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 |
|---|---|---|---|---|---|---|
38,001 | Are there useful applications of SVD that use only the smallest singular values? | Slow feature analysis (SFA) uses the smalles Eigenvalues of the covariance matrix of temporal differences to find the slowest features in a time series,
Minor component analysis (MCA) uses the smallest components in a probabilistic setting--here, not directions of variations are found but constraints,
Extreme component... | Are there useful applications of SVD that use only the smallest singular values? | Slow feature analysis (SFA) uses the smalles Eigenvalues of the covariance matrix of temporal differences to find the slowest features in a time series,
Minor component analysis (MCA) uses the smalles | Are there useful applications of SVD that use only the smallest singular values?
Slow feature analysis (SFA) uses the smalles Eigenvalues of the covariance matrix of temporal differences to find the slowest features in a time series,
Minor component analysis (MCA) uses the smallest components in a probabilistic setting... | Are there useful applications of SVD that use only the smallest singular values?
Slow feature analysis (SFA) uses the smalles Eigenvalues of the covariance matrix of temporal differences to find the slowest features in a time series,
Minor component analysis (MCA) uses the smalles |
38,002 | Are there useful applications of SVD that use only the smallest singular values? | Total Least Squares regression (aka Orthogonal Distance regression) uses the singular vector corresponding to the smallest singular value of the augmented predictor/criterion matrix.
When there is only one dependent variable (i.e., when $k = 1$), both equation 12.3-5 in my Golub & Van Loan (first edition), and the fi... | Are there useful applications of SVD that use only the smallest singular values? | Total Least Squares regression (aka Orthogonal Distance regression) uses the singular vector corresponding to the smallest singular value of the augmented predictor/criterion matrix.
When there is o | Are there useful applications of SVD that use only the smallest singular values?
Total Least Squares regression (aka Orthogonal Distance regression) uses the singular vector corresponding to the smallest singular value of the augmented predictor/criterion matrix.
When there is only one dependent variable (i.e., when ... | Are there useful applications of SVD that use only the smallest singular values?
Total Least Squares regression (aka Orthogonal Distance regression) uses the singular vector corresponding to the smallest singular value of the augmented predictor/criterion matrix.
When there is o |
38,003 | Are there useful applications of SVD that use only the smallest singular values? | Yes, there are. I'm currently working with a professor on a research project where we try to predict very short-term stock market changes based on real-time tweets from Twitter. Unfortunately, the majority of what people say on Twitter about the companies we are tracking is useless rambling. In other words, the largest... | Are there useful applications of SVD that use only the smallest singular values? | Yes, there are. I'm currently working with a professor on a research project where we try to predict very short-term stock market changes based on real-time tweets from Twitter. Unfortunately, the maj | Are there useful applications of SVD that use only the smallest singular values?
Yes, there are. I'm currently working with a professor on a research project where we try to predict very short-term stock market changes based on real-time tweets from Twitter. Unfortunately, the majority of what people say on Twitter abo... | Are there useful applications of SVD that use only the smallest singular values?
Yes, there are. I'm currently working with a professor on a research project where we try to predict very short-term stock market changes based on real-time tweets from Twitter. Unfortunately, the maj |
38,004 | Are there useful applications of SVD that use only the smallest singular values? | It's a bit of a stretch, but consider the portfolio optimization problem: minimize $w^{\top} \Sigma w$ subject to $w^{\top} w \ge 1$. You can think of this as the minimum variance portfolio with an $\ell_2$ constraint. After applying Lagrange Multiplier method, you find that $w$ should be the eigenvector associated wit... | Are there useful applications of SVD that use only the smallest singular values? | It's a bit of a stretch, but consider the portfolio optimization problem: minimize $w^{\top} \Sigma w$ subject to $w^{\top} w \ge 1$. You can think of this as the minimum variance portfolio with an $\ | Are there useful applications of SVD that use only the smallest singular values?
It's a bit of a stretch, but consider the portfolio optimization problem: minimize $w^{\top} \Sigma w$ subject to $w^{\top} w \ge 1$. You can think of this as the minimum variance portfolio with an $\ell_2$ constraint. After applying Lagra... | Are there useful applications of SVD that use only the smallest singular values?
It's a bit of a stretch, but consider the portfolio optimization problem: minimize $w^{\top} \Sigma w$ subject to $w^{\top} w \ge 1$. You can think of this as the minimum variance portfolio with an $\ |
38,005 | Are there useful applications of SVD that use only the smallest singular values? | There is an interesting LSA related paper which concludes that discarding (a lot of) first SVD features can improve results on semantic tests like TOEFL - "Extracting Semantic Represenations from Word Cooccurrence Statistics: Stop-lists, Stemming and SVD" (Bullinaria and Levy, 2012) | Are there useful applications of SVD that use only the smallest singular values? | There is an interesting LSA related paper which concludes that discarding (a lot of) first SVD features can improve results on semantic tests like TOEFL - "Extracting Semantic Represenations from Word | Are there useful applications of SVD that use only the smallest singular values?
There is an interesting LSA related paper which concludes that discarding (a lot of) first SVD features can improve results on semantic tests like TOEFL - "Extracting Semantic Represenations from Word Cooccurrence Statistics: Stop-lists, S... | Are there useful applications of SVD that use only the smallest singular values?
There is an interesting LSA related paper which concludes that discarding (a lot of) first SVD features can improve results on semantic tests like TOEFL - "Extracting Semantic Represenations from Word |
38,006 | Are there useful applications of SVD that use only the smallest singular values? | I'm not aware of any. The smallest singular values correspond to modes that don't contribute much to the reconstruction of the original matrix, or to use the PCA interpretation, don't describe much of the variance in the data. Typically, the modes with smaller singular values are just noise. This doesn't rule out t... | Are there useful applications of SVD that use only the smallest singular values? | I'm not aware of any. The smallest singular values correspond to modes that don't contribute much to the reconstruction of the original matrix, or to use the PCA interpretation, don't describe much o | Are there useful applications of SVD that use only the smallest singular values?
I'm not aware of any. The smallest singular values correspond to modes that don't contribute much to the reconstruction of the original matrix, or to use the PCA interpretation, don't describe much of the variance in the data. Typically,... | Are there useful applications of SVD that use only the smallest singular values?
I'm not aware of any. The smallest singular values correspond to modes that don't contribute much to the reconstruction of the original matrix, or to use the PCA interpretation, don't describe much o |
38,007 | How to test my data against an specific normal distribution? | ks.test in R allows one to adjust the mean and sd of the distribution to be tested against. e.g.
x <- rnorm(1000, 4, 10)
ks.test(x, "pnorm", mean = 4, sd = 10) | How to test my data against an specific normal distribution? | ks.test in R allows one to adjust the mean and sd of the distribution to be tested against. e.g.
x <- rnorm(1000, 4, 10)
ks.test(x, "pnorm", mean = 4, sd = 10) | How to test my data against an specific normal distribution?
ks.test in R allows one to adjust the mean and sd of the distribution to be tested against. e.g.
x <- rnorm(1000, 4, 10)
ks.test(x, "pnorm", mean = 4, sd = 10) | How to test my data against an specific normal distribution?
ks.test in R allows one to adjust the mean and sd of the distribution to be tested against. e.g.
x <- rnorm(1000, 4, 10)
ks.test(x, "pnorm", mean = 4, sd = 10) |
38,008 | How to test my data against an specific normal distribution? | In R, you can just use the function ks.test with the following arguments:
ks.test(your_data, "pnorm", mean=test_mu, sd=test_sd)
Where your_data is your data vector, test_mu is the specific mean of the theoretical normal distribution and test_sd its standard deviation.
To inspect your data graphically, you can use the ... | How to test my data against an specific normal distribution? | In R, you can just use the function ks.test with the following arguments:
ks.test(your_data, "pnorm", mean=test_mu, sd=test_sd)
Where your_data is your data vector, test_mu is the specific mean of th | How to test my data against an specific normal distribution?
In R, you can just use the function ks.test with the following arguments:
ks.test(your_data, "pnorm", mean=test_mu, sd=test_sd)
Where your_data is your data vector, test_mu is the specific mean of the theoretical normal distribution and test_sd its standard ... | How to test my data against an specific normal distribution?
In R, you can just use the function ks.test with the following arguments:
ks.test(your_data, "pnorm", mean=test_mu, sd=test_sd)
Where your_data is your data vector, test_mu is the specific mean of th |
38,009 | How to test my data against an specific normal distribution? | So don't use the default! As already noted, R lets you specify the population mean and standard deviation.
Here is how to do it in MATLAB or anything else that doesn't give you the option to specify:
Standardize your variable by the population parameters: $z_i = \frac{x_i-\mu_0}{\sigma_0}$
... and then test against st... | How to test my data against an specific normal distribution? | So don't use the default! As already noted, R lets you specify the population mean and standard deviation.
Here is how to do it in MATLAB or anything else that doesn't give you the option to specify: | How to test my data against an specific normal distribution?
So don't use the default! As already noted, R lets you specify the population mean and standard deviation.
Here is how to do it in MATLAB or anything else that doesn't give you the option to specify:
Standardize your variable by the population parameters: $z... | How to test my data against an specific normal distribution?
So don't use the default! As already noted, R lets you specify the population mean and standard deviation.
Here is how to do it in MATLAB or anything else that doesn't give you the option to specify: |
38,010 | Regression coefficients by group in R? | data.table also has great tools for solving problems such as this:
library(data.table)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
dat[,coef(lm(y~x)),by=grp]
The first row in each group is the intercept, and the second row is the coefficient:
grp V1
[1,] 1 0.5991761
[2,] ... | Regression coefficients by group in R? | data.table also has great tools for solving problems such as this:
library(data.table)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
dat[,coef(lm(y~x)),by=grp]
The first | Regression coefficients by group in R?
data.table also has great tools for solving problems such as this:
library(data.table)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
dat[,coef(lm(y~x)),by=grp]
The first row in each group is the intercept, and the second row is the coefficient:
g... | Regression coefficients by group in R?
data.table also has great tools for solving problems such as this:
library(data.table)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
dat[,coef(lm(y~x)),by=grp]
The first |
38,011 | Regression coefficients by group in R? | The responses by @Henry and @Zach both work, but I think the most straight-forward way to do what you want is to use lmList in the nlme package:
dat <- data.frame(
GRP = sample(c("A","B","C"), 100, replace=TRUE),
X = runif(100),
Y = runif(100)
)
require(nlme)
lmList(Y ~ X | GRP, data=dat) | Regression coefficients by group in R? | The responses by @Henry and @Zach both work, but I think the most straight-forward way to do what you want is to use lmList in the nlme package:
dat <- data.frame(
GRP = sample(c("A","B","C"), 100, | Regression coefficients by group in R?
The responses by @Henry and @Zach both work, but I think the most straight-forward way to do what you want is to use lmList in the nlme package:
dat <- data.frame(
GRP = sample(c("A","B","C"), 100, replace=TRUE),
X = runif(100),
Y = runif(100)
)
require(nlme)
lmList(Y ~ X ... | Regression coefficients by group in R?
The responses by @Henry and @Zach both work, but I think the most straight-forward way to do what you want is to use lmList in the nlme package:
dat <- data.frame(
GRP = sample(c("A","B","C"), 100, |
38,012 | Regression coefficients by group in R? | Adapting from help("by"), this example may meet your needs
mydf <- data.frame( GRP = rep(c("A","B","C"), each=100), X = rep(1:100,3),
Y = rep(c(2,4,8),each=100) +
rep(c(4,2,1),each=100) * rep(1:100,3) + rnorm(300))
by(mydf, mydf$GRP, function(z) lm(Y ~ X, data = z)) | Regression coefficients by group in R? | Adapting from help("by"), this example may meet your needs
mydf <- data.frame( GRP = rep(c("A","B","C"), each=100), X = rep(1:100,3),
Y = rep(c(2,4,8),each=100) +
| Regression coefficients by group in R?
Adapting from help("by"), this example may meet your needs
mydf <- data.frame( GRP = rep(c("A","B","C"), each=100), X = rep(1:100,3),
Y = rep(c(2,4,8),each=100) +
rep(c(4,2,1),each=100) * rep(1:100,3) + rnorm(300))
by(mydf, mydf$GRP... | Regression coefficients by group in R?
Adapting from help("by"), this example may meet your needs
mydf <- data.frame( GRP = rep(c("A","B","C"), each=100), X = rep(1:100,3),
Y = rep(c(2,4,8),each=100) +
|
38,013 | Regression coefficients by group in R? | If you use package "tidyr" you could do the following
library(data.table)
library(tidyr)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
ncoefs <- 1
dat <- dat[, coef( lm(y ~ x) ), by = grp]
dat[, est := rep( c("intercept", "coef"), .N/(ncoefs + 1)) ]
dat <- dat %>% spread(est, V1)
The resu... | Regression coefficients by group in R? | If you use package "tidyr" you could do the following
library(data.table)
library(tidyr)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
ncoefs <- 1
dat <- dat[, coef( lm(y | Regression coefficients by group in R?
If you use package "tidyr" you could do the following
library(data.table)
library(tidyr)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
ncoefs <- 1
dat <- dat[, coef( lm(y ~ x) ), by = grp]
dat[, est := rep( c("intercept", "coef"), .N/(ncoefs + 1)) ]
d... | Regression coefficients by group in R?
If you use package "tidyr" you could do the following
library(data.table)
library(tidyr)
set.seed(1)
dat <- data.table(x=runif(100), y=runif(100), grp=rep(1:2,50))
ncoefs <- 1
dat <- dat[, coef( lm(y |
38,014 | Creating a uniform prior on the logarithmic scale | It's just a standard change of variables; the (monotone & 1-1) transformation is $y = \exp(x)$ with inverse $x=\log(y)$ and Jacobian $\frac{dx}{dy} = \frac{1}{y}$.
With a uniform prior $p_y(y) \propto 1$ on $\mathbb{R}$ we get $p_x(x) = p_y(x(y)) |\frac{dx}{dy}| \propto \frac{1}{y}$ on $(0, \infty)$.
Edit: Wikipedia h... | Creating a uniform prior on the logarithmic scale | It's just a standard change of variables; the (monotone & 1-1) transformation is $y = \exp(x)$ with inverse $x=\log(y)$ and Jacobian $\frac{dx}{dy} = \frac{1}{y}$.
With a uniform prior $p_y(y) \propto | Creating a uniform prior on the logarithmic scale
It's just a standard change of variables; the (monotone & 1-1) transformation is $y = \exp(x)$ with inverse $x=\log(y)$ and Jacobian $\frac{dx}{dy} = \frac{1}{y}$.
With a uniform prior $p_y(y) \propto 1$ on $\mathbb{R}$ we get $p_x(x) = p_y(x(y)) |\frac{dx}{dy}| \propto... | Creating a uniform prior on the logarithmic scale
It's just a standard change of variables; the (monotone & 1-1) transformation is $y = \exp(x)$ with inverse $x=\log(y)$ and Jacobian $\frac{dx}{dy} = \frac{1}{y}$.
With a uniform prior $p_y(y) \propto |
38,015 | Creating a uniform prior on the logarithmic scale | We are told that the scale parameter is uniform on the logarithmic scale. That means that if x is the scale parameter, then $y=\log(𝑥)$ and the distribution function for $y$ is the one uniform on the logarithmic scale, $p_Y(y) \propto 1$.
Then, applying the Jacobian transformation which comes from the fact that the p... | Creating a uniform prior on the logarithmic scale | We are told that the scale parameter is uniform on the logarithmic scale. That means that if x is the scale parameter, then $y=\log(𝑥)$ and the distribution function for $y$ is the one uniform on the | Creating a uniform prior on the logarithmic scale
We are told that the scale parameter is uniform on the logarithmic scale. That means that if x is the scale parameter, then $y=\log(𝑥)$ and the distribution function for $y$ is the one uniform on the logarithmic scale, $p_Y(y) \propto 1$.
Then, applying the Jacobian t... | Creating a uniform prior on the logarithmic scale
We are told that the scale parameter is uniform on the logarithmic scale. That means that if x is the scale parameter, then $y=\log(𝑥)$ and the distribution function for $y$ is the one uniform on the |
38,016 | Creating a uniform prior on the logarithmic scale | @JMS answer is adequate for the nuts and bolts of changing variables. However, This question may help you a bit with why it is uniform on that scale.
My answer to this question goes through a slightly longer derivation of the "jacobian rule" result given in @JMS's answer. It may help with understanding why the rule a... | Creating a uniform prior on the logarithmic scale | @JMS answer is adequate for the nuts and bolts of changing variables. However, This question may help you a bit with why it is uniform on that scale.
My answer to this question goes through a slightl | Creating a uniform prior on the logarithmic scale
@JMS answer is adequate for the nuts and bolts of changing variables. However, This question may help you a bit with why it is uniform on that scale.
My answer to this question goes through a slightly longer derivation of the "jacobian rule" result given in @JMS's answ... | Creating a uniform prior on the logarithmic scale
@JMS answer is adequate for the nuts and bolts of changing variables. However, This question may help you a bit with why it is uniform on that scale.
My answer to this question goes through a slightl |
38,017 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | I think a simple chi-squared test will do the trick. Do you have 1,000,000 observations for both control and test? If so, your table of observations will be (in R code)
Edit: Woops! Left off a zero!
m <- rbind(c(3, 1000000-3), c(10, 1000000-10))
# [,1] [,2]
# [1,] 3 999997
# [2,] 10 999990
And chi-squared... | Determine if three is statistically different than ten for a very large number of observations (1,00 | I think a simple chi-squared test will do the trick. Do you have 1,000,000 observations for both control and test? If so, your table of observations will be (in R code)
Edit: Woops! Left off a zero!
m | Determine if three is statistically different than ten for a very large number of observations (1,000,000)
I think a simple chi-squared test will do the trick. Do you have 1,000,000 observations for both control and test? If so, your table of observations will be (in R code)
Edit: Woops! Left off a zero!
m <- rbind(c(3... | Determine if three is statistically different than ten for a very large number of observations (1,00
I think a simple chi-squared test will do the trick. Do you have 1,000,000 observations for both control and test? If so, your table of observations will be (in R code)
Edit: Woops! Left off a zero!
m |
38,018 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | The huge denominators throw off one's intuition. Since the sample sizes are identical, and the proportions low, the problem can be recast: 13 events occurred, and were expected (by null hypothesis) to occur equally in both groups. In fact the split was 3 in one group and 10 in the other. How rare is that? The binomial ... | Determine if three is statistically different than ten for a very large number of observations (1,00 | The huge denominators throw off one's intuition. Since the sample sizes are identical, and the proportions low, the problem can be recast: 13 events occurred, and were expected (by null hypothesis) to | Determine if three is statistically different than ten for a very large number of observations (1,000,000)
The huge denominators throw off one's intuition. Since the sample sizes are identical, and the proportions low, the problem can be recast: 13 events occurred, and were expected (by null hypothesis) to occur equall... | Determine if three is statistically different than ten for a very large number of observations (1,00
The huge denominators throw off one's intuition. Since the sample sizes are identical, and the proportions low, the problem can be recast: 13 events occurred, and were expected (by null hypothesis) to |
38,019 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | A (two-sided) Fisher's Exact test gives p-value = 0.092284.
function p = fexact(k, x, m, n)
%FEXACT Fisher's Exact test.
% Y = FEXACT(K, X, M, N) calculates the P-value for Fisher's
% Exact Test.
% K, X, M and N must be nonnegative integer vectors of the same
% length. The following must also hold:
% X <= N ... | Determine if three is statistically different than ten for a very large number of observations (1,00 | A (two-sided) Fisher's Exact test gives p-value = 0.092284.
function p = fexact(k, x, m, n)
%FEXACT Fisher's Exact test.
% Y = FEXACT(K, X, M, N) calculates the P-value for Fisher's
% Exact Test.
| Determine if three is statistically different than ten for a very large number of observations (1,000,000)
A (two-sided) Fisher's Exact test gives p-value = 0.092284.
function p = fexact(k, x, m, n)
%FEXACT Fisher's Exact test.
% Y = FEXACT(K, X, M, N) calculates the P-value for Fisher's
% Exact Test.
% K, X, M a... | Determine if three is statistically different than ten for a very large number of observations (1,00
A (two-sided) Fisher's Exact test gives p-value = 0.092284.
function p = fexact(k, x, m, n)
%FEXACT Fisher's Exact test.
% Y = FEXACT(K, X, M, N) calculates the P-value for Fisher's
% Exact Test.
|
38,020 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | In this case Poisson is good approximation for distribution for number of cases.
There is simple formula to approximate variance of log RR (delta method) .
log RR = 10/3 = 1.2,
se log RR = sqrt(1/3+1/10) = 0.66, so 95%CI = (-0.09; 2.5)
It is not significant difference at 0.05 level using two-sided test.
LR based Chi-s... | Determine if three is statistically different than ten for a very large number of observations (1,00 | In this case Poisson is good approximation for distribution for number of cases.
There is simple formula to approximate variance of log RR (delta method) .
log RR = 10/3 = 1.2,
se log RR = sqrt(1/3+1 | Determine if three is statistically different than ten for a very large number of observations (1,000,000)
In this case Poisson is good approximation for distribution for number of cases.
There is simple formula to approximate variance of log RR (delta method) .
log RR = 10/3 = 1.2,
se log RR = sqrt(1/3+1/10) = 0.66, ... | Determine if three is statistically different than ten for a very large number of observations (1,00
In this case Poisson is good approximation for distribution for number of cases.
There is simple formula to approximate variance of log RR (delta method) .
log RR = 10/3 = 1.2,
se log RR = sqrt(1/3+1 |
38,021 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | If you wanted to check non-parametrically for significance, you could bootstrap the confidence intervals on the ratio, or you could do a permutation test on the two classes. For example, to do the bootstrap, create two arrays: one with 3 ones and 999,997 zeros, and one with 10 ones and 999,990 zeros. Then draw with rep... | Determine if three is statistically different than ten for a very large number of observations (1,00 | If you wanted to check non-parametrically for significance, you could bootstrap the confidence intervals on the ratio, or you could do a permutation test on the two classes. For example, to do the boo | Determine if three is statistically different than ten for a very large number of observations (1,000,000)
If you wanted to check non-parametrically for significance, you could bootstrap the confidence intervals on the ratio, or you could do a permutation test on the two classes. For example, to do the bootstrap, creat... | Determine if three is statistically different than ten for a very large number of observations (1,00
If you wanted to check non-parametrically for significance, you could bootstrap the confidence intervals on the ratio, or you could do a permutation test on the two classes. For example, to do the boo |
38,022 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | I would be really surprised if you find the difference statistically significant. Having said that you may want to use a test for a difference of proportions (3 out of 1M vs 10 out of 1M). | Determine if three is statistically different than ten for a very large number of observations (1,00 | I would be really surprised if you find the difference statistically significant. Having said that you may want to use a test for a difference of proportions (3 out of 1M vs 10 out of 1M). | Determine if three is statistically different than ten for a very large number of observations (1,000,000)
I would be really surprised if you find the difference statistically significant. Having said that you may want to use a test for a difference of proportions (3 out of 1M vs 10 out of 1M). | Determine if three is statistically different than ten for a very large number of observations (1,00
I would be really surprised if you find the difference statistically significant. Having said that you may want to use a test for a difference of proportions (3 out of 1M vs 10 out of 1M). |
38,023 | Determine if three is statistically different than ten for a very large number of observations (1,000,000) | In addition to the other answers:
If you have 1,000,000 observations and when your event comes up only a few times, you are likely to want to look at a lot of different events.
If you look at 100 different events you will run into problems if you work with p<0.05 as criteria for significance. | Determine if three is statistically different than ten for a very large number of observations (1,00 | In addition to the other answers:
If you have 1,000,000 observations and when your event comes up only a few times, you are likely to want to look at a lot of different events.
If you look at 100 diff | Determine if three is statistically different than ten for a very large number of observations (1,000,000)
In addition to the other answers:
If you have 1,000,000 observations and when your event comes up only a few times, you are likely to want to look at a lot of different events.
If you look at 100 different events ... | Determine if three is statistically different than ten for a very large number of observations (1,00
In addition to the other answers:
If you have 1,000,000 observations and when your event comes up only a few times, you are likely to want to look at a lot of different events.
If you look at 100 diff |
38,024 | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | Gauss-hermite (GH) quadrature is a very efficient method compared to the Monte Carlo method (MC). If you care about the trade off between precision versus speed (here measured by the number of function evaluations) GH is far superior to MC sampling for the specific integral here under consideration.
The software I use ... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | Gauss-hermite (GH) quadrature is a very efficient method compared to the Monte Carlo method (MC). If you care about the trade off between precision versus speed (here measured by the number of functio | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
Gauss-hermite (GH) quadrature is a very efficient method compared to the Monte Carlo method (MC). If you care about the trade off between precision versus speed (here measured by the number of function evaluations) GH is far superior to MC sampling for th... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
Gauss-hermite (GH) quadrature is a very efficient method compared to the Monte Carlo method (MC). If you care about the trade off between precision versus speed (here measured by the number of functio |
38,025 | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | Like you mentioned, This is just the average value of $\ln(1+e^x)$, when x is normally distributed with mean μ and variance ν. So all you have to do is:
1) Draw N (large number) of $X_i \sim N(\mu, \nu)$
2) Your estimate $\hat\theta$ is then:
$ \hat\theta = 1/N * \sqrt{2\pi\nu }\sum^N_{i =1}\ln(1+e^{X_i}) $ | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | Like you mentioned, This is just the average value of $\ln(1+e^x)$, when x is normally distributed with mean μ and variance ν. So all you have to do is:
1) Draw N (large number) of $X_i \sim N(\mu, \n | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
Like you mentioned, This is just the average value of $\ln(1+e^x)$, when x is normally distributed with mean μ and variance ν. So all you have to do is:
1) Draw N (large number) of $X_i \sim N(\mu, \nu)$
2) Your estimate $\hat\theta$ is then:
$ \hat\theta... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
Like you mentioned, This is just the average value of $\ln(1+e^x)$, when x is normally distributed with mean μ and variance ν. So all you have to do is:
1) Draw N (large number) of $X_i \sim N(\mu, \n |
38,026 | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | Out of curiosity, I have tried the suggestions of both answers with the following R-code (note that the integrand $\log(1+e^x)$ must be approximated by $x$ for large values to avoid overflow):
f <- function(x, mu, s) {
# approximate integrand for large values
log.part <- ifelse(x<100, log(1+exp(x)), x)
retu... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | Out of curiosity, I have tried the suggestions of both answers with the following R-code (note that the integrand $\log(1+e^x)$ must be approximated by $x$ for large values to avoid overflow):
f <- fu | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
Out of curiosity, I have tried the suggestions of both answers with the following R-code (note that the integrand $\log(1+e^x)$ must be approximated by $x$ for large values to avoid overflow):
f <- function(x, mu, s) {
# approximate integrand for larg... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
Out of curiosity, I have tried the suggestions of both answers with the following R-code (note that the integrand $\log(1+e^x)$ must be approximated by $x$ for large values to avoid overflow):
f <- fu |
38,027 | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | The following diagram compares the desired expectation $E[\ln \left(1+e^x\right)]$ to the 45 degree asymptote $\mu$ (when the variance is 1), and the $x$-axis asymptote when $\mu$ is negative.
That suggests a very simple approximation when $\mu$ is either relatively large negative (0) or relatively large positive ($\m... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed | The following diagram compares the desired expectation $E[\ln \left(1+e^x\right)]$ to the 45 degree asymptote $\mu$ (when the variance is 1), and the $x$-axis asymptote when $\mu$ is negative.
That s | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
The following diagram compares the desired expectation $E[\ln \left(1+e^x\right)]$ to the 45 degree asymptote $\mu$ (when the variance is 1), and the $x$-axis asymptote when $\mu$ is negative.
That suggests a very simple approximation when $\mu$ is eithe... | Expectation of $\ln(1 + e^x)$, where $x$ is normally distributed
The following diagram compares the desired expectation $E[\ln \left(1+e^x\right)]$ to the 45 degree asymptote $\mu$ (when the variance is 1), and the $x$-axis asymptote when $\mu$ is negative.
That s |
38,028 | Why does lasso not converge on a penalization parameter? | I don't know python very well, but I did find one problem with your R code.
You have the 2 lines:
residuals = sum(y_true - y_preds)
mse=residuals^2
Which sums the residuals, then squares them. This is very different from squaring the residuals, then summing them (which it appears that the python code does correctly... | Why does lasso not converge on a penalization parameter? | I don't know python very well, but I did find one problem with your R code.
You have the 2 lines:
residuals = sum(y_true - y_preds)
mse=residuals^2
Which sums the residuals, then squares them. Thi | Why does lasso not converge on a penalization parameter?
I don't know python very well, but I did find one problem with your R code.
You have the 2 lines:
residuals = sum(y_true - y_preds)
mse=residuals^2
Which sums the residuals, then squares them. This is very different from squaring the residuals, then summing t... | Why does lasso not converge on a penalization parameter?
I don't know python very well, but I did find one problem with your R code.
You have the 2 lines:
residuals = sum(y_true - y_preds)
mse=residuals^2
Which sums the residuals, then squares them. Thi |
38,029 | Why does lasso not converge on a penalization parameter? | sklearn has an example that is almost identical to what you're trying to do here: http://scikit-learn.org/stable/auto_examples/exercises/plot_cv_diabetes.html
Indeed this example shows that you do get wildly varying results for alpha for each of the three folds done in that example. This means that you cannot trust the... | Why does lasso not converge on a penalization parameter? | sklearn has an example that is almost identical to what you're trying to do here: http://scikit-learn.org/stable/auto_examples/exercises/plot_cv_diabetes.html
Indeed this example shows that you do get | Why does lasso not converge on a penalization parameter?
sklearn has an example that is almost identical to what you're trying to do here: http://scikit-learn.org/stable/auto_examples/exercises/plot_cv_diabetes.html
Indeed this example shows that you do get wildly varying results for alpha for each of the three folds d... | Why does lasso not converge on a penalization parameter?
sklearn has an example that is almost identical to what you're trying to do here: http://scikit-learn.org/stable/auto_examples/exercises/plot_cv_diabetes.html
Indeed this example shows that you do get |
38,030 | Why does lasso not converge on a penalization parameter? | The multi-collinearity in x1 and x2 is what makes the $\alpha$ value unstable in the Python code. The variance is so small for the distributions that generate these variables, so that the variance of the coefficients are inflated. The variance inflation factor (VIF) could be computed to illustrate this. After the va... | Why does lasso not converge on a penalization parameter? | The multi-collinearity in x1 and x2 is what makes the $\alpha$ value unstable in the Python code. The variance is so small for the distributions that generate these variables, so that the variance of | Why does lasso not converge on a penalization parameter?
The multi-collinearity in x1 and x2 is what makes the $\alpha$ value unstable in the Python code. The variance is so small for the distributions that generate these variables, so that the variance of the coefficients are inflated. The variance inflation factor ... | Why does lasso not converge on a penalization parameter?
The multi-collinearity in x1 and x2 is what makes the $\alpha$ value unstable in the Python code. The variance is so small for the distributions that generate these variables, so that the variance of |
38,031 | Why does lasso not converge on a penalization parameter? | I am going to comment on the R code:
You are resetting variables in the wrong places, i.e., the variables min_mse should be initialized as Inf outside the for loop and
optimal_alpha should be initialized as NULL there. This becomes:
library(glmnet)
library(lars)
library(pracma)
set.seed(1)
k = 2 # number of features... | Why does lasso not converge on a penalization parameter? | I am going to comment on the R code:
You are resetting variables in the wrong places, i.e., the variables min_mse should be initialized as Inf outside the for loop and
optimal_alpha should be initia | Why does lasso not converge on a penalization parameter?
I am going to comment on the R code:
You are resetting variables in the wrong places, i.e., the variables min_mse should be initialized as Inf outside the for loop and
optimal_alpha should be initialized as NULL there. This becomes:
library(glmnet)
library(lars... | Why does lasso not converge on a penalization parameter?
I am going to comment on the R code:
You are resetting variables in the wrong places, i.e., the variables min_mse should be initialized as Inf outside the for loop and
optimal_alpha should be initia |
38,032 | Normal Distribution or not? | For two reasons you picked the wrong kind of plot for visualizing your sample. First, you assume that your data is continuous, so there is no point in counting distinct values. Second, your sample is very small, so even with discrete numbers, in most cases you can expect small counts per value that result with a flat b... | Normal Distribution or not? | For two reasons you picked the wrong kind of plot for visualizing your sample. First, you assume that your data is continuous, so there is no point in counting distinct values. Second, your sample is | Normal Distribution or not?
For two reasons you picked the wrong kind of plot for visualizing your sample. First, you assume that your data is continuous, so there is no point in counting distinct values. Second, your sample is very small, so even with discrete numbers, in most cases you can expect small counts per val... | Normal Distribution or not?
For two reasons you picked the wrong kind of plot for visualizing your sample. First, you assume that your data is continuous, so there is no point in counting distinct values. Second, your sample is |
38,033 | Normal Distribution or not? | Are the following set of values normally distributed? 26, 33, 65, 28, 34, 55, 25, 44, 50, 36, 26, 37, 43, 62, 35, 38, 45, 32, 28, 34
Clearly not; they're integers.
[More properly, it's not a set of observed values that's normally distributed (the ECDF of a set of $n$ known values is discrete, the values themselves are... | Normal Distribution or not? | Are the following set of values normally distributed? 26, 33, 65, 28, 34, 55, 25, 44, 50, 36, 26, 37, 43, 62, 35, 38, 45, 32, 28, 34
Clearly not; they're integers.
[More properly, it's not a set of o | Normal Distribution or not?
Are the following set of values normally distributed? 26, 33, 65, 28, 34, 55, 25, 44, 50, 36, 26, 37, 43, 62, 35, 38, 45, 32, 28, 34
Clearly not; they're integers.
[More properly, it's not a set of observed values that's normally distributed (the ECDF of a set of $n$ known values is discret... | Normal Distribution or not?
Are the following set of values normally distributed? 26, 33, 65, 28, 34, 55, 25, 44, 50, 36, 26, 37, 43, 62, 35, 38, 45, 32, 28, 34
Clearly not; they're integers.
[More properly, it's not a set of o |
38,034 | What are the formulas for exponential, logarithmic, and polynomial trendlines? | I've come to the conclusion that you're probably trying to reproduce what Excel does.
This is not necessarily sensible, but at least it's pretty straightforward.
Here's what Excel does:
linear trendline: ordinary simple regression, fitted by least squares
logarithmic trendline: $y \sim a + b \ln(x)$ -- fitted by takin... | What are the formulas for exponential, logarithmic, and polynomial trendlines? | I've come to the conclusion that you're probably trying to reproduce what Excel does.
This is not necessarily sensible, but at least it's pretty straightforward.
Here's what Excel does:
linear trendl | What are the formulas for exponential, logarithmic, and polynomial trendlines?
I've come to the conclusion that you're probably trying to reproduce what Excel does.
This is not necessarily sensible, but at least it's pretty straightforward.
Here's what Excel does:
linear trendline: ordinary simple regression, fitted b... | What are the formulas for exponential, logarithmic, and polynomial trendlines?
I've come to the conclusion that you're probably trying to reproduce what Excel does.
This is not necessarily sensible, but at least it's pretty straightforward.
Here's what Excel does:
linear trendl |
38,035 | What are the formulas for exponential, logarithmic, and polynomial trendlines? | You can use the forumla for calculating the least squares coefficients of a regression with multiple independent variables. See here for a detailed explanation.
Basically they all reduce down to a single matrix equation given by the formula:
$$\hat\beta = (X'X)^{-1}X'y$$
where
$\beta=\begin{bmatrix}\beta_0 \\ \beta_1 \... | What are the formulas for exponential, logarithmic, and polynomial trendlines? | You can use the forumla for calculating the least squares coefficients of a regression with multiple independent variables. See here for a detailed explanation.
Basically they all reduce down to a sin | What are the formulas for exponential, logarithmic, and polynomial trendlines?
You can use the forumla for calculating the least squares coefficients of a regression with multiple independent variables. See here for a detailed explanation.
Basically they all reduce down to a single matrix equation given by the formula:... | What are the formulas for exponential, logarithmic, and polynomial trendlines?
You can use the forumla for calculating the least squares coefficients of a regression with multiple independent variables. See here for a detailed explanation.
Basically they all reduce down to a sin |
38,036 | What are the formulas for exponential, logarithmic, and polynomial trendlines? | Well, I think that you mean only linear models? Otherwise, as Glen_b has said, the list of models would be uncountable. Just to give you a hint, in statistics we call models linear if they are linear in their parameters. For example, the parameters (in your trend model from the picture) are $m$ and $b$, thus it is a li... | What are the formulas for exponential, logarithmic, and polynomial trendlines? | Well, I think that you mean only linear models? Otherwise, as Glen_b has said, the list of models would be uncountable. Just to give you a hint, in statistics we call models linear if they are linear | What are the formulas for exponential, logarithmic, and polynomial trendlines?
Well, I think that you mean only linear models? Otherwise, as Glen_b has said, the list of models would be uncountable. Just to give you a hint, in statistics we call models linear if they are linear in their parameters. For example, the par... | What are the formulas for exponential, logarithmic, and polynomial trendlines?
Well, I think that you mean only linear models? Otherwise, as Glen_b has said, the list of models would be uncountable. Just to give you a hint, in statistics we call models linear if they are linear |
38,037 | What are the formulas for exponential, logarithmic, and polynomial trendlines? | Use Linearizing transformation.
$$y=m\ln{x}+b$$
All you do is replace $x$ with $\ln{x}$ in all your expressions for $m$ and $b$, e.g.
$$b = \frac{\sum y- m \sum \ln{x}}{n}$$
The same thing with exponential: replace $x$ with $\exp{(x)}$.
Polynomial become a bit more involved, but the same idea. You have to apply ordinar... | What are the formulas for exponential, logarithmic, and polynomial trendlines? | Use Linearizing transformation.
$$y=m\ln{x}+b$$
All you do is replace $x$ with $\ln{x}$ in all your expressions for $m$ and $b$, e.g.
$$b = \frac{\sum y- m \sum \ln{x}}{n}$$
The same thing with expone | What are the formulas for exponential, logarithmic, and polynomial trendlines?
Use Linearizing transformation.
$$y=m\ln{x}+b$$
All you do is replace $x$ with $\ln{x}$ in all your expressions for $m$ and $b$, e.g.
$$b = \frac{\sum y- m \sum \ln{x}}{n}$$
The same thing with exponential: replace $x$ with $\exp{(x)}$.
Poly... | What are the formulas for exponential, logarithmic, and polynomial trendlines?
Use Linearizing transformation.
$$y=m\ln{x}+b$$
All you do is replace $x$ with $\ln{x}$ in all your expressions for $m$ and $b$, e.g.
$$b = \frac{\sum y- m \sum \ln{x}}{n}$$
The same thing with expone |
38,038 | The difference between the three Augmented Dickey–Fuller test (none,drift, trend) | The Wikipedia page states the following:
The testing procedure for the ADF test is the same as for the
Dickey–Fuller test but it is applied to the model
$$ \Delta y_t =
\alpha + \beta t + \gamma y_{t-1} + \delta_1 \Delta y_{t-1} + \cdots +
\delta_{p-1} \Delta y_{t-p+1} + \varepsilon_t
$$
As you very well note... | The difference between the three Augmented Dickey–Fuller test (none,drift, trend) | The Wikipedia page states the following:
The testing procedure for the ADF test is the same as for the
Dickey–Fuller test but it is applied to the model
$$ \Delta y_t =
\alpha + \beta t + \gamm | The difference between the three Augmented Dickey–Fuller test (none,drift, trend)
The Wikipedia page states the following:
The testing procedure for the ADF test is the same as for the
Dickey–Fuller test but it is applied to the model
$$ \Delta y_t =
\alpha + \beta t + \gamma y_{t-1} + \delta_1 \Delta y_{t-1} + ... | The difference between the three Augmented Dickey–Fuller test (none,drift, trend)
The Wikipedia page states the following:
The testing procedure for the ADF test is the same as for the
Dickey–Fuller test but it is applied to the model
$$ \Delta y_t =
\alpha + \beta t + \gamm |
38,039 | The difference between the three Augmented Dickey–Fuller test (none,drift, trend) | The key difference among the three versions of the test are in the specification of the test equation. As a consequence, critical values are different, too.
You want to find the correct specification of the Dickey-Fuller test regression used for testing for a unit root. It means choosing between {no constant, no trend}... | The difference between the three Augmented Dickey–Fuller test (none,drift, trend) | The key difference among the three versions of the test are in the specification of the test equation. As a consequence, critical values are different, too.
You want to find the correct specification | The difference between the three Augmented Dickey–Fuller test (none,drift, trend)
The key difference among the three versions of the test are in the specification of the test equation. As a consequence, critical values are different, too.
You want to find the correct specification of the Dickey-Fuller test regression u... | The difference between the three Augmented Dickey–Fuller test (none,drift, trend)
The key difference among the three versions of the test are in the specification of the test equation. As a consequence, critical values are different, too.
You want to find the correct specification |
38,040 | Plot log-normal distribution in R [closed] | As @ocram stated, the parameters meanlog and sdlog correspond to $\mu$ and $\sqrt{\theta}$ so you do not need to take the log yourself when specifiying the mean and standard deviation of the distribution. Here is some revised code with the true density function plotted (without having to sample as @Glen_b above sugges... | Plot log-normal distribution in R [closed] | As @ocram stated, the parameters meanlog and sdlog correspond to $\mu$ and $\sqrt{\theta}$ so you do not need to take the log yourself when specifiying the mean and standard deviation of the distribut | Plot log-normal distribution in R [closed]
As @ocram stated, the parameters meanlog and sdlog correspond to $\mu$ and $\sqrt{\theta}$ so you do not need to take the log yourself when specifiying the mean and standard deviation of the distribution. Here is some revised code with the true density function plotted (witho... | Plot log-normal distribution in R [closed]
As @ocram stated, the parameters meanlog and sdlog correspond to $\mu$ and $\sqrt{\theta}$ so you do not need to take the log yourself when specifiying the mean and standard deviation of the distribut |
38,041 | Plot log-normal distribution in R [closed] | If $X \sim \text{N}(\mu, \theta)$ and $Y = \exp(X)$, then
$$
\text{E}(X) = \mu, \quad \text{Var}(X) = \theta
$$
and
$$
\text{E}(Y) = \exp(\mu + \tfrac{\theta}{2}), \quad \text{Var}(Y) = (\exp(\theta) - 1) \exp(2 \mu + \theta)
$$
In dlnorm, the paramaters meanlog and sdlog correspond to $\mu$ and $\sqrt{\theta}$. | Plot log-normal distribution in R [closed] | If $X \sim \text{N}(\mu, \theta)$ and $Y = \exp(X)$, then
$$
\text{E}(X) = \mu, \quad \text{Var}(X) = \theta
$$
and
$$
\text{E}(Y) = \exp(\mu + \tfrac{\theta}{2}), \quad \text{Var}(Y) = (\exp(\theta) | Plot log-normal distribution in R [closed]
If $X \sim \text{N}(\mu, \theta)$ and $Y = \exp(X)$, then
$$
\text{E}(X) = \mu, \quad \text{Var}(X) = \theta
$$
and
$$
\text{E}(Y) = \exp(\mu + \tfrac{\theta}{2}), \quad \text{Var}(Y) = (\exp(\theta) - 1) \exp(2 \mu + \theta)
$$
In dlnorm, the paramaters meanlog and sdlog corr... | Plot log-normal distribution in R [closed]
If $X \sim \text{N}(\mu, \theta)$ and $Y = \exp(X)$, then
$$
\text{E}(X) = \mu, \quad \text{Var}(X) = \theta
$$
and
$$
\text{E}(Y) = \exp(\mu + \tfrac{\theta}{2}), \quad \text{Var}(Y) = (\exp(\theta) |
38,042 | Importance of normal distribution | The main reason that the normal distribution is so popular is because it works (is at least good enough in many situations). The reason that it works is really because of the Central Limit Theorem. Rather than trying to look beyond the CLT, I think you (and others) should better appreciate the CLT (I have a cross-sti... | Importance of normal distribution | The main reason that the normal distribution is so popular is because it works (is at least good enough in many situations). The reason that it works is really because of the Central Limit Theorem. | Importance of normal distribution
The main reason that the normal distribution is so popular is because it works (is at least good enough in many situations). The reason that it works is really because of the Central Limit Theorem. Rather than trying to look beyond the CLT, I think you (and others) should better appr... | Importance of normal distribution
The main reason that the normal distribution is so popular is because it works (is at least good enough in many situations). The reason that it works is really because of the Central Limit Theorem. |
38,043 | Importance of normal distribution | The Wikipedia article on the normal distribution contains many reasons. I'll summarize a few of the more useful ones here - but I really do suggest having a read through that article:
It is entirely characterized by two parameters that are easy to estimate
A sum of two jointly normal random variables is also normal
Un... | Importance of normal distribution | The Wikipedia article on the normal distribution contains many reasons. I'll summarize a few of the more useful ones here - but I really do suggest having a read through that article:
It is entirely | Importance of normal distribution
The Wikipedia article on the normal distribution contains many reasons. I'll summarize a few of the more useful ones here - but I really do suggest having a read through that article:
It is entirely characterized by two parameters that are easy to estimate
A sum of two jointly normal ... | Importance of normal distribution
The Wikipedia article on the normal distribution contains many reasons. I'll summarize a few of the more useful ones here - but I really do suggest having a read through that article:
It is entirely |
38,044 | Importance of normal distribution | We all kneel to the central limit theorem.
Here are some slightly less standard reasons why it has become "popular":
Many people never do more than one statistics course or study more than one introductory text. In such courses or texts it is customary to touch on $t$ tests, correlation and regression, for all of wh... | Importance of normal distribution | We all kneel to the central limit theorem.
Here are some slightly less standard reasons why it has become "popular":
Many people never do more than one statistics course or study more than one intr | Importance of normal distribution
We all kneel to the central limit theorem.
Here are some slightly less standard reasons why it has become "popular":
Many people never do more than one statistics course or study more than one introductory text. In such courses or texts it is customary to touch on $t$ tests, correla... | Importance of normal distribution
We all kneel to the central limit theorem.
Here are some slightly less standard reasons why it has become "popular":
Many people never do more than one statistics course or study more than one intr |
38,045 | Importance of normal distribution | I like to view the normal distribution as the curve that approximates (or is the limit of) the sum of many small little random effects. Galton's Bean machines such as the ones in the image below demonstrate this nicely and these simple models make you imagine more easily why and how we see the pattern of the normal cur... | Importance of normal distribution | I like to view the normal distribution as the curve that approximates (or is the limit of) the sum of many small little random effects. Galton's Bean machines such as the ones in the image below demon | Importance of normal distribution
I like to view the normal distribution as the curve that approximates (or is the limit of) the sum of many small little random effects. Galton's Bean machines such as the ones in the image below demonstrate this nicely and these simple models make you imagine more easily why and how we... | Importance of normal distribution
I like to view the normal distribution as the curve that approximates (or is the limit of) the sum of many small little random effects. Galton's Bean machines such as the ones in the image below demon |
38,046 | Importance of normal distribution | Essentially your question is asking about characterisations of the normal distribution. One characterisation of the distribution is that it arises when one takes a second-order Maclaurin approximation to the cumulant generating function of a distribution.
Normal distribution arises by second-order approximation to CG... | Importance of normal distribution | Essentially your question is asking about characterisations of the normal distribution. One characterisation of the distribution is that it arises when one takes a second-order Maclaurin approximatio | Importance of normal distribution
Essentially your question is asking about characterisations of the normal distribution. One characterisation of the distribution is that it arises when one takes a second-order Maclaurin approximation to the cumulant generating function of a distribution.
Normal distribution arises b... | Importance of normal distribution
Essentially your question is asking about characterisations of the normal distribution. One characterisation of the distribution is that it arises when one takes a second-order Maclaurin approximatio |
38,047 | Are Measurements made on the same patient independent? | They are definitely three different data points, but they are also definitely not independent (whether they are same day or different day). What you should do about this depends on the goals of your analysis, but it is likely that a multi-level model is a good choice. Averaging the points is also possible, but it reduc... | Are Measurements made on the same patient independent? | They are definitely three different data points, but they are also definitely not independent (whether they are same day or different day). What you should do about this depends on the goals of your a | Are Measurements made on the same patient independent?
They are definitely three different data points, but they are also definitely not independent (whether they are same day or different day). What you should do about this depends on the goals of your analysis, but it is likely that a multi-level model is a good choi... | Are Measurements made on the same patient independent?
They are definitely three different data points, but they are also definitely not independent (whether they are same day or different day). What you should do about this depends on the goals of your a |
38,048 | Are Measurements made on the same patient independent? | I mostly concur with @PeterFlom's answer. In my opinion, you should not average your data (you are basically throwing away 2/3 of your information, why would you want to do that?), but you should definitely account for the fact that measurements on the same patient will tend to be closer together than measurements on d... | Are Measurements made on the same patient independent? | I mostly concur with @PeterFlom's answer. In my opinion, you should not average your data (you are basically throwing away 2/3 of your information, why would you want to do that?), but you should defi | Are Measurements made on the same patient independent?
I mostly concur with @PeterFlom's answer. In my opinion, you should not average your data (you are basically throwing away 2/3 of your information, why would you want to do that?), but you should definitely account for the fact that measurements on the same patient... | Are Measurements made on the same patient independent?
I mostly concur with @PeterFlom's answer. In my opinion, you should not average your data (you are basically throwing away 2/3 of your information, why would you want to do that?), but you should defi |
38,049 | Are Measurements made on the same patient independent? | The three exams are different data points. Though they are clearly not independent (nor random) observations of all possible exams in your population of interest, at least for any analysis I can imagine.
Others have emphasized that you may do well to include those data points in your analysis (since you already have th... | Are Measurements made on the same patient independent? | The three exams are different data points. Though they are clearly not independent (nor random) observations of all possible exams in your population of interest, at least for any analysis I can imagi | Are Measurements made on the same patient independent?
The three exams are different data points. Though they are clearly not independent (nor random) observations of all possible exams in your population of interest, at least for any analysis I can imagine.
Others have emphasized that you may do well to include those ... | Are Measurements made on the same patient independent?
The three exams are different data points. Though they are clearly not independent (nor random) observations of all possible exams in your population of interest, at least for any analysis I can imagi |
38,050 | Are Measurements made on the same patient independent? | In accordance with the other answers (no, these observations are certainly not independent, so what do you do about it)....
But do you want to use this information to predict other variables? Many of the suggestions so far seem to be assuming you want to use spirometry as a dependent variable, and thus modelling the er... | Are Measurements made on the same patient independent? | In accordance with the other answers (no, these observations are certainly not independent, so what do you do about it)....
But do you want to use this information to predict other variables? Many of | Are Measurements made on the same patient independent?
In accordance with the other answers (no, these observations are certainly not independent, so what do you do about it)....
But do you want to use this information to predict other variables? Many of the suggestions so far seem to be assuming you want to use spirom... | Are Measurements made on the same patient independent?
In accordance with the other answers (no, these observations are certainly not independent, so what do you do about it)....
But do you want to use this information to predict other variables? Many of |
38,051 | Are Measurements made on the same patient independent? | the measurements can be independent or not. if you describe the measured value as $y_t=x_t+\varepsilon_t$, where $x_t$ - true value, and $\varepsilon_t$ - measurement error, then independence means that $cov(\varepsilon_t,\varepsilon_{t-i})=0$ for all times. this may or may not be true. if you have two measurements one... | Are Measurements made on the same patient independent? | the measurements can be independent or not. if you describe the measured value as $y_t=x_t+\varepsilon_t$, where $x_t$ - true value, and $\varepsilon_t$ - measurement error, then independence means th | Are Measurements made on the same patient independent?
the measurements can be independent or not. if you describe the measured value as $y_t=x_t+\varepsilon_t$, where $x_t$ - true value, and $\varepsilon_t$ - measurement error, then independence means that $cov(\varepsilon_t,\varepsilon_{t-i})=0$ for all times. this m... | Are Measurements made on the same patient independent?
the measurements can be independent or not. if you describe the measured value as $y_t=x_t+\varepsilon_t$, where $x_t$ - true value, and $\varepsilon_t$ - measurement error, then independence means th |
38,052 | Where to find mathematical modeling help on low-budget project? | Graduate students. Graduate students, as much as I hate to say this (being one) can be bribed with paper authorships and the like in lieu of actual money. It's important to recognize that said project may get done a little more slowly, as it won't be their first priority if they're also doing something for funding.
Som... | Where to find mathematical modeling help on low-budget project? | Graduate students. Graduate students, as much as I hate to say this (being one) can be bribed with paper authorships and the like in lieu of actual money. It's important to recognize that said project | Where to find mathematical modeling help on low-budget project?
Graduate students. Graduate students, as much as I hate to say this (being one) can be bribed with paper authorships and the like in lieu of actual money. It's important to recognize that said project may get done a little more slowly, as it won't be their... | Where to find mathematical modeling help on low-budget project?
Graduate students. Graduate students, as much as I hate to say this (being one) can be bribed with paper authorships and the like in lieu of actual money. It's important to recognize that said project |
38,053 | Where to find mathematical modeling help on low-budget project? | Not sure if you're a non-profit or not, but Jake Porway has been working on launching a Data Without Borders project, where folks can help out on non-profit projects in need of data analysis skills:
Data Without Borders seeks to match non-profits in need of data
analysis with freelance and pro bono data scientists w... | Where to find mathematical modeling help on low-budget project? | Not sure if you're a non-profit or not, but Jake Porway has been working on launching a Data Without Borders project, where folks can help out on non-profit projects in need of data analysis skills:
| Where to find mathematical modeling help on low-budget project?
Not sure if you're a non-profit or not, but Jake Porway has been working on launching a Data Without Borders project, where folks can help out on non-profit projects in need of data analysis skills:
Data Without Borders seeks to match non-profits in need ... | Where to find mathematical modeling help on low-budget project?
Not sure if you're a non-profit or not, but Jake Porway has been working on launching a Data Without Borders project, where folks can help out on non-profit projects in need of data analysis skills:
|
38,054 | Where to find mathematical modeling help on low-budget project? | You might go to linkedin.com and join some groups that match your needs and ask this question. Alternatively you might approach statistical software developers to see if they could help. | Where to find mathematical modeling help on low-budget project? | You might go to linkedin.com and join some groups that match your needs and ask this question. Alternatively you might approach statistical software developers to see if they could help. | Where to find mathematical modeling help on low-budget project?
You might go to linkedin.com and join some groups that match your needs and ask this question. Alternatively you might approach statistical software developers to see if they could help. | Where to find mathematical modeling help on low-budget project?
You might go to linkedin.com and join some groups that match your needs and ask this question. Alternatively you might approach statistical software developers to see if they could help. |
38,055 | Stacked bar plot | With 60 distinct categories, I feel you may have a hard time making that an effective graphic. You may want to consider a regular bar-chart that is sorted in ascending or descending order. Whether or not these are counts or percentages is up to you. Maybe something like this:
library(ggplot2)
df$names <- reorder(df$nam... | Stacked bar plot | With 60 distinct categories, I feel you may have a hard time making that an effective graphic. You may want to consider a regular bar-chart that is sorted in ascending or descending order. Whether or | Stacked bar plot
With 60 distinct categories, I feel you may have a hard time making that an effective graphic. You may want to consider a regular bar-chart that is sorted in ascending or descending order. Whether or not these are counts or percentages is up to you. Maybe something like this:
library(ggplot2)
df$names ... | Stacked bar plot
With 60 distinct categories, I feel you may have a hard time making that an effective graphic. You may want to consider a regular bar-chart that is sorted in ascending or descending order. Whether or |
38,056 | Stacked bar plot | I doubt you fill find a suitable range of distinct colours with so much categories. Anyway, here are some ideas:
For stacked barchart, you need barplot() with beside=FALSE (which is the default) -- this is in base R (@Chase's solution with ggplot2 is good too)
For generating a color ramp, you can use the RColorBrewer ... | Stacked bar plot | I doubt you fill find a suitable range of distinct colours with so much categories. Anyway, here are some ideas:
For stacked barchart, you need barplot() with beside=FALSE (which is the default) -- t | Stacked bar plot
I doubt you fill find a suitable range of distinct colours with so much categories. Anyway, here are some ideas:
For stacked barchart, you need barplot() with beside=FALSE (which is the default) -- this is in base R (@Chase's solution with ggplot2 is good too)
For generating a color ramp, you can use ... | Stacked bar plot
I doubt you fill find a suitable range of distinct colours with so much categories. Anyway, here are some ideas:
For stacked barchart, you need barplot() with beside=FALSE (which is the default) -- t |
38,057 | Stacked bar plot | For the coloring, either you specify a list of colors or you generate them.
In the latter, I suggest you execute this code
n = 32;
main.name = paste("color palettes; n=",n)
ch.col = c("rainbow(n, start=.7, end=.1)", "heat.colors(n)", "terrain.colors(n)", "topo.colors(n)", "cm.colors(n)");
nt <- length(ch.co... | Stacked bar plot | For the coloring, either you specify a list of colors or you generate them.
In the latter, I suggest you execute this code
n = 32;
main.name = paste("color palettes; n=",n)
ch.col = c("rainbow(n, star | Stacked bar plot
For the coloring, either you specify a list of colors or you generate them.
In the latter, I suggest you execute this code
n = 32;
main.name = paste("color palettes; n=",n)
ch.col = c("rainbow(n, start=.7, end=.1)", "heat.colors(n)", "terrain.colors(n)", "topo.colors(n)", "cm.colors(n)");
n... | Stacked bar plot
For the coloring, either you specify a list of colors or you generate them.
In the latter, I suggest you execute this code
n = 32;
main.name = paste("color palettes; n=",n)
ch.col = c("rainbow(n, star |
38,058 | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of its off-diagonal elements? | Consider the general equi-correlation covariance matrix:
\begin{align}
\Sigma = \begin{bmatrix}
1 & \rho & \cdots & \rho \\
\rho & 1 & \cdots & \rho \\
\vdots & \vdots & \ddots & \vdots \\
\rho & \rho & \cdots & 1
\end{bmatrix} \in \mathbb{R}^{n \times n}. \tag{1}
\end{align}
The sum of all the diagonal elements is $S_... | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of it | Consider the general equi-correlation covariance matrix:
\begin{align}
\Sigma = \begin{bmatrix}
1 & \rho & \cdots & \rho \\
\rho & 1 & \cdots & \rho \\
\vdots & \vdots & \ddots & \vdots \\
\rho & \rho | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of its off-diagonal elements?
Consider the general equi-correlation covariance matrix:
\begin{align}
\Sigma = \begin{bmatrix}
1 & \rho & \cdots & \rho \\
\rho & 1 & \cdots & \rho \\
\vdots & \vdots & \ddots & \vdots \\
\rho &... | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of it
Consider the general equi-correlation covariance matrix:
\begin{align}
\Sigma = \begin{bmatrix}
1 & \rho & \cdots & \rho \\
\rho & 1 & \cdots & \rho \\
\vdots & \vdots & \ddots & \vdots \\
\rho & \rho |
38,059 | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of its off-diagonal elements? | No. Highly-correlated variables will violate this rule:
x <- seq(0, 1, len = 100)
X <- data.frame(x = x, x2 = x^2, x3 = x^3)
X_cor <- cor(X)
sum(X_cor[col(X_cor) != row(X_cor)]) # 5.73822
sum(diag(X_cor)) # 3 | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of it | No. Highly-correlated variables will violate this rule:
x <- seq(0, 1, len = 100)
X <- data.frame(x = x, x2 = x^2, x3 = x^3)
X_cor <- cor(X)
sum(X_cor[col(X_cor) != row(X_cor)]) # 5.73822
sum(diag(X_ | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of its off-diagonal elements?
No. Highly-correlated variables will violate this rule:
x <- seq(0, 1, len = 100)
X <- data.frame(x = x, x2 = x^2, x3 = x^3)
X_cor <- cor(X)
sum(X_cor[col(X_cor) != row(X_cor)]) # 5.73822
sum(di... | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of it
No. Highly-correlated variables will violate this rule:
x <- seq(0, 1, len = 100)
X <- data.frame(x = x, x2 = x^2, x3 = x^3)
X_cor <- cor(X)
sum(X_cor[col(X_cor) != row(X_cor)]) # 5.73822
sum(diag(X_ |
38,060 | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of its off-diagonal elements? | You have been given two good answers. I thought it might be instructive to come at this from a different angle, and suggest how one might realise themselves that the statement is false, by finding a counterexample.
It can often be useful to run simulations (in say R or Python), to test our understanding of things or in... | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of it | You have been given two good answers. I thought it might be instructive to come at this from a different angle, and suggest how one might realise themselves that the statement is false, by finding a c | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of its off-diagonal elements?
You have been given two good answers. I thought it might be instructive to come at this from a different angle, and suggest how one might realise themselves that the statement is false, by findin... | Is the sum of the diagonal elements of a covariance matrix always equal or larger than the sum of it
You have been given two good answers. I thought it might be instructive to come at this from a different angle, and suggest how one might realise themselves that the statement is false, by finding a c |
38,061 | Correcting p-value in multiple regression [duplicate] | I suppose it depends on your intent.
In my day to day, I will regress an outcome on several variables but am really only interested in the effect of one. So I don't need to adjust anything -- those other variables are mainly there to reduce variance or so that I may stratify.
If you're conducting exploratory analyses ... | Correcting p-value in multiple regression [duplicate] | I suppose it depends on your intent.
In my day to day, I will regress an outcome on several variables but am really only interested in the effect of one. So I don't need to adjust anything -- those o | Correcting p-value in multiple regression [duplicate]
I suppose it depends on your intent.
In my day to day, I will regress an outcome on several variables but am really only interested in the effect of one. So I don't need to adjust anything -- those other variables are mainly there to reduce variance or so that I ma... | Correcting p-value in multiple regression [duplicate]
I suppose it depends on your intent.
In my day to day, I will regress an outcome on several variables but am really only interested in the effect of one. So I don't need to adjust anything -- those o |
38,062 | Correcting p-value in multiple regression [duplicate] | I would argue that this depends on why you include multiple regressors in your model. Broadly speaking, two ends of a spectrum come to mind:
A) You are interested in the effect of one regressor and "only" include the others to hopefully avoid omitted variable bias (say, the effect of education on earnings, requiring yo... | Correcting p-value in multiple regression [duplicate] | I would argue that this depends on why you include multiple regressors in your model. Broadly speaking, two ends of a spectrum come to mind:
A) You are interested in the effect of one regressor and "o | Correcting p-value in multiple regression [duplicate]
I would argue that this depends on why you include multiple regressors in your model. Broadly speaking, two ends of a spectrum come to mind:
A) You are interested in the effect of one regressor and "only" include the others to hopefully avoid omitted variable bias (... | Correcting p-value in multiple regression [duplicate]
I would argue that this depends on why you include multiple regressors in your model. Broadly speaking, two ends of a spectrum come to mind:
A) You are interested in the effect of one regressor and "o |
38,063 | Correcting p-value in multiple regression [duplicate] | A reason that people might not be applying corrections is
Because significance cut-off values are arbitrary anyway.
In some field there might be typically some couple of parameters being tested and if that number is not greatly varying then researchers in that field will settle on some value like the typical values as... | Correcting p-value in multiple regression [duplicate] | A reason that people might not be applying corrections is
Because significance cut-off values are arbitrary anyway.
In some field there might be typically some couple of parameters being tested and i | Correcting p-value in multiple regression [duplicate]
A reason that people might not be applying corrections is
Because significance cut-off values are arbitrary anyway.
In some field there might be typically some couple of parameters being tested and if that number is not greatly varying then researchers in that fiel... | Correcting p-value in multiple regression [duplicate]
A reason that people might not be applying corrections is
Because significance cut-off values are arbitrary anyway.
In some field there might be typically some couple of parameters being tested and i |
38,064 | Correcting p-value in multiple regression [duplicate] | Yea, usually you need to because of the problem with multiple comparison (https://en.wikipedia.org/wiki/Multiple_comparisons_problem). Depending on what your are doing (academic or industry) the most conservative and considered the best is the Bonferroni correction (https://en.wikipedia.org/wiki/Bonferroni_correction).... | Correcting p-value in multiple regression [duplicate] | Yea, usually you need to because of the problem with multiple comparison (https://en.wikipedia.org/wiki/Multiple_comparisons_problem). Depending on what your are doing (academic or industry) the most | Correcting p-value in multiple regression [duplicate]
Yea, usually you need to because of the problem with multiple comparison (https://en.wikipedia.org/wiki/Multiple_comparisons_problem). Depending on what your are doing (academic or industry) the most conservative and considered the best is the Bonferroni correction ... | Correcting p-value in multiple regression [duplicate]
Yea, usually you need to because of the problem with multiple comparison (https://en.wikipedia.org/wiki/Multiple_comparisons_problem). Depending on what your are doing (academic or industry) the most |
38,065 | Difference between E(Y) and E(Y|X) in regression | Disclaimer: I only read your question correctly after i wrote this. If $X$ is non random then yeah $E(Y|X) = E(Y)$.
Your mistake comes down to an abuse of notation. Within usual notation $X$ is a random variable which for simplicity, takes values in the reals. $x$ is one such value $X$ could take. Let's sensibly define... | Difference between E(Y) and E(Y|X) in regression | Disclaimer: I only read your question correctly after i wrote this. If $X$ is non random then yeah $E(Y|X) = E(Y)$.
Your mistake comes down to an abuse of notation. Within usual notation $X$ is a rand | Difference between E(Y) and E(Y|X) in regression
Disclaimer: I only read your question correctly after i wrote this. If $X$ is non random then yeah $E(Y|X) = E(Y)$.
Your mistake comes down to an abuse of notation. Within usual notation $X$ is a random variable which for simplicity, takes values in the reals. $x$ is one... | Difference between E(Y) and E(Y|X) in regression
Disclaimer: I only read your question correctly after i wrote this. If $X$ is non random then yeah $E(Y|X) = E(Y)$.
Your mistake comes down to an abuse of notation. Within usual notation $X$ is a rand |
38,066 | Difference between E(Y) and E(Y|X) in regression | If $X$ is not random (i.e. just some constant), then $E[Y|X] = E[Y]$ so it is correct. If $X$ was random then $E[Y] = \beta_0 + \beta_1 E[X] + E[\mu]$ and so, $E[Y|X]$ would not be equal to $E[Y]$ in general | Difference between E(Y) and E(Y|X) in regression | If $X$ is not random (i.e. just some constant), then $E[Y|X] = E[Y]$ so it is correct. If $X$ was random then $E[Y] = \beta_0 + \beta_1 E[X] + E[\mu]$ and so, $E[Y|X]$ would not be equal to $E[Y]$ in | Difference between E(Y) and E(Y|X) in regression
If $X$ is not random (i.e. just some constant), then $E[Y|X] = E[Y]$ so it is correct. If $X$ was random then $E[Y] = \beta_0 + \beta_1 E[X] + E[\mu]$ and so, $E[Y|X]$ would not be equal to $E[Y]$ in general | Difference between E(Y) and E(Y|X) in regression
If $X$ is not random (i.e. just some constant), then $E[Y|X] = E[Y]$ so it is correct. If $X$ was random then $E[Y] = \beta_0 + \beta_1 E[X] + E[\mu]$ and so, $E[Y|X]$ would not be equal to $E[Y]$ in |
38,067 | Difference between E(Y) and E(Y|X) in regression | $\bullet$ You must bear in mind $\mathbb E[\boldsymbol \varepsilon |\mathbf X]=\mathbf 0\implies \mathbb E[\boldsymbol \varepsilon]=\mathbf 0$ but not $\mathbb E[\boldsymbol \varepsilon]=\mathbf 0\implies \mathbb E[\boldsymbol\varepsilon |\mathbf X]=\mathbf 0.$
$\bullet$ For non-stochastic regressors, you can straightf... | Difference between E(Y) and E(Y|X) in regression | $\bullet$ You must bear in mind $\mathbb E[\boldsymbol \varepsilon |\mathbf X]=\mathbf 0\implies \mathbb E[\boldsymbol \varepsilon]=\mathbf 0$ but not $\mathbb E[\boldsymbol \varepsilon]=\mathbf 0\imp | Difference between E(Y) and E(Y|X) in regression
$\bullet$ You must bear in mind $\mathbb E[\boldsymbol \varepsilon |\mathbf X]=\mathbf 0\implies \mathbb E[\boldsymbol \varepsilon]=\mathbf 0$ but not $\mathbb E[\boldsymbol \varepsilon]=\mathbf 0\implies \mathbb E[\boldsymbol\varepsilon |\mathbf X]=\mathbf 0.$
$\bullet$... | Difference between E(Y) and E(Y|X) in regression
$\bullet$ You must bear in mind $\mathbb E[\boldsymbol \varepsilon |\mathbf X]=\mathbf 0\implies \mathbb E[\boldsymbol \varepsilon]=\mathbf 0$ but not $\mathbb E[\boldsymbol \varepsilon]=\mathbf 0\imp |
38,068 | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target value dropped since it's a lone constant? | No. A proper Norm will not allow it to be.
Even the simplest absolute value function as a loss will depend on $t$: $|m(w)-t|’=\pm m’(w)$, here the sign depends on $t$.
TL;DR;
Generally, your loss function will be $L(w|t,X)$, so the first derivative is $\partial L(w|t,X)/\partial w$, and there's no reason for $t$ to dis... | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target | No. A proper Norm will not allow it to be.
Even the simplest absolute value function as a loss will depend on $t$: $|m(w)-t|’=\pm m’(w)$, here the sign depends on $t$.
TL;DR;
Generally, your loss func | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target value dropped since it's a lone constant?
No. A proper Norm will not allow it to be.
Even the simplest absolute value function as a loss will depend on $t$: $|m(w)-t|’=\pm m’(w)$, here the sign depends on $t$.
TL;DR;
Gen... | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target
No. A proper Norm will not allow it to be.
Even the simplest absolute value function as a loss will depend on $t$: $|m(w)-t|’=\pm m’(w)$, here the sign depends on $t$.
TL;DR;
Generally, your loss func |
38,069 | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target value dropped since it's a lone constant? | If we are considering the absolute difference as a norm, that is:
$loss(w) = |m_x(w) - t|$
then $\nabla loss(w)$ is far from simply being equivalent to $\nabla m_x(w)$.
By definition of the derivative for an absolute value (and using the chain rule), we actually get:
$\nabla loss(w) = \frac{m_x(w) - t}{|m_x(w) - t|}. ... | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target | If we are considering the absolute difference as a norm, that is:
$loss(w) = |m_x(w) - t|$
then $\nabla loss(w)$ is far from simply being equivalent to $\nabla m_x(w)$.
By definition of the derivative | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target value dropped since it's a lone constant?
If we are considering the absolute difference as a norm, that is:
$loss(w) = |m_x(w) - t|$
then $\nabla loss(w)$ is far from simply being equivalent to $\nabla m_x(w)$.
By defini... | In GD-optimisation, if the gradient of the error function is w.r.t to the weights, isn't the target
If we are considering the absolute difference as a norm, that is:
$loss(w) = |m_x(w) - t|$
then $\nabla loss(w)$ is far from simply being equivalent to $\nabla m_x(w)$.
By definition of the derivative |
38,070 | How to test whether a correlation is equal to 1? | I would argue that there is not any testing to do. If the sample correlation is not 1, then you reject $H_0: \rho=1$ with certainty.
Having a correlation of 1 means that the points cannot deviate from a diagonal line the way that they can when $\vert \rho \vert < 1$.
EDIT
set.seed(2019)
x <- rexp(1000)
y <- 3*x
plot(x,... | How to test whether a correlation is equal to 1? | I would argue that there is not any testing to do. If the sample correlation is not 1, then you reject $H_0: \rho=1$ with certainty.
Having a correlation of 1 means that the points cannot deviate from | How to test whether a correlation is equal to 1?
I would argue that there is not any testing to do. If the sample correlation is not 1, then you reject $H_0: \rho=1$ with certainty.
Having a correlation of 1 means that the points cannot deviate from a diagonal line the way that they can when $\vert \rho \vert < 1$.
EDI... | How to test whether a correlation is equal to 1?
I would argue that there is not any testing to do. If the sample correlation is not 1, then you reject $H_0: \rho=1$ with certainty.
Having a correlation of 1 means that the points cannot deviate from |
38,071 | How to test whether a correlation is equal to 1? | Using the Fisher Z-transform is one way of doing this (usually used for confidence intervals), bootstrapping would be another.
Here's a brief article for Fisher Z transform for Pearson Product Moment Correlation Coefficient https://www.statisticshowto.datasciencecentral.com/fisher-z/ | How to test whether a correlation is equal to 1? | Using the Fisher Z-transform is one way of doing this (usually used for confidence intervals), bootstrapping would be another.
Here's a brief article for Fisher Z transform for Pearson Product Moment | How to test whether a correlation is equal to 1?
Using the Fisher Z-transform is one way of doing this (usually used for confidence intervals), bootstrapping would be another.
Here's a brief article for Fisher Z transform for Pearson Product Moment Correlation Coefficient https://www.statisticshowto.datasciencecentral.... | How to test whether a correlation is equal to 1?
Using the Fisher Z-transform is one way of doing this (usually used for confidence intervals), bootstrapping would be another.
Here's a brief article for Fisher Z transform for Pearson Product Moment |
38,072 | Calculating confidence in part of a sample (explain like I'm in the humanities please :) ) | ...explain like I'm in the humanities please :)
The paradox of "confidence" relating to inference from sampling is an expression of the hermeneutics of anxiety, tracable to the observations of Kierkegaard. "Statistical Science" puts forward a symbolic synthesis to internalise and "objectify" the antithesis of passion... | Calculating confidence in part of a sample (explain like I'm in the humanities please :) ) | ...explain like I'm in the humanities please :)
The paradox of "confidence" relating to inference from sampling is an expression of the hermeneutics of anxiety, tracable to the observations of Kierke | Calculating confidence in part of a sample (explain like I'm in the humanities please :) )
...explain like I'm in the humanities please :)
The paradox of "confidence" relating to inference from sampling is an expression of the hermeneutics of anxiety, tracable to the observations of Kierkegaard. "Statistical Science"... | Calculating confidence in part of a sample (explain like I'm in the humanities please :) )
...explain like I'm in the humanities please :)
The paradox of "confidence" relating to inference from sampling is an expression of the hermeneutics of anxiety, tracable to the observations of Kierke |
38,073 | Calculating confidence in part of a sample (explain like I'm in the humanities please :) ) | I promised a serious answer (to supplement my gag answer) so here goes. I think at present you are mixing several issues, including considerations relating to descriptive statistics for a sample and some (not clearly specified) population inferences. The main thing I would recommend at this stage is to take some time... | Calculating confidence in part of a sample (explain like I'm in the humanities please :) ) | I promised a serious answer (to supplement my gag answer) so here goes. I think at present you are mixing several issues, including considerations relating to descriptive statistics for a sample and | Calculating confidence in part of a sample (explain like I'm in the humanities please :) )
I promised a serious answer (to supplement my gag answer) so here goes. I think at present you are mixing several issues, including considerations relating to descriptive statistics for a sample and some (not clearly specified) ... | Calculating confidence in part of a sample (explain like I'm in the humanities please :) )
I promised a serious answer (to supplement my gag answer) so here goes. I think at present you are mixing several issues, including considerations relating to descriptive statistics for a sample and |
38,074 | Calculating confidence in part of a sample (explain like I'm in the humanities please :) ) | Reading some of your follow-up comments, it sounds like you want to provide descriptive statistics with confidence intervals.
The uncertainty around the total population size generally isn't a problem. Typically, you'd just assume that the total population is infinite. This is a conservative approach. The you follow th... | Calculating confidence in part of a sample (explain like I'm in the humanities please :) ) | Reading some of your follow-up comments, it sounds like you want to provide descriptive statistics with confidence intervals.
The uncertainty around the total population size generally isn't a problem | Calculating confidence in part of a sample (explain like I'm in the humanities please :) )
Reading some of your follow-up comments, it sounds like you want to provide descriptive statistics with confidence intervals.
The uncertainty around the total population size generally isn't a problem. Typically, you'd just assum... | Calculating confidence in part of a sample (explain like I'm in the humanities please :) )
Reading some of your follow-up comments, it sounds like you want to provide descriptive statistics with confidence intervals.
The uncertainty around the total population size generally isn't a problem |
38,075 | Is Bessel's correction required when calculating mean? | NOPE
When you calculate the variance as $s^2 = \sum_{i=1}^n\Bigg[\dfrac{(x_i-\bar{x})^2}{n-1}\Bigg]$, there is another term in there that you are calculating, the $\bar{x}$.
When you calculate the usual mean, $\bar{x}=\dfrac{\sum_{i=1}^n x_i}{n}$, there is no such other term to calculate, so you do not drop that so-cal... | Is Bessel's correction required when calculating mean? | NOPE
When you calculate the variance as $s^2 = \sum_{i=1}^n\Bigg[\dfrac{(x_i-\bar{x})^2}{n-1}\Bigg]$, there is another term in there that you are calculating, the $\bar{x}$.
When you calculate the usu | Is Bessel's correction required when calculating mean?
NOPE
When you calculate the variance as $s^2 = \sum_{i=1}^n\Bigg[\dfrac{(x_i-\bar{x})^2}{n-1}\Bigg]$, there is another term in there that you are calculating, the $\bar{x}$.
When you calculate the usual mean, $\bar{x}=\dfrac{\sum_{i=1}^n x_i}{n}$, there is no such ... | Is Bessel's correction required when calculating mean?
NOPE
When you calculate the variance as $s^2 = \sum_{i=1}^n\Bigg[\dfrac{(x_i-\bar{x})^2}{n-1}\Bigg]$, there is another term in there that you are calculating, the $\bar{x}$.
When you calculate the usu |
38,076 | Is Bessel's correction required when calculating mean? | No, Bessel's correction is not required for the sample mean
The reason Bessel's correction is required for the sample variance is that estimation of the variance also requires us to use the sample mean to estimate the true mean (around which this variance is formed). This is evident when we compare the the differences... | Is Bessel's correction required when calculating mean? | No, Bessel's correction is not required for the sample mean
The reason Bessel's correction is required for the sample variance is that estimation of the variance also requires us to use the sample mea | Is Bessel's correction required when calculating mean?
No, Bessel's correction is not required for the sample mean
The reason Bessel's correction is required for the sample variance is that estimation of the variance also requires us to use the sample mean to estimate the true mean (around which this variance is formed... | Is Bessel's correction required when calculating mean?
No, Bessel's correction is not required for the sample mean
The reason Bessel's correction is required for the sample variance is that estimation of the variance also requires us to use the sample mea |
38,077 | Is Bessel's correction required when calculating mean? | I'll approach this from the Bessel's correction for variance.
Wikipedia brings in its third proof the expected discrepancy between the true variance $\sigma^2$ and the biased estimate $s_n^2$ is given:
$$
E\left[\sigma^2 - s_n^2\right]
= E\left[\frac{1}{n}\sum_i^n(x_i-\mu)^2-\frac{1}{n}\sum_i^n(x_i-\bar x)^2\right]\\
=... | Is Bessel's correction required when calculating mean? | I'll approach this from the Bessel's correction for variance.
Wikipedia brings in its third proof the expected discrepancy between the true variance $\sigma^2$ and the biased estimate $s_n^2$ is given | Is Bessel's correction required when calculating mean?
I'll approach this from the Bessel's correction for variance.
Wikipedia brings in its third proof the expected discrepancy between the true variance $\sigma^2$ and the biased estimate $s_n^2$ is given:
$$
E\left[\sigma^2 - s_n^2\right]
= E\left[\frac{1}{n}\sum_i^n(... | Is Bessel's correction required when calculating mean?
I'll approach this from the Bessel's correction for variance.
Wikipedia brings in its third proof the expected discrepancy between the true variance $\sigma^2$ and the biased estimate $s_n^2$ is given |
38,078 | Is Bessel's correction required when calculating mean? | No, don't divide by $n - 1$ when estimating the mean.
Suppose I want to estimate how tall the average person is. I randomly select 4 people, and all of them are 6 feet tall. Which is a better estimate of the height of the average person:
$$\frac{6 \text{ feet} + 6 \text{ feet} + 6 \text{ feet} + 6 \text{ feet}}{4} = 6 ... | Is Bessel's correction required when calculating mean? | No, don't divide by $n - 1$ when estimating the mean.
Suppose I want to estimate how tall the average person is. I randomly select 4 people, and all of them are 6 feet tall. Which is a better estimate | Is Bessel's correction required when calculating mean?
No, don't divide by $n - 1$ when estimating the mean.
Suppose I want to estimate how tall the average person is. I randomly select 4 people, and all of them are 6 feet tall. Which is a better estimate of the height of the average person:
$$\frac{6 \text{ feet} + 6 ... | Is Bessel's correction required when calculating mean?
No, don't divide by $n - 1$ when estimating the mean.
Suppose I want to estimate how tall the average person is. I randomly select 4 people, and all of them are 6 feet tall. Which is a better estimate |
38,079 | Dependent and not identically distributed random variables | Next to the "formal" example by Xi'an, a "real-world" example might be height and weight. Already because the two are measured on different scales will they be distributed differently, but they sure are dependent, as taller people tend to be heavier. | Dependent and not identically distributed random variables | Next to the "formal" example by Xi'an, a "real-world" example might be height and weight. Already because the two are measured on different scales will they be distributed differently, but they sure a | Dependent and not identically distributed random variables
Next to the "formal" example by Xi'an, a "real-world" example might be height and weight. Already because the two are measured on different scales will they be distributed differently, but they sure are dependent, as taller people tend to be heavier. | Dependent and not identically distributed random variables
Next to the "formal" example by Xi'an, a "real-world" example might be height and weight. Already because the two are measured on different scales will they be distributed differently, but they sure a |
38,080 | Dependent and not identically distributed random variables | If you randomly draw a card from a deck of playing cards, do not put it back, and draw again. Then, the probability distributions for which card will be drawn in each of the two draws are dependent and not identical.
Otherwise, if the card of first draw is put back and well shuffled before the second draw, then the dis... | Dependent and not identically distributed random variables | If you randomly draw a card from a deck of playing cards, do not put it back, and draw again. Then, the probability distributions for which card will be drawn in each of the two draws are dependent an | Dependent and not identically distributed random variables
If you randomly draw a card from a deck of playing cards, do not put it back, and draw again. Then, the probability distributions for which card will be drawn in each of the two draws are dependent and not identical.
Otherwise, if the card of first draw is put ... | Dependent and not identically distributed random variables
If you randomly draw a card from a deck of playing cards, do not put it back, and draw again. Then, the probability distributions for which card will be drawn in each of the two draws are dependent an |
38,081 | Dependent and not identically distributed random variables | Autorcorrelated processes
A variable in series that 'remembers' its previous values to some degree is not i.i.d.! Any autoregresive value depends on previous values of the variable, and the distribution changes depending on location within the series.
For example, the time series variable $y$, where $t$ indicates perio... | Dependent and not identically distributed random variables | Autorcorrelated processes
A variable in series that 'remembers' its previous values to some degree is not i.i.d.! Any autoregresive value depends on previous values of the variable, and the distributi | Dependent and not identically distributed random variables
Autorcorrelated processes
A variable in series that 'remembers' its previous values to some degree is not i.i.d.! Any autoregresive value depends on previous values of the variable, and the distribution changes depending on location within the series.
For examp... | Dependent and not identically distributed random variables
Autorcorrelated processes
A variable in series that 'remembers' its previous values to some degree is not i.i.d.! Any autoregresive value depends on previous values of the variable, and the distributi |
38,082 | Dependent and not identically distributed random variables | If $\varepsilon_1,\varepsilon_2$ are iid $\mathcal N(0,1)$,
$$X_1=\mu_1+\sigma_1\epsilon_1\qquad X_2=\mu_2+\varrho \epsilon_1 + \sigma_2 \epsilon_2$$
is a pair of dependent RVs that are not identically distributed for most values of the parameters. | Dependent and not identically distributed random variables | If $\varepsilon_1,\varepsilon_2$ are iid $\mathcal N(0,1)$,
$$X_1=\mu_1+\sigma_1\epsilon_1\qquad X_2=\mu_2+\varrho \epsilon_1 + \sigma_2 \epsilon_2$$
is a pair of dependent RVs that are not identicall | Dependent and not identically distributed random variables
If $\varepsilon_1,\varepsilon_2$ are iid $\mathcal N(0,1)$,
$$X_1=\mu_1+\sigma_1\epsilon_1\qquad X_2=\mu_2+\varrho \epsilon_1 + \sigma_2 \epsilon_2$$
is a pair of dependent RVs that are not identically distributed for most values of the parameters. | Dependent and not identically distributed random variables
If $\varepsilon_1,\varepsilon_2$ are iid $\mathcal N(0,1)$,
$$X_1=\mu_1+\sigma_1\epsilon_1\qquad X_2=\mu_2+\varrho \epsilon_1 + \sigma_2 \epsilon_2$$
is a pair of dependent RVs that are not identicall |
38,083 | Dependent and not identically distributed random variables | Some other "real-world"-examples:
Let $(M, F)$ be a pair of measurements on an opposite sex married couple, sampled randomly:
Measurement is height, will have different means.
Measurement is IQ, same mean, different variance.
(But maybe for this example, independence is in doubt ...) Paired data in general can be use... | Dependent and not identically distributed random variables | Some other "real-world"-examples:
Let $(M, F)$ be a pair of measurements on an opposite sex married couple, sampled randomly:
Measurement is height, will have different means.
Measurement is IQ, same | Dependent and not identically distributed random variables
Some other "real-world"-examples:
Let $(M, F)$ be a pair of measurements on an opposite sex married couple, sampled randomly:
Measurement is height, will have different means.
Measurement is IQ, same mean, different variance.
(But maybe for this example, inde... | Dependent and not identically distributed random variables
Some other "real-world"-examples:
Let $(M, F)$ be a pair of measurements on an opposite sex married couple, sampled randomly:
Measurement is height, will have different means.
Measurement is IQ, same |
38,084 | PCA finds a variable to be the most important twice | Your interpretation of PCA components is not correct.
PCA does not tell you which variables account for the most variation in the data, so a statement like
Calcium is the most explanatory of the variance, as well as also being the third most explanatory variable for the variance.
cannot be drawn from a PC analysis.
... | PCA finds a variable to be the most important twice | Your interpretation of PCA components is not correct.
PCA does not tell you which variables account for the most variation in the data, so a statement like
Calcium is the most explanatory of the var | PCA finds a variable to be the most important twice
Your interpretation of PCA components is not correct.
PCA does not tell you which variables account for the most variation in the data, so a statement like
Calcium is the most explanatory of the variance, as well as also being the third most explanatory variable for... | PCA finds a variable to be the most important twice
Your interpretation of PCA components is not correct.
PCA does not tell you which variables account for the most variation in the data, so a statement like
Calcium is the most explanatory of the var |
38,085 | PCA finds a variable to be the most important twice | You aren't interpreting PCA correctly. PCA finds a whole new basis for your data. It's analogous to a change of basis: https://www.math.hmc.edu/calculus/tutorials/changebasis/ but we choose a particular basis
The new basis is not arbitrary: the vectors are selected based on how much variation they account for. That ... | PCA finds a variable to be the most important twice | You aren't interpreting PCA correctly. PCA finds a whole new basis for your data. It's analogous to a change of basis: https://www.math.hmc.edu/calculus/tutorials/changebasis/ but we choose a partic | PCA finds a variable to be the most important twice
You aren't interpreting PCA correctly. PCA finds a whole new basis for your data. It's analogous to a change of basis: https://www.math.hmc.edu/calculus/tutorials/changebasis/ but we choose a particular basis
The new basis is not arbitrary: the vectors are selected ... | PCA finds a variable to be the most important twice
You aren't interpreting PCA correctly. PCA finds a whole new basis for your data. It's analogous to a change of basis: https://www.math.hmc.edu/calculus/tutorials/changebasis/ but we choose a partic |
38,086 | PCA finds a variable to be the most important twice | Correlation is not the same as linear combination with largest variance, which is what PCA finds.
Also the eigenvectors have no particular direction. You can multiply them with $-1$ and those vectors will also be eigenvectors with the same eigenvalue (variance) and then you would get positive $+0.677\cdots$ for third ... | PCA finds a variable to be the most important twice | Correlation is not the same as linear combination with largest variance, which is what PCA finds.
Also the eigenvectors have no particular direction. You can multiply them with $-1$ and those vectors | PCA finds a variable to be the most important twice
Correlation is not the same as linear combination with largest variance, which is what PCA finds.
Also the eigenvectors have no particular direction. You can multiply them with $-1$ and those vectors will also be eigenvectors with the same eigenvalue (variance) and t... | PCA finds a variable to be the most important twice
Correlation is not the same as linear combination with largest variance, which is what PCA finds.
Also the eigenvectors have no particular direction. You can multiply them with $-1$ and those vectors |
38,087 | How is the $\chi^2_1$-distribution not a Gaussian? | As Silverfish said, the problem in your reasoning is that to find the PDF of a squared random variable, or any other transformed random variable, you can't just perform that transformation on the PDF.
If we want to know the actual PDF of a squared random variable we must calculate $P(\chi ^2_1 = x)$. One way to do thi... | How is the $\chi^2_1$-distribution not a Gaussian? | As Silverfish said, the problem in your reasoning is that to find the PDF of a squared random variable, or any other transformed random variable, you can't just perform that transformation on the PDF. | How is the $\chi^2_1$-distribution not a Gaussian?
As Silverfish said, the problem in your reasoning is that to find the PDF of a squared random variable, or any other transformed random variable, you can't just perform that transformation on the PDF.
If we want to know the actual PDF of a squared random variable we m... | How is the $\chi^2_1$-distribution not a Gaussian?
As Silverfish said, the problem in your reasoning is that to find the PDF of a squared random variable, or any other transformed random variable, you can't just perform that transformation on the PDF. |
38,088 | How is the $\chi^2_1$-distribution not a Gaussian? | Normally distributed random variable can take values ranging from $-\infty$ to $\infty$. Squaring any real value would lead to positive values. So how can possibly $Z^2$ be normally distributed?
Squaring probability density function does not make any sense since this would give you "probabilities squared" (more precis... | How is the $\chi^2_1$-distribution not a Gaussian? | Normally distributed random variable can take values ranging from $-\infty$ to $\infty$. Squaring any real value would lead to positive values. So how can possibly $Z^2$ be normally distributed?
Squa | How is the $\chi^2_1$-distribution not a Gaussian?
Normally distributed random variable can take values ranging from $-\infty$ to $\infty$. Squaring any real value would lead to positive values. So how can possibly $Z^2$ be normally distributed?
Squaring probability density function does not make any sense since this ... | How is the $\chi^2_1$-distribution not a Gaussian?
Normally distributed random variable can take values ranging from $-\infty$ to $\infty$. Squaring any real value would lead to positive values. So how can possibly $Z^2$ be normally distributed?
Squa |
38,089 | How is the $\chi^2_1$-distribution not a Gaussian? | Actually, $\chi^2_\infty$ distribution "looks like" normal. The exact relation for $x\sim\chi^2_n$ is $(x-n)/2\sqrt n\sim\mathcal{N}(0,1)$
How come? As you noted a variable from this distribution can be thought of as a sum of squared normals: $Z_1^2+Z_2^2+\dots+Z_n^2$. However, the squared normal random variables are t... | How is the $\chi^2_1$-distribution not a Gaussian? | Actually, $\chi^2_\infty$ distribution "looks like" normal. The exact relation for $x\sim\chi^2_n$ is $(x-n)/2\sqrt n\sim\mathcal{N}(0,1)$
How come? As you noted a variable from this distribution can | How is the $\chi^2_1$-distribution not a Gaussian?
Actually, $\chi^2_\infty$ distribution "looks like" normal. The exact relation for $x\sim\chi^2_n$ is $(x-n)/2\sqrt n\sim\mathcal{N}(0,1)$
How come? As you noted a variable from this distribution can be thought of as a sum of squared normals: $Z_1^2+Z_2^2+\dots+Z_n^2$.... | How is the $\chi^2_1$-distribution not a Gaussian?
Actually, $\chi^2_\infty$ distribution "looks like" normal. The exact relation for $x\sim\chi^2_n$ is $(x-n)/2\sqrt n\sim\mathcal{N}(0,1)$
How come? As you noted a variable from this distribution can |
38,090 | Negative BIC in k-means | Generally, the aim is to minimize BIC, so if you are in a negative territory, a negative number that has the largest modulus (deepest down in the negative territory) indicates the preferred model. Hence, in your plot the best case would appear to be "2".
However, the definition of BIC used in the mclust package happens... | Negative BIC in k-means | Generally, the aim is to minimize BIC, so if you are in a negative territory, a negative number that has the largest modulus (deepest down in the negative territory) indicates the preferred model. Hen | Negative BIC in k-means
Generally, the aim is to minimize BIC, so if you are in a negative territory, a negative number that has the largest modulus (deepest down in the negative territory) indicates the preferred model. Hence, in your plot the best case would appear to be "2".
However, the definition of BIC used in th... | Negative BIC in k-means
Generally, the aim is to minimize BIC, so if you are in a negative territory, a negative number that has the largest modulus (deepest down in the negative territory) indicates the preferred model. Hen |
38,091 | Negative BIC in k-means | This may result useful for someone else.
I was puzzled with the mclust package because I tried Gaussian mixture models to check whether my data followed a uni- or multi-modal Gaussian distribution.
I found that, according to the examples provided in the help, the model that best fitted my data was one with two componen... | Negative BIC in k-means | This may result useful for someone else.
I was puzzled with the mclust package because I tried Gaussian mixture models to check whether my data followed a uni- or multi-modal Gaussian distribution.
I | Negative BIC in k-means
This may result useful for someone else.
I was puzzled with the mclust package because I tried Gaussian mixture models to check whether my data followed a uni- or multi-modal Gaussian distribution.
I found that, according to the examples provided in the help, the model that best fitted my data w... | Negative BIC in k-means
This may result useful for someone else.
I was puzzled with the mclust package because I tried Gaussian mixture models to check whether my data followed a uni- or multi-modal Gaussian distribution.
I |
38,092 | How to use boxplots to find the point where values are more likely to come from different conditions? | @NickCox has presented a good way to visualize your data. I take it you want to find a rule for deciding when to classify a value as condition1 vs condition2.
In an earlier version of your question, you wondered if you should call any value greater than the median of condition1 as a member of condition2. This is no... | How to use boxplots to find the point where values are more likely to come from different conditions | @NickCox has presented a good way to visualize your data. I take it you want to find a rule for deciding when to classify a value as condition1 vs condition2.
In an earlier version of your question | How to use boxplots to find the point where values are more likely to come from different conditions?
@NickCox has presented a good way to visualize your data. I take it you want to find a rule for deciding when to classify a value as condition1 vs condition2.
In an earlier version of your question, you wondered if ... | How to use boxplots to find the point where values are more likely to come from different conditions
@NickCox has presented a good way to visualize your data. I take it you want to find a rule for deciding when to classify a value as condition1 vs condition2.
In an earlier version of your question |
38,093 | How to use boxplots to find the point where values are more likely to come from different conditions? | Here is one of many possibilities. Back in 1979, Emanuel Parzen suggested hybridising the quantile plot and the box plot. Some references are given below. Clearly, the box of the box plot shows median and quartiles, which are just key quantiles. Showing all of the data, namely all the quantiles or order statistics, is ... | How to use boxplots to find the point where values are more likely to come from different conditions | Here is one of many possibilities. Back in 1979, Emanuel Parzen suggested hybridising the quantile plot and the box plot. Some references are given below. Clearly, the box of the box plot shows median | How to use boxplots to find the point where values are more likely to come from different conditions?
Here is one of many possibilities. Back in 1979, Emanuel Parzen suggested hybridising the quantile plot and the box plot. Some references are given below. Clearly, the box of the box plot shows median and quartiles, wh... | How to use boxplots to find the point where values are more likely to come from different conditions
Here is one of many possibilities. Back in 1979, Emanuel Parzen suggested hybridising the quantile plot and the box plot. Some references are given below. Clearly, the box of the box plot shows median |
38,094 | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | You mention in the comments that you are computing the AUC using a 75-25 train-test split, and you are puzzled why AUC is maximized when training your model on only 8 of your 30 regressors. From this you have gotten the impression that AUC is somehow penalizing complexity in your model.
In reality there is something pe... | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | You mention in the comments that you are computing the AUC using a 75-25 train-test split, and you are puzzled why AUC is maximized when training your model on only 8 of your 30 regressors. From this | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
You mention in the comments that you are computing the AUC using a 75-25 train-test split, and you are puzzled why AUC is maximized when training your model on only 8 of your 30 regressors. From this you have gotten the impression that A... | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
You mention in the comments that you are computing the AUC using a 75-25 train-test split, and you are puzzled why AUC is maximized when training your model on only 8 of your 30 regressors. From this |
38,095 | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | There is a good reason why the regression coefficients in logistic regression are estimated by maximizing the likelihood or penalized likelihood. This leads to certain optimality properties. The concordance probability ($c$-index; AUROC) is a useful supplemental measure for describing the final model's predictive dis... | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | There is a good reason why the regression coefficients in logistic regression are estimated by maximizing the likelihood or penalized likelihood. This leads to certain optimality properties. The con | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
There is a good reason why the regression coefficients in logistic regression are estimated by maximizing the likelihood or penalized likelihood. This leads to certain optimality properties. The concordance probability ($c$-index; AURO... | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
There is a good reason why the regression coefficients in logistic regression are estimated by maximizing the likelihood or penalized likelihood. This leads to certain optimality properties. The con |
38,096 | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | This should help clarify a few things, in as few words as possible:
AUC = measure of model's actual predictive performance
BIC = estimate of model's predictive performance
Performance Measures, like AUC, are something you would use to evaluate a model's predictions on data it has never seen before.
Information Crit... | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | This should help clarify a few things, in as few words as possible:
AUC = measure of model's actual predictive performance
BIC = estimate of model's predictive performance
Performance Measures, li | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
This should help clarify a few things, in as few words as possible:
AUC = measure of model's actual predictive performance
BIC = estimate of model's predictive performance
Performance Measures, like AUC, are something you would use t... | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
This should help clarify a few things, in as few words as possible:
AUC = measure of model's actual predictive performance
BIC = estimate of model's predictive performance
Performance Measures, li |
38,097 | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | In logistic regression (I do it univariate for easier typing) you try to explain a binary outome $y_i \in \{0,1\}$ by assuming that it is the outcome of a Bernouilli random variable with a success probability $p_i$ that depends on your explanatory variable $x_i$, i.e. $p_i=P(y_i=1|_{x_i})=f(x_i)$, where $f$ is the logi... | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | In logistic regression (I do it univariate for easier typing) you try to explain a binary outome $y_i \in \{0,1\}$ by assuming that it is the outcome of a Bernouilli random variable with a success pro | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
In logistic regression (I do it univariate for easier typing) you try to explain a binary outome $y_i \in \{0,1\}$ by assuming that it is the outcome of a Bernouilli random variable with a success probability $p_i$ that depends on your e... | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
In logistic regression (I do it univariate for easier typing) you try to explain a binary outome $y_i \in \{0,1\}$ by assuming that it is the outcome of a Bernouilli random variable with a success pro |
38,098 | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | As Marc said, AUC is only a measure of performance, just like missclassification rate. It does not require any information about the model.
Conversely, BIC, AIC, need to know the number of parameters of your model to be evaluated.
There is no good reason, if all of your predictors are relevant, that the missclassific... | Area Under Curve ROC penalizes somehow models with too many explanatory variables? | As Marc said, AUC is only a measure of performance, just like missclassification rate. It does not require any information about the model.
Conversely, BIC, AIC, need to know the number of parameters | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
As Marc said, AUC is only a measure of performance, just like missclassification rate. It does not require any information about the model.
Conversely, BIC, AIC, need to know the number of parameters of your model to be evaluated.
Ther... | Area Under Curve ROC penalizes somehow models with too many explanatory variables?
As Marc said, AUC is only a measure of performance, just like missclassification rate. It does not require any information about the model.
Conversely, BIC, AIC, need to know the number of parameters |
38,099 | Growing number of Gaussians in a mixture | If your goal is to find the maximum-likelihood mixture of size $n+1$, then you can use the existing solution as an initialization, once you have enlarged it to have one more Gaussian. To enlarge it, there are two approaches in the literature. The first approach is to add a new Gaussian in the best possible place, hol... | Growing number of Gaussians in a mixture | If your goal is to find the maximum-likelihood mixture of size $n+1$, then you can use the existing solution as an initialization, once you have enlarged it to have one more Gaussian. To enlarge it, | Growing number of Gaussians in a mixture
If your goal is to find the maximum-likelihood mixture of size $n+1$, then you can use the existing solution as an initialization, once you have enlarged it to have one more Gaussian. To enlarge it, there are two approaches in the literature. The first approach is to add a new... | Growing number of Gaussians in a mixture
If your goal is to find the maximum-likelihood mixture of size $n+1$, then you can use the existing solution as an initialization, once you have enlarged it to have one more Gaussian. To enlarge it, |
38,100 | Growing number of Gaussians in a mixture | Generally it depends on algorithm and software you use. In probably any software you can start your algorithm with some starting values.
However consider that you already fitted some model that is somehow "optimal" for the data. If you'll ask your algorithm for one more Gaussian than you already start in some local opt... | Growing number of Gaussians in a mixture | Generally it depends on algorithm and software you use. In probably any software you can start your algorithm with some starting values.
However consider that you already fitted some model that is som | Growing number of Gaussians in a mixture
Generally it depends on algorithm and software you use. In probably any software you can start your algorithm with some starting values.
However consider that you already fitted some model that is somehow "optimal" for the data. If you'll ask your algorithm for one more Gaussian... | Growing number of Gaussians in a mixture
Generally it depends on algorithm and software you use. In probably any software you can start your algorithm with some starting values.
However consider that you already fitted some model that is som |
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