idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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48,001 | Brewer's method for sampling with unequal probabilities with n>2 | My reading of Brewer's procedure is as follows.
To sample the first unit, set $r=n$ and compute
$$
D_1 = \sum_{i=1}^N\frac{P_i(1-P_i)}{(1-nP_i)}
$$
Then sample the first unit with probability
$$P^{(1)}_i = \frac{P_i (1 - P_i)}{D_1 (1-nP_i)}$$Let the index of the first sampled unit be $I_1$.
To sample the second uni... | Brewer's method for sampling with unequal probabilities with n>2 | My reading of Brewer's procedure is as follows.
To sample the first unit, set $r=n$ and compute
$$
D_1 = \sum_{i=1}^N\frac{P_i(1-P_i)}{(1-nP_i)}
$$
Then sample the first unit with probability
$$P^{ | Brewer's method for sampling with unequal probabilities with n>2
My reading of Brewer's procedure is as follows.
To sample the first unit, set $r=n$ and compute
$$
D_1 = \sum_{i=1}^N\frac{P_i(1-P_i)}{(1-nP_i)}
$$
Then sample the first unit with probability
$$P^{(1)}_i = \frac{P_i (1 - P_i)}{D_1 (1-nP_i)}$$Let the i... | Brewer's method for sampling with unequal probabilities with n>2
My reading of Brewer's procedure is as follows.
To sample the first unit, set $r=n$ and compute
$$
D_1 = \sum_{i=1}^N\frac{P_i(1-P_i)}{(1-nP_i)}
$$
Then sample the first unit with probability
$$P^{ |
48,002 | Brewer's method for sampling with unequal probabilities with n>2 | Update: August 19
I've now had a look at both Brewer's 1963 and 1975 articles. The 1975 method is indeed the generalization for (n>2) as Stas stated, with the same draw probabilities for n = 1 & 2.
This has been a very informative topic. I wondered why there were two versions of "Brewer's Method" in Roberto's question... | Brewer's method for sampling with unequal probabilities with n>2 | Update: August 19
I've now had a look at both Brewer's 1963 and 1975 articles. The 1975 method is indeed the generalization for (n>2) as Stas stated, with the same draw probabilities for n = 1 & 2.
T | Brewer's method for sampling with unequal probabilities with n>2
Update: August 19
I've now had a look at both Brewer's 1963 and 1975 articles. The 1975 method is indeed the generalization for (n>2) as Stas stated, with the same draw probabilities for n = 1 & 2.
This has been a very informative topic. I wondered why t... | Brewer's method for sampling with unequal probabilities with n>2
Update: August 19
I've now had a look at both Brewer's 1963 and 1975 articles. The 1975 method is indeed the generalization for (n>2) as Stas stated, with the same draw probabilities for n = 1 & 2.
T |
48,003 | which statistic should I use for inter-rater agreement on nominal data, multiple responses | You could use a chance-adjusted agreement index (e.g., Cohen's kappa or Scott's pi) for each category separately. Alternatively, you could use the following approach:
Kramer (1980) proposed a method for assessing inter-rater reliability for tasks in which raters could select multiple categories for each object of measu... | which statistic should I use for inter-rater agreement on nominal data, multiple responses | You could use a chance-adjusted agreement index (e.g., Cohen's kappa or Scott's pi) for each category separately. Alternatively, you could use the following approach:
Kramer (1980) proposed a method f | which statistic should I use for inter-rater agreement on nominal data, multiple responses
You could use a chance-adjusted agreement index (e.g., Cohen's kappa or Scott's pi) for each category separately. Alternatively, you could use the following approach:
Kramer (1980) proposed a method for assessing inter-rater reli... | which statistic should I use for inter-rater agreement on nominal data, multiple responses
You could use a chance-adjusted agreement index (e.g., Cohen's kappa or Scott's pi) for each category separately. Alternatively, you could use the following approach:
Kramer (1980) proposed a method f |
48,004 | What do the variables mean in the SVM objective function? | Those two formulae are different things:
$\frac{1}{2} w^T w + C \sum \xi_i$ is one form of the objective function, the function which is minimized over $w$, $b$, and $\xi_i$ (subject to certain constraints, which are where $b$ comes in) to find the best SVM solution.
Once you've found the model (defined by $w$ and $b$... | What do the variables mean in the SVM objective function? | Those two formulae are different things:
$\frac{1}{2} w^T w + C \sum \xi_i$ is one form of the objective function, the function which is minimized over $w$, $b$, and $\xi_i$ (subject to certain const | What do the variables mean in the SVM objective function?
Those two formulae are different things:
$\frac{1}{2} w^T w + C \sum \xi_i$ is one form of the objective function, the function which is minimized over $w$, $b$, and $\xi_i$ (subject to certain constraints, which are where $b$ comes in) to find the best SVM sol... | What do the variables mean in the SVM objective function?
Those two formulae are different things:
$\frac{1}{2} w^T w + C \sum \xi_i$ is one form of the objective function, the function which is minimized over $w$, $b$, and $\xi_i$ (subject to certain const |
48,005 | Is binary hypothesis testing a better statistical term than A/B testing? | The Wikipedia article has accurate information about A/B testing; binary hypothesis testing is another name for A/B testing. A/B testing and split testing are the most widely accepted terms in the business and marketing community. The exact origins of A/B testing are not well known but can be traced back to Google du... | Is binary hypothesis testing a better statistical term than A/B testing? | The Wikipedia article has accurate information about A/B testing; binary hypothesis testing is another name for A/B testing. A/B testing and split testing are the most widely accepted terms in the bu | Is binary hypothesis testing a better statistical term than A/B testing?
The Wikipedia article has accurate information about A/B testing; binary hypothesis testing is another name for A/B testing. A/B testing and split testing are the most widely accepted terms in the business and marketing community. The exact orig... | Is binary hypothesis testing a better statistical term than A/B testing?
The Wikipedia article has accurate information about A/B testing; binary hypothesis testing is another name for A/B testing. A/B testing and split testing are the most widely accepted terms in the bu |
48,006 | Is binary hypothesis testing a better statistical term than A/B testing? | "Binary hypothesis testing" is hypothesis testing when one wants to decide between two hypotheses.
"Two-sample hypothesis testing" is what is known colloquially as A/B testing.
"Paired hypothesis testing" when you compare the same sample before and after an event to find if it had an effect. Similar to A/B testing but ... | Is binary hypothesis testing a better statistical term than A/B testing? | "Binary hypothesis testing" is hypothesis testing when one wants to decide between two hypotheses.
"Two-sample hypothesis testing" is what is known colloquially as A/B testing.
"Paired hypothesis test | Is binary hypothesis testing a better statistical term than A/B testing?
"Binary hypothesis testing" is hypothesis testing when one wants to decide between two hypotheses.
"Two-sample hypothesis testing" is what is known colloquially as A/B testing.
"Paired hypothesis testing" when you compare the same sample before an... | Is binary hypothesis testing a better statistical term than A/B testing?
"Binary hypothesis testing" is hypothesis testing when one wants to decide between two hypotheses.
"Two-sample hypothesis testing" is what is known colloquially as A/B testing.
"Paired hypothesis test |
48,007 | Is binary hypothesis testing a better statistical term than A/B testing? | I agree with the other very good answers. I think, it is mostly engineering backgrounds that prefer the term A/B test and the last years it has become a quite hot term especially within the context of web sites optimisation.
Have in mind that except A/B test you may encounter these terms:
A/B/C tests where you assess... | Is binary hypothesis testing a better statistical term than A/B testing? | I agree with the other very good answers. I think, it is mostly engineering backgrounds that prefer the term A/B test and the last years it has become a quite hot term especially within the context of | Is binary hypothesis testing a better statistical term than A/B testing?
I agree with the other very good answers. I think, it is mostly engineering backgrounds that prefer the term A/B test and the last years it has become a quite hot term especially within the context of web sites optimisation.
Have in mind that exce... | Is binary hypothesis testing a better statistical term than A/B testing?
I agree with the other very good answers. I think, it is mostly engineering backgrounds that prefer the term A/B test and the last years it has become a quite hot term especially within the context of |
48,008 | How to get percentiles from empirical density in R? | The command density(), although very useful for a quick inspection of the KDE, is also very restrictive since it only returns the values on a grid. I prefer to code my own KDE (usually with a Gaussian kernel). This can be obtained as shown below (1-line code):
rm(list=ls())
# Constructing your own KDE
set.seed(123)
sam... | How to get percentiles from empirical density in R? | The command density(), although very useful for a quick inspection of the KDE, is also very restrictive since it only returns the values on a grid. I prefer to code my own KDE (usually with a Gaussian | How to get percentiles from empirical density in R?
The command density(), although very useful for a quick inspection of the KDE, is also very restrictive since it only returns the values on a grid. I prefer to code my own KDE (usually with a Gaussian kernel). This can be obtained as shown below (1-line code):
rm(list... | How to get percentiles from empirical density in R?
The command density(), although very useful for a quick inspection of the KDE, is also very restrictive since it only returns the values on a grid. I prefer to code my own KDE (usually with a Gaussian |
48,009 | How to get percentiles from empirical density in R? | Why re-invent the wheel? I advise you to use the ewcdf function in the spatstat library. If I understand your question correctly, it does exactly what you want:
library(spatstat)
x <- rnorm(100) #data
w <- runif(100) #weights
a1<-ewcdf(x,w) #empricial *weighted* cdf and quantile function
quantile(a1,.2... | How to get percentiles from empirical density in R? | Why re-invent the wheel? I advise you to use the ewcdf function in the spatstat library. If I understand your question correctly, it does exactly what you want:
library(spatstat)
x <- rnorm(100) | How to get percentiles from empirical density in R?
Why re-invent the wheel? I advise you to use the ewcdf function in the spatstat library. If I understand your question correctly, it does exactly what you want:
library(spatstat)
x <- rnorm(100) #data
w <- runif(100) #weights
a1<-ewcdf(x,w) #empricial ... | How to get percentiles from empirical density in R?
Why re-invent the wheel? I advise you to use the ewcdf function in the spatstat library. If I understand your question correctly, it does exactly what you want:
library(spatstat)
x <- rnorm(100) |
48,010 | What are some methods for generating simulated time series data for use in modeling? | One possibility is to fit a time series model to the data you are interested in. Then you can resample the residuals from the fitted model and use them to simulate the data. For example, you can fit an ARIMA model, resample the residuals and then generate new data from the fitted ARIMA model. Instead of generating the ... | What are some methods for generating simulated time series data for use in modeling? | One possibility is to fit a time series model to the data you are interested in. Then you can resample the residuals from the fitted model and use them to simulate the data. For example, you can fit a | What are some methods for generating simulated time series data for use in modeling?
One possibility is to fit a time series model to the data you are interested in. Then you can resample the residuals from the fitted model and use them to simulate the data. For example, you can fit an ARIMA model, resample the residua... | What are some methods for generating simulated time series data for use in modeling?
One possibility is to fit a time series model to the data you are interested in. Then you can resample the residuals from the fitted model and use them to simulate the data. For example, you can fit a |
48,011 | What are some methods for generating simulated time series data for use in modeling? | I suggest you first determine that you wanna predict for one day, one month or one year, next think about situation that effect in your goal variable, and if you could use variable from external database like temperature, and as our friend says use time-series models. I think preparation data is more important modeling... | What are some methods for generating simulated time series data for use in modeling? | I suggest you first determine that you wanna predict for one day, one month or one year, next think about situation that effect in your goal variable, and if you could use variable from external datab | What are some methods for generating simulated time series data for use in modeling?
I suggest you first determine that you wanna predict for one day, one month or one year, next think about situation that effect in your goal variable, and if you could use variable from external database like temperature, and as our fr... | What are some methods for generating simulated time series data for use in modeling?
I suggest you first determine that you wanna predict for one day, one month or one year, next think about situation that effect in your goal variable, and if you could use variable from external datab |
48,012 | Ratio of correlated sample variances (gamma distributed) | If we assume that the underlying normals are jointly normal, then the result is very simple.
This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated chi-square variables (open access).
We have
$$r\equiv \frac{s_y^2}{s_x^2} = \frac {\sigma^2_y}{\sigma^2_x}\frac{(n-1)... | Ratio of correlated sample variances (gamma distributed) | If we assume that the underlying normals are jointly normal, then the result is very simple.
This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated | Ratio of correlated sample variances (gamma distributed)
If we assume that the underlying normals are jointly normal, then the result is very simple.
This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated chi-square variables (open access).
We have
$$r\equiv \frac{... | Ratio of correlated sample variances (gamma distributed)
If we assume that the underlying normals are jointly normal, then the result is very simple.
This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated |
48,013 | Product of Gamma by Beta rv | Given:
$X \sim \text{Beta}(a,b)$ with pdf $f(x)$:
$Y \sim \text{Gamma}(k,\theta)$ with pdf $g(y)$:
Solution: Then, the pdf of the product $Z = X Y$ can be automatically derived via:
where I am using the TransformProduct function from the mathStatica package for Mathematica, and where Hypergeometric1F1 denotes th... | Product of Gamma by Beta rv | Given:
$X \sim \text{Beta}(a,b)$ with pdf $f(x)$:
$Y \sim \text{Gamma}(k,\theta)$ with pdf $g(y)$:
Solution: Then, the pdf of the product $Z = X Y$ can be automatically derived via:
where I am | Product of Gamma by Beta rv
Given:
$X \sim \text{Beta}(a,b)$ with pdf $f(x)$:
$Y \sim \text{Gamma}(k,\theta)$ with pdf $g(y)$:
Solution: Then, the pdf of the product $Z = X Y$ can be automatically derived via:
where I am using the TransformProduct function from the mathStatica package for Mathematica, and where ... | Product of Gamma by Beta rv
Given:
$X \sim \text{Beta}(a,b)$ with pdf $f(x)$:
$Y \sim \text{Gamma}(k,\theta)$ with pdf $g(y)$:
Solution: Then, the pdf of the product $Z = X Y$ can be automatically derived via:
where I am |
48,014 | Product of Gamma by Beta rv | This distribution is called the Gamma-Inverse Beta distribution in this paper. It is available in the R package brr.
nsims <- 1e6
alpha <- 3
beta <- 5
K <- 6
theta <- 4
sims <- rgamma(nsims, shape = K, rate = theta) * rbeta(nsims, alpha, beta)
plot(density(sims, to=3))
curve(brr::dGIB(x, K, beta, alpha, theta), # not... | Product of Gamma by Beta rv | This distribution is called the Gamma-Inverse Beta distribution in this paper. It is available in the R package brr.
nsims <- 1e6
alpha <- 3
beta <- 5
K <- 6
theta <- 4
sims <- rgamma(nsims, shape = | Product of Gamma by Beta rv
This distribution is called the Gamma-Inverse Beta distribution in this paper. It is available in the R package brr.
nsims <- 1e6
alpha <- 3
beta <- 5
K <- 6
theta <- 4
sims <- rgamma(nsims, shape = K, rate = theta) * rbeta(nsims, alpha, beta)
plot(density(sims, to=3))
curve(brr::dGIB(x, K... | Product of Gamma by Beta rv
This distribution is called the Gamma-Inverse Beta distribution in this paper. It is available in the R package brr.
nsims <- 1e6
alpha <- 3
beta <- 5
K <- 6
theta <- 4
sims <- rgamma(nsims, shape = |
48,015 | What's the difference between autoencoders and deep autoencoders? | Autoencoder is basically a technique to find fundamental features representing the input images. A simple autoencoder will have 1 hidden layer between the input and output, wheras a deep autoencoder will have multiple hidden layers (the number of hidden layer depends on your configuration). Refering the first figure be... | What's the difference between autoencoders and deep autoencoders? | Autoencoder is basically a technique to find fundamental features representing the input images. A simple autoencoder will have 1 hidden layer between the input and output, wheras a deep autoencoder w | What's the difference between autoencoders and deep autoencoders?
Autoencoder is basically a technique to find fundamental features representing the input images. A simple autoencoder will have 1 hidden layer between the input and output, wheras a deep autoencoder will have multiple hidden layers (the number of hidden ... | What's the difference between autoencoders and deep autoencoders?
Autoencoder is basically a technique to find fundamental features representing the input images. A simple autoencoder will have 1 hidden layer between the input and output, wheras a deep autoencoder w |
48,016 | How to transform an accuracy distribution for a violin plot | I have various takes on this.
Don't expect too much from transformation. I read your results as saying that the upper quartile (**not* usually called the first quartile) is 1; hence >25% of the values tie at 1 and you have a spike in the distribution. Any one-to-one transformation will inevitably map a spike to a spi... | How to transform an accuracy distribution for a violin plot | I have various takes on this.
Don't expect too much from transformation. I read your results as saying that the upper quartile (**not* usually called the first quartile) is 1; hence >25% of the valu | How to transform an accuracy distribution for a violin plot
I have various takes on this.
Don't expect too much from transformation. I read your results as saying that the upper quartile (**not* usually called the first quartile) is 1; hence >25% of the values tie at 1 and you have a spike in the distribution. Any on... | How to transform an accuracy distribution for a violin plot
I have various takes on this.
Don't expect too much from transformation. I read your results as saying that the upper quartile (**not* usually called the first quartile) is 1; hence >25% of the valu |
48,017 | Statistics of 7 game playoff series | For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game seven, and 20 possible combinations of wins for each of the teams that can win, so 40 possible outcomes. For an N-game seri... | Statistics of 7 game playoff series | For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game sev | Statistics of 7 game playoff series
For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game seven, and 20 possible combinations of wins for each of the teams that can win, so 40 p... | Statistics of 7 game playoff series
For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game sev |
48,018 | Statistics of 7 game playoff series | If p is the probability of winning a single game, and N is the number of games one needs to win the series, then the probability "P" of winning the series is given by:
$$
P = \frac{p^N}{\left(N-1\right)!}\cdot\sum_{i=0}^{N-1}\left[\left(\prod_{j=1}^{N-1}\left(i+j\right)\right)(1-p)^i\right]
$$
Think of the way a game p... | Statistics of 7 game playoff series | If p is the probability of winning a single game, and N is the number of games one needs to win the series, then the probability "P" of winning the series is given by:
$$
P = \frac{p^N}{\left(N-1\righ | Statistics of 7 game playoff series
If p is the probability of winning a single game, and N is the number of games one needs to win the series, then the probability "P" of winning the series is given by:
$$
P = \frac{p^N}{\left(N-1\right)!}\cdot\sum_{i=0}^{N-1}\left[\left(\prod_{j=1}^{N-1}\left(i+j\right)\right)(1-p)^i... | Statistics of 7 game playoff series
If p is the probability of winning a single game, and N is the number of games one needs to win the series, then the probability "P" of winning the series is given by:
$$
P = \frac{p^N}{\left(N-1\righ |
48,019 | Statistics of 7 game playoff series | An alternate way to look at would be a binomial distribution: You need x=3 (exactly 3 successes) in n = 6 (trails) , so if the probability of winning a game is .5 (both teams equally likely) , binomial would say:
P(x=3) = 6C3 * (.5)^3 * (.5)^3 = .3125
This would mean there is 31.25% chance of going to a 7 game series.... | Statistics of 7 game playoff series | An alternate way to look at would be a binomial distribution: You need x=3 (exactly 3 successes) in n = 6 (trails) , so if the probability of winning a game is .5 (both teams equally likely) , binomia | Statistics of 7 game playoff series
An alternate way to look at would be a binomial distribution: You need x=3 (exactly 3 successes) in n = 6 (trails) , so if the probability of winning a game is .5 (both teams equally likely) , binomial would say:
P(x=3) = 6C3 * (.5)^3 * (.5)^3 = .3125
This would mean there is 31.25%... | Statistics of 7 game playoff series
An alternate way to look at would be a binomial distribution: You need x=3 (exactly 3 successes) in n = 6 (trails) , so if the probability of winning a game is .5 (both teams equally likely) , binomia |
48,020 | Statistics of 7 game playoff series | Regarding the discrepancy in real-life results, OP noted "I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams are usually seeded in the first round so that the leading qualifying team plays the team that barely qualified, second pl... | Statistics of 7 game playoff series | Regarding the discrepancy in real-life results, OP noted "I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams a | Statistics of 7 game playoff series
Regarding the discrepancy in real-life results, OP noted "I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams are usually seeded in the first round so that the leading qualifying team plays the t... | Statistics of 7 game playoff series
Regarding the discrepancy in real-life results, OP noted "I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams a |
48,021 | Probability of data point being from distribution in normal mixtures | Let's call the estimates from population 1 and population 2 to be
$\mu_1$ and $\mu_2$ for the means, $\sigma_1$ and $\sigma_2$ for the sd's. Also, let's define $p$ to be the estimated proportion of observations from population 1.
Then, for each observation $x_i$, the estimated probability of belonging to population 1... | Probability of data point being from distribution in normal mixtures | Let's call the estimates from population 1 and population 2 to be
$\mu_1$ and $\mu_2$ for the means, $\sigma_1$ and $\sigma_2$ for the sd's. Also, let's define $p$ to be the estimated proportion of | Probability of data point being from distribution in normal mixtures
Let's call the estimates from population 1 and population 2 to be
$\mu_1$ and $\mu_2$ for the means, $\sigma_1$ and $\sigma_2$ for the sd's. Also, let's define $p$ to be the estimated proportion of observations from population 1.
Then, for each obse... | Probability of data point being from distribution in normal mixtures
Let's call the estimates from population 1 and population 2 to be
$\mu_1$ and $\mu_2$ for the means, $\sigma_1$ and $\sigma_2$ for the sd's. Also, let's define $p$ to be the estimated proportion of |
48,022 | Can the union (or) probability of many non-mutually exclusive events be calculated recursively? | As @HaoYe's comment points out, your recursion via divide-and-conquer is not
quite right: it is not the case that
$$P(A_1\cup A_2\cup A_3\cup A_4) = P( A_1 \cup A_2) + P( A_3 \cup A_4) - P( A_1 \cup A_2) * P( A_3 \cup A_4)$$
but rather that
$$P(A_1\cup A_2\cup A_3\cup A_4) = P( A_1 \cup A_2) + P( A_3 \cup A_4) - P\le... | Can the union (or) probability of many non-mutually exclusive events be calculated recursively? | As @HaoYe's comment points out, your recursion via divide-and-conquer is not
quite right: it is not the case that
$$P(A_1\cup A_2\cup A_3\cup A_4) = P( A_1 \cup A_2) + P( A_3 \cup A_4) - P( A_1 \cup | Can the union (or) probability of many non-mutually exclusive events be calculated recursively?
As @HaoYe's comment points out, your recursion via divide-and-conquer is not
quite right: it is not the case that
$$P(A_1\cup A_2\cup A_3\cup A_4) = P( A_1 \cup A_2) + P( A_3 \cup A_4) - P( A_1 \cup A_2) * P( A_3 \cup A_4)$... | Can the union (or) probability of many non-mutually exclusive events be calculated recursively?
As @HaoYe's comment points out, your recursion via divide-and-conquer is not
quite right: it is not the case that
$$P(A_1\cup A_2\cup A_3\cup A_4) = P( A_1 \cup A_2) + P( A_3 \cup A_4) - P( A_1 \cup |
48,023 | Can the union (or) probability of many non-mutually exclusive events be calculated recursively? | Nothing wrong with this approach. If you use floating point, you will want to or (add up) the two smallest remaining probabilities in each step in order to minimize the effect of rounding errors (floating point is most accurate when the values you add are of similar magnitude).
PS. Since (a or b or c or d) = ((a or b) ... | Can the union (or) probability of many non-mutually exclusive events be calculated recursively? | Nothing wrong with this approach. If you use floating point, you will want to or (add up) the two smallest remaining probabilities in each step in order to minimize the effect of rounding errors (floa | Can the union (or) probability of many non-mutually exclusive events be calculated recursively?
Nothing wrong with this approach. If you use floating point, you will want to or (add up) the two smallest remaining probabilities in each step in order to minimize the effect of rounding errors (floating point is most accur... | Can the union (or) probability of many non-mutually exclusive events be calculated recursively?
Nothing wrong with this approach. If you use floating point, you will want to or (add up) the two smallest remaining probabilities in each step in order to minimize the effect of rounding errors (floa |
48,024 | Dealing with missing data - glmer in lme4 package | Turning my comments into an answer as they seem to have answered your question...
Just exclude the actual missing data. If you format your data with columns ID, Day, environmental variables, response, everything should be fine to just omit the rows where an ID is missing a measurement on a certain day, still keeping th... | Dealing with missing data - glmer in lme4 package | Turning my comments into an answer as they seem to have answered your question...
Just exclude the actual missing data. If you format your data with columns ID, Day, environmental variables, response, | Dealing with missing data - glmer in lme4 package
Turning my comments into an answer as they seem to have answered your question...
Just exclude the actual missing data. If you format your data with columns ID, Day, environmental variables, response, everything should be fine to just omit the rows where an ID is missin... | Dealing with missing data - glmer in lme4 package
Turning my comments into an answer as they seem to have answered your question...
Just exclude the actual missing data. If you format your data with columns ID, Day, environmental variables, response, |
48,025 | Dealing with missing data - glmer in lme4 package | You must create an indicator variable as control in your model: 1 for acoustic receiver in the missing day and 0 for complete data. Repeat the code for the same acoustic receiver in all its observed and unobserved days (in wide format data base, before converting to long format). You can do it for any missing day or on... | Dealing with missing data - glmer in lme4 package | You must create an indicator variable as control in your model: 1 for acoustic receiver in the missing day and 0 for complete data. Repeat the code for the same acoustic receiver in all its observed a | Dealing with missing data - glmer in lme4 package
You must create an indicator variable as control in your model: 1 for acoustic receiver in the missing day and 0 for complete data. Repeat the code for the same acoustic receiver in all its observed and unobserved days (in wide format data base, before converting to lon... | Dealing with missing data - glmer in lme4 package
You must create an indicator variable as control in your model: 1 for acoustic receiver in the missing day and 0 for complete data. Repeat the code for the same acoustic receiver in all its observed a |
48,026 | can an ARMA process with complex unit roots be made stationary by differencing? | Yes, it is possible to render the process stationary regardless of whether the
non-stationary cycles are related to real or complex roots.
For example, the following seasonal random walk:
$$
y_t = y_{t-4} + \epsilon_t \quad \epsilon_t \sim NID(0, \sigma^2) \,,
$$
contains four unit roots: $\pm 1, \pm i$ (i.e., 2 real a... | can an ARMA process with complex unit roots be made stationary by differencing? | Yes, it is possible to render the process stationary regardless of whether the
non-stationary cycles are related to real or complex roots.
For example, the following seasonal random walk:
$$
y_t = y_{ | can an ARMA process with complex unit roots be made stationary by differencing?
Yes, it is possible to render the process stationary regardless of whether the
non-stationary cycles are related to real or complex roots.
For example, the following seasonal random walk:
$$
y_t = y_{t-4} + \epsilon_t \quad \epsilon_t \sim ... | can an ARMA process with complex unit roots be made stationary by differencing?
Yes, it is possible to render the process stationary regardless of whether the
non-stationary cycles are related to real or complex roots.
For example, the following seasonal random walk:
$$
y_t = y_{ |
48,027 | What is this kind of tabular relation diagram called? | I would use the term binary heat map - although I'm not sure if there is a unified term. (There are probably other names used in various fields of application.) Heat maps more often display continuous attributes in cells via a color ramp, but there are plenty of matrix like representations of binary data. For instance ... | What is this kind of tabular relation diagram called? | I would use the term binary heat map - although I'm not sure if there is a unified term. (There are probably other names used in various fields of application.) Heat maps more often display continuous | What is this kind of tabular relation diagram called?
I would use the term binary heat map - although I'm not sure if there is a unified term. (There are probably other names used in various fields of application.) Heat maps more often display continuous attributes in cells via a color ramp, but there are plenty of mat... | What is this kind of tabular relation diagram called?
I would use the term binary heat map - although I'm not sure if there is a unified term. (There are probably other names used in various fields of application.) Heat maps more often display continuous |
48,028 | What is this kind of tabular relation diagram called? | I think this is somewhat harder to answer than it would otherwise be because it's a very simple and commonly used style of chart. My answer may not be universally applicable, and other terms are likely to exist for the exact same layout, but I'm finding a lot of similar results in a Google image search for "comparison ... | What is this kind of tabular relation diagram called? | I think this is somewhat harder to answer than it would otherwise be because it's a very simple and commonly used style of chart. My answer may not be universally applicable, and other terms are likel | What is this kind of tabular relation diagram called?
I think this is somewhat harder to answer than it would otherwise be because it's a very simple and commonly used style of chart. My answer may not be universally applicable, and other terms are likely to exist for the exact same layout, but I'm finding a lot of sim... | What is this kind of tabular relation diagram called?
I think this is somewhat harder to answer than it would otherwise be because it's a very simple and commonly used style of chart. My answer may not be universally applicable, and other terms are likel |
48,029 | How do I use math to predict the next number in the series? | If observations are independent, and if values must either be 1 or 0, with no additional prior information, you may simply assume that the probability that the next value is 1 is equal to the proportion of 1s in the observations.
If you wish to calculate a confidence interval around this estimate, this could reasonably... | How do I use math to predict the next number in the series? | If observations are independent, and if values must either be 1 or 0, with no additional prior information, you may simply assume that the probability that the next value is 1 is equal to the proporti | How do I use math to predict the next number in the series?
If observations are independent, and if values must either be 1 or 0, with no additional prior information, you may simply assume that the probability that the next value is 1 is equal to the proportion of 1s in the observations.
If you wish to calculate a con... | How do I use math to predict the next number in the series?
If observations are independent, and if values must either be 1 or 0, with no additional prior information, you may simply assume that the probability that the next value is 1 is equal to the proporti |
48,030 | p-value for weighted Pearson correlation coefficient | The $P$-value reported for a correlation depends on the sample correlation, the sample size, and a bundle of assumptions not always checked (independence being, in my experience, least checked of all). But there is a difference between a crude $t$-based $P$-value based on a null hypothesis of zero correlation and a mor... | p-value for weighted Pearson correlation coefficient | The $P$-value reported for a correlation depends on the sample correlation, the sample size, and a bundle of assumptions not always checked (independence being, in my experience, least checked of all) | p-value for weighted Pearson correlation coefficient
The $P$-value reported for a correlation depends on the sample correlation, the sample size, and a bundle of assumptions not always checked (independence being, in my experience, least checked of all). But there is a difference between a crude $t$-based $P$-value bas... | p-value for weighted Pearson correlation coefficient
The $P$-value reported for a correlation depends on the sample correlation, the sample size, and a bundle of assumptions not always checked (independence being, in my experience, least checked of all) |
48,031 | p-value for weighted Pearson correlation coefficient | I haven't yet sit down to work the math, but from the few simulations I run it seems that replacing the number of samples $n$ in the formulas with the effective number of samples $n_\text{eff}$ yields very good approximations.
$n_\text{eff} = \exp(H)$, where $H=-\sum_{i=1}^n w_i \ln w_i$ is the entropy of weights (norm... | p-value for weighted Pearson correlation coefficient | I haven't yet sit down to work the math, but from the few simulations I run it seems that replacing the number of samples $n$ in the formulas with the effective number of samples $n_\text{eff}$ yields | p-value for weighted Pearson correlation coefficient
I haven't yet sit down to work the math, but from the few simulations I run it seems that replacing the number of samples $n$ in the formulas with the effective number of samples $n_\text{eff}$ yields very good approximations.
$n_\text{eff} = \exp(H)$, where $H=-\sum... | p-value for weighted Pearson correlation coefficient
I haven't yet sit down to work the math, but from the few simulations I run it seems that replacing the number of samples $n$ in the formulas with the effective number of samples $n_\text{eff}$ yields |
48,032 | Inference with only left-censored data | How much you can conclude depends on the assumptions you are willing to make about the underlying distribution. Even with no assumptions, though, you can still conclude something, although it might not be a whole lot. For example, the sample data are not plausibly consistent with an underlying distribution whose mean... | Inference with only left-censored data | How much you can conclude depends on the assumptions you are willing to make about the underlying distribution. Even with no assumptions, though, you can still conclude something, although it might n | Inference with only left-censored data
How much you can conclude depends on the assumptions you are willing to make about the underlying distribution. Even with no assumptions, though, you can still conclude something, although it might not be a whole lot. For example, the sample data are not plausibly consistent wit... | Inference with only left-censored data
How much you can conclude depends on the assumptions you are willing to make about the underlying distribution. Even with no assumptions, though, you can still conclude something, although it might n |
48,033 | Inference with only left-censored data | I can think of one solution, but I'm open to others.
In the paper linked, the use the formula:
$$F(Z_i) = \prod_{i=1}^j \frac{R_i - n_i}{R_i} $$
The term being multiplied, $\frac{R_i - n_i}{R_i}$, is a pointwise estimate of the rate at time $i$. For low $R_i$ size, or when $n_i = R_i$ or $n_i = 0$, this produces somew... | Inference with only left-censored data | I can think of one solution, but I'm open to others.
In the paper linked, the use the formula:
$$F(Z_i) = \prod_{i=1}^j \frac{R_i - n_i}{R_i} $$
The term being multiplied, $\frac{R_i - n_i}{R_i}$, is | Inference with only left-censored data
I can think of one solution, but I'm open to others.
In the paper linked, the use the formula:
$$F(Z_i) = \prod_{i=1}^j \frac{R_i - n_i}{R_i} $$
The term being multiplied, $\frac{R_i - n_i}{R_i}$, is a pointwise estimate of the rate at time $i$. For low $R_i$ size, or when $n_i =... | Inference with only left-censored data
I can think of one solution, but I'm open to others.
In the paper linked, the use the formula:
$$F(Z_i) = \prod_{i=1}^j \frac{R_i - n_i}{R_i} $$
The term being multiplied, $\frac{R_i - n_i}{R_i}$, is |
48,034 | Does the estimated overdispersion parameter of Negative Binomial depend on mean | I'll begin by noting that there are actually quite a few different estimators of the dispersion parameter of a negative binomial out there, and that my discussion here is limited by those estimators that I am familar with. To answer your question:
In the maximum likelihood case - the answer is yes. For other cases the ... | Does the estimated overdispersion parameter of Negative Binomial depend on mean | I'll begin by noting that there are actually quite a few different estimators of the dispersion parameter of a negative binomial out there, and that my discussion here is limited by those estimators t | Does the estimated overdispersion parameter of Negative Binomial depend on mean
I'll begin by noting that there are actually quite a few different estimators of the dispersion parameter of a negative binomial out there, and that my discussion here is limited by those estimators that I am familar with. To answer your qu... | Does the estimated overdispersion parameter of Negative Binomial depend on mean
I'll begin by noting that there are actually quite a few different estimators of the dispersion parameter of a negative binomial out there, and that my discussion here is limited by those estimators t |
48,035 | Does the estimated overdispersion parameter of Negative Binomial depend on mean | I guess I find the answer via looking at the off diagonal of the Fisher information matrix.
$P(X=x) = \frac{\Gamma(x + \psi)}{\Gamma(\psi)\Gamma(x+1)} \frac{\psi^\psi\mu^x}{(\psi+\mu)^{(\psi+x)}}$
$\log L = n\psi \log\psi - n\log\Gamma(\psi)+\Sigma_{i=1}^n\log\Gamma(x_i+\psi)+\Sigma_{i=1}^nx_i\log\mu-\Sigma_{i=1}^n(\ps... | Does the estimated overdispersion parameter of Negative Binomial depend on mean | I guess I find the answer via looking at the off diagonal of the Fisher information matrix.
$P(X=x) = \frac{\Gamma(x + \psi)}{\Gamma(\psi)\Gamma(x+1)} \frac{\psi^\psi\mu^x}{(\psi+\mu)^{(\psi+x)}}$
$\l | Does the estimated overdispersion parameter of Negative Binomial depend on mean
I guess I find the answer via looking at the off diagonal of the Fisher information matrix.
$P(X=x) = \frac{\Gamma(x + \psi)}{\Gamma(\psi)\Gamma(x+1)} \frac{\psi^\psi\mu^x}{(\psi+\mu)^{(\psi+x)}}$
$\log L = n\psi \log\psi - n\log\Gamma(\psi... | Does the estimated overdispersion parameter of Negative Binomial depend on mean
I guess I find the answer via looking at the off diagonal of the Fisher information matrix.
$P(X=x) = \frac{\Gamma(x + \psi)}{\Gamma(\psi)\Gamma(x+1)} \frac{\psi^\psi\mu^x}{(\psi+\mu)^{(\psi+x)}}$
$\l |
48,036 | When plotting clustering results in the PCA coordinates, does one do PCA or clustering first? | It is hard to see how you could do PCA on clusters; it is quite common to do PCA prior to clustering, particularly when you have a lot of variables. You can then use the PCs as variables.
You might be getting confused between a different two alternatives:
1) Do PCA on the data, then do k-means on the PCs, then plot the... | When plotting clustering results in the PCA coordinates, does one do PCA or clustering first? | It is hard to see how you could do PCA on clusters; it is quite common to do PCA prior to clustering, particularly when you have a lot of variables. You can then use the PCs as variables.
You might be | When plotting clustering results in the PCA coordinates, does one do PCA or clustering first?
It is hard to see how you could do PCA on clusters; it is quite common to do PCA prior to clustering, particularly when you have a lot of variables. You can then use the PCs as variables.
You might be getting confused between ... | When plotting clustering results in the PCA coordinates, does one do PCA or clustering first?
It is hard to see how you could do PCA on clusters; it is quite common to do PCA prior to clustering, particularly when you have a lot of variables. You can then use the PCs as variables.
You might be |
48,037 | Prediction with scikit and an precomputed kernel (SVM) | It seems that gram matrix that you use for predictions is wrong.
Once you fit the SVM its prediction for $x$ is: $ y = \text{sign} \langle w,\phi(x)\rangle $
It sure is possible that one cannot compute $\phi(x)$ but needs to use a pre-computed gram matrix instead. Substitute $ w = \sum_i^m\alpha_i\phi(x_i)$
into the ab... | Prediction with scikit and an precomputed kernel (SVM) | It seems that gram matrix that you use for predictions is wrong.
Once you fit the SVM its prediction for $x$ is: $ y = \text{sign} \langle w,\phi(x)\rangle $
It sure is possible that one cannot comput | Prediction with scikit and an precomputed kernel (SVM)
It seems that gram matrix that you use for predictions is wrong.
Once you fit the SVM its prediction for $x$ is: $ y = \text{sign} \langle w,\phi(x)\rangle $
It sure is possible that one cannot compute $\phi(x)$ but needs to use a pre-computed gram matrix instead. ... | Prediction with scikit and an precomputed kernel (SVM)
It seems that gram matrix that you use for predictions is wrong.
Once you fit the SVM its prediction for $x$ is: $ y = \text{sign} \langle w,\phi(x)\rangle $
It sure is possible that one cannot comput |
48,038 | Prediction with scikit and an precomputed kernel (SVM) | I find that I usually get that error from training/vectorizing outside the kfold loop.
vec.fit_transform(), vec.fit_transform(), clf.fit(), clf.pred() - all four of those function calls has to be inside to loop, to make sure all matricies gets the correct shapes.
Is your testing algo setup something like this?:
k_folds... | Prediction with scikit and an precomputed kernel (SVM) | I find that I usually get that error from training/vectorizing outside the kfold loop.
vec.fit_transform(), vec.fit_transform(), clf.fit(), clf.pred() - all four of those function calls has to be insi | Prediction with scikit and an precomputed kernel (SVM)
I find that I usually get that error from training/vectorizing outside the kfold loop.
vec.fit_transform(), vec.fit_transform(), clf.fit(), clf.pred() - all four of those function calls has to be inside to loop, to make sure all matricies gets the correct shapes.
I... | Prediction with scikit and an precomputed kernel (SVM)
I find that I usually get that error from training/vectorizing outside the kfold loop.
vec.fit_transform(), vec.fit_transform(), clf.fit(), clf.pred() - all four of those function calls has to be insi |
48,039 | CDF and logistic regression | There are several issues here. The biggest is that the thinking behind the code for generating the blue points in the second plot is confused. I can't make any sense of what the logic is supposed to be. The $x$ values are the same as your data, but the $y$ values are the number of wins up to that point on the $x$ ax... | CDF and logistic regression | There are several issues here. The biggest is that the thinking behind the code for generating the blue points in the second plot is confused. I can't make any sense of what the logic is supposed to | CDF and logistic regression
There are several issues here. The biggest is that the thinking behind the code for generating the blue points in the second plot is confused. I can't make any sense of what the logic is supposed to be. The $x$ values are the same as your data, but the $y$ values are the number of wins up... | CDF and logistic regression
There are several issues here. The biggest is that the thinking behind the code for generating the blue points in the second plot is confused. I can't make any sense of what the logic is supposed to |
48,040 | CDF and logistic regression | It would be good to study a standard text on logistic regression. Among other things, you are confusing full conditioning ($Pr[Y=1 | X=x]$ as the logistic model estimates) with partial conditioning ($Pr[Y=1 | X \leq x]$). But more importantly, the model you have fitted is non-stochastic, i.e., the winner is determine... | CDF and logistic regression | It would be good to study a standard text on logistic regression. Among other things, you are confusing full conditioning ($Pr[Y=1 | X=x]$ as the logistic model estimates) with partial conditioning ( | CDF and logistic regression
It would be good to study a standard text on logistic regression. Among other things, you are confusing full conditioning ($Pr[Y=1 | X=x]$ as the logistic model estimates) with partial conditioning ($Pr[Y=1 | X \leq x]$). But more importantly, the model you have fitted is non-stochastic, i... | CDF and logistic regression
It would be good to study a standard text on logistic regression. Among other things, you are confusing full conditioning ($Pr[Y=1 | X=x]$ as the logistic model estimates) with partial conditioning ( |
48,041 | CDF and logistic regression | What the model gives is an estimate of the conditional (population) proportion of times $Y=1$ at each $x$.
If you plot the actual fitted model against the data it was fitted to:
plot(ravenWinNum~ravenScore, data=ravensData)
points(logRegRavens$fitted~ravenScore,ravensData,col=4,pch=16,cex=.8)
... it doesn't seem lik... | CDF and logistic regression | What the model gives is an estimate of the conditional (population) proportion of times $Y=1$ at each $x$.
If you plot the actual fitted model against the data it was fitted to:
plot(ravenWinNum~rave | CDF and logistic regression
What the model gives is an estimate of the conditional (population) proportion of times $Y=1$ at each $x$.
If you plot the actual fitted model against the data it was fitted to:
plot(ravenWinNum~ravenScore, data=ravensData)
points(logRegRavens$fitted~ravenScore,ravensData,col=4,pch=16,cex=.... | CDF and logistic regression
What the model gives is an estimate of the conditional (population) proportion of times $Y=1$ at each $x$.
If you plot the actual fitted model against the data it was fitted to:
plot(ravenWinNum~rave |
48,042 | Residual Vs. Fitted Plot with Outliers | The fitted values can be outlying because of two reasons:
The predictor values are outlying
The predictor values are quite normal compared to the other predictor variables, however the outcome is outlying because of an extreme error term.
In this case I would study the Cooks distance. If it shows the outlying outcom... | Residual Vs. Fitted Plot with Outliers | The fitted values can be outlying because of two reasons:
The predictor values are outlying
The predictor values are quite normal compared to the other predictor variables, however the outcome is ou | Residual Vs. Fitted Plot with Outliers
The fitted values can be outlying because of two reasons:
The predictor values are outlying
The predictor values are quite normal compared to the other predictor variables, however the outcome is outlying because of an extreme error term.
In this case I would study the Cooks di... | Residual Vs. Fitted Plot with Outliers
The fitted values can be outlying because of two reasons:
The predictor values are outlying
The predictor values are quite normal compared to the other predictor variables, however the outcome is ou |
48,043 | Determining similar users from hierarchical clustering | If you have done a hierarchical clustering, it outputs a progressive series of more inclusive clusters. If you want to use the hierarchical clustering to determine which users are similar, you can simply look at the returned dendrogram to see which users are joined at the lowest levels. The cophenetic distance makes ... | Determining similar users from hierarchical clustering | If you have done a hierarchical clustering, it outputs a progressive series of more inclusive clusters. If you want to use the hierarchical clustering to determine which users are similar, you can si | Determining similar users from hierarchical clustering
If you have done a hierarchical clustering, it outputs a progressive series of more inclusive clusters. If you want to use the hierarchical clustering to determine which users are similar, you can simply look at the returned dendrogram to see which users are joine... | Determining similar users from hierarchical clustering
If you have done a hierarchical clustering, it outputs a progressive series of more inclusive clusters. If you want to use the hierarchical clustering to determine which users are similar, you can si |
48,044 | Interpreting Principal Component Analysis output | You appear to be assuming that the largest eigenvalue necessarily can be paired with the largest coefficient within the eigenvectors. That would be wrong.
The question clearly transcends software choice. Here is a fairly silly PCA on five measures of car size using Stata's auto dataset. I used a correlation matrix as ... | Interpreting Principal Component Analysis output | You appear to be assuming that the largest eigenvalue necessarily can be paired with the largest coefficient within the eigenvectors. That would be wrong.
The question clearly transcends software cho | Interpreting Principal Component Analysis output
You appear to be assuming that the largest eigenvalue necessarily can be paired with the largest coefficient within the eigenvectors. That would be wrong.
The question clearly transcends software choice. Here is a fairly silly PCA on five measures of car size using Stat... | Interpreting Principal Component Analysis output
You appear to be assuming that the largest eigenvalue necessarily can be paired with the largest coefficient within the eigenvectors. That would be wrong.
The question clearly transcends software cho |
48,045 | Why is the state-action value function required when the model is unknown? | The $V(s)$'s are sufficient to determine a policy precisely because you have access to the model. In particular you have access to information about the transition structure of the MDP (you know how to look ahead). Without a model you might know you want to get to $s$, but you don't know what actions to take to get the... | Why is the state-action value function required when the model is unknown? | The $V(s)$'s are sufficient to determine a policy precisely because you have access to the model. In particular you have access to information about the transition structure of the MDP (you know how t | Why is the state-action value function required when the model is unknown?
The $V(s)$'s are sufficient to determine a policy precisely because you have access to the model. In particular you have access to information about the transition structure of the MDP (you know how to look ahead). Without a model you might know... | Why is the state-action value function required when the model is unknown?
The $V(s)$'s are sufficient to determine a policy precisely because you have access to the model. In particular you have access to information about the transition structure of the MDP (you know how t |
48,046 | Is this a decent summary of the kernel trick? | Yes. Simply put, $k(x,y) = <\Phi(x), \Phi(y)>$ is the kernel trick. Inner product in the feature space is the evaluation of kernel in the input space. | Is this a decent summary of the kernel trick? | Yes. Simply put, $k(x,y) = <\Phi(x), \Phi(y)>$ is the kernel trick. Inner product in the feature space is the evaluation of kernel in the input space. | Is this a decent summary of the kernel trick?
Yes. Simply put, $k(x,y) = <\Phi(x), \Phi(y)>$ is the kernel trick. Inner product in the feature space is the evaluation of kernel in the input space. | Is this a decent summary of the kernel trick?
Yes. Simply put, $k(x,y) = <\Phi(x), \Phi(y)>$ is the kernel trick. Inner product in the feature space is the evaluation of kernel in the input space. |
48,047 | Is this a decent summary of the kernel trick? | There are plenty of references. For example here or less technical but still nicely explained. It is the other way around, problems that can be expressed in terms of dot products are amenable to be "kernelized", i.e. one can apply the kernel trick yielding to better versions of the algorithms.
A really nice paper revi... | Is this a decent summary of the kernel trick? | There are plenty of references. For example here or less technical but still nicely explained. It is the other way around, problems that can be expressed in terms of dot products are amenable to be " | Is this a decent summary of the kernel trick?
There are plenty of references. For example here or less technical but still nicely explained. It is the other way around, problems that can be expressed in terms of dot products are amenable to be "kernelized", i.e. one can apply the kernel trick yielding to better versio... | Is this a decent summary of the kernel trick?
There are plenty of references. For example here or less technical but still nicely explained. It is the other way around, problems that can be expressed in terms of dot products are amenable to be " |
48,048 | Relationships between t distributions and normal distributions | I'll let someone else address your first question, but regarding the second bullet point, @Glen_b is right: the $t$ distribution is the posterior, not the prior. The problem here is that there is an ambiguity in the text concerning what phrases are being joined by the "and". Consider these two possibilities:
The ... | Relationships between t distributions and normal distributions | I'll let someone else address your first question, but regarding the second bullet point, @Glen_b is right: the $t$ distribution is the posterior, not the prior. The problem here is that there is an | Relationships between t distributions and normal distributions
I'll let someone else address your first question, but regarding the second bullet point, @Glen_b is right: the $t$ distribution is the posterior, not the prior. The problem here is that there is an ambiguity in the text concerning what phrases are being j... | Relationships between t distributions and normal distributions
I'll let someone else address your first question, but regarding the second bullet point, @Glen_b is right: the $t$ distribution is the posterior, not the prior. The problem here is that there is an |
48,049 | Relationships between t distributions and normal distributions | Getting away from the heavy duty math and talking in plain English, I just think of the t distribution as a Normal distribution adjusted for greater uncertainty associated with small samples (low DFs in the table). As a result, the t distribution has wider or fatter tails than the Normal Distribution resulting in much... | Relationships between t distributions and normal distributions | Getting away from the heavy duty math and talking in plain English, I just think of the t distribution as a Normal distribution adjusted for greater uncertainty associated with small samples (low DFs | Relationships between t distributions and normal distributions
Getting away from the heavy duty math and talking in plain English, I just think of the t distribution as a Normal distribution adjusted for greater uncertainty associated with small samples (low DFs in the table). As a result, the t distribution has wider... | Relationships between t distributions and normal distributions
Getting away from the heavy duty math and talking in plain English, I just think of the t distribution as a Normal distribution adjusted for greater uncertainty associated with small samples (low DFs |
48,050 | Why use d-prime instead of percent correct? | D' is a measure of sensitivity, whereas proportion correct is affected by both sensitivity and bias.
In the special case of two balanced classes (same number of signal and noise trials) and zero bias, D' is monotonically mapped to proportion correct, offering no additional insight. However, if the two classes are not ... | Why use d-prime instead of percent correct? | D' is a measure of sensitivity, whereas proportion correct is affected by both sensitivity and bias.
In the special case of two balanced classes (same number of signal and noise trials) and zero bias | Why use d-prime instead of percent correct?
D' is a measure of sensitivity, whereas proportion correct is affected by both sensitivity and bias.
In the special case of two balanced classes (same number of signal and noise trials) and zero bias, D' is monotonically mapped to proportion correct, offering no additional i... | Why use d-prime instead of percent correct?
D' is a measure of sensitivity, whereas proportion correct is affected by both sensitivity and bias.
In the special case of two balanced classes (same number of signal and noise trials) and zero bias |
48,051 | Meta-analysis of means and medians in R? | I thought I would summarize people's suggestions and what I found on my own. It looks like there are meta-analysis methods for analyzing means but not for analyzing medians. These are some sources that are useful for meta-analyzing means:
http://www.stat.rutgers.edu/home/gyang/researches/gmetaRpackage/gmeta.tutorial_2... | Meta-analysis of means and medians in R? | I thought I would summarize people's suggestions and what I found on my own. It looks like there are meta-analysis methods for analyzing means but not for analyzing medians. These are some sources tha | Meta-analysis of means and medians in R?
I thought I would summarize people's suggestions and what I found on my own. It looks like there are meta-analysis methods for analyzing means but not for analyzing medians. These are some sources that are useful for meta-analyzing means:
http://www.stat.rutgers.edu/home/gyang/... | Meta-analysis of means and medians in R?
I thought I would summarize people's suggestions and what I found on my own. It looks like there are meta-analysis methods for analyzing means but not for analyzing medians. These are some sources tha |
48,052 | Meta-analysis of means and medians in R? | If you have medians and range, then you can get the formula for converting to mean and SD from this sentinel paper by Hozo et al., 2005 (http://www.biomedcentral.com/1471-2288/5/13). | Meta-analysis of means and medians in R? | If you have medians and range, then you can get the formula for converting to mean and SD from this sentinel paper by Hozo et al., 2005 (http://www.biomedcentral.com/1471-2288/5/13). | Meta-analysis of means and medians in R?
If you have medians and range, then you can get the formula for converting to mean and SD from this sentinel paper by Hozo et al., 2005 (http://www.biomedcentral.com/1471-2288/5/13). | Meta-analysis of means and medians in R?
If you have medians and range, then you can get the formula for converting to mean and SD from this sentinel paper by Hozo et al., 2005 (http://www.biomedcentral.com/1471-2288/5/13). |
48,053 | Meta-analysis of means and medians in R? | I came through a useful paper by Wan et al, 2014 and I thought it will be very useful for people interested in doing meta-analysis and are reading this post. The paper has formulas for converting median, range and/or IQR to mean and standard deviations (Click here for the paper). The method presented has improvements o... | Meta-analysis of means and medians in R? | I came through a useful paper by Wan et al, 2014 and I thought it will be very useful for people interested in doing meta-analysis and are reading this post. The paper has formulas for converting medi | Meta-analysis of means and medians in R?
I came through a useful paper by Wan et al, 2014 and I thought it will be very useful for people interested in doing meta-analysis and are reading this post. The paper has formulas for converting median, range and/or IQR to mean and standard deviations (Click here for the paper)... | Meta-analysis of means and medians in R?
I came through a useful paper by Wan et al, 2014 and I thought it will be very useful for people interested in doing meta-analysis and are reading this post. The paper has formulas for converting medi |
48,054 | Meta-analysis of means and medians in R? | Since this question was posted and answered there has been more work in this area. In two papers McGrath and colleagues discuss working directly with the medians and compare this with the transformation methods referenced in other answers. Their papers are "One-sample aggregate data meta-analysis of medians" available ... | Meta-analysis of means and medians in R? | Since this question was posted and answered there has been more work in this area. In two papers McGrath and colleagues discuss working directly with the medians and compare this with the transformati | Meta-analysis of means and medians in R?
Since this question was posted and answered there has been more work in this area. In two papers McGrath and colleagues discuss working directly with the medians and compare this with the transformation methods referenced in other answers. Their papers are "One-sample aggregate ... | Meta-analysis of means and medians in R?
Since this question was posted and answered there has been more work in this area. In two papers McGrath and colleagues discuss working directly with the medians and compare this with the transformati |
48,055 | Statistical test for uniform distribution | To test $U(0,10)$ with $n=5$?
Not with much reliability, unless it's extremely nonuniform.
If you're expecting U-shaped alternatives, you might consider say an Anderson-Darling test, (it usually shows up well in power studies, so it's a good one to keep in mind more generally) -- but the power at $n=5$ is going to be ... | Statistical test for uniform distribution | To test $U(0,10)$ with $n=5$?
Not with much reliability, unless it's extremely nonuniform.
If you're expecting U-shaped alternatives, you might consider say an Anderson-Darling test, (it usually show | Statistical test for uniform distribution
To test $U(0,10)$ with $n=5$?
Not with much reliability, unless it's extremely nonuniform.
If you're expecting U-shaped alternatives, you might consider say an Anderson-Darling test, (it usually shows up well in power studies, so it's a good one to keep in mind more generally)... | Statistical test for uniform distribution
To test $U(0,10)$ with $n=5$?
Not with much reliability, unless it's extremely nonuniform.
If you're expecting U-shaped alternatives, you might consider say an Anderson-Darling test, (it usually show |
48,056 | When is a sample proportion p hat instead of x bar | Both questions are essentially applications of the Central Limit Theorem, which says (informally) that "the value of a sum over many samples from a common population will tend to a normal distribution as the number of samples becomes large".
The two questions differ in the type of data that they treat. The "xbar" ques... | When is a sample proportion p hat instead of x bar | Both questions are essentially applications of the Central Limit Theorem, which says (informally) that "the value of a sum over many samples from a common population will tend to a normal distribution | When is a sample proportion p hat instead of x bar
Both questions are essentially applications of the Central Limit Theorem, which says (informally) that "the value of a sum over many samples from a common population will tend to a normal distribution as the number of samples becomes large".
The two questions differ i... | When is a sample proportion p hat instead of x bar
Both questions are essentially applications of the Central Limit Theorem, which says (informally) that "the value of a sum over many samples from a common population will tend to a normal distribution |
48,057 | When is a sample proportion p hat instead of x bar | Below one could be a handy tip. The image clearly distinguishes between sample mean and sample proportions.
Source
Source info: UF Biostatistics Open learning textbook, Module 9, Sampling Distribution of the Sample Mean (in case link dies out in future) | When is a sample proportion p hat instead of x bar | Below one could be a handy tip. The image clearly distinguishes between sample mean and sample proportions.
Source
Source info: UF Biostatistics Open learning textbook, Module 9, Sampling Distributi | When is a sample proportion p hat instead of x bar
Below one could be a handy tip. The image clearly distinguishes between sample mean and sample proportions.
Source
Source info: UF Biostatistics Open learning textbook, Module 9, Sampling Distribution of the Sample Mean (in case link dies out in future) | When is a sample proportion p hat instead of x bar
Below one could be a handy tip. The image clearly distinguishes between sample mean and sample proportions.
Source
Source info: UF Biostatistics Open learning textbook, Module 9, Sampling Distributi |
48,058 | ROC graph interpretation | The accuracy is the ratio of correct results (true positives and true negatives) to total number of tests. If you classify the first six results in the table are "positive" and the rest are "negative", that gives 5 true positives and 9 true negatives. The accuracy is 14/20, which is higher than any other threshold poin... | ROC graph interpretation | The accuracy is the ratio of correct results (true positives and true negatives) to total number of tests. If you classify the first six results in the table are "positive" and the rest are "negative" | ROC graph interpretation
The accuracy is the ratio of correct results (true positives and true negatives) to total number of tests. If you classify the first six results in the table are "positive" and the rest are "negative", that gives 5 true positives and 9 true negatives. The accuracy is 14/20, which is higher than... | ROC graph interpretation
The accuracy is the ratio of correct results (true positives and true negatives) to total number of tests. If you classify the first six results in the table are "positive" and the rest are "negative" |
48,059 | ROC graph interpretation | My understanding is that these are not mathematical assessments. One way to get a best point on the curve is to find the one most protruding in the 10:45 clock direction. But that's assuming you care about fpr equally as tpr, which is not always the case. All subsequent statements are based on this choice. | ROC graph interpretation | My understanding is that these are not mathematical assessments. One way to get a best point on the curve is to find the one most protruding in the 10:45 clock direction. But that's assuming you care | ROC graph interpretation
My understanding is that these are not mathematical assessments. One way to get a best point on the curve is to find the one most protruding in the 10:45 clock direction. But that's assuming you care about fpr equally as tpr, which is not always the case. All subsequent statements are based on... | ROC graph interpretation
My understanding is that these are not mathematical assessments. One way to get a best point on the curve is to find the one most protruding in the 10:45 clock direction. But that's assuming you care |
48,060 | How do i predict with standard errors using betareg package in R? | I've thought about two alternatives to get the prediction interval of the mean of the response:
The paper Ferrari, S.L.P., and Cribari-Neto, F. (2004). Beta Regression for Modeling Rates and Proportions. Journal of Applied Statistics, 31(7), 799–815. (which is in the betareg function references) presents the math form... | How do i predict with standard errors using betareg package in R? | I've thought about two alternatives to get the prediction interval of the mean of the response:
The paper Ferrari, S.L.P., and Cribari-Neto, F. (2004). Beta Regression for Modeling Rates and Proporti | How do i predict with standard errors using betareg package in R?
I've thought about two alternatives to get the prediction interval of the mean of the response:
The paper Ferrari, S.L.P., and Cribari-Neto, F. (2004). Beta Regression for Modeling Rates and Proportions. Journal of Applied Statistics, 31(7), 799–815. (w... | How do i predict with standard errors using betareg package in R?
I've thought about two alternatives to get the prediction interval of the mean of the response:
The paper Ferrari, S.L.P., and Cribari-Neto, F. (2004). Beta Regression for Modeling Rates and Proporti |
48,061 | How do i predict with standard errors using betareg package in R? | I have used the package 'effects' and the function allEffects()
BetaReg <- betareg(Value~x, data = Data)
Effects <- as.data.frame(allEffects(BetaReg, xlevels=list(x=BetaReg$x)))$x
Effects
I think it is correct (looks about right on a graph) :) | How do i predict with standard errors using betareg package in R? | I have used the package 'effects' and the function allEffects()
BetaReg <- betareg(Value~x, data = Data)
Effects <- as.data.frame(allEffects(BetaReg, xlevels=list(x=BetaReg$x)))$x
Effects
I think it | How do i predict with standard errors using betareg package in R?
I have used the package 'effects' and the function allEffects()
BetaReg <- betareg(Value~x, data = Data)
Effects <- as.data.frame(allEffects(BetaReg, xlevels=list(x=BetaReg$x)))$x
Effects
I think it is correct (looks about right on a graph) :) | How do i predict with standard errors using betareg package in R?
I have used the package 'effects' and the function allEffects()
BetaReg <- betareg(Value~x, data = Data)
Effects <- as.data.frame(allEffects(BetaReg, xlevels=list(x=BetaReg$x)))$x
Effects
I think it |
48,062 | Express product as convolution? Or, given $f(s)$, find $g$ satisfying $f(s)=\mathbb{E}[g(X)]$ where $X\sim \mathcal{N}(0,s^2)$ | The short answer is yes and what you want to do is work on the topological group $\mathbb{R}^*$ equipped with the Haar measure $\mu(A) = \int_A \frac{1}{|x|}dx$.
It essentially computes what you computed with log because $e^x : \mathbb{R} \rightarrow \mathbb{R}^*_{\geq 0}$ is a map of topological groups. It takes the... | Express product as convolution? Or, given $f(s)$, find $g$ satisfying $f(s)=\mathbb{E}[g(X)]$ where | The short answer is yes and what you want to do is work on the topological group $\mathbb{R}^*$ equipped with the Haar measure $\mu(A) = \int_A \frac{1}{|x|}dx$.
It essentially computes what you comp | Express product as convolution? Or, given $f(s)$, find $g$ satisfying $f(s)=\mathbb{E}[g(X)]$ where $X\sim \mathcal{N}(0,s^2)$
The short answer is yes and what you want to do is work on the topological group $\mathbb{R}^*$ equipped with the Haar measure $\mu(A) = \int_A \frac{1}{|x|}dx$.
It essentially computes what y... | Express product as convolution? Or, given $f(s)$, find $g$ satisfying $f(s)=\mathbb{E}[g(X)]$ where
The short answer is yes and what you want to do is work on the topological group $\mathbb{R}^*$ equipped with the Haar measure $\mu(A) = \int_A \frac{1}{|x|}dx$.
It essentially computes what you comp |
48,063 | How to conditionally run element of JAGS script based on user supplied variable? | Here is one example of implementing a basic macro substitution system for JAGS scripts.
Explanation of the system
Define a function that takes as arguments any optional elements of the script.
For any aspects of the script that vary across argument values, record a macro token. This should be some unique text. Startin... | How to conditionally run element of JAGS script based on user supplied variable? | Here is one example of implementing a basic macro substitution system for JAGS scripts.
Explanation of the system
Define a function that takes as arguments any optional elements of the script.
For an | How to conditionally run element of JAGS script based on user supplied variable?
Here is one example of implementing a basic macro substitution system for JAGS scripts.
Explanation of the system
Define a function that takes as arguments any optional elements of the script.
For any aspects of the script that vary acros... | How to conditionally run element of JAGS script based on user supplied variable?
Here is one example of implementing a basic macro substitution system for JAGS scripts.
Explanation of the system
Define a function that takes as arguments any optional elements of the script.
For an |
48,064 | How to conditionally run element of JAGS script based on user supplied variable? | In this specific case you could set the beta2 term to zero
for (i in 1:length(y)) {
mu[i] <- beta1 + indicator[i] * beta2 * time[i]
y[i] ~ dnorm(mu[i], tau)
}
where indicator[] is a vector that is one for those data points that you want to model with beta2 and 0 otherwise. You could also use a scalar to chang... | How to conditionally run element of JAGS script based on user supplied variable? | In this specific case you could set the beta2 term to zero
for (i in 1:length(y)) {
mu[i] <- beta1 + indicator[i] * beta2 * time[i]
y[i] ~ dnorm(mu[i], tau)
}
where indicator[] is a vector t | How to conditionally run element of JAGS script based on user supplied variable?
In this specific case you could set the beta2 term to zero
for (i in 1:length(y)) {
mu[i] <- beta1 + indicator[i] * beta2 * time[i]
y[i] ~ dnorm(mu[i], tau)
}
where indicator[] is a vector that is one for those data points that y... | How to conditionally run element of JAGS script based on user supplied variable?
In this specific case you could set the beta2 term to zero
for (i in 1:length(y)) {
mu[i] <- beta1 + indicator[i] * beta2 * time[i]
y[i] ~ dnorm(mu[i], tau)
}
where indicator[] is a vector t |
48,065 | Estimate of variance with the lowest mean square error | Define
$$s^2_d = \frac{\sum_{i=1}^n\left(x_i-\bar{x}\,\right)^2}{d}$$
The statistic $(n - 1)s_{n-1}^2 / \sigma^2$ follows a $\chi^2_{n-1}$ distribution. A $\chi^2_{n-1}$ has mean $n-1$ and variance $2(n-1)$. Hence $\text{E}(s_{n-1}^2) = \sigma^2$, and $\text{Var}(s_{n-1}^2) = 2\frac{\sigma^4 }{ n - 1}$.
Now $s_d^2 = \f... | Estimate of variance with the lowest mean square error | Define
$$s^2_d = \frac{\sum_{i=1}^n\left(x_i-\bar{x}\,\right)^2}{d}$$
The statistic $(n - 1)s_{n-1}^2 / \sigma^2$ follows a $\chi^2_{n-1}$ distribution. A $\chi^2_{n-1}$ has mean $n-1$ and variance $2 | Estimate of variance with the lowest mean square error
Define
$$s^2_d = \frac{\sum_{i=1}^n\left(x_i-\bar{x}\,\right)^2}{d}$$
The statistic $(n - 1)s_{n-1}^2 / \sigma^2$ follows a $\chi^2_{n-1}$ distribution. A $\chi^2_{n-1}$ has mean $n-1$ and variance $2(n-1)$. Hence $\text{E}(s_{n-1}^2) = \sigma^2$, and $\text{Var}(s... | Estimate of variance with the lowest mean square error
Define
$$s^2_d = \frac{\sum_{i=1}^n\left(x_i-\bar{x}\,\right)^2}{d}$$
The statistic $(n - 1)s_{n-1}^2 / \sigma^2$ follows a $\chi^2_{n-1}$ distribution. A $\chi^2_{n-1}$ has mean $n-1$ and variance $2 |
48,066 | Importance of optimizing the correct loss function | Say that I am building a linear regression model p for predicting some values $y_1,…,y_n$.
If the data contains a few extreme outliers in the response - or even just one - the MSE fitted equation can be pulled arbitrarily far away from the MAE one.
Consider the simplest regression model (just an intercept, $\alpha$), ... | Importance of optimizing the correct loss function | Say that I am building a linear regression model p for predicting some values $y_1,…,y_n$.
If the data contains a few extreme outliers in the response - or even just one - the MSE fitted equation can | Importance of optimizing the correct loss function
Say that I am building a linear regression model p for predicting some values $y_1,…,y_n$.
If the data contains a few extreme outliers in the response - or even just one - the MSE fitted equation can be pulled arbitrarily far away from the MAE one.
Consider the simple... | Importance of optimizing the correct loss function
Say that I am building a linear regression model p for predicting some values $y_1,…,y_n$.
If the data contains a few extreme outliers in the response - or even just one - the MSE fitted equation can |
48,067 | Importance of optimizing the correct loss function | I think this question is hard to answer in general and depends heavily on the data that is analyzed.
Simple example: have your model with only the intercept, e.g. we have no predictors and can just estimate the mean and predict that for all new data. If $y$ is normally distributed the sample mean (i.e. the MSE estimate... | Importance of optimizing the correct loss function | I think this question is hard to answer in general and depends heavily on the data that is analyzed.
Simple example: have your model with only the intercept, e.g. we have no predictors and can just es | Importance of optimizing the correct loss function
I think this question is hard to answer in general and depends heavily on the data that is analyzed.
Simple example: have your model with only the intercept, e.g. we have no predictors and can just estimate the mean and predict that for all new data. If $y$ is normally... | Importance of optimizing the correct loss function
I think this question is hard to answer in general and depends heavily on the data that is analyzed.
Simple example: have your model with only the intercept, e.g. we have no predictors and can just es |
48,068 | Importance of optimizing the correct loss function | Often closed form - solution is reason for preferring certain kinds of loss functions. Or it was since now we have computers for solving numerical problems.
If you have loss function with mean absolute errors, that leads to much more difficult problem than error squared as in least squares method.
And problem is al... | Importance of optimizing the correct loss function | Often closed form - solution is reason for preferring certain kinds of loss functions. Or it was since now we have computers for solving numerical problems.
If you have loss function with mean absol | Importance of optimizing the correct loss function
Often closed form - solution is reason for preferring certain kinds of loss functions. Or it was since now we have computers for solving numerical problems.
If you have loss function with mean absolute errors, that leads to much more difficult problem than error squa... | Importance of optimizing the correct loss function
Often closed form - solution is reason for preferring certain kinds of loss functions. Or it was since now we have computers for solving numerical problems.
If you have loss function with mean absol |
48,069 | Conditional Expectation of Poisson Random Variables | Define $S_n=\sum_{i=1}^n X_i$. By symmetry,
$$
\mathrm{E}\left[ X_1 \mid S_n \right] = \mathrm{E}\left[ X_2 \mid S_n \right] = \dots = \mathrm{E}\left[ X_n \mid S_n \right] \quad \textrm{a.s.} \quad (*)
$$
Hence, using $(*)$ and the linearity of the conditional expectation, we have
$$
\mathrm{E}\left[ X_1 \mid S_n ... | Conditional Expectation of Poisson Random Variables | Define $S_n=\sum_{i=1}^n X_i$. By symmetry,
$$
\mathrm{E}\left[ X_1 \mid S_n \right] = \mathrm{E}\left[ X_2 \mid S_n \right] = \dots = \mathrm{E}\left[ X_n \mid S_n \right] \quad \textrm{a.s.} \quad | Conditional Expectation of Poisson Random Variables
Define $S_n=\sum_{i=1}^n X_i$. By symmetry,
$$
\mathrm{E}\left[ X_1 \mid S_n \right] = \mathrm{E}\left[ X_2 \mid S_n \right] = \dots = \mathrm{E}\left[ X_n \mid S_n \right] \quad \textrm{a.s.} \quad (*)
$$
Hence, using $(*)$ and the linearity of the conditional expe... | Conditional Expectation of Poisson Random Variables
Define $S_n=\sum_{i=1}^n X_i$. By symmetry,
$$
\mathrm{E}\left[ X_1 \mid S_n \right] = \mathrm{E}\left[ X_2 \mid S_n \right] = \dots = \mathrm{E}\left[ X_n \mid S_n \right] \quad \textrm{a.s.} \quad |
48,070 | Conditional Expectation of Poisson Random Variables | The OP has apparently found the way, so I am posting an answer.
I will denote $Z \equiv \sum_{i=1}^n X_i$. By linearity of the expected value we have
$$E \left( X_1+X_2+3X_3 |Z \right)= E \left( X_1 |Z \right)+E \left( X_2 |Z \right)+3E \left(X_3 |Z \right)$$
Since the variables are i.i.d. they are also exchangeable,at... | Conditional Expectation of Poisson Random Variables | The OP has apparently found the way, so I am posting an answer.
I will denote $Z \equiv \sum_{i=1}^n X_i$. By linearity of the expected value we have
$$E \left( X_1+X_2+3X_3 |Z \right)= E \left( X_1 | | Conditional Expectation of Poisson Random Variables
The OP has apparently found the way, so I am posting an answer.
I will denote $Z \equiv \sum_{i=1}^n X_i$. By linearity of the expected value we have
$$E \left( X_1+X_2+3X_3 |Z \right)= E \left( X_1 |Z \right)+E \left( X_2 |Z \right)+3E \left(X_3 |Z \right)$$
Since th... | Conditional Expectation of Poisson Random Variables
The OP has apparently found the way, so I am posting an answer.
I will denote $Z \equiv \sum_{i=1}^n X_i$. By linearity of the expected value we have
$$E \left( X_1+X_2+3X_3 |Z \right)= E \left( X_1 | |
48,071 | Pros and Cons: Methods for Detrending Time Series Data | Differenced data
USE: When the series resembles that of a random walk, taking first differences makes it stationary, so that it can be described as linear series representation of autoregressive or moving average terms.
DO NOT USE: When the series appears to randomly fluctuate around its mean.
Log-differenced
USE: Simi... | Pros and Cons: Methods for Detrending Time Series Data | Differenced data
USE: When the series resembles that of a random walk, taking first differences makes it stationary, so that it can be described as linear series representation of autoregressive or mo | Pros and Cons: Methods for Detrending Time Series Data
Differenced data
USE: When the series resembles that of a random walk, taking first differences makes it stationary, so that it can be described as linear series representation of autoregressive or moving average terms.
DO NOT USE: When the series appears to random... | Pros and Cons: Methods for Detrending Time Series Data
Differenced data
USE: When the series resembles that of a random walk, taking first differences makes it stationary, so that it can be described as linear series representation of autoregressive or mo |
48,072 | How To Write Seasonal ARIMA model mathematically | The general $SARIMA(p,d,q)(P,D,Q)_m$ process $X_t$ is the solution of the following equation
$$\Phi(B^m)\phi(B)\nabla_m^D\nabla^d X_t=\Theta(B^m)\theta(B)Z_t,$$
where $Z_t$ is the white noise process. $\nabla_mX_t=X_{t}-X_{t-m}$, $\nabla X_t=X_{t}-X_{t-1}$,
\begin{align}
\Phi(B^m)&=1-\Phi_1B^m -\dots-\Phi_PB^{PM}\\
\ph... | How To Write Seasonal ARIMA model mathematically | The general $SARIMA(p,d,q)(P,D,Q)_m$ process $X_t$ is the solution of the following equation
$$\Phi(B^m)\phi(B)\nabla_m^D\nabla^d X_t=\Theta(B^m)\theta(B)Z_t,$$
where $Z_t$ is the white noise process. | How To Write Seasonal ARIMA model mathematically
The general $SARIMA(p,d,q)(P,D,Q)_m$ process $X_t$ is the solution of the following equation
$$\Phi(B^m)\phi(B)\nabla_m^D\nabla^d X_t=\Theta(B^m)\theta(B)Z_t,$$
where $Z_t$ is the white noise process. $\nabla_mX_t=X_{t}-X_{t-m}$, $\nabla X_t=X_{t}-X_{t-1}$,
\begin{align}... | How To Write Seasonal ARIMA model mathematically
The general $SARIMA(p,d,q)(P,D,Q)_m$ process $X_t$ is the solution of the following equation
$$\Phi(B^m)\phi(B)\nabla_m^D\nabla^d X_t=\Theta(B^m)\theta(B)Z_t,$$
where $Z_t$ is the white noise process. |
48,073 | Struggling with copula theory | It sounds like you are interested in fitting bivariate distributions to data? One way of doing this is to fit a bivariate normal distribution. Unfortunately, many bivariate data sets do not look like a bivariate normal at all. So people have considered more general distributions. One approach is to consider the margina... | Struggling with copula theory | It sounds like you are interested in fitting bivariate distributions to data? One way of doing this is to fit a bivariate normal distribution. Unfortunately, many bivariate data sets do not look like | Struggling with copula theory
It sounds like you are interested in fitting bivariate distributions to data? One way of doing this is to fit a bivariate normal distribution. Unfortunately, many bivariate data sets do not look like a bivariate normal at all. So people have considered more general distributions. One appro... | Struggling with copula theory
It sounds like you are interested in fitting bivariate distributions to data? One way of doing this is to fit a bivariate normal distribution. Unfortunately, many bivariate data sets do not look like |
48,074 | Struggling with copula theory | What convinces you that the Gaussian copula only applies to linear correlation?
This would seem to be a counterexample: a pair of variables with a Gaussian copula, but they're not linearly related:
If that's not what you mean by 'linear correlation', you will need to be more explicit | Struggling with copula theory | What convinces you that the Gaussian copula only applies to linear correlation?
This would seem to be a counterexample: a pair of variables with a Gaussian copula, but they're not linearly related:
I | Struggling with copula theory
What convinces you that the Gaussian copula only applies to linear correlation?
This would seem to be a counterexample: a pair of variables with a Gaussian copula, but they're not linearly related:
If that's not what you mean by 'linear correlation', you will need to be more explicit | Struggling with copula theory
What convinces you that the Gaussian copula only applies to linear correlation?
This would seem to be a counterexample: a pair of variables with a Gaussian copula, but they're not linearly related:
I |
48,075 | How is the number of particles decided in particle filtering? | For this choice I often think about the trade-off between computational cost and the variance of the resulting estimates. As you increase the number of particles or sample size the former increase, while the latter decreases.
Often I do a simple computational experiment:
I create a grid of potential numbers of partic... | How is the number of particles decided in particle filtering? | For this choice I often think about the trade-off between computational cost and the variance of the resulting estimates. As you increase the number of particles or sample size the former increase, wh | How is the number of particles decided in particle filtering?
For this choice I often think about the trade-off between computational cost and the variance of the resulting estimates. As you increase the number of particles or sample size the former increase, while the latter decreases.
Often I do a simple computationa... | How is the number of particles decided in particle filtering?
For this choice I often think about the trade-off between computational cost and the variance of the resulting estimates. As you increase the number of particles or sample size the former increase, wh |
48,076 | Fitting data to a Poisson distribution, what are the errors? | The direct answer to the question - how to deal with small expected counts - is that one might either
(a) combine ranges of cells at the end (a very common approach),
(b) use a different (and perhaps better) goodness of fit test, or
(c) consider dropping the chi-square approximation, and see if one can deal with the... | Fitting data to a Poisson distribution, what are the errors? | The direct answer to the question - how to deal with small expected counts - is that one might either
(a) combine ranges of cells at the end (a very common approach),
(b) use a different (and perhap | Fitting data to a Poisson distribution, what are the errors?
The direct answer to the question - how to deal with small expected counts - is that one might either
(a) combine ranges of cells at the end (a very common approach),
(b) use a different (and perhaps better) goodness of fit test, or
(c) consider dropping t... | Fitting data to a Poisson distribution, what are the errors?
The direct answer to the question - how to deal with small expected counts - is that one might either
(a) combine ranges of cells at the end (a very common approach),
(b) use a different (and perhap |
48,077 | Fitting data to a Poisson distribution, what are the errors? | This doesn't entirely answer your question regarding how to implement the goodness of fit test. But with a sample size that large, your data will almost certainly fail any test comparing it to a Poisson distribution even if your distribution closely resembles a Poisson distribution (unless, of course, you simulated Poi... | Fitting data to a Poisson distribution, what are the errors? | This doesn't entirely answer your question regarding how to implement the goodness of fit test. But with a sample size that large, your data will almost certainly fail any test comparing it to a Poiss | Fitting data to a Poisson distribution, what are the errors?
This doesn't entirely answer your question regarding how to implement the goodness of fit test. But with a sample size that large, your data will almost certainly fail any test comparing it to a Poisson distribution even if your distribution closely resembles... | Fitting data to a Poisson distribution, what are the errors?
This doesn't entirely answer your question regarding how to implement the goodness of fit test. But with a sample size that large, your data will almost certainly fail any test comparing it to a Poiss |
48,078 | Fitting data to a Poisson distribution, what are the errors? | One way to go is by using the "deviance residual" from glms. This is a likelihood ratio for each count. You can easily get these from the glm () function in R. The glm you would fit is one with no intercept, and dummy variables for each year. However with a million observstions this may run into memory problems.
Yo... | Fitting data to a Poisson distribution, what are the errors? | One way to go is by using the "deviance residual" from glms. This is a likelihood ratio for each count. You can easily get these from the glm () function in R. The glm you would fit is one with no | Fitting data to a Poisson distribution, what are the errors?
One way to go is by using the "deviance residual" from glms. This is a likelihood ratio for each count. You can easily get these from the glm () function in R. The glm you would fit is one with no intercept, and dummy variables for each year. However with... | Fitting data to a Poisson distribution, what are the errors?
One way to go is by using the "deviance residual" from glms. This is a likelihood ratio for each count. You can easily get these from the glm () function in R. The glm you would fit is one with no |
48,079 | Fitting data to a Poisson distribution, what are the errors? | You could try fitting an exponential distribution to the inter-arrival times (if you have the times of arrivals in your dataset). In this case, you can do a simple KS Test for the exponential fit. If it fits, then the Poisson fits your count data. See the following reference:
ON THE KOLMOGOROV-SMIRNOV TEST FOR THE EXPO... | Fitting data to a Poisson distribution, what are the errors? | You could try fitting an exponential distribution to the inter-arrival times (if you have the times of arrivals in your dataset). In this case, you can do a simple KS Test for the exponential fit. If | Fitting data to a Poisson distribution, what are the errors?
You could try fitting an exponential distribution to the inter-arrival times (if you have the times of arrivals in your dataset). In this case, you can do a simple KS Test for the exponential fit. If it fits, then the Poisson fits your count data. See the fol... | Fitting data to a Poisson distribution, what are the errors?
You could try fitting an exponential distribution to the inter-arrival times (if you have the times of arrivals in your dataset). In this case, you can do a simple KS Test for the exponential fit. If |
48,080 | Checking assumptions LMM: residual plot with diamond shape | The diamond pattern in the residuals is due to a combination of the "ceiling effect" you have in the observed female data (i.e., a lot of female data points clustered at the maximum value of 18 or so) and the "floor effect" you have in the observed male data (a lot of male data points clustered at the minimum value of ... | Checking assumptions LMM: residual plot with diamond shape | The diamond pattern in the residuals is due to a combination of the "ceiling effect" you have in the observed female data (i.e., a lot of female data points clustered at the maximum value of 18 or so) | Checking assumptions LMM: residual plot with diamond shape
The diamond pattern in the residuals is due to a combination of the "ceiling effect" you have in the observed female data (i.e., a lot of female data points clustered at the maximum value of 18 or so) and the "floor effect" you have in the observed male data (a... | Checking assumptions LMM: residual plot with diamond shape
The diamond pattern in the residuals is due to a combination of the "ceiling effect" you have in the observed female data (i.e., a lot of female data points clustered at the maximum value of 18 or so) |
48,081 | Checking assumptions LMM: residual plot with diamond shape | Diamond (double outward box) distributed residuals are another type of non–monotonic heteroscedasticity that severely violates the homoscedasticity
assumption. For the correction of diamond type of heterocedasticity, the study " Correcting Double Outward Box Distributed Residuals by WCEV"
(http://www.tandfonline.com/d... | Checking assumptions LMM: residual plot with diamond shape | Diamond (double outward box) distributed residuals are another type of non–monotonic heteroscedasticity that severely violates the homoscedasticity
assumption. For the correction of diamond type of he | Checking assumptions LMM: residual plot with diamond shape
Diamond (double outward box) distributed residuals are another type of non–monotonic heteroscedasticity that severely violates the homoscedasticity
assumption. For the correction of diamond type of heterocedasticity, the study " Correcting Double Outward Box Di... | Checking assumptions LMM: residual plot with diamond shape
Diamond (double outward box) distributed residuals are another type of non–monotonic heteroscedasticity that severely violates the homoscedasticity
assumption. For the correction of diamond type of he |
48,082 | Posterior covariance of Normal-Inverse-Wishart not converging properly | The problem was that I was setting the degrees of freedom too low-- it should be at least P+2, where $\Psi$ is a PxP matrix. | Posterior covariance of Normal-Inverse-Wishart not converging properly | The problem was that I was setting the degrees of freedom too low-- it should be at least P+2, where $\Psi$ is a PxP matrix. | Posterior covariance of Normal-Inverse-Wishart not converging properly
The problem was that I was setting the degrees of freedom too low-- it should be at least P+2, where $\Psi$ is a PxP matrix. | Posterior covariance of Normal-Inverse-Wishart not converging properly
The problem was that I was setting the degrees of freedom too low-- it should be at least P+2, where $\Psi$ is a PxP matrix. |
48,083 | Posterior covariance of Normal-Inverse-Wishart not converging properly | Sorry I am slow, but all that changed was setting nu=4?
x = NormalInverseWishartDistribution(np.array([0,0])-3,1,4,np.eye(2)) # nu > 2 + psi.shape[0]
samples = [x.sample() for _ in range(1000)]
data = [np.random.multivariate_normal(mu,cov) for mu,cov in samples]
y = NormalInverseWishartDistribution(np.array([0,0]),1,4,... | Posterior covariance of Normal-Inverse-Wishart not converging properly | Sorry I am slow, but all that changed was setting nu=4?
x = NormalInverseWishartDistribution(np.array([0,0])-3,1,4,np.eye(2)) # nu > 2 + psi.shape[0]
samples = [x.sample() for _ in range(1000)]
data = | Posterior covariance of Normal-Inverse-Wishart not converging properly
Sorry I am slow, but all that changed was setting nu=4?
x = NormalInverseWishartDistribution(np.array([0,0])-3,1,4,np.eye(2)) # nu > 2 + psi.shape[0]
samples = [x.sample() for _ in range(1000)]
data = [np.random.multivariate_normal(mu,cov) for mu,co... | Posterior covariance of Normal-Inverse-Wishart not converging properly
Sorry I am slow, but all that changed was setting nu=4?
x = NormalInverseWishartDistribution(np.array([0,0])-3,1,4,np.eye(2)) # nu > 2 + psi.shape[0]
samples = [x.sample() for _ in range(1000)]
data = |
48,084 | Classify and regress at the same time | Answering the specific question.
Yes, it is possible (but might be undesirable)
Directed graph models (such as ANNs) can cope with that.
You have 5 input variables, and you want to predict gender first, and include this prediction to predict income.
Basically, you need to connect all your inputs to an output, which is ... | Classify and regress at the same time | Answering the specific question.
Yes, it is possible (but might be undesirable)
Directed graph models (such as ANNs) can cope with that.
You have 5 input variables, and you want to predict gender firs | Classify and regress at the same time
Answering the specific question.
Yes, it is possible (but might be undesirable)
Directed graph models (such as ANNs) can cope with that.
You have 5 input variables, and you want to predict gender first, and include this prediction to predict income.
Basically, you need to connect a... | Classify and regress at the same time
Answering the specific question.
Yes, it is possible (but might be undesirable)
Directed graph models (such as ANNs) can cope with that.
You have 5 input variables, and you want to predict gender firs |
48,085 | Classify and regress at the same time | Why do you want to do this? For your proposal to be possible and meaningful, it must be that you have some missing observations on the binary gender predictor (but not all). If that is the case, the usual approach is to use some (multiple) imputation method. See Multiple imputation for missing values or search this s... | Classify and regress at the same time | Why do you want to do this? For your proposal to be possible and meaningful, it must be that you have some missing observations on the binary gender predictor (but not all). If that is the case, the | Classify and regress at the same time
Why do you want to do this? For your proposal to be possible and meaningful, it must be that you have some missing observations on the binary gender predictor (but not all). If that is the case, the usual approach is to use some (multiple) imputation method. See Multiple imputatio... | Classify and regress at the same time
Why do you want to do this? For your proposal to be possible and meaningful, it must be that you have some missing observations on the binary gender predictor (but not all). If that is the case, the |
48,086 | Classify and regress at the same time | As the other answers pointed out, this may not be something you want to do. But in cases where your 2 regression targets are closely related (say in the case of latitude and longitude) it may make sense.
One of the things I really love about neural networks is that they are very flexible, and allow you to define one m... | Classify and regress at the same time | As the other answers pointed out, this may not be something you want to do. But in cases where your 2 regression targets are closely related (say in the case of latitude and longitude) it may make se | Classify and regress at the same time
As the other answers pointed out, this may not be something you want to do. But in cases where your 2 regression targets are closely related (say in the case of latitude and longitude) it may make sense.
One of the things I really love about neural networks is that they are very f... | Classify and regress at the same time
As the other answers pointed out, this may not be something you want to do. But in cases where your 2 regression targets are closely related (say in the case of latitude and longitude) it may make se |
48,087 | Why do PCA and Factor Analysis return different results in this example? | The covariance matrix in this example is given by $$\mathbf C = \left(\begin{array}{c} 1 & \sim 1 & 0 \\ \sim 1 & \sim 1 & 0 \\ 0 & 0 & 100\end{array}\right).$$
To compare PCA and FA, think about how PCA/FA loadings reconstruct the covariance matrix.
The loadings of the first principal component in PCA is a vector $\ma... | Why do PCA and Factor Analysis return different results in this example? | The covariance matrix in this example is given by $$\mathbf C = \left(\begin{array}{c} 1 & \sim 1 & 0 \\ \sim 1 & \sim 1 & 0 \\ 0 & 0 & 100\end{array}\right).$$
To compare PCA and FA, think about how | Why do PCA and Factor Analysis return different results in this example?
The covariance matrix in this example is given by $$\mathbf C = \left(\begin{array}{c} 1 & \sim 1 & 0 \\ \sim 1 & \sim 1 & 0 \\ 0 & 0 & 100\end{array}\right).$$
To compare PCA and FA, think about how PCA/FA loadings reconstruct the covariance matr... | Why do PCA and Factor Analysis return different results in this example?
The covariance matrix in this example is given by $$\mathbf C = \left(\begin{array}{c} 1 & \sim 1 & 0 \\ \sim 1 & \sim 1 & 0 \\ 0 & 0 & 100\end{array}\right).$$
To compare PCA and FA, think about how |
48,088 | Cumulative Distribution Function Inequality (Discrete Distributions) | Try to draw CDF of a Discrete random variable like the (upper) one you have here. Now draw a horizontal line to indicate the level of $a_1$. All you need to do is to find the values of $T$ such that your CDF i.e. $F_T$ satisfies $F_T(T)\leq a_1$. You can move your $a_1$ vertically. Now depending on the level of $a_1$ s... | Cumulative Distribution Function Inequality (Discrete Distributions) | Try to draw CDF of a Discrete random variable like the (upper) one you have here. Now draw a horizontal line to indicate the level of $a_1$. All you need to do is to find the values of $T$ such that y | Cumulative Distribution Function Inequality (Discrete Distributions)
Try to draw CDF of a Discrete random variable like the (upper) one you have here. Now draw a horizontal line to indicate the level of $a_1$. All you need to do is to find the values of $T$ such that your CDF i.e. $F_T$ satisfies $F_T(T)\leq a_1$. You ... | Cumulative Distribution Function Inequality (Discrete Distributions)
Try to draw CDF of a Discrete random variable like the (upper) one you have here. Now draw a horizontal line to indicate the level of $a_1$. All you need to do is to find the values of $T$ such that y |
48,089 | Cumulative Distribution Function Inequality (Discrete Distributions) | Consider a box $\Omega$ filled with tickets. On each ticket $\omega$ is written a number called $X(\omega)$. For any number $x$, whether or not it appears among the tickets, $F_X(x)$ is (defined to be) the proportion of tickets for which $X \le x.$
Let's add some new information to each ticket $\omega$: next to the v... | Cumulative Distribution Function Inequality (Discrete Distributions) | Consider a box $\Omega$ filled with tickets. On each ticket $\omega$ is written a number called $X(\omega)$. For any number $x$, whether or not it appears among the tickets, $F_X(x)$ is (defined to | Cumulative Distribution Function Inequality (Discrete Distributions)
Consider a box $\Omega$ filled with tickets. On each ticket $\omega$ is written a number called $X(\omega)$. For any number $x$, whether or not it appears among the tickets, $F_X(x)$ is (defined to be) the proportion of tickets for which $X \le x.$
... | Cumulative Distribution Function Inequality (Discrete Distributions)
Consider a box $\Omega$ filled with tickets. On each ticket $\omega$ is written a number called $X(\omega)$. For any number $x$, whether or not it appears among the tickets, $F_X(x)$ is (defined to |
48,090 | Why is Pearson's correlation coefficient defined the way it is? | One nice thing you get from dividing by the product of standard deviations is that it guarantees that the correlation coefficient will be between -1 and +1.
If you want to determine if $X$ has a stronger linear relationship with $Y$ or with $Z$ comparing $cov(X,Y)$ with $cov(X,Z)$ directly is not informative, since th... | Why is Pearson's correlation coefficient defined the way it is? | One nice thing you get from dividing by the product of standard deviations is that it guarantees that the correlation coefficient will be between -1 and +1.
If you want to determine if $X$ has a stron | Why is Pearson's correlation coefficient defined the way it is?
One nice thing you get from dividing by the product of standard deviations is that it guarantees that the correlation coefficient will be between -1 and +1.
If you want to determine if $X$ has a stronger linear relationship with $Y$ or with $Z$ comparing ... | Why is Pearson's correlation coefficient defined the way it is?
One nice thing you get from dividing by the product of standard deviations is that it guarantees that the correlation coefficient will be between -1 and +1.
If you want to determine if $X$ has a stron |
48,091 | Is there a way to remove individual trees from a forest in the randomForest package in R? | One idea is, instead of creating one forest with N trees, create N "forests" of 1 tree each by calling randomForest() N times. Then you could manipulate them as you wish. | Is there a way to remove individual trees from a forest in the randomForest package in R? | One idea is, instead of creating one forest with N trees, create N "forests" of 1 tree each by calling randomForest() N times. Then you could manipulate them as you wish. | Is there a way to remove individual trees from a forest in the randomForest package in R?
One idea is, instead of creating one forest with N trees, create N "forests" of 1 tree each by calling randomForest() N times. Then you could manipulate them as you wish. | Is there a way to remove individual trees from a forest in the randomForest package in R?
One idea is, instead of creating one forest with N trees, create N "forests" of 1 tree each by calling randomForest() N times. Then you could manipulate them as you wish. |
48,092 | Is there a way to remove individual trees from a forest in the randomForest package in R? | It depends on which language you are more familiar with.
randomForest package implements the original Fortran version of Breiman's random forest. You should try to modify the Fortran code then.
party packages has everything implemented in C. So, you can try to modify the C code.
WEKA RandomForest is implemented in Ja... | Is there a way to remove individual trees from a forest in the randomForest package in R? | It depends on which language you are more familiar with.
randomForest package implements the original Fortran version of Breiman's random forest. You should try to modify the Fortran code then.
part | Is there a way to remove individual trees from a forest in the randomForest package in R?
It depends on which language you are more familiar with.
randomForest package implements the original Fortran version of Breiman's random forest. You should try to modify the Fortran code then.
party packages has everything impl... | Is there a way to remove individual trees from a forest in the randomForest package in R?
It depends on which language you are more familiar with.
randomForest package implements the original Fortran version of Breiman's random forest. You should try to modify the Fortran code then.
part |
48,093 | How to remove seasonality from daily electricity demand | Modeling daily electricity demand is a data intensive effort. To simplify this, it's easier to start "zoomed out", estimating monthly loads. Here's an article (with a Youtube video) that describes a monthly model that is simple and easy to understand. The article includes R code:
http://revgr.com/2012/11/06/all-fo... | How to remove seasonality from daily electricity demand | Modeling daily electricity demand is a data intensive effort. To simplify this, it's easier to start "zoomed out", estimating monthly loads. Here's an article (with a Youtube video) that describes | How to remove seasonality from daily electricity demand
Modeling daily electricity demand is a data intensive effort. To simplify this, it's easier to start "zoomed out", estimating monthly loads. Here's an article (with a Youtube video) that describes a monthly model that is simple and easy to understand. The art... | How to remove seasonality from daily electricity demand
Modeling daily electricity demand is a data intensive effort. To simplify this, it's easier to start "zoomed out", estimating monthly loads. Here's an article (with a Youtube video) that describes |
48,094 | How to remove seasonality from daily electricity demand | i'm having a fabulous run with ucm. You could model this as daily seasonality & with an annual cycle. You also have a very evident trend.
Post back here if you succeeded with ucm (proc ucm) | How to remove seasonality from daily electricity demand | i'm having a fabulous run with ucm. You could model this as daily seasonality & with an annual cycle. You also have a very evident trend.
Post back here if you succeeded with ucm (proc ucm) | How to remove seasonality from daily electricity demand
i'm having a fabulous run with ucm. You could model this as daily seasonality & with an annual cycle. You also have a very evident trend.
Post back here if you succeeded with ucm (proc ucm) | How to remove seasonality from daily electricity demand
i'm having a fabulous run with ucm. You could model this as daily seasonality & with an annual cycle. You also have a very evident trend.
Post back here if you succeeded with ucm (proc ucm) |
48,095 | Probability of always more heads than tails | For a fixed value of $n$, I'm not sure that a simpler formula exists. However, for $n=\infty$, the probability that there will be more heads than tails permanently is simply equal to $2p-1$. We can show this as follows.
Let $x_k$ denote the probability of success conditional on the current state being $k$ heads. We ... | Probability of always more heads than tails | For a fixed value of $n$, I'm not sure that a simpler formula exists. However, for $n=\infty$, the probability that there will be more heads than tails permanently is simply equal to $2p-1$. We can | Probability of always more heads than tails
For a fixed value of $n$, I'm not sure that a simpler formula exists. However, for $n=\infty$, the probability that there will be more heads than tails permanently is simply equal to $2p-1$. We can show this as follows.
Let $x_k$ denote the probability of success conditiona... | Probability of always more heads than tails
For a fixed value of $n$, I'm not sure that a simpler formula exists. However, for $n=\infty$, the probability that there will be more heads than tails permanently is simply equal to $2p-1$. We can |
48,096 | Why does fixed effect p-value in a mixed model act unintuitively? | Your model corresponds to
$$Y_{ijk}=\beta_0+\beta_i+u_{ij}+e_{ijk}$$
Using the properties of the errors, $e_{ijk}$, The corresponding model for the means is given as:
$$\overline{Y}_{ij}=\beta_0+\beta_i+u_{ij}+\overline{e}_{ij}$$
where the variance of $\overline{e}_{ij}$ is now $\frac{\sigma^2_e}{n_{ij}}$, where $n_{ij... | Why does fixed effect p-value in a mixed model act unintuitively? | Your model corresponds to
$$Y_{ijk}=\beta_0+\beta_i+u_{ij}+e_{ijk}$$
Using the properties of the errors, $e_{ijk}$, The corresponding model for the means is given as:
$$\overline{Y}_{ij}=\beta_0+\beta | Why does fixed effect p-value in a mixed model act unintuitively?
Your model corresponds to
$$Y_{ijk}=\beta_0+\beta_i+u_{ij}+e_{ijk}$$
Using the properties of the errors, $e_{ijk}$, The corresponding model for the means is given as:
$$\overline{Y}_{ij}=\beta_0+\beta_i+u_{ij}+\overline{e}_{ij}$$
where the variance of $\... | Why does fixed effect p-value in a mixed model act unintuitively?
Your model corresponds to
$$Y_{ijk}=\beta_0+\beta_i+u_{ij}+e_{ijk}$$
Using the properties of the errors, $e_{ijk}$, The corresponding model for the means is given as:
$$\overline{Y}_{ij}=\beta_0+\beta |
48,097 | How to calculate estimated proportions and their confidence intervals from a mixed model? | First a note: you can't calculate a decent standard error on the probabilities, you have to do so on a logit scale and use those to construct your confidence intervals. Intervals around probabilities are hardly symmetrical, and definitely not when using a mixed model.
You can easily plot the effects using the package e... | How to calculate estimated proportions and their confidence intervals from a mixed model? | First a note: you can't calculate a decent standard error on the probabilities, you have to do so on a logit scale and use those to construct your confidence intervals. Intervals around probabilities | How to calculate estimated proportions and their confidence intervals from a mixed model?
First a note: you can't calculate a decent standard error on the probabilities, you have to do so on a logit scale and use those to construct your confidence intervals. Intervals around probabilities are hardly symmetrical, and de... | How to calculate estimated proportions and their confidence intervals from a mixed model?
First a note: you can't calculate a decent standard error on the probabilities, you have to do so on a logit scale and use those to construct your confidence intervals. Intervals around probabilities |
48,098 | Why Bayesian logistic (probit) regression instead of standard logistic (probit) regression? | Arguably, Bayesian logistic/probit regression would be better if you had informative prior, or if there was perfect or quasi-perfect separation or if you wanted to fit a hierarchical model.
If you have an informative prior, then use it. And nothing better than use it in a Bayesian approach. If there is perfect separat... | Why Bayesian logistic (probit) regression instead of standard logistic (probit) regression? | Arguably, Bayesian logistic/probit regression would be better if you had informative prior, or if there was perfect or quasi-perfect separation or if you wanted to fit a hierarchical model.
If you ha | Why Bayesian logistic (probit) regression instead of standard logistic (probit) regression?
Arguably, Bayesian logistic/probit regression would be better if you had informative prior, or if there was perfect or quasi-perfect separation or if you wanted to fit a hierarchical model.
If you have an informative prior, the... | Why Bayesian logistic (probit) regression instead of standard logistic (probit) regression?
Arguably, Bayesian logistic/probit regression would be better if you had informative prior, or if there was perfect or quasi-perfect separation or if you wanted to fit a hierarchical model.
If you ha |
48,099 | About Trivariate Normal Distribution | If $X \sim N_k(\mu,\Sigma)$, then $Q=(X-\mu)'\Sigma^{-1}(X-\mu) \sim \chi^2_k$. Further, the level sets of $Q$ are the ellipsoids you refer to. So the 72% you mention comes from a chi-square distribution (these calcs in R):
> pchisq(1.96^2,df=3)
[1] 0.7209157
As do the other numbers:
> pchisq(1.96^2,df=10)
[1] 0.04579... | About Trivariate Normal Distribution | If $X \sim N_k(\mu,\Sigma)$, then $Q=(X-\mu)'\Sigma^{-1}(X-\mu) \sim \chi^2_k$. Further, the level sets of $Q$ are the ellipsoids you refer to. So the 72% you mention comes from a chi-square distribut | About Trivariate Normal Distribution
If $X \sim N_k(\mu,\Sigma)$, then $Q=(X-\mu)'\Sigma^{-1}(X-\mu) \sim \chi^2_k$. Further, the level sets of $Q$ are the ellipsoids you refer to. So the 72% you mention comes from a chi-square distribution (these calcs in R):
> pchisq(1.96^2,df=3)
[1] 0.7209157
As do the other number... | About Trivariate Normal Distribution
If $X \sim N_k(\mu,\Sigma)$, then $Q=(X-\mu)'\Sigma^{-1}(X-\mu) \sim \chi^2_k$. Further, the level sets of $Q$ are the ellipsoids you refer to. So the 72% you mention comes from a chi-square distribut |
48,100 | Relation of slopes of predictors when they are correlated in linear regression | When you omitt $X_2$ from your first regression, your estimate of the coefficient $\beta_{10}$ will be subject to the usual omitted variable bias:
$$plim\: \widehat{\beta}_{10} = \beta_{11} + \beta_{21}\frac{Cov(X_1,X_2)}{Var(X_1)} $$
and the only way that this bias vanishes is if $X_2$ is not useful in predicting $Y$,... | Relation of slopes of predictors when they are correlated in linear regression | When you omitt $X_2$ from your first regression, your estimate of the coefficient $\beta_{10}$ will be subject to the usual omitted variable bias:
$$plim\: \widehat{\beta}_{10} = \beta_{11} + \beta_{2 | Relation of slopes of predictors when they are correlated in linear regression
When you omitt $X_2$ from your first regression, your estimate of the coefficient $\beta_{10}$ will be subject to the usual omitted variable bias:
$$plim\: \widehat{\beta}_{10} = \beta_{11} + \beta_{21}\frac{Cov(X_1,X_2)}{Var(X_1)} $$
and th... | Relation of slopes of predictors when they are correlated in linear regression
When you omitt $X_2$ from your first regression, your estimate of the coefficient $\beta_{10}$ will be subject to the usual omitted variable bias:
$$plim\: \widehat{\beta}_{10} = \beta_{11} + \beta_{2 |
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