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 |
|---|---|---|---|---|---|---|
53,101 | Riddler puzzle - distance from origin after two random jumps of equal length | That puzzle was asking for the mode, not the expected distance. Here is how I solved it.
Without loss of generality, assume that the cricket starts at (1,0) and jumps to the origin. The question is then about the density of distances from (1,0) to any point on the unit circle.
Let $\theta \sim \operatorname{U}(0, 1)$... | Riddler puzzle - distance from origin after two random jumps of equal length | That puzzle was asking for the mode, not the expected distance. Here is how I solved it.
Without loss of generality, assume that the cricket starts at (1,0) and jumps to the origin. The question is | Riddler puzzle - distance from origin after two random jumps of equal length
That puzzle was asking for the mode, not the expected distance. Here is how I solved it.
Without loss of generality, assume that the cricket starts at (1,0) and jumps to the origin. The question is then about the density of distances from (1... | Riddler puzzle - distance from origin after two random jumps of equal length
That puzzle was asking for the mode, not the expected distance. Here is how I solved it.
Without loss of generality, assume that the cricket starts at (1,0) and jumps to the origin. The question is |
53,102 | How to determine by what percent the target variable will change if we change a variable by some percent in Linear Regression? | Let's say we have fitted a model such as:
$$ y = x_1 + x_2 + \epsilon$$
and we obtained esimates giving the following equation:
$$ \hat{y} = 2x_1 + 3x_2 $$
Thus for a 1 unit change in $x_1$ we expect a change of 2 units of $y$.
It is not possible to get the expected % change in $y$ from a % change in $x$ unless the var... | How to determine by what percent the target variable will change if we change a variable by some per | Let's say we have fitted a model such as:
$$ y = x_1 + x_2 + \epsilon$$
and we obtained esimates giving the following equation:
$$ \hat{y} = 2x_1 + 3x_2 $$
Thus for a 1 unit change in $x_1$ we expect | How to determine by what percent the target variable will change if we change a variable by some percent in Linear Regression?
Let's say we have fitted a model such as:
$$ y = x_1 + x_2 + \epsilon$$
and we obtained esimates giving the following equation:
$$ \hat{y} = 2x_1 + 3x_2 $$
Thus for a 1 unit change in $x_1$ we ... | How to determine by what percent the target variable will change if we change a variable by some per
Let's say we have fitted a model such as:
$$ y = x_1 + x_2 + \epsilon$$
and we obtained esimates giving the following equation:
$$ \hat{y} = 2x_1 + 3x_2 $$
Thus for a 1 unit change in $x_1$ we expect |
53,103 | How to determine by what percent the target variable will change if we change a variable by some percent in Linear Regression? | We can back out the answer with a little auxiliary information about the covariates.
Your linear model is probably something like $$E[y \vert x, z] = \hat y= \hat \alpha + \hat \beta \cdot x +\hat \delta \cdot z.$$
What is the change in the expected value of $y$ associated with 1 unit change with $x$? We can easily get... | How to determine by what percent the target variable will change if we change a variable by some per | We can back out the answer with a little auxiliary information about the covariates.
Your linear model is probably something like $$E[y \vert x, z] = \hat y= \hat \alpha + \hat \beta \cdot x +\hat \de | How to determine by what percent the target variable will change if we change a variable by some percent in Linear Regression?
We can back out the answer with a little auxiliary information about the covariates.
Your linear model is probably something like $$E[y \vert x, z] = \hat y= \hat \alpha + \hat \beta \cdot x +\... | How to determine by what percent the target variable will change if we change a variable by some per
We can back out the answer with a little auxiliary information about the covariates.
Your linear model is probably something like $$E[y \vert x, z] = \hat y= \hat \alpha + \hat \beta \cdot x +\hat \de |
53,104 | Autoencoders as dimensionality reduction tools..? | Yes, dimension reduction is one way to use auto-encoders. Consider a feed-forward fully-connected auto-encoder with and input layer, 1 hidden layer with $k$ units, 1 output layer and all linear activation functions. The latent space of this auto-encoder spans the first $k$ principle components of the original data. Thi... | Autoencoders as dimensionality reduction tools..? | Yes, dimension reduction is one way to use auto-encoders. Consider a feed-forward fully-connected auto-encoder with and input layer, 1 hidden layer with $k$ units, 1 output layer and all linear activa | Autoencoders as dimensionality reduction tools..?
Yes, dimension reduction is one way to use auto-encoders. Consider a feed-forward fully-connected auto-encoder with and input layer, 1 hidden layer with $k$ units, 1 output layer and all linear activation functions. The latent space of this auto-encoder spans the first ... | Autoencoders as dimensionality reduction tools..?
Yes, dimension reduction is one way to use auto-encoders. Consider a feed-forward fully-connected auto-encoder with and input layer, 1 hidden layer with $k$ units, 1 output layer and all linear activa |
53,105 | Autoencoders as dimensionality reduction tools..? | Yes, using some form of autoencoder training/pre-training to create good features has been a successful approach in many areas. E.g. for tabular data, using a denoising autoencoder was the winning approach in a recent Kaggle competition. Autoencoder pre-training (recovering masked features) has been used in the TabNet ... | Autoencoders as dimensionality reduction tools..? | Yes, using some form of autoencoder training/pre-training to create good features has been a successful approach in many areas. E.g. for tabular data, using a denoising autoencoder was the winning app | Autoencoders as dimensionality reduction tools..?
Yes, using some form of autoencoder training/pre-training to create good features has been a successful approach in many areas. E.g. for tabular data, using a denoising autoencoder was the winning approach in a recent Kaggle competition. Autoencoder pre-training (recove... | Autoencoders as dimensionality reduction tools..?
Yes, using some form of autoencoder training/pre-training to create good features has been a successful approach in many areas. E.g. for tabular data, using a denoising autoencoder was the winning app |
53,106 | Autoencoders as dimensionality reduction tools..? | As you mentioned, auto-encoders can be used for dimensionality reduction. One of the nice things about them, is they are an unsupervised learning method. If you have a large volume of unlabeled data and a small volume of labeled data, you can train your auto-encoder on the large unlabeled dataset to get a robust repr... | Autoencoders as dimensionality reduction tools..? | As you mentioned, auto-encoders can be used for dimensionality reduction. One of the nice things about them, is they are an unsupervised learning method. If you have a large volume of unlabeled data | Autoencoders as dimensionality reduction tools..?
As you mentioned, auto-encoders can be used for dimensionality reduction. One of the nice things about them, is they are an unsupervised learning method. If you have a large volume of unlabeled data and a small volume of labeled data, you can train your auto-encoder o... | Autoencoders as dimensionality reduction tools..?
As you mentioned, auto-encoders can be used for dimensionality reduction. One of the nice things about them, is they are an unsupervised learning method. If you have a large volume of unlabeled data |
53,107 | Embedding data into a larger dimension space | As Forrest mentioned embedding data into a higher dimension (sometimes called basis expansion) is a common method which allows a linear classifier to observe a non-linear input space. Examples are using the RBF kernel with an SVM or polynomial expansion with linear regression.
However basis expansion isn't always benef... | Embedding data into a larger dimension space | As Forrest mentioned embedding data into a higher dimension (sometimes called basis expansion) is a common method which allows a linear classifier to observe a non-linear input space. Examples are usi | Embedding data into a larger dimension space
As Forrest mentioned embedding data into a higher dimension (sometimes called basis expansion) is a common method which allows a linear classifier to observe a non-linear input space. Examples are using the RBF kernel with an SVM or polynomial expansion with linear regressio... | Embedding data into a larger dimension space
As Forrest mentioned embedding data into a higher dimension (sometimes called basis expansion) is a common method which allows a linear classifier to observe a non-linear input space. Examples are usi |
53,108 | Embedding data into a larger dimension space | are there possible benefits for embedding into a space of larger or same dimension?
In Vector Symbolic Architectures (also known as Hyperdimensional Computing) this is essential. VSAs use algebraic operations (sum, product, permutation) on vectors to encode information in their directions. VSAs can be used to represen... | Embedding data into a larger dimension space | are there possible benefits for embedding into a space of larger or same dimension?
In Vector Symbolic Architectures (also known as Hyperdimensional Computing) this is essential. VSAs use algebraic o | Embedding data into a larger dimension space
are there possible benefits for embedding into a space of larger or same dimension?
In Vector Symbolic Architectures (also known as Hyperdimensional Computing) this is essential. VSAs use algebraic operations (sum, product, permutation) on vectors to encode information in t... | Embedding data into a larger dimension space
are there possible benefits for embedding into a space of larger or same dimension?
In Vector Symbolic Architectures (also known as Hyperdimensional Computing) this is essential. VSAs use algebraic o |
53,109 | Embedding data into a larger dimension space | I suggest that the datapoints of the 5-dimensional dataset are first classified in 13 types of classes(orbits) where these types of classes(orbits) have the following cardinalities: 1,10,32,40,80,160,240,320,480,640,960,1920,3840. These 13 types of classes are generated by the intrinsic properties of the automorphism o... | Embedding data into a larger dimension space | I suggest that the datapoints of the 5-dimensional dataset are first classified in 13 types of classes(orbits) where these types of classes(orbits) have the following cardinalities: 1,10,32,40,80,160, | Embedding data into a larger dimension space
I suggest that the datapoints of the 5-dimensional dataset are first classified in 13 types of classes(orbits) where these types of classes(orbits) have the following cardinalities: 1,10,32,40,80,160,240,320,480,640,960,1920,3840. These 13 types of classes are generated by t... | Embedding data into a larger dimension space
I suggest that the datapoints of the 5-dimensional dataset are first classified in 13 types of classes(orbits) where these types of classes(orbits) have the following cardinalities: 1,10,32,40,80,160, |
53,110 | Programming inverse-transformation sampling for Pareto distribution | I'm assuming you are referecning to Inverse Transform Sampling method. Its very straight forward. Refer Wiki article and this site.
Pareto CDF is given by:
$$
F(X) = 1 -(\frac{k}{x})^\gamma; x\ge k>0 \ and \ \gamma>0
$$
All you do is equate to uniform distribution and solve for $x$
$$
F(X) = U \\
U \sim Uniform(0,1) \\... | Programming inverse-transformation sampling for Pareto distribution | I'm assuming you are referecning to Inverse Transform Sampling method. Its very straight forward. Refer Wiki article and this site.
Pareto CDF is given by:
$$
F(X) = 1 -(\frac{k}{x})^\gamma; x\ge k>0 | Programming inverse-transformation sampling for Pareto distribution
I'm assuming you are referecning to Inverse Transform Sampling method. Its very straight forward. Refer Wiki article and this site.
Pareto CDF is given by:
$$
F(X) = 1 -(\frac{k}{x})^\gamma; x\ge k>0 \ and \ \gamma>0
$$
All you do is equate to uniform ... | Programming inverse-transformation sampling for Pareto distribution
I'm assuming you are referecning to Inverse Transform Sampling method. Its very straight forward. Refer Wiki article and this site.
Pareto CDF is given by:
$$
F(X) = 1 -(\frac{k}{x})^\gamma; x\ge k>0 |
53,111 | Programming inverse-transformation sampling for Pareto distribution | Using the quantile $p=F(x)$ and inverting the CDF equation gives the quantile function:
$$Q(p) = \frac{k}{(1-p)^{1/\gamma}}
\quad \quad \quad \text{for all } 0 \leqslant p \leqslant 1.$$
The corresponding log-quantile function is:
$$\log Q(p) = \log k - \frac{1}{\gamma} \log (1-p)
\quad \quad \quad \text{for all } 0 \l... | Programming inverse-transformation sampling for Pareto distribution | Using the quantile $p=F(x)$ and inverting the CDF equation gives the quantile function:
$$Q(p) = \frac{k}{(1-p)^{1/\gamma}}
\quad \quad \quad \text{for all } 0 \leqslant p \leqslant 1.$$
The correspon | Programming inverse-transformation sampling for Pareto distribution
Using the quantile $p=F(x)$ and inverting the CDF equation gives the quantile function:
$$Q(p) = \frac{k}{(1-p)^{1/\gamma}}
\quad \quad \quad \text{for all } 0 \leqslant p \leqslant 1.$$
The corresponding log-quantile function is:
$$\log Q(p) = \log k ... | Programming inverse-transformation sampling for Pareto distribution
Using the quantile $p=F(x)$ and inverting the CDF equation gives the quantile function:
$$Q(p) = \frac{k}{(1-p)^{1/\gamma}}
\quad \quad \quad \text{for all } 0 \leqslant p \leqslant 1.$$
The correspon |
53,112 | How to choose between an overfit model and a non-overfit model? | First of all, you need to choose before the final test. The purpose of the final test is to measure/estimate generalization error for the already chosen model.
If you choose again based on the test set, you either
need to restrict yourself to not claim any generalization error. I.e. you can say that your optimization ... | How to choose between an overfit model and a non-overfit model? | First of all, you need to choose before the final test. The purpose of the final test is to measure/estimate generalization error for the already chosen model.
If you choose again based on the test se | How to choose between an overfit model and a non-overfit model?
First of all, you need to choose before the final test. The purpose of the final test is to measure/estimate generalization error for the already chosen model.
If you choose again based on the test set, you either
need to restrict yourself to not claim an... | How to choose between an overfit model and a non-overfit model?
First of all, you need to choose before the final test. The purpose of the final test is to measure/estimate generalization error for the already chosen model.
If you choose again based on the test se |
53,113 | How to choose between an overfit model and a non-overfit model? | This is impossible to answer without more information. Class balance, tolerance for false positive/negative results, etc are important factors into deciding if model is for for production.
I've seen models with a very high accuracy score poorly on something like MCC due to most of the predictions being wrong on the min... | How to choose between an overfit model and a non-overfit model? | This is impossible to answer without more information. Class balance, tolerance for false positive/negative results, etc are important factors into deciding if model is for for production.
I've seen m | How to choose between an overfit model and a non-overfit model?
This is impossible to answer without more information. Class balance, tolerance for false positive/negative results, etc are important factors into deciding if model is for for production.
I've seen models with a very high accuracy score poorly on somethin... | How to choose between an overfit model and a non-overfit model?
This is impossible to answer without more information. Class balance, tolerance for false positive/negative results, etc are important factors into deciding if model is for for production.
I've seen m |
53,114 | How to choose between an overfit model and a non-overfit model? | Overfit or not, you should pick the one with the highest test accuracy, conditional on the fact that you have truly kept your test data separate. I would be tempted to find more unseen test data to double check that it has truly generalised well to new data. | How to choose between an overfit model and a non-overfit model? | Overfit or not, you should pick the one with the highest test accuracy, conditional on the fact that you have truly kept your test data separate. I would be tempted to find more unseen test data to do | How to choose between an overfit model and a non-overfit model?
Overfit or not, you should pick the one with the highest test accuracy, conditional on the fact that you have truly kept your test data separate. I would be tempted to find more unseen test data to double check that it has truly generalised well to new dat... | How to choose between an overfit model and a non-overfit model?
Overfit or not, you should pick the one with the highest test accuracy, conditional on the fact that you have truly kept your test data separate. I would be tempted to find more unseen test data to do |
53,115 | How to check the correlation between categorical and numeric independent variable in R? [duplicate] | There are several ways to determine correlation between a categorical and a continuous variable. However, I found only one way to calculate a 'correlation coefficient', and that only works if your categorical variable is dichotomous.
If your categorical variable is dichotomous (only two values), then you can use the po... | How to check the correlation between categorical and numeric independent variable in R? [duplicate] | There are several ways to determine correlation between a categorical and a continuous variable. However, I found only one way to calculate a 'correlation coefficient', and that only works if your cat | How to check the correlation between categorical and numeric independent variable in R? [duplicate]
There are several ways to determine correlation between a categorical and a continuous variable. However, I found only one way to calculate a 'correlation coefficient', and that only works if your categorical variable is... | How to check the correlation between categorical and numeric independent variable in R? [duplicate]
There are several ways to determine correlation between a categorical and a continuous variable. However, I found only one way to calculate a 'correlation coefficient', and that only works if your cat |
53,116 | Generating uniform points inside an $m$-dimensional ball [duplicate] | A simple and efficient method for this problem uses a variation of the well-known Box-Mueller transform, which connects the normal distribution to the uniform distribution on a ball. If we generate a random vector $\mathbf{Z} = (Z_1,...,Z_m)$ composed of IID standard normal random variables and a random variable $U \s... | Generating uniform points inside an $m$-dimensional ball [duplicate] | A simple and efficient method for this problem uses a variation of the well-known Box-Mueller transform, which connects the normal distribution to the uniform distribution on a ball. If we generate a | Generating uniform points inside an $m$-dimensional ball [duplicate]
A simple and efficient method for this problem uses a variation of the well-known Box-Mueller transform, which connects the normal distribution to the uniform distribution on a ball. If we generate a random vector $\mathbf{Z} = (Z_1,...,Z_m)$ compose... | Generating uniform points inside an $m$-dimensional ball [duplicate]
A simple and efficient method for this problem uses a variation of the well-known Box-Mueller transform, which connects the normal distribution to the uniform distribution on a ball. If we generate a |
53,117 | Generating uniform points inside an $m$-dimensional ball [duplicate] | The simplest and least error-prone approach - for low dimensions (see below!) - would still be rejection sampling: pick uniformly distributed points from the $m$-dimensional hypercube circumscribing the sphere, then reject all that fall outside the ball.
runifball <- function(n, centre = 0, center = centre, radius = 1)... | Generating uniform points inside an $m$-dimensional ball [duplicate] | The simplest and least error-prone approach - for low dimensions (see below!) - would still be rejection sampling: pick uniformly distributed points from the $m$-dimensional hypercube circumscribing t | Generating uniform points inside an $m$-dimensional ball [duplicate]
The simplest and least error-prone approach - for low dimensions (see below!) - would still be rejection sampling: pick uniformly distributed points from the $m$-dimensional hypercube circumscribing the sphere, then reject all that fall outside the ba... | Generating uniform points inside an $m$-dimensional ball [duplicate]
The simplest and least error-prone approach - for low dimensions (see below!) - would still be rejection sampling: pick uniformly distributed points from the $m$-dimensional hypercube circumscribing t |
53,118 | Assumption of normality in a sample size of 3 | Honestly, with a sample size of three, there's really not much you can say at all. Those numbers are more likely to come from a normal distribution than, say, $1$, $2$, and $1000$ are, but there are plenty of non-normal distributions that can produce those numbers.
With regards to a standard deviation being relatively... | Assumption of normality in a sample size of 3 | Honestly, with a sample size of three, there's really not much you can say at all. Those numbers are more likely to come from a normal distribution than, say, $1$, $2$, and $1000$ are, but there are p | Assumption of normality in a sample size of 3
Honestly, with a sample size of three, there's really not much you can say at all. Those numbers are more likely to come from a normal distribution than, say, $1$, $2$, and $1000$ are, but there are plenty of non-normal distributions that can produce those numbers.
With re... | Assumption of normality in a sample size of 3
Honestly, with a sample size of three, there's really not much you can say at all. Those numbers are more likely to come from a normal distribution than, say, $1$, $2$, and $1000$ are, but there are p |
53,119 | Assumption of normality in a sample size of 3 | You don't say what null hypothesis you're testing.
If you want to know if your observations indicate that the population mean or median is above $20,$ then a one-sample, one-sided Wilcoxon signed-rank test
cannot give a P-value below $1/8 = 0.125.$
wilcox.test(x, mu=20, alt="g")$p.val
[1] 0.125
By contrast, if you hav... | Assumption of normality in a sample size of 3 | You don't say what null hypothesis you're testing.
If you want to know if your observations indicate that the population mean or median is above $20,$ then a one-sample, one-sided Wilcoxon signed-rank | Assumption of normality in a sample size of 3
You don't say what null hypothesis you're testing.
If you want to know if your observations indicate that the population mean or median is above $20,$ then a one-sample, one-sided Wilcoxon signed-rank test
cannot give a P-value below $1/8 = 0.125.$
wilcox.test(x, mu=20, alt... | Assumption of normality in a sample size of 3
You don't say what null hypothesis you're testing.
If you want to know if your observations indicate that the population mean or median is above $20,$ then a one-sample, one-sided Wilcoxon signed-rank |
53,120 | Assumption of normality in a sample size of 3 | Samples of size 3 cannot be easily used to visually assess whether they might be consistent with having come from a normal population.
Tiny samples actually from a normal distribution may be highly asymmetric.
Symmetry doesn't imply normality. In some cases population symmetry would be very helpful, in others it's fa... | Assumption of normality in a sample size of 3 | Samples of size 3 cannot be easily used to visually assess whether they might be consistent with having come from a normal population.
Tiny samples actually from a normal distribution may be highly a | Assumption of normality in a sample size of 3
Samples of size 3 cannot be easily used to visually assess whether they might be consistent with having come from a normal population.
Tiny samples actually from a normal distribution may be highly asymmetric.
Symmetry doesn't imply normality. In some cases population sym... | Assumption of normality in a sample size of 3
Samples of size 3 cannot be easily used to visually assess whether they might be consistent with having come from a normal population.
Tiny samples actually from a normal distribution may be highly a |
53,121 | Assumption of normality in a sample size of 3 | As Norvia points out, there you cannot make this assumption with a sample size of 3. I would not assume normal distribution. The mann-whitney U test is the better choice. | Assumption of normality in a sample size of 3 | As Norvia points out, there you cannot make this assumption with a sample size of 3. I would not assume normal distribution. The mann-whitney U test is the better choice. | Assumption of normality in a sample size of 3
As Norvia points out, there you cannot make this assumption with a sample size of 3. I would not assume normal distribution. The mann-whitney U test is the better choice. | Assumption of normality in a sample size of 3
As Norvia points out, there you cannot make this assumption with a sample size of 3. I would not assume normal distribution. The mann-whitney U test is the better choice. |
53,122 | Assumption of normality in a sample size of 3 | With classic statistics, I would still do the usual test and arrive at large confidence intervals, for example, which still may be informative in postulating say a prior distribution for any subsequent low data exercise. [EDIT] The rationale is also that it offers a parametric opinion to contrast to the nonparametric/b... | Assumption of normality in a sample size of 3 | With classic statistics, I would still do the usual test and arrive at large confidence intervals, for example, which still may be informative in postulating say a prior distribution for any subsequen | Assumption of normality in a sample size of 3
With classic statistics, I would still do the usual test and arrive at large confidence intervals, for example, which still may be informative in postulating say a prior distribution for any subsequent low data exercise. [EDIT] The rationale is also that it offers a paramet... | Assumption of normality in a sample size of 3
With classic statistics, I would still do the usual test and arrive at large confidence intervals, for example, which still may be informative in postulating say a prior distribution for any subsequen |
53,123 | Finding UMVUE for a function of a Bernoulli parameter | Except when $k=1$, given a finite sequence of i.i.d. Bernoulli
$\mathcal B(θ)$ random variables $X_1,X_2,\ldots,X_m$, there exists no
unbiased estimator of $(1−θ)^{1/k}$, when $k$ is a positive integer.
The reason for this impossibility is that only polynomials in $\theta$ of degree at most $m$ can be unbiasedly e... | Finding UMVUE for a function of a Bernoulli parameter | Except when $k=1$, given a finite sequence of i.i.d. Bernoulli
$\mathcal B(θ)$ random variables $X_1,X_2,\ldots,X_m$, there exists no
unbiased estimator of $(1−θ)^{1/k}$, when $k$ is a positive in | Finding UMVUE for a function of a Bernoulli parameter
Except when $k=1$, given a finite sequence of i.i.d. Bernoulli
$\mathcal B(θ)$ random variables $X_1,X_2,\ldots,X_m$, there exists no
unbiased estimator of $(1−θ)^{1/k}$, when $k$ is a positive integer.
The reason for this impossibility is that only polynomials... | Finding UMVUE for a function of a Bernoulli parameter
Except when $k=1$, given a finite sequence of i.i.d. Bernoulli
$\mathcal B(θ)$ random variables $X_1,X_2,\ldots,X_m$, there exists no
unbiased estimator of $(1−θ)^{1/k}$, when $k$ is a positive in |
53,124 | Finding UMVUE for a function of a Bernoulli parameter | First of all, I'll just point out that it's not enough that $\sum_i X_i$ is sufficient. We need it to be complete-sufficient. Fortunately we know that $\sum_i X_i$ is also a complete statistic by well known properties of the exponential family of distributions.
As you say, we need an estimator $\delta(\cdot)$ based on... | Finding UMVUE for a function of a Bernoulli parameter | First of all, I'll just point out that it's not enough that $\sum_i X_i$ is sufficient. We need it to be complete-sufficient. Fortunately we know that $\sum_i X_i$ is also a complete statistic by wel | Finding UMVUE for a function of a Bernoulli parameter
First of all, I'll just point out that it's not enough that $\sum_i X_i$ is sufficient. We need it to be complete-sufficient. Fortunately we know that $\sum_i X_i$ is also a complete statistic by well known properties of the exponential family of distributions.
As ... | Finding UMVUE for a function of a Bernoulli parameter
First of all, I'll just point out that it's not enough that $\sum_i X_i$ is sufficient. We need it to be complete-sufficient. Fortunately we know that $\sum_i X_i$ is also a complete statistic by wel |
53,125 | stochastic vs. deterministic trend in time series | Deterministic Trend
$$
y_t = \beta_0 + \beta_1 t + \epsilon_t
$$
where $\{\epsilon_t\}$ is white noise, for simplicity. Same discussion applies to the case where $\{\epsilon_t\}$ is a covariance-stationary process (e.g. ARIMA with $d = 0$).
The process is random fluctuations around a deterministic linear trend $\beta_... | stochastic vs. deterministic trend in time series | Deterministic Trend
$$
y_t = \beta_0 + \beta_1 t + \epsilon_t
$$
where $\{\epsilon_t\}$ is white noise, for simplicity. Same discussion applies to the case where $\{\epsilon_t\}$ is a covariance-stati | stochastic vs. deterministic trend in time series
Deterministic Trend
$$
y_t = \beta_0 + \beta_1 t + \epsilon_t
$$
where $\{\epsilon_t\}$ is white noise, for simplicity. Same discussion applies to the case where $\{\epsilon_t\}$ is a covariance-stationary process (e.g. ARIMA with $d = 0$).
The process is random fluctu... | stochastic vs. deterministic trend in time series
Deterministic Trend
$$
y_t = \beta_0 + \beta_1 t + \epsilon_t
$$
where $\{\epsilon_t\}$ is white noise, for simplicity. Same discussion applies to the case where $\{\epsilon_t\}$ is a covariance-stati |
53,126 | Sample size of a "continuous" experimental unit/population instead of "discrete" | All sample size calculations are built on top of a proposed inference that will be made from the data. This might be a confidence interval for an unknown parameter, or a hypothesis test for a set of hypotheses, or a Bayesian posterior inference, etc. Whatever the inference being made, there will be some appropriate m... | Sample size of a "continuous" experimental unit/population instead of "discrete" | All sample size calculations are built on top of a proposed inference that will be made from the data. This might be a confidence interval for an unknown parameter, or a hypothesis test for a set of | Sample size of a "continuous" experimental unit/population instead of "discrete"
All sample size calculations are built on top of a proposed inference that will be made from the data. This might be a confidence interval for an unknown parameter, or a hypothesis test for a set of hypotheses, or a Bayesian posterior inf... | Sample size of a "continuous" experimental unit/population instead of "discrete"
All sample size calculations are built on top of a proposed inference that will be made from the data. This might be a confidence interval for an unknown parameter, or a hypothesis test for a set of |
53,127 | Sample size of a "continuous" experimental unit/population instead of "discrete" | Peter cannot inflate his effective sample size for the purpose of estimating treatment effects just by repeatedly subsampling the same experimental units – this would be the most egregious form of 'pseudo-replication'.
Sample size in the context of a designed experiment is set by the randomization design – since the di... | Sample size of a "continuous" experimental unit/population instead of "discrete" | Peter cannot inflate his effective sample size for the purpose of estimating treatment effects just by repeatedly subsampling the same experimental units – this would be the most egregious form of 'ps | Sample size of a "continuous" experimental unit/population instead of "discrete"
Peter cannot inflate his effective sample size for the purpose of estimating treatment effects just by repeatedly subsampling the same experimental units – this would be the most egregious form of 'pseudo-replication'.
Sample size in the c... | Sample size of a "continuous" experimental unit/population instead of "discrete"
Peter cannot inflate his effective sample size for the purpose of estimating treatment effects just by repeatedly subsampling the same experimental units – this would be the most egregious form of 'ps |
53,128 | Asking nature a single question or a carefully thought out questionnaire | Think of a simple example, and since RA Fisher's primary experience with experimental design was at Rothamsted Experimental station, let us use an agricultural example.
Say you are interested in comparing the effectiveness of various fertilizers. Using the philosophy of vary only one circumstance at a time, you design... | Asking nature a single question or a carefully thought out questionnaire | Think of a simple example, and since RA Fisher's primary experience with experimental design was at Rothamsted Experimental station, let us use an agricultural example.
Say you are interested in comp | Asking nature a single question or a carefully thought out questionnaire
Think of a simple example, and since RA Fisher's primary experience with experimental design was at Rothamsted Experimental station, let us use an agricultural example.
Say you are interested in comparing the effectiveness of various fertilizers.... | Asking nature a single question or a carefully thought out questionnaire
Think of a simple example, and since RA Fisher's primary experience with experimental design was at Rothamsted Experimental station, let us use an agricultural example.
Say you are interested in comp |
53,129 | Asking nature a single question or a carefully thought out questionnaire | Just to add to @kjetilbhalvorsen's answer, the insight Fisher got, or what came out of that is factorial design. This often cited excerpt comes from Arrangement of Field experiment, pg 10 (or 511). Though it seems maybe common sense to us now, Fisher notes most experiments done during that time involves single factors.... | Asking nature a single question or a carefully thought out questionnaire | Just to add to @kjetilbhalvorsen's answer, the insight Fisher got, or what came out of that is factorial design. This often cited excerpt comes from Arrangement of Field experiment, pg 10 (or 511). Th | Asking nature a single question or a carefully thought out questionnaire
Just to add to @kjetilbhalvorsen's answer, the insight Fisher got, or what came out of that is factorial design. This often cited excerpt comes from Arrangement of Field experiment, pg 10 (or 511). Though it seems maybe common sense to us now, Fis... | Asking nature a single question or a carefully thought out questionnaire
Just to add to @kjetilbhalvorsen's answer, the insight Fisher got, or what came out of that is factorial design. This often cited excerpt comes from Arrangement of Field experiment, pg 10 (or 511). Th |
53,130 | Empirical results of Machine Learning/Deep Learning in practice | I don't think it would be possible to answer this question with respect to proprietary models used by private enterprise. But there is a vein of scholarship that focuses on flawed practices, such as in this paper.
Zachary C. Lipton, Jacob Steinhardt "Troubling Trends in Machine Learning Scholarship"
Collectively, mach... | Empirical results of Machine Learning/Deep Learning in practice | I don't think it would be possible to answer this question with respect to proprietary models used by private enterprise. But there is a vein of scholarship that focuses on flawed practices, such as i | Empirical results of Machine Learning/Deep Learning in practice
I don't think it would be possible to answer this question with respect to proprietary models used by private enterprise. But there is a vein of scholarship that focuses on flawed practices, such as in this paper.
Zachary C. Lipton, Jacob Steinhardt "Troub... | Empirical results of Machine Learning/Deep Learning in practice
I don't think it would be possible to answer this question with respect to proprietary models used by private enterprise. But there is a vein of scholarship that focuses on flawed practices, such as i |
53,131 | Can the difference of random variables be uniform distributed? [duplicate] | Consider the following example:
$$
X\sim\text{Unif}(0, 1) \\
Y = 1-X
$$
$X$ and $Y$ are identically distributed as the standard uniform distribution, and $X-Y = 2X-1$, so $X-Y\sim\text{Unif}(-1, 1)$.
Note that this example relied on $X$ and $Y$ being dependent, identically distributed random variables. It is impossible... | Can the difference of random variables be uniform distributed? [duplicate] | Consider the following example:
$$
X\sim\text{Unif}(0, 1) \\
Y = 1-X
$$
$X$ and $Y$ are identically distributed as the standard uniform distribution, and $X-Y = 2X-1$, so $X-Y\sim\text{Unif}(-1, 1)$.
| Can the difference of random variables be uniform distributed? [duplicate]
Consider the following example:
$$
X\sim\text{Unif}(0, 1) \\
Y = 1-X
$$
$X$ and $Y$ are identically distributed as the standard uniform distribution, and $X-Y = 2X-1$, so $X-Y\sim\text{Unif}(-1, 1)$.
Note that this example relied on $X$ and $Y$ ... | Can the difference of random variables be uniform distributed? [duplicate]
Consider the following example:
$$
X\sim\text{Unif}(0, 1) \\
Y = 1-X
$$
$X$ and $Y$ are identically distributed as the standard uniform distribution, and $X-Y = 2X-1$, so $X-Y\sim\text{Unif}(-1, 1)$.
|
53,132 | Can the difference of random variables be uniform distributed? [duplicate] | Independent and identically distributed. If you ask for IID $X$ and $Y$, as others have noted this is not possible. See the answers to this question.
If you are happy to drop either independence between $X$ and $Y$ or them having the same distribution, then there's hope.
Same distribution. If you allow for dependent bu... | Can the difference of random variables be uniform distributed? [duplicate] | Independent and identically distributed. If you ask for IID $X$ and $Y$, as others have noted this is not possible. See the answers to this question.
If you are happy to drop either independence betwe | Can the difference of random variables be uniform distributed? [duplicate]
Independent and identically distributed. If you ask for IID $X$ and $Y$, as others have noted this is not possible. See the answers to this question.
If you are happy to drop either independence between $X$ and $Y$ or them having the same distri... | Can the difference of random variables be uniform distributed? [duplicate]
Independent and identically distributed. If you ask for IID $X$ and $Y$, as others have noted this is not possible. See the answers to this question.
If you are happy to drop either independence betwe |
53,133 | Why decision tree handle unbalanced data well? | Decision trees do not always handle unbalanced data well.
If there is relatively obvious particular partition of our sample space that contains a high-proportion of minority class instances, decision trees can probably find it but that is far from a certainty.
For example, if the minority class is strongly associated... | Why decision tree handle unbalanced data well? | Decision trees do not always handle unbalanced data well.
If there is relatively obvious particular partition of our sample space that contains a high-proportion of minority class instances, decision | Why decision tree handle unbalanced data well?
Decision trees do not always handle unbalanced data well.
If there is relatively obvious particular partition of our sample space that contains a high-proportion of minority class instances, decision trees can probably find it but that is far from a certainty.
For exampl... | Why decision tree handle unbalanced data well?
Decision trees do not always handle unbalanced data well.
If there is relatively obvious particular partition of our sample space that contains a high-proportion of minority class instances, decision |
53,134 | Why decision tree handle unbalanced data well? | I would like to add something to the previous answer - though it's already good (+1).
Decision trees implementations normally use Gini index or Entropy for finding splits. These are functions that are maximized when the classes in a node are perfectly balanced - and therefore reward splits that move away from this bal... | Why decision tree handle unbalanced data well? | I would like to add something to the previous answer - though it's already good (+1).
Decision trees implementations normally use Gini index or Entropy for finding splits. These are functions that ar | Why decision tree handle unbalanced data well?
I would like to add something to the previous answer - though it's already good (+1).
Decision trees implementations normally use Gini index or Entropy for finding splits. These are functions that are maximized when the classes in a node are perfectly balanced - and there... | Why decision tree handle unbalanced data well?
I would like to add something to the previous answer - though it's already good (+1).
Decision trees implementations normally use Gini index or Entropy for finding splits. These are functions that ar |
53,135 | disadvantage of bootstrap (from wiki) | It's wiki, read all wiki with a grain of salt. You should raise a flag as being unclear, opinion-based, or needing a citation because all of those are (partly) true. The recent influx of people in statistics who feel that broad statements can be made and parroted without formal proof need to be reigned in (I include my... | disadvantage of bootstrap (from wiki) | It's wiki, read all wiki with a grain of salt. You should raise a flag as being unclear, opinion-based, or needing a citation because all of those are (partly) true. The recent influx of people in sta | disadvantage of bootstrap (from wiki)
It's wiki, read all wiki with a grain of salt. You should raise a flag as being unclear, opinion-based, or needing a citation because all of those are (partly) true. The recent influx of people in statistics who feel that broad statements can be made and parroted without formal pro... | disadvantage of bootstrap (from wiki)
It's wiki, read all wiki with a grain of salt. You should raise a flag as being unclear, opinion-based, or needing a citation because all of those are (partly) true. The recent influx of people in sta |
53,136 | disadvantage of bootstrap (from wiki) | This may be related to the fact that the bootstrap may sometimes be roughly presented as an "assumption free" procedure that can be used to replace other common e.g. tests when their required assumptions (e.g. normality) are not met. However, bootstrapping is relevant only in certain situations raising assumptions tha... | disadvantage of bootstrap (from wiki) | This may be related to the fact that the bootstrap may sometimes be roughly presented as an "assumption free" procedure that can be used to replace other common e.g. tests when their required assumpti | disadvantage of bootstrap (from wiki)
This may be related to the fact that the bootstrap may sometimes be roughly presented as an "assumption free" procedure that can be used to replace other common e.g. tests when their required assumptions (e.g. normality) are not met. However, bootstrapping is relevant only in cert... | disadvantage of bootstrap (from wiki)
This may be related to the fact that the bootstrap may sometimes be roughly presented as an "assumption free" procedure that can be used to replace other common e.g. tests when their required assumpti |
53,137 | How to apply PCA on 3 dimensional image data in python | In general, your approach may work, and it might even give you something that works somewhat well. However, I would strongly advise against it, or only use something like this as a first step to just get a feel for the problem.
Think about it this way: If you just shift one of the images one pixel to the left, how much... | How to apply PCA on 3 dimensional image data in python | In general, your approach may work, and it might even give you something that works somewhat well. However, I would strongly advise against it, or only use something like this as a first step to just | How to apply PCA on 3 dimensional image data in python
In general, your approach may work, and it might even give you something that works somewhat well. However, I would strongly advise against it, or only use something like this as a first step to just get a feel for the problem.
Think about it this way: If you just ... | How to apply PCA on 3 dimensional image data in python
In general, your approach may work, and it might even give you something that works somewhat well. However, I would strongly advise against it, or only use something like this as a first step to just |
53,138 | How to apply PCA on 3 dimensional image data in python | If you final goal is using SVM, the problem is number of data points instead of the number of dimensions. See following question.
Can support vector machine be used in large data?
In real world SVM will not work very well if you have ~10K data and above.
Your problem is a standard image classification problem using con... | How to apply PCA on 3 dimensional image data in python | If you final goal is using SVM, the problem is number of data points instead of the number of dimensions. See following question.
Can support vector machine be used in large data?
In real world SVM wi | How to apply PCA on 3 dimensional image data in python
If you final goal is using SVM, the problem is number of data points instead of the number of dimensions. See following question.
Can support vector machine be used in large data?
In real world SVM will not work very well if you have ~10K data and above.
Your probl... | How to apply PCA on 3 dimensional image data in python
If you final goal is using SVM, the problem is number of data points instead of the number of dimensions. See following question.
Can support vector machine be used in large data?
In real world SVM wi |
53,139 | P Wasserstein distance in Python | This is implemented in the POT: Python Optimal Transport package, for samples (or, generally, discrete measures): use ot.wasserstein_1d.
If you want to do it for weighted samples (or general discrete distributions with finite support), you can provide the a and b arguments. | P Wasserstein distance in Python | This is implemented in the POT: Python Optimal Transport package, for samples (or, generally, discrete measures): use ot.wasserstein_1d.
If you want to do it for weighted samples (or general discrete | P Wasserstein distance in Python
This is implemented in the POT: Python Optimal Transport package, for samples (or, generally, discrete measures): use ot.wasserstein_1d.
If you want to do it for weighted samples (or general discrete distributions with finite support), you can provide the a and b arguments. | P Wasserstein distance in Python
This is implemented in the POT: Python Optimal Transport package, for samples (or, generally, discrete measures): use ot.wasserstein_1d.
If you want to do it for weighted samples (or general discrete |
53,140 | Controlling variables in causal diagrams | This is more of a comment, in response to comments to Ed Rigdon's answer:
I understand that I shouldn’t control for B because it’s a collider. I want to know how the diagram looks when a variable is controlled for. Then I would be able to see everything explicitly
A good way to do this is by drawing the conditional g... | Controlling variables in causal diagrams | This is more of a comment, in response to comments to Ed Rigdon's answer:
I understand that I shouldn’t control for B because it’s a collider. I want to know how the diagram looks when a variable is | Controlling variables in causal diagrams
This is more of a comment, in response to comments to Ed Rigdon's answer:
I understand that I shouldn’t control for B because it’s a collider. I want to know how the diagram looks when a variable is controlled for. Then I would be able to see everything explicitly
A good way t... | Controlling variables in causal diagrams
This is more of a comment, in response to comments to Ed Rigdon's answer:
I understand that I shouldn’t control for B because it’s a collider. I want to know how the diagram looks when a variable is |
53,141 | Controlling variables in causal diagrams | In the model, B = A + C, and A and C are orthogonal. If you hold B constant, you induce covariance between A and C. Say you hold B constant at 10. Then, if A is 7, C is 3. If A is 6, then C is 4. This creates a confound for the X -> Y relationship, a backdoor path from X through A through the covariance with C to Y.
C ... | Controlling variables in causal diagrams | In the model, B = A + C, and A and C are orthogonal. If you hold B constant, you induce covariance between A and C. Say you hold B constant at 10. Then, if A is 7, C is 3. If A is 6, then C is 4. This | Controlling variables in causal diagrams
In the model, B = A + C, and A and C are orthogonal. If you hold B constant, you induce covariance between A and C. Say you hold B constant at 10. Then, if A is 7, C is 3. If A is 6, then C is 4. This creates a confound for the X -> Y relationship, a backdoor path from X through... | Controlling variables in causal diagrams
In the model, B = A + C, and A and C are orthogonal. If you hold B constant, you induce covariance between A and C. Say you hold B constant at 10. Then, if A is 7, C is 3. If A is 6, then C is 4. This |
53,142 | Controlling variables in causal diagrams | There are seven rules of association. In the first four, $R$ and $T$ are associated with each other:
$$R \rightarrow T$$
$$R \rightarrow S \rightarrow T$$
$$R \leftarrow S \rightarrow T$$
$$R \rightarrow \fbox S \leftarrow T$$
In the second three, $R$ and $T$ are not associated with each other through the path:
$$R \ri... | Controlling variables in causal diagrams | There are seven rules of association. In the first four, $R$ and $T$ are associated with each other:
$$R \rightarrow T$$
$$R \rightarrow S \rightarrow T$$
$$R \leftarrow S \rightarrow T$$
$$R \rightar | Controlling variables in causal diagrams
There are seven rules of association. In the first four, $R$ and $T$ are associated with each other:
$$R \rightarrow T$$
$$R \rightarrow S \rightarrow T$$
$$R \leftarrow S \rightarrow T$$
$$R \rightarrow \fbox S \leftarrow T$$
In the second three, $R$ and $T$ are not associated ... | Controlling variables in causal diagrams
There are seven rules of association. In the first four, $R$ and $T$ are associated with each other:
$$R \rightarrow T$$
$$R \rightarrow S \rightarrow T$$
$$R \leftarrow S \rightarrow T$$
$$R \rightar |
53,143 | In regression analysis, how does one know which transformation to apply to either the response variables or features? | Consider the case of OLS regression:
$$ Y_i = a + b_1X_{i1} + \cdots +b_kX_{ik} + \epsilon_{i}$$
where $Y$ is your response variable, $a$ is the intercept, the $X$s are your independent variables (i.e. predictors), $b_1$ to $b_k$ are the slope coefficients associated with each independent variable, $\epsilon$ represent... | In regression analysis, how does one know which transformation to apply to either the response varia | Consider the case of OLS regression:
$$ Y_i = a + b_1X_{i1} + \cdots +b_kX_{ik} + \epsilon_{i}$$
where $Y$ is your response variable, $a$ is the intercept, the $X$s are your independent variables (i.e | In regression analysis, how does one know which transformation to apply to either the response variables or features?
Consider the case of OLS regression:
$$ Y_i = a + b_1X_{i1} + \cdots +b_kX_{ik} + \epsilon_{i}$$
where $Y$ is your response variable, $a$ is the intercept, the $X$s are your independent variables (i.e. ... | In regression analysis, how does one know which transformation to apply to either the response varia
Consider the case of OLS regression:
$$ Y_i = a + b_1X_{i1} + \cdots +b_kX_{ik} + \epsilon_{i}$$
where $Y$ is your response variable, $a$ is the intercept, the $X$s are your independent variables (i.e |
53,144 | In regression analysis, how does one know which transformation to apply to either the response variables or features? | If your only tool is OLS regression, then you can use methods such as Box-Cox. In the old days, this was true in practice because computers were (first) unavailable and (later) not that powerful or fast.
Nowadays, though, we have very powerful computers and methods that can implement many methods other than OLS, in par... | In regression analysis, how does one know which transformation to apply to either the response varia | If your only tool is OLS regression, then you can use methods such as Box-Cox. In the old days, this was true in practice because computers were (first) unavailable and (later) not that powerful or fa | In regression analysis, how does one know which transformation to apply to either the response variables or features?
If your only tool is OLS regression, then you can use methods such as Box-Cox. In the old days, this was true in practice because computers were (first) unavailable and (later) not that powerful or fast... | In regression analysis, how does one know which transformation to apply to either the response varia
If your only tool is OLS regression, then you can use methods such as Box-Cox. In the old days, this was true in practice because computers were (first) unavailable and (later) not that powerful or fa |
53,145 | In regression analysis, how does one know which transformation to apply to either the response variables or features? | Mosteller and Tukey (1977) offered their bulging rule, suggesting transformations (of X or Y) based on eyeballing the shape of the untransformed relationship.
Mosteller and Tukey suggested looking at the plot of untransformed X and Y, matching the curve that you see to one of the four quadrant curves in the diagram, a... | In regression analysis, how does one know which transformation to apply to either the response varia | Mosteller and Tukey (1977) offered their bulging rule, suggesting transformations (of X or Y) based on eyeballing the shape of the untransformed relationship.
Mosteller and Tukey suggested looking at | In regression analysis, how does one know which transformation to apply to either the response variables or features?
Mosteller and Tukey (1977) offered their bulging rule, suggesting transformations (of X or Y) based on eyeballing the shape of the untransformed relationship.
Mosteller and Tukey suggested looking at t... | In regression analysis, how does one know which transformation to apply to either the response varia
Mosteller and Tukey (1977) offered their bulging rule, suggesting transformations (of X or Y) based on eyeballing the shape of the untransformed relationship.
Mosteller and Tukey suggested looking at |
53,146 | Mixed-effect logistic regression with very large dataset | You do not indicate if you have multiple human subjects in your study who are supposed to mention these preposition phrases. If you do, your model would need to reflect this by also including a random effect for subject.
With your model as it currently is formulated, the odds ratio of 1.26 refers just to a "typical" v... | Mixed-effect logistic regression with very large dataset | You do not indicate if you have multiple human subjects in your study who are supposed to mention these preposition phrases. If you do, your model would need to reflect this by also including a random | Mixed-effect logistic regression with very large dataset
You do not indicate if you have multiple human subjects in your study who are supposed to mention these preposition phrases. If you do, your model would need to reflect this by also including a random effect for subject.
With your model as it currently is formul... | Mixed-effect logistic regression with very large dataset
You do not indicate if you have multiple human subjects in your study who are supposed to mention these preposition phrases. If you do, your model would need to reflect this by also including a random |
53,147 | Mixed-effect logistic regression with very large dataset | As you say, statistical significance is due to the sample size. You should focus on the effect size. 1.26 is not a particularly low odds ratio in many disciplines but in others it is, so it really does depend on what you consider to be low, for your study, in your domain. | Mixed-effect logistic regression with very large dataset | As you say, statistical significance is due to the sample size. You should focus on the effect size. 1.26 is not a particularly low odds ratio in many disciplines but in others it is, so it really doe | Mixed-effect logistic regression with very large dataset
As you say, statistical significance is due to the sample size. You should focus on the effect size. 1.26 is not a particularly low odds ratio in many disciplines but in others it is, so it really does depend on what you consider to be low, for your study, in you... | Mixed-effect logistic regression with very large dataset
As you say, statistical significance is due to the sample size. You should focus on the effect size. 1.26 is not a particularly low odds ratio in many disciplines but in others it is, so it really doe |
53,148 | Forecast accuracy rolling window | Symmetric mape (SMAPE) will be useful for you . Pursue https://stats.stackexchange.com/search?q=SMAPE . | Forecast accuracy rolling window | Symmetric mape (SMAPE) will be useful for you . Pursue https://stats.stackexchange.com/search?q=SMAPE . | Forecast accuracy rolling window
Symmetric mape (SMAPE) will be useful for you . Pursue https://stats.stackexchange.com/search?q=SMAPE . | Forecast accuracy rolling window
Symmetric mape (SMAPE) will be useful for you . Pursue https://stats.stackexchange.com/search?q=SMAPE . |
53,149 | Forecast accuracy rolling window | @IrishStat's suggestion to use SMAPE is probably the most correct.
A log(x+1) transform could also work.
It would be useful to know your application. What is the proportion of 0's? I am curious if you have an intermittent time series.
Intermittent time series are often modeled with Crostons method or the Poisson dist... | Forecast accuracy rolling window | @IrishStat's suggestion to use SMAPE is probably the most correct.
A log(x+1) transform could also work.
It would be useful to know your application. What is the proportion of 0's? I am curious if yo | Forecast accuracy rolling window
@IrishStat's suggestion to use SMAPE is probably the most correct.
A log(x+1) transform could also work.
It would be useful to know your application. What is the proportion of 0's? I am curious if you have an intermittent time series.
Intermittent time series are often modeled with Cr... | Forecast accuracy rolling window
@IrishStat's suggestion to use SMAPE is probably the most correct.
A log(x+1) transform could also work.
It would be useful to know your application. What is the proportion of 0's? I am curious if yo |
53,150 | Forecast accuracy rolling window | Another interesting measure that can deal with zero actuals:
Relative Total Absolute Error: mean of all absolute errors devided by the mean of the absolute actuals.
This measure is a MAPE where the sum is moved into the fraction. An advantage is that the RTAE can deal with zero actuals.
That said, SMAPE and the RTAE a... | Forecast accuracy rolling window | Another interesting measure that can deal with zero actuals:
Relative Total Absolute Error: mean of all absolute errors devided by the mean of the absolute actuals.
This measure is a MAPE where the s | Forecast accuracy rolling window
Another interesting measure that can deal with zero actuals:
Relative Total Absolute Error: mean of all absolute errors devided by the mean of the absolute actuals.
This measure is a MAPE where the sum is moved into the fraction. An advantage is that the RTAE can deal with zero actuals... | Forecast accuracy rolling window
Another interesting measure that can deal with zero actuals:
Relative Total Absolute Error: mean of all absolute errors devided by the mean of the absolute actuals.
This measure is a MAPE where the s |
53,151 | Why doesnt Conditional Variational AutoEncoders(CVAE) cluster data like the vanila VAE? | In a conditional VAE, the approximate posterior is already conditioned on the class -- $q(z|X,c)$, so there is no need for the latent space to separate the class of each input.
How does it know how to produce different classes
The decoder is also conditioned on the class.
How should I interpret what is going on here... | Why doesnt Conditional Variational AutoEncoders(CVAE) cluster data like the vanila VAE? | In a conditional VAE, the approximate posterior is already conditioned on the class -- $q(z|X,c)$, so there is no need for the latent space to separate the class of each input.
How does it know how t | Why doesnt Conditional Variational AutoEncoders(CVAE) cluster data like the vanila VAE?
In a conditional VAE, the approximate posterior is already conditioned on the class -- $q(z|X,c)$, so there is no need for the latent space to separate the class of each input.
How does it know how to produce different classes
The... | Why doesnt Conditional Variational AutoEncoders(CVAE) cluster data like the vanila VAE?
In a conditional VAE, the approximate posterior is already conditioned on the class -- $q(z|X,c)$, so there is no need for the latent space to separate the class of each input.
How does it know how t |
53,152 | What does "large grant" mean in machine learning? | Grants fund research. The joke is that statistics and machine learning receive dramatically different amounts of funding, so what counts as a "large" amount of funding (a "large grant") is different between the two fields. | What does "large grant" mean in machine learning? | Grants fund research. The joke is that statistics and machine learning receive dramatically different amounts of funding, so what counts as a "large" amount of funding (a "large grant") is different b | What does "large grant" mean in machine learning?
Grants fund research. The joke is that statistics and machine learning receive dramatically different amounts of funding, so what counts as a "large" amount of funding (a "large grant") is different between the two fields. | What does "large grant" mean in machine learning?
Grants fund research. The joke is that statistics and machine learning receive dramatically different amounts of funding, so what counts as a "large" amount of funding (a "large grant") is different b |
53,153 | Does the sum of two independent exponentially distributed random variables with different rate parameters follow a gamma distribution? | If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution
\begin{align}
f_Z(z)
&=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx
\\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx
\\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambd... | Does the sum of two independent exponentially distributed random variables with different rate param | If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution
\begin{align}
f_Z(z)
&=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx
\\&=\l | Does the sum of two independent exponentially distributed random variables with different rate parameters follow a gamma distribution?
If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution
\begin{align}
f_Z(z)
&=\int_{-\infty}^\infty f_Y(z-x)... | Does the sum of two independent exponentially distributed random variables with different rate param
If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution
\begin{align}
f_Z(z)
&=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx
\\&=\l |
53,154 | Why do attention models need to choose a maximum sentence length? | A "typical" attention mechanism might assign the weight $w_i$ to one of the source vectors as $w_i \propto \exp(u_i^Tv)$ where $u_i$ is the $i$th "source" vector and $v$ is the query vector. The attention mechanism described in OP from "Pointer Networks" opts for something slightly more involved: $w_i \propto \exp(q^T ... | Why do attention models need to choose a maximum sentence length? | A "typical" attention mechanism might assign the weight $w_i$ to one of the source vectors as $w_i \propto \exp(u_i^Tv)$ where $u_i$ is the $i$th "source" vector and $v$ is the query vector. The atten | Why do attention models need to choose a maximum sentence length?
A "typical" attention mechanism might assign the weight $w_i$ to one of the source vectors as $w_i \propto \exp(u_i^Tv)$ where $u_i$ is the $i$th "source" vector and $v$ is the query vector. The attention mechanism described in OP from "Pointer Networks"... | Why do attention models need to choose a maximum sentence length?
A "typical" attention mechanism might assign the weight $w_i$ to one of the source vectors as $w_i \propto \exp(u_i^Tv)$ where $u_i$ is the $i$th "source" vector and $v$ is the query vector. The atten |
53,155 | Why do attention models need to choose a maximum sentence length? | It is only an efficiency issue. In theory, the attention mechanism can work with arbitrarily long sequences. The reason is that batches must be padded to the same length.
Sentences of the maximum length will use all the attention weights, while shorter sentences will only use the first few.
By this sentence they mean... | Why do attention models need to choose a maximum sentence length? | It is only an efficiency issue. In theory, the attention mechanism can work with arbitrarily long sequences. The reason is that batches must be padded to the same length.
Sentences of the maximum len | Why do attention models need to choose a maximum sentence length?
It is only an efficiency issue. In theory, the attention mechanism can work with arbitrarily long sequences. The reason is that batches must be padded to the same length.
Sentences of the maximum length will use all the attention weights, while shorter ... | Why do attention models need to choose a maximum sentence length?
It is only an efficiency issue. In theory, the attention mechanism can work with arbitrarily long sequences. The reason is that batches must be padded to the same length.
Sentences of the maximum len |
53,156 | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$? | Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. This condition is much stronger than what you actually need to get the result i... | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$? | Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?
Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. This cond... | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?
Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies |
53,157 | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$? | Under the assumption that $X$ and $Y$ are independent and continuous,
\begin{align*}\Bbb P(X<Y)&=\Bbb E^Y[\Bbb I_{X<Y}\mid Y]\\ &=\Bbb E^Y[F_X(Y)]\\&>\Bbb E^Y[F_Y(Y)]\\ &=\int_{\Bbb R} F_Y(y) \, \text{d}F_Y(y) \\&= \frac{1}{2} \int_{\Bbb R} \, \text{d}F_Y^2(y)\\&=\frac{1}{2}F_Y^2(\infty)-\frac{1}{2}F_Y^2(-\infty)\\&=1/... | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$? | Under the assumption that $X$ and $Y$ are independent and continuous,
\begin{align*}\Bbb P(X<Y)&=\Bbb E^Y[\Bbb I_{X<Y}\mid Y]\\ &=\Bbb E^Y[F_X(Y)]\\&>\Bbb E^Y[F_Y(Y)]\\ &=\int_{\Bbb R} F_Y(y) \, \text | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?
Under the assumption that $X$ and $Y$ are independent and continuous,
\begin{align*}\Bbb P(X<Y)&=\Bbb E^Y[\Bbb I_{X<Y}\mid Y]\\ &=\Bbb E^Y[F_X(Y)]\\&>\Bbb E^Y[F_Y(Y)]\\ &=\int_{\Bbb R} F_Y(y) \, \text{d}F_Y(y) \\&= \frac{1}{2} \int_{\Bbb R} \, \text{... | If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?
Under the assumption that $X$ and $Y$ are independent and continuous,
\begin{align*}\Bbb P(X<Y)&=\Bbb E^Y[\Bbb I_{X<Y}\mid Y]\\ &=\Bbb E^Y[F_X(Y)]\\&>\Bbb E^Y[F_Y(Y)]\\ &=\int_{\Bbb R} F_Y(y) \, \text |
53,158 | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variable | Notice first
\begin{eqnarray} \Gamma(1/2) &=& \int_{0}^{\infty}y^{-1/2}e^{-y}dy \\
&=& \int_{0}^{\infty} \sqrt{2}z^{-1} e^{-z^2/2} z \ d z \qquad \text{ (substitute $y=\frac{z^2}{2}$ )} \\
&=& \frac{1}{2} \int_{-\infty}^{\infty} \sqrt{2} e^{-z^2/2} \ d z \qquad \text{ (since even function)} \\
&=& ... | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variab | Notice first
\begin{eqnarray} \Gamma(1/2) &=& \int_{0}^{\infty}y^{-1/2}e^{-y}dy \\
&=& \int_{0}^{\infty} \sqrt{2}z^{-1} e^{-z^2/2} z \ d z \qquad \text{ (substitute $y=\frac{z^2}{2}$ )} \\ | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variable
Notice first
\begin{eqnarray} \Gamma(1/2) &=& \int_{0}^{\infty}y^{-1/2}e^{-y}dy \\
&=& \int_{0}^{\infty} \sqrt{2}z^{-1} e^{-z^2/2} z \ d z \qquad \text{ (substitute $y=\frac{z^2}{2}$ )} \\
&=& \frac{1}{2... | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variab
Notice first
\begin{eqnarray} \Gamma(1/2) &=& \int_{0}^{\infty}y^{-1/2}e^{-y}dy \\
&=& \int_{0}^{\infty} \sqrt{2}z^{-1} e^{-z^2/2} z \ d z \qquad \text{ (substitute $y=\frac{z^2}{2}$ )} \\ |
53,159 | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variable | This is a perfect example of a question-begging mathematical "proof": The kind of "proof" you are trying to construct here is really not terribly useful. If you don't know how to prove the integral:
$$\int \limits_0^\infty y^{-1/2} e^{-y} dy = \sqrt{\pi},$$
then you need to ask yourself why are you willing to assume ... | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variab | This is a perfect example of a question-begging mathematical "proof": The kind of "proof" you are trying to construct here is really not terribly useful. If you don't know how to prove the integral:
| Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variable
This is a perfect example of a question-begging mathematical "proof": The kind of "proof" you are trying to construct here is really not terribly useful. If you don't know how to prove the integral:
$$\int \limits_0^... | Proving $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the expected value of standard normal variab
This is a perfect example of a question-begging mathematical "proof": The kind of "proof" you are trying to construct here is really not terribly useful. If you don't know how to prove the integral:
|
53,160 | Can Adjusted R squared be equal to 1? | Dan and Michael point out the relevant issues. Just for completeness, the relationship between adjusted $R^2$ and $R^2$ is given by (see, e.g., here)
$$
R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-K},
$$
(with $K$ the number of regressors, including the constant). This shows that $R^2_{adjusted}=1$ if $R^2=1$, unless (see bel... | Can Adjusted R squared be equal to 1? | Dan and Michael point out the relevant issues. Just for completeness, the relationship between adjusted $R^2$ and $R^2$ is given by (see, e.g., here)
$$
R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-K},
$$
(wi | Can Adjusted R squared be equal to 1?
Dan and Michael point out the relevant issues. Just for completeness, the relationship between adjusted $R^2$ and $R^2$ is given by (see, e.g., here)
$$
R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-K},
$$
(with $K$ the number of regressors, including the constant). This shows that $R^2_{ad... | Can Adjusted R squared be equal to 1?
Dan and Michael point out the relevant issues. Just for completeness, the relationship between adjusted $R^2$ and $R^2$ is given by (see, e.g., here)
$$
R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-K},
$$
(wi |
53,161 | Can Adjusted R squared be equal to 1? | An adjusted R squared equal to one implies perfect prediction and is an indication of a problem in your model. Adjusted R squared is a penalised version of R square, which is a way of describing the ratio of the residual sum of squares to the total sum of squares - as you approach 1 the implication is that there is no ... | Can Adjusted R squared be equal to 1? | An adjusted R squared equal to one implies perfect prediction and is an indication of a problem in your model. Adjusted R squared is a penalised version of R square, which is a way of describing the r | Can Adjusted R squared be equal to 1?
An adjusted R squared equal to one implies perfect prediction and is an indication of a problem in your model. Adjusted R squared is a penalised version of R square, which is a way of describing the ratio of the residual sum of squares to the total sum of squares - as you approach ... | Can Adjusted R squared be equal to 1?
An adjusted R squared equal to one implies perfect prediction and is an indication of a problem in your model. Adjusted R squared is a penalised version of R square, which is a way of describing the r |
53,162 | How bad is Cholesky decomposition for OLS? | What you are seeing is exactly what one would expect. The condition number of your matrix $X$ is about $10^6$. Double precision floating point calculations give about 18 figures of accuracy so, in double precision, you would expect to get at most $18-6=12$ significant figures for the estimated coefficients from QR and ... | How bad is Cholesky decomposition for OLS? | What you are seeing is exactly what one would expect. The condition number of your matrix $X$ is about $10^6$. Double precision floating point calculations give about 18 figures of accuracy so, in dou | How bad is Cholesky decomposition for OLS?
What you are seeing is exactly what one would expect. The condition number of your matrix $X$ is about $10^6$. Double precision floating point calculations give about 18 figures of accuracy so, in double precision, you would expect to get at most $18-6=12$ significant figures ... | How bad is Cholesky decomposition for OLS?
What you are seeing is exactly what one would expect. The condition number of your matrix $X$ is about $10^6$. Double precision floating point calculations give about 18 figures of accuracy so, in dou |
53,163 | Correlation vs Chi Square | First, in your opening sentence, "affected by" should be "linearly related to". Two variables can be correlated and have not the slightest causal relationship and correlation does not measure all relationships, just linear ones (either on the quantities themselves (Pearson) or their ranks (Spearman)).
So, correlation i... | Correlation vs Chi Square | First, in your opening sentence, "affected by" should be "linearly related to". Two variables can be correlated and have not the slightest causal relationship and correlation does not measure all rela | Correlation vs Chi Square
First, in your opening sentence, "affected by" should be "linearly related to". Two variables can be correlated and have not the slightest causal relationship and correlation does not measure all relationships, just linear ones (either on the quantities themselves (Pearson) or their ranks (Spe... | Correlation vs Chi Square
First, in your opening sentence, "affected by" should be "linearly related to". Two variables can be correlated and have not the slightest causal relationship and correlation does not measure all rela |
53,164 | Correlation vs Chi Square | generally, Chi square is a non-parametric test that is used to show association between two qualitative variables (like gender and eye color) ; while correlation (Pearson coefficient) is used to test the correlation between two quantitative variables (like heart rate and blood pressure) | Correlation vs Chi Square | generally, Chi square is a non-parametric test that is used to show association between two qualitative variables (like gender and eye color) ; while correlation (Pearson coefficient) is used to test | Correlation vs Chi Square
generally, Chi square is a non-parametric test that is used to show association between two qualitative variables (like gender and eye color) ; while correlation (Pearson coefficient) is used to test the correlation between two quantitative variables (like heart rate and blood pressure) | Correlation vs Chi Square
generally, Chi square is a non-parametric test that is used to show association between two qualitative variables (like gender and eye color) ; while correlation (Pearson coefficient) is used to test |
53,165 | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$? | Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly implies $f(x)<f(a)$ and vice versa, i.e. if $f(x)<f(... | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$? | Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $ | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?
Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $f(x)$ is a monotonic (strictly) increasing function, $x<a$ directly ... | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?
Consider the set of $x$, call it $S$, where $x<a$. You seek for the probability, $P(S)$. Any expression that lead to $S$ produces the exact same probability, $b$, irrespective of its decleration. If $ |
53,166 | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$? | If $f$ is strictly increasing then you have:
$$\begin{equation} \begin{aligned}
\{ X < a \}
&= \{ \omega \in \Omega | X(\omega) < a \} \\[6pt]
&= \{ \omega \in \Omega | f(X(\omega)) < f(a) \} \\[6pt]
&= \{ \omega \in \Omega | f(X)(\omega) < f(a) \} \\[6pt]
&= \{ f(X) < f(a) \}, \\[6pt]
\end{aligned} \end{equation}$$
w... | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$? | If $f$ is strictly increasing then you have:
$$\begin{equation} \begin{aligned}
\{ X < a \}
&= \{ \omega \in \Omega | X(\omega) < a \} \\[6pt]
&= \{ \omega \in \Omega | f(X(\omega)) < f(a) \} \\[6pt] | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?
If $f$ is strictly increasing then you have:
$$\begin{equation} \begin{aligned}
\{ X < a \}
&= \{ \omega \in \Omega | X(\omega) < a \} \\[6pt]
&= \{ \omega \in \Omega | f(X(\omega)) < f(a) \} \\[6pt]
&= \{ \omega \in \Omega | f(X)(\omega) < f(a) \} \\[6pt]
&= \{ f(X)... | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?
If $f$ is strictly increasing then you have:
$$\begin{equation} \begin{aligned}
\{ X < a \}
&= \{ \omega \in \Omega | X(\omega) < a \} \\[6pt]
&= \{ \omega \in \Omega | f(X(\omega)) < f(a) \} \\[6pt] |
53,167 | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$? | No.
Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $x\leq y$, one has $f ( x ) \leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.
Then, $P(X<a) = P(X<1) = 1 \neq 0 = P(0<0) = P(f(X) < f(a))$.
Your assumption is... | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$? | No.
Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $x\leq y$, one has $f ( x ) \leq f ( y )$), consider $X$ following a uniform distr | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?
No.
Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $x\leq y$, one has $f ( x ) \leq f ( y )$), consider $X$ following a uniform distribution on $[0, 1]$, $f=0$ and $a=1$.
Then, $P(X<a) = P(X<1) = 1 \ne... | Does $\mathbb{P}(X < a) = \mathbb{P}(f(X) < f(a))$?
No.
Assuming that you call monotonically increasing a function that is non-decreasing (for all $x$ and $y$ such that $x\leq y$, one has $f ( x ) \leq f ( y )$), consider $X$ following a uniform distr |
53,168 | Binomial Regression "logit" vs "cloglog" | If you fit two models, one with logit and one with cloglog, you should report the results of both, and also carry out some type of model comparison technique and report the results of that.
As for the models, this is a great situation in which to use Bayesian multilevel models (see this paper [PDF] by Gelman). We can p... | Binomial Regression "logit" vs "cloglog" | If you fit two models, one with logit and one with cloglog, you should report the results of both, and also carry out some type of model comparison technique and report the results of that.
As for the | Binomial Regression "logit" vs "cloglog"
If you fit two models, one with logit and one with cloglog, you should report the results of both, and also carry out some type of model comparison technique and report the results of that.
As for the models, this is a great situation in which to use Bayesian multilevel models (... | Binomial Regression "logit" vs "cloglog"
If you fit two models, one with logit and one with cloglog, you should report the results of both, and also carry out some type of model comparison technique and report the results of that.
As for the |
53,169 | Measure of smoothness | If this is an image then in my understanding a row should display a gradual change between the adjacent pixels. In such a case autocorrelation (the correlation of data with itself after a shift by one pixel) should work as a measure of smoothness.
However using your example I only get a slight increase in autocorrelati... | Measure of smoothness | If this is an image then in my understanding a row should display a gradual change between the adjacent pixels. In such a case autocorrelation (the correlation of data with itself after a shift by one | Measure of smoothness
If this is an image then in my understanding a row should display a gradual change between the adjacent pixels. In such a case autocorrelation (the correlation of data with itself after a shift by one pixel) should work as a measure of smoothness.
However using your example I only get a slight inc... | Measure of smoothness
If this is an image then in my understanding a row should display a gradual change between the adjacent pixels. In such a case autocorrelation (the correlation of data with itself after a shift by one |
53,170 | Measure of smoothness | The conventional smoothness measures are based on derivatives, and are sometimes called "roughness." For instance, in smoothing splines there's a roughness penalty, which you minimize to get the smooth curve. In case of splines you want continuous first derivative, therefore, the roughness is based on the integral of t... | Measure of smoothness | The conventional smoothness measures are based on derivatives, and are sometimes called "roughness." For instance, in smoothing splines there's a roughness penalty, which you minimize to get the smoot | Measure of smoothness
The conventional smoothness measures are based on derivatives, and are sometimes called "roughness." For instance, in smoothing splines there's a roughness penalty, which you minimize to get the smooth curve. In case of splines you want continuous first derivative, therefore, the roughness is base... | Measure of smoothness
The conventional smoothness measures are based on derivatives, and are sometimes called "roughness." For instance, in smoothing splines there's a roughness penalty, which you minimize to get the smoot |
53,171 | Measure of smoothness | One way to measure non-smoothness is to first smooth the data, subtract it away and compute some measure of how much residuals do you have (i.e. sum squares of all residuals). I.e. you can apply a Laplacian filter and compute the sum of squares of the residuals for both images and compare. | Measure of smoothness | One way to measure non-smoothness is to first smooth the data, subtract it away and compute some measure of how much residuals do you have (i.e. sum squares of all residuals). I.e. you can apply a La | Measure of smoothness
One way to measure non-smoothness is to first smooth the data, subtract it away and compute some measure of how much residuals do you have (i.e. sum squares of all residuals). I.e. you can apply a Laplacian filter and compute the sum of squares of the residuals for both images and compare. | Measure of smoothness
One way to measure non-smoothness is to first smooth the data, subtract it away and compute some measure of how much residuals do you have (i.e. sum squares of all residuals). I.e. you can apply a La |
53,172 | Measure of smoothness | I totally agree with @Tim's comment about variance but I felt motivated to go a step further, as is my want. I took the 20 values and mused as to what AUTOBOX would do with these values essentially revisiting this post How to calculate the standard average of a set excluding outliers? .
AUTOBOX delivered the... | Measure of smoothness | I totally agree with @Tim's comment about variance but I felt motivated to go a step further, as is my want. I took the 20 values and mused as to what AUTOBOX would do with these values essentially | Measure of smoothness
I totally agree with @Tim's comment about variance but I felt motivated to go a step further, as is my want. I took the 20 values and mused as to what AUTOBOX would do with these values essentially revisiting this post How to calculate the standard average of a set excluding outliers? .... | Measure of smoothness
I totally agree with @Tim's comment about variance but I felt motivated to go a step further, as is my want. I took the 20 values and mused as to what AUTOBOX would do with these values essentially |
53,173 | Is Granger causality still relevant? | If you mean econometric theory, I would not be surprised to learn that most of the relevant and needed theoretical results related to the notion of Granger causality have already been derived. After all, the notion of Granger causality is nearly 50 years old. Hence, I do not expect to see much research on the theoretic... | Is Granger causality still relevant? | If you mean econometric theory, I would not be surprised to learn that most of the relevant and needed theoretical results related to the notion of Granger causality have already been derived. After a | Is Granger causality still relevant?
If you mean econometric theory, I would not be surprised to learn that most of the relevant and needed theoretical results related to the notion of Granger causality have already been derived. After all, the notion of Granger causality is nearly 50 years old. Hence, I do not expect ... | Is Granger causality still relevant?
If you mean econometric theory, I would not be surprised to learn that most of the relevant and needed theoretical results related to the notion of Granger causality have already been derived. After a |
53,174 | Is Granger causality still relevant? | "Granger causality" has nothing to do with causality. It is just that the late Sir Clive Granger was a master of marketing his tremendously important work by tremendously catchy names.
We should remember that prof. Granger was heavy into forecasting. And in forecasting, if I can forecast your birthday through your (o... | Is Granger causality still relevant? | "Granger causality" has nothing to do with causality. It is just that the late Sir Clive Granger was a master of marketing his tremendously important work by tremendously catchy names.
We should rem | Is Granger causality still relevant?
"Granger causality" has nothing to do with causality. It is just that the late Sir Clive Granger was a master of marketing his tremendously important work by tremendously catchy names.
We should remember that prof. Granger was heavy into forecasting. And in forecasting, if I can f... | Is Granger causality still relevant?
"Granger causality" has nothing to do with causality. It is just that the late Sir Clive Granger was a master of marketing his tremendously important work by tremendously catchy names.
We should rem |
53,175 | L1 and L2 penalty vs L1 and L2 norms | Norm in mathematics is some function that measures "length" or "size" of a vector. Among the popular norms, there are $\ell_1$, $\ell_2$ and $\ell_p$ norms defined as
$$\begin{align}
\|\boldsymbol{x}\|_1 &= \sum_i | x_i | \\
\| \boldsymbol{x}\|_2 &= \sqrt{ \sum_i |x_i|^2 } \\
\| \boldsymbol{x}\|_p &= \left( \sum_i | x_... | L1 and L2 penalty vs L1 and L2 norms | Norm in mathematics is some function that measures "length" or "size" of a vector. Among the popular norms, there are $\ell_1$, $\ell_2$ and $\ell_p$ norms defined as
$$\begin{align}
\|\boldsymbol{x}\ | L1 and L2 penalty vs L1 and L2 norms
Norm in mathematics is some function that measures "length" or "size" of a vector. Among the popular norms, there are $\ell_1$, $\ell_2$ and $\ell_p$ norms defined as
$$\begin{align}
\|\boldsymbol{x}\|_1 &= \sum_i | x_i | \\
\| \boldsymbol{x}\|_2 &= \sqrt{ \sum_i |x_i|^2 } \\
\| \bo... | L1 and L2 penalty vs L1 and L2 norms
Norm in mathematics is some function that measures "length" or "size" of a vector. Among the popular norms, there are $\ell_1$, $\ell_2$ and $\ell_p$ norms defined as
$$\begin{align}
\|\boldsymbol{x}\ |
53,176 | SE for estimated marginal means | The following code illustrates how this computation is done:
grid <- with(data, expand.grid(f1 = levels(f1), f2 = levels(f2)))
X <- model.matrix(~ f1 * f2, data = grid)
V <- vcov(m)
betas <- fixef(m)
grid$emmean <- c(X %*% betas)
grid$SE <- sqrt(diag(X %*% V %*% t(X)))
grid | SE for estimated marginal means | The following code illustrates how this computation is done:
grid <- with(data, expand.grid(f1 = levels(f1), f2 = levels(f2)))
X <- model.matrix(~ f1 * f2, data = grid)
V <- vcov(m)
betas <- fixef(m) | SE for estimated marginal means
The following code illustrates how this computation is done:
grid <- with(data, expand.grid(f1 = levels(f1), f2 = levels(f2)))
X <- model.matrix(~ f1 * f2, data = grid)
V <- vcov(m)
betas <- fixef(m)
grid$emmean <- c(X %*% betas)
grid$SE <- sqrt(diag(X %*% V %*% t(X)))
grid | SE for estimated marginal means
The following code illustrates how this computation is done:
grid <- with(data, expand.grid(f1 = levels(f1), f2 = levels(f2)))
X <- model.matrix(~ f1 * f2, data = grid)
V <- vcov(m)
betas <- fixef(m) |
53,177 | SE for estimated marginal means | Look at the model summary:
> m
Linear mixed model fit by REML ['lmerMod']
Formula: dep ~ f1 * f2 + (1 | sub)
Data: data
REML criterion at convergence: 371.7578
Random effects:
Groups Name Std.Dev.
sub (Intercept) 1.222
Residual 4.768
Number of obs: 64, groups: sub, 8
Fixed Effects... | SE for estimated marginal means | Look at the model summary:
> m
Linear mixed model fit by REML ['lmerMod']
Formula: dep ~ f1 * f2 + (1 | sub)
Data: data
REML criterion at convergence: 371.7578
Random effects:
Groups Name | SE for estimated marginal means
Look at the model summary:
> m
Linear mixed model fit by REML ['lmerMod']
Formula: dep ~ f1 * f2 + (1 | sub)
Data: data
REML criterion at convergence: 371.7578
Random effects:
Groups Name Std.Dev.
sub (Intercept) 1.222
Residual 4.768
Number of obs: 6... | SE for estimated marginal means
Look at the model summary:
> m
Linear mixed model fit by REML ['lmerMod']
Formula: dep ~ f1 * f2 + (1 | sub)
Data: data
REML criterion at convergence: 371.7578
Random effects:
Groups Name |
53,178 | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$ | To implicitly condition on a variable simply means that we condition on it, but we do not state it as a conditioning variable in the probability statement (i.e., the conditioning is implicit, not explicit). This is usually done for reasons of brevity, particularly if you are making a whole lot of statements where you ... | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$ | To implicitly condition on a variable simply means that we condition on it, but we do not state it as a conditioning variable in the probability statement (i.e., the conditioning is implicit, not expl | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$
To implicitly condition on a variable simply means that we condition on it, but we do not state it as a conditioning variable in the probability statement (i.e., the conditioning is implicit, not explicit). This is usu... | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$
To implicitly condition on a variable simply means that we condition on it, but we do not state it as a conditioning variable in the probability statement (i.e., the conditioning is implicit, not expl |
53,179 | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$ | In addition to the other excellent answer, here I will try to make a more explicit argument. Making the argument explicit is helpful in understanding its underlying assumptions, so we can judge when to use the argument, and when to avoid it. This will be a bayesian version of the argument made in What is the differenc... | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$ | In addition to the other excellent answer, here I will try to make a more explicit argument. Making the argument explicit is helpful in understanding its underlying assumptions, so we can judge when t | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$
In addition to the other excellent answer, here I will try to make a more explicit argument. Making the argument explicit is helpful in understanding its underlying assumptions, so we can judge when to use the argument,... | Bayesian statistical conclusions: we implicitly condition on the known values of any covariates, $x$
In addition to the other excellent answer, here I will try to make a more explicit argument. Making the argument explicit is helpful in understanding its underlying assumptions, so we can judge when t |
53,180 | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat? | The UAT says that a suitable neural network can represent any approximate continuous function, essentially taking an input, passing it through a network of particular weights, and mapping it to an output.
That says nothing about how the particular weights on that network are determined, which is what NFL deals with. Si... | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat? | The UAT says that a suitable neural network can represent any approximate continuous function, essentially taking an input, passing it through a network of particular weights, and mapping it to an out | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat?
The UAT says that a suitable neural network can represent any approximate continuous function, essentially taking an input, passing it through a network of particular weights, and mapping it to an output.
That says nothing about how t... | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat?
The UAT says that a suitable neural network can represent any approximate continuous function, essentially taking an input, passing it through a network of particular weights, and mapping it to an out |
53,181 | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat? | Approximating the optimal decision function for the training data is no guarantee of performance on unseen data. So, UAT and NFLT are not in contradiction, because approximating the training decision surface to arbitrary precision does not mean your algorithm is the best (in fact, it might be worthless). | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat? | Approximating the optimal decision function for the training data is no guarantee of performance on unseen data. So, UAT and NFLT are not in contradiction, because approximating the training decision | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat?
Approximating the optimal decision function for the training data is no guarantee of performance on unseen data. So, UAT and NFLT are not in contradiction, because approximating the training decision surface to arbitrary precision doe... | The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat?
Approximating the optimal decision function for the training data is no guarantee of performance on unseen data. So, UAT and NFLT are not in contradiction, because approximating the training decision |
53,182 | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC? | When you multiply the likelihood by the prior, the resulting function may no longer integrate to $1$, hence why you need to know the normalising constant to analytically solve the posterior.
MCMC samples randomly with frequency proportional to the posterior density (or height of the distribution at any point), without... | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC? | When you multiply the likelihood by the prior, the resulting function may no longer integrate to $1$, hence why you need to know the normalising constant to analytically solve the posterior.
MCMC samp | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC?
When you multiply the likelihood by the prior, the resulting function may no longer integrate to $1$, hence why you need to know the normalising constant to analytically solve the posterior.
MCMC samples randomly with frequency prop... | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC?
When you multiply the likelihood by the prior, the resulting function may no longer integrate to $1$, hence why you need to know the normalising constant to analytically solve the posterior.
MCMC samp |
53,183 | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC? | Not only is that allowed, but the multiple you get is still a valid likelihood function. A valid likelihood function is defined by the requirement that $L_\mathbf{x}(\theta) \propto p(\mathbf{x}|\theta)$ (i.e., it is proportional to the sampling density, with respect to the parameter vector). Multiplication of a like... | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC? | Not only is that allowed, but the multiple you get is still a valid likelihood function. A valid likelihood function is defined by the requirement that $L_\mathbf{x}(\theta) \propto p(\mathbf{x}|\the | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC?
Not only is that allowed, but the multiple you get is still a valid likelihood function. A valid likelihood function is defined by the requirement that $L_\mathbf{x}(\theta) \propto p(\mathbf{x}|\theta)$ (i.e., it is proportional t... | Does multiplying the likelihood by a constant change the Bayesian inference using MCMC?
Not only is that allowed, but the multiple you get is still a valid likelihood function. A valid likelihood function is defined by the requirement that $L_\mathbf{x}(\theta) \propto p(\mathbf{x}|\the |
53,184 | Is a distribution still considered right-skewed if the majority of responses are zero? | This is certainly possible. The most common definition for a distribution to be right skewed is that the skewness
$$ \gamma_1 := E\bigg(\Big(\frac{X-\mu}{\sigma}\Big)^3\bigg) $$
be positive.
For instance, the Poisson distribution with parameter $\lambda$ has skewness $\frac{1}{\sqrt{\lambda}}>0$, so it is always right ... | Is a distribution still considered right-skewed if the majority of responses are zero? | This is certainly possible. The most common definition for a distribution to be right skewed is that the skewness
$$ \gamma_1 := E\bigg(\Big(\frac{X-\mu}{\sigma}\Big)^3\bigg) $$
be positive.
For insta | Is a distribution still considered right-skewed if the majority of responses are zero?
This is certainly possible. The most common definition for a distribution to be right skewed is that the skewness
$$ \gamma_1 := E\bigg(\Big(\frac{X-\mu}{\sigma}\Big)^3\bigg) $$
be positive.
For instance, the Poisson distribution wit... | Is a distribution still considered right-skewed if the majority of responses are zero?
This is certainly possible. The most common definition for a distribution to be right skewed is that the skewness
$$ \gamma_1 := E\bigg(\Big(\frac{X-\mu}{\sigma}\Big)^3\bigg) $$
be positive.
For insta |
53,185 | Is a distribution still considered right-skewed if the majority of responses are zero? | A note from this 11th grade stats class states:
For a right skewed distribution, the mean is typically greater than the median. Also notice that the tail of the distribution on the right hand (positive) side is longer than on the left hand side.
As noted in the comments of this post, the above statement is a generali... | Is a distribution still considered right-skewed if the majority of responses are zero? | A note from this 11th grade stats class states:
For a right skewed distribution, the mean is typically greater than the median. Also notice that the tail of the distribution on the right hand (positi | Is a distribution still considered right-skewed if the majority of responses are zero?
A note from this 11th grade stats class states:
For a right skewed distribution, the mean is typically greater than the median. Also notice that the tail of the distribution on the right hand (positive) side is longer than on the le... | Is a distribution still considered right-skewed if the majority of responses are zero?
A note from this 11th grade stats class states:
For a right skewed distribution, the mean is typically greater than the median. Also notice that the tail of the distribution on the right hand (positi |
53,186 | Should you use an opinion based variable in modelling? (If it predicts well) | I agree with Noah; it is not really a technical statistical question per se. There are several questions you need to have an clear answer.
Do you have a "consistent" subjective rating?
Let say your training data come from an opinion of an existing employee, will the assessment of a new employee going to have the same... | Should you use an opinion based variable in modelling? (If it predicts well) | I agree with Noah; it is not really a technical statistical question per se. There are several questions you need to have an clear answer.
Do you have a "consistent" subjective rating?
Let say your | Should you use an opinion based variable in modelling? (If it predicts well)
I agree with Noah; it is not really a technical statistical question per se. There are several questions you need to have an clear answer.
Do you have a "consistent" subjective rating?
Let say your training data come from an opinion of an ex... | Should you use an opinion based variable in modelling? (If it predicts well)
I agree with Noah; it is not really a technical statistical question per se. There are several questions you need to have an clear answer.
Do you have a "consistent" subjective rating?
Let say your |
53,187 | Should you use an opinion based variable in modelling? (If it predicts well) | If you used stepwise regression, it is possible that you are making a type I error and capitalizing on chance, so be careful about interpreting results from it without a cross-validation sample. In addition, if this variable is highly correlated with another variable in your sample (e.g., wealth), the fact that it emer... | Should you use an opinion based variable in modelling? (If it predicts well) | If you used stepwise regression, it is possible that you are making a type I error and capitalizing on chance, so be careful about interpreting results from it without a cross-validation sample. In ad | Should you use an opinion based variable in modelling? (If it predicts well)
If you used stepwise regression, it is possible that you are making a type I error and capitalizing on chance, so be careful about interpreting results from it without a cross-validation sample. In addition, if this variable is highly correlat... | Should you use an opinion based variable in modelling? (If it predicts well)
If you used stepwise regression, it is possible that you are making a type I error and capitalizing on chance, so be careful about interpreting results from it without a cross-validation sample. In ad |
53,188 | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm? | One very useful fact of the standard uniform distribution is for any $r \in [0,1]$ $P(u \leq r)= r$ for $u \sim U(0,1)$. We're doing a stochastic hill-climbing procedure which means we have two cases to consider: (1) if it's more likely to move $x\to x'$ than $x'\to x$, i.e. the proposed point is "better", we'll alway... | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm? | One very useful fact of the standard uniform distribution is for any $r \in [0,1]$ $P(u \leq r)= r$ for $u \sim U(0,1)$. We're doing a stochastic hill-climbing procedure which means we have two cases | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm?
One very useful fact of the standard uniform distribution is for any $r \in [0,1]$ $P(u \leq r)= r$ for $u \sim U(0,1)$. We're doing a stochastic hill-climbing procedure which means we have two cases to consider: (1) if it's more likely... | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm?
One very useful fact of the standard uniform distribution is for any $r \in [0,1]$ $P(u \leq r)= r$ for $u \sim U(0,1)$. We're doing a stochastic hill-climbing procedure which means we have two cases |
53,189 | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm? | $\alpha$ is the acceptance probability. You don't have to clip it in your code since as you point out $u \le \alpha_{\textrm{clipped}}$ iff $u \le \alpha_{\textrm{unclipped}}$. | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm? | $\alpha$ is the acceptance probability. You don't have to clip it in your code since as you point out $u \le \alpha_{\textrm{clipped}}$ iff $u \le \alpha_{\textrm{unclipped}}$. | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm?
$\alpha$ is the acceptance probability. You don't have to clip it in your code since as you point out $u \le \alpha_{\textrm{clipped}}$ iff $u \le \alpha_{\textrm{unclipped}}$. | Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm?
$\alpha$ is the acceptance probability. You don't have to clip it in your code since as you point out $u \le \alpha_{\textrm{clipped}}$ iff $u \le \alpha_{\textrm{unclipped}}$. |
53,190 | When is it better to use Multiple Linear Regression instead of Polynomial Regression? | First of all note that polynomial regression is a special case of multiple linear regression.
Let's consider three models:
Model1
$Y = \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$
and
Model2
$Y = \beta_1 x_1 + \beta_2 x_1^2 + \beta_3 x_1^3 + \epsilon$
and
Model3
$Y = \beta_1 x_1 + \beta_2 x_1^2 + \beta_3 x_1... | When is it better to use Multiple Linear Regression instead of Polynomial Regression? | First of all note that polynomial regression is a special case of multiple linear regression.
Let's consider three models:
Model1
$Y = \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$
and
Model2
| When is it better to use Multiple Linear Regression instead of Polynomial Regression?
First of all note that polynomial regression is a special case of multiple linear regression.
Let's consider three models:
Model1
$Y = \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$
and
Model2
$Y = \beta_1 x_1 + \beta_2 x_1^2 +... | When is it better to use Multiple Linear Regression instead of Polynomial Regression?
First of all note that polynomial regression is a special case of multiple linear regression.
Let's consider three models:
Model1
$Y = \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$
and
Model2
|
53,191 | Does the interval/ratio distinction ever matter? | As an example, consider a variety of generalized linear models -- for example, gamma, Poisson or inverse Gaussian regression models. Those models (plus some others) correspond to models for data that could be classified as ratio data (the zero is meaningful, 2 really is twice as much as 1, etc).
Further, these models a... | Does the interval/ratio distinction ever matter? | As an example, consider a variety of generalized linear models -- for example, gamma, Poisson or inverse Gaussian regression models. Those models (plus some others) correspond to models for data that | Does the interval/ratio distinction ever matter?
As an example, consider a variety of generalized linear models -- for example, gamma, Poisson or inverse Gaussian regression models. Those models (plus some others) correspond to models for data that could be classified as ratio data (the zero is meaningful, 2 really is ... | Does the interval/ratio distinction ever matter?
As an example, consider a variety of generalized linear models -- for example, gamma, Poisson or inverse Gaussian regression models. Those models (plus some others) correspond to models for data that |
53,192 | Does the interval/ratio distinction ever matter? | Computing percentages or ratios seems to be the easiest statistical operation that is valid for ratio scale but invalid for an interval scale.
Temperature in Celsius or Fahrenheit are examples of interval scale measurement (no fixed 0). For example, it makes no sense to say that 110°C is 10% hotter than 100°C, because ... | Does the interval/ratio distinction ever matter? | Computing percentages or ratios seems to be the easiest statistical operation that is valid for ratio scale but invalid for an interval scale.
Temperature in Celsius or Fahrenheit are examples of inte | Does the interval/ratio distinction ever matter?
Computing percentages or ratios seems to be the easiest statistical operation that is valid for ratio scale but invalid for an interval scale.
Temperature in Celsius or Fahrenheit are examples of interval scale measurement (no fixed 0). For example, it makes no sense to ... | Does the interval/ratio distinction ever matter?
Computing percentages or ratios seems to be the easiest statistical operation that is valid for ratio scale but invalid for an interval scale.
Temperature in Celsius or Fahrenheit are examples of inte |
53,193 | SMOTE data balance - before or during Cross-Validation | The second method should be preferred for exactly the reason that you gave to justify the first. The first method uses the whole data set to synthesize new samples. Cross validation is excluding points from training to give an accurate assessment of the error rate on new data. If you use SMOTE first, information from ... | SMOTE data balance - before or during Cross-Validation | The second method should be preferred for exactly the reason that you gave to justify the first. The first method uses the whole data set to synthesize new samples. Cross validation is excluding point | SMOTE data balance - before or during Cross-Validation
The second method should be preferred for exactly the reason that you gave to justify the first. The first method uses the whole data set to synthesize new samples. Cross validation is excluding points from training to give an accurate assessment of the error rate ... | SMOTE data balance - before or during Cross-Validation
The second method should be preferred for exactly the reason that you gave to justify the first. The first method uses the whole data set to synthesize new samples. Cross validation is excluding point |
53,194 | SMOTE data balance - before or during Cross-Validation | Method 1 should not be used as it leaks information from the test partition into the training set in each fold of the cross-validation. This is because a synthetic example may lie between a real training pattern and a real test pattern or between two real test patterns. Consider a synthetic example generated by rando... | SMOTE data balance - before or during Cross-Validation | Method 1 should not be used as it leaks information from the test partition into the training set in each fold of the cross-validation. This is because a synthetic example may lie between a real trai | SMOTE data balance - before or during Cross-Validation
Method 1 should not be used as it leaks information from the test partition into the training set in each fold of the cross-validation. This is because a synthetic example may lie between a real training pattern and a real test pattern or between two real test pat... | SMOTE data balance - before or during Cross-Validation
Method 1 should not be used as it leaks information from the test partition into the training set in each fold of the cross-validation. This is because a synthetic example may lie between a real trai |
53,195 | Fitting Weibull distribution in R | Here's the fitted pdf and cdf (Weibull) for each of locations 1 to 3:
Let's break down what we need to do here, keeping in mind that the end goal is to estimate the cumulative proportion of area planted with a certain crop at some value for the random variable time $X$:
The first step is to fit a distribution (e.g. $... | Fitting Weibull distribution in R | Here's the fitted pdf and cdf (Weibull) for each of locations 1 to 3:
Let's break down what we need to do here, keeping in mind that the end goal is to estimate the cumulative proportion of area plan | Fitting Weibull distribution in R
Here's the fitted pdf and cdf (Weibull) for each of locations 1 to 3:
Let's break down what we need to do here, keeping in mind that the end goal is to estimate the cumulative proportion of area planted with a certain crop at some value for the random variable time $X$:
The first ste... | Fitting Weibull distribution in R
Here's the fitted pdf and cdf (Weibull) for each of locations 1 to 3:
Let's break down what we need to do here, keeping in mind that the end goal is to estimate the cumulative proportion of area plan |
53,196 | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independent? | Let the random vector $\mathbf{X} \sim N_{p}(\mathbf{\mu},\mathbf{\Sigma})$. If we partition $\mathbf{X}$ as $\left(\begin{array}{c}
\mathbf{X^{(1)}}\\
\mathbf{X^{(2)}}
\end{array}\right)$ and take a non-singular linear transformation to the components of
$\mathbf{X}$ as
\begin{eqnarray*}
\mathbf{Y^{(1)}} &=& \mathbf{... | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independe | Let the random vector $\mathbf{X} \sim N_{p}(\mathbf{\mu},\mathbf{\Sigma})$. If we partition $\mathbf{X}$ as $\left(\begin{array}{c}
\mathbf{X^{(1)}}\\
\mathbf{X^{(2)}}
\end{array}\right)$ and take a | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independent?
Let the random vector $\mathbf{X} \sim N_{p}(\mathbf{\mu},\mathbf{\Sigma})$. If we partition $\mathbf{X}$ as $\left(\begin{array}{c}
\mathbf{X^{(1)}}\\
\mathbf{X^{(2)}}
\end{array}\right)$ and take a non-singular lin... | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independe
Let the random vector $\mathbf{X} \sim N_{p}(\mathbf{\mu},\mathbf{\Sigma})$. If we partition $\mathbf{X}$ as $\left(\begin{array}{c}
\mathbf{X^{(1)}}\\
\mathbf{X^{(2)}}
\end{array}\right)$ and take a |
53,197 | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independent? | Hint:
One can see that $Z$, being a linear combination of jointly normal variables $X$ and $Y$, is itself univariate normal. And two linear combinations (namely, $Z$ and $X$) of jointly normal variables are themselves jointly normal. So one possible way is to find the joint moment generating function of $(Z,X)$ to see ... | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independe | Hint:
One can see that $Z$, being a linear combination of jointly normal variables $X$ and $Y$, is itself univariate normal. And two linear combinations (namely, $Z$ and $X$) of jointly normal variabl | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independent?
Hint:
One can see that $Z$, being a linear combination of jointly normal variables $X$ and $Y$, is itself univariate normal. And two linear combinations (namely, $Z$ and $X$) of jointly normal variables are themselve... | If $(X,Y) \sim \mathcal N(0,\Sigma)$, are $Z = Y - \rho\frac{\sigma_Y}{\sigma_X}X$ and $X$ independe
Hint:
One can see that $Z$, being a linear combination of jointly normal variables $X$ and $Y$, is itself univariate normal. And two linear combinations (namely, $Z$ and $X$) of jointly normal variabl |
53,198 | difference between accuracy and Rand index (R) | Rand index is accuracy computed not in the raw data (which does not work unless you have you data where class 1 is cluster 1).
Instead, it is the accuracy on pairs of points, which is invariant to renaming clusters.
In binary classification, the common definition of accuracy is: (TP+TN)/(TP+FP+FN+TN), that should make ... | difference between accuracy and Rand index (R) | Rand index is accuracy computed not in the raw data (which does not work unless you have you data where class 1 is cluster 1).
Instead, it is the accuracy on pairs of points, which is invariant to ren | difference between accuracy and Rand index (R)
Rand index is accuracy computed not in the raw data (which does not work unless you have you data where class 1 is cluster 1).
Instead, it is the accuracy on pairs of points, which is invariant to renaming clusters.
In binary classification, the common definition of accura... | difference between accuracy and Rand index (R)
Rand index is accuracy computed not in the raw data (which does not work unless you have you data where class 1 is cluster 1).
Instead, it is the accuracy on pairs of points, which is invariant to ren |
53,199 | difference between accuracy and Rand index (R) | The confusion matrix that you use to calculate RI is different than that of accuracy.
Definition of confusion matrix in the Rand Index (RI):
+--------------------------------+--------------------------------------+
| TP: | FN: |
| Same class + same cluster ... | difference between accuracy and Rand index (R) | The confusion matrix that you use to calculate RI is different than that of accuracy.
Definition of confusion matrix in the Rand Index (RI):
+--------------------------------+------------------------ | difference between accuracy and Rand index (R)
The confusion matrix that you use to calculate RI is different than that of accuracy.
Definition of confusion matrix in the Rand Index (RI):
+--------------------------------+--------------------------------------+
| TP: | FN: ... | difference between accuracy and Rand index (R)
The confusion matrix that you use to calculate RI is different than that of accuracy.
Definition of confusion matrix in the Rand Index (RI):
+--------------------------------+------------------------ |
53,200 | difference between accuracy and Rand index (R) | Couple of points to remember:
Rand Index looks at similarity between any two clustering methods. Generally, there are no 'true' labels present here. Whereas for calculating accuracy, you need to compare the true labels with the predicted labels.
Like Has QUIT--Anony-Mousse answered, Rand Index looks for the relationsh... | difference between accuracy and Rand index (R) | Couple of points to remember:
Rand Index looks at similarity between any two clustering methods. Generally, there are no 'true' labels present here. Whereas for calculating accuracy, you need to comp | difference between accuracy and Rand index (R)
Couple of points to remember:
Rand Index looks at similarity between any two clustering methods. Generally, there are no 'true' labels present here. Whereas for calculating accuracy, you need to compare the true labels with the predicted labels.
Like Has QUIT--Anony-Mouss... | difference between accuracy and Rand index (R)
Couple of points to remember:
Rand Index looks at similarity between any two clustering methods. Generally, there are no 'true' labels present here. Whereas for calculating accuracy, you need to comp |
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