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 |
|---|---|---|---|---|---|---|
37,801 | Confidence Intervals for the Parameters of a Logistic Growth Curve | The fitting function growthcurver::SummarizeGrowth is wrapped around Nonlinear Least Squares model minpack.lm::nlsLM, which provides statistics on parameter estimations similar to lm. The NLS model is returned in $mod and is consistent with the returned statistics for K, N0, r.
So K, N0, r are in the sort of standard e... | Confidence Intervals for the Parameters of a Logistic Growth Curve | The fitting function growthcurver::SummarizeGrowth is wrapped around Nonlinear Least Squares model minpack.lm::nlsLM, which provides statistics on parameter estimations similar to lm. The NLS model is | Confidence Intervals for the Parameters of a Logistic Growth Curve
The fitting function growthcurver::SummarizeGrowth is wrapped around Nonlinear Least Squares model minpack.lm::nlsLM, which provides statistics on parameter estimations similar to lm. The NLS model is returned in $mod and is consistent with the returned... | Confidence Intervals for the Parameters of a Logistic Growth Curve
The fitting function growthcurver::SummarizeGrowth is wrapped around Nonlinear Least Squares model minpack.lm::nlsLM, which provides statistics on parameter estimations similar to lm. The NLS model is |
37,802 | Confidence Intervals for the Parameters of a Logistic Growth Curve | If the unknown fixed true parameter such as $r$ is a real number, then the interval using the identity link as you have constructed will work well.
If the unknown fixed true parameter such as $r$ must be a non-negative real number, then you might benefit from using a log link function. In your example your 95% CI for ... | Confidence Intervals for the Parameters of a Logistic Growth Curve | If the unknown fixed true parameter such as $r$ is a real number, then the interval using the identity link as you have constructed will work well.
If the unknown fixed true parameter such as $r$ must | Confidence Intervals for the Parameters of a Logistic Growth Curve
If the unknown fixed true parameter such as $r$ is a real number, then the interval using the identity link as you have constructed will work well.
If the unknown fixed true parameter such as $r$ must be a non-negative real number, then you might benefi... | Confidence Intervals for the Parameters of a Logistic Growth Curve
If the unknown fixed true parameter such as $r$ is a real number, then the interval using the identity link as you have constructed will work well.
If the unknown fixed true parameter such as $r$ must |
37,803 | Feature Engineering : combine a categorical Feature and a continuous Feature | This is probably just a hack that does not solve this kind of problem in general, but may be well-suited for your problem: a person that does not smoke is equivalent to a person that starts smoking at the age of infinity. Hence if you transform your $X2$ into $X2' = 1/X2$, then a person that never smoked should have a ... | Feature Engineering : combine a categorical Feature and a continuous Feature | This is probably just a hack that does not solve this kind of problem in general, but may be well-suited for your problem: a person that does not smoke is equivalent to a person that starts smoking at | Feature Engineering : combine a categorical Feature and a continuous Feature
This is probably just a hack that does not solve this kind of problem in general, but may be well-suited for your problem: a person that does not smoke is equivalent to a person that starts smoking at the age of infinity. Hence if you transfor... | Feature Engineering : combine a categorical Feature and a continuous Feature
This is probably just a hack that does not solve this kind of problem in general, but may be well-suited for your problem: a person that does not smoke is equivalent to a person that starts smoking at |
37,804 | Feature Engineering : combine a categorical Feature and a continuous Feature | Attributes for the smoker class of humans are different than attributes for the non-smoker class of humans. That is okay.
Age started smoking is an attribute of the smoker class. Age started smoking is not an attribute of the non-smoker class. To pretend otherwise, fabricating numeric-looking values, is ill advised a... | Feature Engineering : combine a categorical Feature and a continuous Feature | Attributes for the smoker class of humans are different than attributes for the non-smoker class of humans. That is okay.
Age started smoking is an attribute of the smoker class. Age started smoking | Feature Engineering : combine a categorical Feature and a continuous Feature
Attributes for the smoker class of humans are different than attributes for the non-smoker class of humans. That is okay.
Age started smoking is an attribute of the smoker class. Age started smoking is not an attribute of the non-smoker class... | Feature Engineering : combine a categorical Feature and a continuous Feature
Attributes for the smoker class of humans are different than attributes for the non-smoker class of humans. That is okay.
Age started smoking is an attribute of the smoker class. Age started smoking |
37,805 | Queuing theory for elevators | Not sure I fully understand the problem, but I just think the so-called "soonest on" strategy obviously wins (or at least draws) every time.
Let's say I am currently on the 3rd floor and want to go up to 7th floor. The elevator is on the way down to 1st. At the time when the elevator reaches the 3rd floor on the way do... | Queuing theory for elevators | Not sure I fully understand the problem, but I just think the so-called "soonest on" strategy obviously wins (or at least draws) every time.
Let's say I am currently on the 3rd floor and want to go up | Queuing theory for elevators
Not sure I fully understand the problem, but I just think the so-called "soonest on" strategy obviously wins (or at least draws) every time.
Let's say I am currently on the 3rd floor and want to go up to 7th floor. The elevator is on the way down to 1st. At the time when the elevator reache... | Queuing theory for elevators
Not sure I fully understand the problem, but I just think the so-called "soonest on" strategy obviously wins (or at least draws) every time.
Let's say I am currently on the 3rd floor and want to go up |
37,806 | Why is the type I error (power?!) for shapiro.test on studentized residuals on lm is 10% and for regular residuals is just 5%? | Just for the future, as @glen_b wrote, the answer is that Studentized residuals are not iid. | Why is the type I error (power?!) for shapiro.test on studentized residuals on lm is 10% and for reg | Just for the future, as @glen_b wrote, the answer is that Studentized residuals are not iid. | Why is the type I error (power?!) for shapiro.test on studentized residuals on lm is 10% and for regular residuals is just 5%?
Just for the future, as @glen_b wrote, the answer is that Studentized residuals are not iid. | Why is the type I error (power?!) for shapiro.test on studentized residuals on lm is 10% and for reg
Just for the future, as @glen_b wrote, the answer is that Studentized residuals are not iid. |
37,807 | How can I know when not to answer questions and shut up? | In a little simpler set-up with FAQ answering system, I'm using simple approach with threshold - when alogithm returns probability / similarity below the threshold, I'm falling back into path "I'm not sure" (sligtly below threshold) or "I don't know" (significantly below threshold / below the second threshold). | How can I know when not to answer questions and shut up? | In a little simpler set-up with FAQ answering system, I'm using simple approach with threshold - when alogithm returns probability / similarity below the threshold, I'm falling back into path "I'm not | How can I know when not to answer questions and shut up?
In a little simpler set-up with FAQ answering system, I'm using simple approach with threshold - when alogithm returns probability / similarity below the threshold, I'm falling back into path "I'm not sure" (sligtly below threshold) or "I don't know" (significant... | How can I know when not to answer questions and shut up?
In a little simpler set-up with FAQ answering system, I'm using simple approach with threshold - when alogithm returns probability / similarity below the threshold, I'm falling back into path "I'm not |
37,808 | Statistical analysis of disappearing eagles | I found a way to get to my answer from this comment "look up the hypergeometric distribution" on math.stackexchange:
From Wikipedia's entry on hypergeometric distribution
In probability theory and statistics, the hypergeometric distribution
is a discrete probability distribution that describes the probability
of ... | Statistical analysis of disappearing eagles | I found a way to get to my answer from this comment "look up the hypergeometric distribution" on math.stackexchange:
From Wikipedia's entry on hypergeometric distribution
In probability theory and s | Statistical analysis of disappearing eagles
I found a way to get to my answer from this comment "look up the hypergeometric distribution" on math.stackexchange:
From Wikipedia's entry on hypergeometric distribution
In probability theory and statistics, the hypergeometric distribution
is a discrete probability distr... | Statistical analysis of disappearing eagles
I found a way to get to my answer from this comment "look up the hypergeometric distribution" on math.stackexchange:
From Wikipedia's entry on hypergeometric distribution
In probability theory and s |
37,809 | Does marginalization of some of the latent variables improve convergence in EM? | Let me clean up your question:
First, the equation given by you seems incorrect,
$
p(x|\theta,z_1) \neq \int_{z_1} p(x|\theta, z_1)dz_1,
$
what holds is:
$
p(x|\theta,z_1) = \int_{z_1} p(x, z_1|\theta)dz_1.
$
Second, if you want to connect $p(x|\theta, z_1)$ to $p(x|\theta) $, use the Bayes' theorem:
$
p(x,z_1|\the... | Does marginalization of some of the latent variables improve convergence in EM? | Let me clean up your question:
First, the equation given by you seems incorrect,
$
p(x|\theta,z_1) \neq \int_{z_1} p(x|\theta, z_1)dz_1,
$
what holds is:
$
p(x|\theta,z_1) = \int_{z_1} p(x, z_1|\th | Does marginalization of some of the latent variables improve convergence in EM?
Let me clean up your question:
First, the equation given by you seems incorrect,
$
p(x|\theta,z_1) \neq \int_{z_1} p(x|\theta, z_1)dz_1,
$
what holds is:
$
p(x|\theta,z_1) = \int_{z_1} p(x, z_1|\theta)dz_1.
$
Second, if you want to conne... | Does marginalization of some of the latent variables improve convergence in EM?
Let me clean up your question:
First, the equation given by you seems incorrect,
$
p(x|\theta,z_1) \neq \int_{z_1} p(x|\theta, z_1)dz_1,
$
what holds is:
$
p(x|\theta,z_1) = \int_{z_1} p(x, z_1|\th |
37,810 | Detecting changes in large number of time-series that share seasonality | Transfer Function model identification can be used to develop useful equations that can then be classified or segmented.
"Things" can be simplified for expedience , if that is the goal. Doing this with freely available web resources is a little bit more tricky. | Detecting changes in large number of time-series that share seasonality | Transfer Function model identification can be used to develop useful equations that can then be classified or segmented.
"Things" can be simplified for expedience , if that is the goal. Doing this wi | Detecting changes in large number of time-series that share seasonality
Transfer Function model identification can be used to develop useful equations that can then be classified or segmented.
"Things" can be simplified for expedience , if that is the goal. Doing this with freely available web resources is a little bi... | Detecting changes in large number of time-series that share seasonality
Transfer Function model identification can be used to develop useful equations that can then be classified or segmented.
"Things" can be simplified for expedience , if that is the goal. Doing this wi |
37,811 | Appropriate tests on discrete and paired data | Just to expand on why the t-test is working: this is a result of the Central Limit Theorem, which tells us that the sample mean has a Normal distribution as $n$ grows. Your data is clearly non-normal, taking discrete values only, but the sample mean will be fairly Normally distributed on account of your large sample si... | Appropriate tests on discrete and paired data | Just to expand on why the t-test is working: this is a result of the Central Limit Theorem, which tells us that the sample mean has a Normal distribution as $n$ grows. Your data is clearly non-normal, | Appropriate tests on discrete and paired data
Just to expand on why the t-test is working: this is a result of the Central Limit Theorem, which tells us that the sample mean has a Normal distribution as $n$ grows. Your data is clearly non-normal, taking discrete values only, but the sample mean will be fairly Normally ... | Appropriate tests on discrete and paired data
Just to expand on why the t-test is working: this is a result of the Central Limit Theorem, which tells us that the sample mean has a Normal distribution as $n$ grows. Your data is clearly non-normal, |
37,812 | Bayesian inference on mean of statistic from population | From your description, it sounds like you had two kinds of random variables: time intervals $t_1,\dots,t_n$ and the counts of events $y_1,\dots,y_n$. The occurrence of events depends on length of time intervals given a known, constant rate $\lambda$ according to Poisson distribution
$$
y_i \sim \mathcal{P}(\lambda t_i)... | Bayesian inference on mean of statistic from population | From your description, it sounds like you had two kinds of random variables: time intervals $t_1,\dots,t_n$ and the counts of events $y_1,\dots,y_n$. The occurrence of events depends on length of time | Bayesian inference on mean of statistic from population
From your description, it sounds like you had two kinds of random variables: time intervals $t_1,\dots,t_n$ and the counts of events $y_1,\dots,y_n$. The occurrence of events depends on length of time intervals given a known, constant rate $\lambda$ according to P... | Bayesian inference on mean of statistic from population
From your description, it sounds like you had two kinds of random variables: time intervals $t_1,\dots,t_n$ and the counts of events $y_1,\dots,y_n$. The occurrence of events depends on length of time |
37,813 | How to make Bayesian-style inference for a Poisson process? | If you assume that the rate itself is a Gamma variable, you can easily update its distribution by Bayesian updating (The conjugate prior for a Poisson distribution is a Gamma distribution).
If you use this approach, you'll have to define the event Busy in terms of the Poisson parameter (e.g. $P(\text{Busy}) = P(\lam... | How to make Bayesian-style inference for a Poisson process? | If you assume that the rate itself is a Gamma variable, you can easily update its distribution by Bayesian updating (The conjugate prior for a Poisson distribution is a Gamma distribution).
If you u | How to make Bayesian-style inference for a Poisson process?
If you assume that the rate itself is a Gamma variable, you can easily update its distribution by Bayesian updating (The conjugate prior for a Poisson distribution is a Gamma distribution).
If you use this approach, you'll have to define the event Busy in t... | How to make Bayesian-style inference for a Poisson process?
If you assume that the rate itself is a Gamma variable, you can easily update its distribution by Bayesian updating (The conjugate prior for a Poisson distribution is a Gamma distribution).
If you u |
37,814 | 2-Sample Proportions z-test vs Fisher's Exact Test | Fisher's exact test offers a more exact p-value by running every possible scenario with the given data set and figures out the total number of possible successes and total number of possible failures at the given sample size, then it converts those totals into a p-value. Also note that the total possible outcomes goes ... | 2-Sample Proportions z-test vs Fisher's Exact Test | Fisher's exact test offers a more exact p-value by running every possible scenario with the given data set and figures out the total number of possible successes and total number of possible failures | 2-Sample Proportions z-test vs Fisher's Exact Test
Fisher's exact test offers a more exact p-value by running every possible scenario with the given data set and figures out the total number of possible successes and total number of possible failures at the given sample size, then it converts those totals into a p-valu... | 2-Sample Proportions z-test vs Fisher's Exact Test
Fisher's exact test offers a more exact p-value by running every possible scenario with the given data set and figures out the total number of possible successes and total number of possible failures |
37,815 | Do I still need to include variables used to generate weights in a regression model when applying inverse probability weighting in estimation? | This really depends on your research question. There's no "rule"; it's a theory question. Let's say that one of your variables in $\pi$ is the number of adults in the household ($adults$). This would be fairly common variable to include in an inverse probability weight, because the more adults there are in a household,... | Do I still need to include variables used to generate weights in a regression model when applying in | This really depends on your research question. There's no "rule"; it's a theory question. Let's say that one of your variables in $\pi$ is the number of adults in the household ($adults$). This would | Do I still need to include variables used to generate weights in a regression model when applying inverse probability weighting in estimation?
This really depends on your research question. There's no "rule"; it's a theory question. Let's say that one of your variables in $\pi$ is the number of adults in the household ... | Do I still need to include variables used to generate weights in a regression model when applying in
This really depends on your research question. There's no "rule"; it's a theory question. Let's say that one of your variables in $\pi$ is the number of adults in the household ($adults$). This would |
37,816 | Is ANOVA always more powerful than a two-sample t-test when the data can be blocked? | Using terminology from design of experiments, you seem to have a randomized block experiment (assuming suitable randomization was done) where the blocks are defined by the operators, and you have two treatments (devices). If you ignore the blocks, then the one-way anova will be equivalent to a t-test. Your two questio... | Is ANOVA always more powerful than a two-sample t-test when the data can be blocked? | Using terminology from design of experiments, you seem to have a randomized block experiment (assuming suitable randomization was done) where the blocks are defined by the operators, and you have two | Is ANOVA always more powerful than a two-sample t-test when the data can be blocked?
Using terminology from design of experiments, you seem to have a randomized block experiment (assuming suitable randomization was done) where the blocks are defined by the operators, and you have two treatments (devices). If you ignor... | Is ANOVA always more powerful than a two-sample t-test when the data can be blocked?
Using terminology from design of experiments, you seem to have a randomized block experiment (assuming suitable randomization was done) where the blocks are defined by the operators, and you have two |
37,817 | Confidence Interval for variance given one observation | Viewed through the lens of probability inequalities and connections to the multiple-observation case, this result might not seem so impossible, or, at least, it might seem more plausible.
Let $\renewcommand{\Pr}{\mathbb P}\newcommand{\Ind}[1]{\mathbf 1_{(#1)}}X \sim \mathcal N(\mu,\sigma^2)$ with $\mu$ and $\sigma^2$ u... | Confidence Interval for variance given one observation | Viewed through the lens of probability inequalities and connections to the multiple-observation case, this result might not seem so impossible, or, at least, it might seem more plausible.
Let $\renewc | Confidence Interval for variance given one observation
Viewed through the lens of probability inequalities and connections to the multiple-observation case, this result might not seem so impossible, or, at least, it might seem more plausible.
Let $\renewcommand{\Pr}{\mathbb P}\newcommand{\Ind}[1]{\mathbf 1_{(#1)}}X \si... | Confidence Interval for variance given one observation
Viewed through the lens of probability inequalities and connections to the multiple-observation case, this result might not seem so impossible, or, at least, it might seem more plausible.
Let $\renewc |
37,818 | Confidence Interval for variance given one observation | Time to follow up! Here's the solution I was given:
We will construct a confidence interval of the form $[0,T(X))$, where $T(\cdot)$ is some statistic. By definition this will be a confidence interval with confidence level at least 99% if
$$(\forall \mu \in \mathbb R )(\forall \sigma > 0)\; \mathbb P_{\mu,\sigma_2}(... | Confidence Interval for variance given one observation | Time to follow up! Here's the solution I was given:
We will construct a confidence interval of the form $[0,T(X))$, where $T(\cdot)$ is some statistic. By definition this will be a confidence interva | Confidence Interval for variance given one observation
Time to follow up! Here's the solution I was given:
We will construct a confidence interval of the form $[0,T(X))$, where $T(\cdot)$ is some statistic. By definition this will be a confidence interval with confidence level at least 99% if
$$(\forall \mu \in \mat... | Confidence Interval for variance given one observation
Time to follow up! Here's the solution I was given:
We will construct a confidence interval of the form $[0,T(X))$, where $T(\cdot)$ is some statistic. By definition this will be a confidence interva |
37,819 | Confidence Interval for variance given one observation | The CI's $(0,\infty)$ presumably. | Confidence Interval for variance given one observation | The CI's $(0,\infty)$ presumably. | Confidence Interval for variance given one observation
The CI's $(0,\infty)$ presumably. | Confidence Interval for variance given one observation
The CI's $(0,\infty)$ presumably. |
37,820 | How to fit a Pareto distribution to an observed CDF? | First, you technically have a survival function, which is the complement of the CDF. Second, you must be dealing with some form of censoring from above (insurance policy limits?) as you have plateaus in the CDF and a hard cap at 45,860,334.00. As such, no "pure" Pareto family will fit nicely as these densities are stri... | How to fit a Pareto distribution to an observed CDF? | First, you technically have a survival function, which is the complement of the CDF. Second, you must be dealing with some form of censoring from above (insurance policy limits?) as you have plateaus | How to fit a Pareto distribution to an observed CDF?
First, you technically have a survival function, which is the complement of the CDF. Second, you must be dealing with some form of censoring from above (insurance policy limits?) as you have plateaus in the CDF and a hard cap at 45,860,334.00. As such, no "pure" Pare... | How to fit a Pareto distribution to an observed CDF?
First, you technically have a survival function, which is the complement of the CDF. Second, you must be dealing with some form of censoring from above (insurance policy limits?) as you have plateaus |
37,821 | How to fit a Pareto distribution to an observed CDF? | This Wikipedia page indicates the maximum likelihood estimates have a closed form solution. You can plug these estimates in for the parameter values and plot the resulting parametric CDF to see how well the model fits your data. You can construct tolerance limits (confidence limits for population percentiles) using t... | How to fit a Pareto distribution to an observed CDF? | This Wikipedia page indicates the maximum likelihood estimates have a closed form solution. You can plug these estimates in for the parameter values and plot the resulting parametric CDF to see how w | How to fit a Pareto distribution to an observed CDF?
This Wikipedia page indicates the maximum likelihood estimates have a closed form solution. You can plug these estimates in for the parameter values and plot the resulting parametric CDF to see how well the model fits your data. You can construct tolerance limits (... | How to fit a Pareto distribution to an observed CDF?
This Wikipedia page indicates the maximum likelihood estimates have a closed form solution. You can plug these estimates in for the parameter values and plot the resulting parametric CDF to see how w |
37,822 | Relative advantages of nnet, neuralnet, caret and RSNNS packages [closed] | 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.
Section 7.1 from Bergmeir & Benítez Sánchez (2012) is ... | Relative advantages of nnet, neuralnet, caret and RSNNS packages [closed] | 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.
| Relative advantages of nnet, neuralnet, caret and RSNNS packages [closed]
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.
... | Relative advantages of nnet, neuralnet, caret and RSNNS packages [closed]
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.
|
37,823 | Difference between pointwise and uniform confidence sets | You can see many similar examples in real analysis.
Let's say the $F$ can be indexed as $F_i$(The $\mathfrak{F}$ is countable). The confidence set $C_n$ have those property:
$$
\begin{aligned}
P_{F_i}(\theta \in C_n) & \ge 1-\alpha \quad if \quad i \le n\\
& < 1-\alpha \quad otherwise
\end{aligned}
$$
So $C_n$ is point... | Difference between pointwise and uniform confidence sets | You can see many similar examples in real analysis.
Let's say the $F$ can be indexed as $F_i$(The $\mathfrak{F}$ is countable). The confidence set $C_n$ have those property:
$$
\begin{aligned}
P_{F_i} | Difference between pointwise and uniform confidence sets
You can see many similar examples in real analysis.
Let's say the $F$ can be indexed as $F_i$(The $\mathfrak{F}$ is countable). The confidence set $C_n$ have those property:
$$
\begin{aligned}
P_{F_i}(\theta \in C_n) & \ge 1-\alpha \quad if \quad i \le n\\
& < 1-... | Difference between pointwise and uniform confidence sets
You can see many similar examples in real analysis.
Let's say the $F$ can be indexed as $F_i$(The $\mathfrak{F}$ is countable). The confidence set $C_n$ have those property:
$$
\begin{aligned}
P_{F_i} |
37,824 | Dealing with hierarchical (panel, multi-level) data and fixed effects in LASSO? | First thing to do is to rearrange your data into a standard form. If you've got $n$ samples and $d$ features, that means you want
An $n \times d$ input matrix $X$, in which each column has a mean of $0$ and variance of $1$. This ensures that LASSO's regularization effect treats each dimension "fairly" when deciding wh... | Dealing with hierarchical (panel, multi-level) data and fixed effects in LASSO? | First thing to do is to rearrange your data into a standard form. If you've got $n$ samples and $d$ features, that means you want
An $n \times d$ input matrix $X$, in which each column has a mean of | Dealing with hierarchical (panel, multi-level) data and fixed effects in LASSO?
First thing to do is to rearrange your data into a standard form. If you've got $n$ samples and $d$ features, that means you want
An $n \times d$ input matrix $X$, in which each column has a mean of $0$ and variance of $1$. This ensures th... | Dealing with hierarchical (panel, multi-level) data and fixed effects in LASSO?
First thing to do is to rearrange your data into a standard form. If you've got $n$ samples and $d$ features, that means you want
An $n \times d$ input matrix $X$, in which each column has a mean of |
37,825 | Using an RMSE with derived confidence interval, to generate a prediction interval for an estimate | The CART, as I understand it, does not have homoscedasticity assumptions. If anything it presumes that the variance of each component is independent from the variance of all other components. It doesn't account for correlations in variables either.
The normality assumption is problematic. It is convenient but not ne... | Using an RMSE with derived confidence interval, to generate a prediction interval for an estimate | The CART, as I understand it, does not have homoscedasticity assumptions. If anything it presumes that the variance of each component is independent from the variance of all other components. It doe | Using an RMSE with derived confidence interval, to generate a prediction interval for an estimate
The CART, as I understand it, does not have homoscedasticity assumptions. If anything it presumes that the variance of each component is independent from the variance of all other components. It doesn't account for corre... | Using an RMSE with derived confidence interval, to generate a prediction interval for an estimate
The CART, as I understand it, does not have homoscedasticity assumptions. If anything it presumes that the variance of each component is independent from the variance of all other components. It doe |
37,826 | Fit a VAR model with R [closed] | After you fit your model using:
fit <- VAR(data1,p=6)
You can refine it with refVAR:
fit2 <- refVAR(fit,thres=1.65) | Fit a VAR model with R [closed] | After you fit your model using:
fit <- VAR(data1,p=6)
You can refine it with refVAR:
fit2 <- refVAR(fit,thres=1.65) | Fit a VAR model with R [closed]
After you fit your model using:
fit <- VAR(data1,p=6)
You can refine it with refVAR:
fit2 <- refVAR(fit,thres=1.65) | Fit a VAR model with R [closed]
After you fit your model using:
fit <- VAR(data1,p=6)
You can refine it with refVAR:
fit2 <- refVAR(fit,thres=1.65) |
37,827 | Having a conjugate prior: Deep property or mathematical accident? | It is not by accident. Here you shall find a brief a very nice review on conjugate priors. Concretely, it mentions that if there exist a set of sufficient statistics of fixed dimension for the given likelihood function, then you can construct a conjugate prior to it. Having a set of sufficient statistics means that you... | Having a conjugate prior: Deep property or mathematical accident? | It is not by accident. Here you shall find a brief a very nice review on conjugate priors. Concretely, it mentions that if there exist a set of sufficient statistics of fixed dimension for the given l | Having a conjugate prior: Deep property or mathematical accident?
It is not by accident. Here you shall find a brief a very nice review on conjugate priors. Concretely, it mentions that if there exist a set of sufficient statistics of fixed dimension for the given likelihood function, then you can construct a conjugate... | Having a conjugate prior: Deep property or mathematical accident?
It is not by accident. Here you shall find a brief a very nice review on conjugate priors. Concretely, it mentions that if there exist a set of sufficient statistics of fixed dimension for the given l |
37,828 | Having a conjugate prior: Deep property or mathematical accident? | I am very new to Bayesian statistics, but it seems to me that all of these distributions (and if not all of them then at least those that are useful) share the property that they are described by some limited metric about the observations that define them. I.e., for a normal distribution, you don't need to know every d... | Having a conjugate prior: Deep property or mathematical accident? | I am very new to Bayesian statistics, but it seems to me that all of these distributions (and if not all of them then at least those that are useful) share the property that they are described by some | Having a conjugate prior: Deep property or mathematical accident?
I am very new to Bayesian statistics, but it seems to me that all of these distributions (and if not all of them then at least those that are useful) share the property that they are described by some limited metric about the observations that define the... | Having a conjugate prior: Deep property or mathematical accident?
I am very new to Bayesian statistics, but it seems to me that all of these distributions (and if not all of them then at least those that are useful) share the property that they are described by some |
37,829 | Having a conjugate prior: Deep property or mathematical accident? | What properties are "deep" is a very subjective issue! so the answer depends on your concept of "deep". But, if having conjugate priors is a "deep" property, in some sense, then that sense is mathematical and not statistical. The only reason that (some) statisticians are interested in conjugate priors is that they simp... | Having a conjugate prior: Deep property or mathematical accident? | What properties are "deep" is a very subjective issue! so the answer depends on your concept of "deep". But, if having conjugate priors is a "deep" property, in some sense, then that sense is mathemat | Having a conjugate prior: Deep property or mathematical accident?
What properties are "deep" is a very subjective issue! so the answer depends on your concept of "deep". But, if having conjugate priors is a "deep" property, in some sense, then that sense is mathematical and not statistical. The only reason that (some) ... | Having a conjugate prior: Deep property or mathematical accident?
What properties are "deep" is a very subjective issue! so the answer depends on your concept of "deep". But, if having conjugate priors is a "deep" property, in some sense, then that sense is mathemat |
37,830 | Does Greenhouse-Geisser correction influence the effect size estimation, the statistical significance treshold, or both of them? | Bellow I show part of a code I have written to estimate eta. In the first output I use the normal df, in the second the greenhouse geisser df and in the third the hyndt df. You can get the adjusted df by df*ε. According to what I have written, assuming that everything is correct, eta and partial eta as far as I can see... | Does Greenhouse-Geisser correction influence the effect size estimation, the statistical significanc | Bellow I show part of a code I have written to estimate eta. In the first output I use the normal df, in the second the greenhouse geisser df and in the third the hyndt df. You can get the adjusted df | Does Greenhouse-Geisser correction influence the effect size estimation, the statistical significance treshold, or both of them?
Bellow I show part of a code I have written to estimate eta. In the first output I use the normal df, in the second the greenhouse geisser df and in the third the hyndt df. You can get the ad... | Does Greenhouse-Geisser correction influence the effect size estimation, the statistical significanc
Bellow I show part of a code I have written to estimate eta. In the first output I use the normal df, in the second the greenhouse geisser df and in the third the hyndt df. You can get the adjusted df |
37,831 | Why do Bayesian Networks use acyclicity assumption? | Just to add a little more clarity. this approach is sometimes called Temporal Bayesian Models. I have seen it being used in atleast one other situation of marketing mix models where today's marketing spend influences today's brand & revenue. Today's brand also influences tomorrow's brand & revenue and so on. | Why do Bayesian Networks use acyclicity assumption? | Just to add a little more clarity. this approach is sometimes called Temporal Bayesian Models. I have seen it being used in atleast one other situation of marketing mix models where today's marketing | Why do Bayesian Networks use acyclicity assumption?
Just to add a little more clarity. this approach is sometimes called Temporal Bayesian Models. I have seen it being used in atleast one other situation of marketing mix models where today's marketing spend influences today's brand & revenue. Today's brand also influen... | Why do Bayesian Networks use acyclicity assumption?
Just to add a little more clarity. this approach is sometimes called Temporal Bayesian Models. I have seen it being used in atleast one other situation of marketing mix models where today's marketing |
37,832 | How to analyze and interpret A/B test results via Bootstrap method? | Bootstrapping for a mean does not usually make sense for due to the CLT. Just use the mean and standard error on the mean. This will either give the same result as your bootstrap or your bootstrap would have given a poor result.
If you truly know you want to compare the mean (and it is not clear that you do Better esti... | How to analyze and interpret A/B test results via Bootstrap method? | Bootstrapping for a mean does not usually make sense for due to the CLT. Just use the mean and standard error on the mean. This will either give the same result as your bootstrap or your bootstrap wou | How to analyze and interpret A/B test results via Bootstrap method?
Bootstrapping for a mean does not usually make sense for due to the CLT. Just use the mean and standard error on the mean. This will either give the same result as your bootstrap or your bootstrap would have given a poor result.
If you truly know you w... | How to analyze and interpret A/B test results via Bootstrap method?
Bootstrapping for a mean does not usually make sense for due to the CLT. Just use the mean and standard error on the mean. This will either give the same result as your bootstrap or your bootstrap wou |
37,833 | Difference-in-difference in panel data | You would expect equivalence when T = 2. If you have more than 2 years, use the latter approach (the one that relies on fixed effects). Please see http://econ.lse.ac.uk/staff/spischke/ec533/did.pdf for more. Note: xi: is redundant in the newer versions of Stata. | Difference-in-difference in panel data | You would expect equivalence when T = 2. If you have more than 2 years, use the latter approach (the one that relies on fixed effects). Please see http://econ.lse.ac.uk/staff/spischke/ec533/did.pdf fo | Difference-in-difference in panel data
You would expect equivalence when T = 2. If you have more than 2 years, use the latter approach (the one that relies on fixed effects). Please see http://econ.lse.ac.uk/staff/spischke/ec533/did.pdf for more. Note: xi: is redundant in the newer versions of Stata. | Difference-in-difference in panel data
You would expect equivalence when T = 2. If you have more than 2 years, use the latter approach (the one that relies on fixed effects). Please see http://econ.lse.ac.uk/staff/spischke/ec533/did.pdf fo |
37,834 | Difference-in-difference in panel data | As long as the treatment occurs at the same time for all units, the estimator of DiD is equivalent to the one from panel data (usually called Two-way fixed effects), as shown by Jeffrey Wooldridge in this paper, section 5. To show it, we can generate some random data, apply an effect and estimate with both methods.
np.... | Difference-in-difference in panel data | As long as the treatment occurs at the same time for all units, the estimator of DiD is equivalent to the one from panel data (usually called Two-way fixed effects), as shown by Jeffrey Wooldridge in | Difference-in-difference in panel data
As long as the treatment occurs at the same time for all units, the estimator of DiD is equivalent to the one from panel data (usually called Two-way fixed effects), as shown by Jeffrey Wooldridge in this paper, section 5. To show it, we can generate some random data, apply an eff... | Difference-in-difference in panel data
As long as the treatment occurs at the same time for all units, the estimator of DiD is equivalent to the one from panel data (usually called Two-way fixed effects), as shown by Jeffrey Wooldridge in |
37,835 | Unscented Kalman filter-negative covariance matrix | I also have this problem. However, I realised that depending on the covariance that you fix, it does not happen. If I set an unrealistic covariance matrix, then the error happens. Which makes me think that is perhaps a matter of numerical errors making the covariance matrix non-positive definite.
However, as you say, i... | Unscented Kalman filter-negative covariance matrix | I also have this problem. However, I realised that depending on the covariance that you fix, it does not happen. If I set an unrealistic covariance matrix, then the error happens. Which makes me think | Unscented Kalman filter-negative covariance matrix
I also have this problem. However, I realised that depending on the covariance that you fix, it does not happen. If I set an unrealistic covariance matrix, then the error happens. Which makes me think that is perhaps a matter of numerical errors making the covariance m... | Unscented Kalman filter-negative covariance matrix
I also have this problem. However, I realised that depending on the covariance that you fix, it does not happen. If I set an unrealistic covariance matrix, then the error happens. Which makes me think |
37,836 | Modelling a mixed model in JAGS/BUGS [closed] | Sometimes ifelse doesn't work. Instead
is0X[i] <- ifelse(goalsScored[i, 1]==0, 1, 0)
you should try
is0X[i] <- goalsScored[i, 1]==0
goalsScored[i, 1]==0 returns 1 if True and 0 if False | Modelling a mixed model in JAGS/BUGS [closed] | Sometimes ifelse doesn't work. Instead
is0X[i] <- ifelse(goalsScored[i, 1]==0, 1, 0)
you should try
is0X[i] <- goalsScored[i, 1]==0
goalsScored[i, 1]==0 returns 1 if True and 0 if False | Modelling a mixed model in JAGS/BUGS [closed]
Sometimes ifelse doesn't work. Instead
is0X[i] <- ifelse(goalsScored[i, 1]==0, 1, 0)
you should try
is0X[i] <- goalsScored[i, 1]==0
goalsScored[i, 1]==0 returns 1 if True and 0 if False | Modelling a mixed model in JAGS/BUGS [closed]
Sometimes ifelse doesn't work. Instead
is0X[i] <- ifelse(goalsScored[i, 1]==0, 1, 0)
you should try
is0X[i] <- goalsScored[i, 1]==0
goalsScored[i, 1]==0 returns 1 if True and 0 if False |
37,837 | Modelling a mixed model in JAGS/BUGS [closed] | I do not think you can define zeros[i] ~ dpois(-log(kappa[i]) + C) inside of the model construction.
Try to revise the code to be zeros ~ dpois(-log(kappa[i]) + C) (take out of '[i]').
After defining the model, you re-define the data at zeors:
data$zero=0
Try if this works.
Refer to The zero-crossings trick for JAGS:... | Modelling a mixed model in JAGS/BUGS [closed] | I do not think you can define zeros[i] ~ dpois(-log(kappa[i]) + C) inside of the model construction.
Try to revise the code to be zeros ~ dpois(-log(kappa[i]) + C) (take out of '[i]').
After defining | Modelling a mixed model in JAGS/BUGS [closed]
I do not think you can define zeros[i] ~ dpois(-log(kappa[i]) + C) inside of the model construction.
Try to revise the code to be zeros ~ dpois(-log(kappa[i]) + C) (take out of '[i]').
After defining the model, you re-define the data at zeors:
data$zero=0
Try if this work... | Modelling a mixed model in JAGS/BUGS [closed]
I do not think you can define zeros[i] ~ dpois(-log(kappa[i]) + C) inside of the model construction.
Try to revise the code to be zeros ~ dpois(-log(kappa[i]) + C) (take out of '[i]').
After defining |
37,838 | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? | The bivariate normal distribution is the exception, not the rule!
It is important to recognize that "almost all" joint distributions with normal marginals are not the bivariate normal distribution. That is, the common viewpoint that joint distributions with normal marginals that are not the bivariate normal are somehow... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G | The bivariate normal distribution is the exception, not the rule!
It is important to recognize that "almost all" joint distributions with normal marginals are not the bivariate normal distribution. Th | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?
The bivariate normal distribution is the exception, not the rule!
It is important to recognize that "almost all" joint distributions with normal marginals are not the bivariate normal distribution. That is, the ... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G
The bivariate normal distribution is the exception, not the rule!
It is important to recognize that "almost all" joint distributions with normal marginals are not the bivariate normal distribution. Th |
37,839 | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? | It is true that each element of a multivariate normal vector is itself normally distributed, and you can deduce their means and variances. However, it is not true that any two Guassian random variables are jointly normally distributed. Here is an example:
Edit: In response to the consensus that a random variable that i... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G | It is true that each element of a multivariate normal vector is itself normally distributed, and you can deduce their means and variances. However, it is not true that any two Guassian random variable | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?
It is true that each element of a multivariate normal vector is itself normally distributed, and you can deduce their means and variances. However, it is not true that any two Guassian random variables are joint... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G
It is true that each element of a multivariate normal vector is itself normally distributed, and you can deduce their means and variances. However, it is not true that any two Guassian random variable |
37,840 | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? | The following post contains an outline of a proof, just to give the main ideas and get you started.
Let $z = (Z_1, Z_2)$ be two independent Gaussian random variables and let $x = (X_1, X_2)$ be
$$
x =
\begin{pmatrix}
X_1 \\
X_2
\end{pmatrix}
= \begin{pmatrix}
\alpha_{11} Z_1 + \alpha_{12} Z_2\\
\alpha_{21} Z_1 + \alpha... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G | The following post contains an outline of a proof, just to give the main ideas and get you started.
Let $z = (Z_1, Z_2)$ be two independent Gaussian random variables and let $x = (X_1, X_2)$ be
$$
x = | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?
The following post contains an outline of a proof, just to give the main ideas and get you started.
Let $z = (Z_1, Z_2)$ be two independent Gaussian random variables and let $x = (X_1, X_2)$ be
$$
x =
\begin{pma... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G
The following post contains an outline of a proof, just to give the main ideas and get you started.
Let $z = (Z_1, Z_2)$ be two independent Gaussian random variables and let $x = (X_1, X_2)$ be
$$
x = |
37,841 | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? | I thought it might be worth pointing out a couple of nice examples; one I've mentioned in a couple of older answers here on Cross Validated (e.g. this one) and one rather pretty one which occurred to me the other day.
Here we have two variables, $Y$ and $Z$, that have (uncorrelated) normal distributions, where $Y$ is ... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G | I thought it might be worth pointing out a couple of nice examples; one I've mentioned in a couple of older answers here on Cross Validated (e.g. this one) and one rather pretty one which occurred to | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?
I thought it might be worth pointing out a couple of nice examples; one I've mentioned in a couple of older answers here on Cross Validated (e.g. this one) and one rather pretty one which occurred to me the othe... | Is it possible to have a pair of Gaussian random variables for which the joint distribution is not G
I thought it might be worth pointing out a couple of nice examples; one I've mentioned in a couple of older answers here on Cross Validated (e.g. this one) and one rather pretty one which occurred to |
37,842 | Help with Bayesian derivation of normal model with conjugate prior | [Here is the full text from our book]
The case of a normal distribution with a known variance being quite
unrealistic, we now consider the general case of an iid sample
$$\mathscr{D}_n=(x_1,\ldots,x_n)$$ from the normal distribution
$\mathscr{N}(\mu,\sigma^2)$ and $\theta=(\mu,\sigma^2)$. Keeping the
same pri... | Help with Bayesian derivation of normal model with conjugate prior | [Here is the full text from our book]
The case of a normal distribution with a known variance being quite
unrealistic, we now consider the general case of an iid sample
$$\mathscr{D}_n=(x_1,\ldot | Help with Bayesian derivation of normal model with conjugate prior
[Here is the full text from our book]
The case of a normal distribution with a known variance being quite
unrealistic, we now consider the general case of an iid sample
$$\mathscr{D}_n=(x_1,\ldots,x_n)$$ from the normal distribution
$\mathscr{N}... | Help with Bayesian derivation of normal model with conjugate prior
[Here is the full text from our book]
The case of a normal distribution with a known variance being quite
unrealistic, we now consider the general case of an iid sample
$$\mathscr{D}_n=(x_1,\ldot |
37,843 | What are some use of dense matrices in statistics? | You might find the Java Matrix Benchmark useful. The Matrix Market does not seem to have what you want, although it has many examples. | What are some use of dense matrices in statistics? | You might find the Java Matrix Benchmark useful. The Matrix Market does not seem to have what you want, although it has many examples. | What are some use of dense matrices in statistics?
You might find the Java Matrix Benchmark useful. The Matrix Market does not seem to have what you want, although it has many examples. | What are some use of dense matrices in statistics?
You might find the Java Matrix Benchmark useful. The Matrix Market does not seem to have what you want, although it has many examples. |
37,844 | What are some use of dense matrices in statistics? | Here is large, although I'm not sure if it's dense enough for you. From http://www.grouplens.org/node/73
MovieLens 100k - Consists of 100,000 ratings from 1000 users on 1700
movies.
MovieLens 1M - Consists of 1 million ratings from 6000 users on 4000
movies.
MovieLens 10M - Consists of 10 million ratings and 100,000 t... | What are some use of dense matrices in statistics? | Here is large, although I'm not sure if it's dense enough for you. From http://www.grouplens.org/node/73
MovieLens 100k - Consists of 100,000 ratings from 1000 users on 1700
movies.
MovieLens 1M - Co | What are some use of dense matrices in statistics?
Here is large, although I'm not sure if it's dense enough for you. From http://www.grouplens.org/node/73
MovieLens 100k - Consists of 100,000 ratings from 1000 users on 1700
movies.
MovieLens 1M - Consists of 1 million ratings from 6000 users on 4000
movies.
MovieLens... | What are some use of dense matrices in statistics?
Here is large, although I'm not sure if it's dense enough for you. From http://www.grouplens.org/node/73
MovieLens 100k - Consists of 100,000 ratings from 1000 users on 1700
movies.
MovieLens 1M - Co |
37,845 | What are some use of dense matrices in statistics? | I'm not sure the application you are seeking would make sense in a statistical context. What you're interested in is a linear regression analysis. $A\in R^{m\times n}$ is a matrix of $m$ measurements in which each row is a single measurement of $n$ variables. Two potential applications with possibly $n>5000$ come to my... | What are some use of dense matrices in statistics? | I'm not sure the application you are seeking would make sense in a statistical context. What you're interested in is a linear regression analysis. $A\in R^{m\times n}$ is a matrix of $m$ measurements | What are some use of dense matrices in statistics?
I'm not sure the application you are seeking would make sense in a statistical context. What you're interested in is a linear regression analysis. $A\in R^{m\times n}$ is a matrix of $m$ measurements in which each row is a single measurement of $n$ variables. Two poten... | What are some use of dense matrices in statistics?
I'm not sure the application you are seeking would make sense in a statistical context. What you're interested in is a linear regression analysis. $A\in R^{m\times n}$ is a matrix of $m$ measurements |
37,846 | Assessing the representativeness of population sampling | In survey sampling for commercial and government studies the orthodox approach is as compare the characteristics of the sample with those of the population. For example, comparing the % female, % under 24, etc. The closer the correspondence between the sample and known data for the entire population, the more confide... | Assessing the representativeness of population sampling | In survey sampling for commercial and government studies the orthodox approach is as compare the characteristics of the sample with those of the population. For example, comparing the % female, % und | Assessing the representativeness of population sampling
In survey sampling for commercial and government studies the orthodox approach is as compare the characteristics of the sample with those of the population. For example, comparing the % female, % under 24, etc. The closer the correspondence between the sample an... | Assessing the representativeness of population sampling
In survey sampling for commercial and government studies the orthodox approach is as compare the characteristics of the sample with those of the population. For example, comparing the % female, % und |
37,847 | Feature engineering for sheet music | 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.
Maybe useful in this context, also a little bit too la... | Feature engineering for sheet music | 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.
| Feature engineering for sheet music
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.
Maybe useful in th... | Feature engineering for sheet music
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.
|
37,848 | Train accuracy < Test accuracy with regularization | First, As I know, the training sample size is usually not less than the test sample size. However, it is not the case in your similation. Second, as comment as @gung - Reinstate Monica, how did you train the dataset, the cross-validaiton or other methods? Third, which kind of regularization you used, the LASSO, ridge o... | Train accuracy < Test accuracy with regularization | First, As I know, the training sample size is usually not less than the test sample size. However, it is not the case in your similation. Second, as comment as @gung - Reinstate Monica, how did you tr | Train accuracy < Test accuracy with regularization
First, As I know, the training sample size is usually not less than the test sample size. However, it is not the case in your similation. Second, as comment as @gung - Reinstate Monica, how did you train the dataset, the cross-validaiton or other methods? Third, which ... | Train accuracy < Test accuracy with regularization
First, As I know, the training sample size is usually not less than the test sample size. However, it is not the case in your similation. Second, as comment as @gung - Reinstate Monica, how did you tr |
37,849 | Train accuracy < Test accuracy with regularization | This is only a guess, but I suspect the regularization is interacting with the logistic regression optimizer. In principle, if you can find optimal loss-minimizing parameters, regularization won't increase performance, and instead is likely to lower it (on the training set). However, for large data sets, there are typi... | Train accuracy < Test accuracy with regularization | This is only a guess, but I suspect the regularization is interacting with the logistic regression optimizer. In principle, if you can find optimal loss-minimizing parameters, regularization won't inc | Train accuracy < Test accuracy with regularization
This is only a guess, but I suspect the regularization is interacting with the logistic regression optimizer. In principle, if you can find optimal loss-minimizing parameters, regularization won't increase performance, and instead is likely to lower it (on the training... | Train accuracy < Test accuracy with regularization
This is only a guess, but I suspect the regularization is interacting with the logistic regression optimizer. In principle, if you can find optimal loss-minimizing parameters, regularization won't inc |
37,850 | Train accuracy < Test accuracy with regularization | Did you look at the distribution of the classes... It may most likely be due imbalanced class distibutions. For example, if you sample contain two class labels 'A', 'B' and if 'A' occurs 80% of the times in your dataset. Assume that your classifier almost always classifies any test data as beloning to class 'A'. Then y... | Train accuracy < Test accuracy with regularization | Did you look at the distribution of the classes... It may most likely be due imbalanced class distibutions. For example, if you sample contain two class labels 'A', 'B' and if 'A' occurs 80% of the ti | Train accuracy < Test accuracy with regularization
Did you look at the distribution of the classes... It may most likely be due imbalanced class distibutions. For example, if you sample contain two class labels 'A', 'B' and if 'A' occurs 80% of the times in your dataset. Assume that your classifier almost always classi... | Train accuracy < Test accuracy with regularization
Did you look at the distribution of the classes... It may most likely be due imbalanced class distibutions. For example, if you sample contain two class labels 'A', 'B' and if 'A' occurs 80% of the ti |
37,851 | Using Google Causal Impact package to assess the significance of a planned intervention | About seasonality. From the reference:
. For example, if the data represent daily observations, use 7 for a
day-of-week component. This interface currently only supports up to
one seasonal component.
and
use model.args = list(nseasons = 7, season.duration = 24) to add a
day-of-week component to data with hour... | Using Google Causal Impact package to assess the significance of a planned intervention | About seasonality. From the reference:
. For example, if the data represent daily observations, use 7 for a
day-of-week component. This interface currently only supports up to
one seasonal compo | Using Google Causal Impact package to assess the significance of a planned intervention
About seasonality. From the reference:
. For example, if the data represent daily observations, use 7 for a
day-of-week component. This interface currently only supports up to
one seasonal component.
and
use model.args = lis... | Using Google Causal Impact package to assess the significance of a planned intervention
About seasonality. From the reference:
. For example, if the data represent daily observations, use 7 for a
day-of-week component. This interface currently only supports up to
one seasonal compo |
37,852 | Sum of normal truncated random variables | You could use approximation by saddlepoint methods, for the sum of truncated normals. I will not give the details now, you can look at my answer to General sum of Gamma distributions for hints. What we need is to find the moment-generating function for a truncated normal, which is easy. I will do it here for a stand... | Sum of normal truncated random variables | You could use approximation by saddlepoint methods, for the sum of truncated normals. I will not give the details now, you can look at my answer to General sum of Gamma distributions for hints. Wha | Sum of normal truncated random variables
You could use approximation by saddlepoint methods, for the sum of truncated normals. I will not give the details now, you can look at my answer to General sum of Gamma distributions for hints. What we need is to find the moment-generating function for a truncated normal, whi... | Sum of normal truncated random variables
You could use approximation by saddlepoint methods, for the sum of truncated normals. I will not give the details now, you can look at my answer to General sum of Gamma distributions for hints. Wha |
37,853 | Sum of normal truncated random variables | I'm curious why, but yes, there is a simple way to generate the pdf of this sum of distributions:
## install.packages("truncnorm")
## install.packages("caTools")
library(truncnorm)
x.mu <- c(12, 18, 7)
x.sd <- c(1.5, 2, 0.8)
x.a <- x.mu - 2*x.sd
x.b <- x.mu + 2*x.sd
dmulti <- function(x, a, b, mu, sd)
rowSums(
... | Sum of normal truncated random variables | I'm curious why, but yes, there is a simple way to generate the pdf of this sum of distributions:
## install.packages("truncnorm")
## install.packages("caTools")
library(truncnorm)
x.mu <- c(12, 18, | Sum of normal truncated random variables
I'm curious why, but yes, there is a simple way to generate the pdf of this sum of distributions:
## install.packages("truncnorm")
## install.packages("caTools")
library(truncnorm)
x.mu <- c(12, 18, 7)
x.sd <- c(1.5, 2, 0.8)
x.a <- x.mu - 2*x.sd
x.b <- x.mu + 2*x.sd
dmulti <- ... | Sum of normal truncated random variables
I'm curious why, but yes, there is a simple way to generate the pdf of this sum of distributions:
## install.packages("truncnorm")
## install.packages("caTools")
library(truncnorm)
x.mu <- c(12, 18, |
37,854 | How to subset alternatives in nested multinomial logistic regression? | The general term for the type of model you want is a choice model. Assuming your description of the problem is correct, the choices that you are trying to model are not Fish vs Don't fish, but instead the days of the week.
The second level of your hierarchy is not, I think, really a level of a hierarchy in terms of the... | How to subset alternatives in nested multinomial logistic regression? | The general term for the type of model you want is a choice model. Assuming your description of the problem is correct, the choices that you are trying to model are not Fish vs Don't fish, but instead | How to subset alternatives in nested multinomial logistic regression?
The general term for the type of model you want is a choice model. Assuming your description of the problem is correct, the choices that you are trying to model are not Fish vs Don't fish, but instead the days of the week.
The second level of your hi... | How to subset alternatives in nested multinomial logistic regression?
The general term for the type of model you want is a choice model. Assuming your description of the problem is correct, the choices that you are trying to model are not Fish vs Don't fish, but instead |
37,855 | Interpreting odds ratios less than 1 with 3-category outcome | Given the coeffcient is significant, it means that the cummulative odds for being in a higher food category are .62 times as high for people with less time than for people with no time. Put differently, having more time than no time decreases the odds (and also the probability) for consuming more food (across all categ... | Interpreting odds ratios less than 1 with 3-category outcome | Given the coeffcient is significant, it means that the cummulative odds for being in a higher food category are .62 times as high for people with less time than for people with no time. Put differentl | Interpreting odds ratios less than 1 with 3-category outcome
Given the coeffcient is significant, it means that the cummulative odds for being in a higher food category are .62 times as high for people with less time than for people with no time. Put differently, having more time than no time decreases the odds (and al... | Interpreting odds ratios less than 1 with 3-category outcome
Given the coeffcient is significant, it means that the cummulative odds for being in a higher food category are .62 times as high for people with less time than for people with no time. Put differentl |
37,856 | SVM with quadratic loss | They are different specially in the case of regression. The effect of the quadratic loss is to average errors (so it is more sensitive to outliers). To see this, if you minimize,
$$
L_{2}(y) = \sum_{i}(y-x_{i})^{2}
$$
you get the mean value. The effect of the L1 loss function is to , so that it leads to sparse solution... | SVM with quadratic loss | They are different specially in the case of regression. The effect of the quadratic loss is to average errors (so it is more sensitive to outliers). To see this, if you minimize,
$$
L_{2}(y) = \sum_{i | SVM with quadratic loss
They are different specially in the case of regression. The effect of the quadratic loss is to average errors (so it is more sensitive to outliers). To see this, if you minimize,
$$
L_{2}(y) = \sum_{i}(y-x_{i})^{2}
$$
you get the mean value. The effect of the L1 loss function is to , so that it ... | SVM with quadratic loss
They are different specially in the case of regression. The effect of the quadratic loss is to average errors (so it is more sensitive to outliers). To see this, if you minimize,
$$
L_{2}(y) = \sum_{i |
37,857 | SVM with quadratic loss | "In the dual problem the L2 loss constant can be merged into the kernel."
It can be done in the hinge loss too. You only need to rephrase the formulation.
It is done here, for instance:
A nested heuristic for parameter tuning in Support Vector Machines
E Carrizosa, B Martín-Barragán, D Romero Morales
Computers & Opera... | SVM with quadratic loss | "In the dual problem the L2 loss constant can be merged into the kernel."
It can be done in the hinge loss too. You only need to rephrase the formulation.
It is done here, for instance:
A nested heur | SVM with quadratic loss
"In the dual problem the L2 loss constant can be merged into the kernel."
It can be done in the hinge loss too. You only need to rephrase the formulation.
It is done here, for instance:
A nested heuristic for parameter tuning in Support Vector Machines
E Carrizosa, B Martín-Barragán, D Romero M... | SVM with quadratic loss
"In the dual problem the L2 loss constant can be merged into the kernel."
It can be done in the hinge loss too. You only need to rephrase the formulation.
It is done here, for instance:
A nested heur |
37,858 | Fitting a curve OVER OR UNDER a set of points | As Mark L. Stone comments, your more general problem is indeed known as expectile regression. A nice overview can be found in Waltrup et al. (2015). We need to map your weights of $1$ and $\alpha$ to expectiles that sum to $1$; a little algebra shows that you need $\frac{1}{1+\alpha}$-expectiles.
Again per Mark's comme... | Fitting a curve OVER OR UNDER a set of points | As Mark L. Stone comments, your more general problem is indeed known as expectile regression. A nice overview can be found in Waltrup et al. (2015). We need to map your weights of $1$ and $\alpha$ to | Fitting a curve OVER OR UNDER a set of points
As Mark L. Stone comments, your more general problem is indeed known as expectile regression. A nice overview can be found in Waltrup et al. (2015). We need to map your weights of $1$ and $\alpha$ to expectiles that sum to $1$; a little algebra shows that you need $\frac{1}... | Fitting a curve OVER OR UNDER a set of points
As Mark L. Stone comments, your more general problem is indeed known as expectile regression. A nice overview can be found in Waltrup et al. (2015). We need to map your weights of $1$ and $\alpha$ to |
37,859 | Fitting a curve OVER OR UNDER a set of points | Here is example code using the Python scipy.optimize.differential_evolution genetic algorithm module implementing a "brick wall" that gives a very large error if the genetic algorithm finds parameters that yield any predicted value above or below that of any data point per a code switch. It works in my tests when I fli... | Fitting a curve OVER OR UNDER a set of points | Here is example code using the Python scipy.optimize.differential_evolution genetic algorithm module implementing a "brick wall" that gives a very large error if the genetic algorithm finds parameters | Fitting a curve OVER OR UNDER a set of points
Here is example code using the Python scipy.optimize.differential_evolution genetic algorithm module implementing a "brick wall" that gives a very large error if the genetic algorithm finds parameters that yield any predicted value above or below that of any data point per ... | Fitting a curve OVER OR UNDER a set of points
Here is example code using the Python scipy.optimize.differential_evolution genetic algorithm module implementing a "brick wall" that gives a very large error if the genetic algorithm finds parameters |
37,860 | Asymptotics of the survival function for Anderson Darling distribution? | An asymptotic form for the AD distribution in
C.D. Sinclair and B.D. Spurr, Journal of the American statistical association, 83, p. 1190-1191, (1988) | Asymptotics of the survival function for Anderson Darling distribution? | An asymptotic form for the AD distribution in
C.D. Sinclair and B.D. Spurr, Journal of the American statistical association, 83, p. 1190-1191, (1988) | Asymptotics of the survival function for Anderson Darling distribution?
An asymptotic form for the AD distribution in
C.D. Sinclair and B.D. Spurr, Journal of the American statistical association, 83, p. 1190-1191, (1988) | Asymptotics of the survival function for Anderson Darling distribution?
An asymptotic form for the AD distribution in
C.D. Sinclair and B.D. Spurr, Journal of the American statistical association, 83, p. 1190-1191, (1988) |
37,861 | Ranking of categorical variables in logistic regression | Since you are interested in ranking the categories, you may want to re-code the categorical variables into a number of separate binary variables.
Example: Create a binary variable for express delivery- which would take the value 1 for express delivery cases and 0 otherwise. Similarly, a binary variable for standard de... | Ranking of categorical variables in logistic regression | Since you are interested in ranking the categories, you may want to re-code the categorical variables into a number of separate binary variables.
Example: Create a binary variable for express deliver | Ranking of categorical variables in logistic regression
Since you are interested in ranking the categories, you may want to re-code the categorical variables into a number of separate binary variables.
Example: Create a binary variable for express delivery- which would take the value 1 for express delivery cases and 0... | Ranking of categorical variables in logistic regression
Since you are interested in ranking the categories, you may want to re-code the categorical variables into a number of separate binary variables.
Example: Create a binary variable for express deliver |
37,862 | Ranking of categorical variables in logistic regression | You could fit the logistic regression model using only 1 variable at the time and examine the adjusted R2.
The one explaining most of the variance should have more impact on the model...
I am just guessing, not sure that it is a rigorous solution... | Ranking of categorical variables in logistic regression | You could fit the logistic regression model using only 1 variable at the time and examine the adjusted R2.
The one explaining most of the variance should have more impact on the model...
I am just gu | Ranking of categorical variables in logistic regression
You could fit the logistic regression model using only 1 variable at the time and examine the adjusted R2.
The one explaining most of the variance should have more impact on the model...
I am just guessing, not sure that it is a rigorous solution... | Ranking of categorical variables in logistic regression
You could fit the logistic regression model using only 1 variable at the time and examine the adjusted R2.
The one explaining most of the variance should have more impact on the model...
I am just gu |
37,863 | Ranking of categorical variables in logistic regression | This is a common question with a multitude of answers. The simplest is to use standardized features; the absolute value of coefficients that come back can then loosely be interpreted as 'higher' = 'more influence' on the log(odds). For the most part, using standard scores should not affect your overall results (ROC cur... | Ranking of categorical variables in logistic regression | This is a common question with a multitude of answers. The simplest is to use standardized features; the absolute value of coefficients that come back can then loosely be interpreted as 'higher' = 'mo | Ranking of categorical variables in logistic regression
This is a common question with a multitude of answers. The simplest is to use standardized features; the absolute value of coefficients that come back can then loosely be interpreted as 'higher' = 'more influence' on the log(odds). For the most part, using standar... | Ranking of categorical variables in logistic regression
This is a common question with a multitude of answers. The simplest is to use standardized features; the absolute value of coefficients that come back can then loosely be interpreted as 'higher' = 'mo |
37,864 | Finding the distribution of sample range for a Beta population | edit: please stop down voting this comment, it was a comment and should not have been accepted as the answer by the OP. Please read the comments and read the literature if you are unfamiliar with this technique. It seems that the relationship between the uniform distribution and beta distribution of it's ordered statis... | Finding the distribution of sample range for a Beta population | edit: please stop down voting this comment, it was a comment and should not have been accepted as the answer by the OP. Please read the comments and read the literature if you are unfamiliar with this | Finding the distribution of sample range for a Beta population
edit: please stop down voting this comment, it was a comment and should not have been accepted as the answer by the OP. Please read the comments and read the literature if you are unfamiliar with this technique. It seems that the relationship between the un... | Finding the distribution of sample range for a Beta population
edit: please stop down voting this comment, it was a comment and should not have been accepted as the answer by the OP. Please read the comments and read the literature if you are unfamiliar with this |
37,865 | Anomaly detection on time series | I found this article to be very helpful in my case:
https://mapr.com/blog/deep-learning-tensorflow/
Using this basic RNN structure, I was able to predict the outcome of the next timestep.
By centering all events to the nearest minute, the network was able to recognize the pattern that correlates within the timeline. | Anomaly detection on time series | I found this article to be very helpful in my case:
https://mapr.com/blog/deep-learning-tensorflow/
Using this basic RNN structure, I was able to predict the outcome of the next timestep.
By centeri | Anomaly detection on time series
I found this article to be very helpful in my case:
https://mapr.com/blog/deep-learning-tensorflow/
Using this basic RNN structure, I was able to predict the outcome of the next timestep.
By centering all events to the nearest minute, the network was able to recognize the pattern that... | Anomaly detection on time series
I found this article to be very helpful in my case:
https://mapr.com/blog/deep-learning-tensorflow/
Using this basic RNN structure, I was able to predict the outcome of the next timestep.
By centeri |
37,866 | Anomaly detection on time series | A simple approach would be to treat each event type as independent, and then to build one model per event type. If you expect events to happen on a regular schedule, then an informative feature could be time-since-last-event.
To evaluate the viability of such a model one should do some Exploratory Data Analysis and plo... | Anomaly detection on time series | A simple approach would be to treat each event type as independent, and then to build one model per event type. If you expect events to happen on a regular schedule, then an informative feature could | Anomaly detection on time series
A simple approach would be to treat each event type as independent, and then to build one model per event type. If you expect events to happen on a regular schedule, then an informative feature could be time-since-last-event.
To evaluate the viability of such a model one should do some ... | Anomaly detection on time series
A simple approach would be to treat each event type as independent, and then to build one model per event type. If you expect events to happen on a regular schedule, then an informative feature could |
37,867 | Anomaly detection on time series | There are several ways with which you can tackle this. Before jumping into designing any models standardize your data. Your data seems unlabeled, so initially, what you can do is perform a t-SNE visualization on it which will give you a lot insights to your data. Based on its result you can develop more sensible models... | Anomaly detection on time series | There are several ways with which you can tackle this. Before jumping into designing any models standardize your data. Your data seems unlabeled, so initially, what you can do is perform a t-SNE visua | Anomaly detection on time series
There are several ways with which you can tackle this. Before jumping into designing any models standardize your data. Your data seems unlabeled, so initially, what you can do is perform a t-SNE visualization on it which will give you a lot insights to your data. Based on its result you... | Anomaly detection on time series
There are several ways with which you can tackle this. Before jumping into designing any models standardize your data. Your data seems unlabeled, so initially, what you can do is perform a t-SNE visua |
37,868 | Average time ant needs to get out to the woods | $$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$
$$T/3=9$$
$$T=27$$
From start point, each of 3 paths are equally possible. Two paths lead you back to start point. | Average time ant needs to get out to the woods | $$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$
$$T/3=9$$
$$T=27$$
From start point, each of 3 paths are equally possible. Two paths lead you back to start point. | Average time ant needs to get out to the woods
$$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$
$$T/3=9$$
$$T=27$$
From start point, each of 3 paths are equally possible. Two paths lead you back to start point. | Average time ant needs to get out to the woods
$$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$
$$T/3=9$$
$$T=27$$
From start point, each of 3 paths are equally possible. Two paths lead you back to start point. |
37,869 | Average time ant needs to get out to the woods | To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is,
$$
E[T] = \sum_{\text{path} \in \text{possible paths}} p(\text{path}) T(\text{path})
$$
Every possible path takes Passage $A$ only once, but c... | Average time ant needs to get out to the woods | To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is,
$$
E[T] = \sum_{\text{ | Average time ant needs to get out to the woods
To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is,
$$
E[T] = \sum_{\text{path} \in \text{possible paths}} p(\text{path}) T(\text{path})
$$
Every p... | Average time ant needs to get out to the woods
To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is,
$$
E[T] = \sum_{\text{ |
37,870 | Average time ant needs to get out to the woods | Hello donkordr and welcome to CV.
No, the mean does not answer the question unfortunately.
You can read your problem as a Markov Chain with two states: woods, and ant house.
Now, your transition probabilities are:
From the woods: (does not really matter as it's the goal) $p=1$ to stay in the woods
From the ant hous... | Average time ant needs to get out to the woods | Hello donkordr and welcome to CV.
No, the mean does not answer the question unfortunately.
You can read your problem as a Markov Chain with two states: woods, and ant house.
Now, your transition pro | Average time ant needs to get out to the woods
Hello donkordr and welcome to CV.
No, the mean does not answer the question unfortunately.
You can read your problem as a Markov Chain with two states: woods, and ant house.
Now, your transition probabilities are:
From the woods: (does not really matter as it's the goa... | Average time ant needs to get out to the woods
Hello donkordr and welcome to CV.
No, the mean does not answer the question unfortunately.
You can read your problem as a Markov Chain with two states: woods, and ant house.
Now, your transition pro |
37,871 | Average time ant needs to get out to the woods | Let me join with an explanation which, to me at least, seems even easier:
First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times of $B$ and $C$, i.e. $10$ minutes. This is due to the fact that these paths have the same probability. If they didn't, w... | Average time ant needs to get out to the woods | Let me join with an explanation which, to me at least, seems even easier:
First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times | Average time ant needs to get out to the woods
Let me join with an explanation which, to me at least, seems even easier:
First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times of $B$ and $C$, i.e. $10$ minutes. This is due to the fact that these pat... | Average time ant needs to get out to the woods
Let me join with an explanation which, to me at least, seems even easier:
First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times |
37,872 | Average time ant needs to get out to the woods | Typical way to solve this:
The ant will either take path a and finish or take path b or c and be back in it's starting position.
Let $k$ be the number of times that the ant already has taken path b or c. Let $T_k$ be the expectation value for the time to finish for an ant that took already $k$ times the path. Then the ... | Average time ant needs to get out to the woods | Typical way to solve this:
The ant will either take path a and finish or take path b or c and be back in it's starting position.
Let $k$ be the number of times that the ant already has taken path b or | Average time ant needs to get out to the woods
Typical way to solve this:
The ant will either take path a and finish or take path b or c and be back in it's starting position.
Let $k$ be the number of times that the ant already has taken path b or c. Let $T_k$ be the expectation value for the time to finish for an ant ... | Average time ant needs to get out to the woods
Typical way to solve this:
The ant will either take path a and finish or take path b or c and be back in it's starting position.
Let $k$ be the number of times that the ant already has taken path b or |
37,873 | Average time ant needs to get out to the woods | Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):
Both terms are geometric series:
$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty i\cdot q^i=q \frac{d}{dq}\sum_{i=0}^\infty q^i=\frac{q}{(1-q)^2}, \text{ for } |q|<1$, so for $q=\frac{2}{3}$ this equa... | Average time ant needs to get out to the woods | Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):
Both terms are geometric series:
$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty | Average time ant needs to get out to the woods
Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):
Both terms are geometric series:
$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty i\cdot q^i=q \frac{d}{dq}\sum_{i=0}^\infty q^i=\frac{q}{(1-q)^2}, \text{... | Average time ant needs to get out to the woods
Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):
Both terms are geometric series:
$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty |
37,874 | Average time ant needs to get out to the woods | It always take $7$ min to succeed, which happens once.
It takes an average of $10$ min to fail.
The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries.
So the total time is ($7 + 10\cdot \sum_n (2/3)^n$) min.
Which is $7$ min + $2 \cdot 10$ min $ = 27$. | Average time ant needs to get out to the woods | It always take $7$ min to succeed, which happens once.
It takes an average of $10$ min to fail.
The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries.
So the total time is ($7 + 10 | Average time ant needs to get out to the woods
It always take $7$ min to succeed, which happens once.
It takes an average of $10$ min to fail.
The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries.
So the total time is ($7 + 10\cdot \sum_n (2/3)^n$) min.
Which is $7$ min + $2 \cdot 10$ min $ = 27$. | Average time ant needs to get out to the woods
It always take $7$ min to succeed, which happens once.
It takes an average of $10$ min to fail.
The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries.
So the total time is ($7 + 10 |
37,875 | Is it ever good to increase significance level? | Expanding on @EpiGrad 's answer (which is a good answer):
There are many reasons to ignore p-values altogether: Principally, they answer a question we are very rarely interested in.
If you are going to use p-values, using them as cutoffs often makes little sense.
If you are going to use them as cut-offs, you should de... | Is it ever good to increase significance level? | Expanding on @EpiGrad 's answer (which is a good answer):
There are many reasons to ignore p-values altogether: Principally, they answer a question we are very rarely interested in.
If you are going | Is it ever good to increase significance level?
Expanding on @EpiGrad 's answer (which is a good answer):
There are many reasons to ignore p-values altogether: Principally, they answer a question we are very rarely interested in.
If you are going to use p-values, using them as cutoffs often makes little sense.
If you ... | Is it ever good to increase significance level?
Expanding on @EpiGrad 's answer (which is a good answer):
There are many reasons to ignore p-values altogether: Principally, they answer a question we are very rarely interested in.
If you are going |
37,876 | Is it ever good to increase significance level? | Just if you're not getting significant results? No. Fiddling with the significance level after the experiment is conducted and the results are known is never good practice.
There are circumstances where you might chose a more relaxed p-value, but doing it post hoc is just a bad idea. | Is it ever good to increase significance level? | Just if you're not getting significant results? No. Fiddling with the significance level after the experiment is conducted and the results are known is never good practice.
There are circumstances whe | Is it ever good to increase significance level?
Just if you're not getting significant results? No. Fiddling with the significance level after the experiment is conducted and the results are known is never good practice.
There are circumstances where you might chose a more relaxed p-value, but doing it post hoc is just... | Is it ever good to increase significance level?
Just if you're not getting significant results? No. Fiddling with the significance level after the experiment is conducted and the results are known is never good practice.
There are circumstances whe |
37,877 | Is it ever good to increase significance level? | On the one hand it is somewhat artificial to discount a variable because its p value is higher than 0.01 (that's an unusually stringent criterion). How you get there may be more important than what is the ultimate significance level. A variable that is well grounded on logic or causal links with an acceptable p value... | Is it ever good to increase significance level? | On the one hand it is somewhat artificial to discount a variable because its p value is higher than 0.01 (that's an unusually stringent criterion). How you get there may be more important than what i | Is it ever good to increase significance level?
On the one hand it is somewhat artificial to discount a variable because its p value is higher than 0.01 (that's an unusually stringent criterion). How you get there may be more important than what is the ultimate significance level. A variable that is well grounded on ... | Is it ever good to increase significance level?
On the one hand it is somewhat artificial to discount a variable because its p value is higher than 0.01 (that's an unusually stringent criterion). How you get there may be more important than what i |
37,878 | Is it ever good to increase significance level? | I found myself asking the same question and came here looking for good arguments, and I have to say I'm not convinced. Let me explain why I think it is fine to change significance level after an experiment.
Whatever you calculate using your chosen significance level a, will have the same value whether you choose it be... | Is it ever good to increase significance level? | I found myself asking the same question and came here looking for good arguments, and I have to say I'm not convinced. Let me explain why I think it is fine to change significance level after an exper | Is it ever good to increase significance level?
I found myself asking the same question and came here looking for good arguments, and I have to say I'm not convinced. Let me explain why I think it is fine to change significance level after an experiment.
Whatever you calculate using your chosen significance level a, w... | Is it ever good to increase significance level?
I found myself asking the same question and came here looking for good arguments, and I have to say I'm not convinced. Let me explain why I think it is fine to change significance level after an exper |
37,879 | Why is my regression insignificant when I merge data that produced two significant regressions? | Without seeing your data, this is difficult to answer definitively. One possibility is that your datasets span different ranges of the independent variable. It is well-known that combining data across different groups can sometimes reverse correlations seen in each group individually. This effect is known as Simpson'... | Why is my regression insignificant when I merge data that produced two significant regressions? | Without seeing your data, this is difficult to answer definitively. One possibility is that your datasets span different ranges of the independent variable. It is well-known that combining data acros | Why is my regression insignificant when I merge data that produced two significant regressions?
Without seeing your data, this is difficult to answer definitively. One possibility is that your datasets span different ranges of the independent variable. It is well-known that combining data across different groups can s... | Why is my regression insignificant when I merge data that produced two significant regressions?
Without seeing your data, this is difficult to answer definitively. One possibility is that your datasets span different ranges of the independent variable. It is well-known that combining data acros |
37,880 | Why is my regression insignificant when I merge data that produced two significant regressions? | If your data looks something like this then the reason may be more obvious. Your two original regression lines would be almost parallel and look reasonably plausible but combined they produce a different result which is probably not very helpful.
The data for this chart came from using the R code
exdf <- data.frame... | Why is my regression insignificant when I merge data that produced two significant regressions? | If your data looks something like this then the reason may be more obvious. Your two original regression lines would be almost parallel and look reasonably plausible but combined they produce a diffe | Why is my regression insignificant when I merge data that produced two significant regressions?
If your data looks something like this then the reason may be more obvious. Your two original regression lines would be almost parallel and look reasonably plausible but combined they produce a different result which is pro... | Why is my regression insignificant when I merge data that produced two significant regressions?
If your data looks something like this then the reason may be more obvious. Your two original regression lines would be almost parallel and look reasonably plausible but combined they produce a diffe |
37,881 | Why is my regression insignificant when I merge data that produced two significant regressions? | It's also possible the data points in each dataset may have completely different distributions due to outliers and/or nonlinear relationships between $x$ and $y$, and yet still share nearly identical linear regression coefficients, standard errors, and statistically significant $p$-values. Combining the two datasets co... | Why is my regression insignificant when I merge data that produced two significant regressions? | It's also possible the data points in each dataset may have completely different distributions due to outliers and/or nonlinear relationships between $x$ and $y$, and yet still share nearly identical | Why is my regression insignificant when I merge data that produced two significant regressions?
It's also possible the data points in each dataset may have completely different distributions due to outliers and/or nonlinear relationships between $x$ and $y$, and yet still share nearly identical linear regression coeffi... | Why is my regression insignificant when I merge data that produced two significant regressions?
It's also possible the data points in each dataset may have completely different distributions due to outliers and/or nonlinear relationships between $x$ and $y$, and yet still share nearly identical |
37,882 | Why is my regression insignificant when I merge data that produced two significant regressions? | For more on Simpson's Paradox see Pearl, J., & Mackenzie, D. (2018). Paradoxes Galore! The Book of Why: The New Science of Cause and Effect (Kindle ed., pp. 2843-3283). New York: Basic Books. Also, see Pearl's Causality.
In his book, Pearl gives an example very similar to yours. The problem is that there is a confound... | Why is my regression insignificant when I merge data that produced two significant regressions? | For more on Simpson's Paradox see Pearl, J., & Mackenzie, D. (2018). Paradoxes Galore! The Book of Why: The New Science of Cause and Effect (Kindle ed., pp. 2843-3283). New York: Basic Books. Also, se | Why is my regression insignificant when I merge data that produced two significant regressions?
For more on Simpson's Paradox see Pearl, J., & Mackenzie, D. (2018). Paradoxes Galore! The Book of Why: The New Science of Cause and Effect (Kindle ed., pp. 2843-3283). New York: Basic Books. Also, see Pearl's Causality.
In ... | Why is my regression insignificant when I merge data that produced two significant regressions?
For more on Simpson's Paradox see Pearl, J., & Mackenzie, D. (2018). Paradoxes Galore! The Book of Why: The New Science of Cause and Effect (Kindle ed., pp. 2843-3283). New York: Basic Books. Also, se |
37,883 | Can I interpret logistic regression coefficients and their p-values even if model performance is bad? | 1893 observations is a lot. All kinds of hypothesis tests will yield statistical significance even for tiny effects if the sample size is large (we have many threads on this, here is a similar question on ANOVA I recently had).
Thus, it looks like there is signal in your data, it's just very weak. Some effects in the w... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad | 1893 observations is a lot. All kinds of hypothesis tests will yield statistical significance even for tiny effects if the sample size is large (we have many threads on this, here is a similar questio | Can I interpret logistic regression coefficients and their p-values even if model performance is bad?
1893 observations is a lot. All kinds of hypothesis tests will yield statistical significance even for tiny effects if the sample size is large (we have many threads on this, here is a similar question on ANOVA I recen... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad
1893 observations is a lot. All kinds of hypothesis tests will yield statistical significance even for tiny effects if the sample size is large (we have many threads on this, here is a similar questio |
37,884 | Can I interpret logistic regression coefficients and their p-values even if model performance is bad? | I wanted to add a visual example reproducing results similar to the OP. (I usually understand things when I write the code for them).
Here we have 2000 datapoints. The true relationship between x and y is $y = 0.1x$, but the data is very noisy:
set.seed(1234)
N <- 2000
x <- seq(1, 5, length.out=N)
b <- 0.1
y <- b * x +... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad | I wanted to add a visual example reproducing results similar to the OP. (I usually understand things when I write the code for them).
Here we have 2000 datapoints. The true relationship between x and | Can I interpret logistic regression coefficients and their p-values even if model performance is bad?
I wanted to add a visual example reproducing results similar to the OP. (I usually understand things when I write the code for them).
Here we have 2000 datapoints. The true relationship between x and y is $y = 0.1x$, b... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad
I wanted to add a visual example reproducing results similar to the OP. (I usually understand things when I write the code for them).
Here we have 2000 datapoints. The true relationship between x and |
37,885 | Can I interpret logistic regression coefficients and their p-values even if model performance is bad? | It is routine in many areas to do inference with models that have low performance. I cannot think of any examples with $R^2$ values as low as the pseudo $R^2$ you have, but I know I've read papers in high-end journals that do inference on models with linear regression $R^2$ values around $4\%$, maybe even lower.
A caus... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad | It is routine in many areas to do inference with models that have low performance. I cannot think of any examples with $R^2$ values as low as the pseudo $R^2$ you have, but I know I've read papers in | Can I interpret logistic regression coefficients and their p-values even if model performance is bad?
It is routine in many areas to do inference with models that have low performance. I cannot think of any examples with $R^2$ values as low as the pseudo $R^2$ you have, but I know I've read papers in high-end journals ... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad
It is routine in many areas to do inference with models that have low performance. I cannot think of any examples with $R^2$ values as low as the pseudo $R^2$ you have, but I know I've read papers in |
37,886 | Can I interpret logistic regression coefficients and their p-values even if model performance is bad? | If you are a trader and you can get an pseuo-R² of 0.001 in predicting future financial transactions, you're the richest man in the world. If you are a CPU engineer and your design gets a pseudo-R² of 0.999 at telling if a bit is 0 or 1, you have a useless piece of silicon.
This statement is a copy from the question: D... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad | If you are a trader and you can get an pseuo-R² of 0.001 in predicting future financial transactions, you're the richest man in the world. If you are a CPU engineer and your design gets a pseudo-R² of | Can I interpret logistic regression coefficients and their p-values even if model performance is bad?
If you are a trader and you can get an pseuo-R² of 0.001 in predicting future financial transactions, you're the richest man in the world. If you are a CPU engineer and your design gets a pseudo-R² of 0.999 at telling ... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad
If you are a trader and you can get an pseuo-R² of 0.001 in predicting future financial transactions, you're the richest man in the world. If you are a CPU engineer and your design gets a pseudo-R² of |
37,887 | Can I interpret logistic regression coefficients and their p-values even if model performance is bad? | I would tentatively say, "yes". Remember that pseudo-$R^2$ is an analogue to $R^2$, which in OLS is variance explained. Pseudo-$R^2$ is not that, but it's roughly similar. So your low pseudo-$R^2$ could mean you're missing an explanatory variable, a random effect etc. (many reasons possible). In other words, it's a not... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad | I would tentatively say, "yes". Remember that pseudo-$R^2$ is an analogue to $R^2$, which in OLS is variance explained. Pseudo-$R^2$ is not that, but it's roughly similar. So your low pseudo-$R^2$ cou | Can I interpret logistic regression coefficients and their p-values even if model performance is bad?
I would tentatively say, "yes". Remember that pseudo-$R^2$ is an analogue to $R^2$, which in OLS is variance explained. Pseudo-$R^2$ is not that, but it's roughly similar. So your low pseudo-$R^2$ could mean you're mis... | Can I interpret logistic regression coefficients and their p-values even if model performance is bad
I would tentatively say, "yes". Remember that pseudo-$R^2$ is an analogue to $R^2$, which in OLS is variance explained. Pseudo-$R^2$ is not that, but it's roughly similar. So your low pseudo-$R^2$ cou |
37,888 | regression for binary classification | Intriguing question, I had this question for a while,. Here is my findings
Short Answer
You can create any number of classifier you want, but the point is, you can only prove a few of them to be Bayes/universally-consistent! ( Bayes consistency means that classifier is asymptotically optimal, i.e. with infinite data it... | regression for binary classification | Intriguing question, I had this question for a while,. Here is my findings
Short Answer
You can create any number of classifier you want, but the point is, you can only prove a few of them to be Bayes | regression for binary classification
Intriguing question, I had this question for a while,. Here is my findings
Short Answer
You can create any number of classifier you want, but the point is, you can only prove a few of them to be Bayes/universally-consistent! ( Bayes consistency means that classifier is asymptoticall... | regression for binary classification
Intriguing question, I had this question for a while,. Here is my findings
Short Answer
You can create any number of classifier you want, but the point is, you can only prove a few of them to be Bayes |
37,889 | regression for binary classification | You begin with a misunderstanding. It is important to get the terminology right at the beginning. Logistic regression is not a classifier. It is a direct probability model.
You didn't explain why your problem is an all-or-nothing classification problem vs. a risk estimation problem.
I have to disagree with the answ... | regression for binary classification | You begin with a misunderstanding. It is important to get the terminology right at the beginning. Logistic regression is not a classifier. It is a direct probability model.
You didn't explain why | regression for binary classification
You begin with a misunderstanding. It is important to get the terminology right at the beginning. Logistic regression is not a classifier. It is a direct probability model.
You didn't explain why your problem is an all-or-nothing classification problem vs. a risk estimation prob... | regression for binary classification
You begin with a misunderstanding. It is important to get the terminology right at the beginning. Logistic regression is not a classifier. It is a direct probability model.
You didn't explain why |
37,890 | regression for binary classification | The key question is whether you are likely to need estimates of the probability of class membership, a ranking, or whether you genuinely only interested in a binary classification. In my experience, you often do want the probabilities as the class frequencies, or the misclassification costs are unknown or variable in ... | regression for binary classification | The key question is whether you are likely to need estimates of the probability of class membership, a ranking, or whether you genuinely only interested in a binary classification. In my experience, | regression for binary classification
The key question is whether you are likely to need estimates of the probability of class membership, a ranking, or whether you genuinely only interested in a binary classification. In my experience, you often do want the probabilities as the class frequencies, or the misclassificat... | regression for binary classification
The key question is whether you are likely to need estimates of the probability of class membership, a ranking, or whether you genuinely only interested in a binary classification. In my experience, |
37,891 | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die? | While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for the what the variance is actually "doing." An equivalent expression for the variance in this case is:
$Var(X)$ =$ \sum... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided | While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die?
While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for the what the v... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided
While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for |
37,892 | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die? | If $X$ is the value of the die we already know $\text{E}(X) = 21 / 6$ so we only need to find $\text{E}(X^2)$ since $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)^2$. We can just directly calculate
\begin{align}
\text{E}(X^2) &= \sum_{k=1}^{6} \frac{k^2}{6} \\
&= \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} \\
&= \frac{9... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided | If $X$ is the value of the die we already know $\text{E}(X) = 21 / 6$ so we only need to find $\text{E}(X^2)$ since $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)^2$. We can just directly calculate
\be | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die?
If $X$ is the value of the die we already know $\text{E}(X) = 21 / 6$ so we only need to find $\text{E}(X^2)$ since $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)^2$. We can just directly calculate
\begin{align}
\tex... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided
If $X$ is the value of the die we already know $\text{E}(X) = 21 / 6$ so we only need to find $\text{E}(X^2)$ since $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)^2$. We can just directly calculate
\be |
37,893 | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die? | This is a discrete uniform distribution. So we can use $\frac{(b-a+1)^2-1}{12}$ to solve for the variance. $\frac{(6-1+1)^2-1}{12}$ = $\frac{6^2-1}{12}$ = $\frac{35}{12}$ | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided | This is a discrete uniform distribution. So we can use $\frac{(b-a+1)^2-1}{12}$ to solve for the variance. $\frac{(6-1+1)^2-1}{12}$ = $\frac{6^2-1}{12}$ = $\frac{35}{12}$ | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die?
This is a discrete uniform distribution. So we can use $\frac{(b-a+1)^2-1}{12}$ to solve for the variance. $\frac{(6-1+1)^2-1}{12}$ = $\frac{6^2-1}{12}$ = $\frac{35}{12}$ | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided
This is a discrete uniform distribution. So we can use $\frac{(b-a+1)^2-1}{12}$ to solve for the variance. $\frac{(6-1+1)^2-1}{12}$ = $\frac{6^2-1}{12}$ = $\frac{35}{12}$ |
37,894 | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die? | There are already several good answers posted (as well as one in the comments). My goal here is not to replicate those answers, but rather to try and address an apparent confusion about the "definition of variance".
In your question you say
It seems the variance and standard deviation tacitly ASSUME an a priori normal... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided | There are already several good answers posted (as well as one in the comments). My goal here is not to replicate those answers, but rather to try and address an apparent confusion about the "definitio | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die?
There are already several good answers posted (as well as one in the comments). My goal here is not to replicate those answers, but rather to try and address an apparent confusion about the "definition of variance".... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided
There are already several good answers posted (as well as one in the comments). My goal here is not to replicate those answers, but rather to try and address an apparent confusion about the "definitio |
37,895 | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die? | "That doesn't make sense: if the die starts at zero-based numbering, then the variance changes. But, variance shouldn't depend on the numbering of the die as all faces are equally probable. –
Marcos
Dec 7, 2016 at 20:38"
It seems to me the variance is not the 'variance of the probability' of one side over another, but ... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided | "That doesn't make sense: if the die starts at zero-based numbering, then the variance changes. But, variance shouldn't depend on the numbering of the die as all faces are equally probable. –
Marcos
D | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die?
"That doesn't make sense: if the die starts at zero-based numbering, then the variance changes. But, variance shouldn't depend on the numbering of the die as all faces are equally probable. –
Marcos
Dec 7, 2016 at 2... | What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided
"That doesn't make sense: if the die starts at zero-based numbering, then the variance changes. But, variance shouldn't depend on the numbering of the die as all faces are equally probable. –
Marcos
D |
37,896 | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous variables? [closed] | It is not true. There's similar number of regression algorithms as classification algorithms in machine learning. Most of the classification algorithms have their regression counterparts: there's $k$-NN and $k$-NN regression, SVM's and SVR's for regression, random forest build of classification, or regression trees, XG... | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous | It is not true. There's similar number of regression algorithms as classification algorithms in machine learning. Most of the classification algorithms have their regression counterparts: there's $k$- | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous variables? [closed]
It is not true. There's similar number of regression algorithms as classification algorithms in machine learning. Most of the classification algorithms have their regression counterparts: there's $k$-... | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous
It is not true. There's similar number of regression algorithms as classification algorithms in machine learning. Most of the classification algorithms have their regression counterparts: there's $k$- |
37,897 | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous variables? [closed] | Are there are insights from the machine learning literature that moves away from linear regression using OLS when the goal is predicting a continuous variable?
There are many methods other than standard OLS for modelling continuous data. There are even linear regression methods other than OLS (TLS/Deming regression fo... | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous | Are there are insights from the machine learning literature that moves away from linear regression using OLS when the goal is predicting a continuous variable?
There are many methods other than stand | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous variables? [closed]
Are there are insights from the machine learning literature that moves away from linear regression using OLS when the goal is predicting a continuous variable?
There are many methods other than stand... | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous
Are there are insights from the machine learning literature that moves away from linear regression using OLS when the goal is predicting a continuous variable?
There are many methods other than stand |
37,898 | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous variables? [closed] | IMO, deep learning is under the machine learning umbrella, in that it is deep machine learning, instead of "shallow" machine learning methods (e.g., OLS, KNN, SVM, Random Forest).
Deep learning and artificial neural networks can be used for regression problems, to add another OLS alternative path for you. Keep in mind ... | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous | IMO, deep learning is under the machine learning umbrella, in that it is deep machine learning, instead of "shallow" machine learning methods (e.g., OLS, KNN, SVM, Random Forest).
Deep learning and ar | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous variables? [closed]
IMO, deep learning is under the machine learning umbrella, in that it is deep machine learning, instead of "shallow" machine learning methods (e.g., OLS, KNN, SVM, Random Forest).
Deep learning and ar... | Does machine learning offer alternatives to linear regression (i.e., OLS) for predicting continuous
IMO, deep learning is under the machine learning umbrella, in that it is deep machine learning, instead of "shallow" machine learning methods (e.g., OLS, KNN, SVM, Random Forest).
Deep learning and ar |
37,899 | Can PC1 explain more than 90% of variance? | Certainly that can happen. Below is a simulated example in R with just two original variables. As long as they are strongly enough correlated, they can very well be summarized by the projection to a single line, which means that the first principal component explains an arbitrarily large proportion of total variance:
l... | Can PC1 explain more than 90% of variance? | Certainly that can happen. Below is a simulated example in R with just two original variables. As long as they are strongly enough correlated, they can very well be summarized by the projection to a s | Can PC1 explain more than 90% of variance?
Certainly that can happen. Below is a simulated example in R with just two original variables. As long as they are strongly enough correlated, they can very well be summarized by the projection to a single line, which means that the first principal component explains an arbitr... | Can PC1 explain more than 90% of variance?
Certainly that can happen. Below is a simulated example in R with just two original variables. As long as they are strongly enough correlated, they can very well be summarized by the projection to a s |
37,900 | Can PC1 explain more than 90% of variance? | In case you need a non-simulated example: I did help.search("morpholog", agrep = FALSE) to find morphological data sets in the set of R packages I happen to have installed on my system. A useful one is ade4::tortues, a data set on 48 painted turtles from Jolicoeur and Mosimann 1960.
long, larg, and haut are the length... | Can PC1 explain more than 90% of variance? | In case you need a non-simulated example: I did help.search("morpholog", agrep = FALSE) to find morphological data sets in the set of R packages I happen to have installed on my system. A useful one | Can PC1 explain more than 90% of variance?
In case you need a non-simulated example: I did help.search("morpholog", agrep = FALSE) to find morphological data sets in the set of R packages I happen to have installed on my system. A useful one is ade4::tortues, a data set on 48 painted turtles from Jolicoeur and Mosiman... | Can PC1 explain more than 90% of variance?
In case you need a non-simulated example: I did help.search("morpholog", agrep = FALSE) to find morphological data sets in the set of R packages I happen to have installed on my system. A useful one |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.