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 |
|---|---|---|---|---|---|---|
54,201 | Will removing outliers improve my predictive model? | Turning my comments above into an answer:
If the extreme values are not errors in the data, then you are not helping prediction by removing them. You are instead ignoring data that your model does not explain well. I strongly suggest you retain them in your dataset. Even if you cannot explain them with your model, you... | Will removing outliers improve my predictive model? | Turning my comments above into an answer:
If the extreme values are not errors in the data, then you are not helping prediction by removing them. You are instead ignoring data that your model does no | Will removing outliers improve my predictive model?
Turning my comments above into an answer:
If the extreme values are not errors in the data, then you are not helping prediction by removing them. You are instead ignoring data that your model does not explain well. I strongly suggest you retain them in your dataset. ... | Will removing outliers improve my predictive model?
Turning my comments above into an answer:
If the extreme values are not errors in the data, then you are not helping prediction by removing them. You are instead ignoring data that your model does no |
54,202 | How to make correct predictions of probabilities and their uncertainty in Bayesian logistic regression? | I'm going to start by cleaning up a little notational lack of precision in the first part of the question, then go on to the meat of it.
First, if $y \in \{0,1\}$, $y$ is, more precisely than stated in the question, distributed Bernoulli$(\theta)$, not Binomial$(n,\theta)$ for some arbitrary $n$. The two distributions... | How to make correct predictions of probabilities and their uncertainty in Bayesian logistic regressi | I'm going to start by cleaning up a little notational lack of precision in the first part of the question, then go on to the meat of it.
First, if $y \in \{0,1\}$, $y$ is, more precisely than stated i | How to make correct predictions of probabilities and their uncertainty in Bayesian logistic regression?
I'm going to start by cleaning up a little notational lack of precision in the first part of the question, then go on to the meat of it.
First, if $y \in \{0,1\}$, $y$ is, more precisely than stated in the question, ... | How to make correct predictions of probabilities and their uncertainty in Bayesian logistic regressi
I'm going to start by cleaning up a little notational lack of precision in the first part of the question, then go on to the meat of it.
First, if $y \in \{0,1\}$, $y$ is, more precisely than stated i |
54,203 | Usefulness of convexity of linear regression when there is no closed form solution | When $(X^TX)$ is not invertible there is not one solution but several: an affine subspace. But they are still closed form solutions in a way. They are solutions of the linear system:
$(X^TX)\beta=X^Ty$. Solving this system is not fundamentally more complicated than inverting a matrix I think.
It is also possible to sol... | Usefulness of convexity of linear regression when there is no closed form solution | When $(X^TX)$ is not invertible there is not one solution but several: an affine subspace. But they are still closed form solutions in a way. They are solutions of the linear system:
$(X^TX)\beta=X^Ty | Usefulness of convexity of linear regression when there is no closed form solution
When $(X^TX)$ is not invertible there is not one solution but several: an affine subspace. But they are still closed form solutions in a way. They are solutions of the linear system:
$(X^TX)\beta=X^Ty$. Solving this system is not fundame... | Usefulness of convexity of linear regression when there is no closed form solution
When $(X^TX)$ is not invertible there is not one solution but several: an affine subspace. But they are still closed form solutions in a way. They are solutions of the linear system:
$(X^TX)\beta=X^Ty |
54,204 | How to make predictions with time-dependent covariates with Cox regression | Calculating predicted probabilities using a Cox model
There is a way of obtaining prediction out of a Cox model, as survival probability at time $t$ ($S(t)$) depends on your cox model like so:
$S(t) = e^{-H_0(t) * exp(LP)}$
in this formula $H_0(t)$ is called the baseline hazard at time $t$; and $LP$ is the linear predi... | How to make predictions with time-dependent covariates with Cox regression | Calculating predicted probabilities using a Cox model
There is a way of obtaining prediction out of a Cox model, as survival probability at time $t$ ($S(t)$) depends on your cox model like so:
$S(t) = | How to make predictions with time-dependent covariates with Cox regression
Calculating predicted probabilities using a Cox model
There is a way of obtaining prediction out of a Cox model, as survival probability at time $t$ ($S(t)$) depends on your cox model like so:
$S(t) = e^{-H_0(t) * exp(LP)}$
in this formula $H_0(... | How to make predictions with time-dependent covariates with Cox regression
Calculating predicted probabilities using a Cox model
There is a way of obtaining prediction out of a Cox model, as survival probability at time $t$ ($S(t)$) depends on your cox model like so:
$S(t) = |
54,205 | How to make predictions with time-dependent covariates with Cox regression | How do I know which time-value to use in my calculations?
This isn't a question that your model can answer you. How did you choose the cutoff points?
The result of a Cox-PH model is a survival distribution over time for each patient. Without time dependence, one can look at the overall hazard of each patient to c... | How to make predictions with time-dependent covariates with Cox regression | How do I know which time-value to use in my calculations?
This isn't a question that your model can answer you. How did you choose the cutoff points?
The result of a Cox-PH model is a survival di | How to make predictions with time-dependent covariates with Cox regression
How do I know which time-value to use in my calculations?
This isn't a question that your model can answer you. How did you choose the cutoff points?
The result of a Cox-PH model is a survival distribution over time for each patient. Witho... | How to make predictions with time-dependent covariates with Cox regression
How do I know which time-value to use in my calculations?
This isn't a question that your model can answer you. How did you choose the cutoff points?
The result of a Cox-PH model is a survival di |
54,206 | ARMA - coefficient interpretation | Here Graeme Walsh explains how to better look at the fit than just the coefficients:
ARIMA model interpretation
Here David J Harris describes differences in the model selection parameters AIC, BIC, etc.
AIC,BIC,CIC,DIC,EIC,FIC,GIC,HIC,IIC --- Can I use them interchangeably?
Here an earlier question regarding ARMA mod... | ARMA - coefficient interpretation | Here Graeme Walsh explains how to better look at the fit than just the coefficients:
ARIMA model interpretation
Here David J Harris describes differences in the model selection parameters AIC, BIC, e | ARMA - coefficient interpretation
Here Graeme Walsh explains how to better look at the fit than just the coefficients:
ARIMA model interpretation
Here David J Harris describes differences in the model selection parameters AIC, BIC, etc.
AIC,BIC,CIC,DIC,EIC,FIC,GIC,HIC,IIC --- Can I use them interchangeably?
Here an e... | ARMA - coefficient interpretation
Here Graeme Walsh explains how to better look at the fit than just the coefficients:
ARIMA model interpretation
Here David J Harris describes differences in the model selection parameters AIC, BIC, e |
54,207 | ARMA - coefficient interpretation | You can use the pi weights to help you interpret/decipher/explain the model. The pi weights are obtained by dividing the ma polynomial by the ar polynomial . Presented in this way (i.e. as a pure ar) the model's parameters are simply a weighted average of the past . Bye the way your model is in my experience way and I ... | ARMA - coefficient interpretation | You can use the pi weights to help you interpret/decipher/explain the model. The pi weights are obtained by dividing the ma polynomial by the ar polynomial . Presented in this way (i.e. as a pure ar) | ARMA - coefficient interpretation
You can use the pi weights to help you interpret/decipher/explain the model. The pi weights are obtained by dividing the ma polynomial by the ar polynomial . Presented in this way (i.e. as a pure ar) the model's parameters are simply a weighted average of the past . Bye the way your mo... | ARMA - coefficient interpretation
You can use the pi weights to help you interpret/decipher/explain the model. The pi weights are obtained by dividing the ma polynomial by the ar polynomial . Presented in this way (i.e. as a pure ar) |
54,208 | Choosing a model for my unsupervised machine learning problem | Here are a couple suggestions, given that Gaussian mixture models work well for you in the absence of outliers.
To increase robustness to outliers, you could use a trimmed estimator for Gaussian mixture models instead of fitting with the standard EM algorithm. Some relevant papers:
Neykov et al. (2007). Robust fitting... | Choosing a model for my unsupervised machine learning problem | Here are a couple suggestions, given that Gaussian mixture models work well for you in the absence of outliers.
To increase robustness to outliers, you could use a trimmed estimator for Gaussian mixtu | Choosing a model for my unsupervised machine learning problem
Here are a couple suggestions, given that Gaussian mixture models work well for you in the absence of outliers.
To increase robustness to outliers, you could use a trimmed estimator for Gaussian mixture models instead of fitting with the standard EM algorith... | Choosing a model for my unsupervised machine learning problem
Here are a couple suggestions, given that Gaussian mixture models work well for you in the absence of outliers.
To increase robustness to outliers, you could use a trimmed estimator for Gaussian mixtu |
54,209 | Choosing a model for my unsupervised machine learning problem | Look, this might not be the best idea, but I'll suggest it anyway.
You say that you are dealing with an unsupervised ML problem, but said that:
you can assume that in this 4 dimensional space, your data is a multivariate Gaussian;
your model works nicely when there aren't many outliers
If you can afford not assigning... | Choosing a model for my unsupervised machine learning problem | Look, this might not be the best idea, but I'll suggest it anyway.
You say that you are dealing with an unsupervised ML problem, but said that:
you can assume that in this 4 dimensional space, your d | Choosing a model for my unsupervised machine learning problem
Look, this might not be the best idea, but I'll suggest it anyway.
You say that you are dealing with an unsupervised ML problem, but said that:
you can assume that in this 4 dimensional space, your data is a multivariate Gaussian;
your model works nicely wh... | Choosing a model for my unsupervised machine learning problem
Look, this might not be the best idea, but I'll suggest it anyway.
You say that you are dealing with an unsupervised ML problem, but said that:
you can assume that in this 4 dimensional space, your d |
54,210 | Choosing a model for my unsupervised machine learning problem | I am assuming that the required four clusters will be of somewhat similar density. If that is the case then you can use a density based clustering approach, DBSCAN has worked well for me. You can find all the clusters that you can during training time and have a threshold on the cluster size to preclude the outliers fr... | Choosing a model for my unsupervised machine learning problem | I am assuming that the required four clusters will be of somewhat similar density. If that is the case then you can use a density based clustering approach, DBSCAN has worked well for me. You can find | Choosing a model for my unsupervised machine learning problem
I am assuming that the required four clusters will be of somewhat similar density. If that is the case then you can use a density based clustering approach, DBSCAN has worked well for me. You can find all the clusters that you can during training time and ha... | Choosing a model for my unsupervised machine learning problem
I am assuming that the required four clusters will be of somewhat similar density. If that is the case then you can use a density based clustering approach, DBSCAN has worked well for me. You can find |
54,211 | Choosing a model for my unsupervised machine learning problem | Red flags go off in my head when someone says (paraphrasing heavily)
I can assume property X for my algorithm, but this peroperty X need not be present in the data.
That immediately points to possible performance issues and/ or grounds to revaluate model selection step.
Having said that, I haven't worked with GMM for... | Choosing a model for my unsupervised machine learning problem | Red flags go off in my head when someone says (paraphrasing heavily)
I can assume property X for my algorithm, but this peroperty X need not be present in the data.
That immediately points to possib | Choosing a model for my unsupervised machine learning problem
Red flags go off in my head when someone says (paraphrasing heavily)
I can assume property X for my algorithm, but this peroperty X need not be present in the data.
That immediately points to possible performance issues and/ or grounds to revaluate model s... | Choosing a model for my unsupervised machine learning problem
Red flags go off in my head when someone says (paraphrasing heavily)
I can assume property X for my algorithm, but this peroperty X need not be present in the data.
That immediately points to possib |
54,212 | Choosing a model for my unsupervised machine learning problem | Only 3 dimensions and 30% of your data are outliers? That does not seem to fit with what I normally think of as an outlier. Perhaps you can simply transform your variables to log scale or "cap" the outliers so that they are no more than 3 or 4 standard deviations from the mean. This may create clusters of points wit... | Choosing a model for my unsupervised machine learning problem | Only 3 dimensions and 30% of your data are outliers? That does not seem to fit with what I normally think of as an outlier. Perhaps you can simply transform your variables to log scale or "cap" the | Choosing a model for my unsupervised machine learning problem
Only 3 dimensions and 30% of your data are outliers? That does not seem to fit with what I normally think of as an outlier. Perhaps you can simply transform your variables to log scale or "cap" the outliers so that they are no more than 3 or 4 standard dev... | Choosing a model for my unsupervised machine learning problem
Only 3 dimensions and 30% of your data are outliers? That does not seem to fit with what I normally think of as an outlier. Perhaps you can simply transform your variables to log scale or "cap" the |
54,213 | Why do independent priors for two random variables not result in an independent joint posterior distribution? | When you have an independent prior on $X$ and $Y$, then the posterior might not factor into $X$ and $Y$ pieces just because the likelihood doesn't factor into $X$ and $Y$ pieces.
It's easy to see that
$$
p(x,y|D) \propto p(D|X,Y)p_X(x)p_Y(y).
$$
So in your situation the posterior factors if and only if the likelihood ... | Why do independent priors for two random variables not result in an independent joint posterior dist | When you have an independent prior on $X$ and $Y$, then the posterior might not factor into $X$ and $Y$ pieces just because the likelihood doesn't factor into $X$ and $Y$ pieces.
It's easy to see tha | Why do independent priors for two random variables not result in an independent joint posterior distribution?
When you have an independent prior on $X$ and $Y$, then the posterior might not factor into $X$ and $Y$ pieces just because the likelihood doesn't factor into $X$ and $Y$ pieces.
It's easy to see that
$$
p(x,y... | Why do independent priors for two random variables not result in an independent joint posterior dist
When you have an independent prior on $X$ and $Y$, then the posterior might not factor into $X$ and $Y$ pieces just because the likelihood doesn't factor into $X$ and $Y$ pieces.
It's easy to see tha |
54,214 | Why should the variance equal the mean in Poisson regression? | Because it is a consequence of the functional form of the Poisson distribution that mean and variance are equal. If this condition is not met the model is inadequate and alternatives may be considered such as negative binomial regression (this is called overdispersion). See:
https://en.wikipedia.org/wiki/Poisson_distri... | Why should the variance equal the mean in Poisson regression? | Because it is a consequence of the functional form of the Poisson distribution that mean and variance are equal. If this condition is not met the model is inadequate and alternatives may be considered | Why should the variance equal the mean in Poisson regression?
Because it is a consequence of the functional form of the Poisson distribution that mean and variance are equal. If this condition is not met the model is inadequate and alternatives may be considered such as negative binomial regression (this is called over... | Why should the variance equal the mean in Poisson regression?
Because it is a consequence of the functional form of the Poisson distribution that mean and variance are equal. If this condition is not met the model is inadequate and alternatives may be considered |
54,215 | Why should the variance equal the mean in Poisson regression? | To see this, let's consider the number of crashes for a specific road characteristic. Let's say this number follows a Poisson distribution with a mean $\mu$. This mean is for a certain number of km driven so let's introduce the rate $\lambda$, say 1 crash per km and the total number of km driven $T$. One assumption of ... | Why should the variance equal the mean in Poisson regression? | To see this, let's consider the number of crashes for a specific road characteristic. Let's say this number follows a Poisson distribution with a mean $\mu$. This mean is for a certain number of km dr | Why should the variance equal the mean in Poisson regression?
To see this, let's consider the number of crashes for a specific road characteristic. Let's say this number follows a Poisson distribution with a mean $\mu$. This mean is for a certain number of km driven so let's introduce the rate $\lambda$, say 1 crash pe... | Why should the variance equal the mean in Poisson regression?
To see this, let's consider the number of crashes for a specific road characteristic. Let's say this number follows a Poisson distribution with a mean $\mu$. This mean is for a certain number of km dr |
54,216 | Why should the variance equal the mean in Poisson regression? | Poisson regression allows for inference of regression parameters - generally a parameter vector, $\mathbf{\theta}$ - under the assumption that the errors in the model are distributed according to a Poisson distribution: $\epsilon \sim \operatorname{Poisson}(\lambda)$.
This is appropriate for modelling certain data, a g... | Why should the variance equal the mean in Poisson regression? | Poisson regression allows for inference of regression parameters - generally a parameter vector, $\mathbf{\theta}$ - under the assumption that the errors in the model are distributed according to a Po | Why should the variance equal the mean in Poisson regression?
Poisson regression allows for inference of regression parameters - generally a parameter vector, $\mathbf{\theta}$ - under the assumption that the errors in the model are distributed according to a Poisson distribution: $\epsilon \sim \operatorname{Poisson}(... | Why should the variance equal the mean in Poisson regression?
Poisson regression allows for inference of regression parameters - generally a parameter vector, $\mathbf{\theta}$ - under the assumption that the errors in the model are distributed according to a Po |
54,217 | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0 | The Wald statistic for testing the null hypothesis is:
$$W = \frac{\hat{\theta} - \theta_0}{se(\hat{\theta})}$$
your $\hat{\theta}$ is the point estimate $b$: 0.20483 and $se(\hat{\theta})$ is the SE : 0.06723 used to describe the approximate normal sampling distribution of the test statistic when the null hypothesis... | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0 | The Wald statistic for testing the null hypothesis is:
$$W = \frac{\hat{\theta} - \theta_0}{se(\hat{\theta})}$$
your $\hat{\theta}$ is the point estimate $b$: 0.20483 and $se(\hat{\theta})$ is the S | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0
The Wald statistic for testing the null hypothesis is:
$$W = \frac{\hat{\theta} - \theta_0}{se(\hat{\theta})}$$
your $\hat{\theta}$ is the point estimate $b$: 0.20483 and $se(\hat{\theta})$ is the SE : 0.06723 used to describ... | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0
The Wald statistic for testing the null hypothesis is:
$$W = \frac{\hat{\theta} - \theta_0}{se(\hat{\theta})}$$
your $\hat{\theta}$ is the point estimate $b$: 0.20483 and $se(\hat{\theta})$ is the S |
54,218 | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0 | How about estimating this model:
M = a = nls(y ~ a * exp((b + 0.069) * x), data = d, start = list(a = 1, b = 0))
So b == 0 in this model if and only if b == 0.069 in the original model.
The summary for the new model is:
Formula: y ~ a * exp((b + 0.069) * x)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 0.02... | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0 | How about estimating this model:
M = a = nls(y ~ a * exp((b + 0.069) * x), data = d, start = list(a = 1, b = 0))
So b == 0 in this model if and only if b == 0.069 in the original model.
The summary f | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0
How about estimating this model:
M = a = nls(y ~ a * exp((b + 0.069) * x), data = d, start = list(a = 1, b = 0))
So b == 0 in this model if and only if b == 0.069 in the original model.
The summary for the new model is:
Formul... | How to calculate p-value for a parameter given confidence interval when null hypothesis != 0
How about estimating this model:
M = a = nls(y ~ a * exp((b + 0.069) * x), data = d, start = list(a = 1, b = 0))
So b == 0 in this model if and only if b == 0.069 in the original model.
The summary f |
54,219 | Why does batch normalization use mini-batch statistics instead of the moving averages during training? | There is a follow-up paper by Sergej Ioffe (i.e. Batch Renormalization) which discusses this issue:
https://arxiv.org/abs/1702.03275
A quote from that paper regarding regular batch normalization:
It is natural to ask whether we could simply use the moving averages $\mu, \sigma$ to perform the normalization during tra... | Why does batch normalization use mini-batch statistics instead of the moving averages during trainin | There is a follow-up paper by Sergej Ioffe (i.e. Batch Renormalization) which discusses this issue:
https://arxiv.org/abs/1702.03275
A quote from that paper regarding regular batch normalization:
It | Why does batch normalization use mini-batch statistics instead of the moving averages during training?
There is a follow-up paper by Sergej Ioffe (i.e. Batch Renormalization) which discusses this issue:
https://arxiv.org/abs/1702.03275
A quote from that paper regarding regular batch normalization:
It is natural to as... | Why does batch normalization use mini-batch statistics instead of the moving averages during trainin
There is a follow-up paper by Sergej Ioffe (i.e. Batch Renormalization) which discusses this issue:
https://arxiv.org/abs/1702.03275
A quote from that paper regarding regular batch normalization:
It |
54,220 | How can I test if my observed PDF follows a binomial distribution? | Please note that this question is duplicated, and there is a better answer by Glen_b over here.
Here is an example using the the R package fitdistrplus. The simulated data is strictly binomial, and hence we should fail to reject the null hypothesis.
require("fitdistrplus")
set.seed(10)
n = 25
size = 27
prob = .4
data ... | How can I test if my observed PDF follows a binomial distribution? | Please note that this question is duplicated, and there is a better answer by Glen_b over here.
Here is an example using the the R package fitdistrplus. The simulated data is strictly binomial, and h | How can I test if my observed PDF follows a binomial distribution?
Please note that this question is duplicated, and there is a better answer by Glen_b over here.
Here is an example using the the R package fitdistrplus. The simulated data is strictly binomial, and hence we should fail to reject the null hypothesis.
re... | How can I test if my observed PDF follows a binomial distribution?
Please note that this question is duplicated, and there is a better answer by Glen_b over here.
Here is an example using the the R package fitdistrplus. The simulated data is strictly binomial, and h |
54,221 | Is Weibull distribution a exponential family? | The answer is NO. The logarithm of the weibull density is given by
$$
\log f(x) = \log(k/\lambda) + (k-1)\log(x/\lambda) - (x/\lambda)^k
$$
where $x>0$, $k>0$ (shape parameter), $\lambda>0$ (scale parameter). The problem is the last term. IF $k$ were known (prespecified), this would be a one-parameter exponential ... | Is Weibull distribution a exponential family? | The answer is NO. The logarithm of the weibull density is given by
$$
\log f(x) = \log(k/\lambda) + (k-1)\log(x/\lambda) - (x/\lambda)^k
$$
where $x>0$, $k>0$ (shape parameter), $\lambda>0$ (scale | Is Weibull distribution a exponential family?
The answer is NO. The logarithm of the weibull density is given by
$$
\log f(x) = \log(k/\lambda) + (k-1)\log(x/\lambda) - (x/\lambda)^k
$$
where $x>0$, $k>0$ (shape parameter), $\lambda>0$ (scale parameter). The problem is the last term. IF $k$ were known (prespecifie... | Is Weibull distribution a exponential family?
The answer is NO. The logarithm of the weibull density is given by
$$
\log f(x) = \log(k/\lambda) + (k-1)\log(x/\lambda) - (x/\lambda)^k
$$
where $x>0$, $k>0$ (shape parameter), $\lambda>0$ (scale |
54,222 | Is Weibull distribution a exponential family? | The two parameter Weibull distribution (with $k$ and $\lambda$ as described on wikipedia) is not an exponential family. However, if you fix $k$ to anything, then it is an exponential family having sufficient statistics $x^k$ on support $[0, \infty)$. | Is Weibull distribution a exponential family? | The two parameter Weibull distribution (with $k$ and $\lambda$ as described on wikipedia) is not an exponential family. However, if you fix $k$ to anything, then it is an exponential family having su | Is Weibull distribution a exponential family?
The two parameter Weibull distribution (with $k$ and $\lambda$ as described on wikipedia) is not an exponential family. However, if you fix $k$ to anything, then it is an exponential family having sufficient statistics $x^k$ on support $[0, \infty)$. | Is Weibull distribution a exponential family?
The two parameter Weibull distribution (with $k$ and $\lambda$ as described on wikipedia) is not an exponential family. However, if you fix $k$ to anything, then it is an exponential family having su |
54,223 | Scaling step in Baum-Welch algorithm | Background and Notation
Let $x_1, \ldots, x_T$ be the observations, and $z_1, \ldots, z_T$ be the hidden states. The forward-backward recursions are written in terms of $\alpha(z_n)$ and $\beta(z_n)$.
$\alpha(z_n) = p(x_1,\ldots,x_n,z_n)$ is the joint distribution of all the heretofore observed data and the most recent... | Scaling step in Baum-Welch algorithm | Background and Notation
Let $x_1, \ldots, x_T$ be the observations, and $z_1, \ldots, z_T$ be the hidden states. The forward-backward recursions are written in terms of $\alpha(z_n)$ and $\beta(z_n)$. | Scaling step in Baum-Welch algorithm
Background and Notation
Let $x_1, \ldots, x_T$ be the observations, and $z_1, \ldots, z_T$ be the hidden states. The forward-backward recursions are written in terms of $\alpha(z_n)$ and $\beta(z_n)$.
$\alpha(z_n) = p(x_1,\ldots,x_n,z_n)$ is the joint distribution of all the heretof... | Scaling step in Baum-Welch algorithm
Background and Notation
Let $x_1, \ldots, x_T$ be the observations, and $z_1, \ldots, z_T$ be the hidden states. The forward-backward recursions are written in terms of $\alpha(z_n)$ and $\beta(z_n)$. |
54,224 | Scaling step in Baum-Welch algorithm | I'm not an expert in HMM, but happened to go through the same process of learning and implementing HMM while encountering the same normalization issues. Below is the answer based on my understanding.
- Forward algorithm:
The forward algorithm calculates $\alpha_t(j) \equiv P(z_t = j | x_{1:t})$, i.e., the probability ... | Scaling step in Baum-Welch algorithm | I'm not an expert in HMM, but happened to go through the same process of learning and implementing HMM while encountering the same normalization issues. Below is the answer based on my understanding. | Scaling step in Baum-Welch algorithm
I'm not an expert in HMM, but happened to go through the same process of learning and implementing HMM while encountering the same normalization issues. Below is the answer based on my understanding.
- Forward algorithm:
The forward algorithm calculates $\alpha_t(j) \equiv P(z_t = ... | Scaling step in Baum-Welch algorithm
I'm not an expert in HMM, but happened to go through the same process of learning and implementing HMM while encountering the same normalization issues. Below is the answer based on my understanding. |
54,225 | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0.99$? | This seems to be discussing some symmetric distribution - so that the 0.01 quantile ($q_{0.01}$) is as far from $19.1$ as the 0.99 quantile ($q_{0.99}$). (I've used a normal density in my image but this argument applies to symmetric distributions more generally.)
In the image below we have $P(|X-19.1|\leq a) = 0.98$ (i... | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0 | This seems to be discussing some symmetric distribution - so that the 0.01 quantile ($q_{0.01}$) is as far from $19.1$ as the 0.99 quantile ($q_{0.99}$). (I've used a normal density in my image but th | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0.99$?
This seems to be discussing some symmetric distribution - so that the 0.01 quantile ($q_{0.01}$) is as far from $19.1$ as the 0.99 quantile ($q_{0.99}$). (I've used a normal density in my image but this argument ap... | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0
This seems to be discussing some symmetric distribution - so that the 0.01 quantile ($q_{0.01}$) is as far from $19.1$ as the 0.99 quantile ($q_{0.99}$). (I've used a normal density in my image but th |
54,226 | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0.99$? | This assumes a normal distribution and the first relationship amounts to transform to the standard normal. You can then use 0.99 in the table of the cumulative standard normal distribution to find $a$. The change to 0.99 occurs because the left tail (all values below $-a/1.7$) is left out. | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0 | This assumes a normal distribution and the first relationship amounts to transform to the standard normal. You can then use 0.99 in the table of the cumulative standard normal distribution to find $a$ | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0.99$?
This assumes a normal distribution and the first relationship amounts to transform to the standard normal. You can then use 0.99 in the table of the cumulative standard normal distribution to find $a$. The change t... | If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0
This assumes a normal distribution and the first relationship amounts to transform to the standard normal. You can then use 0.99 in the table of the cumulative standard normal distribution to find $a$ |
54,227 | Can the likelihood function in MLE be equal to zero? | If you have a very inadequate model such that at least one discretely (or continuously) distributed observation has zero probability (or probability density) for any parameter value $\theta\in\Theta$, that is, you essentially observe something that is impossible under that model, then yes, your maximum likelihood would... | Can the likelihood function in MLE be equal to zero? | If you have a very inadequate model such that at least one discretely (or continuously) distributed observation has zero probability (or probability density) for any parameter value $\theta\in\Theta$, | Can the likelihood function in MLE be equal to zero?
If you have a very inadequate model such that at least one discretely (or continuously) distributed observation has zero probability (or probability density) for any parameter value $\theta\in\Theta$, that is, you essentially observe something that is impossible unde... | Can the likelihood function in MLE be equal to zero?
If you have a very inadequate model such that at least one discretely (or continuously) distributed observation has zero probability (or probability density) for any parameter value $\theta\in\Theta$, |
54,228 | Can the likelihood function in MLE be equal to zero? | If you observe a sample that has zero probability density under every possible parameter value then the likelihood function is zero over the parameter space. In this case every parameter value is a legitimate value of the MLE (i.e., the MLE is the whole parameter space) and the likelihood is zero at every point that i... | Can the likelihood function in MLE be equal to zero? | If you observe a sample that has zero probability density under every possible parameter value then the likelihood function is zero over the parameter space. In this case every parameter value is a l | Can the likelihood function in MLE be equal to zero?
If you observe a sample that has zero probability density under every possible parameter value then the likelihood function is zero over the parameter space. In this case every parameter value is a legitimate value of the MLE (i.e., the MLE is the whole parameter sp... | Can the likelihood function in MLE be equal to zero?
If you observe a sample that has zero probability density under every possible parameter value then the likelihood function is zero over the parameter space. In this case every parameter value is a l |
54,229 | Can the likelihood function in MLE be equal to zero? | When you define the MLE as the unique point where the likelihood function has a maximum, and if the parameter space consists of multiple points then the likelihood in the MLE must be non-zero.
The reason is that the likelihood must be non-negative and thus zero is the minimum value of the entire range. If the MLE has z... | Can the likelihood function in MLE be equal to zero? | When you define the MLE as the unique point where the likelihood function has a maximum, and if the parameter space consists of multiple points then the likelihood in the MLE must be non-zero.
The rea | Can the likelihood function in MLE be equal to zero?
When you define the MLE as the unique point where the likelihood function has a maximum, and if the parameter space consists of multiple points then the likelihood in the MLE must be non-zero.
The reason is that the likelihood must be non-negative and thus zero is th... | Can the likelihood function in MLE be equal to zero?
When you define the MLE as the unique point where the likelihood function has a maximum, and if the parameter space consists of multiple points then the likelihood in the MLE must be non-zero.
The rea |
54,230 | Can the likelihood function in MLE be equal to zero? | To add a concrete example to the answer by @Jarle Tufto: If $X_1, X_2, \dotsc, X_n$ are iid $\mathcal{U}(0, \theta)$, then any $\theta$ lesser than the maximum observed value give a zero likelihood, since under this model all observations must be lesser (or equal) to $\theta$. | Can the likelihood function in MLE be equal to zero? | To add a concrete example to the answer by @Jarle Tufto: If $X_1, X_2, \dotsc, X_n$ are iid $\mathcal{U}(0, \theta)$, then any $\theta$ lesser than the maximum observed value give a zero likelihood, s | Can the likelihood function in MLE be equal to zero?
To add a concrete example to the answer by @Jarle Tufto: If $X_1, X_2, \dotsc, X_n$ are iid $\mathcal{U}(0, \theta)$, then any $\theta$ lesser than the maximum observed value give a zero likelihood, since under this model all observations must be lesser (or equal) to... | Can the likelihood function in MLE be equal to zero?
To add a concrete example to the answer by @Jarle Tufto: If $X_1, X_2, \dotsc, X_n$ are iid $\mathcal{U}(0, \theta)$, then any $\theta$ lesser than the maximum observed value give a zero likelihood, s |
54,231 | Updating q-values in q-learning | That's a good idea. What you are thinking of is called eligibility traces. They often improve convergence speed, but this comes at the cost of significantly greater computation and memory cost, as well as increased complexity. The Sutton and Barto book describes them in detail. | Updating q-values in q-learning | That's a good idea. What you are thinking of is called eligibility traces. They often improve convergence speed, but this comes at the cost of significantly greater computation and memory cost, as wel | Updating q-values in q-learning
That's a good idea. What you are thinking of is called eligibility traces. They often improve convergence speed, but this comes at the cost of significantly greater computation and memory cost, as well as increased complexity. The Sutton and Barto book describes them in detail. | Updating q-values in q-learning
That's a good idea. What you are thinking of is called eligibility traces. They often improve convergence speed, but this comes at the cost of significantly greater computation and memory cost, as wel |
54,232 | Updating q-values in q-learning | You are right, in some cases it makes sense to do that.
What you suggest is Monte Carlo, which is the same as TD($\lambda$) with $\lambda = 1$. | Updating q-values in q-learning | You are right, in some cases it makes sense to do that.
What you suggest is Monte Carlo, which is the same as TD($\lambda$) with $\lambda = 1$. | Updating q-values in q-learning
You are right, in some cases it makes sense to do that.
What you suggest is Monte Carlo, which is the same as TD($\lambda$) with $\lambda = 1$. | Updating q-values in q-learning
You are right, in some cases it makes sense to do that.
What you suggest is Monte Carlo, which is the same as TD($\lambda$) with $\lambda = 1$. |
54,233 | Prove that $\Pr\left(A_n\right)=1 \implies \Pr\left(\bigcap_n A_n\right)=1$ | This is easier than you think. We have $\Pr(A_n)=1$ which means that $\Pr(A_n^c)=1-\Pr(A_n)=0$, for all $n \geq 1$. Recall that for probability measures we have countable sub-additivity (also referred to as Boole's inequality), i.e.
$$\Pr\left( \bigcup_{n=1}^{\infty} B_n \right) \leq \sum_{i=1}^{\infty} \Pr(B_n)$$
for ... | Prove that $\Pr\left(A_n\right)=1 \implies \Pr\left(\bigcap_n A_n\right)=1$ | This is easier than you think. We have $\Pr(A_n)=1$ which means that $\Pr(A_n^c)=1-\Pr(A_n)=0$, for all $n \geq 1$. Recall that for probability measures we have countable sub-additivity (also referred | Prove that $\Pr\left(A_n\right)=1 \implies \Pr\left(\bigcap_n A_n\right)=1$
This is easier than you think. We have $\Pr(A_n)=1$ which means that $\Pr(A_n^c)=1-\Pr(A_n)=0$, for all $n \geq 1$. Recall that for probability measures we have countable sub-additivity (also referred to as Boole's inequality), i.e.
$$\Pr\left(... | Prove that $\Pr\left(A_n\right)=1 \implies \Pr\left(\bigcap_n A_n\right)=1$
This is easier than you think. We have $\Pr(A_n)=1$ which means that $\Pr(A_n^c)=1-\Pr(A_n)=0$, for all $n \geq 1$. Recall that for probability measures we have countable sub-additivity (also referred |
54,234 | Is there a way to determine the important features (weight) for an SVM that uses an RBF kernel? | Unfortunately not. Although SVMs are often interpreted as transforming your features into a high-dimensional space and fitting a linear classifier in the new space, the transformation is implicit and cannot be easily retrieved.
In fact, SVMs with the RBF kernel behave more like soft nearest neighbours. To see this, den... | Is there a way to determine the important features (weight) for an SVM that uses an RBF kernel? | Unfortunately not. Although SVMs are often interpreted as transforming your features into a high-dimensional space and fitting a linear classifier in the new space, the transformation is implicit and | Is there a way to determine the important features (weight) for an SVM that uses an RBF kernel?
Unfortunately not. Although SVMs are often interpreted as transforming your features into a high-dimensional space and fitting a linear classifier in the new space, the transformation is implicit and cannot be easily retriev... | Is there a way to determine the important features (weight) for an SVM that uses an RBF kernel?
Unfortunately not. Although SVMs are often interpreted as transforming your features into a high-dimensional space and fitting a linear classifier in the new space, the transformation is implicit and |
54,235 | theta parameter in R's LTM package | Theta estimation according to the expected response pattern:
Theta <- factor.scores(two_pl)
Theta estimation according to the real data response pattern:
Theta <- factor.scores(two_pl, method = "EAP", resp.patterns = data) | theta parameter in R's LTM package | Theta estimation according to the expected response pattern:
Theta <- factor.scores(two_pl)
Theta estimation according to the real data response pattern:
Theta <- factor.scores(two_pl, method = "EAP" | theta parameter in R's LTM package
Theta estimation according to the expected response pattern:
Theta <- factor.scores(two_pl)
Theta estimation according to the real data response pattern:
Theta <- factor.scores(two_pl, method = "EAP", resp.patterns = data) | theta parameter in R's LTM package
Theta estimation according to the expected response pattern:
Theta <- factor.scores(two_pl)
Theta estimation according to the real data response pattern:
Theta <- factor.scores(two_pl, method = "EAP" |
54,236 | theta parameter in R's LTM package | It's the factor.scores function:
WIRStheta <- ltm::factor.scores(two_pl) | theta parameter in R's LTM package | It's the factor.scores function:
WIRStheta <- ltm::factor.scores(two_pl) | theta parameter in R's LTM package
It's the factor.scores function:
WIRStheta <- ltm::factor.scores(two_pl) | theta parameter in R's LTM package
It's the factor.scores function:
WIRStheta <- ltm::factor.scores(two_pl) |
54,237 | Random Effects for Mantel Haenszel | A proper random-effects model extension to the standard Mantel-Haenszel procedure is described by van Houwelingen, Zwinderman, and Stijnen (1993). In essence, one can think of the M-H procedure as a model based on the (non-central) hypergeometric distribution (Mantel & Haenszel, 1959). So, using this as the starting po... | Random Effects for Mantel Haenszel | A proper random-effects model extension to the standard Mantel-Haenszel procedure is described by van Houwelingen, Zwinderman, and Stijnen (1993). In essence, one can think of the M-H procedure as a m | Random Effects for Mantel Haenszel
A proper random-effects model extension to the standard Mantel-Haenszel procedure is described by van Houwelingen, Zwinderman, and Stijnen (1993). In essence, one can think of the M-H procedure as a model based on the (non-central) hypergeometric distribution (Mantel & Haenszel, 1959)... | Random Effects for Mantel Haenszel
A proper random-effects model extension to the standard Mantel-Haenszel procedure is described by van Houwelingen, Zwinderman, and Stijnen (1993). In essence, one can think of the M-H procedure as a m |
54,238 | Random Effects for Mantel Haenszel | Jonathan Deeks and Julian Higgins have a nice document showing all the calculations used in Review Manager. Scroll down to page 8 for DerSimonian and Laird random-effects models that can be used with the Mantel-Haenszel summary models. | Random Effects for Mantel Haenszel | Jonathan Deeks and Julian Higgins have a nice document showing all the calculations used in Review Manager. Scroll down to page 8 for DerSimonian and Laird random-effects models that can be used with | Random Effects for Mantel Haenszel
Jonathan Deeks and Julian Higgins have a nice document showing all the calculations used in Review Manager. Scroll down to page 8 for DerSimonian and Laird random-effects models that can be used with the Mantel-Haenszel summary models. | Random Effects for Mantel Haenszel
Jonathan Deeks and Julian Higgins have a nice document showing all the calculations used in Review Manager. Scroll down to page 8 for DerSimonian and Laird random-effects models that can be used with |
54,239 | How to Compute Bivariate Empirical Distribution? | By definition, the ECDF $F$ at any location $(x,y)$ counts the data points that lie to the left and beneath $(x,y)$. Specifically, writing $(x_i,y_i), i=1, 2, \ldots, n$ for the data points (which may include duplicates),
$$F(x,y) = \frac{1}{n}\times \#\{(x_i, y_i)\mid x_i \le x, \ y_i \le y\}.\tag{1}$$
Equivalently,... | How to Compute Bivariate Empirical Distribution? | By definition, the ECDF $F$ at any location $(x,y)$ counts the data points that lie to the left and beneath $(x,y)$. Specifically, writing $(x_i,y_i), i=1, 2, \ldots, n$ for the data points (which ma | How to Compute Bivariate Empirical Distribution?
By definition, the ECDF $F$ at any location $(x,y)$ counts the data points that lie to the left and beneath $(x,y)$. Specifically, writing $(x_i,y_i), i=1, 2, \ldots, n$ for the data points (which may include duplicates),
$$F(x,y) = \frac{1}{n}\times \#\{(x_i, y_i)\mid... | How to Compute Bivariate Empirical Distribution?
By definition, the ECDF $F$ at any location $(x,y)$ counts the data points that lie to the left and beneath $(x,y)$. Specifically, writing $(x_i,y_i), i=1, 2, \ldots, n$ for the data points (which ma |
54,240 | How to chose the order for polynomial regression? | This question can be generalized for selecting any machine learning algorithm hyper-parameters. For example, number of clusters in K-means, number of Hidden unit in neural networks, etc.
At very high level, there are two ways (not mutually exclusive, in fact combining two ways would be ideal.): Data driven and knowledg... | How to chose the order for polynomial regression? | This question can be generalized for selecting any machine learning algorithm hyper-parameters. For example, number of clusters in K-means, number of Hidden unit in neural networks, etc.
At very high | How to chose the order for polynomial regression?
This question can be generalized for selecting any machine learning algorithm hyper-parameters. For example, number of clusters in K-means, number of Hidden unit in neural networks, etc.
At very high level, there are two ways (not mutually exclusive, in fact combining t... | How to chose the order for polynomial regression?
This question can be generalized for selecting any machine learning algorithm hyper-parameters. For example, number of clusters in K-means, number of Hidden unit in neural networks, etc.
At very high |
54,241 | How to chose the order for polynomial regression? | In a polynomial regression process(gradient descent) try to find the global minima to optimize the cost function.
We choose the degree of polynomial for which the variance as computed by
$\frac{Sr(m)}{n-m-1}$
is a minimum or when there is no significant decrease in its value as the degree of polynomial is increased. In... | How to chose the order for polynomial regression? | In a polynomial regression process(gradient descent) try to find the global minima to optimize the cost function.
We choose the degree of polynomial for which the variance as computed by
$\frac{Sr(m)} | How to chose the order for polynomial regression?
In a polynomial regression process(gradient descent) try to find the global minima to optimize the cost function.
We choose the degree of polynomial for which the variance as computed by
$\frac{Sr(m)}{n-m-1}$
is a minimum or when there is no significant decrease in its ... | How to chose the order for polynomial regression?
In a polynomial regression process(gradient descent) try to find the global minima to optimize the cost function.
We choose the degree of polynomial for which the variance as computed by
$\frac{Sr(m)} |
54,242 | How to perform Validation on Unsupervised learning? | I realize this comes very late, but perhaps it is still useful for anyone looking into the same subject and coming across this question. I don't believe there is a standard method, as you ask. However, I worked on this about two years ago for my MSc thesis in Statistical Science: https://www.universiteitleiden.nl/binar... | How to perform Validation on Unsupervised learning? | I realize this comes very late, but perhaps it is still useful for anyone looking into the same subject and coming across this question. I don't believe there is a standard method, as you ask. However | How to perform Validation on Unsupervised learning?
I realize this comes very late, but perhaps it is still useful for anyone looking into the same subject and coming across this question. I don't believe there is a standard method, as you ask. However, I worked on this about two years ago for my MSc thesis in Statisti... | How to perform Validation on Unsupervised learning?
I realize this comes very late, but perhaps it is still useful for anyone looking into the same subject and coming across this question. I don't believe there is a standard method, as you ask. However |
54,243 | How to perform Validation on Unsupervised learning? | I'm not sure if it will be considered and answer as in fact is a pointer to a possible answer, but at the same time, I don't have enough reputation to add it as a comment. So it will go here, maybe someone with more rights can move it as comment.
I'm struggling with this theme too and today I found this PhD thesis
"CR... | How to perform Validation on Unsupervised learning? | I'm not sure if it will be considered and answer as in fact is a pointer to a possible answer, but at the same time, I don't have enough reputation to add it as a comment. So it will go here, maybe so | How to perform Validation on Unsupervised learning?
I'm not sure if it will be considered and answer as in fact is a pointer to a possible answer, but at the same time, I don't have enough reputation to add it as a comment. So it will go here, maybe someone with more rights can move it as comment.
I'm struggling with ... | How to perform Validation on Unsupervised learning?
I'm not sure if it will be considered and answer as in fact is a pointer to a possible answer, but at the same time, I don't have enough reputation to add it as a comment. So it will go here, maybe so |
54,244 | Architecture of autoencoders | Are there any examples of papers which used architectures consisting of multiple hidden layers?
Yes, e.g. look for "deep autoencoders" a.k.a. "stacked autoencoders", such as {1}:
Hugo Larochelle has the video on it: Neural networks [7.6] : Deep learning - deep autoencoder
Geoffrey Hinton also has a video on it: Lectu... | Architecture of autoencoders | Are there any examples of papers which used architectures consisting of multiple hidden layers?
Yes, e.g. look for "deep autoencoders" a.k.a. "stacked autoencoders", such as {1}:
Hugo Larochelle has | Architecture of autoencoders
Are there any examples of papers which used architectures consisting of multiple hidden layers?
Yes, e.g. look for "deep autoencoders" a.k.a. "stacked autoencoders", such as {1}:
Hugo Larochelle has the video on it: Neural networks [7.6] : Deep learning - deep autoencoder
Geoffrey Hinton ... | Architecture of autoencoders
Are there any examples of papers which used architectures consisting of multiple hidden layers?
Yes, e.g. look for "deep autoencoders" a.k.a. "stacked autoencoders", such as {1}:
Hugo Larochelle has |
54,245 | What to do if you suspect combinations of two distributions? | 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.
Adding to Maarten's answer, there are also the zero-in... | What to do if you suspect combinations of two distributions? | 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.
| What to do if you suspect combinations of two distributions?
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.
... | What to do if you suspect combinations of two distributions?
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.
|
54,246 | What to do if you suspect combinations of two distributions? | Let's say your variable is money spent on meat. In that case your spike at 0 are the vegetarians. You seem to be doing a regression model. Not taking the presence of vegetarians into account would be problematic. One possibility for that kind of data would be a Heckman selection model. | What to do if you suspect combinations of two distributions? | Let's say your variable is money spent on meat. In that case your spike at 0 are the vegetarians. You seem to be doing a regression model. Not taking the presence of vegetarians into account would be | What to do if you suspect combinations of two distributions?
Let's say your variable is money spent on meat. In that case your spike at 0 are the vegetarians. You seem to be doing a regression model. Not taking the presence of vegetarians into account would be problematic. One possibility for that kind of data would be... | What to do if you suspect combinations of two distributions?
Let's say your variable is money spent on meat. In that case your spike at 0 are the vegetarians. You seem to be doing a regression model. Not taking the presence of vegetarians into account would be |
54,247 | What to do if you suspect combinations of two distributions? | Observing a distribution that is actually the convolution of two or more distinct distributions is very common.
In general, you can be observing the same phenomenon for a single population, but observe different causes leading to the need of multiple distributions to describe the data; or you might be observing two di... | What to do if you suspect combinations of two distributions? | Observing a distribution that is actually the convolution of two or more distinct distributions is very common.
In general, you can be observing the same phenomenon for a single population, but obser | What to do if you suspect combinations of two distributions?
Observing a distribution that is actually the convolution of two or more distinct distributions is very common.
In general, you can be observing the same phenomenon for a single population, but observe different causes leading to the need of multiple distrib... | What to do if you suspect combinations of two distributions?
Observing a distribution that is actually the convolution of two or more distinct distributions is very common.
In general, you can be observing the same phenomenon for a single population, but obser |
54,248 | What does this Q-Q plot indicate about my data? | The shape of the plot is consistent with a left-skew, possibly bimodal distribution (with a small mode on the left).
It is possible that there are two groups with similar spread (such as a mixture of two normals with about the same standard deviation, the smaller subpopulation having a lower mean than the rest). This w... | What does this Q-Q plot indicate about my data? | The shape of the plot is consistent with a left-skew, possibly bimodal distribution (with a small mode on the left).
It is possible that there are two groups with similar spread (such as a mixture of | What does this Q-Q plot indicate about my data?
The shape of the plot is consistent with a left-skew, possibly bimodal distribution (with a small mode on the left).
It is possible that there are two groups with similar spread (such as a mixture of two normals with about the same standard deviation, the smaller subpopul... | What does this Q-Q plot indicate about my data?
The shape of the plot is consistent with a left-skew, possibly bimodal distribution (with a small mode on the left).
It is possible that there are two groups with similar spread (such as a mixture of |
54,249 | What does this Q-Q plot indicate about my data? | There are many ways to formally or informally take a sample and check to see if it is approximately normal. The pp or qq plots are usually used as exploratory tools. If that is your intent I would not worry much about error bars. The graph should look like a straight line roughly for normality to be considered a reason... | What does this Q-Q plot indicate about my data? | There are many ways to formally or informally take a sample and check to see if it is approximately normal. The pp or qq plots are usually used as exploratory tools. If that is your intent I would not | What does this Q-Q plot indicate about my data?
There are many ways to formally or informally take a sample and check to see if it is approximately normal. The pp or qq plots are usually used as exploratory tools. If that is your intent I would not worry much about error bars. The graph should look like a straight line... | What does this Q-Q plot indicate about my data?
There are many ways to formally or informally take a sample and check to see if it is approximately normal. The pp or qq plots are usually used as exploratory tools. If that is your intent I would not |
54,250 | How to interpret the classification boundary? | The reason is that you are NOT asking model to provide "a desired boundary", BUT simply ask the model to correctly classify your data.
There are infinite decision boundaries exist, that achieve same classification task with same accuracy.
When we use neural network, the model can chose whatever it wants. In addition, ... | How to interpret the classification boundary? | The reason is that you are NOT asking model to provide "a desired boundary", BUT simply ask the model to correctly classify your data.
There are infinite decision boundaries exist, that achieve same c | How to interpret the classification boundary?
The reason is that you are NOT asking model to provide "a desired boundary", BUT simply ask the model to correctly classify your data.
There are infinite decision boundaries exist, that achieve same classification task with same accuracy.
When we use neural network, the mo... | How to interpret the classification boundary?
The reason is that you are NOT asking model to provide "a desired boundary", BUT simply ask the model to correctly classify your data.
There are infinite decision boundaries exist, that achieve same c |
54,251 | What is central tendency? | You may find it more useful to think of "central tendency" as giving a sense of the distribution's location. This is in contrast to measures of spread (variance, range, etc.), which don't communicate location. From the wikipedia entry for central tendency:
In statistics, a central tendency (or, more commonly, a measur... | What is central tendency? | You may find it more useful to think of "central tendency" as giving a sense of the distribution's location. This is in contrast to measures of spread (variance, range, etc.), which don't communicate | What is central tendency?
You may find it more useful to think of "central tendency" as giving a sense of the distribution's location. This is in contrast to measures of spread (variance, range, etc.), which don't communicate location. From the wikipedia entry for central tendency:
In statistics, a central tendency (o... | What is central tendency?
You may find it more useful to think of "central tendency" as giving a sense of the distribution's location. This is in contrast to measures of spread (variance, range, etc.), which don't communicate |
54,252 | In what cases is it OK to use categorical predictors with many levels in regression? | Nothing is "always ok", as there are always exceptions. For example, logit and probit models get into trouble when one or more categories of your predictor perfectly predict the outcome. This can easily happen regardless of how large your sample size is.
Another case where your model would be somewhat problematic occu... | In what cases is it OK to use categorical predictors with many levels in regression? | Nothing is "always ok", as there are always exceptions. For example, logit and probit models get into trouble when one or more categories of your predictor perfectly predict the outcome. This can easi | In what cases is it OK to use categorical predictors with many levels in regression?
Nothing is "always ok", as there are always exceptions. For example, logit and probit models get into trouble when one or more categories of your predictor perfectly predict the outcome. This can easily happen regardless of how large y... | In what cases is it OK to use categorical predictors with many levels in regression?
Nothing is "always ok", as there are always exceptions. For example, logit and probit models get into trouble when one or more categories of your predictor perfectly predict the outcome. This can easi |
54,253 | In what cases is it OK to use categorical predictors with many levels in regression? | I don't think there is a definite answer. If there are no purely statistical issues (See Maarten Buis' answer) than this is a more theoretical issue.
The way I see it, is while many properties are naturally multi-categorical, there is not always a logical reason of making use of all that data. It can make a model cumb... | In what cases is it OK to use categorical predictors with many levels in regression? | I don't think there is a definite answer. If there are no purely statistical issues (See Maarten Buis' answer) than this is a more theoretical issue.
The way I see it, is while many properties are na | In what cases is it OK to use categorical predictors with many levels in regression?
I don't think there is a definite answer. If there are no purely statistical issues (See Maarten Buis' answer) than this is a more theoretical issue.
The way I see it, is while many properties are naturally multi-categorical, there is... | In what cases is it OK to use categorical predictors with many levels in regression?
I don't think there is a definite answer. If there are no purely statistical issues (See Maarten Buis' answer) than this is a more theoretical issue.
The way I see it, is while many properties are na |
54,254 | Var self-normalised sampling estimator | This is some work showing the Delta Method for approximating the variance of a ratio.
Let $X_1, \ldots, X_n \overset{iid}{\sim} q()$ be samples from your normalized instrumental density $q(\cdot)$. Let $p(\cdot) = C^{-1}p_u(\cdot)$ be your target density. Assume you can only evaluate $p_u$. Call $w_i = w_i(x_i) = p_u(x... | Var self-normalised sampling estimator | This is some work showing the Delta Method for approximating the variance of a ratio.
Let $X_1, \ldots, X_n \overset{iid}{\sim} q()$ be samples from your normalized instrumental density $q(\cdot)$. Le | Var self-normalised sampling estimator
This is some work showing the Delta Method for approximating the variance of a ratio.
Let $X_1, \ldots, X_n \overset{iid}{\sim} q()$ be samples from your normalized instrumental density $q(\cdot)$. Let $p(\cdot) = C^{-1}p_u(\cdot)$ be your target density. Assume you can only evalu... | Var self-normalised sampling estimator
This is some work showing the Delta Method for approximating the variance of a ratio.
Let $X_1, \ldots, X_n \overset{iid}{\sim} q()$ be samples from your normalized instrumental density $q(\cdot)$. Le |
54,255 | Var self-normalised sampling estimator | A very crude approximation to the variance of the self-normalised importance sampling estimator
$$
\hat{\mu}_n = \sum_{i=1}^n {\omega_i x_i \over \sum _{i=1}^n {\omega_i}}
$$
is
$$
\textrm{Var}_q(\hat{\mu}_n) \approx \textrm{Var}_p(\hat{\mu}_n)
(1 + \textrm{Var}_q(W)).
$$
where $p$ is the target distribution and $... | Var self-normalised sampling estimator | A very crude approximation to the variance of the self-normalised importance sampling estimator
$$
\hat{\mu}_n = \sum_{i=1}^n {\omega_i x_i \over \sum _{i=1}^n {\omega_i}}
$$
is
$$
\textrm{Var}_q(\ha | Var self-normalised sampling estimator
A very crude approximation to the variance of the self-normalised importance sampling estimator
$$
\hat{\mu}_n = \sum_{i=1}^n {\omega_i x_i \over \sum _{i=1}^n {\omega_i}}
$$
is
$$
\textrm{Var}_q(\hat{\mu}_n) \approx \textrm{Var}_p(\hat{\mu}_n)
(1 + \textrm{Var}_q(W)).
$$
whe... | Var self-normalised sampling estimator
A very crude approximation to the variance of the self-normalised importance sampling estimator
$$
\hat{\mu}_n = \sum_{i=1}^n {\omega_i x_i \over \sum _{i=1}^n {\omega_i}}
$$
is
$$
\textrm{Var}_q(\ha |
54,256 | Var self-normalised sampling estimator | To get a more accurate estimate than the Delta Method (shown in answer by @Taylor), use bootstrapping https://en.wikipedia.org/wiki/Bootstrapping_(statistics) and https://www.crcpress.com/An-Introduction-to-the-Bootstrap/Efron-Tibshirani/p/book/9780412042317. As a bonus, that will give you an estimate of the entire d... | Var self-normalised sampling estimator | To get a more accurate estimate than the Delta Method (shown in answer by @Taylor), use bootstrapping https://en.wikipedia.org/wiki/Bootstrapping_(statistics) and https://www.crcpress.com/An-Introdu | Var self-normalised sampling estimator
To get a more accurate estimate than the Delta Method (shown in answer by @Taylor), use bootstrapping https://en.wikipedia.org/wiki/Bootstrapping_(statistics) and https://www.crcpress.com/An-Introduction-to-the-Bootstrap/Efron-Tibshirani/p/book/9780412042317. As a bonus, that wi... | Var self-normalised sampling estimator
To get a more accurate estimate than the Delta Method (shown in answer by @Taylor), use bootstrapping https://en.wikipedia.org/wiki/Bootstrapping_(statistics) and https://www.crcpress.com/An-Introdu |
54,257 | How do I avoid computationally singular matrices in R? | Removing highly correlated (or identical) variables by hand can work, but :
It can become unfeasible as the number of variables becomes too large
Selecting the variables by hand is purely arbitrary
With factor variables, it becomes slightly harder to detect correlated variables (unless you look at the predictors with... | How do I avoid computationally singular matrices in R? | Removing highly correlated (or identical) variables by hand can work, but :
It can become unfeasible as the number of variables becomes too large
Selecting the variables by hand is purely arbitrary
| How do I avoid computationally singular matrices in R?
Removing highly correlated (or identical) variables by hand can work, but :
It can become unfeasible as the number of variables becomes too large
Selecting the variables by hand is purely arbitrary
With factor variables, it becomes slightly harder to detect corre... | How do I avoid computationally singular matrices in R?
Removing highly correlated (or identical) variables by hand can work, but :
It can become unfeasible as the number of variables becomes too large
Selecting the variables by hand is purely arbitrary
|
54,258 | How do I avoid computationally singular matrices in R? | RUser4512 has a good answer (+1). I just want to add some comments on Matrix Condition Number, which we can use to check the numerical stability issues. In R the function is kappa.
Here is an example in R. In this example, we create a data with two highly correlated columns. x1 and x2. Note they are not identical but r... | How do I avoid computationally singular matrices in R? | RUser4512 has a good answer (+1). I just want to add some comments on Matrix Condition Number, which we can use to check the numerical stability issues. In R the function is kappa.
Here is an example | How do I avoid computationally singular matrices in R?
RUser4512 has a good answer (+1). I just want to add some comments on Matrix Condition Number, which we can use to check the numerical stability issues. In R the function is kappa.
Here is an example in R. In this example, we create a data with two highly correlate... | How do I avoid computationally singular matrices in R?
RUser4512 has a good answer (+1). I just want to add some comments on Matrix Condition Number, which we can use to check the numerical stability issues. In R the function is kappa.
Here is an example |
54,259 | Does the proportional hazards assumption still matter if the covariate is time-dependent? | You are still assuming that the effect of the value at each covariates/factor at each timepoint is the same, you simply allow the covariate to vary its value over time (but the change in the log-hazard rate associated with a particular value is still exactly the same across all timepoints). Thus, it does not change the... | Does the proportional hazards assumption still matter if the covariate is time-dependent? | You are still assuming that the effect of the value at each covariates/factor at each timepoint is the same, you simply allow the covariate to vary its value over time (but the change in the log-hazar | Does the proportional hazards assumption still matter if the covariate is time-dependent?
You are still assuming that the effect of the value at each covariates/factor at each timepoint is the same, you simply allow the covariate to vary its value over time (but the change in the log-hazard rate associated with a parti... | Does the proportional hazards assumption still matter if the covariate is time-dependent?
You are still assuming that the effect of the value at each covariates/factor at each timepoint is the same, you simply allow the covariate to vary its value over time (but the change in the log-hazar |
54,260 | Does the proportional hazards assumption still matter if the covariate is time-dependent? | I may be wrong but I believe that Björn's answer is not completely correct. The proportional hazards assumption means that the ratio of the hazard for a particular group of observations (determined by the values of the covariates) to the baseline hazard (when all covariates are zero) is constant over time. If there are... | Does the proportional hazards assumption still matter if the covariate is time-dependent? | I may be wrong but I believe that Björn's answer is not completely correct. The proportional hazards assumption means that the ratio of the hazard for a particular group of observations (determined by | Does the proportional hazards assumption still matter if the covariate is time-dependent?
I may be wrong but I believe that Björn's answer is not completely correct. The proportional hazards assumption means that the ratio of the hazard for a particular group of observations (determined by the values of the covariates)... | Does the proportional hazards assumption still matter if the covariate is time-dependent?
I may be wrong but I believe that Björn's answer is not completely correct. The proportional hazards assumption means that the ratio of the hazard for a particular group of observations (determined by |
54,261 | Is adjusted R squared score still appropriate when number of regressors is larger than the sample size? | The adjusted $R^2$ value is specifically for linear regression where it's easy to know the effect of adding many predictors. If you were doing linear regression with more predictors than samples a linear regression would give $R^2=1$ so you must not be using a linear regression model. That means you can't adjust the $R... | Is adjusted R squared score still appropriate when number of regressors is larger than the sample si | The adjusted $R^2$ value is specifically for linear regression where it's easy to know the effect of adding many predictors. If you were doing linear regression with more predictors than samples a lin | Is adjusted R squared score still appropriate when number of regressors is larger than the sample size?
The adjusted $R^2$ value is specifically for linear regression where it's easy to know the effect of adding many predictors. If you were doing linear regression with more predictors than samples a linear regression w... | Is adjusted R squared score still appropriate when number of regressors is larger than the sample si
The adjusted $R^2$ value is specifically for linear regression where it's easy to know the effect of adding many predictors. If you were doing linear regression with more predictors than samples a lin |
54,262 | Is adjusted R squared score still appropriate when number of regressors is larger than the sample size? | To state notation, let $y$ be the $n$-vector of responses, let $X$ be the ($n\times p$) design matrix and let $\beta$ be the $p$-vector of unknown regression coefficients, with $n$ being the sample size. The well known least squares estimate of $\beta$ is $\hat\beta = (X^TX)^{-1} X^Ty$.
The coefficient of determination... | Is adjusted R squared score still appropriate when number of regressors is larger than the sample si | To state notation, let $y$ be the $n$-vector of responses, let $X$ be the ($n\times p$) design matrix and let $\beta$ be the $p$-vector of unknown regression coefficients, with $n$ being the sample si | Is adjusted R squared score still appropriate when number of regressors is larger than the sample size?
To state notation, let $y$ be the $n$-vector of responses, let $X$ be the ($n\times p$) design matrix and let $\beta$ be the $p$-vector of unknown regression coefficients, with $n$ being the sample size. The well kno... | Is adjusted R squared score still appropriate when number of regressors is larger than the sample si
To state notation, let $y$ be the $n$-vector of responses, let $X$ be the ($n\times p$) design matrix and let $\beta$ be the $p$-vector of unknown regression coefficients, with $n$ being the sample si |
54,263 | Is adjusted R squared score still appropriate when number of regressors is larger than the sample size? | The adjusted R-square value is always less than R-square when n>p that means number of observation is greater than the number of parameters. | Is adjusted R squared score still appropriate when number of regressors is larger than the sample si | The adjusted R-square value is always less than R-square when n>p that means number of observation is greater than the number of parameters. | Is adjusted R squared score still appropriate when number of regressors is larger than the sample size?
The adjusted R-square value is always less than R-square when n>p that means number of observation is greater than the number of parameters. | Is adjusted R squared score still appropriate when number of regressors is larger than the sample si
The adjusted R-square value is always less than R-square when n>p that means number of observation is greater than the number of parameters. |
54,264 | What is a practical explanation of affine equivariance and why does it matter for a covariance estimator? | I will first recall the property formally:
Given an $n$ by $p$, $n>p$ data matrix $X$, an affine equivariant estimator of location and scatter $(m(X), S(X))$ is one for which:
$$(0)\quad m(A X)=A m(X)$$
$$(1)\quad S(A X)=A^\top S(X)A$$
for any $p$ by $p$ non singular matrix $A$.
Consider a situation where one would... | What is a practical explanation of affine equivariance and why does it matter for a covariance estim | I will first recall the property formally:
Given an $n$ by $p$, $n>p$ data matrix $X$, an affine equivariant estimator of location and scatter $(m(X), S(X))$ is one for which:
$$(0)\quad m(A X)=A m(X) | What is a practical explanation of affine equivariance and why does it matter for a covariance estimator?
I will first recall the property formally:
Given an $n$ by $p$, $n>p$ data matrix $X$, an affine equivariant estimator of location and scatter $(m(X), S(X))$ is one for which:
$$(0)\quad m(A X)=A m(X)$$
$$(1)\quad... | What is a practical explanation of affine equivariance and why does it matter for a covariance estim
I will first recall the property formally:
Given an $n$ by $p$, $n>p$ data matrix $X$, an affine equivariant estimator of location and scatter $(m(X), S(X))$ is one for which:
$$(0)\quad m(A X)=A m(X) |
54,265 | R and SAS produce the same test-statistics but different p values for normality tests | The actual Kolmogorov-Smirnov, Anderson-Darling and Cramer-von Mises tests are for completely specified distributions. You're estimating the mean and variance of the residuals in your code so you don't have completely specified distributions, which will make your p-values larger than they should be.
There's another tes... | R and SAS produce the same test-statistics but different p values for normality tests | The actual Kolmogorov-Smirnov, Anderson-Darling and Cramer-von Mises tests are for completely specified distributions. You're estimating the mean and variance of the residuals in your code so you don' | R and SAS produce the same test-statistics but different p values for normality tests
The actual Kolmogorov-Smirnov, Anderson-Darling and Cramer-von Mises tests are for completely specified distributions. You're estimating the mean and variance of the residuals in your code so you don't have completely specified distri... | R and SAS produce the same test-statistics but different p values for normality tests
The actual Kolmogorov-Smirnov, Anderson-Darling and Cramer-von Mises tests are for completely specified distributions. You're estimating the mean and variance of the residuals in your code so you don' |
54,266 | With an R function that expects a covariance matrix, can I give it a correlation matrix? | A correlation matrix is a covariance matrix (of standardized variables) so you can do it (a correlation matrix is a valid covariance matrix* after all) -- the question is whether you end up with what you need.
You'll get multivariate normals with unit variance, with the population correlation matrix you supplied as a c... | With an R function that expects a covariance matrix, can I give it a correlation matrix? | A correlation matrix is a covariance matrix (of standardized variables) so you can do it (a correlation matrix is a valid covariance matrix* after all) -- the question is whether you end up with what | With an R function that expects a covariance matrix, can I give it a correlation matrix?
A correlation matrix is a covariance matrix (of standardized variables) so you can do it (a correlation matrix is a valid covariance matrix* after all) -- the question is whether you end up with what you need.
You'll get multivaria... | With an R function that expects a covariance matrix, can I give it a correlation matrix?
A correlation matrix is a covariance matrix (of standardized variables) so you can do it (a correlation matrix is a valid covariance matrix* after all) -- the question is whether you end up with what |
54,267 | With an R function that expects a covariance matrix, can I give it a correlation matrix? | No you cannot do this, those two things are not the same thing. There is no way, in general, you can go from correlation to covariance without knowing the individual variances. As pointed out in the comment below, the correlation will equal the covariance if all the variances are indeed $=1$.
The formula for the corr... | With an R function that expects a covariance matrix, can I give it a correlation matrix? | No you cannot do this, those two things are not the same thing. There is no way, in general, you can go from correlation to covariance without knowing the individual variances. As pointed out in the c | With an R function that expects a covariance matrix, can I give it a correlation matrix?
No you cannot do this, those two things are not the same thing. There is no way, in general, you can go from correlation to covariance without knowing the individual variances. As pointed out in the comment below, the correlation w... | With an R function that expects a covariance matrix, can I give it a correlation matrix?
No you cannot do this, those two things are not the same thing. There is no way, in general, you can go from correlation to covariance without knowing the individual variances. As pointed out in the c |
54,268 | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed] | Questions that are only about how to use R are off topic here; this will be closed.
Regarding the statistical issues involved in this situation, @JeremyMiles has provided a good answer.
For an R-specific response to this, it may help you to read the documentation for ?glm:
For a binomial GLM prior weights are use... | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed] | Questions that are only about how to use R are off topic here; this will be closed.
Regarding the statistical issues involved in this situation, @JeremyMiles has provided a good answer.
For an R-sp | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed]
Questions that are only about how to use R are off topic here; this will be closed.
Regarding the statistical issues involved in this situation, @JeremyMiles has provided a good answer.
For an R-specific response to ... | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed]
Questions that are only about how to use R are off topic here; this will be closed.
Regarding the statistical issues involved in this situation, @JeremyMiles has provided a good answer.
For an R-sp |
54,269 | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed] | First, you don't get the same answer. (And if you think you do, can you provide a reproducible example).
When you use the proportion, you discard information about the level of certainty of the effect. 1 success from two trials is different to 100 successes / 200 trials. | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed] | First, you don't get the same answer. (And if you think you do, can you provide a reproducible example).
When you use the proportion, you discard information about the level of certainty of the effec | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed]
First, you don't get the same answer. (And if you think you do, can you provide a reproducible example).
When you use the proportion, you discard information about the level of certainty of the effect. 1 success from t... | Using proportions directly instead of cbind() in glm() binomial regression is the same [R]? [closed]
First, you don't get the same answer. (And if you think you do, can you provide a reproducible example).
When you use the proportion, you discard information about the level of certainty of the effec |
54,270 | Is the joint probability of two sets equal to their intersection? | Short answer: yes
Long answer: probability is just a measure of the likelihood of some set of events happening (e.g. a coin flip landing heads is the event, and the probability of this event for a fair coin is 0.5). So if you're looking at 2 sets that are exactly the same, namely A intersect B or A,B (although I find t... | Is the joint probability of two sets equal to their intersection? | Short answer: yes
Long answer: probability is just a measure of the likelihood of some set of events happening (e.g. a coin flip landing heads is the event, and the probability of this event for a fai | Is the joint probability of two sets equal to their intersection?
Short answer: yes
Long answer: probability is just a measure of the likelihood of some set of events happening (e.g. a coin flip landing heads is the event, and the probability of this event for a fair coin is 0.5). So if you're looking at 2 sets that ar... | Is the joint probability of two sets equal to their intersection?
Short answer: yes
Long answer: probability is just a measure of the likelihood of some set of events happening (e.g. a coin flip landing heads is the event, and the probability of this event for a fai |
54,271 | Is the joint probability of two sets equal to their intersection? | Yes, commas generally are used to denote intersection even by those who are otherwise very careful to avoid the possibility of their writings being misinterpreted. The most common usage is $P(X\leq
x, Y \leq y)$ for the more prolix and correct $P\big ( (X \leq x)\cap (Y \leq y)\big)$ to denote the value of the joint C... | Is the joint probability of two sets equal to their intersection? | Yes, commas generally are used to denote intersection even by those who are otherwise very careful to avoid the possibility of their writings being misinterpreted. The most common usage is $P(X\leq
x | Is the joint probability of two sets equal to their intersection?
Yes, commas generally are used to denote intersection even by those who are otherwise very careful to avoid the possibility of their writings being misinterpreted. The most common usage is $P(X\leq
x, Y \leq y)$ for the more prolix and correct $P\big ( ... | Is the joint probability of two sets equal to their intersection?
Yes, commas generally are used to denote intersection even by those who are otherwise very careful to avoid the possibility of their writings being misinterpreted. The most common usage is $P(X\leq
x |
54,272 | Applying L1, L2 and Tikhonov Regularization to Neural Nets: Possible Misconceptions | If you are interested in pursuing neural networks (or other machine learning), you would be well served to spend some time on the mathematical fundamentals. I started skimming this trendy new treatise, which IMO is not too difficult mathematically (although I have seen some complain otherwise, and I was a math major, s... | Applying L1, L2 and Tikhonov Regularization to Neural Nets: Possible Misconceptions | If you are interested in pursuing neural networks (or other machine learning), you would be well served to spend some time on the mathematical fundamentals. I started skimming this trendy new treatise | Applying L1, L2 and Tikhonov Regularization to Neural Nets: Possible Misconceptions
If you are interested in pursuing neural networks (or other machine learning), you would be well served to spend some time on the mathematical fundamentals. I started skimming this trendy new treatise, which IMO is not too difficult mat... | Applying L1, L2 and Tikhonov Regularization to Neural Nets: Possible Misconceptions
If you are interested in pursuing neural networks (or other machine learning), you would be well served to spend some time on the mathematical fundamentals. I started skimming this trendy new treatise |
54,273 | How to classify data which is spiral in shape? | You could use SVM with an RBF kernel. Example:
import numpy as np
import matplotlib.pyplot as plt
import mlpy # sudo pip install mlpy
f = np.loadtxt("spiral.data")
x, y = f[:, :2], f[:, 2]
svm = mlpy.LibSvm(svm_type='c_svc', kernel_type='rbf', gamma=100)
svm.learn(x, y)
xmin, xmax = x[:,0].min()-0.1, x[:,0].max()+0.1
y... | How to classify data which is spiral in shape? | You could use SVM with an RBF kernel. Example:
import numpy as np
import matplotlib.pyplot as plt
import mlpy # sudo pip install mlpy
f = np.loadtxt("spiral.data")
x, y = f[:, :2], f[:, 2]
svm = mlpy. | How to classify data which is spiral in shape?
You could use SVM with an RBF kernel. Example:
import numpy as np
import matplotlib.pyplot as plt
import mlpy # sudo pip install mlpy
f = np.loadtxt("spiral.data")
x, y = f[:, :2], f[:, 2]
svm = mlpy.LibSvm(svm_type='c_svc', kernel_type='rbf', gamma=100)
svm.learn(x, y)
xm... | How to classify data which is spiral in shape?
You could use SVM with an RBF kernel. Example:
import numpy as np
import matplotlib.pyplot as plt
import mlpy # sudo pip install mlpy
f = np.loadtxt("spiral.data")
x, y = f[:, :2], f[:, 2]
svm = mlpy. |
54,274 | How to classify data which is spiral in shape? | I had similar experiments comparing to Franck's answer. Please check this post.
Do all machine learning algorithms separate data linearly?
In the post we use tree, boosting and K nearest neighbor on spiral data.
KNN is most intuitive one, it make the classification according to a given point's neighbors. So, spira... | How to classify data which is spiral in shape? | I had similar experiments comparing to Franck's answer. Please check this post.
Do all machine learning algorithms separate data linearly?
In the post we use tree, boosting and K nearest neighbor o | How to classify data which is spiral in shape?
I had similar experiments comparing to Franck's answer. Please check this post.
Do all machine learning algorithms separate data linearly?
In the post we use tree, boosting and K nearest neighbor on spiral data.
KNN is most intuitive one, it make the classification ac... | How to classify data which is spiral in shape?
I had similar experiments comparing to Franck's answer. Please check this post.
Do all machine learning algorithms separate data linearly?
In the post we use tree, boosting and K nearest neighbor o |
54,275 | How to classify data which is spiral in shape? | For this dummy problem you can increase the number of features. One particular way that I found to work is using extreme learning machines. Basically, you create a random matrix $K$ with columns equal to number of old features, $d$, and rows equal to number of new features $d'$(I had to use $d'=300d$). Also, create a r... | How to classify data which is spiral in shape? | For this dummy problem you can increase the number of features. One particular way that I found to work is using extreme learning machines. Basically, you create a random matrix $K$ with columns equal | How to classify data which is spiral in shape?
For this dummy problem you can increase the number of features. One particular way that I found to work is using extreme learning machines. Basically, you create a random matrix $K$ with columns equal to number of old features, $d$, and rows equal to number of new features... | How to classify data which is spiral in shape?
For this dummy problem you can increase the number of features. One particular way that I found to work is using extreme learning machines. Basically, you create a random matrix $K$ with columns equal |
54,276 | How to classify data which is spiral in shape? | You won't experience a "spiral" in the real world. But it is one of the more complex yet easy to visualize non-linear datasets. The playground in your question is for building intuition with neural networks. The other answers gave solutions that work but in my opinion miss the point of what can be learned here.
In the ... | How to classify data which is spiral in shape? | You won't experience a "spiral" in the real world. But it is one of the more complex yet easy to visualize non-linear datasets. The playground in your question is for building intuition with neural ne | How to classify data which is spiral in shape?
You won't experience a "spiral" in the real world. But it is one of the more complex yet easy to visualize non-linear datasets. The playground in your question is for building intuition with neural networks. The other answers gave solutions that work but in my opinion miss... | How to classify data which is spiral in shape?
You won't experience a "spiral" in the real world. But it is one of the more complex yet easy to visualize non-linear datasets. The playground in your question is for building intuition with neural ne |
54,277 | How to evaluate effect size from a regression output | Maybe an example will be helpful. This very simple example is from Gelman and Hill (2006, p.31-34). We want to predict cognitive test scores of children (kid.score) given their mothers' education (mom.hs) and IQ (mom.iq): mom.hs is a binary predictor indicating whether mother graduated from high school (1) or not (0), ... | How to evaluate effect size from a regression output | Maybe an example will be helpful. This very simple example is from Gelman and Hill (2006, p.31-34). We want to predict cognitive test scores of children (kid.score) given their mothers' education (mom | How to evaluate effect size from a regression output
Maybe an example will be helpful. This very simple example is from Gelman and Hill (2006, p.31-34). We want to predict cognitive test scores of children (kid.score) given their mothers' education (mom.hs) and IQ (mom.iq): mom.hs is a binary predictor indicating wheth... | How to evaluate effect size from a regression output
Maybe an example will be helpful. This very simple example is from Gelman and Hill (2006, p.31-34). We want to predict cognitive test scores of children (kid.score) given their mothers' education (mom |
54,278 | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distribution is provided? | You can compute pairwise KL divergence as a function of parameters in closed form for two Gaussian distributions $p$ and $q$. The uni-variate case:
$KL(p||q) = \log \frac{\sigma_2}{\sigma_1} + \frac{\sigma_{1}^{2} + (\mu_1-\mu_2)^2}{2\sigma_{2}^{2}} - \frac{1}{2}$
and the multi-variate case:
$KL(p||q) = \frac{1}{2}\lef... | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distributio | You can compute pairwise KL divergence as a function of parameters in closed form for two Gaussian distributions $p$ and $q$. The uni-variate case:
$KL(p||q) = \log \frac{\sigma_2}{\sigma_1} + \frac{\ | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distribution is provided?
You can compute pairwise KL divergence as a function of parameters in closed form for two Gaussian distributions $p$ and $q$. The uni-variate case:
$KL(p||q) = \log \frac{\sigma_2}{\sigma_1} + \frac{\sigma... | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distributio
You can compute pairwise KL divergence as a function of parameters in closed form for two Gaussian distributions $p$ and $q$. The uni-variate case:
$KL(p||q) = \log \frac{\sigma_2}{\sigma_1} + \frac{\ |
54,279 | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distribution is provided? | To complete the answer given by Vadim, there are also many approximation of the Kullback-Leibler divergence between mixtures of Gaussian distributions.
These approximations are surprisingly easy to compute and implement. These paper by Hershey & Olsen proposes 7 or 8 different approximations and advise the use of the v... | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distributio | To complete the answer given by Vadim, there are also many approximation of the Kullback-Leibler divergence between mixtures of Gaussian distributions.
These approximations are surprisingly easy to co | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distribution is provided?
To complete the answer given by Vadim, there are also many approximation of the Kullback-Leibler divergence between mixtures of Gaussian distributions.
These approximations are surprisingly easy to compute... | How to use Kullback-leibler divergence if mean and standard deviation of of two Gaussian Distributio
To complete the answer given by Vadim, there are also many approximation of the Kullback-Leibler divergence between mixtures of Gaussian distributions.
These approximations are surprisingly easy to co |
54,280 | Regression. Interaction term correlated with the variables | Keep it. It's one of those choices between unbiasedness and precision.
The negative effect of having correlated independent variables is that they inflate each other's variance (the statistics that quantifies this phenomenon is called variance inflation factor). The results are enlarged standard error, which leads to l... | Regression. Interaction term correlated with the variables | Keep it. It's one of those choices between unbiasedness and precision.
The negative effect of having correlated independent variables is that they inflate each other's variance (the statistics that qu | Regression. Interaction term correlated with the variables
Keep it. It's one of those choices between unbiasedness and precision.
The negative effect of having correlated independent variables is that they inflate each other's variance (the statistics that quantifies this phenomenon is called variance inflation factor)... | Regression. Interaction term correlated with the variables
Keep it. It's one of those choices between unbiasedness and precision.
The negative effect of having correlated independent variables is that they inflate each other's variance (the statistics that qu |
54,281 | Regression. Interaction term correlated with the variables | Consider (approximately) centreing your $X$ and $Y$ variables which will minimise the correlation between them and their interaction.
Although it is true that interpretation of the results of a regression can be easier if the predictors are independent it is not essential. After all the reason we do it is to see the ef... | Regression. Interaction term correlated with the variables | Consider (approximately) centreing your $X$ and $Y$ variables which will minimise the correlation between them and their interaction.
Although it is true that interpretation of the results of a regres | Regression. Interaction term correlated with the variables
Consider (approximately) centreing your $X$ and $Y$ variables which will minimise the correlation between them and their interaction.
Although it is true that interpretation of the results of a regression can be easier if the predictors are independent it is no... | Regression. Interaction term correlated with the variables
Consider (approximately) centreing your $X$ and $Y$ variables which will minimise the correlation between them and their interaction.
Although it is true that interpretation of the results of a regres |
54,282 | Proof of recurrence between cumulants and central-moments | The equation given by Wikipedia connects cumulants to moments (generally).
A proof of a formula connecting cumulants to central moments is found in A Recursive Formulation of the Old Problem of Obtaining Moments from Cumulants and Vice Versa
Letting $K(t)$ be the cumulant-generating function, and $M(t)$ the moment-gene... | Proof of recurrence between cumulants and central-moments | The equation given by Wikipedia connects cumulants to moments (generally).
A proof of a formula connecting cumulants to central moments is found in A Recursive Formulation of the Old Problem of Obtain | Proof of recurrence between cumulants and central-moments
The equation given by Wikipedia connects cumulants to moments (generally).
A proof of a formula connecting cumulants to central moments is found in A Recursive Formulation of the Old Problem of Obtaining Moments from Cumulants and Vice Versa
Letting $K(t)$ be th... | Proof of recurrence between cumulants and central-moments
The equation given by Wikipedia connects cumulants to moments (generally).
A proof of a formula connecting cumulants to central moments is found in A Recursive Formulation of the Old Problem of Obtain |
54,283 | Proof of recurrence between cumulants and central-moments | The paper mentioned and the formula cited by Itronneberg still refer to "raw" (non-central, "at the origin") moments. To verify this, take n=3: you get $\kappa_3 = \theta_3 - \kappa_1\theta_2$. (EDIT: fixed according to comment by Gâteau-Gallois) Hence, $\theta_n$ clearly denotes raw moments: the cumulant and the centr... | Proof of recurrence between cumulants and central-moments | The paper mentioned and the formula cited by Itronneberg still refer to "raw" (non-central, "at the origin") moments. To verify this, take n=3: you get $\kappa_3 = \theta_3 - \kappa_1\theta_2$. (EDIT: | Proof of recurrence between cumulants and central-moments
The paper mentioned and the formula cited by Itronneberg still refer to "raw" (non-central, "at the origin") moments. To verify this, take n=3: you get $\kappa_3 = \theta_3 - \kappa_1\theta_2$. (EDIT: fixed according to comment by Gâteau-Gallois) Hence, $\theta_... | Proof of recurrence between cumulants and central-moments
The paper mentioned and the formula cited by Itronneberg still refer to "raw" (non-central, "at the origin") moments. To verify this, take n=3: you get $\kappa_3 = \theta_3 - \kappa_1\theta_2$. (EDIT: |
54,284 | Multi parameter Metropolis-Hastings | You actually have a single joint prior, which is a function of the parameter vector $\theta = [a_1, ..., a_d]$. If the parameters are treated independently, the prior factorizes into a product of the 'individual priors' that you mentioned. That is:
$$p(\theta) = \prod_{i = 1}^d p(a_i)$$
Classic Metropolis-Hastings look... | Multi parameter Metropolis-Hastings | You actually have a single joint prior, which is a function of the parameter vector $\theta = [a_1, ..., a_d]$. If the parameters are treated independently, the prior factorizes into a product of the | Multi parameter Metropolis-Hastings
You actually have a single joint prior, which is a function of the parameter vector $\theta = [a_1, ..., a_d]$. If the parameters are treated independently, the prior factorizes into a product of the 'individual priors' that you mentioned. That is:
$$p(\theta) = \prod_{i = 1}^d p(a_i... | Multi parameter Metropolis-Hastings
You actually have a single joint prior, which is a function of the parameter vector $\theta = [a_1, ..., a_d]$. If the parameters are treated independently, the prior factorizes into a product of the |
54,285 | Multi parameter Metropolis-Hastings | To add to the existing thread, you could think of doing this using what are called piecewise or blockwise updates.
If you're considering the gaussian case and block updates you would have candidate values that are generated using a mean vector and a co-variance matrix and everything is accepted or rejected at once.
I... | Multi parameter Metropolis-Hastings | To add to the existing thread, you could think of doing this using what are called piecewise or blockwise updates.
If you're considering the gaussian case and block updates you would have candidate v | Multi parameter Metropolis-Hastings
To add to the existing thread, you could think of doing this using what are called piecewise or blockwise updates.
If you're considering the gaussian case and block updates you would have candidate values that are generated using a mean vector and a co-variance matrix and everything... | Multi parameter Metropolis-Hastings
To add to the existing thread, you could think of doing this using what are called piecewise or blockwise updates.
If you're considering the gaussian case and block updates you would have candidate v |
54,286 | Mean Absolute Deviation of Student t distributions? | General solution
Let $f$ be the density of a symmetrical distribution with zero expectation (such as any Student $t$ distribution with $\nu \gt 1$ degrees of freedom) and $F$ be its distribution function. Integrating by parts for $\mu \ge 0$, observe that
$$\int_\mu^\infty t f(t) dt = (t(1-F(t)))\big|_\mu^\infty + \in... | Mean Absolute Deviation of Student t distributions? | General solution
Let $f$ be the density of a symmetrical distribution with zero expectation (such as any Student $t$ distribution with $\nu \gt 1$ degrees of freedom) and $F$ be its distribution funct | Mean Absolute Deviation of Student t distributions?
General solution
Let $f$ be the density of a symmetrical distribution with zero expectation (such as any Student $t$ distribution with $\nu \gt 1$ degrees of freedom) and $F$ be its distribution function. Integrating by parts for $\mu \ge 0$, observe that
$$\int_\mu^... | Mean Absolute Deviation of Student t distributions?
General solution
Let $f$ be the density of a symmetrical distribution with zero expectation (such as any Student $t$ distribution with $\nu \gt 1$ degrees of freedom) and $F$ be its distribution funct |
54,287 | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation | It depends on your model and its probabilistic structure. Here are some options:
Composite/Quasi/Pseudo likelihood methods:
Varin, Cristiano, Nancy Reid, and David Firth. "An overview of composite likelihood methods." Statistica Sinica (2011): 5-42.
This paper contains an overview of composite likelihood methods, w... | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation | It depends on your model and its probabilistic structure. Here are some options:
Composite/Quasi/Pseudo likelihood methods:
Varin, Cristiano, Nancy Reid, and David Firth. "An overview of composite | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation
It depends on your model and its probabilistic structure. Here are some options:
Composite/Quasi/Pseudo likelihood methods:
Varin, Cristiano, Nancy Reid, and David Firth. "An overview of composite likelihood methods."... | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation
It depends on your model and its probabilistic structure. Here are some options:
Composite/Quasi/Pseudo likelihood methods:
Varin, Cristiano, Nancy Reid, and David Firth. "An overview of composite |
54,288 | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation | You could consider the method of simulated moments, see for instance:
http://www.stat.columbia.edu/~gelman/research/published/moments.pdf
Otherwise, synthetic likelihood might be another option:
http://www.nature.com/nature/journal/v466/n7310/full/nature09319.html | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation | You could consider the method of simulated moments, see for instance:
http://www.stat.columbia.edu/~gelman/research/published/moments.pdf
Otherwise, synthetic likelihood might be another option:
http | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation
You could consider the method of simulated moments, see for instance:
http://www.stat.columbia.edu/~gelman/research/published/moments.pdf
Otherwise, synthetic likelihood might be another option:
http://www.nature.com/na... | Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation
You could consider the method of simulated moments, see for instance:
http://www.stat.columbia.edu/~gelman/research/published/moments.pdf
Otherwise, synthetic likelihood might be another option:
http |
54,289 | The distribution of the AUC | AUC can be viewed as Wilcoxon-Mann-Whitney Test. And here is some demo, where for the R code I posted, I first calculate AUC, then use Wilcoxon-Mann-Whitney Test to calculate the number. Then verify both numbers are the same which is 0.911332. For a hypothesis testing, it is not hard to derive confidence interval. Righ... | The distribution of the AUC | AUC can be viewed as Wilcoxon-Mann-Whitney Test. And here is some demo, where for the R code I posted, I first calculate AUC, then use Wilcoxon-Mann-Whitney Test to calculate the number. Then verify b | The distribution of the AUC
AUC can be viewed as Wilcoxon-Mann-Whitney Test. And here is some demo, where for the R code I posted, I first calculate AUC, then use Wilcoxon-Mann-Whitney Test to calculate the number. Then verify both numbers are the same which is 0.911332. For a hypothesis testing, it is not hard to deri... | The distribution of the AUC
AUC can be viewed as Wilcoxon-Mann-Whitney Test. And here is some demo, where for the R code I posted, I first calculate AUC, then use Wilcoxon-Mann-Whitney Test to calculate the number. Then verify b |
54,290 | The distribution of the AUC | As hxd1011 said, the AUC is equivalent to the U statistic calculated for the Mann-Whitney-U aka Wilcoxon-rank-sum test, [normalized to [0;1] by the product of observations in each group]. The original paper by Mann and Whitney (1947) contains a proof for approximate normality and the U statistic is approximately Normal... | The distribution of the AUC | As hxd1011 said, the AUC is equivalent to the U statistic calculated for the Mann-Whitney-U aka Wilcoxon-rank-sum test, [normalized to [0;1] by the product of observations in each group]. The original | The distribution of the AUC
As hxd1011 said, the AUC is equivalent to the U statistic calculated for the Mann-Whitney-U aka Wilcoxon-rank-sum test, [normalized to [0;1] by the product of observations in each group]. The original paper by Mann and Whitney (1947) contains a proof for approximate normality and the U stati... | The distribution of the AUC
As hxd1011 said, the AUC is equivalent to the U statistic calculated for the Mann-Whitney-U aka Wilcoxon-rank-sum test, [normalized to [0;1] by the product of observations in each group]. The original |
54,291 | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution | Updated with full solution since OP has now solved it.
For a complete solution, one needs to first show that $ Y_i:= X_i - \bar{X}$ is a Gaussian random variable, whence it suffices to find its mean and variance to characterize the distribution. Knowing something about Gaussian random vectors makes this straight-forwa... | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution | Updated with full solution since OP has now solved it.
For a complete solution, one needs to first show that $ Y_i:= X_i - \bar{X}$ is a Gaussian random variable, whence it suffices to find its mean | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution
Updated with full solution since OP has now solved it.
For a complete solution, one needs to first show that $ Y_i:= X_i - \bar{X}$ is a Gaussian random variable, whence it suffices to find its mean and variance to characterize the d... | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution
Updated with full solution since OP has now solved it.
For a complete solution, one needs to first show that $ Y_i:= X_i - \bar{X}$ is a Gaussian random variable, whence it suffices to find its mean |
54,292 | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution | Student001 already gave a solution for the question and accepted.
Micheal M's comments are also very helpful, so I will post another solution following Micheal's suggestion without using matrix notation.
$X_1-\bar{X}=X_1-\frac{X_1+X_2+...+X_n}{n}\\=X_1-\frac{X_1}{n}-\frac{X_2+X_3+...+X_n}{n}\\=(1-\frac{1}{n})X_1-\frac{... | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution | Student001 already gave a solution for the question and accepted.
Micheal M's comments are also very helpful, so I will post another solution following Micheal's suggestion without using matrix notati | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution
Student001 already gave a solution for the question and accepted.
Micheal M's comments are also very helpful, so I will post another solution following Micheal's suggestion without using matrix notation.
$X_1-\bar{X}=X_1-\frac{X_1+X_2... | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution
Student001 already gave a solution for the question and accepted.
Micheal M's comments are also very helpful, so I will post another solution following Micheal's suggestion without using matrix notati |
54,293 | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution | Staying in the univarite case, since $X_i, i=1$ ,$\dots ,N$ are iid Normally distributed with mean $\mu$ and variance $\sigma^2$, we have, as you mentioned,
$E[X_i -\bar X] = E[X_i- \frac{1}{n}\sum_j^N X_j] = E[X_i] - \frac{1}{n}\sum_j^N E[X_j] = \mu - \frac{1}{n}n\mu =0$
For the variance, notice that
$Var[X_i-\bar X] ... | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution | Staying in the univarite case, since $X_i, i=1$ ,$\dots ,N$ are iid Normally distributed with mean $\mu$ and variance $\sigma^2$, we have, as you mentioned,
$E[X_i -\bar X] = E[X_i- \frac{1}{n}\sum_j^ | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution
Staying in the univarite case, since $X_i, i=1$ ,$\dots ,N$ are iid Normally distributed with mean $\mu$ and variance $\sigma^2$, we have, as you mentioned,
$E[X_i -\bar X] = E[X_i- \frac{1}{n}\sum_j^N X_j] = E[X_i] - \frac{1}{n}\sum_... | What is the distribution of $X_i-\bar{X}$ when $X_i$ has $N(\mu,\sigma)$ distribution
Staying in the univarite case, since $X_i, i=1$ ,$\dots ,N$ are iid Normally distributed with mean $\mu$ and variance $\sigma^2$, we have, as you mentioned,
$E[X_i -\bar X] = E[X_i- \frac{1}{n}\sum_j^ |
54,294 | Categorizing Continuous Random Variable in Logistic Regression | Instead of throwing away data by categorizing, you could consider fitting your continuous predictor as a spline function with a specified number of knots or with the number of knots chosen by cross-validation. That will use up no more degrees of freedom than categorization. If you are willing to envision up to 8 catego... | Categorizing Continuous Random Variable in Logistic Regression | Instead of throwing away data by categorizing, you could consider fitting your continuous predictor as a spline function with a specified number of knots or with the number of knots chosen by cross-va | Categorizing Continuous Random Variable in Logistic Regression
Instead of throwing away data by categorizing, you could consider fitting your continuous predictor as a spline function with a specified number of knots or with the number of knots chosen by cross-validation. That will use up no more degrees of freedom tha... | Categorizing Continuous Random Variable in Logistic Regression
Instead of throwing away data by categorizing, you could consider fitting your continuous predictor as a spline function with a specified number of knots or with the number of knots chosen by cross-va |
54,295 | Categorizing Continuous Random Variable in Logistic Regression | Since it seems like "ease of interpretation" is important to you, I think you would be interested to learn about nomograms, which are essentially a model represented in a diagrammatical way. Instead of relying on some ad hoc categorization procedure, you can fit ornate trends using statistically principled methods such... | Categorizing Continuous Random Variable in Logistic Regression | Since it seems like "ease of interpretation" is important to you, I think you would be interested to learn about nomograms, which are essentially a model represented in a diagrammatical way. Instead o | Categorizing Continuous Random Variable in Logistic Regression
Since it seems like "ease of interpretation" is important to you, I think you would be interested to learn about nomograms, which are essentially a model represented in a diagrammatical way. Instead of relying on some ad hoc categorization procedure, you ca... | Categorizing Continuous Random Variable in Logistic Regression
Since it seems like "ease of interpretation" is important to you, I think you would be interested to learn about nomograms, which are essentially a model represented in a diagrammatical way. Instead o |
54,296 | Categorizing Continuous Random Variable in Logistic Regression | I always think you can do most task by two approaches: knowledge driven and data driven include binning your continuous features.
By knowledge driven, you can think about what binning will make sense from what the actual feature represents. For example, if you are binning a household income, you definitely can find so... | Categorizing Continuous Random Variable in Logistic Regression | I always think you can do most task by two approaches: knowledge driven and data driven include binning your continuous features.
By knowledge driven, you can think about what binning will make sense | Categorizing Continuous Random Variable in Logistic Regression
I always think you can do most task by two approaches: knowledge driven and data driven include binning your continuous features.
By knowledge driven, you can think about what binning will make sense from what the actual feature represents. For example, if... | Categorizing Continuous Random Variable in Logistic Regression
I always think you can do most task by two approaches: knowledge driven and data driven include binning your continuous features.
By knowledge driven, you can think about what binning will make sense |
54,297 | Categorizing Continuous Random Variable in Logistic Regression | Thanks to those who tried to answer it. However, I don't think either of these answers are that much helpful to me. In fact there is a phd thesis written on this available here. There are also some R packages e.g. CatPredi that can be used as well. | Categorizing Continuous Random Variable in Logistic Regression | Thanks to those who tried to answer it. However, I don't think either of these answers are that much helpful to me. In fact there is a phd thesis written on this available here. There are also some R | Categorizing Continuous Random Variable in Logistic Regression
Thanks to those who tried to answer it. However, I don't think either of these answers are that much helpful to me. In fact there is a phd thesis written on this available here. There are also some R packages e.g. CatPredi that can be used as well. | Categorizing Continuous Random Variable in Logistic Regression
Thanks to those who tried to answer it. However, I don't think either of these answers are that much helpful to me. In fact there is a phd thesis written on this available here. There are also some R |
54,298 | A bar graph and its look - should I add titles, where should values go etc | I agree with EdM's point that "bar plots simply have too much ink for the information conveyed." Here's a ggplot2 version of his answer:
library(ggplot2)
df <- data.frame(years=c(1991, 1993, 1997, 2001, 2005, 2007, 2011, 2015),
freq=c(43.20, 52.13, 47.93, 46.29, 40.57, 53.88, 48.92, 50.92))
p <- (gg... | A bar graph and its look - should I add titles, where should values go etc | I agree with EdM's point that "bar plots simply have too much ink for the information conveyed." Here's a ggplot2 version of his answer:
library(ggplot2)
df <- data.frame(years=c(1991, 1993, 1997, 2 | A bar graph and its look - should I add titles, where should values go etc
I agree with EdM's point that "bar plots simply have too much ink for the information conveyed." Here's a ggplot2 version of his answer:
library(ggplot2)
df <- data.frame(years=c(1991, 1993, 1997, 2001, 2005, 2007, 2011, 2015),
... | A bar graph and its look - should I add titles, where should values go etc
I agree with EdM's point that "bar plots simply have too much ink for the information conveyed." Here's a ggplot2 version of his answer:
library(ggplot2)
df <- data.frame(years=c(1991, 1993, 1997, 2 |
54,299 | A bar graph and its look - should I add titles, where should values go etc | Maybe "Tufte`rize" your plot:
ggplot(dat, aes(years, freq)) +
geom_bar(stat = "identity", width=0.55, fill="grey")+
scale_y_continuous(breaks = seq(0,50,10)) +
geom_hline(yintercept= seq(0,50,10), col="white") +
theme_classic(base_size = 16) + theme(axis.ticks=element_blank()) +
labs(x=NULL, y=NULL) +
... | A bar graph and its look - should I add titles, where should values go etc | Maybe "Tufte`rize" your plot:
ggplot(dat, aes(years, freq)) +
geom_bar(stat = "identity", width=0.55, fill="grey")+
scale_y_continuous(breaks = seq(0,50,10)) +
geom_hline(yintercept= seq(0,50 | A bar graph and its look - should I add titles, where should values go etc
Maybe "Tufte`rize" your plot:
ggplot(dat, aes(years, freq)) +
geom_bar(stat = "identity", width=0.55, fill="grey")+
scale_y_continuous(breaks = seq(0,50,10)) +
geom_hline(yintercept= seq(0,50,10), col="white") +
theme_classic(base_si... | A bar graph and its look - should I add titles, where should values go etc
Maybe "Tufte`rize" your plot:
ggplot(dat, aes(years, freq)) +
geom_bar(stat = "identity", width=0.55, fill="grey")+
scale_y_continuous(breaks = seq(0,50,10)) +
geom_hline(yintercept= seq(0,50 |
54,300 | A bar graph and its look - should I add titles, where should values go etc | This is not a good application of a bar plot. Tufte would not be pleased, even with the allegedly "Tufte`rized" bar plots recommended in another answer. Bar plots simply have too much ink used for the information conveyed. See Tufte's website and books, and if possible attend one of his seminars on how to display data ... | A bar graph and its look - should I add titles, where should values go etc | This is not a good application of a bar plot. Tufte would not be pleased, even with the allegedly "Tufte`rized" bar plots recommended in another answer. Bar plots simply have too much ink used for the | A bar graph and its look - should I add titles, where should values go etc
This is not a good application of a bar plot. Tufte would not be pleased, even with the allegedly "Tufte`rized" bar plots recommended in another answer. Bar plots simply have too much ink used for the information conveyed. See Tufte's website an... | A bar graph and its look - should I add titles, where should values go etc
This is not a good application of a bar plot. Tufte would not be pleased, even with the allegedly "Tufte`rized" bar plots recommended in another answer. Bar plots simply have too much ink used for the |
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