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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38,901 | Does adjusting for superfluous variables bias OLS estimates? | There's no contradiction.
The first paragraph there talks about superfluous variables.
If Simpson's paradox applies, the variables are not superfluous. | Does adjusting for superfluous variables bias OLS estimates? | There's no contradiction.
The first paragraph there talks about superfluous variables.
If Simpson's paradox applies, the variables are not superfluous. | Does adjusting for superfluous variables bias OLS estimates?
There's no contradiction.
The first paragraph there talks about superfluous variables.
If Simpson's paradox applies, the variables are not superfluous. | Does adjusting for superfluous variables bias OLS estimates?
There's no contradiction.
The first paragraph there talks about superfluous variables.
If Simpson's paradox applies, the variables are not superfluous. |
38,902 | Does adjusting for superfluous variables bias OLS estimates? | Consider a postulated linear regression model
$$y_i = b_0 + b_1X_{1i} + b_2X_{2i} + u_i,\;\; i=1,...,n$$
As a matter of algebra (and not any stochastic assumptions), the OLS estimator in matrix notation is
$$\hat b = b + \left(\mathbf X'\mathbf X\right)^{-1}\mathbf X'\mathbf u$$
Its expected value conditional on the re... | Does adjusting for superfluous variables bias OLS estimates? | Consider a postulated linear regression model
$$y_i = b_0 + b_1X_{1i} + b_2X_{2i} + u_i,\;\; i=1,...,n$$
As a matter of algebra (and not any stochastic assumptions), the OLS estimator in matrix notati | Does adjusting for superfluous variables bias OLS estimates?
Consider a postulated linear regression model
$$y_i = b_0 + b_1X_{1i} + b_2X_{2i} + u_i,\;\; i=1,...,n$$
As a matter of algebra (and not any stochastic assumptions), the OLS estimator in matrix notation is
$$\hat b = b + \left(\mathbf X'\mathbf X\right)^{-1}\... | Does adjusting for superfluous variables bias OLS estimates?
Consider a postulated linear regression model
$$y_i = b_0 + b_1X_{1i} + b_2X_{2i} + u_i,\;\; i=1,...,n$$
As a matter of algebra (and not any stochastic assumptions), the OLS estimator in matrix notati |
38,903 | How can I calculate t-score without knowing true population mean? | As far as I know μ is used to define true population mean.
Not quite, and here's the rub. μ represents whatever the true mean is. It's defined by the problem for which this little bit of statistical inference is the analysis, not by the data itself (that would make it an estimate, not a hypothesis)
So in formula a... | How can I calculate t-score without knowing true population mean? | As far as I know μ is used to define true population mean.
Not quite, and here's the rub. μ represents whatever the true mean is. It's defined by the problem for which this little bit of statistic | How can I calculate t-score without knowing true population mean?
As far as I know μ is used to define true population mean.
Not quite, and here's the rub. μ represents whatever the true mean is. It's defined by the problem for which this little bit of statistical inference is the analysis, not by the data itself (... | How can I calculate t-score without knowing true population mean?
As far as I know μ is used to define true population mean.
Not quite, and here's the rub. μ represents whatever the true mean is. It's defined by the problem for which this little bit of statistic |
38,904 | How can I calculate t-score without knowing true population mean? | There are two different $\mu$'s involved here:
the hypothesized mean that you use in the numerator of your
t-statistic for a t-test (sometimes denoted as $\mu_0$), and
the true population mean, $\mu$.
The t-test is actually to see whether the true population mean differs from the hypothesized mean -- that is, it's a ... | How can I calculate t-score without knowing true population mean? | There are two different $\mu$'s involved here:
the hypothesized mean that you use in the numerator of your
t-statistic for a t-test (sometimes denoted as $\mu_0$), and
the true population mean, $\mu$ | How can I calculate t-score without knowing true population mean?
There are two different $\mu$'s involved here:
the hypothesized mean that you use in the numerator of your
t-statistic for a t-test (sometimes denoted as $\mu_0$), and
the true population mean, $\mu$.
The t-test is actually to see whether the true popu... | How can I calculate t-score without knowing true population mean?
There are two different $\mu$'s involved here:
the hypothesized mean that you use in the numerator of your
t-statistic for a t-test (sometimes denoted as $\mu_0$), and
the true population mean, $\mu$ |
38,905 | Approximate the distribution of the sum of ind. Beta r.v | If you want better approximations than what you get from the central limit theorem, there is results in a book dedicated exclusively to the beta distribution: http://www.amazon.com/Handbook-Beta-Distribution-Applications-Statistics/dp/0824753968/ref=sr_1_1?s=books&ie=UTF8&qid=1403444915&sr=1-1&keywords=beta+distribu... | Approximate the distribution of the sum of ind. Beta r.v | If you want better approximations than what you get from the central limit theorem, there is results in a book dedicated exclusively to the beta distribution: http://www.amazon.com/Handbook-Beta-Di | Approximate the distribution of the sum of ind. Beta r.v
If you want better approximations than what you get from the central limit theorem, there is results in a book dedicated exclusively to the beta distribution: http://www.amazon.com/Handbook-Beta-Distribution-Applications-Statistics/dp/0824753968/ref=sr_1_1?s=b... | Approximate the distribution of the sum of ind. Beta r.v
If you want better approximations than what you get from the central limit theorem, there is results in a book dedicated exclusively to the beta distribution: http://www.amazon.com/Handbook-Beta-Di |
38,906 | Approximate the distribution of the sum of ind. Beta r.v | Since you cannot share details, we cannot know to what extend you can use transformations of your variables, and still get what you need. For what is worth, a usual transformation here is the following:
$$X_i \sim \beta(1,K) \Rightarrow Z_i=-\ln(1-X_i) \sim \text{Exp}(K)$$
and
$$S_z = \sum_{i=1}^NZ_i \sim \text{Erlang}... | Approximate the distribution of the sum of ind. Beta r.v | Since you cannot share details, we cannot know to what extend you can use transformations of your variables, and still get what you need. For what is worth, a usual transformation here is the followin | Approximate the distribution of the sum of ind. Beta r.v
Since you cannot share details, we cannot know to what extend you can use transformations of your variables, and still get what you need. For what is worth, a usual transformation here is the following:
$$X_i \sim \beta(1,K) \Rightarrow Z_i=-\ln(1-X_i) \sim \text... | Approximate the distribution of the sum of ind. Beta r.v
Since you cannot share details, we cannot know to what extend you can use transformations of your variables, and still get what you need. For what is worth, a usual transformation here is the followin |
38,907 | Approximate the distribution of the sum of ind. Beta r.v | CLT: $(S_n-n\mu)/\sigma\sqrt{n}$ is approximately distribured as $\mathrm{N}(0,1)$ for large $n$.
To determine $\mu$ and $\sigma$ use this and this. | Approximate the distribution of the sum of ind. Beta r.v | CLT: $(S_n-n\mu)/\sigma\sqrt{n}$ is approximately distribured as $\mathrm{N}(0,1)$ for large $n$.
To determine $\mu$ and $\sigma$ use this and this. | Approximate the distribution of the sum of ind. Beta r.v
CLT: $(S_n-n\mu)/\sigma\sqrt{n}$ is approximately distribured as $\mathrm{N}(0,1)$ for large $n$.
To determine $\mu$ and $\sigma$ use this and this. | Approximate the distribution of the sum of ind. Beta r.v
CLT: $(S_n-n\mu)/\sigma\sqrt{n}$ is approximately distribured as $\mathrm{N}(0,1)$ for large $n$.
To determine $\mu$ and $\sigma$ use this and this. |
38,908 | False discovery rate of multiple regressions models | @Dian breathe easy, it's pretty much not too difficult. So let's work from familiar territory to false discovery rate (FDR).
First, I see that you have a bunch of outcomes, with a varying number of predictors. Someone who is more familiar with multivariate regression (i.e. multiple dependent variables, assuming possibl... | False discovery rate of multiple regressions models | @Dian breathe easy, it's pretty much not too difficult. So let's work from familiar territory to false discovery rate (FDR).
First, I see that you have a bunch of outcomes, with a varying number of pr | False discovery rate of multiple regressions models
@Dian breathe easy, it's pretty much not too difficult. So let's work from familiar territory to false discovery rate (FDR).
First, I see that you have a bunch of outcomes, with a varying number of predictors. Someone who is more familiar with multivariate regression ... | False discovery rate of multiple regressions models
@Dian breathe easy, it's pretty much not too difficult. So let's work from familiar territory to false discovery rate (FDR).
First, I see that you have a bunch of outcomes, with a varying number of pr |
38,909 | False discovery rate of multiple regressions models | What you're doing is a type of step-wise regression, so FDR in this context serves to guide the model selection: the p-value would be that of the goodness-of-fit (coefficient of determination), not of any individual explanatory variable. | False discovery rate of multiple regressions models | What you're doing is a type of step-wise regression, so FDR in this context serves to guide the model selection: the p-value would be that of the goodness-of-fit (coefficient of determination), not of | False discovery rate of multiple regressions models
What you're doing is a type of step-wise regression, so FDR in this context serves to guide the model selection: the p-value would be that of the goodness-of-fit (coefficient of determination), not of any individual explanatory variable. | False discovery rate of multiple regressions models
What you're doing is a type of step-wise regression, so FDR in this context serves to guide the model selection: the p-value would be that of the goodness-of-fit (coefficient of determination), not of |
38,910 | How to get only desirable comparisons from post-hoc | Suppose that we have one response variable and one explanatory variable (5 levels).
con.data <- data.frame(x = c(rnorm(75), c(rnorm(75)+5)),
category = rep(c("A","B1","B2","C1","C2"), each=30))
If we do ANOVA followed by classical TukeyHSD post-hoc...
m1 <- aov(con.data$x ~ con.data$category); su... | How to get only desirable comparisons from post-hoc | Suppose that we have one response variable and one explanatory variable (5 levels).
con.data <- data.frame(x = c(rnorm(75), c(rnorm(75)+5)),
category = rep(c("A","B1","B2","C1"," | How to get only desirable comparisons from post-hoc
Suppose that we have one response variable and one explanatory variable (5 levels).
con.data <- data.frame(x = c(rnorm(75), c(rnorm(75)+5)),
category = rep(c("A","B1","B2","C1","C2"), each=30))
If we do ANOVA followed by classical TukeyHSD post-... | How to get only desirable comparisons from post-hoc
Suppose that we have one response variable and one explanatory variable (5 levels).
con.data <- data.frame(x = c(rnorm(75), c(rnorm(75)+5)),
category = rep(c("A","B1","B2","C1"," |
38,911 | How to get only desirable comparisons from post-hoc | Running step on your model will also work.
# Full model
m1.0 = glm (logICPMS ~ SITE*ISLAND*ORG*ORGL*CAGE*Metal,
data = M.m, na.action = na.exclude)
# this removes each interacting factor to give you the MAM model
step(m1.0)
# MAM model
m1.1 = glm(formula = logICPMS ~ ISLAND + ORG + CAGE + Metal + ISLAND:O... | How to get only desirable comparisons from post-hoc | Running step on your model will also work.
# Full model
m1.0 = glm (logICPMS ~ SITE*ISLAND*ORG*ORGL*CAGE*Metal,
data = M.m, na.action = na.exclude)
# this removes each interacting factor | How to get only desirable comparisons from post-hoc
Running step on your model will also work.
# Full model
m1.0 = glm (logICPMS ~ SITE*ISLAND*ORG*ORGL*CAGE*Metal,
data = M.m, na.action = na.exclude)
# this removes each interacting factor to give you the MAM model
step(m1.0)
# MAM model
m1.1 = glm(formula... | How to get only desirable comparisons from post-hoc
Running step on your model will also work.
# Full model
m1.0 = glm (logICPMS ~ SITE*ISLAND*ORG*ORGL*CAGE*Metal,
data = M.m, na.action = na.exclude)
# this removes each interacting factor |
38,912 | How useful is the CLT in applications? | The CLT certainly informs applications all the time, since we deal with distributions of averages or sums very frequently (including in cases that may not be always obvious; for example, $s_n^2$ - the sample variance with denominator $n$ - is an average, and so the ordinary sample variance is just a slightly rescaled a... | How useful is the CLT in applications? | The CLT certainly informs applications all the time, since we deal with distributions of averages or sums very frequently (including in cases that may not be always obvious; for example, $s_n^2$ - the | How useful is the CLT in applications?
The CLT certainly informs applications all the time, since we deal with distributions of averages or sums very frequently (including in cases that may not be always obvious; for example, $s_n^2$ - the sample variance with denominator $n$ - is an average, and so the ordinary sample... | How useful is the CLT in applications?
The CLT certainly informs applications all the time, since we deal with distributions of averages or sums very frequently (including in cases that may not be always obvious; for example, $s_n^2$ - the |
38,913 | How useful is the CLT in applications? | Notwithstanding the problems with CLT described by Glen_b, if we are talking about econometric analysis, then practicaly all results are asymptotic (except when Bayesian analysis is used). Hence any application which is based on econometrics is based on CLT. For example Lars Hansen got the Nobel prize last year for w... | How useful is the CLT in applications? | Notwithstanding the problems with CLT described by Glen_b, if we are talking about econometric analysis, then practicaly all results are asymptotic (except when Bayesian analysis is used). Hence any | How useful is the CLT in applications?
Notwithstanding the problems with CLT described by Glen_b, if we are talking about econometric analysis, then practicaly all results are asymptotic (except when Bayesian analysis is used). Hence any application which is based on econometrics is based on CLT. For example Lars Han... | How useful is the CLT in applications?
Notwithstanding the problems with CLT described by Glen_b, if we are talking about econometric analysis, then practicaly all results are asymptotic (except when Bayesian analysis is used). Hence any |
38,914 | Should control variables be included in model if statistically insignificant? | One reason to include control variables is precisely because they can affect other variables. In this case, the statistical significance of the control variable is completely irrelevant.
However, you may run into journal editors who disagree. | Should control variables be included in model if statistically insignificant? | One reason to include control variables is precisely because they can affect other variables. In this case, the statistical significance of the control variable is completely irrelevant.
However, you | Should control variables be included in model if statistically insignificant?
One reason to include control variables is precisely because they can affect other variables. In this case, the statistical significance of the control variable is completely irrelevant.
However, you may run into journal editors who disagree. | Should control variables be included in model if statistically insignificant?
One reason to include control variables is precisely because they can affect other variables. In this case, the statistical significance of the control variable is completely irrelevant.
However, you |
38,915 | Should control variables be included in model if statistically insignificant? | Just a short comment: your p-values should reflect the number of models you are "trying out". In some ways your approach of trying models with and without subsets of variables is one aspect of p-hacking. Your research question alone (not the data) should determine what is a control variable and what is a variable of in... | Should control variables be included in model if statistically insignificant? | Just a short comment: your p-values should reflect the number of models you are "trying out". In some ways your approach of trying models with and without subsets of variables is one aspect of p-hacki | Should control variables be included in model if statistically insignificant?
Just a short comment: your p-values should reflect the number of models you are "trying out". In some ways your approach of trying models with and without subsets of variables is one aspect of p-hacking. Your research question alone (not the ... | Should control variables be included in model if statistically insignificant?
Just a short comment: your p-values should reflect the number of models you are "trying out". In some ways your approach of trying models with and without subsets of variables is one aspect of p-hacki |
38,916 | Is there a word for the phenomenon that the old are generally less affected by risk factors? | Survival bias and competing risks. Also, elderly having a high value of a risk factor who have not been affected by that risk factor have demonstrated a robustness to that particular factor in general. This is why age $\times$ risk factor interactions can be important to pre-specify in a model. | Is there a word for the phenomenon that the old are generally less affected by risk factors? | Survival bias and competing risks. Also, elderly having a high value of a risk factor who have not been affected by that risk factor have demonstrated a robustness to that particular factor in genera | Is there a word for the phenomenon that the old are generally less affected by risk factors?
Survival bias and competing risks. Also, elderly having a high value of a risk factor who have not been affected by that risk factor have demonstrated a robustness to that particular factor in general. This is why age $\times... | Is there a word for the phenomenon that the old are generally less affected by risk factors?
Survival bias and competing risks. Also, elderly having a high value of a risk factor who have not been affected by that risk factor have demonstrated a robustness to that particular factor in genera |
38,917 | Is there a word for the phenomenon that the old are generally less affected by risk factors? | Age $\times$ risk factor interaction are, as mentioned, important. Some risk factors may have a different form of association in later life. Low values of some risk factors can be associated with mortality due to them being possible indicators of poor health in later life, leading to a u-shaped relationship. There is a... | Is there a word for the phenomenon that the old are generally less affected by risk factors? | Age $\times$ risk factor interaction are, as mentioned, important. Some risk factors may have a different form of association in later life. Low values of some risk factors can be associated with mort | Is there a word for the phenomenon that the old are generally less affected by risk factors?
Age $\times$ risk factor interaction are, as mentioned, important. Some risk factors may have a different form of association in later life. Low values of some risk factors can be associated with mortality due to them being pos... | Is there a word for the phenomenon that the old are generally less affected by risk factors?
Age $\times$ risk factor interaction are, as mentioned, important. Some risk factors may have a different form of association in later life. Low values of some risk factors can be associated with mort |
38,918 | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | A $I(0)$ and a $I(1)$ timeseries can not be cointegrated. There is no linear combination of the timeseries that is stationary. And the definition of cointegration is if there is a combination of them that is stationary, they're cointegrated.
I think you should fit a VAR with the stationary variable in levels and the n... | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | A $I(0)$ and a $I(1)$ timeseries can not be cointegrated. There is no linear combination of the timeseries that is stationary. And the definition of cointegration is if there is a combination of them | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
A $I(0)$ and a $I(1)$ timeseries can not be cointegrated. There is no linear combination of the timeseries that is stationary. And the definition of cointegration is if there is a combination of them that is stationary, they're c... | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
A $I(0)$ and a $I(1)$ timeseries can not be cointegrated. There is no linear combination of the timeseries that is stationary. And the definition of cointegration is if there is a combination of them |
38,919 | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | If johansen test result is not significant, meaning no cointegration, then take the 1st difference of the other variable to ensure stationarity. In some cases you may need to take the ln of the 1 st difference. | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | If johansen test result is not significant, meaning no cointegration, then take the 1st difference of the other variable to ensure stationarity. In some cases you may need to take the ln of the 1 st d | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
If johansen test result is not significant, meaning no cointegration, then take the 1st difference of the other variable to ensure stationarity. In some cases you may need to take the ln of the 1 st difference. | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
If johansen test result is not significant, meaning no cointegration, then take the 1st difference of the other variable to ensure stationarity. In some cases you may need to take the ln of the 1 st d |
38,920 | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | In case you have a mix of I(0) and I(1) variables, you can apply the tests proposed by Pesaran et al (2001), where you can test for cointegration. The link to the paper is:
http://onlinelibrary.wiley.com/doi/10.1002/jae.616/pdf
All the best | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | In case you have a mix of I(0) and I(1) variables, you can apply the tests proposed by Pesaran et al (2001), where you can test for cointegration. The link to the paper is:
http://onlinelibrary.wiley. | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
In case you have a mix of I(0) and I(1) variables, you can apply the tests proposed by Pesaran et al (2001), where you can test for cointegration. The link to the paper is:
http://onlinelibrary.wiley.com/doi/10.1002/jae.616/pdf
A... | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
In case you have a mix of I(0) and I(1) variables, you can apply the tests proposed by Pesaran et al (2001), where you can test for cointegration. The link to the paper is:
http://onlinelibrary.wiley. |
38,921 | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | ARDL model approach described by Pesaran is the only way to find the cointegration among the variables having different orders I(0) and I(1) but keeping in mind none of the variable should stationery at I(2) | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration? | ARDL model approach described by Pesaran is the only way to find the cointegration among the variables having different orders I(0) and I(1) but keeping in mind none of the variable should stationery | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
ARDL model approach described by Pesaran is the only way to find the cointegration among the variables having different orders I(0) and I(1) but keeping in mind none of the variable should stationery at I(2) | Time series: one I(1) and one I(0) variable, should I use VAR/VEC, test for cointegration?
ARDL model approach described by Pesaran is the only way to find the cointegration among the variables having different orders I(0) and I(1) but keeping in mind none of the variable should stationery |
38,922 | Why will a statistic be significant with sufficiently large samples unless the population effect is exactly zero? | With increasing sample size, the statistical power (see below) to detect even the smallest effect size is also increasing and these tiny effect sizes are then found to be statistically significant, even though they bear no relevance at all. Just as a thought experiment to illustrate it further: What if you could includ... | Why will a statistic be significant with sufficiently large samples unless the population effect is | With increasing sample size, the statistical power (see below) to detect even the smallest effect size is also increasing and these tiny effect sizes are then found to be statistically significant, ev | Why will a statistic be significant with sufficiently large samples unless the population effect is exactly zero?
With increasing sample size, the statistical power (see below) to detect even the smallest effect size is also increasing and these tiny effect sizes are then found to be statistically significant, even tho... | Why will a statistic be significant with sufficiently large samples unless the population effect is
With increasing sample size, the statistical power (see below) to detect even the smallest effect size is also increasing and these tiny effect sizes are then found to be statistically significant, ev |
38,923 | Why will a statistic be significant with sufficiently large samples unless the population effect is exactly zero? | For concreteness, imagine a one sample test of means (large sample, on something where the population mean and variance exists to make the argument a little simpler).
Let the difference between the true mean and the hypothesized sample mean be any nonzero $\delta$. Then the sampling distribution of the sample mean mi... | Why will a statistic be significant with sufficiently large samples unless the population effect is | For concreteness, imagine a one sample test of means (large sample, on something where the population mean and variance exists to make the argument a little simpler).
Let the difference between the t | Why will a statistic be significant with sufficiently large samples unless the population effect is exactly zero?
For concreteness, imagine a one sample test of means (large sample, on something where the population mean and variance exists to make the argument a little simpler).
Let the difference between the true me... | Why will a statistic be significant with sufficiently large samples unless the population effect is
For concreteness, imagine a one sample test of means (large sample, on something where the population mean and variance exists to make the argument a little simpler).
Let the difference between the t |
38,924 | Regression with ARIMA(0,0,0) errors different from linear regression | As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, possibly including the constant, using the AICc. With one covariate, the model is
$$y_t = \beta_0 x_t + n_t$$
where $n_t$ is an ARIMA process. Note that the intercept is shifted to t... | Regression with ARIMA(0,0,0) errors different from linear regression | As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, possibly including the constant, using the AICc. With one cova | Regression with ARIMA(0,0,0) errors different from linear regression
As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, possibly including the constant, using the AICc. With one covariate, the model is
$$y_t = \beta_0 x_t + n_t$$
whe... | Regression with ARIMA(0,0,0) errors different from linear regression
As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, possibly including the constant, using the AICc. With one cova |
38,925 | Definition of "precision" of an estimator | Precision is typically defined as the reciprocal of the variance of an estimator. A high variance estimator has low precision and vice versa. | Definition of "precision" of an estimator | Precision is typically defined as the reciprocal of the variance of an estimator. A high variance estimator has low precision and vice versa. | Definition of "precision" of an estimator
Precision is typically defined as the reciprocal of the variance of an estimator. A high variance estimator has low precision and vice versa. | Definition of "precision" of an estimator
Precision is typically defined as the reciprocal of the variance of an estimator. A high variance estimator has low precision and vice versa. |
38,926 | Definition of "precision" of an estimator | Precision can be defined as 1/variance, or 1/standard deviation. See Wikipedia's article on Normal distribution (section "Alternative formulations") and references provided therein. | Definition of "precision" of an estimator | Precision can be defined as 1/variance, or 1/standard deviation. See Wikipedia's article on Normal distribution (section "Alternative formulations") and references provided therein. | Definition of "precision" of an estimator
Precision can be defined as 1/variance, or 1/standard deviation. See Wikipedia's article on Normal distribution (section "Alternative formulations") and references provided therein. | Definition of "precision" of an estimator
Precision can be defined as 1/variance, or 1/standard deviation. See Wikipedia's article on Normal distribution (section "Alternative formulations") and references provided therein. |
38,927 | What's the correct distribution for page reading time? | It helps to have data. Jakob Nielsen has measured reading times for web pages ("How Little Do Users Read", 2008), which gives some strong hints:
The data show that the variation in reading time is directly proportional to the number of words on a page.
The variation therefore should be expressed as a time relative t... | What's the correct distribution for page reading time? | It helps to have data. Jakob Nielsen has measured reading times for web pages ("How Little Do Users Read", 2008), which gives some strong hints:
The data show that the variation in reading time is d | What's the correct distribution for page reading time?
It helps to have data. Jakob Nielsen has measured reading times for web pages ("How Little Do Users Read", 2008), which gives some strong hints:
The data show that the variation in reading time is directly proportional to the number of words on a page.
The varia... | What's the correct distribution for page reading time?
It helps to have data. Jakob Nielsen has measured reading times for web pages ("How Little Do Users Read", 2008), which gives some strong hints:
The data show that the variation in reading time is d |
38,928 | What's the correct distribution for page reading time? | You could use a gamma distribution. Check it out on Wikipedia.
Gamma distributions are commonly used to model waiting times, like the situation you have here. | What's the correct distribution for page reading time? | You could use a gamma distribution. Check it out on Wikipedia.
Gamma distributions are commonly used to model waiting times, like the situation you have here. | What's the correct distribution for page reading time?
You could use a gamma distribution. Check it out on Wikipedia.
Gamma distributions are commonly used to model waiting times, like the situation you have here. | What's the correct distribution for page reading time?
You could use a gamma distribution. Check it out on Wikipedia.
Gamma distributions are commonly used to model waiting times, like the situation you have here. |
38,929 | Regression on a non-normal dependent variable | The normality assumption is a convenient property of model's residuals, since it enables correct inferences about the estimated parameters and critical values of many other tests are also dependent on this assumption (therefore some corrections should be made, or you may roughly take more strict rule-of-thumb criteria,... | Regression on a non-normal dependent variable | The normality assumption is a convenient property of model's residuals, since it enables correct inferences about the estimated parameters and critical values of many other tests are also dependent on | Regression on a non-normal dependent variable
The normality assumption is a convenient property of model's residuals, since it enables correct inferences about the estimated parameters and critical values of many other tests are also dependent on this assumption (therefore some corrections should be made, or you may ro... | Regression on a non-normal dependent variable
The normality assumption is a convenient property of model's residuals, since it enables correct inferences about the estimated parameters and critical values of many other tests are also dependent on |
38,930 | Regression on a non-normal dependent variable | Ordinal regression is not affected by empty cells of Y. Quantile grouping is not required unless you just want to reduce computational burden. Proportional odds or continuation ratio ordinal logistic models are likely to be able to handle the distribution of Y you plotted (with no grouping of Y). | Regression on a non-normal dependent variable | Ordinal regression is not affected by empty cells of Y. Quantile grouping is not required unless you just want to reduce computational burden. Proportional odds or continuation ratio ordinal logisti | Regression on a non-normal dependent variable
Ordinal regression is not affected by empty cells of Y. Quantile grouping is not required unless you just want to reduce computational burden. Proportional odds or continuation ratio ordinal logistic models are likely to be able to handle the distribution of Y you plotted... | Regression on a non-normal dependent variable
Ordinal regression is not affected by empty cells of Y. Quantile grouping is not required unless you just want to reduce computational burden. Proportional odds or continuation ratio ordinal logisti |
38,931 | Origin of strange formula for equilibrium standard deviation | The authors are providing a simple means for estimating the parameters of a mean-reverting Orstein-Uhlenbeck process via a regression on returns at discretized points in time.
The model they are considering has a representation as a stochastic differential equation of the form [pg. 16, Eq. (12)]
$$
\newcommand{\rd}{\m... | Origin of strange formula for equilibrium standard deviation | The authors are providing a simple means for estimating the parameters of a mean-reverting Orstein-Uhlenbeck process via a regression on returns at discretized points in time.
The model they are consi | Origin of strange formula for equilibrium standard deviation
The authors are providing a simple means for estimating the parameters of a mean-reverting Orstein-Uhlenbeck process via a regression on returns at discretized points in time.
The model they are considering has a representation as a stochastic differential eq... | Origin of strange formula for equilibrium standard deviation
The authors are providing a simple means for estimating the parameters of a mean-reverting Orstein-Uhlenbeck process via a regression on returns at discretized points in time.
The model they are consi |
38,932 | Distance between empirically generated distributions (in R) | You can do a Kolmogorov-Smirnov test using the ks.test function. See ?ks.test.
In general, when you are looking for a function in R (and you don't know its name) try using ??. For instance, ??"Kolmogorov Smirnov". If nothing comes up RSiteSearch("whatever you're looking for") should help :) | Distance between empirically generated distributions (in R) | You can do a Kolmogorov-Smirnov test using the ks.test function. See ?ks.test.
In general, when you are looking for a function in R (and you don't know its name) try using ??. For instance, ??"Kolmogo | Distance between empirically generated distributions (in R)
You can do a Kolmogorov-Smirnov test using the ks.test function. See ?ks.test.
In general, when you are looking for a function in R (and you don't know its name) try using ??. For instance, ??"Kolmogorov Smirnov". If nothing comes up RSiteSearch("whatever you'... | Distance between empirically generated distributions (in R)
You can do a Kolmogorov-Smirnov test using the ks.test function. See ?ks.test.
In general, when you are looking for a function in R (and you don't know its name) try using ??. For instance, ??"Kolmogo |
38,933 | Distance between empirically generated distributions (in R) | A standard way to compare distributions is to use the Kullback-Leibler divergence. As usual, there's an R package that does this for you! From the ?KLdiv help page in the flexmix package, we get the following bit of code:
## Gaussian and Student t are much closer to each other than
## to the uniform:
> library(flexmix)... | Distance between empirically generated distributions (in R) | A standard way to compare distributions is to use the Kullback-Leibler divergence. As usual, there's an R package that does this for you! From the ?KLdiv help page in the flexmix package, we get the f | Distance between empirically generated distributions (in R)
A standard way to compare distributions is to use the Kullback-Leibler divergence. As usual, there's an R package that does this for you! From the ?KLdiv help page in the flexmix package, we get the following bit of code:
## Gaussian and Student t are much clo... | Distance between empirically generated distributions (in R)
A standard way to compare distributions is to use the Kullback-Leibler divergence. As usual, there's an R package that does this for you! From the ?KLdiv help page in the flexmix package, we get the f |
38,934 | Distance between empirically generated distributions (in R) | First thing: define "distance". That sounds like a stupid question, but what do you mean by distance? Is the data paired? Then -and only then- it makes sense to look at the sum of (squared) differences to decide about the distance between two datasets. If not, you have to resort to other means.
The next question is: is... | Distance between empirically generated distributions (in R) | First thing: define "distance". That sounds like a stupid question, but what do you mean by distance? Is the data paired? Then -and only then- it makes sense to look at the sum of (squared) difference | Distance between empirically generated distributions (in R)
First thing: define "distance". That sounds like a stupid question, but what do you mean by distance? Is the data paired? Then -and only then- it makes sense to look at the sum of (squared) differences to decide about the distance between two datasets. If not,... | Distance between empirically generated distributions (in R)
First thing: define "distance". That sounds like a stupid question, but what do you mean by distance? Is the data paired? Then -and only then- it makes sense to look at the sum of (squared) difference |
38,935 | How to test group differences on a five point variable? | What you are looking for seems to be a test for comparing two groups where observations are kind of ordinal data. In this case, I would suggest to apply a trend test to see if there are any differences between the CTL and TRT group.
Using a t-test would not acknowledge the fact your data are discrete, and the Gaussian... | How to test group differences on a five point variable? | What you are looking for seems to be a test for comparing two groups where observations are kind of ordinal data. In this case, I would suggest to apply a trend test to see if there are any difference | How to test group differences on a five point variable?
What you are looking for seems to be a test for comparing two groups where observations are kind of ordinal data. In this case, I would suggest to apply a trend test to see if there are any differences between the CTL and TRT group.
Using a t-test would not ackno... | How to test group differences on a five point variable?
What you are looking for seems to be a test for comparing two groups where observations are kind of ordinal data. In this case, I would suggest to apply a trend test to see if there are any difference |
38,936 | How to test group differences on a five point variable? | Three things come to mind:
Contingency table analysis using Fisher's exact test or Chi Square (but will only tell you that somewhere in the table there is a difference that is significant. You'd have to visualize your data or do post-hoc tests to know where this difference is.) Not my preferred solution.
A non-paramet... | How to test group differences on a five point variable? | Three things come to mind:
Contingency table analysis using Fisher's exact test or Chi Square (but will only tell you that somewhere in the table there is a difference that is significant. You'd have | How to test group differences on a five point variable?
Three things come to mind:
Contingency table analysis using Fisher's exact test or Chi Square (but will only tell you that somewhere in the table there is a difference that is significant. You'd have to visualize your data or do post-hoc tests to know where this ... | How to test group differences on a five point variable?
Three things come to mind:
Contingency table analysis using Fisher's exact test or Chi Square (but will only tell you that somewhere in the table there is a difference that is significant. You'd have |
38,937 | How to test group differences on a five point variable? | This question is a little unusual because the nature of "different" is unspecified. This response is formulated in the spirit of trying to detect as many kinds of differences as possible, not just changes of location ("trend").
One approach that might have more power than most, while remaining agnostic about the relat... | How to test group differences on a five point variable? | This question is a little unusual because the nature of "different" is unspecified. This response is formulated in the spirit of trying to detect as many kinds of differences as possible, not just ch | How to test group differences on a five point variable?
This question is a little unusual because the nature of "different" is unspecified. This response is formulated in the spirit of trying to detect as many kinds of differences as possible, not just changes of location ("trend").
One approach that might have more p... | How to test group differences on a five point variable?
This question is a little unusual because the nature of "different" is unspecified. This response is formulated in the spirit of trying to detect as many kinds of differences as possible, not just ch |
38,938 | Precedent for Bootstrap-like procedure with "invented" data? | I don't know.
What you've shown is a legitimate Monte Carlo simulation
Bootstrap is also a Monte Carlo method, but it is more about estimating distributions.
In general yes, especially if imputation is giving poor results. In special cases when imputation works great, no. In simple words, it will be as good as strongl... | Precedent for Bootstrap-like procedure with "invented" data? | I don't know.
What you've shown is a legitimate Monte Carlo simulation
Bootstrap is also a Monte Carlo method, but it is more about estimating distributions.
In general yes, especially if imputation | Precedent for Bootstrap-like procedure with "invented" data?
I don't know.
What you've shown is a legitimate Monte Carlo simulation
Bootstrap is also a Monte Carlo method, but it is more about estimating distributions.
In general yes, especially if imputation is giving poor results. In special cases when imputation wo... | Precedent for Bootstrap-like procedure with "invented" data?
I don't know.
What you've shown is a legitimate Monte Carlo simulation
Bootstrap is also a Monte Carlo method, but it is more about estimating distributions.
In general yes, especially if imputation |
38,939 | Precedent for Bootstrap-like procedure with "invented" data? | An alternative approach would be to impute the missing raw data using a missing data replacement procedure. You could then run the PCA on the correlation matrix that resulted from the imputed dataset (see also multiple imputation).
Here are a few links on missing data imputation in R:
Gelman on missing data imputation... | Precedent for Bootstrap-like procedure with "invented" data? | An alternative approach would be to impute the missing raw data using a missing data replacement procedure. You could then run the PCA on the correlation matrix that resulted from the imputed dataset | Precedent for Bootstrap-like procedure with "invented" data?
An alternative approach would be to impute the missing raw data using a missing data replacement procedure. You could then run the PCA on the correlation matrix that resulted from the imputed dataset (see also multiple imputation).
Here are a few links on mis... | Precedent for Bootstrap-like procedure with "invented" data?
An alternative approach would be to impute the missing raw data using a missing data replacement procedure. You could then run the PCA on the correlation matrix that resulted from the imputed dataset |
38,940 | Precedent for Bootstrap-like procedure with "invented" data? | I think we need to know more about the nature of the data to make recommendations on how to deal with the missing values. An exploratory task that jumps out to me is to look at the behavior of variables A through H when I is present, versus A through H when J is present. Is there anything interesting to take into accou... | Precedent for Bootstrap-like procedure with "invented" data? | I think we need to know more about the nature of the data to make recommendations on how to deal with the missing values. An exploratory task that jumps out to me is to look at the behavior of variabl | Precedent for Bootstrap-like procedure with "invented" data?
I think we need to know more about the nature of the data to make recommendations on how to deal with the missing values. An exploratory task that jumps out to me is to look at the behavior of variables A through H when I is present, versus A through H when J... | Precedent for Bootstrap-like procedure with "invented" data?
I think we need to know more about the nature of the data to make recommendations on how to deal with the missing values. An exploratory task that jumps out to me is to look at the behavior of variabl |
38,941 | Free treemapping software | Flowing Data has a tutorial on how to use the map.market function in the portfolio package in R. | Free treemapping software | Flowing Data has a tutorial on how to use the map.market function in the portfolio package in R. | Free treemapping software
Flowing Data has a tutorial on how to use the map.market function in the portfolio package in R. | Free treemapping software
Flowing Data has a tutorial on how to use the map.market function in the portfolio package in R. |
38,942 | Free treemapping software | Protovis has an nice treemap layout. I could try to add this into webvis if you want to create it from R, but it isn't currently an option. | Free treemapping software | Protovis has an nice treemap layout. I could try to add this into webvis if you want to create it from R, but it isn't currently an option. | Free treemapping software
Protovis has an nice treemap layout. I could try to add this into webvis if you want to create it from R, but it isn't currently an option. | Free treemapping software
Protovis has an nice treemap layout. I could try to add this into webvis if you want to create it from R, but it isn't currently an option. |
38,943 | Free treemapping software | While I haven't used either of these, I have these two projects bookmarked:
Treeviz
JTreeMap
Both are implemented in Java, are free and open source (MIT and Apache 2.0 licenses, respectively). Both appear to be actually libraries, but seem to come with an example application. | Free treemapping software | While I haven't used either of these, I have these two projects bookmarked:
Treeviz
JTreeMap
Both are implemented in Java, are free and open source (MIT and Apache 2.0 licenses, respectively). Both | Free treemapping software
While I haven't used either of these, I have these two projects bookmarked:
Treeviz
JTreeMap
Both are implemented in Java, are free and open source (MIT and Apache 2.0 licenses, respectively). Both appear to be actually libraries, but seem to come with an example application. | Free treemapping software
While I haven't used either of these, I have these two projects bookmarked:
Treeviz
JTreeMap
Both are implemented in Java, are free and open source (MIT and Apache 2.0 licenses, respectively). Both |
38,944 | Free treemapping software | While commercial, Macrofocus provides a free 30-days evaluation of their TreeMap software. | Free treemapping software | While commercial, Macrofocus provides a free 30-days evaluation of their TreeMap software. | Free treemapping software
While commercial, Macrofocus provides a free 30-days evaluation of their TreeMap software. | Free treemapping software
While commercial, Macrofocus provides a free 30-days evaluation of their TreeMap software. |
38,945 | Free treemapping software | The treemap package for R is superior - better look, much more flexible - to the map.market function in package portfolio featured in FlowingData's tutorial. | Free treemapping software | The treemap package for R is superior - better look, much more flexible - to the map.market function in package portfolio featured in FlowingData's tutorial. | Free treemapping software
The treemap package for R is superior - better look, much more flexible - to the map.market function in package portfolio featured in FlowingData's tutorial. | Free treemapping software
The treemap package for R is superior - better look, much more flexible - to the map.market function in package portfolio featured in FlowingData's tutorial. |
38,946 | What is the name of this normalization (which yields mean 0 and sd 1)? | The quantity $z = \frac{X - \mu}{\sigma}$ is a standard score. So, standardization is a common way to refer to it. | What is the name of this normalization (which yields mean 0 and sd 1)? | The quantity $z = \frac{X - \mu}{\sigma}$ is a standard score. So, standardization is a common way to refer to it. | What is the name of this normalization (which yields mean 0 and sd 1)?
The quantity $z = \frac{X - \mu}{\sigma}$ is a standard score. So, standardization is a common way to refer to it. | What is the name of this normalization (which yields mean 0 and sd 1)?
The quantity $z = \frac{X - \mu}{\sigma}$ is a standard score. So, standardization is a common way to refer to it. |
38,947 | What is the name of this normalization (which yields mean 0 and sd 1)? | I think it is just called z-score.
[@ttnphns remark: that is correct, however "z-score" also has other meanings in statistics.
z-standardization, z standard value - probably the most widely used terms for the linear transform to mean 0 and sd 1] | What is the name of this normalization (which yields mean 0 and sd 1)? | I think it is just called z-score.
[@ttnphns remark: that is correct, however "z-score" also has other meanings in statistics.
z-standardization, z standard value - probably the most widely used terms | What is the name of this normalization (which yields mean 0 and sd 1)?
I think it is just called z-score.
[@ttnphns remark: that is correct, however "z-score" also has other meanings in statistics.
z-standardization, z standard value - probably the most widely used terms for the linear transform to mean 0 and sd 1] | What is the name of this normalization (which yields mean 0 and sd 1)?
I think it is just called z-score.
[@ttnphns remark: that is correct, however "z-score" also has other meanings in statistics.
z-standardization, z standard value - probably the most widely used terms |
38,948 | What is the name of this normalization (which yields mean 0 and sd 1)? | As ars' answer states, standardization is the transformation that involves:
mean-centering, and
rescaling to unit variance.
A generalization of standardization is whitening or sphering whereby a set of one or more variables is linearly-transformed (typically after mean-centering) so that the covariance matrix is the ... | What is the name of this normalization (which yields mean 0 and sd 1)? | As ars' answer states, standardization is the transformation that involves:
mean-centering, and
rescaling to unit variance.
A generalization of standardization is whitening or sphering whereby a set | What is the name of this normalization (which yields mean 0 and sd 1)?
As ars' answer states, standardization is the transformation that involves:
mean-centering, and
rescaling to unit variance.
A generalization of standardization is whitening or sphering whereby a set of one or more variables is linearly-transformed... | What is the name of this normalization (which yields mean 0 and sd 1)?
As ars' answer states, standardization is the transformation that involves:
mean-centering, and
rescaling to unit variance.
A generalization of standardization is whitening or sphering whereby a set |
38,949 | What is the name of this normalization (which yields mean 0 and sd 1)? | You have to say it was "mean-zero standardized." Dividing by the original calculated standard deviation, $\sigma$, results in a standardized variate with a standard deviation of 1. It has mean zero since it was "centered" (the average was subtracted as well). It's also not normalizing since normalizing implies the f... | What is the name of this normalization (which yields mean 0 and sd 1)? | You have to say it was "mean-zero standardized." Dividing by the original calculated standard deviation, $\sigma$, results in a standardized variate with a standard deviation of 1. It has mean zero | What is the name of this normalization (which yields mean 0 and sd 1)?
You have to say it was "mean-zero standardized." Dividing by the original calculated standard deviation, $\sigma$, results in a standardized variate with a standard deviation of 1. It has mean zero since it was "centered" (the average was subtract... | What is the name of this normalization (which yields mean 0 and sd 1)?
You have to say it was "mean-zero standardized." Dividing by the original calculated standard deviation, $\sigma$, results in a standardized variate with a standard deviation of 1. It has mean zero |
38,950 | Chi squared for goodnes of fit test always rejects my fits | You are keeping bins with 0 counts and assigning them an uncertainty of 0. When you measure 0 counts, 0 is a poor estimate of the uncertainty. 0 is a valid outcome even when you expect a few counts. Common practices are to remove those bins or assign them a non-zero value for the std (I would use 1.14 since it's when y... | Chi squared for goodnes of fit test always rejects my fits | You are keeping bins with 0 counts and assigning them an uncertainty of 0. When you measure 0 counts, 0 is a poor estimate of the uncertainty. 0 is a valid outcome even when you expect a few counts. C | Chi squared for goodnes of fit test always rejects my fits
You are keeping bins with 0 counts and assigning them an uncertainty of 0. When you measure 0 counts, 0 is a poor estimate of the uncertainty. 0 is a valid outcome even when you expect a few counts. Common practices are to remove those bins or assign them a non... | Chi squared for goodnes of fit test always rejects my fits
You are keeping bins with 0 counts and assigning them an uncertainty of 0. When you measure 0 counts, 0 is a poor estimate of the uncertainty. 0 is a valid outcome even when you expect a few counts. C |
38,951 | Chi squared for goodnes of fit test always rejects my fits | In a way, the "eye test" measures an effect size but not a p-value. Looking at the histogram versus the fit line can give you a sense of "how far" the two are apart, but the p-value will depend very much on the sample size represented in the graph, which is not as immediately obvious. With sufficient sample size, arbit... | Chi squared for goodnes of fit test always rejects my fits | In a way, the "eye test" measures an effect size but not a p-value. Looking at the histogram versus the fit line can give you a sense of "how far" the two are apart, but the p-value will depend very m | Chi squared for goodnes of fit test always rejects my fits
In a way, the "eye test" measures an effect size but not a p-value. Looking at the histogram versus the fit line can give you a sense of "how far" the two are apart, but the p-value will depend very much on the sample size represented in the graph, which is not... | Chi squared for goodnes of fit test always rejects my fits
In a way, the "eye test" measures an effect size but not a p-value. Looking at the histogram versus the fit line can give you a sense of "how far" the two are apart, but the p-value will depend very m |
38,952 | Chi squared for goodnes of fit test always rejects my fits | Your fit is pretty good. It is not perfect. In some sense, you are correct that your eye calibration is off, but that is not such a terrible issue. Basically every model is at least a little bit wrong. With sufficient data, the slightest of deviations can be caught. However, the eye test is often a powerful and useful ... | Chi squared for goodnes of fit test always rejects my fits | Your fit is pretty good. It is not perfect. In some sense, you are correct that your eye calibration is off, but that is not such a terrible issue. Basically every model is at least a little bit wrong | Chi squared for goodnes of fit test always rejects my fits
Your fit is pretty good. It is not perfect. In some sense, you are correct that your eye calibration is off, but that is not such a terrible issue. Basically every model is at least a little bit wrong. With sufficient data, the slightest of deviations can be ca... | Chi squared for goodnes of fit test always rejects my fits
Your fit is pretty good. It is not perfect. In some sense, you are correct that your eye calibration is off, but that is not such a terrible issue. Basically every model is at least a little bit wrong |
38,953 | Probability of one horse finishing ahead of another [duplicate] | This is similar to Christian Hennig's answer.
If you make the strong assumption that the probabilities are in effect weights, and the first horse is sampled with probabilities proportional to the weights of all the horses, the second horse sampled with probabilities proportional to the weights of all the horses except ... | Probability of one horse finishing ahead of another [duplicate] | This is similar to Christian Hennig's answer.
If you make the strong assumption that the probabilities are in effect weights, and the first horse is sampled with probabilities proportional to the weig | Probability of one horse finishing ahead of another [duplicate]
This is similar to Christian Hennig's answer.
If you make the strong assumption that the probabilities are in effect weights, and the first horse is sampled with probabilities proportional to the weights of all the horses, the second horse sampled with pro... | Probability of one horse finishing ahead of another [duplicate]
This is similar to Christian Hennig's answer.
If you make the strong assumption that the probabilities are in effect weights, and the first horse is sampled with probabilities proportional to the weig |
38,954 | Probability of one horse finishing ahead of another [duplicate] | The given information is not enough to compute these probabilities in general, because it may be that for example Horse C is of a kind that it either wins or gets frustrated and finishes last. But it may also just be the best horse and if it doesn't win it may be very likely that it comes second. Which of these is the ... | Probability of one horse finishing ahead of another [duplicate] | The given information is not enough to compute these probabilities in general, because it may be that for example Horse C is of a kind that it either wins or gets frustrated and finishes last. But it | Probability of one horse finishing ahead of another [duplicate]
The given information is not enough to compute these probabilities in general, because it may be that for example Horse C is of a kind that it either wins or gets frustrated and finishes last. But it may also just be the best horse and if it doesn't win it... | Probability of one horse finishing ahead of another [duplicate]
The given information is not enough to compute these probabilities in general, because it may be that for example Horse C is of a kind that it either wins or gets frustrated and finishes last. But it |
38,955 | Probability of one horse finishing ahead of another [duplicate] | Let $p_i$, $i=1,2,\dots,n$ denote the probabilities that each horse wins given in the problem statement. A model that leads to simple calculations is to assume that some monotonically decreasing transformation $Z_i$ of the race times $T_i$ follow exponential distributions with rate parameters $\lambda_i=p_i$. For ins... | Probability of one horse finishing ahead of another [duplicate] | Let $p_i$, $i=1,2,\dots,n$ denote the probabilities that each horse wins given in the problem statement. A model that leads to simple calculations is to assume that some monotonically decreasing tran | Probability of one horse finishing ahead of another [duplicate]
Let $p_i$, $i=1,2,\dots,n$ denote the probabilities that each horse wins given in the problem statement. A model that leads to simple calculations is to assume that some monotonically decreasing transformation $Z_i$ of the race times $T_i$ follow exponent... | Probability of one horse finishing ahead of another [duplicate]
Let $p_i$, $i=1,2,\dots,n$ denote the probabilities that each horse wins given in the problem statement. A model that leads to simple calculations is to assume that some monotonically decreasing tran |
38,956 | How to obtain the model coefficients of GAM in R? | The model you fitted with {mgcv} is
$$
y_i = \alpha + f(x_i) + \varepsilon_i
$$
not
$$
y_i = \alpha + \beta \cdot f(x_i) + \varepsilon_i
$$
So that makes me a little unsure as to what $\beta$ you want. In the {mgcv} formulation, the smooth function $f$ is parameterised as a penalised spline and represented as a sum of ... | How to obtain the model coefficients of GAM in R? | The model you fitted with {mgcv} is
$$
y_i = \alpha + f(x_i) + \varepsilon_i
$$
not
$$
y_i = \alpha + \beta \cdot f(x_i) + \varepsilon_i
$$
So that makes me a little unsure as to what $\beta$ you want | How to obtain the model coefficients of GAM in R?
The model you fitted with {mgcv} is
$$
y_i = \alpha + f(x_i) + \varepsilon_i
$$
not
$$
y_i = \alpha + \beta \cdot f(x_i) + \varepsilon_i
$$
So that makes me a little unsure as to what $\beta$ you want. In the {mgcv} formulation, the smooth function $f$ is parameterised ... | How to obtain the model coefficients of GAM in R?
The model you fitted with {mgcv} is
$$
y_i = \alpha + f(x_i) + \varepsilon_i
$$
not
$$
y_i = \alpha + \beta \cdot f(x_i) + \varepsilon_i
$$
So that makes me a little unsure as to what $\beta$ you want |
38,957 | Why does Hutchinson's trace estimator reduce computation complexity? | You are right that for calculating the trace of a matrix this does not reduce cost vs a simple calculation...but this trick is very useful when we need to compute the trace of a function of a matrix, $tr(f(A))$.
For example, consider estimating a log determinant, which can be rewritten as $tr( log(A) )$, the trace of t... | Why does Hutchinson's trace estimator reduce computation complexity? | You are right that for calculating the trace of a matrix this does not reduce cost vs a simple calculation...but this trick is very useful when we need to compute the trace of a function of a matrix, | Why does Hutchinson's trace estimator reduce computation complexity?
You are right that for calculating the trace of a matrix this does not reduce cost vs a simple calculation...but this trick is very useful when we need to compute the trace of a function of a matrix, $tr(f(A))$.
For example, consider estimating a log ... | Why does Hutchinson's trace estimator reduce computation complexity?
You are right that for calculating the trace of a matrix this does not reduce cost vs a simple calculation...but this trick is very useful when we need to compute the trace of a function of a matrix, |
38,958 | Why does Hutchinson's trace estimator reduce computation complexity? | I have never heard of this trick before, and it does look strange ... since the trace is a sum of the $n$ diagonal terms, while the quadratic form needs a double sum over all the $n^2$ terms. So I guess the "trick" must be used only as one idea in combination with others, for some more involved problem than just a trac... | Why does Hutchinson's trace estimator reduce computation complexity? | I have never heard of this trick before, and it does look strange ... since the trace is a sum of the $n$ diagonal terms, while the quadratic form needs a double sum over all the $n^2$ terms. So I gue | Why does Hutchinson's trace estimator reduce computation complexity?
I have never heard of this trick before, and it does look strange ... since the trace is a sum of the $n$ diagonal terms, while the quadratic form needs a double sum over all the $n^2$ terms. So I guess the "trick" must be used only as one idea in com... | Why does Hutchinson's trace estimator reduce computation complexity?
I have never heard of this trick before, and it does look strange ... since the trace is a sum of the $n$ diagonal terms, while the quadratic form needs a double sum over all the $n^2$ terms. So I gue |
38,959 | Why does Hutchinson's trace estimator reduce computation complexity? | It depends a lot on what accuracy you need as well as the form of the matrix. In the application I had, there was a matrix $A=GP$ that was the product of a large (say $N\times M$) sparse matrix $G$ of genotypes and a projection matrix $P$ on to the residual space of a linear regression. This means $A$ wasn't itself sp... | Why does Hutchinson's trace estimator reduce computation complexity? | It depends a lot on what accuracy you need as well as the form of the matrix. In the application I had, there was a matrix $A=GP$ that was the product of a large (say $N\times M$) sparse matrix $G$ o | Why does Hutchinson's trace estimator reduce computation complexity?
It depends a lot on what accuracy you need as well as the form of the matrix. In the application I had, there was a matrix $A=GP$ that was the product of a large (say $N\times M$) sparse matrix $G$ of genotypes and a projection matrix $P$ on to the r... | Why does Hutchinson's trace estimator reduce computation complexity?
It depends a lot on what accuracy you need as well as the form of the matrix. In the application I had, there was a matrix $A=GP$ that was the product of a large (say $N\times M$) sparse matrix $G$ o |
38,960 | what is the difference between a two-sample t-test and a paired t-test | In a two-sample test, you have two independent samples. Of course, by independence, we expect the two sets of measurements will not be correlated. Sample sizes for the two samples need not be equal. (But it often makes sense for them to be approximately equal.)
In a paired test you have one sample of pairs. Typically, ... | what is the difference between a two-sample t-test and a paired t-test | In a two-sample test, you have two independent samples. Of course, by independence, we expect the two sets of measurements will not be correlated. Sample sizes for the two samples need not be equal. ( | what is the difference between a two-sample t-test and a paired t-test
In a two-sample test, you have two independent samples. Of course, by independence, we expect the two sets of measurements will not be correlated. Sample sizes for the two samples need not be equal. (But it often makes sense for them to be approxima... | what is the difference between a two-sample t-test and a paired t-test
In a two-sample test, you have two independent samples. Of course, by independence, we expect the two sets of measurements will not be correlated. Sample sizes for the two samples need not be equal. ( |
38,961 | what is the difference between a two-sample t-test and a paired t-test | When you use a paired T-test, you are essentially doing a one-sample test, where your one sample consists of the paired differences between outcomes in two groups. If you create a new sample of these difference values and then apply the formula for a one-sample T-test, you will see that this is equivalent to the paire... | what is the difference between a two-sample t-test and a paired t-test | When you use a paired T-test, you are essentially doing a one-sample test, where your one sample consists of the paired differences between outcomes in two groups. If you create a new sample of these | what is the difference between a two-sample t-test and a paired t-test
When you use a paired T-test, you are essentially doing a one-sample test, where your one sample consists of the paired differences between outcomes in two groups. If you create a new sample of these difference values and then apply the formula for... | what is the difference between a two-sample t-test and a paired t-test
When you use a paired T-test, you are essentially doing a one-sample test, where your one sample consists of the paired differences between outcomes in two groups. If you create a new sample of these |
38,962 | Applying Wilks' theorem to uniform distribution | From the Wikipedia page for Wilks' theorem:
The theorem no longer applies when any one of the estimated parameters is at its upper or lower limit: Wilks’ theorem assumes that the ‘true’ but unknown values of the estimated parameters lie within the interior of the supported parameter space. The likelihood maximum may n... | Applying Wilks' theorem to uniform distribution | From the Wikipedia page for Wilks' theorem:
The theorem no longer applies when any one of the estimated parameters is at its upper or lower limit: Wilks’ theorem assumes that the ‘true’ but unknown v | Applying Wilks' theorem to uniform distribution
From the Wikipedia page for Wilks' theorem:
The theorem no longer applies when any one of the estimated parameters is at its upper or lower limit: Wilks’ theorem assumes that the ‘true’ but unknown values of the estimated parameters lie within the interior of the support... | Applying Wilks' theorem to uniform distribution
From the Wikipedia page for Wilks' theorem:
The theorem no longer applies when any one of the estimated parameters is at its upper or lower limit: Wilks’ theorem assumes that the ‘true’ but unknown v |
38,963 | Applying Wilks' theorem to uniform distribution | Wilk's theorem assumes that the support of the distribution, i.e., the sample space, does not depend on the unknown parameter. Here, the parameter $b$ determines the width of the sample space so Wilk's theorem cannot be applied.
All the standard asymptotic tests associated with likelihood theory (Wald tests, likelihood... | Applying Wilks' theorem to uniform distribution | Wilk's theorem assumes that the support of the distribution, i.e., the sample space, does not depend on the unknown parameter. Here, the parameter $b$ determines the width of the sample space so Wilk' | Applying Wilks' theorem to uniform distribution
Wilk's theorem assumes that the support of the distribution, i.e., the sample space, does not depend on the unknown parameter. Here, the parameter $b$ determines the width of the sample space so Wilk's theorem cannot be applied.
All the standard asymptotic tests associate... | Applying Wilks' theorem to uniform distribution
Wilk's theorem assumes that the support of the distribution, i.e., the sample space, does not depend on the unknown parameter. Here, the parameter $b$ determines the width of the sample space so Wilk' |
38,964 | Applying Wilks' theorem to uniform distribution | Wilks' theorem does not apply here
Before getting to Wilks' theorem, one problem here (as with many questions about uniform distributions on this site) is that you have not taken account of the support of the uniform distribution in your mathematics, and this leads you to an incorrect form for the likelihood-ratio stat... | Applying Wilks' theorem to uniform distribution | Wilks' theorem does not apply here
Before getting to Wilks' theorem, one problem here (as with many questions about uniform distributions on this site) is that you have not taken account of the suppor | Applying Wilks' theorem to uniform distribution
Wilks' theorem does not apply here
Before getting to Wilks' theorem, one problem here (as with many questions about uniform distributions on this site) is that you have not taken account of the support of the uniform distribution in your mathematics, and this leads you to... | Applying Wilks' theorem to uniform distribution
Wilks' theorem does not apply here
Before getting to Wilks' theorem, one problem here (as with many questions about uniform distributions on this site) is that you have not taken account of the suppor |
38,965 | Is (covariance) stationarity preserved under log or exponential transformation? | The underlying idea is that equality of moments does not survive nonlinear transformations. In particular, when two variables $X$ and $Y$ have the same moments (up to some finite order) and $f$ is a nonlinear transformation, there is no assurance that $f(X)$ and $f(Y)$ will have any of the same moments.
Thus, when the... | Is (covariance) stationarity preserved under log or exponential transformation? | The underlying idea is that equality of moments does not survive nonlinear transformations. In particular, when two variables $X$ and $Y$ have the same moments (up to some finite order) and $f$ is a | Is (covariance) stationarity preserved under log or exponential transformation?
The underlying idea is that equality of moments does not survive nonlinear transformations. In particular, when two variables $X$ and $Y$ have the same moments (up to some finite order) and $f$ is a nonlinear transformation, there is no as... | Is (covariance) stationarity preserved under log or exponential transformation?
The underlying idea is that equality of moments does not survive nonlinear transformations. In particular, when two variables $X$ and $Y$ have the same moments (up to some finite order) and $f$ is a |
38,966 | Is (covariance) stationarity preserved under log or exponential transformation? | Weak stationarity is generally not preserved under exponential or logarithmic transformations.
Exponential transformation
Covariance (or weak) stationarity requires the second moment to be finite. If a random variable has a finite second moment, it is not guaranteed that the second (or even first) moment of its exponen... | Is (covariance) stationarity preserved under log or exponential transformation? | Weak stationarity is generally not preserved under exponential or logarithmic transformations.
Exponential transformation
Covariance (or weak) stationarity requires the second moment to be finite. If | Is (covariance) stationarity preserved under log or exponential transformation?
Weak stationarity is generally not preserved under exponential or logarithmic transformations.
Exponential transformation
Covariance (or weak) stationarity requires the second moment to be finite. If a random variable has a finite second mo... | Is (covariance) stationarity preserved under log or exponential transformation?
Weak stationarity is generally not preserved under exponential or logarithmic transformations.
Exponential transformation
Covariance (or weak) stationarity requires the second moment to be finite. If |
38,967 | Distribution of the ratio of a Normal distribution divided by Lognormal distribution | Finding a closed-form distribution for the product or ratio looks rather challenging, as even simple cases (zero mean and unit standard deviation) do not appear to produce closed forms.
The other part of the question ... to find the moments of $Z = X/Y$ ... is readily solvable, assuming that $X$ and $Y$ are independen... | Distribution of the ratio of a Normal distribution divided by Lognormal distribution | Finding a closed-form distribution for the product or ratio looks rather challenging, as even simple cases (zero mean and unit standard deviation) do not appear to produce closed forms.
The other part | Distribution of the ratio of a Normal distribution divided by Lognormal distribution
Finding a closed-form distribution for the product or ratio looks rather challenging, as even simple cases (zero mean and unit standard deviation) do not appear to produce closed forms.
The other part of the question ... to find the mo... | Distribution of the ratio of a Normal distribution divided by Lognormal distribution
Finding a closed-form distribution for the product or ratio looks rather challenging, as even simple cases (zero mean and unit standard deviation) do not appear to produce closed forms.
The other part |
38,968 | Distribution of the ratio of a Normal distribution divided by Lognormal distribution | The density of $Z$ is
$$f(z)=\frac{1}{2\pi\sigma_x\sigma_y}\int_0^\infty \exp\{ -[zy-\mu_x]^2/2\sigma_x^2-[\log(y)-\mu_y]^2/2\sigma_y^2\}~\text dy$$
since the Jacobian of turning $(x,y)$ into $(zy,y)$ is $y$. | Distribution of the ratio of a Normal distribution divided by Lognormal distribution | The density of $Z$ is
$$f(z)=\frac{1}{2\pi\sigma_x\sigma_y}\int_0^\infty \exp\{ -[zy-\mu_x]^2/2\sigma_x^2-[\log(y)-\mu_y]^2/2\sigma_y^2\}~\text dy$$
since the Jacobian of turning $(x,y)$ into $(zy,y)$ | Distribution of the ratio of a Normal distribution divided by Lognormal distribution
The density of $Z$ is
$$f(z)=\frac{1}{2\pi\sigma_x\sigma_y}\int_0^\infty \exp\{ -[zy-\mu_x]^2/2\sigma_x^2-[\log(y)-\mu_y]^2/2\sigma_y^2\}~\text dy$$
since the Jacobian of turning $(x,y)$ into $(zy,y)$ is $y$. | Distribution of the ratio of a Normal distribution divided by Lognormal distribution
The density of $Z$ is
$$f(z)=\frac{1}{2\pi\sigma_x\sigma_y}\int_0^\infty \exp\{ -[zy-\mu_x]^2/2\sigma_x^2-[\log(y)-\mu_y]^2/2\sigma_y^2\}~\text dy$$
since the Jacobian of turning $(x,y)$ into $(zy,y)$ |
38,969 | Using two sensors instead of one: Statistically meaningful? | Intuitively, if the noise and measurement errors from your sensors are independent and have zero mean, then using multiple sensors will improve the signal to noise ratio because the noise and errors will tend to cancel each other out as you increase the number of sensors.
There is an interesting parallel to be made wit... | Using two sensors instead of one: Statistically meaningful? | Intuitively, if the noise and measurement errors from your sensors are independent and have zero mean, then using multiple sensors will improve the signal to noise ratio because the noise and errors w | Using two sensors instead of one: Statistically meaningful?
Intuitively, if the noise and measurement errors from your sensors are independent and have zero mean, then using multiple sensors will improve the signal to noise ratio because the noise and errors will tend to cancel each other out as you increase the number... | Using two sensors instead of one: Statistically meaningful?
Intuitively, if the noise and measurement errors from your sensors are independent and have zero mean, then using multiple sensors will improve the signal to noise ratio because the noise and errors w |
38,970 | Using two sensors instead of one: Statistically meaningful? | Sensor fusion via a Kalman filter takes 2 (or more) separate measurements, combines them within the filter calculations, and outputs a smoother/more accurate measurement of the underlying.
Some good stuff online about this, e.g. https://simondlevy.academic.wlu.edu/kalman-tutorial/the-extended-kalman-filter-an-interacti... | Using two sensors instead of one: Statistically meaningful? | Sensor fusion via a Kalman filter takes 2 (or more) separate measurements, combines them within the filter calculations, and outputs a smoother/more accurate measurement of the underlying.
Some good s | Using two sensors instead of one: Statistically meaningful?
Sensor fusion via a Kalman filter takes 2 (or more) separate measurements, combines them within the filter calculations, and outputs a smoother/more accurate measurement of the underlying.
Some good stuff online about this, e.g. https://simondlevy.academic.wlu... | Using two sensors instead of one: Statistically meaningful?
Sensor fusion via a Kalman filter takes 2 (or more) separate measurements, combines them within the filter calculations, and outputs a smoother/more accurate measurement of the underlying.
Some good s |
38,971 | What is the variance of Y = AX where A is a matrix? | This (linear transform) is typically listed as a property of covariance, but easy to show as well:
$$\begin{align}\operatorname{cov}(AX)&=\mathbb E[AXX^TA^T]-\mathbb E[AX]\mathbb E[X^TA^T]\\&=A\mathbb E[XX^T]A^T-A\mathbb E[X]\mathbb E[X^T]A^T\\&=A(\mathbb E[XX^T]-\mathbb E[X]\mathbb E[X^T])A^T\\&=ADA^T\end{align}$$ | What is the variance of Y = AX where A is a matrix? | This (linear transform) is typically listed as a property of covariance, but easy to show as well:
$$\begin{align}\operatorname{cov}(AX)&=\mathbb E[AXX^TA^T]-\mathbb E[AX]\mathbb E[X^TA^T]\\&=A\mathbb | What is the variance of Y = AX where A is a matrix?
This (linear transform) is typically listed as a property of covariance, but easy to show as well:
$$\begin{align}\operatorname{cov}(AX)&=\mathbb E[AXX^TA^T]-\mathbb E[AX]\mathbb E[X^TA^T]\\&=A\mathbb E[XX^T]A^T-A\mathbb E[X]\mathbb E[X^T]A^T\\&=A(\mathbb E[XX^T]-\mat... | What is the variance of Y = AX where A is a matrix?
This (linear transform) is typically listed as a property of covariance, but easy to show as well:
$$\begin{align}\operatorname{cov}(AX)&=\mathbb E[AXX^TA^T]-\mathbb E[AX]\mathbb E[X^TA^T]\\&=A\mathbb |
38,972 | For a time series problem, Why is it preferrable to use a time series model over a model without an explicit time component? | Yes, in principle your "classical" approaches would catch periodicities and autocorrelations as well. Fitting an AR time series model is not that much different from OLS regressing the actuals on lagged values of the actuals, after all. However:
Suppose you run a standard linear regression with day, month and year as ... | For a time series problem, Why is it preferrable to use a time series model over a model without an | Yes, in principle your "classical" approaches would catch periodicities and autocorrelations as well. Fitting an AR time series model is not that much different from OLS regressing the actuals on lagg | For a time series problem, Why is it preferrable to use a time series model over a model without an explicit time component?
Yes, in principle your "classical" approaches would catch periodicities and autocorrelations as well. Fitting an AR time series model is not that much different from OLS regressing the actuals on... | For a time series problem, Why is it preferrable to use a time series model over a model without an
Yes, in principle your "classical" approaches would catch periodicities and autocorrelations as well. Fitting an AR time series model is not that much different from OLS regressing the actuals on lagg |
38,973 | For a time series problem, Why is it preferrable to use a time series model over a model without an explicit time component? | Superficially, the time series approach is more convenient mathematically, whereas the human date is more convenient for presenting the data/results.
On a deeper level, as you correctly noted, one could use day, month and year as three independent variables, particularly, especially if there are reasons to think that t... | For a time series problem, Why is it preferrable to use a time series model over a model without an | Superficially, the time series approach is more convenient mathematically, whereas the human date is more convenient for presenting the data/results.
On a deeper level, as you correctly noted, one cou | For a time series problem, Why is it preferrable to use a time series model over a model without an explicit time component?
Superficially, the time series approach is more convenient mathematically, whereas the human date is more convenient for presenting the data/results.
On a deeper level, as you correctly noted, on... | For a time series problem, Why is it preferrable to use a time series model over a model without an
Superficially, the time series approach is more convenient mathematically, whereas the human date is more convenient for presenting the data/results.
On a deeper level, as you correctly noted, one cou |
38,974 | For a time series problem, Why is it preferrable to use a time series model over a model without an explicit time component? | If you think that the impact of X on y changes over time regression will not work (well, OLS won't). Autocorrelation is another problem inherent in time series data although special SE have been developed to address that. If Y influences itself over time then linear regression (non-time series) is not going to work cor... | For a time series problem, Why is it preferrable to use a time series model over a model without an | If you think that the impact of X on y changes over time regression will not work (well, OLS won't). Autocorrelation is another problem inherent in time series data although special SE have been devel | For a time series problem, Why is it preferrable to use a time series model over a model without an explicit time component?
If you think that the impact of X on y changes over time regression will not work (well, OLS won't). Autocorrelation is another problem inherent in time series data although special SE have been ... | For a time series problem, Why is it preferrable to use a time series model over a model without an
If you think that the impact of X on y changes over time regression will not work (well, OLS won't). Autocorrelation is another problem inherent in time series data although special SE have been devel |
38,975 | What math classes are relevant for machine learning? [closed] | A great professional of machine learning should know a mix of the ingredients that make up the Computer Scientist and Statistician. Furthermore, It should also know some math to be able to realy learn these subjects.
Math
1) Calculus in $\mathbb{R}^n$ and Optimization (diferential calculus, interior optimization, prope... | What math classes are relevant for machine learning? [closed] | A great professional of machine learning should know a mix of the ingredients that make up the Computer Scientist and Statistician. Furthermore, It should also know some math to be able to realy learn | What math classes are relevant for machine learning? [closed]
A great professional of machine learning should know a mix of the ingredients that make up the Computer Scientist and Statistician. Furthermore, It should also know some math to be able to realy learn these subjects.
Math
1) Calculus in $\mathbb{R}^n$ and Op... | What math classes are relevant for machine learning? [closed]
A great professional of machine learning should know a mix of the ingredients that make up the Computer Scientist and Statistician. Furthermore, It should also know some math to be able to realy learn |
38,976 | Right-skewed distribution with mean equals to mode? | Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $\times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution... | Right-skewed distribution with mean equals to mode? | Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success | Right-skewed distribution with mean equals to mode?
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $\times$ 0.1 = 1, and 1 also is the mode (a... | Right-skewed distribution with mean equals to mode?
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success |
38,977 | Right-skewed distribution with mean equals to mode? | If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1. | Right-skewed distribution with mean equals to mode? | If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is r | Right-skewed distribution with mean equals to mode?
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1. | Right-skewed distribution with mean equals to mode?
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is r |
38,978 | What are the benefits of layer-specific learning rates? | Here is a great article on this topic:
Basically, it usually makes training faster. The first layers are usually good enough, so the learning rate for them could be lower, but the last layers need to be tuned for out dataset, so lr should be higher. | What are the benefits of layer-specific learning rates? | Here is a great article on this topic:
Basically, it usually makes training faster. The first layers are usually good enough, so the learning rate for them could be lower, but the last layers need to | What are the benefits of layer-specific learning rates?
Here is a great article on this topic:
Basically, it usually makes training faster. The first layers are usually good enough, so the learning rate for them could be lower, but the last layers need to be tuned for out dataset, so lr should be higher. | What are the benefits of layer-specific learning rates?
Here is a great article on this topic:
Basically, it usually makes training faster. The first layers are usually good enough, so the learning rate for them could be lower, but the last layers need to |
38,979 | What are the benefits of layer-specific learning rates? | The layer-specific learning rates help in overcoming the slow learning (thus slow training) problem in deep neural networks. As stated in the paper Layer-Specific Adaptive Learning Rates for Deep Networks:
When the gradient descent methods are used to train deep networks, additional problems are introduced. As the num... | What are the benefits of layer-specific learning rates? | The layer-specific learning rates help in overcoming the slow learning (thus slow training) problem in deep neural networks. As stated in the paper Layer-Specific Adaptive Learning Rates for Deep Netw | What are the benefits of layer-specific learning rates?
The layer-specific learning rates help in overcoming the slow learning (thus slow training) problem in deep neural networks. As stated in the paper Layer-Specific Adaptive Learning Rates for Deep Networks:
When the gradient descent methods are used to train deep ... | What are the benefits of layer-specific learning rates?
The layer-specific learning rates help in overcoming the slow learning (thus slow training) problem in deep neural networks. As stated in the paper Layer-Specific Adaptive Learning Rates for Deep Netw |
38,980 | What does the angle bracket mean in variance formula? | It's the expected value of $(X-\mu)^2$, i.e., it's the same as $\sigma^2=E[(X-\mu)^2]$. | What does the angle bracket mean in variance formula? | It's the expected value of $(X-\mu)^2$, i.e., it's the same as $\sigma^2=E[(X-\mu)^2]$. | What does the angle bracket mean in variance formula?
It's the expected value of $(X-\mu)^2$, i.e., it's the same as $\sigma^2=E[(X-\mu)^2]$. | What does the angle bracket mean in variance formula?
It's the expected value of $(X-\mu)^2$, i.e., it's the same as $\sigma^2=E[(X-\mu)^2]$. |
38,981 | What does the angle bracket mean in variance formula? | It means an inner product for the multi-dimensional case. When $X \in \mathbb{R}^n$ and $n \geq 2$ and want to define variance, the definition of the variance is related to the inner product of $X-\mu$ to itself, and denoted as $\langle X-\mu, X-\mu\rangle$ | What does the angle bracket mean in variance formula? | It means an inner product for the multi-dimensional case. When $X \in \mathbb{R}^n$ and $n \geq 2$ and want to define variance, the definition of the variance is related to the inner product of $X-\mu | What does the angle bracket mean in variance formula?
It means an inner product for the multi-dimensional case. When $X \in \mathbb{R}^n$ and $n \geq 2$ and want to define variance, the definition of the variance is related to the inner product of $X-\mu$ to itself, and denoted as $\langle X-\mu, X-\mu\rangle$ | What does the angle bracket mean in variance formula?
It means an inner product for the multi-dimensional case. When $X \in \mathbb{R}^n$ and $n \geq 2$ and want to define variance, the definition of the variance is related to the inner product of $X-\mu |
38,982 | How do you multiply two conditional probabilities? | Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $\times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two... | How do you multiply two conditional probabilities? | Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $\times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are sca | How do you multiply two conditional probabilities?
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $\times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply mu... | How do you multiply two conditional probabilities?
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $\times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are sca |
38,983 | How do you multiply two conditional probabilities? | If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) =... | How do you multiply two conditional probabilities? | If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for | How do you multiply two conditional probabilities?
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b... | How do you multiply two conditional probabilities?
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for |
38,984 | Subtracting very small probabilities - How to compute? [duplicate] | To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:
$$\begin{equation} \begin{aligned}
\exp(\ell_1) - \exp(\ell_2)
&= \exp(\ell_1) (1 - \exp(-(\ell_1 - \ell_2))). \\[6pt]
\end{aligned} \end{equation}$$
This result converts the difference ... | Subtracting very small probabilities - How to compute? [duplicate] | To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:
$$\begin{equation} \begin{aligned}
\exp(\ell_1) - \exp(\ell_2)
&= | Subtracting very small probabilities - How to compute? [duplicate]
To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:
$$\begin{equation} \begin{aligned}
\exp(\ell_1) - \exp(\ell_2)
&= \exp(\ell_1) (1 - \exp(-(\ell_1 - \ell_2))). \\[6pt]
... | Subtracting very small probabilities - How to compute? [duplicate]
To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:
$$\begin{equation} \begin{aligned}
\exp(\ell_1) - \exp(\ell_2)
&= |
38,985 | Subtracting very small probabilities - How to compute? [duplicate] | The following workaround is often very useful for these sorts of problems:
subtract the smaller of l1 and l2 from each of l1 and l2 (effectively, we have multiplied the probabilities by some constant z = exp(min(l1,l2)))
Now compute the sum using standard functions (you can use log1p and expm1 if you want).
Afterwards... | Subtracting very small probabilities - How to compute? [duplicate] | The following workaround is often very useful for these sorts of problems:
subtract the smaller of l1 and l2 from each of l1 and l2 (effectively, we have multiplied the probabilities by some constant | Subtracting very small probabilities - How to compute? [duplicate]
The following workaround is often very useful for these sorts of problems:
subtract the smaller of l1 and l2 from each of l1 and l2 (effectively, we have multiplied the probabilities by some constant z = exp(min(l1,l2)))
Now compute the sum using stand... | Subtracting very small probabilities - How to compute? [duplicate]
The following workaround is often very useful for these sorts of problems:
subtract the smaller of l1 and l2 from each of l1 and l2 (effectively, we have multiplied the probabilities by some constant |
38,986 | Likelihood function when $X\sim U(0,\theta)$ | The fact that all the observable values $X_1,...,X_n$ are all non-negative is generally considered to be implicit throughout the working (since this is the support of the original distribution) and so it is not stated explicitly. You can certainly add it in explicitly if you wish, which is what you are doing in the c... | Likelihood function when $X\sim U(0,\theta)$ | The fact that all the observable values $X_1,...,X_n$ are all non-negative is generally considered to be implicit throughout the working (since this is the support of the original distribution) and s | Likelihood function when $X\sim U(0,\theta)$
The fact that all the observable values $X_1,...,X_n$ are all non-negative is generally considered to be implicit throughout the working (since this is the support of the original distribution) and so it is not stated explicitly. You can certainly add it in explicitly if y... | Likelihood function when $X\sim U(0,\theta)$
The fact that all the observable values $X_1,...,X_n$ are all non-negative is generally considered to be implicit throughout the working (since this is the support of the original distribution) and s |
38,987 | Likelihood function when $X\sim U(0,\theta)$ | You are correct that the third equality could use the event $\{\min(X_1, \ldots, X_n)\geq 0, \max(X_1, \ldots, X_n)\leq\theta\}$.
The more compact notation is used because for any value of $\theta$, $\min(X_1, \ldots, X_n)\geq 0$ almost surely, so that event is not informative. The only useful event is $\max(X_1, \ldot... | Likelihood function when $X\sim U(0,\theta)$ | You are correct that the third equality could use the event $\{\min(X_1, \ldots, X_n)\geq 0, \max(X_1, \ldots, X_n)\leq\theta\}$.
The more compact notation is used because for any value of $\theta$, $ | Likelihood function when $X\sim U(0,\theta)$
You are correct that the third equality could use the event $\{\min(X_1, \ldots, X_n)\geq 0, \max(X_1, \ldots, X_n)\leq\theta\}$.
The more compact notation is used because for any value of $\theta$, $\min(X_1, \ldots, X_n)\geq 0$ almost surely, so that event is not informati... | Likelihood function when $X\sim U(0,\theta)$
You are correct that the third equality could use the event $\{\min(X_1, \ldots, X_n)\geq 0, \max(X_1, \ldots, X_n)\leq\theta\}$.
The more compact notation is used because for any value of $\theta$, $ |
38,988 | Likelihood function when $X\sim U(0,\theta)$ | Just to stress a point, namely that the notion of likelihood is one of a function of $\theta$ indexed by the sample $(x_1,\ldots,x_n)$, instead of the density as a function of $(x_1,\ldots,x_n)$ indexed by the parameter $\theta$. This means that, as a function of $\theta$, and for a given sample $(x_1,\ldots,x_n)$ fro... | Likelihood function when $X\sim U(0,\theta)$ | Just to stress a point, namely that the notion of likelihood is one of a function of $\theta$ indexed by the sample $(x_1,\ldots,x_n)$, instead of the density as a function of $(x_1,\ldots,x_n)$ index | Likelihood function when $X\sim U(0,\theta)$
Just to stress a point, namely that the notion of likelihood is one of a function of $\theta$ indexed by the sample $(x_1,\ldots,x_n)$, instead of the density as a function of $(x_1,\ldots,x_n)$ indexed by the parameter $\theta$. This means that, as a function of $\theta$, ... | Likelihood function when $X\sim U(0,\theta)$
Just to stress a point, namely that the notion of likelihood is one of a function of $\theta$ indexed by the sample $(x_1,\ldots,x_n)$, instead of the density as a function of $(x_1,\ldots,x_n)$ index |
38,989 | Is there a difference between Bayesian and Classical sufficiency? | In addition to the other excellent answer: The question of equivalence between Bayesian (B-) sufficiency and Classical (F-sufficiency) is answered in the abstract by NOT. But, this is based on abstract measure-theoretic definitions, and proofs are technical using measure theory. But, counterexamples to equivalence are... | Is there a difference between Bayesian and Classical sufficiency? | In addition to the other excellent answer: The question of equivalence between Bayesian (B-) sufficiency and Classical (F-sufficiency) is answered in the abstract by NOT. But, this is based on abstra | Is there a difference between Bayesian and Classical sufficiency?
In addition to the other excellent answer: The question of equivalence between Bayesian (B-) sufficiency and Classical (F-sufficiency) is answered in the abstract by NOT. But, this is based on abstract measure-theoretic definitions, and proofs are techn... | Is there a difference between Bayesian and Classical sufficiency?
In addition to the other excellent answer: The question of equivalence between Bayesian (B-) sufficiency and Classical (F-sufficiency) is answered in the abstract by NOT. But, this is based on abstra |
38,990 | Is there a difference between Bayesian and Classical sufficiency? | Here is one example of differentiation between classical and Bayesian statistics: when comparing two models $\mathcal{M}_1$ and $\mathcal{M}_2$, a statistic $S(\cdot)$ may be sufficient for both models, hence sufficient in a classical sense, but insufficient for model comparison as e.g. in Bayes factors, when the condi... | Is there a difference between Bayesian and Classical sufficiency? | Here is one example of differentiation between classical and Bayesian statistics: when comparing two models $\mathcal{M}_1$ and $\mathcal{M}_2$, a statistic $S(\cdot)$ may be sufficient for both model | Is there a difference between Bayesian and Classical sufficiency?
Here is one example of differentiation between classical and Bayesian statistics: when comparing two models $\mathcal{M}_1$ and $\mathcal{M}_2$, a statistic $S(\cdot)$ may be sufficient for both models, hence sufficient in a classical sense, but insuffic... | Is there a difference between Bayesian and Classical sufficiency?
Here is one example of differentiation between classical and Bayesian statistics: when comparing two models $\mathcal{M}_1$ and $\mathcal{M}_2$, a statistic $S(\cdot)$ may be sufficient for both model |
38,991 | MLE for a homogeneous Poisson process? | You can use Maximum Likelihood Estimation, either with synchronous data (time-binned data) or asynchronous data (time-stamped data). The likelihood function changes accordingly.
For time-binned (or synchronous) data, you can simply use the joint Poisson probability mass function for your observed counts as the likeliho... | MLE for a homogeneous Poisson process? | You can use Maximum Likelihood Estimation, either with synchronous data (time-binned data) or asynchronous data (time-stamped data). The likelihood function changes accordingly.
For time-binned (or sy | MLE for a homogeneous Poisson process?
You can use Maximum Likelihood Estimation, either with synchronous data (time-binned data) or asynchronous data (time-stamped data). The likelihood function changes accordingly.
For time-binned (or synchronous) data, you can simply use the joint Poisson probability mass function f... | MLE for a homogeneous Poisson process?
You can use Maximum Likelihood Estimation, either with synchronous data (time-binned data) or asynchronous data (time-stamped data). The likelihood function changes accordingly.
For time-binned (or sy |
38,992 | MLE for a homogeneous Poisson process? | For the homogeneous Poisson process with rate $\lambda$ the likelihood function can be written as
$L(\lambda)=\prod\limits_{n=1}^{N}\dfrac{\left(\lambda.(t_n-t_{n-1})\right)^1.e^{-\lambda(t_n-t_{n-1})}}{1!}$, with $t_0=0$, s.t., the log-likelihood
$l(\lambda)=\sum\limits_{n=1}^{N}ln(\lambda) + ln(t_n-t_{n-1})-\lambda(t... | MLE for a homogeneous Poisson process? | For the homogeneous Poisson process with rate $\lambda$ the likelihood function can be written as
$L(\lambda)=\prod\limits_{n=1}^{N}\dfrac{\left(\lambda.(t_n-t_{n-1})\right)^1.e^{-\lambda(t_n-t_{n-1}) | MLE for a homogeneous Poisson process?
For the homogeneous Poisson process with rate $\lambda$ the likelihood function can be written as
$L(\lambda)=\prod\limits_{n=1}^{N}\dfrac{\left(\lambda.(t_n-t_{n-1})\right)^1.e^{-\lambda(t_n-t_{n-1})}}{1!}$, with $t_0=0$, s.t., the log-likelihood
$l(\lambda)=\sum\limits_{n=1}^{N}... | MLE for a homogeneous Poisson process?
For the homogeneous Poisson process with rate $\lambda$ the likelihood function can be written as
$L(\lambda)=\prod\limits_{n=1}^{N}\dfrac{\left(\lambda.(t_n-t_{n-1})\right)^1.e^{-\lambda(t_n-t_{n-1}) |
38,993 | MLE for a homogeneous Poisson process? | Expanding from @sandipan-dey's answer for homogenous process: | MLE for a homogeneous Poisson process? | Expanding from @sandipan-dey's answer for homogenous process: | MLE for a homogeneous Poisson process?
Expanding from @sandipan-dey's answer for homogenous process: | MLE for a homogeneous Poisson process?
Expanding from @sandipan-dey's answer for homogenous process: |
38,994 | Why does glm() provide estimates and standard errors on the link scale? | Hard to know for sure, but there are a few reasons the link scale is useful.
Using standard errors as a summary of uncertainty is generally more reliable on the link scale, where the domain of the parameters is unbounded and where the assumption that the likelihood surface is approximately quadratic ($\leftrightarrow$... | Why does glm() provide estimates and standard errors on the link scale? | Hard to know for sure, but there are a few reasons the link scale is useful.
Using standard errors as a summary of uncertainty is generally more reliable on the link scale, where the domain of the pa | Why does glm() provide estimates and standard errors on the link scale?
Hard to know for sure, but there are a few reasons the link scale is useful.
Using standard errors as a summary of uncertainty is generally more reliable on the link scale, where the domain of the parameters is unbounded and where the assumption t... | Why does glm() provide estimates and standard errors on the link scale?
Hard to know for sure, but there are a few reasons the link scale is useful.
Using standard errors as a summary of uncertainty is generally more reliable on the link scale, where the domain of the pa |
38,995 | Finding the slope at different points in a sigmoid curve | Your question is very broad. There are many ways to do this, even without assuming a specific function. For the following I assume that you have a good reason to use the Gompertz model.
First let's fit the model:
y <- c(0.5,3.0,22.2,46.0,77.3,97.0,98.9,100.0)
x <- seq_along(y)
plot(x, y)
fit <- nls(y ~ SSgompertz(x, A... | Finding the slope at different points in a sigmoid curve | Your question is very broad. There are many ways to do this, even without assuming a specific function. For the following I assume that you have a good reason to use the Gompertz model.
First let's fi | Finding the slope at different points in a sigmoid curve
Your question is very broad. There are many ways to do this, even without assuming a specific function. For the following I assume that you have a good reason to use the Gompertz model.
First let's fit the model:
y <- c(0.5,3.0,22.2,46.0,77.3,97.0,98.9,100.0)
x <... | Finding the slope at different points in a sigmoid curve
Your question is very broad. There are many ways to do this, even without assuming a specific function. For the following I assume that you have a good reason to use the Gompertz model.
First let's fi |
38,996 | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic? | No, posterior distribution is not a statistic if you define it strictly and agree that there is a distinction between statistic and estimator. First of all, it is an estimator of the posterior distribution (see What is the difference between an estimator and a statistic?). Second, as already noticed in the comment by @... | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic? | No, posterior distribution is not a statistic if you define it strictly and agree that there is a distinction between statistic and estimator. First of all, it is an estimator of the posterior distrib | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic?
No, posterior distribution is not a statistic if you define it strictly and agree that there is a distinction between statistic and estimator. First of all, it is an estimator of the posterior distribution (see What is the difference between an estimator... | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic?
No, posterior distribution is not a statistic if you define it strictly and agree that there is a distinction between statistic and estimator. First of all, it is an estimator of the posterior distrib |
38,997 | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic? | Yes, a posterior is a statistic. As you point out a statistic is any function of the data and the posterior is a function of the data and thus a statistic. Yes, Bayesians use the notion of statistic but typically not describing the posterior as a statistic. We are certainly interested in sufficient statistics since a p... | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic? | Yes, a posterior is a statistic. As you point out a statistic is any function of the data and the posterior is a function of the data and thus a statistic. Yes, Bayesians use the notion of statistic b | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic?
Yes, a posterior is a statistic. As you point out a statistic is any function of the data and the posterior is a function of the data and thus a statistic. Yes, Bayesians use the notion of statistic but typically not describing the posterior as a statist... | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic?
Yes, a posterior is a statistic. As you point out a statistic is any function of the data and the posterior is a function of the data and thus a statistic. Yes, Bayesians use the notion of statistic b |
38,998 | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic? | I don't agree with the view that the posterior is not a statistic because it is an estimator (e.g., in the accepted answer to this question, and the accompanying answer to the related question on alleged differences between a statistic and an estimator). But for other reasons, set out in detail below, I would say that... | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic? | I don't agree with the view that the posterior is not a statistic because it is an estimator (e.g., in the accepted answer to this question, and the accompanying answer to the related question on alle | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic?
I don't agree with the view that the posterior is not a statistic because it is an estimator (e.g., in the accepted answer to this question, and the accompanying answer to the related question on alleged differences between a statistic and an estimator).... | Is the posterior distribution $P(\theta|\mathbf{X})$ a statistic?
I don't agree with the view that the posterior is not a statistic because it is an estimator (e.g., in the accepted answer to this question, and the accompanying answer to the related question on alle |
38,999 | Given a GLM using Tweedie, how do I find the coefficients? | Are you familiar with generalized linear models in R? If so, you can fit Tweedie glms just like any other glms.
The glm family definition necessary to make this happen is provided by the statmod R package from CRAN.
Tweedie glms assume that the variance function is a power function:
$${\rm var}(y)=V(\mu)\phi=\mu^\alpha... | Given a GLM using Tweedie, how do I find the coefficients? | Are you familiar with generalized linear models in R? If so, you can fit Tweedie glms just like any other glms.
The glm family definition necessary to make this happen is provided by the statmod R pac | Given a GLM using Tweedie, how do I find the coefficients?
Are you familiar with generalized linear models in R? If so, you can fit Tweedie glms just like any other glms.
The glm family definition necessary to make this happen is provided by the statmod R package from CRAN.
Tweedie glms assume that the variance functio... | Given a GLM using Tweedie, how do I find the coefficients?
Are you familiar with generalized linear models in R? If so, you can fit Tweedie glms just like any other glms.
The glm family definition necessary to make this happen is provided by the statmod R pac |
39,000 | Multicollinearity between ln(x) and ln(x)^2 | Except for very small counts, $\log(x)^2$ is essentially a linear function of $\log(x)$:
The colored lines are least squares fits to $\log(x)^2$ vs $\log(x)$ for various ranges of counts $x$. They are extremely good once $x$ exceeds $10$ (and still awfully good even when $x\gt 4$ or so).
Introducing the square of a v... | Multicollinearity between ln(x) and ln(x)^2 | Except for very small counts, $\log(x)^2$ is essentially a linear function of $\log(x)$:
The colored lines are least squares fits to $\log(x)^2$ vs $\log(x)$ for various ranges of counts $x$. They a | Multicollinearity between ln(x) and ln(x)^2
Except for very small counts, $\log(x)^2$ is essentially a linear function of $\log(x)$:
The colored lines are least squares fits to $\log(x)^2$ vs $\log(x)$ for various ranges of counts $x$. They are extremely good once $x$ exceeds $10$ (and still awfully good even when $x... | Multicollinearity between ln(x) and ln(x)^2
Except for very small counts, $\log(x)^2$ is essentially a linear function of $\log(x)$:
The colored lines are least squares fits to $\log(x)^2$ vs $\log(x)$ for various ranges of counts $x$. They a |
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