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 |
|---|---|---|---|---|---|---|
39,001 | Multicollinearity between ln(x) and ln(x)^2 | The source of the collinearity is that $f(x) = x^2$. One way to reduce the correlation between $x$ and $x^2$ is to center $x$. Let $z=x-E(x)$ and compute $z^2$. Because the low end of the scale now has large absolute values, its square becomes large, making the relationship between $z$ and $z^2$ less linear than that b... | Multicollinearity between ln(x) and ln(x)^2 | The source of the collinearity is that $f(x) = x^2$. One way to reduce the correlation between $x$ and $x^2$ is to center $x$. Let $z=x-E(x)$ and compute $z^2$. Because the low end of the scale now ha | Multicollinearity between ln(x) and ln(x)^2
The source of the collinearity is that $f(x) = x^2$. One way to reduce the correlation between $x$ and $x^2$ is to center $x$. Let $z=x-E(x)$ and compute $z^2$. Because the low end of the scale now has large absolute values, its square becomes large, making the relationship b... | Multicollinearity between ln(x) and ln(x)^2
The source of the collinearity is that $f(x) = x^2$. One way to reduce the correlation between $x$ and $x^2$ is to center $x$. Let $z=x-E(x)$ and compute $z^2$. Because the low end of the scale now ha |
39,002 | Instrumental variable exclusion restriction | There are two criteria for good instruments:
The instrument $z$ is correlated with the endogenous variable $x$ (relevance).
The instrument $z$ affects dependent variable $y$ only through $x$. In other words, $z$ itself does not cause $y$. This is the exclusion restriction.
You can check 1 (relevance) statistically, b... | Instrumental variable exclusion restriction | There are two criteria for good instruments:
The instrument $z$ is correlated with the endogenous variable $x$ (relevance).
The instrument $z$ affects dependent variable $y$ only through $x$. In othe | Instrumental variable exclusion restriction
There are two criteria for good instruments:
The instrument $z$ is correlated with the endogenous variable $x$ (relevance).
The instrument $z$ affects dependent variable $y$ only through $x$. In other words, $z$ itself does not cause $y$. This is the exclusion restriction.
... | Instrumental variable exclusion restriction
There are two criteria for good instruments:
The instrument $z$ is correlated with the endogenous variable $x$ (relevance).
The instrument $z$ affects dependent variable $y$ only through $x$. In othe |
39,003 | Log odds ratio - what happens if linearity fails? | If the functional relationship between the exposure and average response is not an S-shaped logistic curve, there are still reasons why we might consider an S-shaped logistic curve as a meaningful summary of those data.
As an example, we might have omitted a prognostic factor from a model, meaning that the true margin... | Log odds ratio - what happens if linearity fails? | If the functional relationship between the exposure and average response is not an S-shaped logistic curve, there are still reasons why we might consider an S-shaped logistic curve as a meaningful sum | Log odds ratio - what happens if linearity fails?
If the functional relationship between the exposure and average response is not an S-shaped logistic curve, there are still reasons why we might consider an S-shaped logistic curve as a meaningful summary of those data.
As an example, we might have omitted a prognostic... | Log odds ratio - what happens if linearity fails?
If the functional relationship between the exposure and average response is not an S-shaped logistic curve, there are still reasons why we might consider an S-shaped logistic curve as a meaningful sum |
39,004 | Log odds ratio - what happens if linearity fails? | You get biased and inconsistent coefficient estimates, and biased standard errors. Bias in standard errors can be in both directions and the probability of types I and II errors could increase.
You can tackle non-linearity by introducing different functional forms of the predictor that had a non-linear relationship wit... | Log odds ratio - what happens if linearity fails? | You get biased and inconsistent coefficient estimates, and biased standard errors. Bias in standard errors can be in both directions and the probability of types I and II errors could increase.
You ca | Log odds ratio - what happens if linearity fails?
You get biased and inconsistent coefficient estimates, and biased standard errors. Bias in standard errors can be in both directions and the probability of types I and II errors could increase.
You can tackle non-linearity by introducing different functional forms of th... | Log odds ratio - what happens if linearity fails?
You get biased and inconsistent coefficient estimates, and biased standard errors. Bias in standard errors can be in both directions and the probability of types I and II errors could increase.
You ca |
39,005 | Log odds ratio - what happens if linearity fails? | The assumption that your target probability can be modelled as a linear combination of log-odds ratios scaled by your inputs is equivalent to assuming that it is a combination of independent pieces of Bernoulli evidence. When that's not the case, you typically build a more complex model with cross terms.
Seeing the lo... | Log odds ratio - what happens if linearity fails? | The assumption that your target probability can be modelled as a linear combination of log-odds ratios scaled by your inputs is equivalent to assuming that it is a combination of independent pieces of | Log odds ratio - what happens if linearity fails?
The assumption that your target probability can be modelled as a linear combination of log-odds ratios scaled by your inputs is equivalent to assuming that it is a combination of independent pieces of Bernoulli evidence. When that's not the case, you typically build a ... | Log odds ratio - what happens if linearity fails?
The assumption that your target probability can be modelled as a linear combination of log-odds ratios scaled by your inputs is equivalent to assuming that it is a combination of independent pieces of |
39,006 | Kruskal-Wallis test: assumption testing and interpretation of the results | The KW test (also the Mann-Whitney U-test) is essentially always a test for stochastic dominance. What that means is it is testing to see if there exists at least one group such that you would typically get a larger (lesser) value from it than the rest if you drew a value at random from each.
People assume this mean... | Kruskal-Wallis test: assumption testing and interpretation of the results | The KW test (also the Mann-Whitney U-test) is essentially always a test for stochastic dominance. What that means is it is testing to see if there exists at least one group such that you would typica | Kruskal-Wallis test: assumption testing and interpretation of the results
The KW test (also the Mann-Whitney U-test) is essentially always a test for stochastic dominance. What that means is it is testing to see if there exists at least one group such that you would typically get a larger (lesser) value from it than t... | Kruskal-Wallis test: assumption testing and interpretation of the results
The KW test (also the Mann-Whitney U-test) is essentially always a test for stochastic dominance. What that means is it is testing to see if there exists at least one group such that you would typica |
39,007 | Kruskal-Wallis test: assumption testing and interpretation of the results | This question (Non-normal distribution even with Kruskal-Wallis test) has a nice summary on the difference between inferring stochastic dominance vs. equality of medians.
To answer your specific questions:
The K-W test for chickwts gives
Kruskal-Wallis rank sum test
data: weight by feed
Kruskal-Wallis chi-squared =... | Kruskal-Wallis test: assumption testing and interpretation of the results | This question (Non-normal distribution even with Kruskal-Wallis test) has a nice summary on the difference between inferring stochastic dominance vs. equality of medians.
To answer your specific quest | Kruskal-Wallis test: assumption testing and interpretation of the results
This question (Non-normal distribution even with Kruskal-Wallis test) has a nice summary on the difference between inferring stochastic dominance vs. equality of medians.
To answer your specific questions:
The K-W test for chickwts gives
Kruska... | Kruskal-Wallis test: assumption testing and interpretation of the results
This question (Non-normal distribution even with Kruskal-Wallis test) has a nice summary on the difference between inferring stochastic dominance vs. equality of medians.
To answer your specific quest |
39,008 | Is it possible to get fitted values 0 or 1 in logistic regression when the fitting algorithm converges? | R is giving you two different warnings because these really are two distinct issues.
Very loosely, the algorithm that fits a logistic regression model (typically some version of Newton-Raphson) looks around for the coefficient estimates that will maximize the log likelihood. It will estimate the model at a given poi... | Is it possible to get fitted values 0 or 1 in logistic regression when the fitting algorithm converg | R is giving you two different warnings because these really are two distinct issues.
Very loosely, the algorithm that fits a logistic regression model (typically some version of Newton-Raphson) look | Is it possible to get fitted values 0 or 1 in logistic regression when the fitting algorithm converges?
R is giving you two different warnings because these really are two distinct issues.
Very loosely, the algorithm that fits a logistic regression model (typically some version of Newton-Raphson) looks around for the... | Is it possible to get fitted values 0 or 1 in logistic regression when the fitting algorithm converg
R is giving you two different warnings because these really are two distinct issues.
Very loosely, the algorithm that fits a logistic regression model (typically some version of Newton-Raphson) look |
39,009 | How to simulate Poisson arrival times if the rate varies with time? | There are already several reasonable answers, but to try and add some clarity, here I will try to synthesize these, and do some verification on the resulting algorithm.
Homogeneous Process
For a homogeneous 1D Poisson process with constant rate $\lambda$, the number of events $n$ in a time interval $T$ will follow a P... | How to simulate Poisson arrival times if the rate varies with time? | There are already several reasonable answers, but to try and add some clarity, here I will try to synthesize these, and do some verification on the resulting algorithm.
Homogeneous Process
For a homo | How to simulate Poisson arrival times if the rate varies with time?
There are already several reasonable answers, but to try and add some clarity, here I will try to synthesize these, and do some verification on the resulting algorithm.
Homogeneous Process
For a homogeneous 1D Poisson process with constant rate $\lamb... | How to simulate Poisson arrival times if the rate varies with time?
There are already several reasonable answers, but to try and add some clarity, here I will try to synthesize these, and do some verification on the resulting algorithm.
Homogeneous Process
For a homo |
39,010 | How to simulate Poisson arrival times if the rate varies with time? | There are several methods. For simple rate functions like yours I find the inversion approach to be the easiest. See algorithm 3 of this paper; it's 5 lines. Edit: and the algorithm is:
rates = c(1,2,3)
s = sum(rates)
cs = c(0, cumsum(rates))
R = function(t) {
L = length(rates)
t0 = floor(t)
d = t0 %% L + 1
ret... | How to simulate Poisson arrival times if the rate varies with time? | There are several methods. For simple rate functions like yours I find the inversion approach to be the easiest. See algorithm 3 of this paper; it's 5 lines. Edit: and the algorithm is:
rates = c(1,2, | How to simulate Poisson arrival times if the rate varies with time?
There are several methods. For simple rate functions like yours I find the inversion approach to be the easiest. See algorithm 3 of this paper; it's 5 lines. Edit: and the algorithm is:
rates = c(1,2,3)
s = sum(rates)
cs = c(0, cumsum(rates))
R = funct... | How to simulate Poisson arrival times if the rate varies with time?
There are several methods. For simple rate functions like yours I find the inversion approach to be the easiest. See algorithm 3 of this paper; it's 5 lines. Edit: and the algorithm is:
rates = c(1,2, |
39,011 | How to simulate Poisson arrival times if the rate varies with time? | It's pretty straightforward, thanks to the fact that the interarrival times of a Poisson process are exponentially distributed and the exponential distribution has the memoryless property. (As an aside, note that if you are generating the time points, you are generating the interarrival times too, as they are just the... | How to simulate Poisson arrival times if the rate varies with time? | It's pretty straightforward, thanks to the fact that the interarrival times of a Poisson process are exponentially distributed and the exponential distribution has the memoryless property. (As an asi | How to simulate Poisson arrival times if the rate varies with time?
It's pretty straightforward, thanks to the fact that the interarrival times of a Poisson process are exponentially distributed and the exponential distribution has the memoryless property. (As an aside, note that if you are generating the time points,... | How to simulate Poisson arrival times if the rate varies with time?
It's pretty straightforward, thanks to the fact that the interarrival times of a Poisson process are exponentially distributed and the exponential distribution has the memoryless property. (As an asi |
39,012 | How to simulate Poisson arrival times if the rate varies with time? | Your intensities are so low that it's impossible to ignore the boundary effects. By that I mean that a time between the last event in the first hour and the first event in the next hour is covered by two Poisson (exponential) processes.
Method 1:
So, the simplest way to treat this (certainly, not the most efficient) i... | How to simulate Poisson arrival times if the rate varies with time? | Your intensities are so low that it's impossible to ignore the boundary effects. By that I mean that a time between the last event in the first hour and the first event in the next hour is covered by | How to simulate Poisson arrival times if the rate varies with time?
Your intensities are so low that it's impossible to ignore the boundary effects. By that I mean that a time between the last event in the first hour and the first event in the next hour is covered by two Poisson (exponential) processes.
Method 1:
So, ... | How to simulate Poisson arrival times if the rate varies with time?
Your intensities are so low that it's impossible to ignore the boundary effects. By that I mean that a time between the last event in the first hour and the first event in the next hour is covered by |
39,013 | How to simulate Poisson arrival times if the rate varies with time? | If Python is not a problem for you, I advise you to have a look at this new open-source library: tick. Among others, it provides simulation and inference methods for point processes. You can find an example explaining the simulation of an inhomogeneous Poisson process here. | How to simulate Poisson arrival times if the rate varies with time? | If Python is not a problem for you, I advise you to have a look at this new open-source library: tick. Among others, it provides simulation and inference methods for point processes. You can find an e | How to simulate Poisson arrival times if the rate varies with time?
If Python is not a problem for you, I advise you to have a look at this new open-source library: tick. Among others, it provides simulation and inference methods for point processes. You can find an example explaining the simulation of an inhomogeneous... | How to simulate Poisson arrival times if the rate varies with time?
If Python is not a problem for you, I advise you to have a look at this new open-source library: tick. Among others, it provides simulation and inference methods for point processes. You can find an e |
39,014 | Computing Confidence Intervals for Coefficients in Logistic Regression [duplicate] | I just discovered that someone answered this question in another post. The answer is, confint uses profile confidence intervals, whereas I was computing a Wald confidence interval (which can equivalently be computed using confint.default). | Computing Confidence Intervals for Coefficients in Logistic Regression [duplicate] | I just discovered that someone answered this question in another post. The answer is, confint uses profile confidence intervals, whereas I was computing a Wald confidence interval (which can equivalen | Computing Confidence Intervals for Coefficients in Logistic Regression [duplicate]
I just discovered that someone answered this question in another post. The answer is, confint uses profile confidence intervals, whereas I was computing a Wald confidence interval (which can equivalently be computed using confint.default... | Computing Confidence Intervals for Coefficients in Logistic Regression [duplicate]
I just discovered that someone answered this question in another post. The answer is, confint uses profile confidence intervals, whereas I was computing a Wald confidence interval (which can equivalen |
39,015 | Consistency in mean square vs. "normal" consistency | ADDENDUM 30-3-2017
An important clarification: None of the derivations below guarantees that "$\mu$" is the "true value" we are attempting to estimate. What we only show is that if $\theta_n$ converges in $L^2$ to some constant, then this constant is also its probability limit.
Whether the probability limit of the est... | Consistency in mean square vs. "normal" consistency | ADDENDUM 30-3-2017
An important clarification: None of the derivations below guarantees that "$\mu$" is the "true value" we are attempting to estimate. What we only show is that if $\theta_n$ converge | Consistency in mean square vs. "normal" consistency
ADDENDUM 30-3-2017
An important clarification: None of the derivations below guarantees that "$\mu$" is the "true value" we are attempting to estimate. What we only show is that if $\theta_n$ converges in $L^2$ to some constant, then this constant is also its probabil... | Consistency in mean square vs. "normal" consistency
ADDENDUM 30-3-2017
An important clarification: None of the derivations below guarantees that "$\mu$" is the "true value" we are attempting to estimate. What we only show is that if $\theta_n$ converge |
39,016 | Consistency in mean square vs. "normal" consistency | This seems to really be more of a question about mathematical probability, so I will ignore the statistical context of your question in this answer. Please let me know if this is not helpful enough.
1. Yes, it is true that converge in mean square, also called convergence in $L^2$, implies convergence in probability.
Th... | Consistency in mean square vs. "normal" consistency | This seems to really be more of a question about mathematical probability, so I will ignore the statistical context of your question in this answer. Please let me know if this is not helpful enough.
1 | Consistency in mean square vs. "normal" consistency
This seems to really be more of a question about mathematical probability, so I will ignore the statistical context of your question in this answer. Please let me know if this is not helpful enough.
1. Yes, it is true that converge in mean square, also called converge... | Consistency in mean square vs. "normal" consistency
This seems to really be more of a question about mathematical probability, so I will ignore the statistical context of your question in this answer. Please let me know if this is not helpful enough.
1 |
39,017 | Consistency in mean square vs. "normal" consistency | I have a very straight forward answer to this from a machine learning perspective. So given $\theta$ is mean square consistent, a reformulation is (Bias - Variance tradeoff):
\begin{equation}
\mathbb{E}\left[(\theta-\mu)^2\right] \to 0 \Leftrightarrow Var\left[\theta\right] + Bias\left(\theta,\mu\right)^2 \to 0
\end{e... | Consistency in mean square vs. "normal" consistency | I have a very straight forward answer to this from a machine learning perspective. So given $\theta$ is mean square consistent, a reformulation is (Bias - Variance tradeoff):
\begin{equation}
\mathbb{ | Consistency in mean square vs. "normal" consistency
I have a very straight forward answer to this from a machine learning perspective. So given $\theta$ is mean square consistent, a reformulation is (Bias - Variance tradeoff):
\begin{equation}
\mathbb{E}\left[(\theta-\mu)^2\right] \to 0 \Leftrightarrow Var\left[\theta... | Consistency in mean square vs. "normal" consistency
I have a very straight forward answer to this from a machine learning perspective. So given $\theta$ is mean square consistent, a reformulation is (Bias - Variance tradeoff):
\begin{equation}
\mathbb{ |
39,018 | Choosing the number of bootstrap resamples | A bootstrap sample is usually taken to mean that the sample size of the resample is equal to the original sample size. What you are doing is to take resamples from the original sample with larger and larger (re)sample sizes. There is no reason to believe that this will represent the properties of the (original) sampl... | Choosing the number of bootstrap resamples | A bootstrap sample is usually taken to mean that the sample size of the resample is equal to the original sample size. What you are doing is to take resamples from the original sample with larger and | Choosing the number of bootstrap resamples
A bootstrap sample is usually taken to mean that the sample size of the resample is equal to the original sample size. What you are doing is to take resamples from the original sample with larger and larger (re)sample sizes. There is no reason to believe that this will repre... | Choosing the number of bootstrap resamples
A bootstrap sample is usually taken to mean that the sample size of the resample is equal to the original sample size. What you are doing is to take resamples from the original sample with larger and |
39,019 | Can Median Absolute Deviation (MAD)/SD be used to determine if a distribution is normal or not? | Considered as a formal test of normality: If $M$ = (sample) median absolute deviation from the median and $s$ = standard deviation, then you could indeed use a measure like $R = M/s$ (or its reciprocal) as a test statistic for a test of normality.
Note however, that such tests cannot tell you something is normal, onl... | Can Median Absolute Deviation (MAD)/SD be used to determine if a distribution is normal or not? | Considered as a formal test of normality: If $M$ = (sample) median absolute deviation from the median and $s$ = standard deviation, then you could indeed use a measure like $R = M/s$ (or its reciproc | Can Median Absolute Deviation (MAD)/SD be used to determine if a distribution is normal or not?
Considered as a formal test of normality: If $M$ = (sample) median absolute deviation from the median and $s$ = standard deviation, then you could indeed use a measure like $R = M/s$ (or its reciprocal) as a test statistic ... | Can Median Absolute Deviation (MAD)/SD be used to determine if a distribution is normal or not?
Considered as a formal test of normality: If $M$ = (sample) median absolute deviation from the median and $s$ = standard deviation, then you could indeed use a measure like $R = M/s$ (or its reciproc |
39,020 | What's the name of this correlation/association measure between binary variables? | Using a,b,c,d convention of the 4-fold table, as here,
Y
1 0
-------
1 | a | b |
X -------
0 | c | d |
-------
a = number of cases on which both X and Y are 1
b = number of cases where X is 1 and Y is 0
c = number of cases where X is 0 an... | What's the name of this correlation/association measure between binary variables? | Using a,b,c,d convention of the 4-fold table, as here,
Y
1 0
-------
1 | a | b |
X -------
0 | c | d |
-------
a = num | What's the name of this correlation/association measure between binary variables?
Using a,b,c,d convention of the 4-fold table, as here,
Y
1 0
-------
1 | a | b |
X -------
0 | c | d |
-------
a = number of cases on which both X and Y are... | What's the name of this correlation/association measure between binary variables?
Using a,b,c,d convention of the 4-fold table, as here,
Y
1 0
-------
1 | a | b |
X -------
0 | c | d |
-------
a = num |
39,021 | What's the name of this correlation/association measure between binary variables? | Hubalek, Z. Coefficients of association and similarity, based on binary (presence-absence) data: an evaluation (Biol. Rev., 1982)
reviews and ranks 42 different correlation coefficients for binary data. Only 3 of them meet basic statistical desiderata. Unfortunately, the issue of PRE (proportionate reduction of error) ... | What's the name of this correlation/association measure between binary variables? | Hubalek, Z. Coefficients of association and similarity, based on binary (presence-absence) data: an evaluation (Biol. Rev., 1982)
reviews and ranks 42 different correlation coefficients for binary dat | What's the name of this correlation/association measure between binary variables?
Hubalek, Z. Coefficients of association and similarity, based on binary (presence-absence) data: an evaluation (Biol. Rev., 1982)
reviews and ranks 42 different correlation coefficients for binary data. Only 3 of them meet basic statistic... | What's the name of this correlation/association measure between binary variables?
Hubalek, Z. Coefficients of association and similarity, based on binary (presence-absence) data: an evaluation (Biol. Rev., 1982)
reviews and ranks 42 different correlation coefficients for binary dat |
39,022 | Why use differencing and Box-Cox in time series? | The Box-Cox transformation is a family of power transformations indexed by a parameter lambda. Whenever you use it the parameter needs to be estimated from the data. In time series the process could have a non-constant variance. If the variance changes with time, the process is nonstationary. It is often desirable t... | Why use differencing and Box-Cox in time series? | The Box-Cox transformation is a family of power transformations indexed by a parameter lambda. Whenever you use it the parameter needs to be estimated from the data. In time series the process could | Why use differencing and Box-Cox in time series?
The Box-Cox transformation is a family of power transformations indexed by a parameter lambda. Whenever you use it the parameter needs to be estimated from the data. In time series the process could have a non-constant variance. If the variance changes with time, the p... | Why use differencing and Box-Cox in time series?
The Box-Cox transformation is a family of power transformations indexed by a parameter lambda. Whenever you use it the parameter needs to be estimated from the data. In time series the process could |
39,023 | Are Random Forests more powerful than generalized linear models? | You should try lots of models. The 'no free lunch' theorem states that there is no one best model - every situation is different. Logistic regression for example is desirable when it works because parameters are very interpretable. Random forests are great because they can deal with very difficult patterns, but forget... | Are Random Forests more powerful than generalized linear models? | You should try lots of models. The 'no free lunch' theorem states that there is no one best model - every situation is different. Logistic regression for example is desirable when it works because par | Are Random Forests more powerful than generalized linear models?
You should try lots of models. The 'no free lunch' theorem states that there is no one best model - every situation is different. Logistic regression for example is desirable when it works because parameters are very interpretable. Random forests are gre... | Are Random Forests more powerful than generalized linear models?
You should try lots of models. The 'no free lunch' theorem states that there is no one best model - every situation is different. Logistic regression for example is desirable when it works because par |
39,024 | Are Random Forests more powerful than generalized linear models? | One point to consider is are you interested in making predictions or understanding associations and carrying out inference (confidence intervals around effects). Although random forests provide a variable-importance summary, this technique is primarily aimed at prediction; there is no inference. Many researchers think ... | Are Random Forests more powerful than generalized linear models? | One point to consider is are you interested in making predictions or understanding associations and carrying out inference (confidence intervals around effects). Although random forests provide a vari | Are Random Forests more powerful than generalized linear models?
One point to consider is are you interested in making predictions or understanding associations and carrying out inference (confidence intervals around effects). Although random forests provide a variable-importance summary, this technique is primarily ai... | Are Random Forests more powerful than generalized linear models?
One point to consider is are you interested in making predictions or understanding associations and carrying out inference (confidence intervals around effects). Although random forests provide a vari |
39,025 | Are Random Forests more powerful than generalized linear models? | Let's have some simple examples to show the differences. Our example have a single independent variable x and a single dependent variable - either real y or categorical z:
x y z
...
0 0.01 A
1 1.98 A
2.01 4.02 B
2.99 6.01 B
...
One can see that y grows as x grows and that z=B for values of x grater than so... | Are Random Forests more powerful than generalized linear models? | Let's have some simple examples to show the differences. Our example have a single independent variable x and a single dependent variable - either real y or categorical z:
x y z
...
0 0.01 A
| Are Random Forests more powerful than generalized linear models?
Let's have some simple examples to show the differences. Our example have a single independent variable x and a single dependent variable - either real y or categorical z:
x y z
...
0 0.01 A
1 1.98 A
2.01 4.02 B
2.99 6.01 B
...
One can see th... | Are Random Forests more powerful than generalized linear models?
Let's have some simple examples to show the differences. Our example have a single independent variable x and a single dependent variable - either real y or categorical z:
x y z
...
0 0.01 A
|
39,026 | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y) | You should present Mean (X/Y) if X/Y is a useful measure and a mean is a useful way to summarize it. By Jensen's Inequality we know that the ratio of the mean is never equal to the mean of the ratio except under some special circumstances. | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y) | You should present Mean (X/Y) if X/Y is a useful measure and a mean is a useful way to summarize it. By Jensen's Inequality we know that the ratio of the mean is never equal to the mean of the ratio e | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y)
You should present Mean (X/Y) if X/Y is a useful measure and a mean is a useful way to summarize it. By Jensen's Inequality we know that the ratio of the mean is never equal to the mean of the ratio except under some special circumstances. | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y)
You should present Mean (X/Y) if X/Y is a useful measure and a mean is a useful way to summarize it. By Jensen's Inequality we know that the ratio of the mean is never equal to the mean of the ratio e |
39,027 | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y) | $Z=Y/X$ may be meaningful for individual users as their individual average volume per upload, but $\text{Mean}(Y/X)$ does not look meaningful in aggregate as some users use the system more than others.
If you took a weighted mean of $Z=Y/X$ to account for this, the natural weights would be the numbers of uploads $X$ ... | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y) | $Z=Y/X$ may be meaningful for individual users as their individual average volume per upload, but $\text{Mean}(Y/X)$ does not look meaningful in aggregate as some users use the system more than others | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y)
$Z=Y/X$ may be meaningful for individual users as their individual average volume per upload, but $\text{Mean}(Y/X)$ does not look meaningful in aggregate as some users use the system more than others.
If you took a weighted mean of $Z=Y/X$ to account for this, the natural we... | Usages of Mean(X/Y) vs. Mean(X) / Mean(Y)
$Z=Y/X$ may be meaningful for individual users as their individual average volume per upload, but $\text{Mean}(Y/X)$ does not look meaningful in aggregate as some users use the system more than others |
39,028 | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate] | First, there are more Markov chain Monte Carlo (MCMC) samplers aside from the Metropolis-Hastings (MH), but I will focus on MH.
It is not that a Markov chain is needed, it is that using an MH algorithm the samples obtained themselves form a Markov chain that converge to the stationary distribution indicated in the acce... | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate] | First, there are more Markov chain Monte Carlo (MCMC) samplers aside from the Metropolis-Hastings (MH), but I will focus on MH.
It is not that a Markov chain is needed, it is that using an MH algorith | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate]
First, there are more Markov chain Monte Carlo (MCMC) samplers aside from the Metropolis-Hastings (MH), but I will focus on MH.
It is not that a Markov chain is needed, it is that using an MH algorithm the samples obtained themselves form a ... | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate]
First, there are more Markov chain Monte Carlo (MCMC) samplers aside from the Metropolis-Hastings (MH), but I will focus on MH.
It is not that a Markov chain is needed, it is that using an MH algorith |
39,029 | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate] | This is just confusion with the terminology. It's not that you need to "use" a Markov chain to draw Metropolis-Hastings samples, it's that the sequence of Metropolis-Hastings draws is a Markov chain. This is an innate part of the algorithm design.
"Markov chain Monte Carlo" is just a qualifier indicating that the draws... | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate] | This is just confusion with the terminology. It's not that you need to "use" a Markov chain to draw Metropolis-Hastings samples, it's that the sequence of Metropolis-Hastings draws is a Markov chain. | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate]
This is just confusion with the terminology. It's not that you need to "use" a Markov chain to draw Metropolis-Hastings samples, it's that the sequence of Metropolis-Hastings draws is a Markov chain. This is an innate part of the algorithm d... | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate]
This is just confusion with the terminology. It's not that you need to "use" a Markov chain to draw Metropolis-Hastings samples, it's that the sequence of Metropolis-Hastings draws is a Markov chain. |
39,030 | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate] | MCMC in general and Metropolis-Hastings in particular is quite distinct from rejection sampling.
Note that rejection sampling is independent from one generated value to the next -- it doesn't matter what value you just generated, the distribution of the next one doesn't consider it. MCMC involves a series of moves. At... | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate] | MCMC in general and Metropolis-Hastings in particular is quite distinct from rejection sampling.
Note that rejection sampling is independent from one generated value to the next -- it doesn't matter | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate]
MCMC in general and Metropolis-Hastings in particular is quite distinct from rejection sampling.
Note that rejection sampling is independent from one generated value to the next -- it doesn't matter what value you just generated, the distri... | Does "Markov Chain Monte Carlo" method really need "Markov Chain"? [duplicate]
MCMC in general and Metropolis-Hastings in particular is quite distinct from rejection sampling.
Note that rejection sampling is independent from one generated value to the next -- it doesn't matter |
39,031 | Are linear classifiers (SVM, Logistic Regression) deterministic? | First, your definition of "deterministic" and "linear classifier" are not clear to me. For example, are you asking if the model building deterministic or model prediction deterministic? In addition, most people will think SVM is not a linear model but you treat it is linear.
I am trying to guess what you want to ask f... | Are linear classifiers (SVM, Logistic Regression) deterministic? | First, your definition of "deterministic" and "linear classifier" are not clear to me. For example, are you asking if the model building deterministic or model prediction deterministic? In addition, m | Are linear classifiers (SVM, Logistic Regression) deterministic?
First, your definition of "deterministic" and "linear classifier" are not clear to me. For example, are you asking if the model building deterministic or model prediction deterministic? In addition, most people will think SVM is not a linear model but you... | Are linear classifiers (SVM, Logistic Regression) deterministic?
First, your definition of "deterministic" and "linear classifier" are not clear to me. For example, are you asking if the model building deterministic or model prediction deterministic? In addition, m |
39,032 | Are linear classifiers (SVM, Logistic Regression) deterministic? | I would think that any algorithm that can prove it reaches a global error minimum (linear/logistic regression, support vector machines) should stay the same, except for maybe a few trailing decimal places. Models that do not make this guarantee that could get stuck in a local minimum, like neural networks or random for... | Are linear classifiers (SVM, Logistic Regression) deterministic? | I would think that any algorithm that can prove it reaches a global error minimum (linear/logistic regression, support vector machines) should stay the same, except for maybe a few trailing decimal pl | Are linear classifiers (SVM, Logistic Regression) deterministic?
I would think that any algorithm that can prove it reaches a global error minimum (linear/logistic regression, support vector machines) should stay the same, except for maybe a few trailing decimal places. Models that do not make this guarantee that could... | Are linear classifiers (SVM, Logistic Regression) deterministic?
I would think that any algorithm that can prove it reaches a global error minimum (linear/logistic regression, support vector machines) should stay the same, except for maybe a few trailing decimal pl |
39,033 | Square of the Sample Mean as estimator of the variance | You have $X_1, X_2, \dots, X_n$ are iid from an unknown distribution with mean (say) $\mu$ and variance (say) $\sigma^2$.
$\bar{X}$ is an unbiased estimator of the mean, and thus $E(\bar{X}) = \mu$. Also, $Var(\bar{X}) = \sigma^2/n$. Thus since,
\begin{align*}
E[\bar{X}^2] & = Var(\bar{X}) + E[\bar{X}]^2\\
& = \dfrac{... | Square of the Sample Mean as estimator of the variance | You have $X_1, X_2, \dots, X_n$ are iid from an unknown distribution with mean (say) $\mu$ and variance (say) $\sigma^2$.
$\bar{X}$ is an unbiased estimator of the mean, and thus $E(\bar{X}) = \mu$. | Square of the Sample Mean as estimator of the variance
You have $X_1, X_2, \dots, X_n$ are iid from an unknown distribution with mean (say) $\mu$ and variance (say) $\sigma^2$.
$\bar{X}$ is an unbiased estimator of the mean, and thus $E(\bar{X}) = \mu$. Also, $Var(\bar{X}) = \sigma^2/n$. Thus since,
\begin{align*}
E[\... | Square of the Sample Mean as estimator of the variance
You have $X_1, X_2, \dots, X_n$ are iid from an unknown distribution with mean (say) $\mu$ and variance (say) $\sigma^2$.
$\bar{X}$ is an unbiased estimator of the mean, and thus $E(\bar{X}) = \mu$. |
39,034 | Square of the Sample Mean as estimator of the variance | Here is a solution using the 'Moment of Moment' functions in the mathStatica package for Mathematica. In particular, let $s_1$ denote the sample sum, i.e. $s_1 = \sum_{i=1}^nX_i$. Then, you seek
$$E\big[{\big(\frac{s_1}{n}\big)}^2\big]$$
which is the $1^{\text{st}}$ Raw Moment of $(\frac{s_1}{n})^2$, expressed here in... | Square of the Sample Mean as estimator of the variance | Here is a solution using the 'Moment of Moment' functions in the mathStatica package for Mathematica. In particular, let $s_1$ denote the sample sum, i.e. $s_1 = \sum_{i=1}^nX_i$. Then, you seek
$$E\ | Square of the Sample Mean as estimator of the variance
Here is a solution using the 'Moment of Moment' functions in the mathStatica package for Mathematica. In particular, let $s_1$ denote the sample sum, i.e. $s_1 = \sum_{i=1}^nX_i$. Then, you seek
$$E\big[{\big(\frac{s_1}{n}\big)}^2\big]$$
which is the $1^{\text{st}... | Square of the Sample Mean as estimator of the variance
Here is a solution using the 'Moment of Moment' functions in the mathStatica package for Mathematica. In particular, let $s_1$ denote the sample sum, i.e. $s_1 = \sum_{i=1}^nX_i$. Then, you seek
$$E\ |
39,035 | Square of the Sample Mean as estimator of the variance | Do you actually mean something like "$\frac{1}{n-1} \sum_i \left(x_i - \bar{x} \right)^2$, where $\bar{x}$ is the sample mean, is an unbiased estimator of the population variance?" Or perhaps, "Is $\frac{1}{n} \sum_i x_i^2 - \bar{x}^2$ an unbiased estimator of the population variance?"
Trivial counterexample for what y... | Square of the Sample Mean as estimator of the variance | Do you actually mean something like "$\frac{1}{n-1} \sum_i \left(x_i - \bar{x} \right)^2$, where $\bar{x}$ is the sample mean, is an unbiased estimator of the population variance?" Or perhaps, "Is $\f | Square of the Sample Mean as estimator of the variance
Do you actually mean something like "$\frac{1}{n-1} \sum_i \left(x_i - \bar{x} \right)^2$, where $\bar{x}$ is the sample mean, is an unbiased estimator of the population variance?" Or perhaps, "Is $\frac{1}{n} \sum_i x_i^2 - \bar{x}^2$ an unbiased estimator of the ... | Square of the Sample Mean as estimator of the variance
Do you actually mean something like "$\frac{1}{n-1} \sum_i \left(x_i - \bar{x} \right)^2$, where $\bar{x}$ is the sample mean, is an unbiased estimator of the population variance?" Or perhaps, "Is $\f |
39,036 | Why are over identified models preferred over just identified models in Structural Equation Modeling? | The point of running an structural equation model is to be able to be wrong - and that's only true if it's over-identified (i.e. has degrees of freedom greater than zero). You can specify a multiple regression model as a structural equation model, you'll get the same answer, and the model will be just identified, so it... | Why are over identified models preferred over just identified models in Structural Equation Modeling | The point of running an structural equation model is to be able to be wrong - and that's only true if it's over-identified (i.e. has degrees of freedom greater than zero). You can specify a multiple r | Why are over identified models preferred over just identified models in Structural Equation Modeling?
The point of running an structural equation model is to be able to be wrong - and that's only true if it's over-identified (i.e. has degrees of freedom greater than zero). You can specify a multiple regression model as... | Why are over identified models preferred over just identified models in Structural Equation Modeling
The point of running an structural equation model is to be able to be wrong - and that's only true if it's over-identified (i.e. has degrees of freedom greater than zero). You can specify a multiple r |
39,037 | Regression variable has no meaning for one category | This will happen naturally, with no intervention on your part.
Consider, for instance, dummy coding. This system uses vectors of zeros and ones to indicate the categorical variables in a way that allows straightforward interpretation of the coefficients. A variable with $k$ categories is represented by $k-1$ terms (a... | Regression variable has no meaning for one category | This will happen naturally, with no intervention on your part.
Consider, for instance, dummy coding. This system uses vectors of zeros and ones to indicate the categorical variables in a way that all | Regression variable has no meaning for one category
This will happen naturally, with no intervention on your part.
Consider, for instance, dummy coding. This system uses vectors of zeros and ones to indicate the categorical variables in a way that allows straightforward interpretation of the coefficients. A variable ... | Regression variable has no meaning for one category
This will happen naturally, with no intervention on your part.
Consider, for instance, dummy coding. This system uses vectors of zeros and ones to indicate the categorical variables in a way that all |
39,038 | Regression variable has no meaning for one category | If I understand this, what you are saying is that when there is only one person, the question of whether communication exists is meaningless.
You can solve this issue by making the two IVs into one:
One person
Two people, no communication
Two people, communication
Three+ people, communication
Three+ people, no comm... | Regression variable has no meaning for one category | If I understand this, what you are saying is that when there is only one person, the question of whether communication exists is meaningless.
You can solve this issue by making the two IVs into one: | Regression variable has no meaning for one category
If I understand this, what you are saying is that when there is only one person, the question of whether communication exists is meaningless.
You can solve this issue by making the two IVs into one:
One person
Two people, no communication
Two people, communication
... | Regression variable has no meaning for one category
If I understand this, what you are saying is that when there is only one person, the question of whether communication exists is meaningless.
You can solve this issue by making the two IVs into one: |
39,039 | What's the Kendall Tau's distance between these 2 rankings? | The Kendall tau distance in your case is, indeed, 1.
See the following python code:
import itertools
def kendallTau(A, B):
pairs = itertools.combinations(range(0, len(A)), 2)
distance = 0
for x, y in pairs:
a = A[x] - A[y]
b = B[x] - B[y]
# if discordant (different signs)
... | What's the Kendall Tau's distance between these 2 rankings? | The Kendall tau distance in your case is, indeed, 1.
See the following python code:
import itertools
def kendallTau(A, B):
pairs = itertools.combinations(range(0, len(A)), 2)
distance = 0
| What's the Kendall Tau's distance between these 2 rankings?
The Kendall tau distance in your case is, indeed, 1.
See the following python code:
import itertools
def kendallTau(A, B):
pairs = itertools.combinations(range(0, len(A)), 2)
distance = 0
for x, y in pairs:
a = A[x] - A[y]
b = B[... | What's the Kendall Tau's distance between these 2 rankings?
The Kendall tau distance in your case is, indeed, 1.
See the following python code:
import itertools
def kendallTau(A, B):
pairs = itertools.combinations(range(0, len(A)), 2)
distance = 0
|
39,040 | What's the Kendall Tau's distance between these 2 rankings? | The wrong answer got accepted! The correct answer is 3.
From Wikipedia:
Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort algorithm would take to place one list in the same order as the other list.
import itertools
def kendall_tau_distance(orde... | What's the Kendall Tau's distance between these 2 rankings? | The wrong answer got accepted! The correct answer is 3.
From Wikipedia:
Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort alg | What's the Kendall Tau's distance between these 2 rankings?
The wrong answer got accepted! The correct answer is 3.
From Wikipedia:
Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort algorithm would take to place one list in the same order as the... | What's the Kendall Tau's distance between these 2 rankings?
The wrong answer got accepted! The correct answer is 3.
From Wikipedia:
Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort alg |
39,041 | What's the Kendall Tau's distance between these 2 rankings? | The Kendall tau distance in this instance is 3. It is also known as Kemeny distance. See here and here.
In some fields rankings are also allowed to have ties, therefore the Kemeny could be considered as a distance of 6 instead of 3. That's a confusion that arises quite often. But in your situation it is 3 because ties ... | What's the Kendall Tau's distance between these 2 rankings? | The Kendall tau distance in this instance is 3. It is also known as Kemeny distance. See here and here.
In some fields rankings are also allowed to have ties, therefore the Kemeny could be considered | What's the Kendall Tau's distance between these 2 rankings?
The Kendall tau distance in this instance is 3. It is also known as Kemeny distance. See here and here.
In some fields rankings are also allowed to have ties, therefore the Kemeny could be considered as a distance of 6 instead of 3. That's a confusion that ari... | What's the Kendall Tau's distance between these 2 rankings?
The Kendall tau distance in this instance is 3. It is also known as Kemeny distance. See here and here.
In some fields rankings are also allowed to have ties, therefore the Kemeny could be considered |
39,042 | What's the Kendall Tau's distance between these 2 rankings? | Kendell tau distance is the number of swaps which needs to be done for making the two lists the same. It can also be considered as a variant of Insertion Sort, where each swap adds +1 to Kendell distance. | What's the Kendall Tau's distance between these 2 rankings? | Kendell tau distance is the number of swaps which needs to be done for making the two lists the same. It can also be considered as a variant of Insertion Sort, where each swap adds +1 to Kendell dista | What's the Kendall Tau's distance between these 2 rankings?
Kendell tau distance is the number of swaps which needs to be done for making the two lists the same. It can also be considered as a variant of Insertion Sort, where each swap adds +1 to Kendell distance. | What's the Kendall Tau's distance between these 2 rankings?
Kendell tau distance is the number of swaps which needs to be done for making the two lists the same. It can also be considered as a variant of Insertion Sort, where each swap adds +1 to Kendell dista |
39,043 | Interpretation of $\mathbf{y}^T(\mathbf{I}-\mathbf{H})\mathbf{y}$ in OLS | It's the sum of squared residuals. As you say (no boldface in my notation), $(I-H)y=r$ gives the residuals, as we subtract the fitted values $Hy$ from the dependent variable $y$. Also, $I-H$ is symmetric and idempotent, i.e., $(I-H)=(I-H)'$ and $(I-H)^2=I-H$.
This follows from $H=X(X'X)^{-1}X'$, which is symmetric and... | Interpretation of $\mathbf{y}^T(\mathbf{I}-\mathbf{H})\mathbf{y}$ in OLS | It's the sum of squared residuals. As you say (no boldface in my notation), $(I-H)y=r$ gives the residuals, as we subtract the fitted values $Hy$ from the dependent variable $y$. Also, $I-H$ is symmet | Interpretation of $\mathbf{y}^T(\mathbf{I}-\mathbf{H})\mathbf{y}$ in OLS
It's the sum of squared residuals. As you say (no boldface in my notation), $(I-H)y=r$ gives the residuals, as we subtract the fitted values $Hy$ from the dependent variable $y$. Also, $I-H$ is symmetric and idempotent, i.e., $(I-H)=(I-H)'$ and $(... | Interpretation of $\mathbf{y}^T(\mathbf{I}-\mathbf{H})\mathbf{y}$ in OLS
It's the sum of squared residuals. As you say (no boldface in my notation), $(I-H)y=r$ gives the residuals, as we subtract the fitted values $Hy$ from the dependent variable $y$. Also, $I-H$ is symmet |
39,044 | White's test for heteroskedasticity in R | The whites.htest() function implements White's test for heteroskedasticity for vector autoregressions (VAR). It requires a varest object as input. However, from your description it seems that your model is not a VAR (vector autoregression) but a simple linear model.
Hence, the model should be estimated by lm() as previ... | White's test for heteroskedasticity in R | The whites.htest() function implements White's test for heteroskedasticity for vector autoregressions (VAR). It requires a varest object as input. However, from your description it seems that your mod | White's test for heteroskedasticity in R
The whites.htest() function implements White's test for heteroskedasticity for vector autoregressions (VAR). It requires a varest object as input. However, from your description it seems that your model is not a VAR (vector autoregression) but a simple linear model.
Hence, the m... | White's test for heteroskedasticity in R
The whites.htest() function implements White's test for heteroskedasticity for vector autoregressions (VAR). It requires a varest object as input. However, from your description it seems that your mod |
39,045 | KL divergence between two univariate Poisson distributions | This is fairly straight-foward. The log ratio of the densities is equal to:
$$ \log\left (\frac {f_1}{f_2}\right)=x\log\left (\frac {\lambda_1}{\lambda_2}\right)+\lambda_2-\lambda_1$$
Then you take expectation of this expression wrt $ f_1$, which simply replaces $ x $ with its expectation (in this case). So you have... | KL divergence between two univariate Poisson distributions | This is fairly straight-foward. The log ratio of the densities is equal to:
$$ \log\left (\frac {f_1}{f_2}\right)=x\log\left (\frac {\lambda_1}{\lambda_2}\right)+\lambda_2-\lambda_1$$
Then you take e | KL divergence between two univariate Poisson distributions
This is fairly straight-foward. The log ratio of the densities is equal to:
$$ \log\left (\frac {f_1}{f_2}\right)=x\log\left (\frac {\lambda_1}{\lambda_2}\right)+\lambda_2-\lambda_1$$
Then you take expectation of this expression wrt $ f_1$, which simply replac... | KL divergence between two univariate Poisson distributions
This is fairly straight-foward. The log ratio of the densities is equal to:
$$ \log\left (\frac {f_1}{f_2}\right)=x\log\left (\frac {\lambda_1}{\lambda_2}\right)+\lambda_2-\lambda_1$$
Then you take e |
39,046 | Significance in beta regression and glm binomial | The binomial is for modeling Bernoulli variables (i.e., binary) or binomial variables (i.e., the number of successes from a certain number of independent trials). So this should not be applied to computed rates (successes divided by trials) directly but glm() wants you to supply a matrix with successes and failures. Co... | Significance in beta regression and glm binomial | The binomial is for modeling Bernoulli variables (i.e., binary) or binomial variables (i.e., the number of successes from a certain number of independent trials). So this should not be applied to comp | Significance in beta regression and glm binomial
The binomial is for modeling Bernoulli variables (i.e., binary) or binomial variables (i.e., the number of successes from a certain number of independent trials). So this should not be applied to computed rates (successes divided by trials) directly but glm() wants you t... | Significance in beta regression and glm binomial
The binomial is for modeling Bernoulli variables (i.e., binary) or binomial variables (i.e., the number of successes from a certain number of independent trials). So this should not be applied to comp |
39,047 | Should the fumble rate of NFL teams be a normal distribution? | Theoretically, a normal distribution has a nonzero probability of negative numbers. So that's right out. A normal also has fully continuous distribution, whereas fumble rates would be discrete or rational.
It could be very close, and good enough, for example, the sum of many binomials (had a fumble or didnt with x% ch... | Should the fumble rate of NFL teams be a normal distribution? | Theoretically, a normal distribution has a nonzero probability of negative numbers. So that's right out. A normal also has fully continuous distribution, whereas fumble rates would be discrete or rati | Should the fumble rate of NFL teams be a normal distribution?
Theoretically, a normal distribution has a nonzero probability of negative numbers. So that's right out. A normal also has fully continuous distribution, whereas fumble rates would be discrete or rational.
It could be very close, and good enough, for exampl... | Should the fumble rate of NFL teams be a normal distribution?
Theoretically, a normal distribution has a nonzero probability of negative numbers. So that's right out. A normal also has fully continuous distribution, whereas fumble rates would be discrete or rati |
39,048 | Should the fumble rate of NFL teams be a normal distribution? | would distribution of season fumbles (y) by NFL team (x) be a normal distribution?
Under no circumstances is a non-negative and discrete random variable actually normal.
In some circumstances (discreteness aside) it might not be terrible as an approximation, but it wouldn't be the first approximation I'd look at.
"... | Should the fumble rate of NFL teams be a normal distribution? | would distribution of season fumbles (y) by NFL team (x) be a normal distribution?
Under no circumstances is a non-negative and discrete random variable actually normal.
In some circumstances (disc | Should the fumble rate of NFL teams be a normal distribution?
would distribution of season fumbles (y) by NFL team (x) be a normal distribution?
Under no circumstances is a non-negative and discrete random variable actually normal.
In some circumstances (discreteness aside) it might not be terrible as an approximati... | Should the fumble rate of NFL teams be a normal distribution?
would distribution of season fumbles (y) by NFL team (x) be a normal distribution?
Under no circumstances is a non-negative and discrete random variable actually normal.
In some circumstances (disc |
39,049 | Which residuals to analyse when dependent variable is transformed? | For residual analysis, you should use the residuals obtained directly from the regression. No back-transformation is needed. This is because you want to make sure that your regression is valid (that it satisfies the underlying assumptions) which is sort of a "mechanical" issue, not subject-matter issue. Thus you look a... | Which residuals to analyse when dependent variable is transformed? | For residual analysis, you should use the residuals obtained directly from the regression. No back-transformation is needed. This is because you want to make sure that your regression is valid (that i | Which residuals to analyse when dependent variable is transformed?
For residual analysis, you should use the residuals obtained directly from the regression. No back-transformation is needed. This is because you want to make sure that your regression is valid (that it satisfies the underlying assumptions) which is sort... | Which residuals to analyse when dependent variable is transformed?
For residual analysis, you should use the residuals obtained directly from the regression. No back-transformation is needed. This is because you want to make sure that your regression is valid (that i |
39,050 | Which residuals to analyse when dependent variable is transformed? | You might be better off fitting a generalized linear model instead of a "plain" linear model, and analyzing the residuals of the GLM instead. This procedure and a few good reasons for doing so are laid out in this answer. GLMs have more than one kind of residual, but there is a large literature on analyzing them.
In ca... | Which residuals to analyse when dependent variable is transformed? | You might be better off fitting a generalized linear model instead of a "plain" linear model, and analyzing the residuals of the GLM instead. This procedure and a few good reasons for doing so are lai | Which residuals to analyse when dependent variable is transformed?
You might be better off fitting a generalized linear model instead of a "plain" linear model, and analyzing the residuals of the GLM instead. This procedure and a few good reasons for doing so are laid out in this answer. GLMs have more than one kind of... | Which residuals to analyse when dependent variable is transformed?
You might be better off fitting a generalized linear model instead of a "plain" linear model, and analyzing the residuals of the GLM instead. This procedure and a few good reasons for doing so are lai |
39,051 | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables | why can't I expand the $(X + Y)^2$ to get $E(X^2 + 2XY + Y^2)$
You can.
and then use linearity to get $E(X^2) + 2E(X)E(Y) + E(Y^2)$
You can
Then since $X$ and $Y$ are independent,
right ... but you already used independence in the previous step when you wrote $E(XY)$ as $E(X)\,E(Y)$.
this would give $E(X)^2 + 2(... | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables | why can't I expand the $(X + Y)^2$ to get $E(X^2 + 2XY + Y^2)$
You can.
and then use linearity to get $E(X^2) + 2E(X)E(Y) + E(Y^2)$
You can
Then since $X$ and $Y$ are independent,
right ... but | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables
why can't I expand the $(X + Y)^2$ to get $E(X^2 + 2XY + Y^2)$
You can.
and then use linearity to get $E(X^2) + 2E(X)E(Y) + E(Y^2)$
You can
Then since $X$ and $Y$ are independent,
right ... but you already used independence in t... | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables
why can't I expand the $(X + Y)^2$ to get $E(X^2 + 2XY + Y^2)$
You can.
and then use linearity to get $E(X^2) + 2E(X)E(Y) + E(Y^2)$
You can
Then since $X$ and $Y$ are independent,
right ... but |
39,052 | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables | You cannot say $E(X^2)=E(X)^2$ since $X$ is not independent of $X$.
For this problem you can set $Z=X_1+X_2 \sim \text{Poisson}$ with mean $2$. Than you can find $E(Z^2)$ easily. | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables | You cannot say $E(X^2)=E(X)^2$ since $X$ is not independent of $X$.
For this problem you can set $Z=X_1+X_2 \sim \text{Poisson}$ with mean $2$. Than you can find $E(Z^2)$ easily. | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables
You cannot say $E(X^2)=E(X)^2$ since $X$ is not independent of $X$.
For this problem you can set $Z=X_1+X_2 \sim \text{Poisson}$ with mean $2$. Than you can find $E(Z^2)$ easily. | Expectation of $(X + Y)^2$ where $X$ and $Y$ are independent Poisson random variables
You cannot say $E(X^2)=E(X)^2$ since $X$ is not independent of $X$.
For this problem you can set $Z=X_1+X_2 \sim \text{Poisson}$ with mean $2$. Than you can find $E(Z^2)$ easily. |
39,053 | In convolutional neural network, what does fully-connected layer mean? | Every neuron from the previous layer is connected to every neuron on the next layer1. | In convolutional neural network, what does fully-connected layer mean? | Every neuron from the previous layer is connected to every neuron on the next layer1. | In convolutional neural network, what does fully-connected layer mean?
Every neuron from the previous layer is connected to every neuron on the next layer1. | In convolutional neural network, what does fully-connected layer mean?
Every neuron from the previous layer is connected to every neuron on the next layer1. |
39,054 | In convolutional neural network, what does fully-connected layer mean? | The convolutional and the Pooling layers create a feature space, and the flatten FUlly connected layer can be thought as a cheap way of learning a linear function out of the feature space. For the convenience of understanding think of it as a PCA that selects the good features among the feature space created by the Con... | In convolutional neural network, what does fully-connected layer mean? | The convolutional and the Pooling layers create a feature space, and the flatten FUlly connected layer can be thought as a cheap way of learning a linear function out of the feature space. For the con | In convolutional neural network, what does fully-connected layer mean?
The convolutional and the Pooling layers create a feature space, and the flatten FUlly connected layer can be thought as a cheap way of learning a linear function out of the feature space. For the convenience of understanding think of it as a PCA th... | In convolutional neural network, what does fully-connected layer mean?
The convolutional and the Pooling layers create a feature space, and the flatten FUlly connected layer can be thought as a cheap way of learning a linear function out of the feature space. For the con |
39,055 | How do I enter a continuous variable as a random effect in a linear mixed effects model? | I'll elaborate on what I think Sergio meant in his comment.
A random effect is always associated with a categorical variable. This categorical variable will most often divide the observations into different observational units (this could for instance be Dam in your data set as it seems reasonable to assume that observ... | How do I enter a continuous variable as a random effect in a linear mixed effects model? | I'll elaborate on what I think Sergio meant in his comment.
A random effect is always associated with a categorical variable. This categorical variable will most often divide the observations into dif | How do I enter a continuous variable as a random effect in a linear mixed effects model?
I'll elaborate on what I think Sergio meant in his comment.
A random effect is always associated with a categorical variable. This categorical variable will most often divide the observations into different observational units (thi... | How do I enter a continuous variable as a random effect in a linear mixed effects model?
I'll elaborate on what I think Sergio meant in his comment.
A random effect is always associated with a categorical variable. This categorical variable will most often divide the observations into dif |
39,056 | Linearity between predictors and dependent variable in a linear model | To add to AdamO's answer, I was taught to base my decisions regarding model assumptions more on whether failing to correct the assumption in some way causes me to misrepresent my data. For a concrete example of what I mean, I simulated some data in R and created some plots and ran some diagnostics using these data.
# ... | Linearity between predictors and dependent variable in a linear model | To add to AdamO's answer, I was taught to base my decisions regarding model assumptions more on whether failing to correct the assumption in some way causes me to misrepresent my data. For a concrete | Linearity between predictors and dependent variable in a linear model
To add to AdamO's answer, I was taught to base my decisions regarding model assumptions more on whether failing to correct the assumption in some way causes me to misrepresent my data. For a concrete example of what I mean, I simulated some data in ... | Linearity between predictors and dependent variable in a linear model
To add to AdamO's answer, I was taught to base my decisions regarding model assumptions more on whether failing to correct the assumption in some way causes me to misrepresent my data. For a concrete |
39,057 | Linearity between predictors and dependent variable in a linear model | The emphasis upon so called assumptions in linear regression modeling is evidence that pedagogy in the applied fields is not in line with traditional statistical theory. In particular, the aforementioned "straight line assumption" (or non-existence of higher order effects) is completely erroneous depending upon your i... | Linearity between predictors and dependent variable in a linear model | The emphasis upon so called assumptions in linear regression modeling is evidence that pedagogy in the applied fields is not in line with traditional statistical theory. In particular, the aforementi | Linearity between predictors and dependent variable in a linear model
The emphasis upon so called assumptions in linear regression modeling is evidence that pedagogy in the applied fields is not in line with traditional statistical theory. In particular, the aforementioned "straight line assumption" (or non-existence ... | Linearity between predictors and dependent variable in a linear model
The emphasis upon so called assumptions in linear regression modeling is evidence that pedagogy in the applied fields is not in line with traditional statistical theory. In particular, the aforementi |
39,058 | Linear regression with constrained coefficient | General constrained OLS problem
Recall that the OLS problem, subject to linear constraints can be written as
$$
\begin{align}
\arg\min_{\boldsymbol{\beta}}\boldsymbol{Y}'\boldsymbol{Y} - \boldsymbol{Y}'\mathbf{X}\boldsymbol{\beta} - \boldsymbol{\beta}'\mathbf{X}'\boldsymbol{Y} + \boldsymbol{\beta}'\mathbf{X}'\mathbf{X}... | Linear regression with constrained coefficient | General constrained OLS problem
Recall that the OLS problem, subject to linear constraints can be written as
$$
\begin{align}
\arg\min_{\boldsymbol{\beta}}\boldsymbol{Y}'\boldsymbol{Y} - \boldsymbol{Y | Linear regression with constrained coefficient
General constrained OLS problem
Recall that the OLS problem, subject to linear constraints can be written as
$$
\begin{align}
\arg\min_{\boldsymbol{\beta}}\boldsymbol{Y}'\boldsymbol{Y} - \boldsymbol{Y}'\mathbf{X}\boldsymbol{\beta} - \boldsymbol{\beta}'\mathbf{X}'\boldsymbo... | Linear regression with constrained coefficient
General constrained OLS problem
Recall that the OLS problem, subject to linear constraints can be written as
$$
\begin{align}
\arg\min_{\boldsymbol{\beta}}\boldsymbol{Y}'\boldsymbol{Y} - \boldsymbol{Y |
39,059 | Linear regression with constrained coefficient | This problem can be formulated as a standard errors-in-variables problem.
Write:
\begin{equation}
y - x_2 = a(x_1-x_2) + c
\end{equation}
now call $z = y-x_2$ and $w = x_1 -x_2$. Then you have the following problem
\begin{equation}
z = aw + c
\end{equation}
Be careful, Error have changed and are now correlated.... | Linear regression with constrained coefficient | This problem can be formulated as a standard errors-in-variables problem.
Write:
\begin{equation}
y - x_2 = a(x_1-x_2) + c
\end{equation}
now call $z = y-x_2$ and $w = x_1 -x_2$. Then you have the | Linear regression with constrained coefficient
This problem can be formulated as a standard errors-in-variables problem.
Write:
\begin{equation}
y - x_2 = a(x_1-x_2) + c
\end{equation}
now call $z = y-x_2$ and $w = x_1 -x_2$. Then you have the following problem
\begin{equation}
z = aw + c
\end{equation}
Be care... | Linear regression with constrained coefficient
This problem can be formulated as a standard errors-in-variables problem.
Write:
\begin{equation}
y - x_2 = a(x_1-x_2) + c
\end{equation}
now call $z = y-x_2$ and $w = x_1 -x_2$. Then you have the |
39,060 | Linear regression with constrained coefficient | This is an old question, but it may help you.
You can use ConsReg package.
See the example below:
Imagine you want the following constraints in your parameters:
All coefficients will be less than 1 and greater than -1
$x_4 < 0.2$
The coefficient of $x_3$ and $x_3^2$ must satisfied: $(x_3 + x_3^2 > 0.01$)
Your can put... | Linear regression with constrained coefficient | This is an old question, but it may help you.
You can use ConsReg package.
See the example below:
Imagine you want the following constraints in your parameters:
All coefficients will be less than 1 a | Linear regression with constrained coefficient
This is an old question, but it may help you.
You can use ConsReg package.
See the example below:
Imagine you want the following constraints in your parameters:
All coefficients will be less than 1 and greater than -1
$x_4 < 0.2$
The coefficient of $x_3$ and $x_3^2$ must ... | Linear regression with constrained coefficient
This is an old question, but it may help you.
You can use ConsReg package.
See the example below:
Imagine you want the following constraints in your parameters:
All coefficients will be less than 1 a |
39,061 | Main idea of Bagging | Bootstrapping is a concept in statistics of approximating the sampling distribution of a statistic by repeatedly sampling from a given sample of size $n$. We construct $B$ samples, each of size $n$, by sampling with replacement from the original sample. The statistic of interest is calculated for each of the $B$ sample... | Main idea of Bagging | Bootstrapping is a concept in statistics of approximating the sampling distribution of a statistic by repeatedly sampling from a given sample of size $n$. We construct $B$ samples, each of size $n$, b | Main idea of Bagging
Bootstrapping is a concept in statistics of approximating the sampling distribution of a statistic by repeatedly sampling from a given sample of size $n$. We construct $B$ samples, each of size $n$, by sampling with replacement from the original sample. The statistic of interest is calculated for e... | Main idea of Bagging
Bootstrapping is a concept in statistics of approximating the sampling distribution of a statistic by repeatedly sampling from a given sample of size $n$. We construct $B$ samples, each of size $n$, b |
39,062 | Model fitting when errors take a Cauchy distribution | The least squares estimates for the regression coefficients are only equal to the maximum-likelihood estimates when the errors have a normal distribution (see here for the proof).
If you really wanted maximum likelihood estimates for regression parameters with Cauchy errors, just look at that likelihood:
$$L(\beta,\sig... | Model fitting when errors take a Cauchy distribution | The least squares estimates for the regression coefficients are only equal to the maximum-likelihood estimates when the errors have a normal distribution (see here for the proof).
If you really wanted | Model fitting when errors take a Cauchy distribution
The least squares estimates for the regression coefficients are only equal to the maximum-likelihood estimates when the errors have a normal distribution (see here for the proof).
If you really wanted maximum likelihood estimates for regression parameters with Cauchy... | Model fitting when errors take a Cauchy distribution
The least squares estimates for the regression coefficients are only equal to the maximum-likelihood estimates when the errors have a normal distribution (see here for the proof).
If you really wanted |
39,063 | Model fitting when errors take a Cauchy distribution | The Jeffreys (posterior) distribution is quite nice to do inference in a linear regression model with location-scale errors. Inference based on the Jeffreys distribution achieves very good frequentist properties: it provides confidence intervals whose coverage is close to the nominal coverage, even for very small sampl... | Model fitting when errors take a Cauchy distribution | The Jeffreys (posterior) distribution is quite nice to do inference in a linear regression model with location-scale errors. Inference based on the Jeffreys distribution achieves very good frequentist | Model fitting when errors take a Cauchy distribution
The Jeffreys (posterior) distribution is quite nice to do inference in a linear regression model with location-scale errors. Inference based on the Jeffreys distribution achieves very good frequentist properties: it provides confidence intervals whose coverage is clo... | Model fitting when errors take a Cauchy distribution
The Jeffreys (posterior) distribution is quite nice to do inference in a linear regression model with location-scale errors. Inference based on the Jeffreys distribution achieves very good frequentist |
39,064 | Model fitting when errors take a Cauchy distribution | GraphPad Prism can do nonlinear regression assuming a Cauchy distribution. That is our robust method. The mathematical details are explained in detail on pages 11-14 of BMC Bioinformatics 2006, 7:123 doi:10.1186/1471-2105-7-123, Detecting outliers when fitting data with nonlinear regression – a
new method based on rob... | Model fitting when errors take a Cauchy distribution | GraphPad Prism can do nonlinear regression assuming a Cauchy distribution. That is our robust method. The mathematical details are explained in detail on pages 11-14 of BMC Bioinformatics 2006, 7:123 | Model fitting when errors take a Cauchy distribution
GraphPad Prism can do nonlinear regression assuming a Cauchy distribution. That is our robust method. The mathematical details are explained in detail on pages 11-14 of BMC Bioinformatics 2006, 7:123 doi:10.1186/1471-2105-7-123, Detecting outliers when fitting data w... | Model fitting when errors take a Cauchy distribution
GraphPad Prism can do nonlinear regression assuming a Cauchy distribution. That is our robust method. The mathematical details are explained in detail on pages 11-14 of BMC Bioinformatics 2006, 7:123 |
39,065 | Model fitting when errors take a Cauchy distribution | A bit too late, but it may be useful for others in the future.
Scortchi wrote the likelihood as if the errors where independent and there is no simple answer in that case. To complement what he said in the end about OLS, note that if you instead, specify the joint likelihood with dependent errors, you will end up with ... | Model fitting when errors take a Cauchy distribution | A bit too late, but it may be useful for others in the future.
Scortchi wrote the likelihood as if the errors where independent and there is no simple answer in that case. To complement what he said i | Model fitting when errors take a Cauchy distribution
A bit too late, but it may be useful for others in the future.
Scortchi wrote the likelihood as if the errors where independent and there is no simple answer in that case. To complement what he said in the end about OLS, note that if you instead, specify the joint li... | Model fitting when errors take a Cauchy distribution
A bit too late, but it may be useful for others in the future.
Scortchi wrote the likelihood as if the errors where independent and there is no simple answer in that case. To complement what he said i |
39,066 | Model fitting when errors take a Cauchy distribution | The use of Cauchy errors IS NOT a robust method. It leads to a model that can capture outliers, but if there are no outliers, then the resulting model becomes very restrictive since it is being assumed that the distribution of the errors is heavy tailed with a specific tail behaviour. Because the Cauchy distribution is... | Model fitting when errors take a Cauchy distribution | The use of Cauchy errors IS NOT a robust method. It leads to a model that can capture outliers, but if there are no outliers, then the resulting model becomes very restrictive since it is being assume | Model fitting when errors take a Cauchy distribution
The use of Cauchy errors IS NOT a robust method. It leads to a model that can capture outliers, but if there are no outliers, then the resulting model becomes very restrictive since it is being assumed that the distribution of the errors is heavy tailed with a specif... | Model fitting when errors take a Cauchy distribution
The use of Cauchy errors IS NOT a robust method. It leads to a model that can capture outliers, but if there are no outliers, then the resulting model becomes very restrictive since it is being assume |
39,067 | coxph ran out of iterations and did not converge | This may be a case where, as the coxph() documentation page puts it, "the actual MLE estimate of a coefficient is infinity" so that "the associated coefficient grows at a steady pace and a race condition will exist in the fitting routine." In particular, close interrelations of the start / end times with the total_usag... | coxph ran out of iterations and did not converge | This may be a case where, as the coxph() documentation page puts it, "the actual MLE estimate of a coefficient is infinity" so that "the associated coefficient grows at a steady pace and a race condit | coxph ran out of iterations and did not converge
This may be a case where, as the coxph() documentation page puts it, "the actual MLE estimate of a coefficient is infinity" so that "the associated coefficient grows at a steady pace and a race condition will exist in the fitting routine." In particular, close interrelat... | coxph ran out of iterations and did not converge
This may be a case where, as the coxph() documentation page puts it, "the actual MLE estimate of a coefficient is infinity" so that "the associated coefficient grows at a steady pace and a race condit |
39,068 | coxph ran out of iterations and did not converge | The result:
f_cox
Call:
coxph(formula = f_sur ~ total_usage + cluster(fault_id), data = ff_usage)
coef exp(coef) se(coef) robust se z p
total_usage -0.00407 0.996 0 0 -Inf 0
This is the warning message. Notice it is not a failure to converge:
Warning message:
In fitter(X, Y, str... | coxph ran out of iterations and did not converge | The result:
f_cox
Call:
coxph(formula = f_sur ~ total_usage + cluster(fault_id), data = ff_usage)
coef exp(coef) se(coef) robust se z p
total_usage -0.00407 0.996 0 | coxph ran out of iterations and did not converge
The result:
f_cox
Call:
coxph(formula = f_sur ~ total_usage + cluster(fault_id), data = ff_usage)
coef exp(coef) se(coef) robust se z p
total_usage -0.00407 0.996 0 0 -Inf 0
This is the warning message. Notice it is not a failure ... | coxph ran out of iterations and did not converge
The result:
f_cox
Call:
coxph(formula = f_sur ~ total_usage + cluster(fault_id), data = ff_usage)
coef exp(coef) se(coef) robust se z p
total_usage -0.00407 0.996 0 |
39,069 | How to determine trend strength from linear regression slope? | A couple notes: Usually people write the equation as either $y = b_0 + b_1X$ or $y = a + bx$ Your version is OK, but might be confusing when you see other versions.
In your equation, $a$ is a measure of how much $f(x)$ is expected to rise for a 1 unit increase in $x$. If $a$ is positive, then $f(x)$ is expected to rise... | How to determine trend strength from linear regression slope? | A couple notes: Usually people write the equation as either $y = b_0 + b_1X$ or $y = a + bx$ Your version is OK, but might be confusing when you see other versions.
In your equation, $a$ is a measure | How to determine trend strength from linear regression slope?
A couple notes: Usually people write the equation as either $y = b_0 + b_1X$ or $y = a + bx$ Your version is OK, but might be confusing when you see other versions.
In your equation, $a$ is a measure of how much $f(x)$ is expected to rise for a 1 unit increa... | How to determine trend strength from linear regression slope?
A couple notes: Usually people write the equation as either $y = b_0 + b_1X$ or $y = a + bx$ Your version is OK, but might be confusing when you see other versions.
In your equation, $a$ is a measure |
39,070 | How to determine trend strength from linear regression slope? | R^2 is a scaled measure of the error in the fit. Here is some more information on it.
http://mathworld.wolfram.com/CorrelationCoefficient.html
Although R^2 is useful, there is no perfect measure. They each have strengths and weaknesses. I find that I use measures like the Akaike Information Criterion much more becau... | How to determine trend strength from linear regression slope? | R^2 is a scaled measure of the error in the fit. Here is some more information on it.
http://mathworld.wolfram.com/CorrelationCoefficient.html
Although R^2 is useful, there is no perfect measure. Th | How to determine trend strength from linear regression slope?
R^2 is a scaled measure of the error in the fit. Here is some more information on it.
http://mathworld.wolfram.com/CorrelationCoefficient.html
Although R^2 is useful, there is no perfect measure. They each have strengths and weaknesses. I find that I use ... | How to determine trend strength from linear regression slope?
R^2 is a scaled measure of the error in the fit. Here is some more information on it.
http://mathworld.wolfram.com/CorrelationCoefficient.html
Although R^2 is useful, there is no perfect measure. Th |
39,071 | How to determine trend strength from linear regression slope? | One way to make comparisons on the relative "strength" of 2 independent variables in the same linear model is to divide them by their respective standard deviations (i.e. use z-scores). It's still not an apples to apples comparison, but it can be useful to see if, for example, a 1 SD increase in X results in a greater ... | How to determine trend strength from linear regression slope? | One way to make comparisons on the relative "strength" of 2 independent variables in the same linear model is to divide them by their respective standard deviations (i.e. use z-scores). It's still not | How to determine trend strength from linear regression slope?
One way to make comparisons on the relative "strength" of 2 independent variables in the same linear model is to divide them by their respective standard deviations (i.e. use z-scores). It's still not an apples to apples comparison, but it can be useful to s... | How to determine trend strength from linear regression slope?
One way to make comparisons on the relative "strength" of 2 independent variables in the same linear model is to divide them by their respective standard deviations (i.e. use z-scores). It's still not |
39,072 | How to interpret a logistic regression model with all negative coefficient? | Yes, it is possible. Couple of things here. The direction of your predictors are critical to the interpretation; if they were scaled in the opposite direction, they would be positive. Second, it would seem on the face that your predictors lower the log odds given a unit change, which in itself might be good if say the ... | How to interpret a logistic regression model with all negative coefficient? | Yes, it is possible. Couple of things here. The direction of your predictors are critical to the interpretation; if they were scaled in the opposite direction, they would be positive. Second, it would | How to interpret a logistic regression model with all negative coefficient?
Yes, it is possible. Couple of things here. The direction of your predictors are critical to the interpretation; if they were scaled in the opposite direction, they would be positive. Second, it would seem on the face that your predictors lower... | How to interpret a logistic regression model with all negative coefficient?
Yes, it is possible. Couple of things here. The direction of your predictors are critical to the interpretation; if they were scaled in the opposite direction, they would be positive. Second, it would |
39,073 | How to interpret a logistic regression model with all negative coefficient? | From your description I see nothing out of the ordinary.
That the intercept is negative corresponds to that the estimated probability of the response is less than 50% when all model covariates equal zero.
If the coefficients of the model covariates are negative, then yes, the corresponding odds ratios are smaller tha... | How to interpret a logistic regression model with all negative coefficient? | From your description I see nothing out of the ordinary.
That the intercept is negative corresponds to that the estimated probability of the response is less than 50% when all model covariates equal | How to interpret a logistic regression model with all negative coefficient?
From your description I see nothing out of the ordinary.
That the intercept is negative corresponds to that the estimated probability of the response is less than 50% when all model covariates equal zero.
If the coefficients of the model cova... | How to interpret a logistic regression model with all negative coefficient?
From your description I see nothing out of the ordinary.
That the intercept is negative corresponds to that the estimated probability of the response is less than 50% when all model covariates equal |
39,074 | How to interpret a logistic regression model with all negative coefficient? | I didn't really get logistic regression until I thought about it this way:
Picture the S curve (the logistic) going from -3 to 3
Look at the coefficient estimate on your constant term. Mark it on the x axis of the S curve.
Each coefficient moves you $\beta$ units along the X axis of the S curve. If you want to know ... | How to interpret a logistic regression model with all negative coefficient? | I didn't really get logistic regression until I thought about it this way:
Picture the S curve (the logistic) going from -3 to 3
Look at the coefficient estimate on your constant term. Mark it on th | How to interpret a logistic regression model with all negative coefficient?
I didn't really get logistic regression until I thought about it this way:
Picture the S curve (the logistic) going from -3 to 3
Look at the coefficient estimate on your constant term. Mark it on the x axis of the S curve.
Each coefficient mo... | How to interpret a logistic regression model with all negative coefficient?
I didn't really get logistic regression until I thought about it this way:
Picture the S curve (the logistic) going from -3 to 3
Look at the coefficient estimate on your constant term. Mark it on th |
39,075 | ANOVA results do not match post-hoc Tukey test, how to proceed? | This is not an area where there is universal agreement.
My view is that 1) the two tests answer different questions, so it's not surprising that they get different answers. 2) This discrepancy is more a demonstration of the problems of p values and, especially, using cutoff values like p < .05. 3) It also gets at the p... | ANOVA results do not match post-hoc Tukey test, how to proceed? | This is not an area where there is universal agreement.
My view is that 1) the two tests answer different questions, so it's not surprising that they get different answers. 2) This discrepancy is more | ANOVA results do not match post-hoc Tukey test, how to proceed?
This is not an area where there is universal agreement.
My view is that 1) the two tests answer different questions, so it's not surprising that they get different answers. 2) This discrepancy is more a demonstration of the problems of p values and, especi... | ANOVA results do not match post-hoc Tukey test, how to proceed?
This is not an area where there is universal agreement.
My view is that 1) the two tests answer different questions, so it's not surprising that they get different answers. 2) This discrepancy is more |
39,076 | ANOVA results do not match post-hoc Tukey test, how to proceed? | It's possible that you could have no significant differences among all of the individual means with the most liberal of tests—even a planned comparison. I was just tasking my students to develop simulated data with just this feature.
You do an ANOVA to test the pattern of your data. If the test is passed then, in mos... | ANOVA results do not match post-hoc Tukey test, how to proceed? | It's possible that you could have no significant differences among all of the individual means with the most liberal of tests—even a planned comparison. I was just tasking my students to develop simu | ANOVA results do not match post-hoc Tukey test, how to proceed?
It's possible that you could have no significant differences among all of the individual means with the most liberal of tests—even a planned comparison. I was just tasking my students to develop simulated data with just this feature.
You do an ANOVA to te... | ANOVA results do not match post-hoc Tukey test, how to proceed?
It's possible that you could have no significant differences among all of the individual means with the most liberal of tests—even a planned comparison. I was just tasking my students to develop simu |
39,077 | ANOVA results do not match post-hoc Tukey test, how to proceed? | I used SPSS and I used to have the same problem and I have tried different tests in the PostHoc. For equal variance assumed, I suggest you use Dunnet test in which you can have different results if you change the selection in Control category (First or Last) and sometimes in the Test (2-sided, < Control, > Control). Fo... | ANOVA results do not match post-hoc Tukey test, how to proceed? | I used SPSS and I used to have the same problem and I have tried different tests in the PostHoc. For equal variance assumed, I suggest you use Dunnet test in which you can have different results if yo | ANOVA results do not match post-hoc Tukey test, how to proceed?
I used SPSS and I used to have the same problem and I have tried different tests in the PostHoc. For equal variance assumed, I suggest you use Dunnet test in which you can have different results if you change the selection in Control category (First or Las... | ANOVA results do not match post-hoc Tukey test, how to proceed?
I used SPSS and I used to have the same problem and I have tried different tests in the PostHoc. For equal variance assumed, I suggest you use Dunnet test in which you can have different results if yo |
39,078 | Multiple imputation with the Amelia package | Amelia assumes that the data follow a multivariate normal distribution, so all information about the relations in the data can be summarized by just means and covariances. When data are incomplete, Amelia uses the well-known EM algorithm to find corrected estimates of the means and covariances. See Little and Rubin (20... | Multiple imputation with the Amelia package | Amelia assumes that the data follow a multivariate normal distribution, so all information about the relations in the data can be summarized by just means and covariances. When data are incomplete, Am | Multiple imputation with the Amelia package
Amelia assumes that the data follow a multivariate normal distribution, so all information about the relations in the data can be summarized by just means and covariances. When data are incomplete, Amelia uses the well-known EM algorithm to find corrected estimates of the mea... | Multiple imputation with the Amelia package
Amelia assumes that the data follow a multivariate normal distribution, so all information about the relations in the data can be summarized by just means and covariances. When data are incomplete, Am |
39,079 | Probability distribution of income | Typically lognormal distributions or sometimes pareto distributions are used to model the distribution of income. Here you can find information how well these distrubtions fit real data for Germany, UK and the US:
http://ideas.repec.org/p/wpa/wuwpmi/0505006.html
Here is a proposal to use a generalized lognormal distrib... | Probability distribution of income | Typically lognormal distributions or sometimes pareto distributions are used to model the distribution of income. Here you can find information how well these distrubtions fit real data for Germany, U | Probability distribution of income
Typically lognormal distributions or sometimes pareto distributions are used to model the distribution of income. Here you can find information how well these distrubtions fit real data for Germany, UK and the US:
http://ideas.repec.org/p/wpa/wuwpmi/0505006.html
Here is a proposal to ... | Probability distribution of income
Typically lognormal distributions or sometimes pareto distributions are used to model the distribution of income. Here you can find information how well these distrubtions fit real data for Germany, U |
39,080 | Probability distribution of income | If you want to include heavy tails while maintaining most of the remaining features of the lognormal, might I suggest the log-Cauchy or, if you need finite moments, the log-Student? | Probability distribution of income | If you want to include heavy tails while maintaining most of the remaining features of the lognormal, might I suggest the log-Cauchy or, if you need finite moments, the log-Student? | Probability distribution of income
If you want to include heavy tails while maintaining most of the remaining features of the lognormal, might I suggest the log-Cauchy or, if you need finite moments, the log-Student? | Probability distribution of income
If you want to include heavy tails while maintaining most of the remaining features of the lognormal, might I suggest the log-Cauchy or, if you need finite moments, the log-Student? |
39,081 | Skills & coursework needed to be a data analyst | Aside from the technical skills, like R or SAS, SQL will be important, and a few other higher level skills, including:
Data Manipulation: To be able to analyse data, you will frequently have to spend some considerable time acquiring the data and manipulating it into a form which can be analysed. Many statisticians wil... | Skills & coursework needed to be a data analyst | Aside from the technical skills, like R or SAS, SQL will be important, and a few other higher level skills, including:
Data Manipulation: To be able to analyse data, you will frequently have to spend | Skills & coursework needed to be a data analyst
Aside from the technical skills, like R or SAS, SQL will be important, and a few other higher level skills, including:
Data Manipulation: To be able to analyse data, you will frequently have to spend some considerable time acquiring the data and manipulating it into a for... | Skills & coursework needed to be a data analyst
Aside from the technical skills, like R or SAS, SQL will be important, and a few other higher level skills, including:
Data Manipulation: To be able to analyse data, you will frequently have to spend |
39,082 | Skills & coursework needed to be a data analyst | SAS is important in the pharmaceutical industry but not necessarily in other disciplines. in business and marketing time series analysis and survey sampling are particularly important. Yes big pharma and business use SAS a lot but it is expensive and makes more economical sense with multiple users that you would find... | Skills & coursework needed to be a data analyst | SAS is important in the pharmaceutical industry but not necessarily in other disciplines. in business and marketing time series analysis and survey sampling are particularly important. Yes big pharm | Skills & coursework needed to be a data analyst
SAS is important in the pharmaceutical industry but not necessarily in other disciplines. in business and marketing time series analysis and survey sampling are particularly important. Yes big pharma and business use SAS a lot but it is expensive and makes more economic... | Skills & coursework needed to be a data analyst
SAS is important in the pharmaceutical industry but not necessarily in other disciplines. in business and marketing time series analysis and survey sampling are particularly important. Yes big pharm |
39,083 | Skills & coursework needed to be a data analyst | I would recommend that you take some microeconometrics courses. Business analysts spend an inordinate amounts of time patching up analyses that don't make much sense once you've done this. One common example is running regressions of revenue or profit on price and believing that these estimate some sort of demand elast... | Skills & coursework needed to be a data analyst | I would recommend that you take some microeconometrics courses. Business analysts spend an inordinate amounts of time patching up analyses that don't make much sense once you've done this. One common | Skills & coursework needed to be a data analyst
I would recommend that you take some microeconometrics courses. Business analysts spend an inordinate amounts of time patching up analyses that don't make much sense once you've done this. One common example is running regressions of revenue or profit on price and believi... | Skills & coursework needed to be a data analyst
I would recommend that you take some microeconometrics courses. Business analysts spend an inordinate amounts of time patching up analyses that don't make much sense once you've done this. One common |
39,084 | Skills & coursework needed to be a data analyst | In addition to the many great suggestions, I'd add that you should try to learn some soft/people/analysis skills and perhaps at least a bit about one field where you might want to apply your skills as well.
In the real world, no one hands you a clean set of well-documented data with a precise question to answer. You wi... | Skills & coursework needed to be a data analyst | In addition to the many great suggestions, I'd add that you should try to learn some soft/people/analysis skills and perhaps at least a bit about one field where you might want to apply your skills as | Skills & coursework needed to be a data analyst
In addition to the many great suggestions, I'd add that you should try to learn some soft/people/analysis skills and perhaps at least a bit about one field where you might want to apply your skills as well.
In the real world, no one hands you a clean set of well-documente... | Skills & coursework needed to be a data analyst
In addition to the many great suggestions, I'd add that you should try to learn some soft/people/analysis skills and perhaps at least a bit about one field where you might want to apply your skills as |
39,085 | Skills & coursework needed to be a data analyst | 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.
I recommend reading these two before doing anything. H... | Skills & coursework needed to be a data analyst | 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.
| Skills & coursework needed to be a data analyst
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.
I reco... | Skills & coursework needed to be a data analyst
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.
|
39,086 | Inter-rater statistic for skewed rankings | A measure that is low when highly skewed raters agree is actually highly desirable. Gwet's AC1 specifically assumes that chance agreement should be at most 50%, but if both raters vote +ve 90% of the time, Cohen and Fleiss/Scott says that chance agreement is 81% on the positives and 1% on the negatives for a total of ... | Inter-rater statistic for skewed rankings | A measure that is low when highly skewed raters agree is actually highly desirable. Gwet's AC1 specifically assumes that chance agreement should be at most 50%, but if both raters vote +ve 90% of the | Inter-rater statistic for skewed rankings
A measure that is low when highly skewed raters agree is actually highly desirable. Gwet's AC1 specifically assumes that chance agreement should be at most 50%, but if both raters vote +ve 90% of the time, Cohen and Fleiss/Scott says that chance agreement is 81% on the positiv... | Inter-rater statistic for skewed rankings
A measure that is low when highly skewed raters agree is actually highly desirable. Gwet's AC1 specifically assumes that chance agreement should be at most 50%, but if both raters vote +ve 90% of the |
39,087 | Inter-rater statistic for skewed rankings | Since skewness is a problem in your case, you might want to use the AC1 interrater reliability statistic proposed by Gwet (2001, 2002). See e.g. Gwet 2008. It is a "more robust chance-corrected statistic that consistently yields reliable results" as compared to $\kappa$.
The $\kappa$ statistics can be problematic, bec... | Inter-rater statistic for skewed rankings | Since skewness is a problem in your case, you might want to use the AC1 interrater reliability statistic proposed by Gwet (2001, 2002). See e.g. Gwet 2008. It is a "more robust chance-corrected statis | Inter-rater statistic for skewed rankings
Since skewness is a problem in your case, you might want to use the AC1 interrater reliability statistic proposed by Gwet (2001, 2002). See e.g. Gwet 2008. It is a "more robust chance-corrected statistic that consistently yields reliable results" as compared to $\kappa$.
The $... | Inter-rater statistic for skewed rankings
Since skewness is a problem in your case, you might want to use the AC1 interrater reliability statistic proposed by Gwet (2001, 2002). See e.g. Gwet 2008. It is a "more robust chance-corrected statis |
39,088 | Inter-rater statistic for skewed rankings | I think most of them test concordance versus discordance and so they stress the degree with which the raters agree and so the fact that they will tend to vote yes a 10% of the time is not a factor. Sample size could be though because if the sample size is small you won't have many yeses to compare among the voters. T... | Inter-rater statistic for skewed rankings | I think most of them test concordance versus discordance and so they stress the degree with which the raters agree and so the fact that they will tend to vote yes a 10% of the time is not a factor. S | Inter-rater statistic for skewed rankings
I think most of them test concordance versus discordance and so they stress the degree with which the raters agree and so the fact that they will tend to vote yes a 10% of the time is not a factor. Sample size could be though because if the sample size is small you won't have ... | Inter-rater statistic for skewed rankings
I think most of them test concordance versus discordance and so they stress the degree with which the raters agree and so the fact that they will tend to vote yes a 10% of the time is not a factor. S |
39,089 | Evaluating recommender systems with (implicit) binary ratings only | I suggest to use Expected Utility or R-score. Assume your model has created an ordered list of recommendations where the first item is the one the user is most likely and the last is the one the user is least likely interested in. Let's say that this recommendations are specified by $r_i$ where $i$ is the position in t... | Evaluating recommender systems with (implicit) binary ratings only | I suggest to use Expected Utility or R-score. Assume your model has created an ordered list of recommendations where the first item is the one the user is most likely and the last is the one the user | Evaluating recommender systems with (implicit) binary ratings only
I suggest to use Expected Utility or R-score. Assume your model has created an ordered list of recommendations where the first item is the one the user is most likely and the last is the one the user is least likely interested in. Let's say that this re... | Evaluating recommender systems with (implicit) binary ratings only
I suggest to use Expected Utility or R-score. Assume your model has created an ordered list of recommendations where the first item is the one the user is most likely and the last is the one the user |
39,090 | Association or relationship | Under the definitions you've listed, "association" and "relationship" would not be interchangeable. However, I would argue that a better use of the term "relationship" would make them fairly synonymous in this application. I think that your teacher was making an important, and correct, point about correlation and reg... | Association or relationship | Under the definitions you've listed, "association" and "relationship" would not be interchangeable. However, I would argue that a better use of the term "relationship" would make them fairly synonymo | Association or relationship
Under the definitions you've listed, "association" and "relationship" would not be interchangeable. However, I would argue that a better use of the term "relationship" would make them fairly synonymous in this application. I think that your teacher was making an important, and correct, poi... | Association or relationship
Under the definitions you've listed, "association" and "relationship" would not be interchangeable. However, I would argue that a better use of the term "relationship" would make them fairly synonymo |
39,091 | Association or relationship | Yes. You are basically correct. Regression is used when you want to show how a dependent variable $Y$ is related to one or more independent variables. When we refer to correlation we are taking about an association. Regression is often used to predict future responses for $y$ based on given values for $x$. In leas... | Association or relationship | Yes. You are basically correct. Regression is used when you want to show how a dependent variable $Y$ is related to one or more independent variables. When we refer to correlation we are taking abo | Association or relationship
Yes. You are basically correct. Regression is used when you want to show how a dependent variable $Y$ is related to one or more independent variables. When we refer to correlation we are taking about an association. Regression is often used to predict future responses for $y$ based on gi... | Association or relationship
Yes. You are basically correct. Regression is used when you want to show how a dependent variable $Y$ is related to one or more independent variables. When we refer to correlation we are taking abo |
39,092 | How would you fit ARIMA model with lots of autocorrelations? | I think the issue here is whether a hypothesis test of the residuals is appropriate. You have 60000 observations, so any model will fail a residual test as there is so much data. That doesn't make the model bad, it just means that you have enough data to be able to tell that the model is an inaccurate representation of... | How would you fit ARIMA model with lots of autocorrelations? | I think the issue here is whether a hypothesis test of the residuals is appropriate. You have 60000 observations, so any model will fail a residual test as there is so much data. That doesn't make the | How would you fit ARIMA model with lots of autocorrelations?
I think the issue here is whether a hypothesis test of the residuals is appropriate. You have 60000 observations, so any model will fail a residual test as there is so much data. That doesn't make the model bad, it just means that you have enough data to be a... | How would you fit ARIMA model with lots of autocorrelations?
I think the issue here is whether a hypothesis test of the residuals is appropriate. You have 60000 observations, so any model will fail a residual test as there is so much data. That doesn't make the |
39,093 | How would you fit ARIMA model with lots of autocorrelations? | We are working with data like this for a major fast food franchise. The series represents the demand for tacos in 15 minute intervals for the last 5 years (180,000 observations) . This series can be treated by building 96 separate models (4x24) for each 15 minute interval a daily model reflecting overall trends,level s... | How would you fit ARIMA model with lots of autocorrelations? | We are working with data like this for a major fast food franchise. The series represents the demand for tacos in 15 minute intervals for the last 5 years (180,000 observations) . This series can be t | How would you fit ARIMA model with lots of autocorrelations?
We are working with data like this for a major fast food franchise. The series represents the demand for tacos in 15 minute intervals for the last 5 years (180,000 observations) . This series can be treated by building 96 separate models (4x24) for each 15 mi... | How would you fit ARIMA model with lots of autocorrelations?
We are working with data like this for a major fast food franchise. The series represents the demand for tacos in 15 minute intervals for the last 5 years (180,000 observations) . This series can be t |
39,094 | Detect outliers in mixture of Gaussians | I have suggested, in comments, that an "outlier" in this situation might be defined as a member of a "small" cluster centered at an "extreme" value. The meanings of the quoted terms need to be quantified, but apparently they can be: "small" would be a cluster of less than 10 values and "extreme" can be determined as o... | Detect outliers in mixture of Gaussians | I have suggested, in comments, that an "outlier" in this situation might be defined as a member of a "small" cluster centered at an "extreme" value. The meanings of the quoted terms need to be quanti | Detect outliers in mixture of Gaussians
I have suggested, in comments, that an "outlier" in this situation might be defined as a member of a "small" cluster centered at an "extreme" value. The meanings of the quoted terms need to be quantified, but apparently they can be: "small" would be a cluster of less than 10 val... | Detect outliers in mixture of Gaussians
I have suggested, in comments, that an "outlier" in this situation might be defined as a member of a "small" cluster centered at an "extreme" value. The meanings of the quoted terms need to be quanti |
39,095 | Detect outliers in mixture of Gaussians | I'm not sure I understand the issue here, but the MAD-Median rule:
$\frac{|X-M|}{MADN}>2.24$, where $M$ is the median and $MADN$ is the $\frac{\text{median absolute deviation from the median}}{0.6745}$
is pretty commonly used. Wilcox's WRS package in R has an out() function that fits this and returns the cases to keep... | Detect outliers in mixture of Gaussians | I'm not sure I understand the issue here, but the MAD-Median rule:
$\frac{|X-M|}{MADN}>2.24$, where $M$ is the median and $MADN$ is the $\frac{\text{median absolute deviation from the median}}{0.6745} | Detect outliers in mixture of Gaussians
I'm not sure I understand the issue here, but the MAD-Median rule:
$\frac{|X-M|}{MADN}>2.24$, where $M$ is the median and $MADN$ is the $\frac{\text{median absolute deviation from the median}}{0.6745}$
is pretty commonly used. Wilcox's WRS package in R has an out() function that... | Detect outliers in mixture of Gaussians
I'm not sure I understand the issue here, but the MAD-Median rule:
$\frac{|X-M|}{MADN}>2.24$, where $M$ is the median and $MADN$ is the $\frac{\text{median absolute deviation from the median}}{0.6745} |
39,096 | Detect outliers in mixture of Gaussians | If your range of possible distributions of non-outliers is so broad, I don't think you can have any outliers. But perhaps you can impose some restrictions on the mixture?
For example, if N = 10,000 and it's a mixture of $\mathcal{N}~(9900, 10, 10)$ and $\mathcal{N}~(100, 50, 100)$ then some very large values would be n... | Detect outliers in mixture of Gaussians | If your range of possible distributions of non-outliers is so broad, I don't think you can have any outliers. But perhaps you can impose some restrictions on the mixture?
For example, if N = 10,000 an | Detect outliers in mixture of Gaussians
If your range of possible distributions of non-outliers is so broad, I don't think you can have any outliers. But perhaps you can impose some restrictions on the mixture?
For example, if N = 10,000 and it's a mixture of $\mathcal{N}~(9900, 10, 10)$ and $\mathcal{N}~(100, 50, 100)... | Detect outliers in mixture of Gaussians
If your range of possible distributions of non-outliers is so broad, I don't think you can have any outliers. But perhaps you can impose some restrictions on the mixture?
For example, if N = 10,000 an |
39,097 | Detect outliers in mixture of Gaussians | The most elegant solution I can think of is a mixture of Gaussians model, in which you have k Gaussians corresponding to your signal (with a prior encouraging their variances to be reasonably small), and 1 diffuse Gaussian capturing the outliers ("diffuse" means huge variance), where you specify the prior proportion of... | Detect outliers in mixture of Gaussians | The most elegant solution I can think of is a mixture of Gaussians model, in which you have k Gaussians corresponding to your signal (with a prior encouraging their variances to be reasonably small), | Detect outliers in mixture of Gaussians
The most elegant solution I can think of is a mixture of Gaussians model, in which you have k Gaussians corresponding to your signal (with a prior encouraging their variances to be reasonably small), and 1 diffuse Gaussian capturing the outliers ("diffuse" means huge variance), w... | Detect outliers in mixture of Gaussians
The most elegant solution I can think of is a mixture of Gaussians model, in which you have k Gaussians corresponding to your signal (with a prior encouraging their variances to be reasonably small), |
39,098 | Joint distribution of two gamma random variables | As stated the problem does not make sense, because a joint distribution cannot be found from the marginal distributions! The only meaningful case (as an homework) is to assume independence. In which case the density of the joint distribution is obviously the product of both densities... | Joint distribution of two gamma random variables | As stated the problem does not make sense, because a joint distribution cannot be found from the marginal distributions! The only meaningful case (as an homework) is to assume independence. In which c | Joint distribution of two gamma random variables
As stated the problem does not make sense, because a joint distribution cannot be found from the marginal distributions! The only meaningful case (as an homework) is to assume independence. In which case the density of the joint distribution is obviously the product of b... | Joint distribution of two gamma random variables
As stated the problem does not make sense, because a joint distribution cannot be found from the marginal distributions! The only meaningful case (as an homework) is to assume independence. In which c |
39,099 | Joint distribution of two gamma random variables | OP notrockstar knows the solution for the case when the random variables are independent but presumably cannot use it since a solution without the
independence assumption is being sought. Perhaps the OP has posted only
a simplified version of the question, and what has been left out makes
a solution possible. For exa... | Joint distribution of two gamma random variables | OP notrockstar knows the solution for the case when the random variables are independent but presumably cannot use it since a solution without the
independence assumption is being sought. Perhaps the | Joint distribution of two gamma random variables
OP notrockstar knows the solution for the case when the random variables are independent but presumably cannot use it since a solution without the
independence assumption is being sought. Perhaps the OP has posted only
a simplified version of the question, and what has ... | Joint distribution of two gamma random variables
OP notrockstar knows the solution for the case when the random variables are independent but presumably cannot use it since a solution without the
independence assumption is being sought. Perhaps the |
39,100 | Kruskal-Wallis or Fligner test to check homogeneity of variances? | If I understand correctly, you have one predictor (explanatory variable $x$) and one criterion (predicted variable $y$) in a simple linear regression. The significance tests rests on the model assumption that for each observation $i$
$$
y_{i} = \beta_{0} + \beta_{1} x_{i} + \epsilon_{i}
$$
where $\beta_{0}, \beta_{1}$... | Kruskal-Wallis or Fligner test to check homogeneity of variances? | If I understand correctly, you have one predictor (explanatory variable $x$) and one criterion (predicted variable $y$) in a simple linear regression. The significance tests rests on the model assumpt | Kruskal-Wallis or Fligner test to check homogeneity of variances?
If I understand correctly, you have one predictor (explanatory variable $x$) and one criterion (predicted variable $y$) in a simple linear regression. The significance tests rests on the model assumption that for each observation $i$
$$
y_{i} = \beta_{0... | Kruskal-Wallis or Fligner test to check homogeneity of variances?
If I understand correctly, you have one predictor (explanatory variable $x$) and one criterion (predicted variable $y$) in a simple linear regression. The significance tests rests on the model assumpt |
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