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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39,301 | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | Framing the question
You are asking an applied and subjective question, and thus, any answer needs to be infused with applied and subjective considerations.
From a purely statistical perspective, the mean and median both provide different information about the central tendency of a sample of data. Thus, neither is cor... | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | Framing the question
You are asking an applied and subjective question, and thus, any answer needs to be infused with applied and subjective considerations.
From a purely statistical perspective, the | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
Framing the question
You are asking an applied and subjective question, and thus, any answer needs to be infused with applied and subjective considerations.
From a purely statistical perspective, the mean and median both pro... | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
Framing the question
You are asking an applied and subjective question, and thus, any answer needs to be infused with applied and subjective considerations.
From a purely statistical perspective, the |
39,302 | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | You can read about measures of central tendency here: http://en.wikipedia.org/wiki/Central_tendency .
Generally, you analyse a sample in order to tell something about a (much larger) population. Often you know more about the population than merely the data in your sample, usually something that motivated you to take a ... | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | You can read about measures of central tendency here: http://en.wikipedia.org/wiki/Central_tendency .
Generally, you analyse a sample in order to tell something about a (much larger) population. Often | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
You can read about measures of central tendency here: http://en.wikipedia.org/wiki/Central_tendency .
Generally, you analyse a sample in order to tell something about a (much larger) population. Often you know more about the ... | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
You can read about measures of central tendency here: http://en.wikipedia.org/wiki/Central_tendency .
Generally, you analyse a sample in order to tell something about a (much larger) population. Often |
39,303 | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | There are no hard and fast rules. They convey different information and have different properties. You select the statistic that best conveys what you want to convey. Or better yet, select statistics that best describe the data. Keep this same thing in mind when you're selecting a measure of central tendency to ana... | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | There are no hard and fast rules. They convey different information and have different properties. You select the statistic that best conveys what you want to convey. Or better yet, select statisti | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
There are no hard and fast rules. They convey different information and have different properties. You select the statistic that best conveys what you want to convey. Or better yet, select statistics that best describe the... | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
There are no hard and fast rules. They convey different information and have different properties. You select the statistic that best conveys what you want to convey. Or better yet, select statisti |
39,304 | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | Be careful with medians: they are biased estimators and the degree of bias can change depending on the skew of the distribution and the sample size (see Miller, 1988). This means that if you are comparing two conditions that have either different skew or different sample sizes, you may find a difference that is in fact... | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | Be careful with medians: they are biased estimators and the degree of bias can change depending on the skew of the distribution and the sample size (see Miller, 1988). This means that if you are compa | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
Be careful with medians: they are biased estimators and the degree of bias can change depending on the skew of the distribution and the sample size (see Miller, 1988). This means that if you are comparing two conditions that ... | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
Be careful with medians: they are biased estimators and the degree of bias can change depending on the skew of the distribution and the sample size (see Miller, 1988). This means that if you are compa |
39,305 | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | "deviations in the data are the devil" is just not true I think - well I don't agree with it at least. I'd say its more like "chilli" than the "devil" - as much as you can reasonably handle is good, but it can get nasty if there is too much.
The most general procedure I know of to "choose a statistic" to report your d... | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | "deviations in the data are the devil" is just not true I think - well I don't agree with it at least. I'd say its more like "chilli" than the "devil" - as much as you can reasonably handle is good, | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
"deviations in the data are the devil" is just not true I think - well I don't agree with it at least. I'd say its more like "chilli" than the "devil" - as much as you can reasonably handle is good, but it can get nasty if t... | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
"deviations in the data are the devil" is just not true I think - well I don't agree with it at least. I'd say its more like "chilli" than the "devil" - as much as you can reasonably handle is good, |
39,306 | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | In your examples you seem to make the underlying assumption that you are "blind" to how the data you are going to analyze were generated. In other words, whether mean or median is the right tool for say something robust about central tendency of a sample of data largely depends on your (prior) knowledge of the data gen... | When does the amount of skew or prevalence of outliers make the median preferable to the mean? | In your examples you seem to make the underlying assumption that you are "blind" to how the data you are going to analyze were generated. In other words, whether mean or median is the right tool for s | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
In your examples you seem to make the underlying assumption that you are "blind" to how the data you are going to analyze were generated. In other words, whether mean or median is the right tool for say something robust about... | When does the amount of skew or prevalence of outliers make the median preferable to the mean?
In your examples you seem to make the underlying assumption that you are "blind" to how the data you are going to analyze were generated. In other words, whether mean or median is the right tool for s |
39,307 | Forecasting beyond one season using Holt-Winters' exponential smoothing | I am not very familiar with Holt-Winters, however I have this excellent book by @Rob Hyndman. The package forecast (which is based on the book) of statistical package R gives the following result on your data:
> hw<-read.table("~/R/stackoverflow/hw.txt")
> tt<-ts(hw[,3],start=c(1999,1),freq=12)
> aa<-forecast(tt)
> pl... | Forecasting beyond one season using Holt-Winters' exponential smoothing | I am not very familiar with Holt-Winters, however I have this excellent book by @Rob Hyndman. The package forecast (which is based on the book) of statistical package R gives the following result on y | Forecasting beyond one season using Holt-Winters' exponential smoothing
I am not very familiar with Holt-Winters, however I have this excellent book by @Rob Hyndman. The package forecast (which is based on the book) of statistical package R gives the following result on your data:
> hw<-read.table("~/R/stackoverflow/hw... | Forecasting beyond one season using Holt-Winters' exponential smoothing
I am not very familiar with Holt-Winters, however I have this excellent book by @Rob Hyndman. The package forecast (which is based on the book) of statistical package R gives the following result on y |
39,308 | Forecasting beyond one season using Holt-Winters' exponential smoothing | The formulae for Holt-Winters' method include forecasting the seasonal component. You don't need $\gamma=0$. See a forecasting textbook for the details. | Forecasting beyond one season using Holt-Winters' exponential smoothing | The formulae for Holt-Winters' method include forecasting the seasonal component. You don't need $\gamma=0$. See a forecasting textbook for the details. | Forecasting beyond one season using Holt-Winters' exponential smoothing
The formulae for Holt-Winters' method include forecasting the seasonal component. You don't need $\gamma=0$. See a forecasting textbook for the details. | Forecasting beyond one season using Holt-Winters' exponential smoothing
The formulae for Holt-Winters' method include forecasting the seasonal component. You don't need $\gamma=0$. See a forecasting textbook for the details. |
39,309 | Forecasting beyond one season using Holt-Winters' exponential smoothing | 1: http://i.stack.imgur.com/kxU4t.jpg reflects a questioning of the highly unusual Oct 2009 value 130 Oct-09 2301.41 . Time series analysis actually challenges the data rather than fitting a presumed set of models. The residuals from the following model
1: http://i.stack.imgur.com/Q4W5h.jpg more closely exhibit the re... | Forecasting beyond one season using Holt-Winters' exponential smoothing | 1: http://i.stack.imgur.com/kxU4t.jpg reflects a questioning of the highly unusual Oct 2009 value 130 Oct-09 2301.41 . Time series analysis actually challenges the data rather than fitting a presumed | Forecasting beyond one season using Holt-Winters' exponential smoothing
1: http://i.stack.imgur.com/kxU4t.jpg reflects a questioning of the highly unusual Oct 2009 value 130 Oct-09 2301.41 . Time series analysis actually challenges the data rather than fitting a presumed set of models. The residuals from the following ... | Forecasting beyond one season using Holt-Winters' exponential smoothing
1: http://i.stack.imgur.com/kxU4t.jpg reflects a questioning of the highly unusual Oct 2009 value 130 Oct-09 2301.41 . Time series analysis actually challenges the data rather than fitting a presumed |
39,310 | How to add horizontal lines to ggplot2 boxplot? | Found solution myself. Maybe someone could use it:
#step 1: preparing data
ageMetaData <- ddply(data,~group,summarise,
mean=mean(age),
sd=sd(age),
min=min(age),
max=max(age),
median=median(age),
Q1=summary(age)['1st Qu.'],
Q3=summary(ag... | How to add horizontal lines to ggplot2 boxplot? | Found solution myself. Maybe someone could use it:
#step 1: preparing data
ageMetaData <- ddply(data,~group,summarise,
mean=mean(age),
sd=sd(age),
min=min(age),
| How to add horizontal lines to ggplot2 boxplot?
Found solution myself. Maybe someone could use it:
#step 1: preparing data
ageMetaData <- ddply(data,~group,summarise,
mean=mean(age),
sd=sd(age),
min=min(age),
max=max(age),
median=median(age),
Q1=su... | How to add horizontal lines to ggplot2 boxplot?
Found solution myself. Maybe someone could use it:
#step 1: preparing data
ageMetaData <- ddply(data,~group,summarise,
mean=mean(age),
sd=sd(age),
min=min(age),
|
39,311 | How to add horizontal lines to ggplot2 boxplot? | There is a simpler solution using stat_boxplot(geom ='errorbar')
I provide an example:
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))
bp + geom_boxplot() + stat_boxplot(geom ='errorbar')
Result: | How to add horizontal lines to ggplot2 boxplot? | There is a simpler solution using stat_boxplot(geom ='errorbar')
I provide an example:
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))
bp + geom_boxplot() + stat_boxplot(geom ='e | How to add horizontal lines to ggplot2 boxplot?
There is a simpler solution using stat_boxplot(geom ='errorbar')
I provide an example:
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))
bp + geom_boxplot() + stat_boxplot(geom ='errorbar')
Result: | How to add horizontal lines to ggplot2 boxplot?
There is a simpler solution using stat_boxplot(geom ='errorbar')
I provide an example:
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))
bp + geom_boxplot() + stat_boxplot(geom ='e |
39,312 | How to add horizontal lines to ggplot2 boxplot? | I think that it looks better if stat_boxplot(geom ='errorbar') on the first line as it hides the vertical line.
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))+ stat_boxplot(geom ='errorbar')
bp + geom_boxplot() | How to add horizontal lines to ggplot2 boxplot? | I think that it looks better if stat_boxplot(geom ='errorbar') on the first line as it hides the vertical line.
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))+ stat_boxp | How to add horizontal lines to ggplot2 boxplot?
I think that it looks better if stat_boxplot(geom ='errorbar') on the first line as it hides the vertical line.
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))+ stat_boxplot(geom ='errorbar')
bp + geom_boxplot() | How to add horizontal lines to ggplot2 boxplot?
I think that it looks better if stat_boxplot(geom ='errorbar') on the first line as it hides the vertical line.
bp <- ggplot(iris, aes(factor(Species), Sepal.Width, fill = Species))+ stat_boxp |
39,313 | Why are fitted.values not part the R object returned from arima? | Use fitted() function from the forecast package. Since arima object saves residuals it is easy to compute fitted values from it. | Why are fitted.values not part the R object returned from arima? | Use fitted() function from the forecast package. Since arima object saves residuals it is easy to compute fitted values from it. | Why are fitted.values not part the R object returned from arima?
Use fitted() function from the forecast package. Since arima object saves residuals it is easy to compute fitted values from it. | Why are fitted.values not part the R object returned from arima?
Use fitted() function from the forecast package. Since arima object saves residuals it is easy to compute fitted values from it. |
39,314 | Interactive decision trees | Try Orange Canvas, it will give you option to build interactive decision tree. | Interactive decision trees | Try Orange Canvas, it will give you option to build interactive decision tree. | Interactive decision trees
Try Orange Canvas, it will give you option to build interactive decision tree. | Interactive decision trees
Try Orange Canvas, it will give you option to build interactive decision tree. |
39,315 | Interactive decision trees | Try the examples under dendrogram. You can make it as interactive as you want.
require(graphics); require(utils)
hc <- hclust(dist(USArrests), "ave")
(dend1 <- as.dendrogram(hc)) # "print()" method
str(dend1) # "str()" method
str(dend1, max = 2) # only the first two sub-levels
op <- par(mfrow= c(2,2), mar ... | Interactive decision trees | Try the examples under dendrogram. You can make it as interactive as you want.
require(graphics); require(utils)
hc <- hclust(dist(USArrests), "ave")
(dend1 <- as.dendrogram(hc)) # "print()" method | Interactive decision trees
Try the examples under dendrogram. You can make it as interactive as you want.
require(graphics); require(utils)
hc <- hclust(dist(USArrests), "ave")
(dend1 <- as.dendrogram(hc)) # "print()" method
str(dend1) # "str()" method
str(dend1, max = 2) # only the first two sub-levels
op... | Interactive decision trees
Try the examples under dendrogram. You can make it as interactive as you want.
require(graphics); require(utils)
hc <- hclust(dist(USArrests), "ave")
(dend1 <- as.dendrogram(hc)) # "print()" method |
39,316 | How do you solve a Poisson process problem | The two most important characteristics of a Poisson process with rate $\lambda$ are
For any interval $(s, t)$, the number of arrivals within the interval follows a Poisson distribution with mean $\lambda (t-s)$.
The number of arrivals in disjoint intervals are independent of one another. So, if $s_1 < t_1 < s_2 < t_2$... | How do you solve a Poisson process problem | The two most important characteristics of a Poisson process with rate $\lambda$ are
For any interval $(s, t)$, the number of arrivals within the interval follows a Poisson distribution with mean $\la | How do you solve a Poisson process problem
The two most important characteristics of a Poisson process with rate $\lambda$ are
For any interval $(s, t)$, the number of arrivals within the interval follows a Poisson distribution with mean $\lambda (t-s)$.
The number of arrivals in disjoint intervals are independent of ... | How do you solve a Poisson process problem
The two most important characteristics of a Poisson process with rate $\lambda$ are
For any interval $(s, t)$, the number of arrivals within the interval follows a Poisson distribution with mean $\la |
39,317 | How do you solve a Poisson process problem | If I were his instructor I'd prefer the following explanation, which obviously is equivalent to that given by @cardinal.
Let $N_t$ be the Poisson process of counting the number of errors on consequtive pages, of rate $\lambda=1.6/page$. One supposes that we observe the process
at integer moments t (here t being assimi... | How do you solve a Poisson process problem | If I were his instructor I'd prefer the following explanation, which obviously is equivalent to that given by @cardinal.
Let $N_t$ be the Poisson process of counting the number of errors on consequtiv | How do you solve a Poisson process problem
If I were his instructor I'd prefer the following explanation, which obviously is equivalent to that given by @cardinal.
Let $N_t$ be the Poisson process of counting the number of errors on consequtive pages, of rate $\lambda=1.6/page$. One supposes that we observe the proces... | How do you solve a Poisson process problem
If I were his instructor I'd prefer the following explanation, which obviously is equivalent to that given by @cardinal.
Let $N_t$ be the Poisson process of counting the number of errors on consequtiv |
39,318 | How do extreme values scale with sample size? | Assume that the random variables $x_k$ are i.i.d., nonnegative, integer valued, bounded by $n$, and such that $P(x_k=0)$ and $P(x_k=1)$ are both positive. For every $N\ge1$, let
$$
X_N= \min\{x_1,\ldots,x_N\}.
$$
Then, when $N\to+\infty$,
$$
E(X_N)=c^N(1+o(1)),
$$
where $c<1$ is independent of $N$ and given by
$... | How do extreme values scale with sample size? | Assume that the random variables $x_k$ are i.i.d., nonnegative, integer valued, bounded by $n$, and such that $P(x_k=0)$ and $P(x_k=1)$ are both positive. For every $N\ge1$, let
$$
X_N= \min\{x_1,\ld | How do extreme values scale with sample size?
Assume that the random variables $x_k$ are i.i.d., nonnegative, integer valued, bounded by $n$, and such that $P(x_k=0)$ and $P(x_k=1)$ are both positive. For every $N\ge1$, let
$$
X_N= \min\{x_1,\ldots,x_N\}.
$$
Then, when $N\to+\infty$,
$$
E(X_N)=c^N(1+o(1)),
$$
whe... | How do extreme values scale with sample size?
Assume that the random variables $x_k$ are i.i.d., nonnegative, integer valued, bounded by $n$, and such that $P(x_k=0)$ and $P(x_k=1)$ are both positive. For every $N\ge1$, let
$$
X_N= \min\{x_1,\ld |
39,319 | How do extreme values scale with sample size? | The table in this page of this book might help you. The explicit formulas for the expectation of minimum of sample of binomial distributions is given in the page before. | How do extreme values scale with sample size? | The table in this page of this book might help you. The explicit formulas for the expectation of minimum of sample of binomial distributions is given in the page before. | How do extreme values scale with sample size?
The table in this page of this book might help you. The explicit formulas for the expectation of minimum of sample of binomial distributions is given in the page before. | How do extreme values scale with sample size?
The table in this page of this book might help you. The explicit formulas for the expectation of minimum of sample of binomial distributions is given in the page before. |
39,320 | How do extreme values scale with sample size? | The distribution of the minimum of any set of N iid random variables is:
$$f_{min}(x)=Nf(x)[1-F(x)]^{N-1}$$
Where $f(x)$ is the pdf and $F(x)$ is the cdf (this is sometime called a $Beta-F$ distribution, because it is a compound of a Beta distribution and an arbitrary distribution). Hence the expectation (in this part... | How do extreme values scale with sample size? | The distribution of the minimum of any set of N iid random variables is:
$$f_{min}(x)=Nf(x)[1-F(x)]^{N-1}$$
Where $f(x)$ is the pdf and $F(x)$ is the cdf (this is sometime called a $Beta-F$ distributi | How do extreme values scale with sample size?
The distribution of the minimum of any set of N iid random variables is:
$$f_{min}(x)=Nf(x)[1-F(x)]^{N-1}$$
Where $f(x)$ is the pdf and $F(x)$ is the cdf (this is sometime called a $Beta-F$ distribution, because it is a compound of a Beta distribution and an arbitrary distr... | How do extreme values scale with sample size?
The distribution of the minimum of any set of N iid random variables is:
$$f_{min}(x)=Nf(x)[1-F(x)]^{N-1}$$
Where $f(x)$ is the pdf and $F(x)$ is the cdf (this is sometime called a $Beta-F$ distributi |
39,321 | Logistic Regression: Classification Tables a la SPSS in R | I'm not aware of a specific command, but this might be a start:
# generate some data
> N <- 100
> X <- rnorm(N, 175, 7)
> Y <- 0.4*X + 10 + rnorm(N, 0, 3)
# dichotomize Y
> Yfac <- cut(Y, breaks=c(-Inf, median(Y), Inf), labels=c("lo", "hi"))
# logistic regression
> glmFit <- glm(Yfac ~ X, family=binomial(link="logit"... | Logistic Regression: Classification Tables a la SPSS in R | I'm not aware of a specific command, but this might be a start:
# generate some data
> N <- 100
> X <- rnorm(N, 175, 7)
> Y <- 0.4*X + 10 + rnorm(N, 0, 3)
# dichotomize Y
> Yfac <- cut(Y, breaks=c(-I | Logistic Regression: Classification Tables a la SPSS in R
I'm not aware of a specific command, but this might be a start:
# generate some data
> N <- 100
> X <- rnorm(N, 175, 7)
> Y <- 0.4*X + 10 + rnorm(N, 0, 3)
# dichotomize Y
> Yfac <- cut(Y, breaks=c(-Inf, median(Y), Inf), labels=c("lo", "hi"))
# logistic regress... | Logistic Regression: Classification Tables a la SPSS in R
I'm not aware of a specific command, but this might be a start:
# generate some data
> N <- 100
> X <- rnorm(N, 175, 7)
> Y <- 0.4*X + 10 + rnorm(N, 0, 3)
# dichotomize Y
> Yfac <- cut(Y, breaks=c(-I |
39,322 | Logistic Regression: Classification Tables a la SPSS in R | Thomas D. Fletcher has a function called ClassLog() (for "classification analysis for a logistic regression model") in his QuantPsyc package. However, I like @caracal's response because it is self-made and easily customizable. | Logistic Regression: Classification Tables a la SPSS in R | Thomas D. Fletcher has a function called ClassLog() (for "classification analysis for a logistic regression model") in his QuantPsyc package. However, I like @caracal's response because it is self-mad | Logistic Regression: Classification Tables a la SPSS in R
Thomas D. Fletcher has a function called ClassLog() (for "classification analysis for a logistic regression model") in his QuantPsyc package. However, I like @caracal's response because it is self-made and easily customizable. | Logistic Regression: Classification Tables a la SPSS in R
Thomas D. Fletcher has a function called ClassLog() (for "classification analysis for a logistic regression model") in his QuantPsyc package. However, I like @caracal's response because it is self-mad |
39,323 | How to rank the results of questions with categorical answers? | If all your questions have the same response scale and they are standard Likert items, scaling the item 1,2,3,4,5 and taking the mean is generally fine.
You can investigate the robustness of the rank ordering by experimenting with different scaling procedures (e.g., 0, 0, 0, 1, 1 is common where you want to assess the ... | How to rank the results of questions with categorical answers? | If all your questions have the same response scale and they are standard Likert items, scaling the item 1,2,3,4,5 and taking the mean is generally fine.
You can investigate the robustness of the rank | How to rank the results of questions with categorical answers?
If all your questions have the same response scale and they are standard Likert items, scaling the item 1,2,3,4,5 and taking the mean is generally fine.
You can investigate the robustness of the rank ordering by experimenting with different scaling procedur... | How to rank the results of questions with categorical answers?
If all your questions have the same response scale and they are standard Likert items, scaling the item 1,2,3,4,5 and taking the mean is generally fine.
You can investigate the robustness of the rank |
39,324 | How to rank the results of questions with categorical answers? | Recoding your data with numerical values seems ok, provided the assumption of an ordinal scale holds. This is often the case for Likert-type item, but see these related questions:
Is it appropriate to treat n-point Likert scale data as n trials from a binomial process?
Under what conditions should Likert scales be use... | How to rank the results of questions with categorical answers? | Recoding your data with numerical values seems ok, provided the assumption of an ordinal scale holds. This is often the case for Likert-type item, but see these related questions:
Is it appropriate t | How to rank the results of questions with categorical answers?
Recoding your data with numerical values seems ok, provided the assumption of an ordinal scale holds. This is often the case for Likert-type item, but see these related questions:
Is it appropriate to treat n-point Likert scale data as n trials from a bino... | How to rank the results of questions with categorical answers?
Recoding your data with numerical values seems ok, provided the assumption of an ordinal scale holds. This is often the case for Likert-type item, but see these related questions:
Is it appropriate t |
39,325 | How to rank the results of questions with categorical answers? | If I plot the distribution of answers for each question, I can identify which questions have lots of 'good' answers (distribution is negatively skewed) or those with lots of 'bad' answers (positively skewed histogram). So picking the extremes is easy but this is also dependent on the data.
Is an absolute ranking neces... | How to rank the results of questions with categorical answers? | If I plot the distribution of answers for each question, I can identify which questions have lots of 'good' answers (distribution is negatively skewed) or those with lots of 'bad' answers (positively | How to rank the results of questions with categorical answers?
If I plot the distribution of answers for each question, I can identify which questions have lots of 'good' answers (distribution is negatively skewed) or those with lots of 'bad' answers (positively skewed histogram). So picking the extremes is easy but th... | How to rank the results of questions with categorical answers?
If I plot the distribution of answers for each question, I can identify which questions have lots of 'good' answers (distribution is negatively skewed) or those with lots of 'bad' answers (positively |
39,326 | How to rank the results of questions with categorical answers? | A small additional point to the answers already given:
Without assuming your ordinal data is interval, you can compare any convenient quantiles - e.g. medians.
Or, when comparing X vs Y which are both ordered categorical, you can estimate something like P(Y>X) - P(X>Y) or P(Y>X) + 0.5 * P(Y=X) (etc.), where you estimat... | How to rank the results of questions with categorical answers? | A small additional point to the answers already given:
Without assuming your ordinal data is interval, you can compare any convenient quantiles - e.g. medians.
Or, when comparing X vs Y which are both | How to rank the results of questions with categorical answers?
A small additional point to the answers already given:
Without assuming your ordinal data is interval, you can compare any convenient quantiles - e.g. medians.
Or, when comparing X vs Y which are both ordered categorical, you can estimate something like P(Y... | How to rank the results of questions with categorical answers?
A small additional point to the answers already given:
Without assuming your ordinal data is interval, you can compare any convenient quantiles - e.g. medians.
Or, when comparing X vs Y which are both |
39,327 | How to rank the results of questions with categorical answers? | You can rank ordinal distributions by means of an intuitive dominance criterion: the answers to one question are better than the answers to another when it is more likely than not that a randomly chosen answer to the first will be better than a randomly chosen answer to the second.
In more detail: put all the answers t... | How to rank the results of questions with categorical answers? | You can rank ordinal distributions by means of an intuitive dominance criterion: the answers to one question are better than the answers to another when it is more likely than not that a randomly chos | How to rank the results of questions with categorical answers?
You can rank ordinal distributions by means of an intuitive dominance criterion: the answers to one question are better than the answers to another when it is more likely than not that a randomly chosen answer to the first will be better than a randomly cho... | How to rank the results of questions with categorical answers?
You can rank ordinal distributions by means of an intuitive dominance criterion: the answers to one question are better than the answers to another when it is more likely than not that a randomly chos |
39,328 | What practical implications/interpretations are there of a kurtotic distribution? | The kurtosis also indicates the "fat tailedness" of the distribution. A distribution with high kurtosis will have many extreme events (events far away from the center) and many "typical" events (events near the center). A distribution with low kurtosis will have events a moderate distance from the center.
This picture... | What practical implications/interpretations are there of a kurtotic distribution? | The kurtosis also indicates the "fat tailedness" of the distribution. A distribution with high kurtosis will have many extreme events (events far away from the center) and many "typical" events (event | What practical implications/interpretations are there of a kurtotic distribution?
The kurtosis also indicates the "fat tailedness" of the distribution. A distribution with high kurtosis will have many extreme events (events far away from the center) and many "typical" events (events near the center). A distribution wit... | What practical implications/interpretations are there of a kurtotic distribution?
The kurtosis also indicates the "fat tailedness" of the distribution. A distribution with high kurtosis will have many extreme events (events far away from the center) and many "typical" events (event |
39,329 | What practical implications/interpretations are there of a kurtotic distribution? | I seem to remember that the median has a smaller standard error than the mean when the samples are drawn from a leptokurtic distribution, but the mean has a smaller standard error when the distribution is platykurtic. I think I read this in one of Wilcox' books. Thus the kurtosis may dictate which kinds of locational t... | What practical implications/interpretations are there of a kurtotic distribution? | I seem to remember that the median has a smaller standard error than the mean when the samples are drawn from a leptokurtic distribution, but the mean has a smaller standard error when the distributio | What practical implications/interpretations are there of a kurtotic distribution?
I seem to remember that the median has a smaller standard error than the mean when the samples are drawn from a leptokurtic distribution, but the mean has a smaller standard error when the distribution is platykurtic. I think I read this ... | What practical implications/interpretations are there of a kurtotic distribution?
I seem to remember that the median has a smaller standard error than the mean when the samples are drawn from a leptokurtic distribution, but the mean has a smaller standard error when the distributio |
39,330 | What practical implications/interpretations are there of a kurtotic distribution? | Haven't got an example of a dataset in mind with which to answer your question about the interpretation, but this answer to a related question indicates that a practical implication of kurtosis is biased variance estimates.
In considering the interpretive difference rationally, I think there's relevant information in t... | What practical implications/interpretations are there of a kurtotic distribution? | Haven't got an example of a dataset in mind with which to answer your question about the interpretation, but this answer to a related question indicates that a practical implication of kurtosis is bia | What practical implications/interpretations are there of a kurtotic distribution?
Haven't got an example of a dataset in mind with which to answer your question about the interpretation, but this answer to a related question indicates that a practical implication of kurtosis is biased variance estimates.
In considering... | What practical implications/interpretations are there of a kurtotic distribution?
Haven't got an example of a dataset in mind with which to answer your question about the interpretation, but this answer to a related question indicates that a practical implication of kurtosis is bia |
39,331 | What practical implications/interpretations are there of a kurtotic distribution? | There is the Kurtosis risk which isn't explained fantastically well at that link.
In general, measures of normality (or deviation therefrom) are crucial if you are using analyses that assume normality. For example, the standard workhorse Pearson-r correlation coefficient is severely sensitive to outliers and becomes e... | What practical implications/interpretations are there of a kurtotic distribution? | There is the Kurtosis risk which isn't explained fantastically well at that link.
In general, measures of normality (or deviation therefrom) are crucial if you are using analyses that assume normalit | What practical implications/interpretations are there of a kurtotic distribution?
There is the Kurtosis risk which isn't explained fantastically well at that link.
In general, measures of normality (or deviation therefrom) are crucial if you are using analyses that assume normality. For example, the standard workhorse... | What practical implications/interpretations are there of a kurtotic distribution?
There is the Kurtosis risk which isn't explained fantastically well at that link.
In general, measures of normality (or deviation therefrom) are crucial if you are using analyses that assume normalit |
39,332 | Continuity of the location | The expectation does not have to be a continuous function of the parameter. The following example illustrates what can go wrong.
For any number $0 \le p \le 1,$ let $g(x,p)=0$ when $x \le 1$ and otherwise
$$g(x,p) = (p+1)x^{-(p+2)} = \frac{\mathrm{d}}{\mathrm{d}x}\left(1 - x^{-(p+1)} \right) = \frac{\mathrm{d}}{\mat... | Continuity of the location | The expectation does not have to be a continuous function of the parameter. The following example illustrates what can go wrong.
For any number $0 \le p \le 1,$ let $g(x,p)=0$ when $x \le 1$ and oth | Continuity of the location
The expectation does not have to be a continuous function of the parameter. The following example illustrates what can go wrong.
For any number $0 \le p \le 1,$ let $g(x,p)=0$ when $x \le 1$ and otherwise
$$g(x,p) = (p+1)x^{-(p+2)} = \frac{\mathrm{d}}{\mathrm{d}x}\left(1 - x^{-(p+1)} \right... | Continuity of the location
The expectation does not have to be a continuous function of the parameter. The following example illustrates what can go wrong.
For any number $0 \le p \le 1,$ let $g(x,p)=0$ when $x \le 1$ and oth |
39,333 | Discrimination/Slope Item Response Theory Models | A steeper slope means a stronger relationship between ability and the question. It means that the item (and therefore the test) is more reliable - and reliability is the inverse of measurement error. Less error = better discrimination.
You can get higher reliability / discrimination / less measurement error by having m... | Discrimination/Slope Item Response Theory Models | A steeper slope means a stronger relationship between ability and the question. It means that the item (and therefore the test) is more reliable - and reliability is the inverse of measurement error. | Discrimination/Slope Item Response Theory Models
A steeper slope means a stronger relationship between ability and the question. It means that the item (and therefore the test) is more reliable - and reliability is the inverse of measurement error. Less error = better discrimination.
You can get higher reliability / di... | Discrimination/Slope Item Response Theory Models
A steeper slope means a stronger relationship between ability and the question. It means that the item (and therefore the test) is more reliable - and reliability is the inverse of measurement error. |
39,334 | Discrimination/Slope Item Response Theory Models | Intuitions are useful, but it is worth understanding the math behind the model. We model the probability of answering correctly to $i$-th question by a person with the ability $\theta$, $p_i({\theta})$. The model can have up to three parameters $a_i, b_i, c_i$ and uses a logistic function.
$$
p_i({\theta})=c_i + \frac{... | Discrimination/Slope Item Response Theory Models | Intuitions are useful, but it is worth understanding the math behind the model. We model the probability of answering correctly to $i$-th question by a person with the ability $\theta$, $p_i({\theta}) | Discrimination/Slope Item Response Theory Models
Intuitions are useful, but it is worth understanding the math behind the model. We model the probability of answering correctly to $i$-th question by a person with the ability $\theta$, $p_i({\theta})$. The model can have up to three parameters $a_i, b_i, c_i$ and uses a... | Discrimination/Slope Item Response Theory Models
Intuitions are useful, but it is worth understanding the math behind the model. We model the probability of answering correctly to $i$-th question by a person with the ability $\theta$, $p_i({\theta}) |
39,335 | XGBoost Classifier not capturing extreme probabilities | This is what I would focus on.
Limitations on max_depth might cause terminal nodes to group together observations with very small probabilities with other observations where probabilities aren't that small, so the effect is to move the leaf weights away from extreme values. Likewise, something similar with large proba... | XGBoost Classifier not capturing extreme probabilities | This is what I would focus on.
Limitations on max_depth might cause terminal nodes to group together observations with very small probabilities with other observations where probabilities aren't that | XGBoost Classifier not capturing extreme probabilities
This is what I would focus on.
Limitations on max_depth might cause terminal nodes to group together observations with very small probabilities with other observations where probabilities aren't that small, so the effect is to move the leaf weights away from extre... | XGBoost Classifier not capturing extreme probabilities
This is what I would focus on.
Limitations on max_depth might cause terminal nodes to group together observations with very small probabilities with other observations where probabilities aren't that |
39,336 | XGBoost Classifier not capturing extreme probabilities | If you want precise estimates of probabilities, don't use algorithms based on decision trees. To get a probability estimate of a decision tree, you would count the class occurrences in the node and divide by the node size. This would never lead to a smooth function describing the probabilities, same as regression trees... | XGBoost Classifier not capturing extreme probabilities | If you want precise estimates of probabilities, don't use algorithms based on decision trees. To get a probability estimate of a decision tree, you would count the class occurrences in the node and di | XGBoost Classifier not capturing extreme probabilities
If you want precise estimates of probabilities, don't use algorithms based on decision trees. To get a probability estimate of a decision tree, you would count the class occurrences in the node and divide by the node size. This would never lead to a smooth function... | XGBoost Classifier not capturing extreme probabilities
If you want precise estimates of probabilities, don't use algorithms based on decision trees. To get a probability estimate of a decision tree, you would count the class occurrences in the node and di |
39,337 | XGBoost Classifier not capturing extreme probabilities | As @Sycorax and @BenReiniger pointed out, the problem is that the probabilities are not calibrated (or not as calibrated as well as you'd prefer). Here is how you could calibrate the XGBoost probabilities. Use the following model:
P(y|x) = 1/(1+exp(-(a+x)))
where x is the logit function of the original probabilities pr... | XGBoost Classifier not capturing extreme probabilities | As @Sycorax and @BenReiniger pointed out, the problem is that the probabilities are not calibrated (or not as calibrated as well as you'd prefer). Here is how you could calibrate the XGBoost probabili | XGBoost Classifier not capturing extreme probabilities
As @Sycorax and @BenReiniger pointed out, the problem is that the probabilities are not calibrated (or not as calibrated as well as you'd prefer). Here is how you could calibrate the XGBoost probabilities. Use the following model:
P(y|x) = 1/(1+exp(-(a+x)))
where x... | XGBoost Classifier not capturing extreme probabilities
As @Sycorax and @BenReiniger pointed out, the problem is that the probabilities are not calibrated (or not as calibrated as well as you'd prefer). Here is how you could calibrate the XGBoost probabili |
39,338 | How come model prediction accuracy high but model does not generalise well | It's hard to say without digging deeply into your model and your data. However, it seems like you have been doing a lot of cross-validation, model tuning, cross-validation, model tuning and so forth. That, together with bad out-of-sample performance, suggests that you are overfitting to your test set. That is harder th... | How come model prediction accuracy high but model does not generalise well | It's hard to say without digging deeply into your model and your data. However, it seems like you have been doing a lot of cross-validation, model tuning, cross-validation, model tuning and so forth. | How come model prediction accuracy high but model does not generalise well
It's hard to say without digging deeply into your model and your data. However, it seems like you have been doing a lot of cross-validation, model tuning, cross-validation, model tuning and so forth. That, together with bad out-of-sample perform... | How come model prediction accuracy high but model does not generalise well
It's hard to say without digging deeply into your model and your data. However, it seems like you have been doing a lot of cross-validation, model tuning, cross-validation, model tuning and so forth. |
39,339 | How come model prediction accuracy high but model does not generalise well | I second Stephan's answer that the likely culprit is overfitting the entire dataset. That said, another thing to validate is that there are no differences between data processing pipelines in your training vs. production code. E.g. are you normalizing the features before training? If so, do you record the means and sta... | How come model prediction accuracy high but model does not generalise well | I second Stephan's answer that the likely culprit is overfitting the entire dataset. That said, another thing to validate is that there are no differences between data processing pipelines in your tra | How come model prediction accuracy high but model does not generalise well
I second Stephan's answer that the likely culprit is overfitting the entire dataset. That said, another thing to validate is that there are no differences between data processing pipelines in your training vs. production code. E.g. are you norma... | How come model prediction accuracy high but model does not generalise well
I second Stephan's answer that the likely culprit is overfitting the entire dataset. That said, another thing to validate is that there are no differences between data processing pipelines in your tra |
39,340 | How come model prediction accuracy high but model does not generalise well | To me it seems to be a data problem. You are splitting the data 70:30, but are all data from the 70% prior to data from the 30% set?
It can be a problem if you mix older and newer data in the training set. If time is involved in the generation of data, which seems to be the case as you have live data, test set should n... | How come model prediction accuracy high but model does not generalise well | To me it seems to be a data problem. You are splitting the data 70:30, but are all data from the 70% prior to data from the 30% set?
It can be a problem if you mix older and newer data in the training | How come model prediction accuracy high but model does not generalise well
To me it seems to be a data problem. You are splitting the data 70:30, but are all data from the 70% prior to data from the 30% set?
It can be a problem if you mix older and newer data in the training set. If time is involved in the generation o... | How come model prediction accuracy high but model does not generalise well
To me it seems to be a data problem. You are splitting the data 70:30, but are all data from the 70% prior to data from the 30% set?
It can be a problem if you mix older and newer data in the training |
39,341 | Trying to calculate confidence intervals for a Monte-Carlo estimate of Pi. What am I doing wrong? | The standard error that you want is the standard deviation of the estimate $\hat\pi$ at a fixed point in the sequence, over multiple experiments. This standard error will give you a confidence interval that includes the actual value $\pi$ in 95% of experiments. You don't need a huge sample size to get reasonable esti... | Trying to calculate confidence intervals for a Monte-Carlo estimate of Pi. What am I doing wrong? | The standard error that you want is the standard deviation of the estimate $\hat\pi$ at a fixed point in the sequence, over multiple experiments. This standard error will give you a confidence interv | Trying to calculate confidence intervals for a Monte-Carlo estimate of Pi. What am I doing wrong?
The standard error that you want is the standard deviation of the estimate $\hat\pi$ at a fixed point in the sequence, over multiple experiments. This standard error will give you a confidence interval that includes the a... | Trying to calculate confidence intervals for a Monte-Carlo estimate of Pi. What am I doing wrong?
The standard error that you want is the standard deviation of the estimate $\hat\pi$ at a fixed point in the sequence, over multiple experiments. This standard error will give you a confidence interv |
39,342 | Circular reasoning in Harrell BBR 18-19? | That argument seems circular to me; we have to assume that c-index and $\chi^2$ are superior metrics to show the flaw in accuracy as a metric
In your comments, you mention that you agree and understand why we should prefer c-index and likelihood ratio chi square (LHRCS) to classification accuracy. In your post, you a... | Circular reasoning in Harrell BBR 18-19? | That argument seems circular to me; we have to assume that c-index and $\chi^2$ are superior metrics to show the flaw in accuracy as a metric
In your comments, you mention that you agree and understa | Circular reasoning in Harrell BBR 18-19?
That argument seems circular to me; we have to assume that c-index and $\chi^2$ are superior metrics to show the flaw in accuracy as a metric
In your comments, you mention that you agree and understand why we should prefer c-index and likelihood ratio chi square (LHRCS) to clas... | Circular reasoning in Harrell BBR 18-19?
That argument seems circular to me; we have to assume that c-index and $\chi^2$ are superior metrics to show the flaw in accuracy as a metric
In your comments, you mention that you agree and understa |
39,343 | how to generate data from cdf which is not in closed form? | Solution
By exploiting the principles explained at https://stats.stackexchange.com/a/495347/919, we can examine the expression for
$$f_X(x,\beta,\alpha) \propto \frac{(x/\alpha)^{\beta+1}}{\left(1 + (x/\alpha)^\beta\right)^2}\frac{\mathrm{d}x}{x}$$
and see that if it is to make any sense at all (a question I will presc... | how to generate data from cdf which is not in closed form? | Solution
By exploiting the principles explained at https://stats.stackexchange.com/a/495347/919, we can examine the expression for
$$f_X(x,\beta,\alpha) \propto \frac{(x/\alpha)^{\beta+1}}{\left(1 + ( | how to generate data from cdf which is not in closed form?
Solution
By exploiting the principles explained at https://stats.stackexchange.com/a/495347/919, we can examine the expression for
$$f_X(x,\beta,\alpha) \propto \frac{(x/\alpha)^{\beta+1}}{\left(1 + (x/\alpha)^\beta\right)^2}\frac{\mathrm{d}x}{x}$$
and see that... | how to generate data from cdf which is not in closed form?
Solution
By exploiting the principles explained at https://stats.stackexchange.com/a/495347/919, we can examine the expression for
$$f_X(x,\beta,\alpha) \propto \frac{(x/\alpha)^{\beta+1}}{\left(1 + ( |
39,344 | how to generate data from cdf which is not in closed form? | Searching for something related
Step 1 use a transformation to simplify te distribution. With $t = (x/\alpha)^\beta$ or $x=\alpha t^{1/\beta}$ it becomes
$$f(t) \propto \frac{t^{1/\beta}}{(1+t)^{2}}$$
Step 2. Compare with other distributions. We can use the following quotient to see if this is a Pearson distribution... | how to generate data from cdf which is not in closed form? | Searching for something related
Step 1 use a transformation to simplify te distribution. With $t = (x/\alpha)^\beta$ or $x=\alpha t^{1/\beta}$ it becomes
$$f(t) \propto \frac{t^{1/\beta}}{(1+t)^{2}} | how to generate data from cdf which is not in closed form?
Searching for something related
Step 1 use a transformation to simplify te distribution. With $t = (x/\alpha)^\beta$ or $x=\alpha t^{1/\beta}$ it becomes
$$f(t) \propto \frac{t^{1/\beta}}{(1+t)^{2}}$$
Step 2. Compare with other distributions. We can use the ... | how to generate data from cdf which is not in closed form?
Searching for something related
Step 1 use a transformation to simplify te distribution. With $t = (x/\alpha)^\beta$ or $x=\alpha t^{1/\beta}$ it becomes
$$f(t) \propto \frac{t^{1/\beta}}{(1+t)^{2}} |
39,345 | how to generate data from cdf which is not in closed form? | I cannot make sense of your PDF. So I will show an acceptance-rejection method using a PDF that does make sense to me.
The method is similar
to that of @Ertxiem (+1), except that I guess mine works better if your
(correct) density function has support $(0,\infty).$ Both methods are
general ones, and I hope you can use ... | how to generate data from cdf which is not in closed form? | I cannot make sense of your PDF. So I will show an acceptance-rejection method using a PDF that does make sense to me.
The method is similar
to that of @Ertxiem (+1), except that I guess mine works be | how to generate data from cdf which is not in closed form?
I cannot make sense of your PDF. So I will show an acceptance-rejection method using a PDF that does make sense to me.
The method is similar
to that of @Ertxiem (+1), except that I guess mine works better if your
(correct) density function has support $(0,\inft... | how to generate data from cdf which is not in closed form?
I cannot make sense of your PDF. So I will show an acceptance-rejection method using a PDF that does make sense to me.
The method is similar
to that of @Ertxiem (+1), except that I guess mine works be |
39,346 | how to generate data from cdf which is not in closed form? | You may try to use a Monte-Carlo method.
Generate a random value for $(x, \alpha, \beta)$ and $f_e$. If $f_e < f(x, \alpha, \beta)$ accept that triple, otherwise, reject it.
This is easier to make work if $(x, \alpha, \beta)$ and $f$ are bounded than if they're unbounded.
Edit:
I will assume that $x \in [x_L; x_u]$, $... | how to generate data from cdf which is not in closed form? | You may try to use a Monte-Carlo method.
Generate a random value for $(x, \alpha, \beta)$ and $f_e$. If $f_e < f(x, \alpha, \beta)$ accept that triple, otherwise, reject it.
This is easier to make wor | how to generate data from cdf which is not in closed form?
You may try to use a Monte-Carlo method.
Generate a random value for $(x, \alpha, \beta)$ and $f_e$. If $f_e < f(x, \alpha, \beta)$ accept that triple, otherwise, reject it.
This is easier to make work if $(x, \alpha, \beta)$ and $f$ are bounded than if they're... | how to generate data from cdf which is not in closed form?
You may try to use a Monte-Carlo method.
Generate a random value for $(x, \alpha, \beta)$ and $f_e$. If $f_e < f(x, \alpha, \beta)$ accept that triple, otherwise, reject it.
This is easier to make wor |
39,347 | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ implies $\lim_{n\to\infty} P(X_n=k) = P(X=k)$ | $F$ actually is continuous at all non-integers (where, of course, its graph is horizontal, reflecting zero chance that $X$ will be nonintegral). Consequently, in particular, the convergence in distribution of the sequence implies convergence at (say) the values $k\pm 1/2,$ from which you may instantly conclude for any... | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ imp | $F$ actually is continuous at all non-integers (where, of course, its graph is horizontal, reflecting zero chance that $X$ will be nonintegral). Consequently, in particular, the convergence in distri | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ implies $\lim_{n\to\infty} P(X_n=k) = P(X=k)$
$F$ actually is continuous at all non-integers (where, of course, its graph is horizontal, reflecting zero chance that $X$ will be nonintegral). Consequently, in particular, th... | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ imp
$F$ actually is continuous at all non-integers (where, of course, its graph is horizontal, reflecting zero chance that $X$ will be nonintegral). Consequently, in particular, the convergence in distri |
39,348 | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ implies $\lim_{n\to\infty} P(X_n=k) = P(X=k)$ | Following on from whuber's excellent answer, we can also see that this theory applies for any set of random variables with countable support, even if they are not positive integers. To see this, consider a sequence of random variables with distribution on a countable support $\mathscr{X} \subset \mathbb{R}$. For any ... | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ imp | Following on from whuber's excellent answer, we can also see that this theory applies for any set of random variables with countable support, even if they are not positive integers. To see this, cons | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ implies $\lim_{n\to\infty} P(X_n=k) = P(X=k)$
Following on from whuber's excellent answer, we can also see that this theory applies for any set of random variables with countable support, even if they are not positive integ... | Let $X,X_1,X_2,X_3,...$ be positive integer random variables. Show that $X_n \overset{d}{\to} X$ imp
Following on from whuber's excellent answer, we can also see that this theory applies for any set of random variables with countable support, even if they are not positive integers. To see this, cons |
39,349 | Why is the chi-square test more popular than the G-test? | The Pearson test is popular because it's simple to compute - it's amenable to hand-calculation even without a calculator (or historically, even without log-tables) - and yet generally has good power compared to alternatives; the simplicity means it continues to be taught in the most basic subjects. There might be argue... | Why is the chi-square test more popular than the G-test? | The Pearson test is popular because it's simple to compute - it's amenable to hand-calculation even without a calculator (or historically, even without log-tables) - and yet generally has good power c | Why is the chi-square test more popular than the G-test?
The Pearson test is popular because it's simple to compute - it's amenable to hand-calculation even without a calculator (or historically, even without log-tables) - and yet generally has good power compared to alternatives; the simplicity means it continues to b... | Why is the chi-square test more popular than the G-test?
The Pearson test is popular because it's simple to compute - it's amenable to hand-calculation even without a calculator (or historically, even without log-tables) - and yet generally has good power c |
39,350 | Idea for an using of mixed model. Used variable with all possible category as an random effect | This is an interesting question. Indeed, understanding random effects $u_i$ as draws from a larger population is well in line with the underlying model, which treats them as a random variable, e.g.:
$u_i \sim_{iid} Normal(0, \sigma^2)$
But there are arguments for treating entities as coming from a larger population eve... | Idea for an using of mixed model. Used variable with all possible category as an random effect | This is an interesting question. Indeed, understanding random effects $u_i$ as draws from a larger population is well in line with the underlying model, which treats them as a random variable, e.g.:
$ | Idea for an using of mixed model. Used variable with all possible category as an random effect
This is an interesting question. Indeed, understanding random effects $u_i$ as draws from a larger population is well in line with the underlying model, which treats them as a random variable, e.g.:
$u_i \sim_{iid} Normal(0, ... | Idea for an using of mixed model. Used variable with all possible category as an random effect
This is an interesting question. Indeed, understanding random effects $u_i$ as draws from a larger population is well in line with the underlying model, which treats them as a random variable, e.g.:
$ |
39,351 | Idea for an using of mixed model. Used variable with all possible category as an random effect | Yes I think your intuition is correct and you can fit random intercepts. There are several criteria for assessing if a factor should be treated as random and being a sample from a population is only one. Often different criteria point in different directions and it is a matter of judgement. In this case if you really w... | Idea for an using of mixed model. Used variable with all possible category as an random effect | Yes I think your intuition is correct and you can fit random intercepts. There are several criteria for assessing if a factor should be treated as random and being a sample from a population is only o | Idea for an using of mixed model. Used variable with all possible category as an random effect
Yes I think your intuition is correct and you can fit random intercepts. There are several criteria for assessing if a factor should be treated as random and being a sample from a population is only one. Often different crite... | Idea for an using of mixed model. Used variable with all possible category as an random effect
Yes I think your intuition is correct and you can fit random intercepts. There are several criteria for assessing if a factor should be treated as random and being a sample from a population is only o |
39,352 | Why is the convergence rate important? | Perhaps the two most familiar and most used limit theorems are
the Central Limit Theorem (CLT) and the Law of Large Numbers (LLN). Both are useful for proving other theoretical theorems. Here I discuss a few kinds of practical applications in which one hopes sample size is large enough to use CLT and LLN to make useful... | Why is the convergence rate important? | Perhaps the two most familiar and most used limit theorems are
the Central Limit Theorem (CLT) and the Law of Large Numbers (LLN). Both are useful for proving other theoretical theorems. Here I discus | Why is the convergence rate important?
Perhaps the two most familiar and most used limit theorems are
the Central Limit Theorem (CLT) and the Law of Large Numbers (LLN). Both are useful for proving other theoretical theorems. Here I discuss a few kinds of practical applications in which one hopes sample size is large e... | Why is the convergence rate important?
Perhaps the two most familiar and most used limit theorems are
the Central Limit Theorem (CLT) and the Law of Large Numbers (LLN). Both are useful for proving other theoretical theorems. Here I discus |
39,353 | Why is the convergence rate important? | Here is an example of how to apply such theorems:
Let us say that we want to fit a function $g$ to some data observed and let us assume that the setting is 'good' in the sense that the data really comes from a true function $f$ and the observed data points really come from IID random variables or so (assumptions that w... | Why is the convergence rate important? | Here is an example of how to apply such theorems:
Let us say that we want to fit a function $g$ to some data observed and let us assume that the setting is 'good' in the sense that the data really com | Why is the convergence rate important?
Here is an example of how to apply such theorems:
Let us say that we want to fit a function $g$ to some data observed and let us assume that the setting is 'good' in the sense that the data really comes from a true function $f$ and the observed data points really come from IID ran... | Why is the convergence rate important?
Here is an example of how to apply such theorems:
Let us say that we want to fit a function $g$ to some data observed and let us assume that the setting is 'good' in the sense that the data really com |
39,354 | What does $N(x|\mu, \sigma^2)$ mean? | Here, it means the normal PDF:
$$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$
The $\mu,\sigma^2$ in given side means that you can treat them as known quantities. | What does $N(x|\mu, \sigma^2)$ mean? | Here, it means the normal PDF:
$$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$
The $\mu,\sigma^2$ in given side means that you can treat them as known quantities. | What does $N(x|\mu, \sigma^2)$ mean?
Here, it means the normal PDF:
$$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$
The $\mu,\sigma^2$ in given side means that you can treat them as known quantities. | What does $N(x|\mu, \sigma^2)$ mean?
Here, it means the normal PDF:
$$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$
The $\mu,\sigma^2$ in given side means that you can treat them as known quantities. |
39,355 | What does $N(x|\mu, \sigma^2)$ mean? | $N(x|\mu, \sigma^2)$ combines the two notations: $x \sim N(\mu, \sigma^2)$ and $p(x| \mu, \sigma^2)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma^2$. | What does $N(x|\mu, \sigma^2)$ mean? | $N(x|\mu, \sigma^2)$ combines the two notations: $x \sim N(\mu, \sigma^2)$ and $p(x| \mu, \sigma^2)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma^2$. | What does $N(x|\mu, \sigma^2)$ mean?
$N(x|\mu, \sigma^2)$ combines the two notations: $x \sim N(\mu, \sigma^2)$ and $p(x| \mu, \sigma^2)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma^2$. | What does $N(x|\mu, \sigma^2)$ mean?
$N(x|\mu, \sigma^2)$ combines the two notations: $x \sim N(\mu, \sigma^2)$ and $p(x| \mu, \sigma^2)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma^2$. |
39,356 | probability that random draws from the same pool will collectively select 90% of the pool | The answer is $n=17$.
I can't see an easy analytic solution to this question. Instead, we will develop an analytic solution to a closely related problem, and then find the answer to your exact question via simulation.
Clarification:
Since the question is slightly vague, let me re-state the problem. There are $200$ name... | probability that random draws from the same pool will collectively select 90% of the pool | The answer is $n=17$.
I can't see an easy analytic solution to this question. Instead, we will develop an analytic solution to a closely related problem, and then find the answer to your exact questio | probability that random draws from the same pool will collectively select 90% of the pool
The answer is $n=17$.
I can't see an easy analytic solution to this question. Instead, we will develop an analytic solution to a closely related problem, and then find the answer to your exact question via simulation.
Clarificatio... | probability that random draws from the same pool will collectively select 90% of the pool
The answer is $n=17$.
I can't see an easy analytic solution to this question. Instead, we will develop an analytic solution to a closely related problem, and then find the answer to your exact questio |
39,357 | probability that random draws from the same pool will collectively select 90% of the pool | Here is a general analytic solution --- does not require simulation
This is a variation on the classical occupancy problem, where you are sampling lots of thirty names at each sampling point, instead of sampling individual names. The simplest way to compute this result is by framing the problem as a Markov chain, and ... | probability that random draws from the same pool will collectively select 90% of the pool | Here is a general analytic solution --- does not require simulation
This is a variation on the classical occupancy problem, where you are sampling lots of thirty names at each sampling point, instead | probability that random draws from the same pool will collectively select 90% of the pool
Here is a general analytic solution --- does not require simulation
This is a variation on the classical occupancy problem, where you are sampling lots of thirty names at each sampling point, instead of sampling individual names. ... | probability that random draws from the same pool will collectively select 90% of the pool
Here is a general analytic solution --- does not require simulation
This is a variation on the classical occupancy problem, where you are sampling lots of thirty names at each sampling point, instead |
39,358 | probability that random draws from the same pool will collectively select 90% of the pool | The answer is $n = 17$, with $P(N_{30}\ge180)=0.976388$.
The approach I took to calculate the probability after 30 draws was to determine the probability of drawing seen vs. unseen names at each round.
When drawing $n$ names out of $p=200$ after having seen $s$ of them, let's call $U_s$ the number of names out of those... | probability that random draws from the same pool will collectively select 90% of the pool | The answer is $n = 17$, with $P(N_{30}\ge180)=0.976388$.
The approach I took to calculate the probability after 30 draws was to determine the probability of drawing seen vs. unseen names at each round | probability that random draws from the same pool will collectively select 90% of the pool
The answer is $n = 17$, with $P(N_{30}\ge180)=0.976388$.
The approach I took to calculate the probability after 30 draws was to determine the probability of drawing seen vs. unseen names at each round.
When drawing $n$ names out o... | probability that random draws from the same pool will collectively select 90% of the pool
The answer is $n = 17$, with $P(N_{30}\ge180)=0.976388$.
The approach I took to calculate the probability after 30 draws was to determine the probability of drawing seen vs. unseen names at each round |
39,359 | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$? | If $X$ is a Gamma $\mathcal G(p,1)$ variate, its density is
$$f_X(x)= \frac{x^{p-1}}{(p-1)!}e^{-x}\mathbb I_{\mathbb R^*_+}(x)$$
Therefore, the probability density of $Y_1=\sqrt{X_1}$ is
$$f_{Y_1}(y)=\frac{y^{2(p-1)}}{(p-1)!}e^{-y^2}\mathbb I_{\mathbb R^*_+}(y)\times\overbrace{\left|\frac{\text{d}x}{\text{d}y}\right|}^... | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$? | If $X$ is a Gamma $\mathcal G(p,1)$ variate, its density is
$$f_X(x)= \frac{x^{p-1}}{(p-1)!}e^{-x}\mathbb I_{\mathbb R^*_+}(x)$$
Therefore, the probability density of $Y_1=\sqrt{X_1}$ is
$$f_{Y_1}(y)= | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$?
If $X$ is a Gamma $\mathcal G(p,1)$ variate, its density is
$$f_X(x)= \frac{x^{p-1}}{(p-1)!}e^{-x}\mathbb I_{\mathbb R^*_+}(x)$$
Therefore, the probability density of $Y_1=\sqrt{X_1}$ is
$$f_{Y_1}(y)=\frac{y^{2(p-1)}}{(p-1)!}e^{-y^2}\mathbb I_{\mathbb R^*_+}(y)\ti... | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$?
If $X$ is a Gamma $\mathcal G(p,1)$ variate, its density is
$$f_X(x)= \frac{x^{p-1}}{(p-1)!}e^{-x}\mathbb I_{\mathbb R^*_+}(x)$$
Therefore, the probability density of $Y_1=\sqrt{X_1}$ is
$$f_{Y_1}(y)= |
39,360 | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$? | I think the moment generating function approach works fine, but easier if we consider MGF of $\ln Y$.
Assuming of course $\mathsf{Gamma}(p,1)$ refers to shape $p$ parameterization, i.e. with density $$f(x)=\frac{e^{-x}x^{p-1}}{\Gamma(p)}\mathbf1_{x>0}$$ with $p>0$ as in @Xi'an's answer.
We have
\begin{align}
E\left[e^{... | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$? | I think the moment generating function approach works fine, but easier if we consider MGF of $\ln Y$.
Assuming of course $\mathsf{Gamma}(p,1)$ refers to shape $p$ parameterization, i.e. with density $ | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$?
I think the moment generating function approach works fine, but easier if we consider MGF of $\ln Y$.
Assuming of course $\mathsf{Gamma}(p,1)$ refers to shape $p$ parameterization, i.e. with density $$f(x)=\frac{e^{-x}x^{p-1}}{\Gamma(p)}\mathbf1_{x>0}$$ with $p>0$... | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$?
I think the moment generating function approach works fine, but easier if we consider MGF of $\ln Y$.
Assuming of course $\mathsf{Gamma}(p,1)$ refers to shape $p$ parameterization, i.e. with density $ |
39,361 | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$? | ... Comment continued: R code for the simple simulation is as shown
below. Unfortunately, the simulation gives no clue how to work the
problem. (See @Xi'an's Answer.)
set.seed(1023)
p = 2; m = 10^6
x1 = rgamma(m,p,1); x2 = rgamma(m,p+.5,1)
y = 2*sqrt(x1*x2)
hist(y, br=60, prob=T, col="skyblue2")
curve(dgamma(x,2*p,1... | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$? | ... Comment continued: R code for the simple simulation is as shown
below. Unfortunately, the simulation gives no clue how to work the
problem. (See @Xi'an's Answer.)
set.seed(1023)
p = 2; m = 10^6
x | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$?
... Comment continued: R code for the simple simulation is as shown
below. Unfortunately, the simulation gives no clue how to work the
problem. (See @Xi'an's Answer.)
set.seed(1023)
p = 2; m = 10^6
x1 = rgamma(m,p,1); x2 = rgamma(m,p+.5,1)
y = 2*sqrt(x1*x2)
hist... | How do I show that $Y=2\sqrt{X_1X_2}\sim$Gamma$(2p,1)$?
... Comment continued: R code for the simple simulation is as shown
below. Unfortunately, the simulation gives no clue how to work the
problem. (See @Xi'an's Answer.)
set.seed(1023)
p = 2; m = 10^6
x |
39,362 | Model ensembling - averaging of probabilities | From the law of total probability we know that for disjoint events $H_n$, we can calculate:
$$P(A) = \sum_n P(A|H_n) * P(H_n)$$
Basically if $P(A|H_n), n:1,...,N$ are different networks emitting probabilities, and $H_n$ is a disjoint hypothesis space then the result is a probability.
When doing simple averaging they ar... | Model ensembling - averaging of probabilities | From the law of total probability we know that for disjoint events $H_n$, we can calculate:
$$P(A) = \sum_n P(A|H_n) * P(H_n)$$
Basically if $P(A|H_n), n:1,...,N$ are different networks emitting proba | Model ensembling - averaging of probabilities
From the law of total probability we know that for disjoint events $H_n$, we can calculate:
$$P(A) = \sum_n P(A|H_n) * P(H_n)$$
Basically if $P(A|H_n), n:1,...,N$ are different networks emitting probabilities, and $H_n$ is a disjoint hypothesis space then the result is a pr... | Model ensembling - averaging of probabilities
From the law of total probability we know that for disjoint events $H_n$, we can calculate:
$$P(A) = \sum_n P(A|H_n) * P(H_n)$$
Basically if $P(A|H_n), n:1,...,N$ are different networks emitting proba |
39,363 | Model ensembling - averaging of probabilities | As I already noticed in my comment, you can find a partial answer to your question and further details in my other answer and the references that were provided.
What you seem to be asking, is "how do we know that the average of probability forecasts is a valid probability?", at least this is how I understand it. Your ... | Model ensembling - averaging of probabilities | As I already noticed in my comment, you can find a partial answer to your question and further details in my other answer and the references that were provided.
What you seem to be asking, is "how do | Model ensembling - averaging of probabilities
As I already noticed in my comment, you can find a partial answer to your question and further details in my other answer and the references that were provided.
What you seem to be asking, is "how do we know that the average of probability forecasts is a valid probability?... | Model ensembling - averaging of probabilities
As I already noticed in my comment, you can find a partial answer to your question and further details in my other answer and the references that were provided.
What you seem to be asking, is "how do |
39,364 | Model ensembling - averaging of probabilities | Yes, there is theoretical basis, and No, we don't know why it works. Look up "forecast combination puzzle" in interweb, e.g. this presentation, p.20. Somehow, a simple average of multiple models appears to outperform single model forecast and weighted average forecasts in practice. There are many hypotheses of why this... | Model ensembling - averaging of probabilities | Yes, there is theoretical basis, and No, we don't know why it works. Look up "forecast combination puzzle" in interweb, e.g. this presentation, p.20. Somehow, a simple average of multiple models appea | Model ensembling - averaging of probabilities
Yes, there is theoretical basis, and No, we don't know why it works. Look up "forecast combination puzzle" in interweb, e.g. this presentation, p.20. Somehow, a simple average of multiple models appears to outperform single model forecast and weighted average forecasts in p... | Model ensembling - averaging of probabilities
Yes, there is theoretical basis, and No, we don't know why it works. Look up "forecast combination puzzle" in interweb, e.g. this presentation, p.20. Somehow, a simple average of multiple models appea |
39,365 | Model ensembling - averaging of probabilities | Yes there is a theory for it, it's called ensemble learning.
The method of bagging (bootstrap aggregating) relies on it. This is used for example in Random Forests.
The intuitive idea is that by averaging models that have a very low bias but a high variance, you can reduce that variance while still keeping the bias low... | Model ensembling - averaging of probabilities | Yes there is a theory for it, it's called ensemble learning.
The method of bagging (bootstrap aggregating) relies on it. This is used for example in Random Forests.
The intuitive idea is that by avera | Model ensembling - averaging of probabilities
Yes there is a theory for it, it's called ensemble learning.
The method of bagging (bootstrap aggregating) relies on it. This is used for example in Random Forests.
The intuitive idea is that by averaging models that have a very low bias but a high variance, you can reduce ... | Model ensembling - averaging of probabilities
Yes there is a theory for it, it's called ensemble learning.
The method of bagging (bootstrap aggregating) relies on it. This is used for example in Random Forests.
The intuitive idea is that by avera |
39,366 | Removing duplicates before train test split | Interesting question.
The effect of duplicates in training data is slightly different than the effect of duplicates in the test data.
If an element is duplicated in the training data, it is effectively the same as having its 'weight' doubled. That element becomes twice as important when the classifier is fitting your d... | Removing duplicates before train test split | Interesting question.
The effect of duplicates in training data is slightly different than the effect of duplicates in the test data.
If an element is duplicated in the training data, it is effectivel | Removing duplicates before train test split
Interesting question.
The effect of duplicates in training data is slightly different than the effect of duplicates in the test data.
If an element is duplicated in the training data, it is effectively the same as having its 'weight' doubled. That element becomes twice as imp... | Removing duplicates before train test split
Interesting question.
The effect of duplicates in training data is slightly different than the effect of duplicates in the test data.
If an element is duplicated in the training data, it is effectivel |
39,367 | Removing duplicates before train test split | I'd like to add 2 points to @Ingolifs nice answer:
The main idea behind recommending to deduplicate or not is to think what that amounts to wrt. your application. Both have their point, but wrt. testing a slightly different kind of generalization ability:
If you want to test the ability to correctly predict new (say,... | Removing duplicates before train test split | I'd like to add 2 points to @Ingolifs nice answer:
The main idea behind recommending to deduplicate or not is to think what that amounts to wrt. your application. Both have their point, but wrt. test | Removing duplicates before train test split
I'd like to add 2 points to @Ingolifs nice answer:
The main idea behind recommending to deduplicate or not is to think what that amounts to wrt. your application. Both have their point, but wrt. testing a slightly different kind of generalization ability:
If you want to tes... | Removing duplicates before train test split
I'd like to add 2 points to @Ingolifs nice answer:
The main idea behind recommending to deduplicate or not is to think what that amounts to wrt. your application. Both have their point, but wrt. test |
39,368 | Controlling for confounding variables in linear mixed effects models (lmer) | You have stated that you believe Time is a confounding variable in this analysis. If so then you should include Time as a covariate in the analysis.
However, before doing so, it is important to ensure that the variable is indeed a (potential) confounder, or a competing exposure.
To be a confounder, it must be a cause, ... | Controlling for confounding variables in linear mixed effects models (lmer) | You have stated that you believe Time is a confounding variable in this analysis. If so then you should include Time as a covariate in the analysis.
However, before doing so, it is important to ensure | Controlling for confounding variables in linear mixed effects models (lmer)
You have stated that you believe Time is a confounding variable in this analysis. If so then you should include Time as a covariate in the analysis.
However, before doing so, it is important to ensure that the variable is indeed a (potential) c... | Controlling for confounding variables in linear mixed effects models (lmer)
You have stated that you believe Time is a confounding variable in this analysis. If so then you should include Time as a covariate in the analysis.
However, before doing so, it is important to ensure |
39,369 | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.
Another way to think of it is that there'... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B d | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the ... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B d |
39,370 | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a \cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a \cdot b$ pairs, in half of which A comes before B. So the proportion of pair... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a \cdot 2b$ pairs of delivery times. The window in which A | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a \cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver tim... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a \cdot 2b$ pairs of delivery times. The window in which A |
39,371 | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | I propose another way of looking at it, only if you had a pc during the interview of course.
We can simulate the process with R, for example.
Let's simulate 1000 values from A and the same from B, we know that both are uniforms, are independent.
a <- runif(1000, 8, 10) # A deliveries
b <- runif(1000, 9, 11) # B deliver... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | I propose another way of looking at it, only if you had a pc during the interview of course.
We can simulate the process with R, for example.
Let's simulate 1000 values from A and the same from B, we | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
I propose another way of looking at it, only if you had a pc during the interview of course.
We can simulate the process with R, for example.
Let's simulate 1000 values from A and the same from B, we know that both are uniforms, ... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
I propose another way of looking at it, only if you had a pc during the interview of course.
We can simulate the process with R, for example.
Let's simulate 1000 values from A and the same from B, we |
39,372 | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery betwee... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no const | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, su... | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no const |
39,373 | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | It is zero:
if the B truck has at least one delivery in the period of [9-11] at least one delivery is made after (or equal to ) 10
and that delivery is not before the deliveries of A (which are all before 10) | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A? | It is zero:
if the B truck has at least one delivery in the period of [9-11] at least one delivery is made after (or equal to ) 10
and that delivery is not before the deliveries of A (which are all be | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
It is zero:
if the B truck has at least one delivery in the period of [9-11] at least one delivery is made after (or equal to ) 10
and that delivery is not before the deliveries of A (which are all before 10) | If A is distributed uniformly on [8,10] and B on [9,11], what is the probability that B<A?
It is zero:
if the B truck has at least one delivery in the period of [9-11] at least one delivery is made after (or equal to ) 10
and that delivery is not before the deliveries of A (which are all be |
39,374 | Resources on Explainable AI | I would recommend a great book by Christoph Molnar: Interpretable Machine Learning - A Guide for Making Black Box Models Explainable.
It touches upon both Interpretable Models, e.g. Linear/Logistic Regression, Generalized Linear Models (GLMs), Generative Additive Models (GAMs), Decision Trees and Model-Agnostic Methods... | Resources on Explainable AI | I would recommend a great book by Christoph Molnar: Interpretable Machine Learning - A Guide for Making Black Box Models Explainable.
It touches upon both Interpretable Models, e.g. Linear/Logistic Re | Resources on Explainable AI
I would recommend a great book by Christoph Molnar: Interpretable Machine Learning - A Guide for Making Black Box Models Explainable.
It touches upon both Interpretable Models, e.g. Linear/Logistic Regression, Generalized Linear Models (GLMs), Generative Additive Models (GAMs), Decision Tree... | Resources on Explainable AI
I would recommend a great book by Christoph Molnar: Interpretable Machine Learning - A Guide for Making Black Box Models Explainable.
It touches upon both Interpretable Models, e.g. Linear/Logistic Re |
39,375 | Resources on Explainable AI | Longer papers which I found when I recently started exploring this topic are:
Explanation in Artificial Intelligence: Insights from the Social Sciences by Tim Miller with lots of references
The Mythos of Model Interpretability by Zachary Lipton
For a more applied perspective and descriptions of concrete applications... | Resources on Explainable AI | Longer papers which I found when I recently started exploring this topic are:
Explanation in Artificial Intelligence: Insights from the Social Sciences by Tim Miller with lots of references
The Mytho | Resources on Explainable AI
Longer papers which I found when I recently started exploring this topic are:
Explanation in Artificial Intelligence: Insights from the Social Sciences by Tim Miller with lots of references
The Mythos of Model Interpretability by Zachary Lipton
For a more applied perspective and descripti... | Resources on Explainable AI
Longer papers which I found when I recently started exploring this topic are:
Explanation in Artificial Intelligence: Insights from the Social Sciences by Tim Miller with lots of references
The Mytho |
39,376 | Resources on Explainable AI | There is a lot of content coming up everyday. Banking, Pharma, Fraud detection are already using explaibale AI in many ways.
You can get examples/main people( from IBM, Google, Microsoft, FB)/research paper from the github page of these framework.
we had done very exhaustive literature study to build our own XAI framew... | Resources on Explainable AI | There is a lot of content coming up everyday. Banking, Pharma, Fraud detection are already using explaibale AI in many ways.
You can get examples/main people( from IBM, Google, Microsoft, FB)/research | Resources on Explainable AI
There is a lot of content coming up everyday. Banking, Pharma, Fraud detection are already using explaibale AI in many ways.
You can get examples/main people( from IBM, Google, Microsoft, FB)/research paper from the github page of these framework.
we had done very exhaustive literature study... | Resources on Explainable AI
There is a lot of content coming up everyday. Banking, Pharma, Fraud detection are already using explaibale AI in many ways.
You can get examples/main people( from IBM, Google, Microsoft, FB)/research |
39,377 | Resources on Explainable AI | If you very new to Explainable AI, this book on Applied Machine Learning Explainability techniques would be the best book for you. It has wonderful code examples and very practical scenarios and dataset presented in the tutorials, where you can gain hands on knowledge on XAI.
Book link - https://amzn.to/3f6v6H3
GitHub ... | Resources on Explainable AI | If you very new to Explainable AI, this book on Applied Machine Learning Explainability techniques would be the best book for you. It has wonderful code examples and very practical scenarios and datas | Resources on Explainable AI
If you very new to Explainable AI, this book on Applied Machine Learning Explainability techniques would be the best book for you. It has wonderful code examples and very practical scenarios and dataset presented in the tutorials, where you can gain hands on knowledge on XAI.
Book link - htt... | Resources on Explainable AI
If you very new to Explainable AI, this book on Applied Machine Learning Explainability techniques would be the best book for you. It has wonderful code examples and very practical scenarios and datas |
39,378 | a fast uniform order statistic generator | The R code means returning
$$(E_1,E_1+E_2,\ldots,E_1+\cdots+E_n)\Big/\sum_{i=1}^{n+1} E_i$$ and the result follows from checking that the differences between the cumulated sums of exponentials renormalised by the overall sum has the same distribution as the differences between order statistics for a uniform sample, $S_... | a fast uniform order statistic generator | The R code means returning
$$(E_1,E_1+E_2,\ldots,E_1+\cdots+E_n)\Big/\sum_{i=1}^{n+1} E_i$$ and the result follows from checking that the differences between the cumulated sums of exponentials renorma | a fast uniform order statistic generator
The R code means returning
$$(E_1,E_1+E_2,\ldots,E_1+\cdots+E_n)\Big/\sum_{i=1}^{n+1} E_i$$ and the result follows from checking that the differences between the cumulated sums of exponentials renormalised by the overall sum has the same distribution as the differences between o... | a fast uniform order statistic generator
The R code means returning
$$(E_1,E_1+E_2,\ldots,E_1+\cdots+E_n)\Big/\sum_{i=1}^{n+1} E_i$$ and the result follows from checking that the differences between the cumulated sums of exponentials renorma |
39,379 | Satterthwaite degrees of freedom in a mixed model change drastically depending on the DV | The Satterthwaite method depends on the dependent variable through the Hessian of the (reml) log-likelihood/deviance function with respect to the variance-parameters (there is also a gradient involved), so it is not surprising if the denominator degrees of freedom change with a change of the dependent variable.
In thi... | Satterthwaite degrees of freedom in a mixed model change drastically depending on the DV | The Satterthwaite method depends on the dependent variable through the Hessian of the (reml) log-likelihood/deviance function with respect to the variance-parameters (there is also a gradient involved | Satterthwaite degrees of freedom in a mixed model change drastically depending on the DV
The Satterthwaite method depends on the dependent variable through the Hessian of the (reml) log-likelihood/deviance function with respect to the variance-parameters (there is also a gradient involved), so it is not surprising if t... | Satterthwaite degrees of freedom in a mixed model change drastically depending on the DV
The Satterthwaite method depends on the dependent variable through the Hessian of the (reml) log-likelihood/deviance function with respect to the variance-parameters (there is also a gradient involved |
39,380 | Wasserstein Loss is very sensitive to model architecture | Usually, the same architecture and parameters would not be good for training both GAN and WGAN.
In a typical GAN, you want to avoid making the discriminator more powerful than the generator, and you want to avoid training the discriminator so much that it "overpowers" the generator and always finds the fakes.
In WGAN... | Wasserstein Loss is very sensitive to model architecture | Usually, the same architecture and parameters would not be good for training both GAN and WGAN.
In a typical GAN, you want to avoid making the discriminator more powerful than the generator, and you | Wasserstein Loss is very sensitive to model architecture
Usually, the same architecture and parameters would not be good for training both GAN and WGAN.
In a typical GAN, you want to avoid making the discriminator more powerful than the generator, and you want to avoid training the discriminator so much that it "overp... | Wasserstein Loss is very sensitive to model architecture
Usually, the same architecture and parameters would not be good for training both GAN and WGAN.
In a typical GAN, you want to avoid making the discriminator more powerful than the generator, and you |
39,381 | Wasserstein Loss is very sensitive to model architecture | Try to substitute gradient clipping with gradient penalty in WGAN, if you haven't done so yet. The important thing is that you should NOT use batch normalization in WGAN discriminator, as it breaks the whole idea. The authors of the WGAN paper suggest to use layer norm. | Wasserstein Loss is very sensitive to model architecture | Try to substitute gradient clipping with gradient penalty in WGAN, if you haven't done so yet. The important thing is that you should NOT use batch normalization in WGAN discriminator, as it breaks th | Wasserstein Loss is very sensitive to model architecture
Try to substitute gradient clipping with gradient penalty in WGAN, if you haven't done so yet. The important thing is that you should NOT use batch normalization in WGAN discriminator, as it breaks the whole idea. The authors of the WGAN paper suggest to use laye... | Wasserstein Loss is very sensitive to model architecture
Try to substitute gradient clipping with gradient penalty in WGAN, if you haven't done so yet. The important thing is that you should NOT use batch normalization in WGAN discriminator, as it breaks th |
39,382 | What is distribution parameterization? | Reparameterization means the substitution of a function for a parameter, where the parameters are the coefficients of a distribution. References on this do not help much. Parameterization is the explicit form for a distribution. For example, the gamma distribution has two different parameterizations that are in common ... | What is distribution parameterization? | Reparameterization means the substitution of a function for a parameter, where the parameters are the coefficients of a distribution. References on this do not help much. Parameterization is the expli | What is distribution parameterization?
Reparameterization means the substitution of a function for a parameter, where the parameters are the coefficients of a distribution. References on this do not help much. Parameterization is the explicit form for a distribution. For example, the gamma distribution has two differen... | What is distribution parameterization?
Reparameterization means the substitution of a function for a parameter, where the parameters are the coefficients of a distribution. References on this do not help much. Parameterization is the expli |
39,383 | What is distribution parameterization? | It means to use a parameter or a set of parameters to describe a probability distribution.
The easiest example would be the Bernoulli distribution with one parameter $p$: Suppose we want to have a probability distribution on the discrete outcome of a coin flip. We use $p$ to represent the probability of getting a HEAD(... | What is distribution parameterization? | It means to use a parameter or a set of parameters to describe a probability distribution.
The easiest example would be the Bernoulli distribution with one parameter $p$: Suppose we want to have a pro | What is distribution parameterization?
It means to use a parameter or a set of parameters to describe a probability distribution.
The easiest example would be the Bernoulli distribution with one parameter $p$: Suppose we want to have a probability distribution on the discrete outcome of a coin flip. We use $p$ to repre... | What is distribution parameterization?
It means to use a parameter or a set of parameters to describe a probability distribution.
The easiest example would be the Bernoulli distribution with one parameter $p$: Suppose we want to have a pro |
39,384 | Simple method of forecasting number of guests given current and historical data | There are simple methods to use but they are probably profoundly wrong as daily data presents a ton of opportunities . Simply striking daily averages is both simple and useless BUT I guess if there is no analytics around it is probably better than the overall average . The absolute LAST approach would be to use the ove... | Simple method of forecasting number of guests given current and historical data | There are simple methods to use but they are probably profoundly wrong as daily data presents a ton of opportunities . Simply striking daily averages is both simple and useless BUT I guess if there is | Simple method of forecasting number of guests given current and historical data
There are simple methods to use but they are probably profoundly wrong as daily data presents a ton of opportunities . Simply striking daily averages is both simple and useless BUT I guess if there is no analytics around it is probably bett... | Simple method of forecasting number of guests given current and historical data
There are simple methods to use but they are probably profoundly wrong as daily data presents a ton of opportunities . Simply striking daily averages is both simple and useless BUT I guess if there is |
39,385 | Simple method of forecasting number of guests given current and historical data | This is not a trivial task as there are many ways to approach this problem, as well as things to take into consideration. Instead of proposing a model, I will give you some general advice.
What you are describing is a time-series and predicting the number of guests in the future is a forecasting problem. A time-series... | Simple method of forecasting number of guests given current and historical data | This is not a trivial task as there are many ways to approach this problem, as well as things to take into consideration. Instead of proposing a model, I will give you some general advice.
What you a | Simple method of forecasting number of guests given current and historical data
This is not a trivial task as there are many ways to approach this problem, as well as things to take into consideration. Instead of proposing a model, I will give you some general advice.
What you are describing is a time-series and predi... | Simple method of forecasting number of guests given current and historical data
This is not a trivial task as there are many ways to approach this problem, as well as things to take into consideration. Instead of proposing a model, I will give you some general advice.
What you a |
39,386 | What type of chart is this? | It is a chord diagram or radial network diagram. | What type of chart is this? | It is a chord diagram or radial network diagram. | What type of chart is this?
It is a chord diagram or radial network diagram. | What type of chart is this?
It is a chord diagram or radial network diagram. |
39,387 | What type of chart is this? | It's a circos plot, see here. | What type of chart is this? | It's a circos plot, see here. | What type of chart is this?
It's a circos plot, see here. | What type of chart is this?
It's a circos plot, see here. |
39,388 | How to simulate a uniform distribution of a triangular area? | The uniform distribution on the simplex $y_1+y_2+y_3=1$, all $y_i\ge 0$, is known as the Dirichlet$(1,1,1)$ distribution. By setting $x_i=(1-3\times 0.1)y_i + 0.1$ you will achieve a uniform distribution on the simplex $x_1+x_2+x_3=0.7$, because it shrinks everything with a constant scale factor and therefore preserve... | How to simulate a uniform distribution of a triangular area? | The uniform distribution on the simplex $y_1+y_2+y_3=1$, all $y_i\ge 0$, is known as the Dirichlet$(1,1,1)$ distribution. By setting $x_i=(1-3\times 0.1)y_i + 0.1$ you will achieve a uniform distribu | How to simulate a uniform distribution of a triangular area?
The uniform distribution on the simplex $y_1+y_2+y_3=1$, all $y_i\ge 0$, is known as the Dirichlet$(1,1,1)$ distribution. By setting $x_i=(1-3\times 0.1)y_i + 0.1$ you will achieve a uniform distribution on the simplex $x_1+x_2+x_3=0.7$, because it shrinks e... | How to simulate a uniform distribution of a triangular area?
The uniform distribution on the simplex $y_1+y_2+y_3=1$, all $y_i\ge 0$, is known as the Dirichlet$(1,1,1)$ distribution. By setting $x_i=(1-3\times 0.1)y_i + 0.1$ you will achieve a uniform distribu |
39,389 | How to simulate a uniform distribution of a triangular area? | For Mathematica users, an easy way to do what the OP asks is to define the right-angled isosceles triangle on the Cartesian plane:
R = Triangle[{{.1, .1}, {.1, .8}, {.8, .1}}];
... and then draw (Uniform) random numbers from it:
pts = RandomPoint[R, 10^4];
All done.
To visualise both the triangle R and the sample... | How to simulate a uniform distribution of a triangular area? | For Mathematica users, an easy way to do what the OP asks is to define the right-angled isosceles triangle on the Cartesian plane:
R = Triangle[{{.1, .1}, {.1, .8}, {.8, .1}}];
... and then draw ( | How to simulate a uniform distribution of a triangular area?
For Mathematica users, an easy way to do what the OP asks is to define the right-angled isosceles triangle on the Cartesian plane:
R = Triangle[{{.1, .1}, {.1, .8}, {.8, .1}}];
... and then draw (Uniform) random numbers from it:
pts = RandomPoint[R, 10^4]... | How to simulate a uniform distribution of a triangular area?
For Mathematica users, an easy way to do what the OP asks is to define the right-angled isosceles triangle on the Cartesian plane:
R = Triangle[{{.1, .1}, {.1, .8}, {.8, .1}}];
... and then draw ( |
39,390 | What is the difference between the Poisson distribution and the uniform distribution? | @scortchi has the right answer. To summarize:
The arrival time stamps are uniformly distributed.
The inter-arrival times are exponentially distributed.
The count of arrivals per uniform time period is Poisson distributed. | What is the difference between the Poisson distribution and the uniform distribution? | @scortchi has the right answer. To summarize:
The arrival time stamps are uniformly distributed.
The inter-arrival times are exponentially distributed.
The count of arrivals per uniform time period i | What is the difference between the Poisson distribution and the uniform distribution?
@scortchi has the right answer. To summarize:
The arrival time stamps are uniformly distributed.
The inter-arrival times are exponentially distributed.
The count of arrivals per uniform time period is Poisson distributed. | What is the difference between the Poisson distribution and the uniform distribution?
@scortchi has the right answer. To summarize:
The arrival time stamps are uniformly distributed.
The inter-arrival times are exponentially distributed.
The count of arrivals per uniform time period i |
39,391 | For regression with time varying parameters, SGD or Kalman filter? | Both of these things can be used in an online manner, but they do this in different ways. So they are not competitors.
The Kalman filter has two purposes. First, for a batch of data, it will yield the log-likelihood of all your observed data, assuming you are estimating a Linear-Gaussian state space model. The log-like... | For regression with time varying parameters, SGD or Kalman filter? | Both of these things can be used in an online manner, but they do this in different ways. So they are not competitors.
The Kalman filter has two purposes. First, for a batch of data, it will yield the | For regression with time varying parameters, SGD or Kalman filter?
Both of these things can be used in an online manner, but they do this in different ways. So they are not competitors.
The Kalman filter has two purposes. First, for a batch of data, it will yield the log-likelihood of all your observed data, assuming y... | For regression with time varying parameters, SGD or Kalman filter?
Both of these things can be used in an online manner, but they do this in different ways. So they are not competitors.
The Kalman filter has two purposes. First, for a batch of data, it will yield the |
39,392 | For regression with time varying parameters, SGD or Kalman filter? | The Kalman filter is a model based optimization algorithm that assumes linear dynamics and Gaussian noise. If these assumptions hold, it is guaranteed to converge to the optimum and should be used instead of SGD.
SGD is a model free heuristic which (hopefully) converges to a local optimum. It 'works' for non-linear dy... | For regression with time varying parameters, SGD or Kalman filter? | The Kalman filter is a model based optimization algorithm that assumes linear dynamics and Gaussian noise. If these assumptions hold, it is guaranteed to converge to the optimum and should be used ins | For regression with time varying parameters, SGD or Kalman filter?
The Kalman filter is a model based optimization algorithm that assumes linear dynamics and Gaussian noise. If these assumptions hold, it is guaranteed to converge to the optimum and should be used instead of SGD.
SGD is a model free heuristic which (ho... | For regression with time varying parameters, SGD or Kalman filter?
The Kalman filter is a model based optimization algorithm that assumes linear dynamics and Gaussian noise. If these assumptions hold, it is guaranteed to converge to the optimum and should be used ins |
39,393 | For regression with time varying parameters, SGD or Kalman filter? | With time varying parameters, only the Kalman filter can be used. (Unless you find an innovative way to use SGD).
Let's look at a situation where both algorithms would make sense: linear regression $Y=\beta X+\epsilon$. Call $n$ the length of vector $X$.
SGD (for MLE) assumes a fixed $\beta$ and will just find it. It i... | For regression with time varying parameters, SGD or Kalman filter? | With time varying parameters, only the Kalman filter can be used. (Unless you find an innovative way to use SGD).
Let's look at a situation where both algorithms would make sense: linear regression $Y | For regression with time varying parameters, SGD or Kalman filter?
With time varying parameters, only the Kalman filter can be used. (Unless you find an innovative way to use SGD).
Let's look at a situation where both algorithms would make sense: linear regression $Y=\beta X+\epsilon$. Call $n$ the length of vector $X$... | For regression with time varying parameters, SGD or Kalman filter?
With time varying parameters, only the Kalman filter can be used. (Unless you find an innovative way to use SGD).
Let's look at a situation where both algorithms would make sense: linear regression $Y |
39,394 | For regression with time varying parameters, SGD or Kalman filter? | Stochastic gradient descent is an optimization algorithm. It is a variant of gradient descent. It is used to find minima or maxima of functions. The difference between SGD and vanilla gradient descent is that SGD works on samples of the objective function, while vanilla gradient descent works on the exact objective fun... | For regression with time varying parameters, SGD or Kalman filter? | Stochastic gradient descent is an optimization algorithm. It is a variant of gradient descent. It is used to find minima or maxima of functions. The difference between SGD and vanilla gradient descent | For regression with time varying parameters, SGD or Kalman filter?
Stochastic gradient descent is an optimization algorithm. It is a variant of gradient descent. It is used to find minima or maxima of functions. The difference between SGD and vanilla gradient descent is that SGD works on samples of the objective functi... | For regression with time varying parameters, SGD or Kalman filter?
Stochastic gradient descent is an optimization algorithm. It is a variant of gradient descent. It is used to find minima or maxima of functions. The difference between SGD and vanilla gradient descent |
39,395 | For regression with time varying parameters, SGD or Kalman filter? | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
There is an alternative at all theses methods which u... | For regression with time varying parameters, SGD or Kalman filter? | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| For regression with time varying parameters, SGD or Kalman filter?
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
... | For regression with time varying parameters, SGD or Kalman filter?
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
|
39,396 | Who is the father(or mother) of the linear least squares analysis as we know it? | I highly recommend Prof. Stephen Stigler's A History of Statistics: Measurement of Uncertainty before 1900. Chapter 1 discusses your question in depth. (A link is available here.)
Legendre first developed the method of least squares and derived the normal equations in 1805 as a way to solve an overdetermined system of ... | Who is the father(or mother) of the linear least squares analysis as we know it? | I highly recommend Prof. Stephen Stigler's A History of Statistics: Measurement of Uncertainty before 1900. Chapter 1 discusses your question in depth. (A link is available here.)
Legendre first devel | Who is the father(or mother) of the linear least squares analysis as we know it?
I highly recommend Prof. Stephen Stigler's A History of Statistics: Measurement of Uncertainty before 1900. Chapter 1 discusses your question in depth. (A link is available here.)
Legendre first developed the method of least squares and de... | Who is the father(or mother) of the linear least squares analysis as we know it?
I highly recommend Prof. Stephen Stigler's A History of Statistics: Measurement of Uncertainty before 1900. Chapter 1 discusses your question in depth. (A link is available here.)
Legendre first devel |
39,397 | Gradient boosting - extreme predictions vs predictions close to 0.5 | I have prepared a short script to show what I think should be the right intuition.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn import ensemble
from sklearn.model_selection import train_test_split
def create_dataset(location, scale, N):
class_zero = pd.DataFrame({
'x... | Gradient boosting - extreme predictions vs predictions close to 0.5 | I have prepared a short script to show what I think should be the right intuition.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn import ensemble
from sklearn.mode | Gradient boosting - extreme predictions vs predictions close to 0.5
I have prepared a short script to show what I think should be the right intuition.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn import ensemble
from sklearn.model_selection import train_test_split
def create_dat... | Gradient boosting - extreme predictions vs predictions close to 0.5
I have prepared a short script to show what I think should be the right intuition.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn import ensemble
from sklearn.mode |
39,398 | Gradient boosting - extreme predictions vs predictions close to 0.5 | First I would suggest to use one data set instead of two to explore the output probability predictions. The reason is simple: if we change the data no one knows what will happen. As demoed in @IcannotFixThis's answer the exact same model will have different probability output, if the data changes from overlapping to mo... | Gradient boosting - extreme predictions vs predictions close to 0.5 | First I would suggest to use one data set instead of two to explore the output probability predictions. The reason is simple: if we change the data no one knows what will happen. As demoed in @Icannot | Gradient boosting - extreme predictions vs predictions close to 0.5
First I would suggest to use one data set instead of two to explore the output probability predictions. The reason is simple: if we change the data no one knows what will happen. As demoed in @IcannotFixThis's answer the exact same model will have diff... | Gradient boosting - extreme predictions vs predictions close to 0.5
First I would suggest to use one data set instead of two to explore the output probability predictions. The reason is simple: if we change the data no one knows what will happen. As demoed in @Icannot |
39,399 | Gradient boosting - extreme predictions vs predictions close to 0.5 | The predictions usually depend on your model. Decision trees in general yield quite "calibrated" outputs which can nearly be interpreted as a probability. Some like SVM for example, don't. But this also highly depends on over-fitting/under-fitting. Or on the number of features (no, it does not have to be "better" neces... | Gradient boosting - extreme predictions vs predictions close to 0.5 | The predictions usually depend on your model. Decision trees in general yield quite "calibrated" outputs which can nearly be interpreted as a probability. Some like SVM for example, don't. But this al | Gradient boosting - extreme predictions vs predictions close to 0.5
The predictions usually depend on your model. Decision trees in general yield quite "calibrated" outputs which can nearly be interpreted as a probability. Some like SVM for example, don't. But this also highly depends on over-fitting/under-fitting. Or ... | Gradient boosting - extreme predictions vs predictions close to 0.5
The predictions usually depend on your model. Decision trees in general yield quite "calibrated" outputs which can nearly be interpreted as a probability. Some like SVM for example, don't. But this al |
39,400 | Gradient boosting - extreme predictions vs predictions close to 0.5 | The first scenario may be due to over fitting of the training data. In-sample and out of sample performance also depends on what evaluation metric you are using (or is applicable to the problem).
Besides comparing the metrics, try checking the confusion matrices as well to check the misclassifications.
Using metrics li... | Gradient boosting - extreme predictions vs predictions close to 0.5 | The first scenario may be due to over fitting of the training data. In-sample and out of sample performance also depends on what evaluation metric you are using (or is applicable to the problem).
Besi | Gradient boosting - extreme predictions vs predictions close to 0.5
The first scenario may be due to over fitting of the training data. In-sample and out of sample performance also depends on what evaluation metric you are using (or is applicable to the problem).
Besides comparing the metrics, try checking the confusio... | Gradient boosting - extreme predictions vs predictions close to 0.5
The first scenario may be due to over fitting of the training data. In-sample and out of sample performance also depends on what evaluation metric you are using (or is applicable to the problem).
Besi |
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