random number of random variables (Cambridge University Worksheet) | The easy way is to use the law of total variance:
$$\text{Var}(S) = E_N\left[\text{Var}(S|N)\right] + \text{Var}_N\left[E(S|N)\right] =\text{E}_N\left[N\cdot \text{Var}(X)\right] + \text{Var}_N\left[N\cdot\text{E}(X)\right]$$
Can you do it from there? It's pretty much just substitution (well, that and really basic prop... | Variance of sum of random number of random variables (Cambridge University Worksheet) | The easy way is to use the law of total variance:
$$\text{Var}(S) = E_N\left[\text{Var}(S|N)\right] + \text{Var}_N\left[E(S|N)\right] =\text{E}_N\left[N\cdot \text{Var}(X)\right] + \text{Var}_N\left[N | Variance of sum of random number of random variables (Cambridge University Worksheet)
The easy way is to use the law of total variance:
$$\text{Var}(S) = E_N\left[\text{Var}(S|N)\right] + \text{Var}_N\left[E(S|N)\right] =\text{E}_N\left[N\cdot \text{Var}(X)\right] + \text{Var}_N\left[N\cdot\text{E}(X)\right]$$
Can you ... | Variance of sum of random number of random variables (Cambridge University Worksheet)
The easy way is to use the law of total variance:
$$\text{Var}(S) = E_N\left[\text{Var}(S|N)\right] + \text{Var}_N\left[E(S|N)\right] =\text{E}_N\left[N\cdot \text{Var}(X)\right] + \text{Var}_N\left[N |
40,077 | Variance of sum of random number of random variables (Cambridge University Worksheet) | The law of total variance is the easiest way to do this. But there are several occasions when we don't know how many random variables we are dealing with (e.g. branching processes such as Galton-Watson, birth-death processes, queues) where probability-generating functions are a useful technique. It is possible to deriv... | Variance of sum of random number of random variables (Cambridge University Worksheet) | The law of total variance is the easiest way to do this. But there are several occasions when we don't know how many random variables we are dealing with (e.g. branching processes such as Galton-Watso | Variance of sum of random number of random variables (Cambridge University Worksheet)
The law of total variance is the easiest way to do this. But there are several occasions when we don't know how many random variables we are dealing with (e.g. branching processes such as Galton-Watson, birth-death processes, queues) ... | Variance of sum of random number of random variables (Cambridge University Worksheet)
The law of total variance is the easiest way to do this. But there are several occasions when we don't know how many random variables we are dealing with (e.g. branching processes such as Galton-Watso |
40,078 | Variance of sum of random number of random variables (Cambridge University Worksheet) | The solution to calculating the mean:
$\mathbb{E}(S_N) = 0.P(N=0) + \mathbb{E}(X_1).P(N=1) + \mathbb{E}(X_1+X_2).P(N=2) + . . .$
$= 0 + \mu_2P(N=1) + 2\mu_2P(N=2) + . . . = \mu_2\sum_{i=0}^\infty i.P(N=i)$
and the infinite sum above is just equal to the expectation of $N$, hence:
$\mathbb{E}(S_N) = \mu_2.\mathbb{E}(N) ... | Variance of sum of random number of random variables (Cambridge University Worksheet) | The solution to calculating the mean:
$\mathbb{E}(S_N) = 0.P(N=0) + \mathbb{E}(X_1).P(N=1) + \mathbb{E}(X_1+X_2).P(N=2) + . . .$
$= 0 + \mu_2P(N=1) + 2\mu_2P(N=2) + . . . = \mu_2\sum_{i=0}^\infty i.P( | Variance of sum of random number of random variables (Cambridge University Worksheet)
The solution to calculating the mean:
$\mathbb{E}(S_N) = 0.P(N=0) + \mathbb{E}(X_1).P(N=1) + \mathbb{E}(X_1+X_2).P(N=2) + . . .$
$= 0 + \mu_2P(N=1) + 2\mu_2P(N=2) + . . . = \mu_2\sum_{i=0}^\infty i.P(N=i)$
and the infinite sum above i... | Variance of sum of random number of random variables (Cambridge University Worksheet)
The solution to calculating the mean:
$\mathbb{E}(S_N) = 0.P(N=0) + \mathbb{E}(X_1).P(N=1) + \mathbb{E}(X_1+X_2).P(N=2) + . . .$
$= 0 + \mu_2P(N=1) + 2\mu_2P(N=2) + . . . = \mu_2\sum_{i=0}^\infty i.P( |
40,079 | Converting 2nd order Markov chain to the 1st order equivalent | Here's a way to do it:
(I may be writing my state vectors and transition matrices transposed relative to the way you might have learned them, or even the way they're usually done. If that's the case you'll need to translate back.)
The probability model gives you probabilities for 4 output states at time $t$ in terms of... | Converting 2nd order Markov chain to the 1st order equivalent | Here's a way to do it:
(I may be writing my state vectors and transition matrices transposed relative to the way you might have learned them, or even the way they're usually done. If that's the case y | Converting 2nd order Markov chain to the 1st order equivalent
Here's a way to do it:
(I may be writing my state vectors and transition matrices transposed relative to the way you might have learned them, or even the way they're usually done. If that's the case you'll need to translate back.)
The probability model gives... | Converting 2nd order Markov chain to the 1st order equivalent
Here's a way to do it:
(I may be writing my state vectors and transition matrices transposed relative to the way you might have learned them, or even the way they're usually done. If that's the case y |
40,080 | Converting 2nd order Markov chain to the 1st order equivalent | The first order transition matrix: $T^1$ is of size $[k*k]$. And the second order transition matrix: $T^2$ is of size $[k^2*k]$. So you want to reduce the number of rows from $k^2$ to $k$ by merging.
An example is given on the Wikipedia link, you should be able to convert $T^2$ to $T^1$ simply by marginalising over the... | Converting 2nd order Markov chain to the 1st order equivalent | The first order transition matrix: $T^1$ is of size $[k*k]$. And the second order transition matrix: $T^2$ is of size $[k^2*k]$. So you want to reduce the number of rows from $k^2$ to $k$ by merging.
| Converting 2nd order Markov chain to the 1st order equivalent
The first order transition matrix: $T^1$ is of size $[k*k]$. And the second order transition matrix: $T^2$ is of size $[k^2*k]$. So you want to reduce the number of rows from $k^2$ to $k$ by merging.
An example is given on the Wikipedia link, you should be a... | Converting 2nd order Markov chain to the 1st order equivalent
The first order transition matrix: $T^1$ is of size $[k*k]$. And the second order transition matrix: $T^2$ is of size $[k^2*k]$. So you want to reduce the number of rows from $k^2$ to $k$ by merging.
|
40,081 | How to think of reduced dimensions in PCA on facial images (eigenfaces)? | Just a hint, after reading your comment. Each image (face) is represented as a stacked vector of length $N$. The different faces make up a dataset stored in a matrix $X$ of size $K\times N$. You might be confused about the fact that you use the PCA to obtain a set of eigenvectors (eigenfaces) $I = \{u_1, u_2, \ldots, u... | How to think of reduced dimensions in PCA on facial images (eigenfaces)? | Just a hint, after reading your comment. Each image (face) is represented as a stacked vector of length $N$. The different faces make up a dataset stored in a matrix $X$ of size $K\times N$. You might | How to think of reduced dimensions in PCA on facial images (eigenfaces)?
Just a hint, after reading your comment. Each image (face) is represented as a stacked vector of length $N$. The different faces make up a dataset stored in a matrix $X$ of size $K\times N$. You might be confused about the fact that you use the PC... | How to think of reduced dimensions in PCA on facial images (eigenfaces)?
Just a hint, after reading your comment. Each image (face) is represented as a stacked vector of length $N$. The different faces make up a dataset stored in a matrix $X$ of size $K\times N$. You might |
40,082 | How to think of reduced dimensions in PCA on facial images (eigenfaces)? | To clarify a bit more: in your original high-dimensional space, pixels are the dimensions. In the new space, each image is represented as a linear combination of a relatively small number of basis images, the eigenfaces. So in the new space, the eigenfaces are the dimensions. | How to think of reduced dimensions in PCA on facial images (eigenfaces)? | To clarify a bit more: in your original high-dimensional space, pixels are the dimensions. In the new space, each image is represented as a linear combination of a relatively small number of basis ima | How to think of reduced dimensions in PCA on facial images (eigenfaces)?
To clarify a bit more: in your original high-dimensional space, pixels are the dimensions. In the new space, each image is represented as a linear combination of a relatively small number of basis images, the eigenfaces. So in the new space, the e... | How to think of reduced dimensions in PCA on facial images (eigenfaces)?
To clarify a bit more: in your original high-dimensional space, pixels are the dimensions. In the new space, each image is represented as a linear combination of a relatively small number of basis ima |
40,083 | Asymptotic Theory in Economics | Ferguson's A Course in Large Sample Theory is the best concise introduction to the topic, and it is written in a nice didactic way of having an equivalent of a week's lecture course material in a chapter followed by a strong set of exercises. (Ferguson introduced GMM in 1968 under the name of the minimum $\chi^2$, and ... | Asymptotic Theory in Economics | Ferguson's A Course in Large Sample Theory is the best concise introduction to the topic, and it is written in a nice didactic way of having an equivalent of a week's lecture course material in a chap | Asymptotic Theory in Economics
Ferguson's A Course in Large Sample Theory is the best concise introduction to the topic, and it is written in a nice didactic way of having an equivalent of a week's lecture course material in a chapter followed by a strong set of exercises. (Ferguson introduced GMM in 1968 under the nam... | Asymptotic Theory in Economics
Ferguson's A Course in Large Sample Theory is the best concise introduction to the topic, and it is written in a nice didactic way of having an equivalent of a week's lecture course material in a chap |
40,084 | Asymptotic Theory in Economics | Since you mention Greene's book, I assume you are interested in more in-depth understanding of asymptotic statistics. Then I can recommend A. van der Vaart's "Asymptotic statistics" and H. White's "Asymptotic theory for econometricians". Also J. Wooldridge's "Econometric Analysis of Cross Section and Panel Data" has ni... | Asymptotic Theory in Economics | Since you mention Greene's book, I assume you are interested in more in-depth understanding of asymptotic statistics. Then I can recommend A. van der Vaart's "Asymptotic statistics" and H. White's "As | Asymptotic Theory in Economics
Since you mention Greene's book, I assume you are interested in more in-depth understanding of asymptotic statistics. Then I can recommend A. van der Vaart's "Asymptotic statistics" and H. White's "Asymptotic theory for econometricians". Also J. Wooldridge's "Econometric Analysis of Cross... | Asymptotic Theory in Economics
Since you mention Greene's book, I assume you are interested in more in-depth understanding of asymptotic statistics. Then I can recommend A. van der Vaart's "Asymptotic statistics" and H. White's "As |
40,085 | Asymptotic Theory in Economics | "Asymptotic theory for econometricians" by Halbert White. "Asymptotic Theory of Statistics and Probability", by Anirban DasGupta. | Asymptotic Theory in Economics | "Asymptotic theory for econometricians" by Halbert White. "Asymptotic Theory of Statistics and Probability", by Anirban DasGupta. | Asymptotic Theory in Economics
"Asymptotic theory for econometricians" by Halbert White. "Asymptotic Theory of Statistics and Probability", by Anirban DasGupta. | Asymptotic Theory in Economics
"Asymptotic theory for econometricians" by Halbert White. "Asymptotic Theory of Statistics and Probability", by Anirban DasGupta. |
40,086 | Fitting a Poisson distribution with lme4 and nlme [closed] | You are correct; there is no way to specify the family because the nlme package is only for linear mixed models or non-linear mixed models, which assume Gaussian errors. The range of models fitted by nlme does not include the generalised linear mixed model (GLMM).
That lmer() takes a family argument is unfortunate, and... | Fitting a Poisson distribution with lme4 and nlme [closed] | You are correct; there is no way to specify the family because the nlme package is only for linear mixed models or non-linear mixed models, which assume Gaussian errors. The range of models fitted by | Fitting a Poisson distribution with lme4 and nlme [closed]
You are correct; there is no way to specify the family because the nlme package is only for linear mixed models or non-linear mixed models, which assume Gaussian errors. The range of models fitted by nlme does not include the generalised linear mixed model (GLM... | Fitting a Poisson distribution with lme4 and nlme [closed]
You are correct; there is no way to specify the family because the nlme package is only for linear mixed models or non-linear mixed models, which assume Gaussian errors. The range of models fitted by |
40,087 | Which distributions are parameterization invariant when based on the Jeffreys prior? | Yes. And actually this is the interesting invariance property: it means that two Bayesians using a different parameterization of the model but both using the Jeffreys prior, obtain the same posterior distribution (up to change-of-variables) to draw inference.
Conceptually, there's no prior predictive distribution based... | Which distributions are parameterization invariant when based on the Jeffreys prior? | Yes. And actually this is the interesting invariance property: it means that two Bayesians using a different parameterization of the model but both using the Jeffreys prior, obtain the same posterior | Which distributions are parameterization invariant when based on the Jeffreys prior?
Yes. And actually this is the interesting invariance property: it means that two Bayesians using a different parameterization of the model but both using the Jeffreys prior, obtain the same posterior distribution (up to change-of-varia... | Which distributions are parameterization invariant when based on the Jeffreys prior?
Yes. And actually this is the interesting invariance property: it means that two Bayesians using a different parameterization of the model but both using the Jeffreys prior, obtain the same posterior |
40,088 | Does $\sum_{i\neq j} \text{Cov}(X_i, X_j) = 0$ imply $\text{Cov}(X_i, X_j) = 0, \,\forall\,i \neq j$ | Consider three random variables with covariance matrix
$$\left[\begin{matrix}1 & a & 0\\a & 1 & -a\\ 0 & -a & 1\end{matrix}\right]$$
which has leading principal minors $1$, $1-a^2$, and $1-2a^2$ and thus is positive definite
(as all covariance matrices must be) provided that $\vert a\vert < \frac{1}{\sqrt{2}}$.
Obvious... | Does $\sum_{i\neq j} \text{Cov}(X_i, X_j) = 0$ imply $\text{Cov}(X_i, X_j) = 0, \,\forall\,i \neq j$ | Consider three random variables with covariance matrix
$$\left[\begin{matrix}1 & a & 0\\a & 1 & -a\\ 0 & -a & 1\end{matrix}\right]$$
which has leading principal minors $1$, $1-a^2$, and $1-2a^2$ and t | Does $\sum_{i\neq j} \text{Cov}(X_i, X_j) = 0$ imply $\text{Cov}(X_i, X_j) = 0, \,\forall\,i \neq j$
Consider three random variables with covariance matrix
$$\left[\begin{matrix}1 & a & 0\\a & 1 & -a\\ 0 & -a & 1\end{matrix}\right]$$
which has leading principal minors $1$, $1-a^2$, and $1-2a^2$ and thus is positive def... | Does $\sum_{i\neq j} \text{Cov}(X_i, X_j) = 0$ imply $\text{Cov}(X_i, X_j) = 0, \,\forall\,i \neq j$
Consider three random variables with covariance matrix
$$\left[\begin{matrix}1 & a & 0\\a & 1 & -a\\ 0 & -a & 1\end{matrix}\right]$$
which has leading principal minors $1$, $1-a^2$, and $1-2a^2$ and t |
40,089 | Chi-square independence test for A/B split testing | Here's one way to lay out the chi-squared test -- you don't necessarily need all this detail. The test doesn't care which way around you have rows and columns, so I'll do it
your way around.
Comparing proportions via Pearson's Chi-squared test of independence
at the 5% level #(for this example, at least - you cho... | Chi-square independence test for A/B split testing | Here's one way to lay out the chi-squared test -- you don't necessarily need all this detail. The test doesn't care which way around you have rows and columns, so I'll do it
your way around.
Comparin | Chi-square independence test for A/B split testing
Here's one way to lay out the chi-squared test -- you don't necessarily need all this detail. The test doesn't care which way around you have rows and columns, so I'll do it
your way around.
Comparing proportions via Pearson's Chi-squared test of independence
at the 5... | Chi-square independence test for A/B split testing
Here's one way to lay out the chi-squared test -- you don't necessarily need all this detail. The test doesn't care which way around you have rows and columns, so I'll do it
your way around.
Comparin |
40,090 | Chi-square independence test for A/B split testing | The chi-square test of independence only tests for whether there is a relationship between the variables, though, not that there is no difference between CTR:
H0: Click-through and interface are independent.
Ha: Click-through and interface are not independent (that is, something interesting is going on; one of your int... | Chi-square independence test for A/B split testing | The chi-square test of independence only tests for whether there is a relationship between the variables, though, not that there is no difference between CTR:
H0: Click-through and interface are indep | Chi-square independence test for A/B split testing
The chi-square test of independence only tests for whether there is a relationship between the variables, though, not that there is no difference between CTR:
H0: Click-through and interface are independent.
Ha: Click-through and interface are not independent (that is,... | Chi-square independence test for A/B split testing
The chi-square test of independence only tests for whether there is a relationship between the variables, though, not that there is no difference between CTR:
H0: Click-through and interface are indep |
40,091 | Value at $D_\max$ from Kolmogorov-Smirnov test in R | Something like that? Dmax occurs at the value max.at.
set.seed(12345)
x <- rnorm(10000, 5, 5)
y <- rnorm(10000, 7, 6.5)
# remove any missings from the data
x <- x[!is.na(x)]
y <- y[!is.na(y)]
ecdf.x <- ecdf(x)
ecdf.y <- ecdf(y)
plot(ecdf.x, xlim=c(min(c(x,y)), max(c(x,y))), verticals=T, cex.lab=1.2, cex.axis=1.3,
... | Value at $D_\max$ from Kolmogorov-Smirnov test in R | Something like that? Dmax occurs at the value max.at.
set.seed(12345)
x <- rnorm(10000, 5, 5)
y <- rnorm(10000, 7, 6.5)
# remove any missings from the data
x <- x[!is.na(x)]
y <- y[!is.na(y)]
ecdf | Value at $D_\max$ from Kolmogorov-Smirnov test in R
Something like that? Dmax occurs at the value max.at.
set.seed(12345)
x <- rnorm(10000, 5, 5)
y <- rnorm(10000, 7, 6.5)
# remove any missings from the data
x <- x[!is.na(x)]
y <- y[!is.na(y)]
ecdf.x <- ecdf(x)
ecdf.y <- ecdf(y)
plot(ecdf.x, xlim=c(min(c(x,y)), ma... | Value at $D_\max$ from Kolmogorov-Smirnov test in R
Something like that? Dmax occurs at the value max.at.
set.seed(12345)
x <- rnorm(10000, 5, 5)
y <- rnorm(10000, 7, 6.5)
# remove any missings from the data
x <- x[!is.na(x)]
y <- y[!is.na(y)]
ecdf |
40,092 | Value at $D_\max$ from Kolmogorov-Smirnov test in R | You could also use @COOLSerdash's answer plus environments to make the ks.test function output the value directly, like this:
ks.test.2 <- function(x, y, ..., alternative = c("two.sided", "less", "greater"),
exact = NULL)
{
e <- new.env()
ks.test.2 <- ks.test
environment(ks.test.2) <- e
e$C_pkstwo <- stats:::C_pk... | Value at $D_\max$ from Kolmogorov-Smirnov test in R | You could also use @COOLSerdash's answer plus environments to make the ks.test function output the value directly, like this:
ks.test.2 <- function(x, y, ..., alternative = c("two.sided", "less", "gre | Value at $D_\max$ from Kolmogorov-Smirnov test in R
You could also use @COOLSerdash's answer plus environments to make the ks.test function output the value directly, like this:
ks.test.2 <- function(x, y, ..., alternative = c("two.sided", "less", "greater"),
exact = NULL)
{
e <- new.env()
ks.test.2 <- ks.test
en... | Value at $D_\max$ from Kolmogorov-Smirnov test in R
You could also use @COOLSerdash's answer plus environments to make the ks.test function output the value directly, like this:
ks.test.2 <- function(x, y, ..., alternative = c("two.sided", "less", "gre |
40,093 | How to explain how I divided a bimodal distribution based on kernel density estimation | You could fit a two-component mixture model using http://cran.r-project.org/web/packages/mixtools/index.html. Try using normalmixEM. You could then follow Erich Schubert's suggestions and find the region where Pr[data point was generated from the component with the smaller mean] >= 0.50.
Edit: example R code:
library... | How to explain how I divided a bimodal distribution based on kernel density estimation | You could fit a two-component mixture model using http://cran.r-project.org/web/packages/mixtools/index.html. Try using normalmixEM. You could then follow Erich Schubert's suggestions and find the r | How to explain how I divided a bimodal distribution based on kernel density estimation
You could fit a two-component mixture model using http://cran.r-project.org/web/packages/mixtools/index.html. Try using normalmixEM. You could then follow Erich Schubert's suggestions and find the region where Pr[data point was gen... | How to explain how I divided a bimodal distribution based on kernel density estimation
You could fit a two-component mixture model using http://cran.r-project.org/web/packages/mixtools/index.html. Try using normalmixEM. You could then follow Erich Schubert's suggestions and find the r |
40,094 | How to explain how I divided a bimodal distribution based on kernel density estimation | It would probably make more sense if you also estimated the "height" (actually, more of a weight) of both, and then set the threshold to the tipping point.
I.e. model the data as $$p_1 \cdot pdf(x, \mu_1, \sigma_1) + p_2 \cdot pdf(x, \mu_2, \sigma_2)$$
and set the threshold to $x$ where $$p_1 \cdot pdf(x, \mu_1, \sigma... | How to explain how I divided a bimodal distribution based on kernel density estimation | It would probably make more sense if you also estimated the "height" (actually, more of a weight) of both, and then set the threshold to the tipping point.
I.e. model the data as $$p_1 \cdot pdf(x, \m | How to explain how I divided a bimodal distribution based on kernel density estimation
It would probably make more sense if you also estimated the "height" (actually, more of a weight) of both, and then set the threshold to the tipping point.
I.e. model the data as $$p_1 \cdot pdf(x, \mu_1, \sigma_1) + p_2 \cdot pdf(x,... | How to explain how I divided a bimodal distribution based on kernel density estimation
It would probably make more sense if you also estimated the "height" (actually, more of a weight) of both, and then set the threshold to the tipping point.
I.e. model the data as $$p_1 \cdot pdf(x, \m |
40,095 | How to explain how I divided a bimodal distribution based on kernel density estimation | I am using this example and I sometimes get this error
Error in uniroot(f = f, lower = -10, upper = 10) :
f() values at end points not of opposite sign
and so i changed the lower to -1 and for some of the datasets it fixed it, but still errors out on others. Not sure if that can be dynamically set based on the input... | How to explain how I divided a bimodal distribution based on kernel density estimation | I am using this example and I sometimes get this error
Error in uniroot(f = f, lower = -10, upper = 10) :
f() values at end points not of opposite sign
and so i changed the lower to -1 and for some | How to explain how I divided a bimodal distribution based on kernel density estimation
I am using this example and I sometimes get this error
Error in uniroot(f = f, lower = -10, upper = 10) :
f() values at end points not of opposite sign
and so i changed the lower to -1 and for some of the datasets it fixed it, but... | How to explain how I divided a bimodal distribution based on kernel density estimation
I am using this example and I sometimes get this error
Error in uniroot(f = f, lower = -10, upper = 10) :
f() values at end points not of opposite sign
and so i changed the lower to -1 and for some |
40,096 | Is it valid to use a difference score as an independent variable in a regression analysis | Difference scores as independent variables are fine, but they impose a functionally more restrictive form on the equation. Consider;
$y = \beta_{11}(X_2) - \beta_{21}(X_1) + e_1$
Versus the equation;
$y = \beta_{12}(\Delta X) + e_2$
Where $\Delta X = X_2 - X_1$. You can see the second equation is a special case of the ... | Is it valid to use a difference score as an independent variable in a regression analysis | Difference scores as independent variables are fine, but they impose a functionally more restrictive form on the equation. Consider;
$y = \beta_{11}(X_2) - \beta_{21}(X_1) + e_1$
Versus the equation;
| Is it valid to use a difference score as an independent variable in a regression analysis
Difference scores as independent variables are fine, but they impose a functionally more restrictive form on the equation. Consider;
$y = \beta_{11}(X_2) - \beta_{21}(X_1) + e_1$
Versus the equation;
$y = \beta_{12}(\Delta X) + e_... | Is it valid to use a difference score as an independent variable in a regression analysis
Difference scores as independent variables are fine, but they impose a functionally more restrictive form on the equation. Consider;
$y = \beta_{11}(X_2) - \beta_{21}(X_1) + e_1$
Versus the equation;
|
40,097 | Is it valid to use a difference score as an independent variable in a regression analysis | As far as I know, there is no reason you can't use a difference score as an independent variable in a regression. It violates no assumptions.
Your second question is more complex. Your idea of using T1 measures as covariates is often done. People also sometimes use difference scores as a DV (as you probably know from ... | Is it valid to use a difference score as an independent variable in a regression analysis | As far as I know, there is no reason you can't use a difference score as an independent variable in a regression. It violates no assumptions.
Your second question is more complex. Your idea of using | Is it valid to use a difference score as an independent variable in a regression analysis
As far as I know, there is no reason you can't use a difference score as an independent variable in a regression. It violates no assumptions.
Your second question is more complex. Your idea of using T1 measures as covariates is o... | Is it valid to use a difference score as an independent variable in a regression analysis
As far as I know, there is no reason you can't use a difference score as an independent variable in a regression. It violates no assumptions.
Your second question is more complex. Your idea of using |
40,098 | Is it valid to use a difference score as an independent variable in a regression analysis | Using a difference score as a predictor in multiple regression will usually lead to some loss of model fit, i.e. R-squared will be less than what it could be if you leave both variables in the difference score to be their own predictors with their own slopes. That is, if you have a model like this:
$$
y' = a + b_1 d, ... | Is it valid to use a difference score as an independent variable in a regression analysis | Using a difference score as a predictor in multiple regression will usually lead to some loss of model fit, i.e. R-squared will be less than what it could be if you leave both variables in the differe | Is it valid to use a difference score as an independent variable in a regression analysis
Using a difference score as a predictor in multiple regression will usually lead to some loss of model fit, i.e. R-squared will be less than what it could be if you leave both variables in the difference score to be their own pred... | Is it valid to use a difference score as an independent variable in a regression analysis
Using a difference score as a predictor in multiple regression will usually lead to some loss of model fit, i.e. R-squared will be less than what it could be if you leave both variables in the differe |
40,099 | What is coskewness and how can it be calculated? | In this paper coskewness is defined as
$$
coskew_{i,m} = \frac{COV(r_i,(r_m-\mu_m)^2) }{E[(r_m-\mu_m)^3]}.
$$
You can calculate it by using the standard moment estimators - that's what I would do.
Thus, given a sample for market returns $(r_m^j)_{j=1}^N$ and asset returns $(r_i^j)_{j=1}^N$ you calculate the quantities ... | What is coskewness and how can it be calculated? | In this paper coskewness is defined as
$$
coskew_{i,m} = \frac{COV(r_i,(r_m-\mu_m)^2) }{E[(r_m-\mu_m)^3]}.
$$
You can calculate it by using the standard moment estimators - that's what I would do.
Thu | What is coskewness and how can it be calculated?
In this paper coskewness is defined as
$$
coskew_{i,m} = \frac{COV(r_i,(r_m-\mu_m)^2) }{E[(r_m-\mu_m)^3]}.
$$
You can calculate it by using the standard moment estimators - that's what I would do.
Thus, given a sample for market returns $(r_m^j)_{j=1}^N$ and asset return... | What is coskewness and how can it be calculated?
In this paper coskewness is defined as
$$
coskew_{i,m} = \frac{COV(r_i,(r_m-\mu_m)^2) }{E[(r_m-\mu_m)^3]}.
$$
You can calculate it by using the standard moment estimators - that's what I would do.
Thu |
40,100 | What is coskewness and how can it be calculated? | Is there a standard definition?
Yes, these quantities were defined in
Kraus, A., Litzenberger, R.H., 1976. Skewness preference and the valuation of risk assets.
Journal of Finance 31, 1085-1100.
How to calculate it?
Check page 6 of the following document
http://asianfa2012.mcu.edu.tw/fullpaper/10312.pdf
what are m... | What is coskewness and how can it be calculated? | Is there a standard definition?
Yes, these quantities were defined in
Kraus, A., Litzenberger, R.H., 1976. Skewness preference and the valuation of risk assets.
Journal of Finance 31, 1085-1100.
| What is coskewness and how can it be calculated?
Is there a standard definition?
Yes, these quantities were defined in
Kraus, A., Litzenberger, R.H., 1976. Skewness preference and the valuation of risk assets.
Journal of Finance 31, 1085-1100.
How to calculate it?
Check page 6 of the following document
http://asia... | What is coskewness and how can it be calculated?
Is there a standard definition?
Yes, these quantities were defined in
Kraus, A., Litzenberger, R.H., 1976. Skewness preference and the valuation of risk assets.
Journal of Finance 31, 1085-1100.
|
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