idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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38,101 | Linear regression with strongly non-normal response variable | The distribution of the response is irrelevant. Inference based on small samples requires the errors to be approximately normal (better look at the QQ-plot of the residuals than at its density because the tails are important). If you are only interested in descriptive results or if the sample size is not too small, you... | Linear regression with strongly non-normal response variable | The distribution of the response is irrelevant. Inference based on small samples requires the errors to be approximately normal (better look at the QQ-plot of the residuals than at its density because | Linear regression with strongly non-normal response variable
The distribution of the response is irrelevant. Inference based on small samples requires the errors to be approximately normal (better look at the QQ-plot of the residuals than at its density because the tails are important). If you are only interested in de... | Linear regression with strongly non-normal response variable
The distribution of the response is irrelevant. Inference based on small samples requires the errors to be approximately normal (better look at the QQ-plot of the residuals than at its density because |
38,102 | Linear regression with strongly non-normal response variable | Your distribution is not beta if your density plot is to be taken at face value. A beta distribution cannot have two modes within (0, 1). However, no density plot for a bounded variable (at a guess here from some kernel density estimation procedure) can be taken at face value unless the estimation includes adjustments ... | Linear regression with strongly non-normal response variable | Your distribution is not beta if your density plot is to be taken at face value. A beta distribution cannot have two modes within (0, 1). However, no density plot for a bounded variable (at a guess he | Linear regression with strongly non-normal response variable
Your distribution is not beta if your density plot is to be taken at face value. A beta distribution cannot have two modes within (0, 1). However, no density plot for a bounded variable (at a guess here from some kernel density estimation procedure) can be ta... | Linear regression with strongly non-normal response variable
Your distribution is not beta if your density plot is to be taken at face value. A beta distribution cannot have two modes within (0, 1). However, no density plot for a bounded variable (at a guess he |
38,103 | Linear regression with strongly non-normal response variable | Strictly speaking, the normality of residuals assumption is not needed for OLS to work, it becomes an issue especially in hypothesis testing. Since your residuals actually seem to be normally distributed, you're fine even in this area. Additionally, OLS does not assume anything about the distribution of variables so yo... | Linear regression with strongly non-normal response variable | Strictly speaking, the normality of residuals assumption is not needed for OLS to work, it becomes an issue especially in hypothesis testing. Since your residuals actually seem to be normally distribu | Linear regression with strongly non-normal response variable
Strictly speaking, the normality of residuals assumption is not needed for OLS to work, it becomes an issue especially in hypothesis testing. Since your residuals actually seem to be normally distributed, you're fine even in this area. Additionally, OLS does ... | Linear regression with strongly non-normal response variable
Strictly speaking, the normality of residuals assumption is not needed for OLS to work, it becomes an issue especially in hypothesis testing. Since your residuals actually seem to be normally distribu |
38,104 | Linear regression with strongly non-normal response variable | Although the other answers have already addressed the question, I would like to add another powerful option that would solve most of the problems related to the distribution-assumptions: quantile regression.
Depending on the research interests, this method can be extremely powerful.
As someone has already said before, ... | Linear regression with strongly non-normal response variable | Although the other answers have already addressed the question, I would like to add another powerful option that would solve most of the problems related to the distribution-assumptions: quantile regr | Linear regression with strongly non-normal response variable
Although the other answers have already addressed the question, I would like to add another powerful option that would solve most of the problems related to the distribution-assumptions: quantile regression.
Depending on the research interests, this method ca... | Linear regression with strongly non-normal response variable
Although the other answers have already addressed the question, I would like to add another powerful option that would solve most of the problems related to the distribution-assumptions: quantile regr |
38,105 | Mean has lower standard error than 5% trimmed mean? | If the underlying population is normally distributed without contamination then the sample mean is the best unbiased estimate (in the sense of the lowest mean square error) of the centre of the population distribution.
This is not always the case with other distributions, which might include those with contamination.... | Mean has lower standard error than 5% trimmed mean? | If the underlying population is normally distributed without contamination then the sample mean is the best unbiased estimate (in the sense of the lowest mean square error) of the centre of the popula | Mean has lower standard error than 5% trimmed mean?
If the underlying population is normally distributed without contamination then the sample mean is the best unbiased estimate (in the sense of the lowest mean square error) of the centre of the population distribution.
This is not always the case with other distribu... | Mean has lower standard error than 5% trimmed mean?
If the underlying population is normally distributed without contamination then the sample mean is the best unbiased estimate (in the sense of the lowest mean square error) of the centre of the popula |
38,106 | Mean has lower standard error than 5% trimmed mean? | This does seem surprising at first sight, but here is a guess at what is happening.
Focus on what a bootstrap sample is, namely a sample with replacement. So, every now and again some of these samples will include repetitions of the outliers or wild values. Those samples will be trimmed, but in some cases the trimming... | Mean has lower standard error than 5% trimmed mean? | This does seem surprising at first sight, but here is a guess at what is happening.
Focus on what a bootstrap sample is, namely a sample with replacement. So, every now and again some of these sample | Mean has lower standard error than 5% trimmed mean?
This does seem surprising at first sight, but here is a guess at what is happening.
Focus on what a bootstrap sample is, namely a sample with replacement. So, every now and again some of these samples will include repetitions of the outliers or wild values. Those sam... | Mean has lower standard error than 5% trimmed mean?
This does seem surprising at first sight, but here is a guess at what is happening.
Focus on what a bootstrap sample is, namely a sample with replacement. So, every now and again some of these sample |
38,107 | Mean has lower standard error than 5% trimmed mean? | The efficiency of the trimmed mean depends on the shape of the distribution.
If the underlying distribution is very asymmetric -- say, Exponential -- then trimming will bias your mean in the negative direction.
Or, say, if the distribution is a mixture of two distributions with different means, trimming could remove mo... | Mean has lower standard error than 5% trimmed mean? | The efficiency of the trimmed mean depends on the shape of the distribution.
If the underlying distribution is very asymmetric -- say, Exponential -- then trimming will bias your mean in the negative | Mean has lower standard error than 5% trimmed mean?
The efficiency of the trimmed mean depends on the shape of the distribution.
If the underlying distribution is very asymmetric -- say, Exponential -- then trimming will bias your mean in the negative direction.
Or, say, if the distribution is a mixture of two distribu... | Mean has lower standard error than 5% trimmed mean?
The efficiency of the trimmed mean depends on the shape of the distribution.
If the underlying distribution is very asymmetric -- say, Exponential -- then trimming will bias your mean in the negative |
38,108 | Excel, Heatmap & Data Visualization without add-ins | The short answer is no, there is no easy way to create most of the graphics you mention. But in any graphics environment where you can draw line segments (such as the pen plotter drivers from the 60's, 70's, and 80's), you can construct workable visualizations. So one method is to focus on joined scatterplots (which is... | Excel, Heatmap & Data Visualization without add-ins | The short answer is no, there is no easy way to create most of the graphics you mention. But in any graphics environment where you can draw line segments (such as the pen plotter drivers from the 60's | Excel, Heatmap & Data Visualization without add-ins
The short answer is no, there is no easy way to create most of the graphics you mention. But in any graphics environment where you can draw line segments (such as the pen plotter drivers from the 60's, 70's, and 80's), you can construct workable visualizations. So one... | Excel, Heatmap & Data Visualization without add-ins
The short answer is no, there is no easy way to create most of the graphics you mention. But in any graphics environment where you can draw line segments (such as the pen plotter drivers from the 60's |
38,109 | Excel, Heatmap & Data Visualization without add-ins | As the way I've found to plot heatmaps in graphs is very simple and can be used to plot almost everything, I think it will interet some people. For curious people (and those who have time to lose).
My idea is to create a "screen" pixel by pixel.
Create a table with the same size of what you want to plot, filled with on... | Excel, Heatmap & Data Visualization without add-ins | As the way I've found to plot heatmaps in graphs is very simple and can be used to plot almost everything, I think it will interet some people. For curious people (and those who have time to lose).
My | Excel, Heatmap & Data Visualization without add-ins
As the way I've found to plot heatmaps in graphs is very simple and can be used to plot almost everything, I think it will interet some people. For curious people (and those who have time to lose).
My idea is to create a "screen" pixel by pixel.
Create a table with th... | Excel, Heatmap & Data Visualization without add-ins
As the way I've found to plot heatmaps in graphs is very simple and can be used to plot almost everything, I think it will interet some people. For curious people (and those who have time to lose).
My |
38,110 | Excel, Heatmap & Data Visualization without add-ins | Jon Peltier has a really great site with a lot of guides on how to bend Excel to make some more exotic charts. For example, Box plots and Heat maps. You don't need his plug-in to execute a lot of them.
Excel can't do a lot of things, and sometimes the things that you can make it do are hard to maintain and update, ... | Excel, Heatmap & Data Visualization without add-ins | Jon Peltier has a really great site with a lot of guides on how to bend Excel to make some more exotic charts. For example, Box plots and Heat maps. You don't need his plug-in to execute a lot of th | Excel, Heatmap & Data Visualization without add-ins
Jon Peltier has a really great site with a lot of guides on how to bend Excel to make some more exotic charts. For example, Box plots and Heat maps. You don't need his plug-in to execute a lot of them.
Excel can't do a lot of things, and sometimes the things that ... | Excel, Heatmap & Data Visualization without add-ins
Jon Peltier has a really great site with a lot of guides on how to bend Excel to make some more exotic charts. For example, Box plots and Heat maps. You don't need his plug-in to execute a lot of th |
38,111 | Excel, Heatmap & Data Visualization without add-ins | There a few tools out there to create treemaps in Excel. Try Treemap or Sparklines | Excel, Heatmap & Data Visualization without add-ins | There a few tools out there to create treemaps in Excel. Try Treemap or Sparklines | Excel, Heatmap & Data Visualization without add-ins
There a few tools out there to create treemaps in Excel. Try Treemap or Sparklines | Excel, Heatmap & Data Visualization without add-ins
There a few tools out there to create treemaps in Excel. Try Treemap or Sparklines |
38,112 | Are misses in my data distributed completely at random? | As @Dirk Eddelbuettel already mentioned, your questions is not very clear. In fact, I think you are asking two questions. The first question is related to your M(C)AR assumption. The second question is about (an) appropriate R package(s).
(1) "Testing" for MAR
To test if age has an effect on the missingness of your sc... | Are misses in my data distributed completely at random? | As @Dirk Eddelbuettel already mentioned, your questions is not very clear. In fact, I think you are asking two questions. The first question is related to your M(C)AR assumption. The second question i | Are misses in my data distributed completely at random?
As @Dirk Eddelbuettel already mentioned, your questions is not very clear. In fact, I think you are asking two questions. The first question is related to your M(C)AR assumption. The second question is about (an) appropriate R package(s).
(1) "Testing" for MAR
To... | Are misses in my data distributed completely at random?
As @Dirk Eddelbuettel already mentioned, your questions is not very clear. In fact, I think you are asking two questions. The first question is related to your M(C)AR assumption. The second question i |
38,113 | Are misses in my data distributed completely at random? | As far as I understand your question, you want to investigate if missing values in your data appear due to some pattern. In this case, you don't need any "missing value analysis" -- this is the same problem as checking whether the score is bigger than 0.7 or whatever. Just convert your dataset into two-class factor (mi... | Are misses in my data distributed completely at random? | As far as I understand your question, you want to investigate if missing values in your data appear due to some pattern. In this case, you don't need any "missing value analysis" -- this is the same p | Are misses in my data distributed completely at random?
As far as I understand your question, you want to investigate if missing values in your data appear due to some pattern. In this case, you don't need any "missing value analysis" -- this is the same problem as checking whether the score is bigger than 0.7 or whate... | Are misses in my data distributed completely at random?
As far as I understand your question, you want to investigate if missing values in your data appear due to some pattern. In this case, you don't need any "missing value analysis" -- this is the same p |
38,114 | Are misses in my data distributed completely at random? | Your question is a little difficult to decipher. One approach for dealing with missing data is imputation -- and there is a substantial literature on this and an already large and growing set of packages at CRAN so you may want to start there. | Are misses in my data distributed completely at random? | Your question is a little difficult to decipher. One approach for dealing with missing data is imputation -- and there is a substantial literature on this and an already large and growing set of pack | Are misses in my data distributed completely at random?
Your question is a little difficult to decipher. One approach for dealing with missing data is imputation -- and there is a substantial literature on this and an already large and growing set of packages at CRAN so you may want to start there. | Are misses in my data distributed completely at random?
Your question is a little difficult to decipher. One approach for dealing with missing data is imputation -- and there is a substantial literature on this and an already large and growing set of pack |
38,115 | What does Φ mean? [closed] | The Phyrexian or Greek letter Φ is being used to describe the CDF (cumulative distribution function) of the standard normal.
The lowercase Greek letter φ is being used to describe the PDF (probability density function) of the standard normal distribution.
And as noted in the other answer as well (I copy it to make thi... | What does Φ mean? [closed] | The Phyrexian or Greek letter Φ is being used to describe the CDF (cumulative distribution function) of the standard normal.
The lowercase Greek letter φ is being used to describe the PDF (probability | What does Φ mean? [closed]
The Phyrexian or Greek letter Φ is being used to describe the CDF (cumulative distribution function) of the standard normal.
The lowercase Greek letter φ is being used to describe the PDF (probability density function) of the standard normal distribution.
And as noted in the other answer as ... | What does Φ mean? [closed]
The Phyrexian or Greek letter Φ is being used to describe the CDF (cumulative distribution function) of the standard normal.
The lowercase Greek letter φ is being used to describe the PDF (probability |
38,116 | What does Φ mean? [closed] | As the previos answer noticed, $\Phi$ is just a short notation to denote the CDF of random variable which has standard normal distribution. But as I see, in METAL documentation you need $\Phi^{-1}$ (normal quantile function) so these links to Wiki can help you.
https://en.wikipedia.org/wiki/Error_function#Related_funct... | What does Φ mean? [closed] | As the previos answer noticed, $\Phi$ is just a short notation to denote the CDF of random variable which has standard normal distribution. But as I see, in METAL documentation you need $\Phi^{-1}$ (n | What does Φ mean? [closed]
As the previos answer noticed, $\Phi$ is just a short notation to denote the CDF of random variable which has standard normal distribution. But as I see, in METAL documentation you need $\Phi^{-1}$ (normal quantile function) so these links to Wiki can help you.
https://en.wikipedia.org/wiki/E... | What does Φ mean? [closed]
As the previos answer noticed, $\Phi$ is just a short notation to denote the CDF of random variable which has standard normal distribution. But as I see, in METAL documentation you need $\Phi^{-1}$ (n |
38,117 | When to use Mean(X/Y) versus Mean(X)/Mean(Y)? | If $x_i =$ number of items consumed on active days (for person $i$) and $y=$ number of active days (for person $i$), then...
$\text{Mean}(x/y)$ is the average number of items consumed per person on a day in which that person is active. This gives equal weight to each person's data.
$\frac{\text{Mean}(x)}{\text{Mean}(y... | When to use Mean(X/Y) versus Mean(X)/Mean(Y)? | If $x_i =$ number of items consumed on active days (for person $i$) and $y=$ number of active days (for person $i$), then...
$\text{Mean}(x/y)$ is the average number of items consumed per person on a | When to use Mean(X/Y) versus Mean(X)/Mean(Y)?
If $x_i =$ number of items consumed on active days (for person $i$) and $y=$ number of active days (for person $i$), then...
$\text{Mean}(x/y)$ is the average number of items consumed per person on a day in which that person is active. This gives equal weight to each perso... | When to use Mean(X/Y) versus Mean(X)/Mean(Y)?
If $x_i =$ number of items consumed on active days (for person $i$) and $y=$ number of active days (for person $i$), then...
$\text{Mean}(x/y)$ is the average number of items consumed per person on a |
38,118 | When to use Mean(X/Y) versus Mean(X)/Mean(Y)? | I first consider $mean(x/y)$ versus $mean(x)/mean(y)$.
As Eoin and dariober suggest, the more practically relevant quantity should be used, which is often the former, although the latter is not that outlandish.
To illustrate both, let's modify the OP's example and imagine opening a bakery with bakers A, B and C.
... | When to use Mean(X/Y) versus Mean(X)/Mean(Y)? | I first consider $mean(x/y)$ versus $mean(x)/mean(y)$.
As Eoin and dariober suggest, the more practically relevant quantity should be used, which is often the former, although the latter is not t | When to use Mean(X/Y) versus Mean(X)/Mean(Y)?
I first consider $mean(x/y)$ versus $mean(x)/mean(y)$.
As Eoin and dariober suggest, the more practically relevant quantity should be used, which is often the former, although the latter is not that outlandish.
To illustrate both, let's modify the OP's example and imag... | When to use Mean(X/Y) versus Mean(X)/Mean(Y)?
I first consider $mean(x/y)$ versus $mean(x)/mean(y)$.
As Eoin and dariober suggest, the more practically relevant quantity should be used, which is often the former, although the latter is not t |
38,119 | When to use Mean(X/Y) versus Mean(X)/Mean(Y)? | In my opinion mean(X/Y) is more meaningful because your experimental unit (not sure this is the correct term) is the individual, not the aggregate. Let's try to see this with a contrived example. Consider this data set:
individual cons act ratio
A 5 2 2.5
B 5 2 2.5
C 5 ... | When to use Mean(X/Y) versus Mean(X)/Mean(Y)? | In my opinion mean(X/Y) is more meaningful because your experimental unit (not sure this is the correct term) is the individual, not the aggregate. Let's try to see this with a contrived example. Cons | When to use Mean(X/Y) versus Mean(X)/Mean(Y)?
In my opinion mean(X/Y) is more meaningful because your experimental unit (not sure this is the correct term) is the individual, not the aggregate. Let's try to see this with a contrived example. Consider this data set:
individual cons act ratio
A 5 2 2.5
... | When to use Mean(X/Y) versus Mean(X)/Mean(Y)?
In my opinion mean(X/Y) is more meaningful because your experimental unit (not sure this is the correct term) is the individual, not the aggregate. Let's try to see this with a contrived example. Cons |
38,120 | Multiply, add, or condition on probability? | Let's follow up on GlenB's advice and make those Venn diagrams. We do this below with the heterosexual stereotype colours representing mother sick with red/pink and dad sick with blue.
With the two variables mother and father you can create 4 different disjoint situations.
father sick and not mom sick
not fath... | Multiply, add, or condition on probability? | Let's follow up on GlenB's advice and make those Venn diagrams. We do this below with the heterosexual stereotype colours representing mother sick with red/pink and dad sick with blue.
With the two v | Multiply, add, or condition on probability?
Let's follow up on GlenB's advice and make those Venn diagrams. We do this below with the heterosexual stereotype colours representing mother sick with red/pink and dad sick with blue.
With the two variables mother and father you can create 4 different disjoint situations.
... | Multiply, add, or condition on probability?
Let's follow up on GlenB's advice and make those Venn diagrams. We do this below with the heterosexual stereotype colours representing mother sick with red/pink and dad sick with blue.
With the two v |
38,121 | Multiply, add, or condition on probability? | My first question is that: I find it particularly difficult to differentiate between addition or multiplication rule when it comes to probabilities from independent events.
That's not a question (you don't ask anything), but the answer to what I assume is your implied question is simple: there isn't an addition rule f... | Multiply, add, or condition on probability? | My first question is that: I find it particularly difficult to differentiate between addition or multiplication rule when it comes to probabilities from independent events.
That's not a question (you | Multiply, add, or condition on probability?
My first question is that: I find it particularly difficult to differentiate between addition or multiplication rule when it comes to probabilities from independent events.
That's not a question (you don't ask anything), but the answer to what I assume is your implied questi... | Multiply, add, or condition on probability?
My first question is that: I find it particularly difficult to differentiate between addition or multiplication rule when it comes to probabilities from independent events.
That's not a question (you |
38,122 | Can I remove sample outliers using standard deviation? | The use of a standard deviation-based threshold for "outlier" detection is generally not a good idea. I think there may be some confusion as to what "normal data" mean as this refers to the statistical distribution of the data, not a qualitative description of the population from which they derive. Your post suggests t... | Can I remove sample outliers using standard deviation? | The use of a standard deviation-based threshold for "outlier" detection is generally not a good idea. I think there may be some confusion as to what "normal data" mean as this refers to the statistica | Can I remove sample outliers using standard deviation?
The use of a standard deviation-based threshold for "outlier" detection is generally not a good idea. I think there may be some confusion as to what "normal data" mean as this refers to the statistical distribution of the data, not a qualitative description of the ... | Can I remove sample outliers using standard deviation?
The use of a standard deviation-based threshold for "outlier" detection is generally not a good idea. I think there may be some confusion as to what "normal data" mean as this refers to the statistica |
38,123 | Can I remove sample outliers using standard deviation? | The answer from @Billy (+1) gets to the critical points of the question you posed. These are just a few further thoughts on your modeling strategy that are too extensive to fit into comments.
First, from what you describe it's not clear what you will gain with elastic net. With 6000 cases and what seems to be an outcom... | Can I remove sample outliers using standard deviation? | The answer from @Billy (+1) gets to the critical points of the question you posed. These are just a few further thoughts on your modeling strategy that are too extensive to fit into comments.
First, f | Can I remove sample outliers using standard deviation?
The answer from @Billy (+1) gets to the critical points of the question you posed. These are just a few further thoughts on your modeling strategy that are too extensive to fit into comments.
First, from what you describe it's not clear what you will gain with elas... | Can I remove sample outliers using standard deviation?
The answer from @Billy (+1) gets to the critical points of the question you posed. These are just a few further thoughts on your modeling strategy that are too extensive to fit into comments.
First, f |
38,124 | Is there any classification algorithm that doesn't give probability? | TLDR: probabilities are not required to build a ROC curve, only a numerical scale supporting the decision.
I'm studying the ROC Curve, and I was wondering if there is any
classification algorithm that doesn't return the output class as a
result of a certain threshold from the probabilities of the algo?
I previou... | Is there any classification algorithm that doesn't give probability? | TLDR: probabilities are not required to build a ROC curve, only a numerical scale supporting the decision.
I'm studying the ROC Curve, and I was wondering if there is any
classification algorithm | Is there any classification algorithm that doesn't give probability?
TLDR: probabilities are not required to build a ROC curve, only a numerical scale supporting the decision.
I'm studying the ROC Curve, and I was wondering if there is any
classification algorithm that doesn't return the output class as a
result ... | Is there any classification algorithm that doesn't give probability?
TLDR: probabilities are not required to build a ROC curve, only a numerical scale supporting the decision.
I'm studying the ROC Curve, and I was wondering if there is any
classification algorithm |
38,125 | Is there any classification algorithm that doesn't give probability? | Support Vector Machines and $k$-Nearest Neighbors come to mind.
(See here for a motivation for short answers. Longer answers are always welcome.) | Is there any classification algorithm that doesn't give probability? | Support Vector Machines and $k$-Nearest Neighbors come to mind.
(See here for a motivation for short answers. Longer answers are always welcome.) | Is there any classification algorithm that doesn't give probability?
Support Vector Machines and $k$-Nearest Neighbors come to mind.
(See here for a motivation for short answers. Longer answers are always welcome.) | Is there any classification algorithm that doesn't give probability?
Support Vector Machines and $k$-Nearest Neighbors come to mind.
(See here for a motivation for short answers. Longer answers are always welcome.) |
38,126 | Independence of $X+Y$ and $X-Y$ | They're not: If $X+Y=12$ then both rolls were sixes, so $X-Y=0$. So you have:
$$1 = \mathbb{P}(X-Y =0|X+Y=12) \neq \mathbb{P}(X-Y =0) = \frac{1}{6}.$$ | Independence of $X+Y$ and $X-Y$ | They're not: If $X+Y=12$ then both rolls were sixes, so $X-Y=0$. So you have:
$$1 = \mathbb{P}(X-Y =0|X+Y=12) \neq \mathbb{P}(X-Y =0) = \frac{1}{6}.$$ | Independence of $X+Y$ and $X-Y$
They're not: If $X+Y=12$ then both rolls were sixes, so $X-Y=0$. So you have:
$$1 = \mathbb{P}(X-Y =0|X+Y=12) \neq \mathbb{P}(X-Y =0) = \frac{1}{6}.$$ | Independence of $X+Y$ and $X-Y$
They're not: If $X+Y=12$ then both rolls were sixes, so $X-Y=0$. So you have:
$$1 = \mathbb{P}(X-Y =0|X+Y=12) \neq \mathbb{P}(X-Y =0) = \frac{1}{6}.$$ |
38,127 | Independence of $X+Y$ and $X-Y$ | When checking if the random variables are independent or not, the first thing needing to check is the range of the random variables. If the range of one random variable varies according to values of other random variables, then they are not independent, and stop. Otherwise, need to check further.
In this problem, when ... | Independence of $X+Y$ and $X-Y$ | When checking if the random variables are independent or not, the first thing needing to check is the range of the random variables. If the range of one random variable varies according to values of o | Independence of $X+Y$ and $X-Y$
When checking if the random variables are independent or not, the first thing needing to check is the range of the random variables. If the range of one random variable varies according to values of other random variables, then they are not independent, and stop. Otherwise, need to check... | Independence of $X+Y$ and $X-Y$
When checking if the random variables are independent or not, the first thing needing to check is the range of the random variables. If the range of one random variable varies according to values of o |
38,128 | How to handle predictors that are highly correlated to the response | It sounds like what you have is a powerfully predictive variable, and there is no reason to remove it.
What you have to watch out in situations like this is what is called leakage. Leakage is when you have a predictor that is just some version of your response in disguise.
For example, suppose that you have a system a... | How to handle predictors that are highly correlated to the response | It sounds like what you have is a powerfully predictive variable, and there is no reason to remove it.
What you have to watch out in situations like this is what is called leakage. Leakage is when yo | How to handle predictors that are highly correlated to the response
It sounds like what you have is a powerfully predictive variable, and there is no reason to remove it.
What you have to watch out in situations like this is what is called leakage. Leakage is when you have a predictor that is just some version of your... | How to handle predictors that are highly correlated to the response
It sounds like what you have is a powerfully predictive variable, and there is no reason to remove it.
What you have to watch out in situations like this is what is called leakage. Leakage is when yo |
38,129 | How to handle predictors that are highly correlated to the response | In addition to @Matthew Drury's answer, you could train different models for different transaction methods (cheques, non-cheques). This way the features of people using cheques would be highlighted and the column will also remain in the data.
See if the tool you are using allows grouping when fitting models. This coul... | How to handle predictors that are highly correlated to the response | In addition to @Matthew Drury's answer, you could train different models for different transaction methods (cheques, non-cheques). This way the features of people using cheques would be highlighted an | How to handle predictors that are highly correlated to the response
In addition to @Matthew Drury's answer, you could train different models for different transaction methods (cheques, non-cheques). This way the features of people using cheques would be highlighted and the column will also remain in the data.
See if t... | How to handle predictors that are highly correlated to the response
In addition to @Matthew Drury's answer, you could train different models for different transaction methods (cheques, non-cheques). This way the features of people using cheques would be highlighted an |
38,130 | How to handle predictors that are highly correlated to the response | Imbalance
By far the cleanest* way to (50%-50%) balance your training set is to use observation weights. Say 1% of cases is fraudulent, 99% is not, then give each non-fraudulent case weight 1, and each fraudulent transaction weight 99. This is equivalent to adding 98 copies of each fraudulent cases back into the traini... | How to handle predictors that are highly correlated to the response | Imbalance
By far the cleanest* way to (50%-50%) balance your training set is to use observation weights. Say 1% of cases is fraudulent, 99% is not, then give each non-fraudulent case weight 1, and eac | How to handle predictors that are highly correlated to the response
Imbalance
By far the cleanest* way to (50%-50%) balance your training set is to use observation weights. Say 1% of cases is fraudulent, 99% is not, then give each non-fraudulent case weight 1, and each fraudulent transaction weight 99. This is equivale... | How to handle predictors that are highly correlated to the response
Imbalance
By far the cleanest* way to (50%-50%) balance your training set is to use observation weights. Say 1% of cases is fraudulent, 99% is not, then give each non-fraudulent case weight 1, and eac |
38,131 | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? | (Started as a comment, but it's much too long)
Let's consider this a different way. A more general version of the question is --- can we use reasoning involving conditional probabilities when the thing we condition on is false?
It's not simply permissible -- it's necessary.
Consider this in the context of Bayes theorem... | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? | (Started as a comment, but it's much too long)
Let's consider this a different way. A more general version of the question is --- can we use reasoning involving conditional probabilities when the thin | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true?
(Started as a comment, but it's much too long)
Let's consider this a different way. A more general version of the question is --- can we use reasoning involving conditional probabilities when the thing we condition on is false?
It's not sim... | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true?
(Started as a comment, but it's much too long)
Let's consider this a different way. A more general version of the question is --- can we use reasoning involving conditional probabilities when the thin |
38,132 | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? | You have a misunderstanding. The 'alternative' hypothesis ($H_1$) is simply the negation of the null hypothesis. When conducting, say, a power analysis, we will specify a specific sampling distribution around a point estimate (for example a mean treatment effect) that we believe in, but rejecting the null does not ma... | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? | You have a misunderstanding. The 'alternative' hypothesis ($H_1$) is simply the negation of the null hypothesis. When conducting, say, a power analysis, we will specify a specific sampling distribut | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true?
You have a misunderstanding. The 'alternative' hypothesis ($H_1$) is simply the negation of the null hypothesis. When conducting, say, a power analysis, we will specify a specific sampling distribution around a point estimate (for example... | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true?
You have a misunderstanding. The 'alternative' hypothesis ($H_1$) is simply the negation of the null hypothesis. When conducting, say, a power analysis, we will specify a specific sampling distribut |
38,133 | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? | The principle is like a "fuzzy" version of the contraposition principle (or reductio ad absurdum principle, I'm not sure).
Consider that every dog has four legs. Then if you sample an animal with two legs you are sure it is not a dog.
Now only consider that every dog has a high probability to have four legs (in other ... | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? | The principle is like a "fuzzy" version of the contraposition principle (or reductio ad absurdum principle, I'm not sure).
Consider that every dog has four legs. Then if you sample an animal with two | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true?
The principle is like a "fuzzy" version of the contraposition principle (or reductio ad absurdum principle, I'm not sure).
Consider that every dog has four legs. Then if you sample an animal with two legs you are sure it is not a dog.
Now ... | If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true?
The principle is like a "fuzzy" version of the contraposition principle (or reductio ad absurdum principle, I'm not sure).
Consider that every dog has four legs. Then if you sample an animal with two |
38,134 | Is kNN best for classification? | There is no such thing as the best classifier, it always depends on the context, what kind of data/problem is at hand. As you mention, kNN is slow when you have a lot of observations, since it does not generalize over data in advance, it scans historical database each time a prediction is needed.
With kNN you need to ... | Is kNN best for classification? | There is no such thing as the best classifier, it always depends on the context, what kind of data/problem is at hand. As you mention, kNN is slow when you have a lot of observations, since it does no | Is kNN best for classification?
There is no such thing as the best classifier, it always depends on the context, what kind of data/problem is at hand. As you mention, kNN is slow when you have a lot of observations, since it does not generalize over data in advance, it scans historical database each time a prediction i... | Is kNN best for classification?
There is no such thing as the best classifier, it always depends on the context, what kind of data/problem is at hand. As you mention, kNN is slow when you have a lot of observations, since it does no |
38,135 | Is kNN best for classification? | What you're referring to is called Bias.
Since kNN is not model based, it has low Bias, but that also means it can have high Variance. This is called the Bias-Variance tradeoff.
Basically, there's no guarantee that just because it has low Bias it will have a good "testing performance". Quite the contrary, it could easi... | Is kNN best for classification? | What you're referring to is called Bias.
Since kNN is not model based, it has low Bias, but that also means it can have high Variance. This is called the Bias-Variance tradeoff.
Basically, there's no | Is kNN best for classification?
What you're referring to is called Bias.
Since kNN is not model based, it has low Bias, but that also means it can have high Variance. This is called the Bias-Variance tradeoff.
Basically, there's no guarantee that just because it has low Bias it will have a good "testing performance". Q... | Is kNN best for classification?
What you're referring to is called Bias.
Since kNN is not model based, it has low Bias, but that also means it can have high Variance. This is called the Bias-Variance tradeoff.
Basically, there's no |
38,136 | Is kNN best for classification? | Do you know $k$? If $k$ is unknown all bets are off.
How do you define 'best'? In a statistical sense, best implies minimizing the risk with a squared error loss function. If this isn't the case, and even if this is the case how are you going to compare methods?
As addressed by inzl there is no best classifier. ... | Is kNN best for classification? | Do you know $k$? If $k$ is unknown all bets are off.
How do you define 'best'? In a statistical sense, best implies minimizing the risk with a squared error loss function. If this isn't the case | Is kNN best for classification?
Do you know $k$? If $k$ is unknown all bets are off.
How do you define 'best'? In a statistical sense, best implies minimizing the risk with a squared error loss function. If this isn't the case, and even if this is the case how are you going to compare methods?
As addressed by inz... | Is kNN best for classification?
Do you know $k$? If $k$ is unknown all bets are off.
How do you define 'best'? In a statistical sense, best implies minimizing the risk with a squared error loss function. If this isn't the case |
38,137 | Is kNN best for classification? | Given infinite data, k-NN is guaranteed to approach the Bayes error rate under ideal conditions. You probably don't have infinite data, and your k is probably not large enough (it has to approach infinity).
In practice, there's no reason k-NN should be the best classifier given finite data! | Is kNN best for classification? | Given infinite data, k-NN is guaranteed to approach the Bayes error rate under ideal conditions. You probably don't have infinite data, and your k is probably not large enough (it has to approach infi | Is kNN best for classification?
Given infinite data, k-NN is guaranteed to approach the Bayes error rate under ideal conditions. You probably don't have infinite data, and your k is probably not large enough (it has to approach infinity).
In practice, there's no reason k-NN should be the best classifier given finite da... | Is kNN best for classification?
Given infinite data, k-NN is guaranteed to approach the Bayes error rate under ideal conditions. You probably don't have infinite data, and your k is probably not large enough (it has to approach infi |
38,138 | Is kNN best for classification? | I would atleast consider NaiveBayes along with knn.
You can do cross validation with both knn and Naive Bayes on your training data and select the best one. | Is kNN best for classification? | I would atleast consider NaiveBayes along with knn.
You can do cross validation with both knn and Naive Bayes on your training data and select the best one. | Is kNN best for classification?
I would atleast consider NaiveBayes along with knn.
You can do cross validation with both knn and Naive Bayes on your training data and select the best one. | Is kNN best for classification?
I would atleast consider NaiveBayes along with knn.
You can do cross validation with both knn and Naive Bayes on your training data and select the best one. |
38,139 | A naive question about the Kolmogorov Smirnov test | You can standardize the exponential distribution easily enough multiplying variates by the rate parameter (it's a reciprocal scale parameter). But if you're estimating the rate parameter from the data, the Kolmogorov–Smirnov statistic doesn't have the same distribution as when the exponential distribution is completely... | A naive question about the Kolmogorov Smirnov test | You can standardize the exponential distribution easily enough multiplying variates by the rate parameter (it's a reciprocal scale parameter). But if you're estimating the rate parameter from the data | A naive question about the Kolmogorov Smirnov test
You can standardize the exponential distribution easily enough multiplying variates by the rate parameter (it's a reciprocal scale parameter). But if you're estimating the rate parameter from the data, the Kolmogorov–Smirnov statistic doesn't have the same distribution... | A naive question about the Kolmogorov Smirnov test
You can standardize the exponential distribution easily enough multiplying variates by the rate parameter (it's a reciprocal scale parameter). But if you're estimating the rate parameter from the data |
38,140 | A naive question about the Kolmogorov Smirnov test | You don't need to normalize, but can get the p-value for a goodness-of-fit test by simulation. Here is some sample R code, taken from Greg Snow's answer to a similar question (KS test - R, Minitab (and Systat)):
data <- c(7.2,10.5,10.67,0.15,3.92,3.28,0.89,2.29,13.82,0.43)
simp <- replicate(100000, {x <- rexp(length(d... | A naive question about the Kolmogorov Smirnov test | You don't need to normalize, but can get the p-value for a goodness-of-fit test by simulation. Here is some sample R code, taken from Greg Snow's answer to a similar question (KS test - R, Minitab (an | A naive question about the Kolmogorov Smirnov test
You don't need to normalize, but can get the p-value for a goodness-of-fit test by simulation. Here is some sample R code, taken from Greg Snow's answer to a similar question (KS test - R, Minitab (and Systat)):
data <- c(7.2,10.5,10.67,0.15,3.92,3.28,0.89,2.29,13.82,0... | A naive question about the Kolmogorov Smirnov test
You don't need to normalize, but can get the p-value for a goodness-of-fit test by simulation. Here is some sample R code, taken from Greg Snow's answer to a similar question (KS test - R, Minitab (an |
38,141 | A naive question about the Kolmogorov Smirnov test | No, you don't need to normalise your data since the KS statistic is defined in terms of the raw data (actually in terms of the empirical distribution of these data):
http://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Smirnov_test#Kolmogorov.E2.80.93Smirnov_statistic
I don't know Python, but in R you can conduct this test ... | A naive question about the Kolmogorov Smirnov test | No, you don't need to normalise your data since the KS statistic is defined in terms of the raw data (actually in terms of the empirical distribution of these data):
http://en.wikipedia.org/wiki/Kolmo | A naive question about the Kolmogorov Smirnov test
No, you don't need to normalise your data since the KS statistic is defined in terms of the raw data (actually in terms of the empirical distribution of these data):
http://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Smirnov_test#Kolmogorov.E2.80.93Smirnov_statistic
I don... | A naive question about the Kolmogorov Smirnov test
No, you don't need to normalise your data since the KS statistic is defined in terms of the raw data (actually in terms of the empirical distribution of these data):
http://en.wikipedia.org/wiki/Kolmo |
38,142 | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$? | Clearly not.
An easy counterexample (here done in R), that I think satisfies all your constraints:
set.seed(239843)
x=rnorm(100,100,1)
y=rep(c(0.01,0.99),times=50)
z=x*y
var(x)
[1] 0.8413043
var(y)
[1] 0.2425253
var(z)
[1] 2425.296
What's going on:
x is a series with mean 100 and sd 1.
y alternates between 0.01... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v | Clearly not.
An easy counterexample (here done in R), that I think satisfies all your constraints:
set.seed(239843)
x=rnorm(100,100,1)
y=rep(c(0.01,0.99),times=50)
z=x*y
var(x)
[1] 0.8413043
var( | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$?
Clearly not.
An easy counterexample (here done in R), that I think satisfies all your constraints:
set.seed(239843)
x=rnorm(100,100,1)
y=rep(c(0.01,0.99),times=50)
z=x*y
var(x)
[1] 0.8413043
var(y)
[1] 0.24... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v
Clearly not.
An easy counterexample (here done in R), that I think satisfies all your constraints:
set.seed(239843)
x=rnorm(100,100,1)
y=rep(c(0.01,0.99),times=50)
z=x*y
var(x)
[1] 0.8413043
var( |
38,143 | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$? | I do not think Var$(Z)\le $Var$(X)$. Imagine that $X$ is a time series that meanders about values near 100, almost always between 98 and 102. Now imagine that $Z$ meanders between 0 and 100, but is always less than $X$. The variance of $Z$ is clearly going to be larger in such a case than the variance of $X$. This is a... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v | I do not think Var$(Z)\le $Var$(X)$. Imagine that $X$ is a time series that meanders about values near 100, almost always between 98 and 102. Now imagine that $Z$ meanders between 0 and 100, but is al | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$?
I do not think Var$(Z)\le $Var$(X)$. Imagine that $X$ is a time series that meanders about values near 100, almost always between 98 and 102. Now imagine that $Z$ meanders between 0 and 100, but is always less t... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v
I do not think Var$(Z)\le $Var$(X)$. Imagine that $X$ is a time series that meanders about values near 100, almost always between 98 and 102. Now imagine that $Z$ meanders between 0 and 100, but is al |
38,144 | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$? | For the general case, the answer is no. For the specific cases, it is also no.
A simple counter example is take $ y\sim U (0,1) $ and take $ x\sim Gamma (a, a) $ such that we have $ E (x)=1 $ and $ var (x)=a^{-1}$ . Take $ x $ and $ y $ as independent, and we have:
$$ var (z) =E [var (z|y)]+var [E (z|y)]=E [y^2a^{-1}... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v | For the general case, the answer is no. For the specific cases, it is also no.
A simple counter example is take $ y\sim U (0,1) $ and take $ x\sim Gamma (a, a) $ such that we have $ E (x)=1 $ and $ v | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$?
For the general case, the answer is no. For the specific cases, it is also no.
A simple counter example is take $ y\sim U (0,1) $ and take $ x\sim Gamma (a, a) $ such that we have $ E (x)=1 $ and $ var (x)=a^{-... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v
For the general case, the answer is no. For the specific cases, it is also no.
A simple counter example is take $ y\sim U (0,1) $ and take $ x\sim Gamma (a, a) $ such that we have $ E (x)=1 $ and $ v |
38,145 | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$? | Let us be clear that the "variance" under discussion appears to be a random variable derived from a finite portion of a time series. Specifically, the raw $k^\text{th}$ moment of $\mathrm{X} =(X_1, X_2, \ldots, X_N)$ is
$$\mu_k(\mathrm{X}) = (X_1^k+X_2^k+\cdots+X_N^k)/N,$$
which is a random variable, and the variance ... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v | Let us be clear that the "variance" under discussion appears to be a random variable derived from a finite portion of a time series. Specifically, the raw $k^\text{th}$ moment of $\mathrm{X} =(X_1, X | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$?
Let us be clear that the "variance" under discussion appears to be a random variable derived from a finite portion of a time series. Specifically, the raw $k^\text{th}$ moment of $\mathrm{X} =(X_1, X_2, \ldots,... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v
Let us be clear that the "variance" under discussion appears to be a random variable derived from a finite portion of a time series. Specifically, the raw $k^\text{th}$ moment of $\mathrm{X} =(X_1, X |
38,146 | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$? | Assume that the processes $\{X\}$ and $\{Y\}$ are ergodic/stationary with finite moments, and independent. Then $\{XY\}$ is also ergodic and
$$\operatorname{Var(XY)} = E(X^2Y^2) - [E(XY)]^2 = E(X^2)E(Y^2) - [E(X)]^2[E(Y)]^2$$
the break-up of expected values due to independence.
You are asking
$$E(X^2)E(Y^2) - [E(X)]^... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v | Assume that the processes $\{X\}$ and $\{Y\}$ are ergodic/stationary with finite moments, and independent. Then $\{XY\}$ is also ergodic and
$$\operatorname{Var(XY)} = E(X^2Y^2) - [E(XY)]^2 = E(X^2)E( | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{var}(X)$?
Assume that the processes $\{X\}$ and $\{Y\}$ are ergodic/stationary with finite moments, and independent. Then $\{XY\}$ is also ergodic and
$$\operatorname{Var(XY)} = E(X^2Y^2) - [E(XY)]^2 = E(X^2)E(Y^2) - [E(X... | If two time series $X$ and $Z$ follow $0 \leq Z \leq X$, can we say that $\text{var}(Z) \leq \text{v
Assume that the processes $\{X\}$ and $\{Y\}$ are ergodic/stationary with finite moments, and independent. Then $\{XY\}$ is also ergodic and
$$\operatorname{Var(XY)} = E(X^2Y^2) - [E(XY)]^2 = E(X^2)E( |
38,147 | Quadratic models with R. The use of poly(..) and I(..) functions (R-language) | You have the first model all sorts of wrong; that model
var1 ~ var2 * var3
says the variance in var1 is explained by the main effects of factor var2 and continuous covariate var3 and their interaction. In other words, the model is one where each level of var2 has a separate intercept and slope for the regressions line... | Quadratic models with R. The use of poly(..) and I(..) functions (R-language) | You have the first model all sorts of wrong; that model
var1 ~ var2 * var3
says the variance in var1 is explained by the main effects of factor var2 and continuous covariate var3 and their interactio | Quadratic models with R. The use of poly(..) and I(..) functions (R-language)
You have the first model all sorts of wrong; that model
var1 ~ var2 * var3
says the variance in var1 is explained by the main effects of factor var2 and continuous covariate var3 and their interaction. In other words, the model is one where ... | Quadratic models with R. The use of poly(..) and I(..) functions (R-language)
You have the first model all sorts of wrong; that model
var1 ~ var2 * var3
says the variance in var1 is explained by the main effects of factor var2 and continuous covariate var3 and their interactio |
38,148 | Why does the bootstrapped correlation revolve around zero while the original correlation $\approx 0.52$? | There is an obvious reason for that: You are sampling from both series separately, thus destroying any correlation. You probably want to sample pairs, not observations in each series, e.g.
index <- sample(132,132, replace=TRUE)
euro.nzd.corr[i] = cor(euro[index], nzd[index])
Fixing your code should allow you to recove... | Why does the bootstrapped correlation revolve around zero while the original correlation $\approx 0. | There is an obvious reason for that: You are sampling from both series separately, thus destroying any correlation. You probably want to sample pairs, not observations in each series, e.g.
index <- sa | Why does the bootstrapped correlation revolve around zero while the original correlation $\approx 0.52$?
There is an obvious reason for that: You are sampling from both series separately, thus destroying any correlation. You probably want to sample pairs, not observations in each series, e.g.
index <- sample(132,132, r... | Why does the bootstrapped correlation revolve around zero while the original correlation $\approx 0.
There is an obvious reason for that: You are sampling from both series separately, thus destroying any correlation. You probably want to sample pairs, not observations in each series, e.g.
index <- sa |
38,149 | Standard errors for covariance estimate in R | In response to whuber's follow-up, I would advocate that an all-purpose black-box approach would be using a non-parametric bootstrap.
The basic pseudocode is:
Jointly resample from observed rows of data, allowing for replications and holding the sample size fixed.
Re-estimate covariance in the resampled data.
Repeat 1... | Standard errors for covariance estimate in R | In response to whuber's follow-up, I would advocate that an all-purpose black-box approach would be using a non-parametric bootstrap.
The basic pseudocode is:
Jointly resample from observed rows of d | Standard errors for covariance estimate in R
In response to whuber's follow-up, I would advocate that an all-purpose black-box approach would be using a non-parametric bootstrap.
The basic pseudocode is:
Jointly resample from observed rows of data, allowing for replications and holding the sample size fixed.
Re-estima... | Standard errors for covariance estimate in R
In response to whuber's follow-up, I would advocate that an all-purpose black-box approach would be using a non-parametric bootstrap.
The basic pseudocode is:
Jointly resample from observed rows of d |
38,150 | Standard errors for covariance estimate in R | This is not an answer to the original question, but to your request to AdamO.
(As far as I'm concerned he's covered the original question.)
I'd make it a comment but I think it's too long
Would you be able to derive a closed form solution assuming the variables are normal for example?
see
http://en.wikipedia.org/wiki... | Standard errors for covariance estimate in R | This is not an answer to the original question, but to your request to AdamO.
(As far as I'm concerned he's covered the original question.)
I'd make it a comment but I think it's too long
Would you b | Standard errors for covariance estimate in R
This is not an answer to the original question, but to your request to AdamO.
(As far as I'm concerned he's covered the original question.)
I'd make it a comment but I think it's too long
Would you be able to derive a closed form solution assuming the variables are normal f... | Standard errors for covariance estimate in R
This is not an answer to the original question, but to your request to AdamO.
(As far as I'm concerned he's covered the original question.)
I'd make it a comment but I think it's too long
Would you b |
38,151 | Standard errors for covariance estimate in R | The OP's question does not define what formula R uses as sample covariance estimator. However, following the link in Glen's answer, I assume that R is using:
$$m_{11} = \frac{1}{n} \sum _{i=1}^n \left(X_i-\bar{X}\right) \left(Y_i-\bar{Y}\right)$$
... also known as the $m_{11}$ sample central moment, which can be expre... | Standard errors for covariance estimate in R | The OP's question does not define what formula R uses as sample covariance estimator. However, following the link in Glen's answer, I assume that R is using:
$$m_{11} = \frac{1}{n} \sum _{i=1}^n \lef | Standard errors for covariance estimate in R
The OP's question does not define what formula R uses as sample covariance estimator. However, following the link in Glen's answer, I assume that R is using:
$$m_{11} = \frac{1}{n} \sum _{i=1}^n \left(X_i-\bar{X}\right) \left(Y_i-\bar{Y}\right)$$
... also known as the $m_{1... | Standard errors for covariance estimate in R
The OP's question does not define what formula R uses as sample covariance estimator. However, following the link in Glen's answer, I assume that R is using:
$$m_{11} = \frac{1}{n} \sum _{i=1}^n \lef |
38,152 | Standard errors for covariance estimate in R | There is some confusion in the discussion above.
Simple algebra shows that the expression ascribed to Richardson:
Var$(S_{XY})=\frac{(n−1)^2}{n^3}(\mu_{22}−\mu^2_{11})+\frac{(n−1)}{n^3}(\mu^2_{11}+\mu_{20}\mu_{02})$
is identical to that obtained by wolfies using MathStatica.
Both expressions clearly agree on the coef... | Standard errors for covariance estimate in R | There is some confusion in the discussion above.
Simple algebra shows that the expression ascribed to Richardson:
Var$(S_{XY})=\frac{(n−1)^2}{n^3}(\mu_{22}−\mu^2_{11})+\frac{(n−1)}{n^3}(\mu^2_{11}+\m | Standard errors for covariance estimate in R
There is some confusion in the discussion above.
Simple algebra shows that the expression ascribed to Richardson:
Var$(S_{XY})=\frac{(n−1)^2}{n^3}(\mu_{22}−\mu^2_{11})+\frac{(n−1)}{n^3}(\mu^2_{11}+\mu_{20}\mu_{02})$
is identical to that obtained by wolfies using MathStatica... | Standard errors for covariance estimate in R
There is some confusion in the discussion above.
Simple algebra shows that the expression ascribed to Richardson:
Var$(S_{XY})=\frac{(n−1)^2}{n^3}(\mu_{22}−\mu^2_{11})+\frac{(n−1)}{n^3}(\mu^2_{11}+\m |
38,153 | Which performance measure to use when using SVM: MSE or MAE? | Actually, looking at both MAE and RMSE gives you additional information about the distribution of the errors:
$\mathrm{MAE} \leq \mathrm{RMSE} \leq \mathrm{MAE}^2$ (for regression)
if $\mathrm{RMSE}$ is close to $\mathrm{MAE}$, the model makes many relatively small errors
if $\mathrm{RMSE}$ is close to $\mathrm{MAE}^2... | Which performance measure to use when using SVM: MSE or MAE? | Actually, looking at both MAE and RMSE gives you additional information about the distribution of the errors:
$\mathrm{MAE} \leq \mathrm{RMSE} \leq \mathrm{MAE}^2$ (for regression)
if $\mathrm{RMSE}$ | Which performance measure to use when using SVM: MSE or MAE?
Actually, looking at both MAE and RMSE gives you additional information about the distribution of the errors:
$\mathrm{MAE} \leq \mathrm{RMSE} \leq \mathrm{MAE}^2$ (for regression)
if $\mathrm{RMSE}$ is close to $\mathrm{MAE}$, the model makes many relativel... | Which performance measure to use when using SVM: MSE or MAE?
Actually, looking at both MAE and RMSE gives you additional information about the distribution of the errors:
$\mathrm{MAE} \leq \mathrm{RMSE} \leq \mathrm{MAE}^2$ (for regression)
if $\mathrm{RMSE}$ |
38,154 | Which performance measure to use when using SVM: MSE or MAE? | The choice of performance metric depends on what is important for the application that you are interested in. The MSE is a good performance metric for many applications as there is good reason to suppose that noise process is Gaussian. Sometimes it is better to use the MAE if you don't want your performance metric to... | Which performance measure to use when using SVM: MSE or MAE? | The choice of performance metric depends on what is important for the application that you are interested in. The MSE is a good performance metric for many applications as there is good reason to sup | Which performance measure to use when using SVM: MSE or MAE?
The choice of performance metric depends on what is important for the application that you are interested in. The MSE is a good performance metric for many applications as there is good reason to suppose that noise process is Gaussian. Sometimes it is bette... | Which performance measure to use when using SVM: MSE or MAE?
The choice of performance metric depends on what is important for the application that you are interested in. The MSE is a good performance metric for many applications as there is good reason to sup |
38,155 | Which performance measure to use when using SVM: MSE or MAE? | MAE is more intuitive than MSE to simply evaluate the overall error.
MSE is easier to handle mathematically for variance analysis. For example, MSE is used to calculate the error variance $s_e^2$, which is a recurring value in regression statistics. | Which performance measure to use when using SVM: MSE or MAE? | MAE is more intuitive than MSE to simply evaluate the overall error.
MSE is easier to handle mathematically for variance analysis. For example, MSE is used to calculate the error variance $s_e^2$, whi | Which performance measure to use when using SVM: MSE or MAE?
MAE is more intuitive than MSE to simply evaluate the overall error.
MSE is easier to handle mathematically for variance analysis. For example, MSE is used to calculate the error variance $s_e^2$, which is a recurring value in regression statistics. | Which performance measure to use when using SVM: MSE or MAE?
MAE is more intuitive than MSE to simply evaluate the overall error.
MSE is easier to handle mathematically for variance analysis. For example, MSE is used to calculate the error variance $s_e^2$, whi |
38,156 | Whether to use EFA or PCA to assess dimensionality of a set of Likert items | EFA versus PCA
In a previous question on the differences between EFA and PCA, I state:
Principal components analysis involves extracting linear composites of observed variables.
Factor analysis is based on a formal model predicting observed variables from theoretical latent factors.
I find that typically within the c... | Whether to use EFA or PCA to assess dimensionality of a set of Likert items | EFA versus PCA
In a previous question on the differences between EFA and PCA, I state:
Principal components analysis involves extracting linear composites of observed variables.
Factor analysis is ba | Whether to use EFA or PCA to assess dimensionality of a set of Likert items
EFA versus PCA
In a previous question on the differences between EFA and PCA, I state:
Principal components analysis involves extracting linear composites of observed variables.
Factor analysis is based on a formal model predicting observed va... | Whether to use EFA or PCA to assess dimensionality of a set of Likert items
EFA versus PCA
In a previous question on the differences between EFA and PCA, I state:
Principal components analysis involves extracting linear composites of observed variables.
Factor analysis is ba |
38,157 | Whether to use EFA or PCA to assess dimensionality of a set of Likert items | Firstly, neither PCA or EFA will give you an estimate of the dimension of the scale. They are both essentially data reduction techniques. That being said, EFA is probably better for this purpose as it tells you how much of the variance in each question is accounted for in the model (the communality).
To estimate dimen... | Whether to use EFA or PCA to assess dimensionality of a set of Likert items | Firstly, neither PCA or EFA will give you an estimate of the dimension of the scale. They are both essentially data reduction techniques. That being said, EFA is probably better for this purpose as it | Whether to use EFA or PCA to assess dimensionality of a set of Likert items
Firstly, neither PCA or EFA will give you an estimate of the dimension of the scale. They are both essentially data reduction techniques. That being said, EFA is probably better for this purpose as it tells you how much of the variance in each ... | Whether to use EFA or PCA to assess dimensionality of a set of Likert items
Firstly, neither PCA or EFA will give you an estimate of the dimension of the scale. They are both essentially data reduction techniques. That being said, EFA is probably better for this purpose as it |
38,158 | Whether to use EFA or PCA to assess dimensionality of a set of Likert items | Two things not mentioned so far:
One: With only 6 items, you are going to have a hard time finding a lot of dimensions.
Two: If you do EFA, rather than look at scree plots or eigenvalues or some other numeric test, examine several solutions and see which makes sense. Ideally, you'll be able to follow @richiemorrisroe... | Whether to use EFA or PCA to assess dimensionality of a set of Likert items | Two things not mentioned so far:
One: With only 6 items, you are going to have a hard time finding a lot of dimensions.
Two: If you do EFA, rather than look at scree plots or eigenvalues or some oth | Whether to use EFA or PCA to assess dimensionality of a set of Likert items
Two things not mentioned so far:
One: With only 6 items, you are going to have a hard time finding a lot of dimensions.
Two: If you do EFA, rather than look at scree plots or eigenvalues or some other numeric test, examine several solutions a... | Whether to use EFA or PCA to assess dimensionality of a set of Likert items
Two things not mentioned so far:
One: With only 6 items, you are going to have a hard time finding a lot of dimensions.
Two: If you do EFA, rather than look at scree plots or eigenvalues or some oth |
38,159 | Interpreting two-sided, two-sample, Welch T-Test | (1a) You don't need the Welch test to cope with different sample sizes. That's automatically handled by the Student t-test.
(1b) If you think there's a real chance the variances in the two populations are strongly different, then you are assuming a priori that the two populations differ. It might not be a difference ... | Interpreting two-sided, two-sample, Welch T-Test | (1a) You don't need the Welch test to cope with different sample sizes. That's automatically handled by the Student t-test.
(1b) If you think there's a real chance the variances in the two population | Interpreting two-sided, two-sample, Welch T-Test
(1a) You don't need the Welch test to cope with different sample sizes. That's automatically handled by the Student t-test.
(1b) If you think there's a real chance the variances in the two populations are strongly different, then you are assuming a priori that the two p... | Interpreting two-sided, two-sample, Welch T-Test
(1a) You don't need the Welch test to cope with different sample sizes. That's automatically handled by the Student t-test.
(1b) If you think there's a real chance the variances in the two population |
38,160 | Interpreting two-sided, two-sample, Welch T-Test | Dividing by 2 is for p-values. If you compare critical values the division by 2 is not necessary. The function getCriticalValue should be the quantile function of Student's t distribution. Thus it should take 2 values, the probability and the degrees of freedom. If you want 2-sided hypothesis as indicated by your code,... | Interpreting two-sided, two-sample, Welch T-Test | Dividing by 2 is for p-values. If you compare critical values the division by 2 is not necessary. The function getCriticalValue should be the quantile function of Student's t distribution. Thus it sho | Interpreting two-sided, two-sample, Welch T-Test
Dividing by 2 is for p-values. If you compare critical values the division by 2 is not necessary. The function getCriticalValue should be the quantile function of Student's t distribution. Thus it should take 2 values, the probability and the degrees of freedom. If you w... | Interpreting two-sided, two-sample, Welch T-Test
Dividing by 2 is for p-values. If you compare critical values the division by 2 is not necessary. The function getCriticalValue should be the quantile function of Student's t distribution. Thus it sho |
38,161 | Interpreting two-sided, two-sample, Welch T-Test | It's not absolutely necessary to round the degrees of freedom to an integer. Student's t-distribution can be defined for all positive real values of this parameter. Restricting it to a positive integer may make the critical value easier to calculate though, depending on how you're doing that. And it will make very litt... | Interpreting two-sided, two-sample, Welch T-Test | It's not absolutely necessary to round the degrees of freedom to an integer. Student's t-distribution can be defined for all positive real values of this parameter. Restricting it to a positive intege | Interpreting two-sided, two-sample, Welch T-Test
It's not absolutely necessary to round the degrees of freedom to an integer. Student's t-distribution can be defined for all positive real values of this parameter. Restricting it to a positive integer may make the critical value easier to calculate though, depending on ... | Interpreting two-sided, two-sample, Welch T-Test
It's not absolutely necessary to round the degrees of freedom to an integer. Student's t-distribution can be defined for all positive real values of this parameter. Restricting it to a positive intege |
38,162 | Interpreting two-sided, two-sample, Welch T-Test | I'm working with the OP on the benchmarking project and wanted to thank you all for clearing some things up. Also I wanted to provide a bit more information in case that affects the advice.
The sample size ranges from 5 - 700+ (as many as can be completed in 8 seconds or until the margin of error is at or below 1%. The... | Interpreting two-sided, two-sample, Welch T-Test | I'm working with the OP on the benchmarking project and wanted to thank you all for clearing some things up. Also I wanted to provide a bit more information in case that affects the advice.
The sample | Interpreting two-sided, two-sample, Welch T-Test
I'm working with the OP on the benchmarking project and wanted to thank you all for clearing some things up. Also I wanted to provide a bit more information in case that affects the advice.
The sample size ranges from 5 - 700+ (as many as can be completed in 8 seconds or... | Interpreting two-sided, two-sample, Welch T-Test
I'm working with the OP on the benchmarking project and wanted to thank you all for clearing some things up. Also I wanted to provide a bit more information in case that affects the advice.
The sample |
38,163 | SAS/IML compared to R | Full disclosure: I work at SAS.
The IML blog is http://blogs.sas.com/iml.
Both languages are matrix-vector languages with a rich run-time library and the ability to write your own functions. For data analysis tasks and matrix computations, they both provide the neccessary tools to help you analyze your data.
The SAS/IM... | SAS/IML compared to R | Full disclosure: I work at SAS.
The IML blog is http://blogs.sas.com/iml.
Both languages are matrix-vector languages with a rich run-time library and the ability to write your own functions. For data | SAS/IML compared to R
Full disclosure: I work at SAS.
The IML blog is http://blogs.sas.com/iml.
Both languages are matrix-vector languages with a rich run-time library and the ability to write your own functions. For data analysis tasks and matrix computations, they both provide the neccessary tools to help you analyze... | SAS/IML compared to R
Full disclosure: I work at SAS.
The IML blog is http://blogs.sas.com/iml.
Both languages are matrix-vector languages with a rich run-time library and the ability to write your own functions. For data |
38,164 | SAS/IML compared to R | You might want to pick up (or look at) a copy of Rick Wicklin's book: Statistical Programming with SAS IML software
https://support.sas.com/content/dam/SAS/support/en/books/statistical-programming-with-sas-iml-software/63119_excerpt.pdf
He also has a blog about IML.
And, on SAS' site, there is a section about IML:
http... | SAS/IML compared to R | You might want to pick up (or look at) a copy of Rick Wicklin's book: Statistical Programming with SAS IML software
https://support.sas.com/content/dam/SAS/support/en/books/statistical-programming-wit | SAS/IML compared to R
You might want to pick up (or look at) a copy of Rick Wicklin's book: Statistical Programming with SAS IML software
https://support.sas.com/content/dam/SAS/support/en/books/statistical-programming-with-sas-iml-software/63119_excerpt.pdf
He also has a blog about IML.
And, on SAS' site, there is a s... | SAS/IML compared to R
You might want to pick up (or look at) a copy of Rick Wicklin's book: Statistical Programming with SAS IML software
https://support.sas.com/content/dam/SAS/support/en/books/statistical-programming-wit |
38,165 | SAS/IML compared to R | I have never used it, but I know for new versions of IML, you can call R routines. Maybe start by looking at http://support.sas.com/rnd/app/studio/statr.pdf. | SAS/IML compared to R | I have never used it, but I know for new versions of IML, you can call R routines. Maybe start by looking at http://support.sas.com/rnd/app/studio/statr.pdf. | SAS/IML compared to R
I have never used it, but I know for new versions of IML, you can call R routines. Maybe start by looking at http://support.sas.com/rnd/app/studio/statr.pdf. | SAS/IML compared to R
I have never used it, but I know for new versions of IML, you can call R routines. Maybe start by looking at http://support.sas.com/rnd/app/studio/statr.pdf. |
38,166 | Binomial test for a binary variable | You cannot determine this through a statistical test, for a trivial reason and a profound reason.
The trivial reason is that your data consist of $k$ ones and $n-k$ zeros with $n$ about 100k. These data conform extremely closely to a Bernoulli($k/n$) distribution. No testing is necessary.
The profound reason is that ... | Binomial test for a binary variable | You cannot determine this through a statistical test, for a trivial reason and a profound reason.
The trivial reason is that your data consist of $k$ ones and $n-k$ zeros with $n$ about 100k. These d | Binomial test for a binary variable
You cannot determine this through a statistical test, for a trivial reason and a profound reason.
The trivial reason is that your data consist of $k$ ones and $n-k$ zeros with $n$ about 100k. These data conform extremely closely to a Bernoulli($k/n$) distribution. No testing is nec... | Binomial test for a binary variable
You cannot determine this through a statistical test, for a trivial reason and a profound reason.
The trivial reason is that your data consist of $k$ ones and $n-k$ zeros with $n$ about 100k. These d |
38,167 | Binomial test for a binary variable | I completely agree with @whuber -- just wanted to add:
If you were to try to transform the data. How would you go about doing so? You would map 0 to some number say, -5 and 1 to some other number?, say 5?
So now instead of having:
0 0 0 1 0 1 1 0 1 0 1
You have:
-5 -5 -5 5 -5 5 5 -5 5 -5 5
This cannot possible be n... | Binomial test for a binary variable | I completely agree with @whuber -- just wanted to add:
If you were to try to transform the data. How would you go about doing so? You would map 0 to some number say, -5 and 1 to some other number?, | Binomial test for a binary variable
I completely agree with @whuber -- just wanted to add:
If you were to try to transform the data. How would you go about doing so? You would map 0 to some number say, -5 and 1 to some other number?, say 5?
So now instead of having:
0 0 0 1 0 1 1 0 1 0 1
You have:
-5 -5 -5 5 -5 5 5 ... | Binomial test for a binary variable
I completely agree with @whuber -- just wanted to add:
If you were to try to transform the data. How would you go about doing so? You would map 0 to some number say, -5 and 1 to some other number?, |
38,168 | Binomial test for a binary variable | ALL binary variables have the binomial distribution, provided that the probability of success (probability to observe 1) does not change and that all their instances are independent. A rule of thumb says that binomial distribution can be fairly approximated by normal distribution when n*p>30, with n=number of instances... | Binomial test for a binary variable | ALL binary variables have the binomial distribution, provided that the probability of success (probability to observe 1) does not change and that all their instances are independent. A rule of thumb s | Binomial test for a binary variable
ALL binary variables have the binomial distribution, provided that the probability of success (probability to observe 1) does not change and that all their instances are independent. A rule of thumb says that binomial distribution can be fairly approximated by normal distribution whe... | Binomial test for a binary variable
ALL binary variables have the binomial distribution, provided that the probability of success (probability to observe 1) does not change and that all their instances are independent. A rule of thumb s |
38,169 | Conditional Expectation as a function of X | Without getting very much into measure theory,consider the random vector $(X,Y)$ with density $f_{X,Y}(\cdot,\cdot)$ (wrt a dominating measure $\text d\mu(x,y)$) decomposed as
$$f_{X,Y}(x,y)=f_Y(y)\times f_{X|Y}(x|y)$$
where
$f_Y(\cdot)$ is a probability density (wrt the appropriate dominating measure $\text d\mu_2(y)... | Conditional Expectation as a function of X | Without getting very much into measure theory,consider the random vector $(X,Y)$ with density $f_{X,Y}(\cdot,\cdot)$ (wrt a dominating measure $\text d\mu(x,y)$) decomposed as
$$f_{X,Y}(x,y)=f_Y(y)\ti | Conditional Expectation as a function of X
Without getting very much into measure theory,consider the random vector $(X,Y)$ with density $f_{X,Y}(\cdot,\cdot)$ (wrt a dominating measure $\text d\mu(x,y)$) decomposed as
$$f_{X,Y}(x,y)=f_Y(y)\times f_{X|Y}(x|y)$$
where
$f_Y(\cdot)$ is a probability density (wrt the appr... | Conditional Expectation as a function of X
Without getting very much into measure theory,consider the random vector $(X,Y)$ with density $f_{X,Y}(\cdot,\cdot)$ (wrt a dominating measure $\text d\mu(x,y)$) decomposed as
$$f_{X,Y}(x,y)=f_Y(y)\ti |
38,170 | Conditional Expectation as a function of X | If you take the expectation of $X$, it's not a function of $X$. You integrate $xf(x)$ over $x$, and so $x$ is gone, producing only a number.
If you take the expectation of $X|Y=y$, it's not a function of $X$. You integrate $xf(x|y)$ over $x$ (think of it as $xf(x)$ with a different $f(x)$ for different $y$), and so $x$... | Conditional Expectation as a function of X | If you take the expectation of $X$, it's not a function of $X$. You integrate $xf(x)$ over $x$, and so $x$ is gone, producing only a number.
If you take the expectation of $X|Y=y$, it's not a function | Conditional Expectation as a function of X
If you take the expectation of $X$, it's not a function of $X$. You integrate $xf(x)$ over $x$, and so $x$ is gone, producing only a number.
If you take the expectation of $X|Y=y$, it's not a function of $X$. You integrate $xf(x|y)$ over $x$ (think of it as $xf(x)$ with a diff... | Conditional Expectation as a function of X
If you take the expectation of $X$, it's not a function of $X$. You integrate $xf(x)$ over $x$, and so $x$ is gone, producing only a number.
If you take the expectation of $X|Y=y$, it's not a function |
38,171 | Can you say that you reject the null at the 95% level? | Sure. This is just a case of sloppy language.
You can either say "I reject the null hypothesis with 95% confidence" or "I reject the null at a significance level of .05." Both of these statements are shorthand for this much longer statement:
"If the null hypotheses were true, and I repeated this experiment/survey/analy... | Can you say that you reject the null at the 95% level? | Sure. This is just a case of sloppy language.
You can either say "I reject the null hypothesis with 95% confidence" or "I reject the null at a significance level of .05." Both of these statements are | Can you say that you reject the null at the 95% level?
Sure. This is just a case of sloppy language.
You can either say "I reject the null hypothesis with 95% confidence" or "I reject the null at a significance level of .05." Both of these statements are shorthand for this much longer statement:
"If the null hypotheses... | Can you say that you reject the null at the 95% level?
Sure. This is just a case of sloppy language.
You can either say "I reject the null hypothesis with 95% confidence" or "I reject the null at a significance level of .05." Both of these statements are |
38,172 | Can you say that you reject the null at the 95% level? | You can* but should not.
You're conflating language of confidence intervals with hypothesis tests. The level of a test is the significance level. "Confidence" levels (coverage) is a property of confidence intervals. There are connections between them but they are NOT the same thing and you should not mix terminology be... | Can you say that you reject the null at the 95% level? | You can* but should not.
You're conflating language of confidence intervals with hypothesis tests. The level of a test is the significance level. "Confidence" levels (coverage) is a property of confid | Can you say that you reject the null at the 95% level?
You can* but should not.
You're conflating language of confidence intervals with hypothesis tests. The level of a test is the significance level. "Confidence" levels (coverage) is a property of confidence intervals. There are connections between them but they are N... | Can you say that you reject the null at the 95% level?
You can* but should not.
You're conflating language of confidence intervals with hypothesis tests. The level of a test is the significance level. "Confidence" levels (coverage) is a property of confid |
38,173 | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form? | Your estimator is what's known as an M-estimator of $\rho$-type, where in this case $\rho = -f$.
If your function $f$ is differentiable, then it is known that under some (fairly strong) conditions, the M-estimator is consistent for the true maximizer of $f$, and is in fact asymptotically normal. See Chapter 7 of Boos &... | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form? | Your estimator is what's known as an M-estimator of $\rho$-type, where in this case $\rho = -f$.
If your function $f$ is differentiable, then it is known that under some (fairly strong) conditions, th | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form?
Your estimator is what's known as an M-estimator of $\rho$-type, where in this case $\rho = -f$.
If your function $f$ is differentiable, then it is known that under some (fairly strong) conditions, the M-estimator is consis... | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form?
Your estimator is what's known as an M-estimator of $\rho$-type, where in this case $\rho = -f$.
If your function $f$ is differentiable, then it is known that under some (fairly strong) conditions, th |
38,174 | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form? | On a general basis, $\hat\theta(X)$ is biased, due to the fact that it is equivariant under reparameterisation, i.e., if$$\eta=h(\theta)$$ is another parameterisation of the model, with $h$ a bijection, then $$\hat\eta(X)=h(\hat\theta(X))$$ while unbiasedness does not carry under arbitrary transforms $h$.
Note also tha... | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form? | On a general basis, $\hat\theta(X)$ is biased, due to the fact that it is equivariant under reparameterisation, i.e., if$$\eta=h(\theta)$$ is another parameterisation of the model, with $h$ a bijectio | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form?
On a general basis, $\hat\theta(X)$ is biased, due to the fact that it is equivariant under reparameterisation, i.e., if$$\eta=h(\theta)$$ is another parameterisation of the model, with $h$ a bijection, then $$\hat\eta(X)=h... | How to prove unbiasedness/consistency/normality of an estimator that doesn't have a closed form?
On a general basis, $\hat\theta(X)$ is biased, due to the fact that it is equivariant under reparameterisation, i.e., if$$\eta=h(\theta)$$ is another parameterisation of the model, with $h$ a bijectio |
38,175 | Why the second term is transposed, but not the first one? | If you use the convention that $(\boldsymbol{x} - \boldsymbol{\mu})$ is a column vector, i.e. $(\boldsymbol{x} - \boldsymbol{\mu}) = \begin{bmatrix}
x_{1} - \mu_1\\
x_{2} - \mu_2\\
\vdots \\
x_{m}- \mu_m
\end{bmatrix}$, then $(\boldsymbol{x} - \boldsymbol{\mu})^T$ i... | Why the second term is transposed, but not the first one? | If you use the convention that $(\boldsymbol{x} - \boldsymbol{\mu})$ is a column vector, i.e. $(\boldsymbol{x} - \boldsymbol{\mu}) = \begin{bmatrix}
x_{1} - \mu_1\\
x_{2} - \mu_ | Why the second term is transposed, but not the first one?
If you use the convention that $(\boldsymbol{x} - \boldsymbol{\mu})$ is a column vector, i.e. $(\boldsymbol{x} - \boldsymbol{\mu}) = \begin{bmatrix}
x_{1} - \mu_1\\
x_{2} - \mu_2\\
\vdots \\
x_{m}- \mu_m
\end... | Why the second term is transposed, but not the first one?
If you use the convention that $(\boldsymbol{x} - \boldsymbol{\mu})$ is a column vector, i.e. $(\boldsymbol{x} - \boldsymbol{\mu}) = \begin{bmatrix}
x_{1} - \mu_1\\
x_{2} - \mu_ |
38,176 | Why the second term is transposed, but not the first one? | When you multiply matrices, the adjacent dimensions need to match, so you can multiply the (n, k) matrix by (k, m) matrix, or (m, k) by (k, n), but not any other way around. Where you would see the transpose symbol it depends on if the data is stored row-wise or column-wise. If you take something like a dot product of ... | Why the second term is transposed, but not the first one? | When you multiply matrices, the adjacent dimensions need to match, so you can multiply the (n, k) matrix by (k, m) matrix, or (m, k) by (k, n), but not any other way around. Where you would see the tr | Why the second term is transposed, but not the first one?
When you multiply matrices, the adjacent dimensions need to match, so you can multiply the (n, k) matrix by (k, m) matrix, or (m, k) by (k, n), but not any other way around. Where you would see the transpose symbol it depends on if the data is stored row-wise or... | Why the second term is transposed, but not the first one?
When you multiply matrices, the adjacent dimensions need to match, so you can multiply the (n, k) matrix by (k, m) matrix, or (m, k) by (k, n), but not any other way around. Where you would see the tr |
38,177 | Why the second term is transposed, but not the first one? | For any column vector $x$ (eg $x \in \mathbb R^{n \times 1}$)
$x^Tx$ is (a 1x1 matrix and thus 'is isomorphic to' (*)) a scalar.
$xx^T$ is a matrix. (and if it's 1x1, then it could be treated as a scalar similarly.)
(*) in your case 'is isomorphic to' just means 'can be treated as' | Why the second term is transposed, but not the first one? | For any column vector $x$ (eg $x \in \mathbb R^{n \times 1}$)
$x^Tx$ is (a 1x1 matrix and thus 'is isomorphic to' (*)) a scalar.
$xx^T$ is a matrix. (and if it's 1x1, then it could be treated as a s | Why the second term is transposed, but not the first one?
For any column vector $x$ (eg $x \in \mathbb R^{n \times 1}$)
$x^Tx$ is (a 1x1 matrix and thus 'is isomorphic to' (*)) a scalar.
$xx^T$ is a matrix. (and if it's 1x1, then it could be treated as a scalar similarly.)
(*) in your case 'is isomorphic to' just m... | Why the second term is transposed, but not the first one?
For any column vector $x$ (eg $x \in \mathbb R^{n \times 1}$)
$x^Tx$ is (a 1x1 matrix and thus 'is isomorphic to' (*)) a scalar.
$xx^T$ is a matrix. (and if it's 1x1, then it could be treated as a s |
38,178 | How to simulate standard deviation | Standard error decreases as the sample size increases. Standard deviation is a related concept but perhaps not related enough to warrant such similar terminology that confuses everyone who is starting to learn statistics.
A sampling distribution is the distribution of values you would get if you repeatedly sampled from... | How to simulate standard deviation | Standard error decreases as the sample size increases. Standard deviation is a related concept but perhaps not related enough to warrant such similar terminology that confuses everyone who is starting | How to simulate standard deviation
Standard error decreases as the sample size increases. Standard deviation is a related concept but perhaps not related enough to warrant such similar terminology that confuses everyone who is starting to learn statistics.
A sampling distribution is the distribution of values you would... | How to simulate standard deviation
Standard error decreases as the sample size increases. Standard deviation is a related concept but perhaps not related enough to warrant such similar terminology that confuses everyone who is starting |
38,179 | How to simulate standard deviation | Standard deviation does not decrease with sample size. The bigger your sample is, the closer the standard deviation should be to the standard deviation of the population. What follows, with larger sample size the spread of the standard deviations estimated on larger vs smaller samples would decrease, because based on l... | How to simulate standard deviation | Standard deviation does not decrease with sample size. The bigger your sample is, the closer the standard deviation should be to the standard deviation of the population. What follows, with larger sam | How to simulate standard deviation
Standard deviation does not decrease with sample size. The bigger your sample is, the closer the standard deviation should be to the standard deviation of the population. What follows, with larger sample size the spread of the standard deviations estimated on larger vs smaller samples... | How to simulate standard deviation
Standard deviation does not decrease with sample size. The bigger your sample is, the closer the standard deviation should be to the standard deviation of the population. What follows, with larger sam |
38,180 | How to simulate standard deviation | You specifically ask about simulation. Following @Dave's Answer (+1), here are a couple of simulations in R.
Suppose I take a million samples of size $n = 16$ from a population distributed
as $\mathsf{Gamma}(\mathrm{shape} = 4,\, \mathrm{rate}=.1),$ so that the population mean is $\mu = 40$
the population variance is $... | How to simulate standard deviation | You specifically ask about simulation. Following @Dave's Answer (+1), here are a couple of simulations in R.
Suppose I take a million samples of size $n = 16$ from a population distributed
as $\mathsf | How to simulate standard deviation
You specifically ask about simulation. Following @Dave's Answer (+1), here are a couple of simulations in R.
Suppose I take a million samples of size $n = 16$ from a population distributed
as $\mathsf{Gamma}(\mathrm{shape} = 4,\, \mathrm{rate}=.1),$ so that the population mean is $\mu... | How to simulate standard deviation
You specifically ask about simulation. Following @Dave's Answer (+1), here are a couple of simulations in R.
Suppose I take a million samples of size $n = 16$ from a population distributed
as $\mathsf |
38,181 | Why is power of a hypothesis test a concern when we can bootstrap any representative sample to make n approach infinity? | The amount of information relating to the hypotheses that you have is simply the information in the original data.
Resampling that information, whether bootstrapping, permutation testing or any other resampling, cannot add information that wasn't already there.
The point of bootstrapping is to estimate the sampling dis... | Why is power of a hypothesis test a concern when we can bootstrap any representative sample to make | The amount of information relating to the hypotheses that you have is simply the information in the original data.
Resampling that information, whether bootstrapping, permutation testing or any other | Why is power of a hypothesis test a concern when we can bootstrap any representative sample to make n approach infinity?
The amount of information relating to the hypotheses that you have is simply the information in the original data.
Resampling that information, whether bootstrapping, permutation testing or any other... | Why is power of a hypothesis test a concern when we can bootstrap any representative sample to make
The amount of information relating to the hypotheses that you have is simply the information in the original data.
Resampling that information, whether bootstrapping, permutation testing or any other |
38,182 | Distribution of Y from distribution of X | For a general function $h$, there is no direct formula to get the pdf of the random variable $Y=h(X)$ knowing the pdf of $X$. There is a formula in case when $h$ is a differentiable one-to-one mapping from the range (the support, I should say) of $X$ to the range of $Y$. I guess from your question that you don't know t... | Distribution of Y from distribution of X | For a general function $h$, there is no direct formula to get the pdf of the random variable $Y=h(X)$ knowing the pdf of $X$. There is a formula in case when $h$ is a differentiable one-to-one mapping | Distribution of Y from distribution of X
For a general function $h$, there is no direct formula to get the pdf of the random variable $Y=h(X)$ knowing the pdf of $X$. There is a formula in case when $h$ is a differentiable one-to-one mapping from the range (the support, I should say) of $X$ to the range of $Y$. I guess... | Distribution of Y from distribution of X
For a general function $h$, there is no direct formula to get the pdf of the random variable $Y=h(X)$ knowing the pdf of $X$. There is a formula in case when $h$ is a differentiable one-to-one mapping |
38,183 | Distribution of Y from distribution of X | Suppose $X$ has a standard normal distribution, then pdf of $X$ is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$
Suppose the mathematical relationship between the random variable $Y$ and $X$ is $Y=2X$. It will be easy to find the pdf of $Y$.
Basically there are two methods to find the pdf of Y
1. use the CDF then tak... | Distribution of Y from distribution of X | Suppose $X$ has a standard normal distribution, then pdf of $X$ is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$
Suppose the mathematical relationship between the random variable $Y$ and $X$ is $Y=2 | Distribution of Y from distribution of X
Suppose $X$ has a standard normal distribution, then pdf of $X$ is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$
Suppose the mathematical relationship between the random variable $Y$ and $X$ is $Y=2X$. It will be easy to find the pdf of $Y$.
Basically there are two methods to ... | Distribution of Y from distribution of X
Suppose $X$ has a standard normal distribution, then pdf of $X$ is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$
Suppose the mathematical relationship between the random variable $Y$ and $X$ is $Y=2 |
38,184 | Distribution of Y from distribution of X | Yes, it is possible. Let's see how.
When $Y = f(X)$, the conditional P(Y|X) can be written as
$$P(Y=y|x) = \delta(y - f(x))$$
where $\delta$ is the Dirac delta.
The PDF of Y can be obtained by marginalising over X:
$$ P(Y=y) = \int P(y|x) P(x)dx = \int \delta(y - f(x))P(x)dx $$
To compute such integral, we can use that... | Distribution of Y from distribution of X | Yes, it is possible. Let's see how.
When $Y = f(X)$, the conditional P(Y|X) can be written as
$$P(Y=y|x) = \delta(y - f(x))$$
where $\delta$ is the Dirac delta.
The PDF of Y can be obtained by margina | Distribution of Y from distribution of X
Yes, it is possible. Let's see how.
When $Y = f(X)$, the conditional P(Y|X) can be written as
$$P(Y=y|x) = \delta(y - f(x))$$
where $\delta$ is the Dirac delta.
The PDF of Y can be obtained by marginalising over X:
$$ P(Y=y) = \int P(y|x) P(x)dx = \int \delta(y - f(x))P(x)dx $$
... | Distribution of Y from distribution of X
Yes, it is possible. Let's see how.
When $Y = f(X)$, the conditional P(Y|X) can be written as
$$P(Y=y|x) = \delta(y - f(x))$$
where $\delta$ is the Dirac delta.
The PDF of Y can be obtained by margina |
38,185 | Distribution of Y from distribution of X | As I understand your question, you ask about situation where we have random variable $X$, its density function $f(x)$, and you are interested of finding probability density function of random variable $Y$ defined as $Y = f(X)$.
Such relationship is pretty straightforward for cumulative distribution function, but does n... | Distribution of Y from distribution of X | As I understand your question, you ask about situation where we have random variable $X$, its density function $f(x)$, and you are interested of finding probability density function of random variable | Distribution of Y from distribution of X
As I understand your question, you ask about situation where we have random variable $X$, its density function $f(x)$, and you are interested of finding probability density function of random variable $Y$ defined as $Y = f(X)$.
Such relationship is pretty straightforward for cum... | Distribution of Y from distribution of X
As I understand your question, you ask about situation where we have random variable $X$, its density function $f(x)$, and you are interested of finding probability density function of random variable |
38,186 | Nonlinear effect in an interaction term | This is exactly why I switched from Stata to R and Frank's rms package (called Design back then) a few years ago.
Anyway, this somewhat hack-ish code will at least get you started. The syntax is a little outdated and there may be better ways to code this (haven't used Stata in a while), but here it goes
EDIT: Re-writte... | Nonlinear effect in an interaction term | This is exactly why I switched from Stata to R and Frank's rms package (called Design back then) a few years ago.
Anyway, this somewhat hack-ish code will at least get you started. The syntax is a lit | Nonlinear effect in an interaction term
This is exactly why I switched from Stata to R and Frank's rms package (called Design back then) a few years ago.
Anyway, this somewhat hack-ish code will at least get you started. The syntax is a little outdated and there may be better ways to code this (haven't used Stata in a ... | Nonlinear effect in an interaction term
This is exactly why I switched from Stata to R and Frank's rms package (called Design back then) a few years ago.
Anyway, this somewhat hack-ish code will at least get you started. The syntax is a lit |
38,187 | Nonlinear effect in an interaction term | The following uses the R rms package using ordinary least squares modeling, and models the nonlinear effect smoothly using a restricted cubic spline with 4 knots at default knot locations. This generates one linear component and 2 nonlinear components for a total of 3 parameters per treatment group.
require(rms)
dd <-... | Nonlinear effect in an interaction term | The following uses the R rms package using ordinary least squares modeling, and models the nonlinear effect smoothly using a restricted cubic spline with 4 knots at default knot locations. This gener | Nonlinear effect in an interaction term
The following uses the R rms package using ordinary least squares modeling, and models the nonlinear effect smoothly using a restricted cubic spline with 4 knots at default knot locations. This generates one linear component and 2 nonlinear components for a total of 3 parameters... | Nonlinear effect in an interaction term
The following uses the R rms package using ordinary least squares modeling, and models the nonlinear effect smoothly using a restricted cubic spline with 4 knots at default knot locations. This gener |
38,188 | Nonlinear effect in an interaction term | Have you considered using a generalized additive model? Wikipedia link here
Basically the model would be
$$
g(y) = X'\beta+\displaystyle\sum_j f_j(Z_j)+\epsilon
$$
or in your specific case
$$
B = logit\left(f(S,T)\right)
$$
In R, you could use the mgcv package, and run something like
library(mgcv)
m = gam(B~te(S,T),f... | Nonlinear effect in an interaction term | Have you considered using a generalized additive model? Wikipedia link here
Basically the model would be
$$
g(y) = X'\beta+\displaystyle\sum_j f_j(Z_j)+\epsilon
$$
or in your specific case
$$
B = lo | Nonlinear effect in an interaction term
Have you considered using a generalized additive model? Wikipedia link here
Basically the model would be
$$
g(y) = X'\beta+\displaystyle\sum_j f_j(Z_j)+\epsilon
$$
or in your specific case
$$
B = logit\left(f(S,T)\right)
$$
In R, you could use the mgcv package, and run somethin... | Nonlinear effect in an interaction term
Have you considered using a generalized additive model? Wikipedia link here
Basically the model would be
$$
g(y) = X'\beta+\displaystyle\sum_j f_j(Z_j)+\epsilon
$$
or in your specific case
$$
B = lo |
38,189 | Nonlinear effect in an interaction term | You may argue a non-linear effect by showing that a non-linear model fits better. For example you could implement a piecewise linear model to take into account changes in the influence of S. Dependent on your hypothesis, you could also linearise your factors. For example, a log transform of factors may reduce your resi... | Nonlinear effect in an interaction term | You may argue a non-linear effect by showing that a non-linear model fits better. For example you could implement a piecewise linear model to take into account changes in the influence of S. Dependent | Nonlinear effect in an interaction term
You may argue a non-linear effect by showing that a non-linear model fits better. For example you could implement a piecewise linear model to take into account changes in the influence of S. Dependent on your hypothesis, you could also linearise your factors. For example, a log t... | Nonlinear effect in an interaction term
You may argue a non-linear effect by showing that a non-linear model fits better. For example you could implement a piecewise linear model to take into account changes in the influence of S. Dependent |
38,190 | How to interpret coefficients of $x$ and $x^2$ in same regression | Such an equation describes a curved relationship between $y$ and $x$ - a parabola:
(This particular set of parameters correspond to a minimum at $x= -\frac{_5}{^6}$, just off the left margin of this plot.)
Consequently, you should keep all terms in the same x-variable together, since they describe the way $y$ is relat... | How to interpret coefficients of $x$ and $x^2$ in same regression | Such an equation describes a curved relationship between $y$ and $x$ - a parabola:
(This particular set of parameters correspond to a minimum at $x= -\frac{_5}{^6}$, just off the left margin of this | How to interpret coefficients of $x$ and $x^2$ in same regression
Such an equation describes a curved relationship between $y$ and $x$ - a parabola:
(This particular set of parameters correspond to a minimum at $x= -\frac{_5}{^6}$, just off the left margin of this plot.)
Consequently, you should keep all terms in the ... | How to interpret coefficients of $x$ and $x^2$ in same regression
Such an equation describes a curved relationship between $y$ and $x$ - a parabola:
(This particular set of parameters correspond to a minimum at $x= -\frac{_5}{^6}$, just off the left margin of this |
38,191 | How to interpret coefficients of $x$ and $x^2$ in same regression | It doesn't pay to interpret them separately. They are connected. The formula for the vertex in a quadratic equation $y = a x^{2} + b x + c$ is $-\frac{b}{2a}$. The effect of changing $x$ from $s$ to $t$ is $a (t^{2} - s^{2}) + b (t - s)$. In a regression setting I often set $s$ to the first quartile of $x$ and $t$ ... | How to interpret coefficients of $x$ and $x^2$ in same regression | It doesn't pay to interpret them separately. They are connected. The formula for the vertex in a quadratic equation $y = a x^{2} + b x + c$ is $-\frac{b}{2a}$. The effect of changing $x$ from $s$ t | How to interpret coefficients of $x$ and $x^2$ in same regression
It doesn't pay to interpret them separately. They are connected. The formula for the vertex in a quadratic equation $y = a x^{2} + b x + c$ is $-\frac{b}{2a}$. The effect of changing $x$ from $s$ to $t$ is $a (t^{2} - s^{2}) + b (t - s)$. In a regres... | How to interpret coefficients of $x$ and $x^2$ in same regression
It doesn't pay to interpret them separately. They are connected. The formula for the vertex in a quadratic equation $y = a x^{2} + b x + c$ is $-\frac{b}{2a}$. The effect of changing $x$ from $s$ t |
38,192 | How to interpret coefficients of $x$ and $x^2$ in same regression | The most straightforward way to interpret is through multivariate Taylor expansion. If you don't know what is it, then forget what I just wrote.
If you take a derivative of the model specification, you'll see that your coefficients are the Taylor series coefficients. | How to interpret coefficients of $x$ and $x^2$ in same regression | The most straightforward way to interpret is through multivariate Taylor expansion. If you don't know what is it, then forget what I just wrote.
If you take a derivative of the model specification, yo | How to interpret coefficients of $x$ and $x^2$ in same regression
The most straightforward way to interpret is through multivariate Taylor expansion. If you don't know what is it, then forget what I just wrote.
If you take a derivative of the model specification, you'll see that your coefficients are the Taylor series ... | How to interpret coefficients of $x$ and $x^2$ in same regression
The most straightforward way to interpret is through multivariate Taylor expansion. If you don't know what is it, then forget what I just wrote.
If you take a derivative of the model specification, yo |
38,193 | Sign of coefficients in linear regression vs. the sign of correlation | Expanding on @Maarten 's answer:
Suppose you are predicting the damage done by a fire. First, you look at one IV: Number of firefighters called to the scene. To your surprise, you find a strong positive relationship: More firefighters, more damage. Then you think of adding "size of fire" and add that to the equation... | Sign of coefficients in linear regression vs. the sign of correlation | Expanding on @Maarten 's answer:
Suppose you are predicting the damage done by a fire. First, you look at one IV: Number of firefighters called to the scene. To your surprise, you find a strong posi | Sign of coefficients in linear regression vs. the sign of correlation
Expanding on @Maarten 's answer:
Suppose you are predicting the damage done by a fire. First, you look at one IV: Number of firefighters called to the scene. To your surprise, you find a strong positive relationship: More firefighters, more damage.... | Sign of coefficients in linear regression vs. the sign of correlation
Expanding on @Maarten 's answer:
Suppose you are predicting the damage done by a fire. First, you look at one IV: Number of firefighters called to the scene. To your surprise, you find a strong posi |
38,194 | Sign of coefficients in linear regression vs. the sign of correlation | The statement is true iff you included only one explanatory variable. I suppose you know that the regression coefficent in such a regression is just $r_{x,y}\frac{s_y}{s_x}$, where $r_{x,y}$ is the correlation coefficient, and $s_y$ and $s_x$ are the standard deviations of $y$ and $x$ respectively. Since standard devia... | Sign of coefficients in linear regression vs. the sign of correlation | The statement is true iff you included only one explanatory variable. I suppose you know that the regression coefficent in such a regression is just $r_{x,y}\frac{s_y}{s_x}$, where $r_{x,y}$ is the co | Sign of coefficients in linear regression vs. the sign of correlation
The statement is true iff you included only one explanatory variable. I suppose you know that the regression coefficent in such a regression is just $r_{x,y}\frac{s_y}{s_x}$, where $r_{x,y}$ is the correlation coefficient, and $s_y$ and $s_x$ are the... | Sign of coefficients in linear regression vs. the sign of correlation
The statement is true iff you included only one explanatory variable. I suppose you know that the regression coefficent in such a regression is just $r_{x,y}\frac{s_y}{s_x}$, where $r_{x,y}$ is the co |
38,195 | Sign of coefficients in linear regression vs. the sign of correlation | Assume you have a regression with two regressors (plus a constant),
$$y_i = a + b_1x_{1i}+b_2x_{2i} + u_i$$
Then if you work the normal equations (a bit tedious) you will find that
$$\hat b_1 = \frac {\operatorname {\hat Var}(X_2)\cdot \operatorname {\hat Cov}(Y,X_1) - \operatorname {\hat Cov}(X_1,X_2)\cdot \operat... | Sign of coefficients in linear regression vs. the sign of correlation | Assume you have a regression with two regressors (plus a constant),
$$y_i = a + b_1x_{1i}+b_2x_{2i} + u_i$$
Then if you work the normal equations (a bit tedious) you will find that
$$\hat b_1 = \frac | Sign of coefficients in linear regression vs. the sign of correlation
Assume you have a regression with two regressors (plus a constant),
$$y_i = a + b_1x_{1i}+b_2x_{2i} + u_i$$
Then if you work the normal equations (a bit tedious) you will find that
$$\hat b_1 = \frac {\operatorname {\hat Var}(X_2)\cdot \operatornam... | Sign of coefficients in linear regression vs. the sign of correlation
Assume you have a regression with two regressors (plus a constant),
$$y_i = a + b_1x_{1i}+b_2x_{2i} + u_i$$
Then if you work the normal equations (a bit tedious) you will find that
$$\hat b_1 = \frac |
38,196 | Is " independent and identically distributed" an assumption or a fact ? | In practice being independent and identically distributed is an assumption; it may sometimes be a good approximation, but it's next to impossible to demonstrate that it actually holds.
Generally, the best you can do is show that it doesn't fail too badly.
This is what diagnostics, and to some extent hypothesis tests a... | Is " independent and identically distributed" an assumption or a fact ? | In practice being independent and identically distributed is an assumption; it may sometimes be a good approximation, but it's next to impossible to demonstrate that it actually holds.
Generally, the | Is " independent and identically distributed" an assumption or a fact ?
In practice being independent and identically distributed is an assumption; it may sometimes be a good approximation, but it's next to impossible to demonstrate that it actually holds.
Generally, the best you can do is show that it doesn't fail to... | Is " independent and identically distributed" an assumption or a fact ?
In practice being independent and identically distributed is an assumption; it may sometimes be a good approximation, but it's next to impossible to demonstrate that it actually holds.
Generally, the |
38,197 | Is " independent and identically distributed" an assumption or a fact ? | Just to add to the discussion, this is mostly an assumption that simplifies the mathematics of inference.
To take a concrete example, I am in the field of image processing and usually most algorithms will assume that the noise in the image is IID. This is hardly ever the case because most of the time we do some pre pro... | Is " independent and identically distributed" an assumption or a fact ? | Just to add to the discussion, this is mostly an assumption that simplifies the mathematics of inference.
To take a concrete example, I am in the field of image processing and usually most algorithms | Is " independent and identically distributed" an assumption or a fact ?
Just to add to the discussion, this is mostly an assumption that simplifies the mathematics of inference.
To take a concrete example, I am in the field of image processing and usually most algorithms will assume that the noise in the image is IID. ... | Is " independent and identically distributed" an assumption or a fact ?
Just to add to the discussion, this is mostly an assumption that simplifies the mathematics of inference.
To take a concrete example, I am in the field of image processing and usually most algorithms |
38,198 | Is " independent and identically distributed" an assumption or a fact ? | De Finetti would say that conditional independence is a logical consequence of your assumption that the sequence of random variables whose values you can observe in your experiment is exchangeable. | Is " independent and identically distributed" an assumption or a fact ? | De Finetti would say that conditional independence is a logical consequence of your assumption that the sequence of random variables whose values you can observe in your experiment is exchangeable. | Is " independent and identically distributed" an assumption or a fact ?
De Finetti would say that conditional independence is a logical consequence of your assumption that the sequence of random variables whose values you can observe in your experiment is exchangeable. | Is " independent and identically distributed" an assumption or a fact ?
De Finetti would say that conditional independence is a logical consequence of your assumption that the sequence of random variables whose values you can observe in your experiment is exchangeable. |
38,199 | Is " independent and identically distributed" an assumption or a fact ? | It depends on the problem but iid is usually an assumption based on two random variables being approximately independent and identically distributed (or at least we have good reason to believe they are). In most cases where we assume iid, we can't make the claim of perfect independence or that the distributions of the... | Is " independent and identically distributed" an assumption or a fact ? | It depends on the problem but iid is usually an assumption based on two random variables being approximately independent and identically distributed (or at least we have good reason to believe they ar | Is " independent and identically distributed" an assumption or a fact ?
It depends on the problem but iid is usually an assumption based on two random variables being approximately independent and identically distributed (or at least we have good reason to believe they are). In most cases where we assume iid, we can't... | Is " independent and identically distributed" an assumption or a fact ?
It depends on the problem but iid is usually an assumption based on two random variables being approximately independent and identically distributed (or at least we have good reason to believe they ar |
38,200 | Why does statistical significance increase with data, BUT the effects may not be meaningful? | I think this comes from the fact that in the real world you don't really expect the standard null hypothesis to be true. If you're comparing the means of two populations, the null hypothesis says that $\mu_1 = \mu_2$, that is the two means are exactly equal. In many situations however a more accurate null hypothesis wo... | Why does statistical significance increase with data, BUT the effects may not be meaningful? | I think this comes from the fact that in the real world you don't really expect the standard null hypothesis to be true. If you're comparing the means of two populations, the null hypothesis says that | Why does statistical significance increase with data, BUT the effects may not be meaningful?
I think this comes from the fact that in the real world you don't really expect the standard null hypothesis to be true. If you're comparing the means of two populations, the null hypothesis says that $\mu_1 = \mu_2$, that is t... | Why does statistical significance increase with data, BUT the effects may not be meaningful?
I think this comes from the fact that in the real world you don't really expect the standard null hypothesis to be true. If you're comparing the means of two populations, the null hypothesis says that |
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