idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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54,301 | Expectation of rational formula | I symbolically evaluated the double integral for the desired expectation in MAPLE as follows (note that I executed expand just so that I could get the image to display better).
expand(int(int(a*x^2*y^2/(1+b*x^2)*1/(2*Pi)*exp(-1/2*x^2)*exp(-1/2*y^2),x=-infinity..infinity),y=-infinity..infinity));
Here is the result:
I... | Expectation of rational formula | I symbolically evaluated the double integral for the desired expectation in MAPLE as follows (note that I executed expand just so that I could get the image to display better).
expand(int(int(a*x^2*y^ | Expectation of rational formula
I symbolically evaluated the double integral for the desired expectation in MAPLE as follows (note that I executed expand just so that I could get the image to display better).
expand(int(int(a*x^2*y^2/(1+b*x^2)*1/(2*Pi)*exp(-1/2*x^2)*exp(-1/2*y^2),x=-infinity..infinity),y=-infinity..inf... | Expectation of rational formula
I symbolically evaluated the double integral for the desired expectation in MAPLE as follows (note that I executed expand just so that I could get the image to display better).
expand(int(int(a*x^2*y^ |
54,302 | Expectation of rational formula | This has an elementary solution. It employs a technique often found to be useful when integrating exponentials: a fraction can be expressed in terms of an integral of an exponential function.
Because $X$ and $Y$ are independent, the expectation splits into the product of $\mathbb{E}(Y^2)=1$ and
$$\mathbb{E}\left(\fr... | Expectation of rational formula | This has an elementary solution. It employs a technique often found to be useful when integrating exponentials: a fraction can be expressed in terms of an integral of an exponential function.
Becaus | Expectation of rational formula
This has an elementary solution. It employs a technique often found to be useful when integrating exponentials: a fraction can be expressed in terms of an integral of an exponential function.
Because $X$ and $Y$ are independent, the expectation splits into the product of $\mathbb{E}(Y^... | Expectation of rational formula
This has an elementary solution. It employs a technique often found to be useful when integrating exponentials: a fraction can be expressed in terms of an integral of an exponential function.
Becaus |
54,303 | Expectation of rational formula | No real answer, but it makes things more simple:
$$\mathbb{E}Z=\mathbb{E}\frac{aX^{2}Y^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}Y^{2}\mathbb{E}\frac{bX^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}\left[1-\frac{1}{1+bX^{2}}\right]=\frac{a}{b}-\frac{a}{b}\mathbb{E}\frac{1}{1+bX^{2}}$$
So actually to be found is $$\mathbb{E}\frac{1}{1... | Expectation of rational formula | No real answer, but it makes things more simple:
$$\mathbb{E}Z=\mathbb{E}\frac{aX^{2}Y^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}Y^{2}\mathbb{E}\frac{bX^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}\left[1-\frac{1}{ | Expectation of rational formula
No real answer, but it makes things more simple:
$$\mathbb{E}Z=\mathbb{E}\frac{aX^{2}Y^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}Y^{2}\mathbb{E}\frac{bX^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}\left[1-\frac{1}{1+bX^{2}}\right]=\frac{a}{b}-\frac{a}{b}\mathbb{E}\frac{1}{1+bX^{2}}$$
So actually to be... | Expectation of rational formula
No real answer, but it makes things more simple:
$$\mathbb{E}Z=\mathbb{E}\frac{aX^{2}Y^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}Y^{2}\mathbb{E}\frac{bX^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}\left[1-\frac{1}{ |
54,304 | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC | Slice sampling is a special case of Gibbs sampling, to the point that in Monte Carlo Statistical Methods, we started our chapters on Gibbs sampling with a first chapter on slice sampling (Chapter 8).
To quote verbatim from the book (p.326):
Slice sampling relies upon the decomposition of the density $f(x)$ as
$$ f(x... | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC | Slice sampling is a special case of Gibbs sampling, to the point that in Monte Carlo Statistical Methods, we started our chapters on Gibbs sampling with a first chapter on slice sampling (Chapter 8).
| The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC
Slice sampling is a special case of Gibbs sampling, to the point that in Monte Carlo Statistical Methods, we started our chapters on Gibbs sampling with a first chapter on slice sampling (Chapter 8).
To quote verbatim from the book (p.326):
Slice... | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC
Slice sampling is a special case of Gibbs sampling, to the point that in Monte Carlo Statistical Methods, we started our chapters on Gibbs sampling with a first chapter on slice sampling (Chapter 8).
|
54,305 | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC | I prove that the stepping-out and shrinkage procedure satisfies detailed balance, but this is of course not enough to show irreducibility or ergodicity. And it's easy to construct examples in which there are regions with zero probability density in which a slice sampler that looks only at the part of the slice found b... | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC | I prove that the stepping-out and shrinkage procedure satisfies detailed balance, but this is of course not enough to show irreducibility or ergodicity. And it's easy to construct examples in which t | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC
I prove that the stepping-out and shrinkage procedure satisfies detailed balance, but this is of course not enough to show irreducibility or ergodicity. And it's easy to construct examples in which there are regions with zero probability density ... | The Harris recurrence of a stepping-out slice-sampling-within-Gibbs MCMC
I prove that the stepping-out and shrinkage procedure satisfies detailed balance, but this is of course not enough to show irreducibility or ergodicity. And it's easy to construct examples in which t |
54,306 | Do conjugate priors just lead to a posterior that is a modification of the parameters of the prior? | This question is actually somewhat subtle, and it brings to attention an interesting quirk of usage that I hadn't noticed before.
For every practical definition of conjugate distributions that I'm familiar with, it is the case that the posterior of a model using a conjugate prior is a modified form of the prior. The wi... | Do conjugate priors just lead to a posterior that is a modification of the parameters of the prior? | This question is actually somewhat subtle, and it brings to attention an interesting quirk of usage that I hadn't noticed before.
For every practical definition of conjugate distributions that I'm fam | Do conjugate priors just lead to a posterior that is a modification of the parameters of the prior?
This question is actually somewhat subtle, and it brings to attention an interesting quirk of usage that I hadn't noticed before.
For every practical definition of conjugate distributions that I'm familiar with, it is th... | Do conjugate priors just lead to a posterior that is a modification of the parameters of the prior?
This question is actually somewhat subtle, and it brings to attention an interesting quirk of usage that I hadn't noticed before.
For every practical definition of conjugate distributions that I'm fam |
54,307 | Feature subsampling with gradient boosting | A large motivation in restricting the number of predictors available to each learner in a random forest is to encourage variance between the trees. Because each tree has the same starting point, tricks like row and column subsampling are necessary to ensure that you don't have the same tree multiple times. This isn't n... | Feature subsampling with gradient boosting | A large motivation in restricting the number of predictors available to each learner in a random forest is to encourage variance between the trees. Because each tree has the same starting point, trick | Feature subsampling with gradient boosting
A large motivation in restricting the number of predictors available to each learner in a random forest is to encourage variance between the trees. Because each tree has the same starting point, tricks like row and column subsampling are necessary to ensure that you don't have... | Feature subsampling with gradient boosting
A large motivation in restricting the number of predictors available to each learner in a random forest is to encourage variance between the trees. Because each tree has the same starting point, trick |
54,308 | Why is computing the Bayesian Evidence difficult? | I like @ShijiaBian's answer. I would add the following.
The normalizing constant is important because without it, (1) you won't have a valid probability distribution and (2) you can't assess relative probabilities of values of the parameter. For example, if you modeled data $x_t$ as Gaussian conditional on a mean $\the... | Why is computing the Bayesian Evidence difficult? | I like @ShijiaBian's answer. I would add the following.
The normalizing constant is important because without it, (1) you won't have a valid probability distribution and (2) you can't assess relative | Why is computing the Bayesian Evidence difficult?
I like @ShijiaBian's answer. I would add the following.
The normalizing constant is important because without it, (1) you won't have a valid probability distribution and (2) you can't assess relative probabilities of values of the parameter. For example, if you modeled ... | Why is computing the Bayesian Evidence difficult?
I like @ShijiaBian's answer. I would add the following.
The normalizing constant is important because without it, (1) you won't have a valid probability distribution and (2) you can't assess relative |
54,309 | Why is computing the Bayesian Evidence difficult? | The posterior distribution has no relationship with the law of total probability, even though they are similar looking.
The given $P_X(x)$ is the normalizing constant. The reason that this is hard to compute is because 1). the conjugate property only can be applied for some specific distributions; 2). The prior and th... | Why is computing the Bayesian Evidence difficult? | The posterior distribution has no relationship with the law of total probability, even though they are similar looking.
The given $P_X(x)$ is the normalizing constant. The reason that this is hard to | Why is computing the Bayesian Evidence difficult?
The posterior distribution has no relationship with the law of total probability, even though they are similar looking.
The given $P_X(x)$ is the normalizing constant. The reason that this is hard to compute is because 1). the conjugate property only can be applied for... | Why is computing the Bayesian Evidence difficult?
The posterior distribution has no relationship with the law of total probability, even though they are similar looking.
The given $P_X(x)$ is the normalizing constant. The reason that this is hard to |
54,310 | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go) | As I wrote in the comments,
I suspect that this distribution doesn't have a name, because it represents the outcome of a non-independent random process. But the description of the problem is binomial-flavored, so it's probably related to the binomial and Poisson distributions. In general I think these kinds of questio... | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go) | As I wrote in the comments,
I suspect that this distribution doesn't have a name, because it represents the outcome of a non-independent random process. But the description of the problem is binomial | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go)
As I wrote in the comments,
I suspect that this distribution doesn't have a name, because it represents the outcome of a non-independent random process. But the description of the problem is binomial-flavored, so it's probably r... | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go)
As I wrote in the comments,
I suspect that this distribution doesn't have a name, because it represents the outcome of a non-independent random process. But the description of the problem is binomial |
54,311 | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go) | To simplify the issue: let the board be one-dimensional of size $1 \times n$.
Equivalent problem
For each $k$-th step number we can ask the following equivalent question:
In how many ways can we distribute the $k$ stones such that only 2 or 3 are touching together (more or bigger groups are not a valid end-states). Th... | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go) | To simplify the issue: let the board be one-dimensional of size $1 \times n$.
Equivalent problem
For each $k$-th step number we can ask the following equivalent question:
In how many ways can we dist | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go)
To simplify the issue: let the board be one-dimensional of size $1 \times n$.
Equivalent problem
For each $k$-th step number we can ask the following equivalent question:
In how many ways can we distribute the $k$ stones such th... | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go)
To simplify the issue: let the board be one-dimensional of size $1 \times n$.
Equivalent problem
For each $k$-th step number we can ask the following equivalent question:
In how many ways can we dist |
54,312 | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go) | Your second graph looks log-normal to me. See sigma=1, mu=0 here: https://en.wikipedia.org/wiki/Log-normal_distribution.
Can you perform a regression of your data to log normal and report the results? | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go) | Your second graph looks log-normal to me. See sigma=1, mu=0 here: https://en.wikipedia.org/wiki/Log-normal_distribution.
Can you perform a regression of your data to log normal and report the resul | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go)
Your second graph looks log-normal to me. See sigma=1, mu=0 here: https://en.wikipedia.org/wiki/Log-normal_distribution.
Can you perform a regression of your data to log normal and report the results? | What kind of distribution is this? (Number of stones until 2 are adjacent in a game of Go)
Your second graph looks log-normal to me. See sigma=1, mu=0 here: https://en.wikipedia.org/wiki/Log-normal_distribution.
Can you perform a regression of your data to log normal and report the resul |
54,313 | Using the central limit theorem | The binomial distribution of size $n$ and probability $p$ has probability mass function $P(Y_n=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Setting $p=2/3$ gives:
$$P(Y_n=k)=\binom{n}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{n-k}=\frac{1}{3^n}\binom{n}{k}2^k.$$
It follows that the left-hand-side of your equation sneaki... | Using the central limit theorem | The binomial distribution of size $n$ and probability $p$ has probability mass function $P(Y_n=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Setting $p=2/3$ gives:
$$P(Y_n=k)=\binom{n}{k}\left(\frac{2}{3}\right)^k\ | Using the central limit theorem
The binomial distribution of size $n$ and probability $p$ has probability mass function $P(Y_n=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Setting $p=2/3$ gives:
$$P(Y_n=k)=\binom{n}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{n-k}=\frac{1}{3^n}\binom{n}{k}2^k.$$
It follows that the left-h... | Using the central limit theorem
The binomial distribution of size $n$ and probability $p$ has probability mass function $P(Y_n=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Setting $p=2/3$ gives:
$$P(Y_n=k)=\binom{n}{k}\left(\frac{2}{3}\right)^k\ |
54,314 | dropout regularization in gbm | Check out this paper: DART: Dropouts meet Multiple Additive Regression Trees (Arxiv PDF).
Their interpertation of dropout is this: instead of developing the next tree from the residual of all previous trees, develop the next tree from the residual of a sample of previous trees. The effect on the model is similar in tha... | dropout regularization in gbm | Check out this paper: DART: Dropouts meet Multiple Additive Regression Trees (Arxiv PDF).
Their interpertation of dropout is this: instead of developing the next tree from the residual of all previous | dropout regularization in gbm
Check out this paper: DART: Dropouts meet Multiple Additive Regression Trees (Arxiv PDF).
Their interpertation of dropout is this: instead of developing the next tree from the residual of all previous trees, develop the next tree from the residual of a sample of previous trees. The effect ... | dropout regularization in gbm
Check out this paper: DART: Dropouts meet Multiple Additive Regression Trees (Arxiv PDF).
Their interpertation of dropout is this: instead of developing the next tree from the residual of all previous |
54,315 | Relation between Bayesian analysis and Bayesian hierarchical analysis? | In my view, hierarchical modeling in a Bayesian setting mainly refers to the building of a complex prior structure. Consider a parameter of interest $\theta_{0}$ and your observation $(x_i)$.
Now, consider for example that you are adding a supplemental layer to your model $p(\theta_0|\theta_1)$ through hyperprior $p(\... | Relation between Bayesian analysis and Bayesian hierarchical analysis? | In my view, hierarchical modeling in a Bayesian setting mainly refers to the building of a complex prior structure. Consider a parameter of interest $\theta_{0}$ and your observation $(x_i)$.
Now, co | Relation between Bayesian analysis and Bayesian hierarchical analysis?
In my view, hierarchical modeling in a Bayesian setting mainly refers to the building of a complex prior structure. Consider a parameter of interest $\theta_{0}$ and your observation $(x_i)$.
Now, consider for example that you are adding a suppleme... | Relation between Bayesian analysis and Bayesian hierarchical analysis?
In my view, hierarchical modeling in a Bayesian setting mainly refers to the building of a complex prior structure. Consider a parameter of interest $\theta_{0}$ and your observation $(x_i)$.
Now, co |
54,316 | Choosing words in a topic, which cut-off for LDA topics? | It is important to remember that topic models such as LDA were primarily developed for unsupervised text summarization. So often, there is not a "best" choice for how many top words to show. Most research papers on topic models tend to use the top 5-20 words. If you use more than 20 words, then you start to defeat the ... | Choosing words in a topic, which cut-off for LDA topics? | It is important to remember that topic models such as LDA were primarily developed for unsupervised text summarization. So often, there is not a "best" choice for how many top words to show. Most rese | Choosing words in a topic, which cut-off for LDA topics?
It is important to remember that topic models such as LDA were primarily developed for unsupervised text summarization. So often, there is not a "best" choice for how many top words to show. Most research papers on topic models tend to use the top 5-20 words. If ... | Choosing words in a topic, which cut-off for LDA topics?
It is important to remember that topic models such as LDA were primarily developed for unsupervised text summarization. So often, there is not a "best" choice for how many top words to show. Most rese |
54,317 | Simple kNN example | Using the Anderson's Iris data set available in the iris {datasets} in R, I worked on a makeshift function (simply to make sure I got the idea) to predict three species of Iris based on different botanical measurements in the dataset:
We want to predict the actual species Iris (Iris setosa, Iris virginica and Iris ver... | Simple kNN example | Using the Anderson's Iris data set available in the iris {datasets} in R, I worked on a makeshift function (simply to make sure I got the idea) to predict three species of Iris based on different bota | Simple kNN example
Using the Anderson's Iris data set available in the iris {datasets} in R, I worked on a makeshift function (simply to make sure I got the idea) to predict three species of Iris based on different botanical measurements in the dataset:
We want to predict the actual species Iris (Iris setosa, Iris vir... | Simple kNN example
Using the Anderson's Iris data set available in the iris {datasets} in R, I worked on a makeshift function (simply to make sure I got the idea) to predict three species of Iris based on different bota |
54,318 | Simple kNN example | I assume you've read the Wikipedia kNN Entry? (It has a diagram illustrating how it works in 2 dimensions.)
As a simple example, assume you're looking to classify homes by "Well-maintained" or "Not well-maintained". You have a map and you are able to stick pins on this map: green for "Well-maintained" and red for "Not ... | Simple kNN example | I assume you've read the Wikipedia kNN Entry? (It has a diagram illustrating how it works in 2 dimensions.)
As a simple example, assume you're looking to classify homes by "Well-maintained" or "Not we | Simple kNN example
I assume you've read the Wikipedia kNN Entry? (It has a diagram illustrating how it works in 2 dimensions.)
As a simple example, assume you're looking to classify homes by "Well-maintained" or "Not well-maintained". You have a map and you are able to stick pins on this map: green for "Well-maintained... | Simple kNN example
I assume you've read the Wikipedia kNN Entry? (It has a diagram illustrating how it works in 2 dimensions.)
As a simple example, assume you're looking to classify homes by "Well-maintained" or "Not we |
54,319 | GLM coefficient estimates distribution | Assuming $\hat{\beta}$ is the MLE, we do not know the distribution of $\hat{\beta}$ but we do know the asymptotic distribution of $\sqrt n (\hat{\beta} - \beta)$ to be $N\bigg(0, \bigg(E\Big[-\frac{\partial^2 \mathcal{L}(\beta)}{\partial \beta \partial \beta^T}\Big]\bigg)^{-1}\bigg)$.
Rearranging terms we say $\hat\be... | GLM coefficient estimates distribution | Assuming $\hat{\beta}$ is the MLE, we do not know the distribution of $\hat{\beta}$ but we do know the asymptotic distribution of $\sqrt n (\hat{\beta} - \beta)$ to be $N\bigg(0, \bigg(E\Big[-\frac{\p | GLM coefficient estimates distribution
Assuming $\hat{\beta}$ is the MLE, we do not know the distribution of $\hat{\beta}$ but we do know the asymptotic distribution of $\sqrt n (\hat{\beta} - \beta)$ to be $N\bigg(0, \bigg(E\Big[-\frac{\partial^2 \mathcal{L}(\beta)}{\partial \beta \partial \beta^T}\Big]\bigg)^{-1}\big... | GLM coefficient estimates distribution
Assuming $\hat{\beta}$ is the MLE, we do not know the distribution of $\hat{\beta}$ but we do know the asymptotic distribution of $\sqrt n (\hat{\beta} - \beta)$ to be $N\bigg(0, \bigg(E\Big[-\frac{\p |
54,320 | GLM coefficient estimates distribution | Your question is, imho, not directly related to GLMs but can be answered using more simple examples that do not require likelihood approaches:
Take $X_i\sim iid (\mu,1)$, i.e., the $X_i$ are independently and identically distributed from a distribution with some mean $\mu$ and a variance of 1 (for simplicity).
Then, b... | GLM coefficient estimates distribution | Your question is, imho, not directly related to GLMs but can be answered using more simple examples that do not require likelihood approaches:
Take $X_i\sim iid (\mu,1)$, i.e., the $X_i$ are independe | GLM coefficient estimates distribution
Your question is, imho, not directly related to GLMs but can be answered using more simple examples that do not require likelihood approaches:
Take $X_i\sim iid (\mu,1)$, i.e., the $X_i$ are independently and identically distributed from a distribution with some mean $\mu$ and a v... | GLM coefficient estimates distribution
Your question is, imho, not directly related to GLMs but can be answered using more simple examples that do not require likelihood approaches:
Take $X_i\sim iid (\mu,1)$, i.e., the $X_i$ are independe |
54,321 | Class Balancing in Deep Neural Network | Yes, they need to compute the weights ones but not assigned it to the whole loss. Instead each pixel in the loss (before summing it in both directions) should take a weight. So overall 'weights' is a tensor just like the others. Lets say there are only two classes, and frequency of $ C_1 $ is twofold of $ C_2 $. One of... | Class Balancing in Deep Neural Network | Yes, they need to compute the weights ones but not assigned it to the whole loss. Instead each pixel in the loss (before summing it in both directions) should take a weight. So overall 'weights' is a | Class Balancing in Deep Neural Network
Yes, they need to compute the weights ones but not assigned it to the whole loss. Instead each pixel in the loss (before summing it in both directions) should take a weight. So overall 'weights' is a tensor just like the others. Lets say there are only two classes, and frequency o... | Class Balancing in Deep Neural Network
Yes, they need to compute the weights ones but not assigned it to the whole loss. Instead each pixel in the loss (before summing it in both directions) should take a weight. So overall 'weights' is a |
54,322 | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2)`? | Here is Cheng & Feast Gamma generator code, on which R rbeta and rgamma functions are based:
function x=gamrnd_cheng(alpha)
% Gamma(alpha,1) generator using Cheng--Feast method
% Algorithm 4.35
c1=alpha-1; c2=(alpha-1/(6*alpha))/c1; c3=2/c1; c4=1+c3;
c5=1/sqrt(alpha);
flag=0;
while flag==0;
U1=rand; U2=rand;
if... | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2 | Here is Cheng & Feast Gamma generator code, on which R rbeta and rgamma functions are based:
function x=gamrnd_cheng(alpha)
% Gamma(alpha,1) generator using Cheng--Feast method
% Algorithm 4.35
c1=alp | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2)`?
Here is Cheng & Feast Gamma generator code, on which R rbeta and rgamma functions are based:
function x=gamrnd_cheng(alpha)
% Gamma(alpha,1) generator using Cheng--Feast method
% Algorithm 4.35
c1=alpha-1; c2=(alpha-... | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2
Here is Cheng & Feast Gamma generator code, on which R rbeta and rgamma functions are based:
function x=gamrnd_cheng(alpha)
% Gamma(alpha,1) generator using Cheng--Feast method
% Algorithm 4.35
c1=alp |
54,323 | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2)`? | Following up on Xi'an's suggestion, the result in the question indeed seems to be an artifact of the particular (small) parameters chosen. In particular, when choosing higher $\alpha$ and $\beta$, such that the beta density becomes more concentrated with a correspondingly higher density at the mode, such that a larger ... | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2 | Following up on Xi'an's suggestion, the result in the question indeed seems to be an artifact of the particular (small) parameters chosen. In particular, when choosing higher $\alpha$ and $\beta$, suc | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2)`?
Following up on Xi'an's suggestion, the result in the question indeed seems to be an artifact of the particular (small) parameters chosen. In particular, when choosing higher $\alpha$ and $\beta$, such that the beta ... | Why is rejection sampling with acceptance probability 2/3 for Beta(2,2) not slower than `rbeta(N,2,2
Following up on Xi'an's suggestion, the result in the question indeed seems to be an artifact of the particular (small) parameters chosen. In particular, when choosing higher $\alpha$ and $\beta$, suc |
54,324 | Why does SGD and back propagation work with ReLUs? | At x = 0, the ReLU function is no longer differentiable, however it is sub-differentiable and any value in the range [0,1] is a valid choice of sub-gradient. You may see some implementations simply use 0 sub-gradient at the x = 0 singularity. For further details see the Wikipedia article: Subdervative. | Why does SGD and back propagation work with ReLUs? | At x = 0, the ReLU function is no longer differentiable, however it is sub-differentiable and any value in the range [0,1] is a valid choice of sub-gradient. You may see some implementations simply us | Why does SGD and back propagation work with ReLUs?
At x = 0, the ReLU function is no longer differentiable, however it is sub-differentiable and any value in the range [0,1] is a valid choice of sub-gradient. You may see some implementations simply use 0 sub-gradient at the x = 0 singularity. For further details see th... | Why does SGD and back propagation work with ReLUs?
At x = 0, the ReLU function is no longer differentiable, however it is sub-differentiable and any value in the range [0,1] is a valid choice of sub-gradient. You may see some implementations simply us |
54,325 | What is the purpose of the scaling factor used in dropout? | If a p=0.5 dropout is used, only half on the neurons are activate during training, while if we activate them all at test time, the output of the dropout layer would get "doubled", so in this regard it makes sense to multiply the output by a factor 1-p to neutralize that effect.
Here's a quote from the dropout paper htt... | What is the purpose of the scaling factor used in dropout? | If a p=0.5 dropout is used, only half on the neurons are activate during training, while if we activate them all at test time, the output of the dropout layer would get "doubled", so in this regard it | What is the purpose of the scaling factor used in dropout?
If a p=0.5 dropout is used, only half on the neurons are activate during training, while if we activate them all at test time, the output of the dropout layer would get "doubled", so in this regard it makes sense to multiply the output by a factor 1-p to neutra... | What is the purpose of the scaling factor used in dropout?
If a p=0.5 dropout is used, only half on the neurons are activate during training, while if we activate them all at test time, the output of the dropout layer would get "doubled", so in this regard it |
54,326 | Diagonal in ROC plot? | Assume that you have the following result:
score label
1.000 positive
0.900 negative
0.900 positive
0.900 negative
0.500 negative
0.200 positive
Manually plot the ROC curve for the possible thresholds of 1.0, 0.9, 0.5, 0.2 and you do get a sloped part.
The reason are duplicate scores.
Beware, there are some poor imple... | Diagonal in ROC plot? | Assume that you have the following result:
score label
1.000 positive
0.900 negative
0.900 positive
0.900 negative
0.500 negative
0.200 positive
Manually plot the ROC curve for the possible threshold | Diagonal in ROC plot?
Assume that you have the following result:
score label
1.000 positive
0.900 negative
0.900 positive
0.900 negative
0.500 negative
0.200 positive
Manually plot the ROC curve for the possible thresholds of 1.0, 0.9, 0.5, 0.2 and you do get a sloped part.
The reason are duplicate scores.
Beware, the... | Diagonal in ROC plot?
Assume that you have the following result:
score label
1.000 positive
0.900 negative
0.900 positive
0.900 negative
0.500 negative
0.200 positive
Manually plot the ROC curve for the possible threshold |
54,327 | Diagonal in ROC plot? | Yes, this is "legal". If the jump from one threshold to the next raises the amount of false positives and false negatives together the result is a diagonal line.
Two reasons that might happen:
You have 2 observations with same threshold but with different ground truth
The resolution between 2 thresholds is large enoug... | Diagonal in ROC plot? | Yes, this is "legal". If the jump from one threshold to the next raises the amount of false positives and false negatives together the result is a diagonal line.
Two reasons that might happen:
You ha | Diagonal in ROC plot?
Yes, this is "legal". If the jump from one threshold to the next raises the amount of false positives and false negatives together the result is a diagonal line.
Two reasons that might happen:
You have 2 observations with same threshold but with different ground truth
The resolution between 2 thr... | Diagonal in ROC plot?
Yes, this is "legal". If the jump from one threshold to the next raises the amount of false positives and false negatives together the result is a diagonal line.
Two reasons that might happen:
You ha |
54,328 | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate] | The thing is $Y$ is not that continuous to begin with. To be continuous, the distribution function of $Y$ must be absolutely continuous (see definition 1.32, page 10 of link http://math.arizona.edu/~jwatkins/probnotes.pdf by @fcop). You see the distribution of Y has a half impulse (Dirac delta) function at zero. When y... | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate | The thing is $Y$ is not that continuous to begin with. To be continuous, the distribution function of $Y$ must be absolutely continuous (see definition 1.32, page 10 of link http://math.arizona.edu/~j | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate]
The thing is $Y$ is not that continuous to begin with. To be continuous, the distribution function of $Y$ must be absolutely continuous (see definition 1.32, page 10 of link http://math.arizona.edu/~jwatkins/probnotes.... | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate
The thing is $Y$ is not that continuous to begin with. To be continuous, the distribution function of $Y$ must be absolutely continuous (see definition 1.32, page 10 of link http://math.arizona.edu/~j |
54,329 | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate] | Let's just go with a simple intuitive explanation. But it can get really mathy really fast (if you prefer).
A continuous distribution is a line between two points A and B. On this line there are infinitely many points, no matter if the distance between the points (A, B) is extremely small. If all of those infinite poin... | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate | Let's just go with a simple intuitive explanation. But it can get really mathy really fast (if you prefer).
A continuous distribution is a line between two points A and B. On this line there are infin | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate]
Let's just go with a simple intuitive explanation. But it can get really mathy really fast (if you prefer).
A continuous distribution is a line between two points A and B. On this line there are infinitely many points,... | Why is "the probability that a continuous random variable equals some value always zero"? [duplicate
Let's just go with a simple intuitive explanation. But it can get really mathy really fast (if you prefer).
A continuous distribution is a line between two points A and B. On this line there are infin |
54,330 | Can I use linear model on each variable to determine which variables are important? | No. The proposed method seems very unlikely to produce useful results. The problem I would anticipate is lots of false positive results.
For example, suppose we wish to predict $y$ and have two predictors $x$ and $z$. Let
$y = x + \epsilon_1$
$z = x + \epsilon_2$
Then the proposed method is likely to select both $x$ a... | Can I use linear model on each variable to determine which variables are important? | No. The proposed method seems very unlikely to produce useful results. The problem I would anticipate is lots of false positive results.
For example, suppose we wish to predict $y$ and have two predi | Can I use linear model on each variable to determine which variables are important?
No. The proposed method seems very unlikely to produce useful results. The problem I would anticipate is lots of false positive results.
For example, suppose we wish to predict $y$ and have two predictors $x$ and $z$. Let
$y = x + \eps... | Can I use linear model on each variable to determine which variables are important?
No. The proposed method seems very unlikely to produce useful results. The problem I would anticipate is lots of false positive results.
For example, suppose we wish to predict $y$ and have two predi |
54,331 | Can I use linear model on each variable to determine which variables are important? | If your predictors (biomarkers) are colinear, univariate regressions may grossly over / underestimate effect sizes, depending on sign of the colinearity and the sign of the product of their effect sizes. This is known as Simpson's paradox, or in general as omitted-variable bias, as mentioned above. I would therefore no... | Can I use linear model on each variable to determine which variables are important? | If your predictors (biomarkers) are colinear, univariate regressions may grossly over / underestimate effect sizes, depending on sign of the colinearity and the sign of the product of their effect siz | Can I use linear model on each variable to determine which variables are important?
If your predictors (biomarkers) are colinear, univariate regressions may grossly over / underestimate effect sizes, depending on sign of the colinearity and the sign of the product of their effect sizes. This is known as Simpson's parad... | Can I use linear model on each variable to determine which variables are important?
If your predictors (biomarkers) are colinear, univariate regressions may grossly over / underestimate effect sizes, depending on sign of the colinearity and the sign of the product of their effect siz |
54,332 | Can I use linear model on each variable to determine which variables are important? | Aristotle said that, “The whole is greater than the sum of its parts.” Each simple linear regression is merely testing a part. However, I imagine that many diseases are associated with combinations of markers (the whole). What you really care about are the combination of markers. As a result, your algorithm may not w... | Can I use linear model on each variable to determine which variables are important? | Aristotle said that, “The whole is greater than the sum of its parts.” Each simple linear regression is merely testing a part. However, I imagine that many diseases are associated with combinations o | Can I use linear model on each variable to determine which variables are important?
Aristotle said that, “The whole is greater than the sum of its parts.” Each simple linear regression is merely testing a part. However, I imagine that many diseases are associated with combinations of markers (the whole). What you real... | Can I use linear model on each variable to determine which variables are important?
Aristotle said that, “The whole is greater than the sum of its parts.” Each simple linear regression is merely testing a part. However, I imagine that many diseases are associated with combinations o |
54,333 | What is UV decomposition? | If $A$ is a matrix of rank $k$ and size $m$ by $n$, $A$ can be written as
$A=UV^{T}$
where $U$ is of size $m$ by $k$ and $V$ is of size $n$ by $k$. The columns of $U$ and $V$ need not necessarily be orthogonal.
If you have the SVD of $A$, then it's easy to compute this low rank factorization from the SVD. Given t... | What is UV decomposition? | If $A$ is a matrix of rank $k$ and size $m$ by $n$, $A$ can be written as
$A=UV^{T}$
where $U$ is of size $m$ by $k$ and $V$ is of size $n$ by $k$. The columns of $U$ and $V$ need not necessarily be | What is UV decomposition?
If $A$ is a matrix of rank $k$ and size $m$ by $n$, $A$ can be written as
$A=UV^{T}$
where $U$ is of size $m$ by $k$ and $V$ is of size $n$ by $k$. The columns of $U$ and $V$ need not necessarily be orthogonal.
If you have the SVD of $A$, then it's easy to compute this low rank factorizat... | What is UV decomposition?
If $A$ is a matrix of rank $k$ and size $m$ by $n$, $A$ can be written as
$A=UV^{T}$
where $U$ is of size $m$ by $k$ and $V$ is of size $n$ by $k$. The columns of $U$ and $V$ need not necessarily be |
54,334 | Calculating the expression for the derivative of a Gaussian process | For simplicity I assume x has one dimension.
$\frac{\partial f}{\partial x}$ is normally distributed with expectation:
$E[\frac{\partial f}{\partial x}] = \frac{\partial}{\partial x}E[f]$
And Covariance:
$\text{Cov}(\frac{\partial f_1}{\partial x_1},\frac{\partial f_2}{\partial x_2})$ = $\frac{\partial^2 }{\partial x_2... | Calculating the expression for the derivative of a Gaussian process | For simplicity I assume x has one dimension.
$\frac{\partial f}{\partial x}$ is normally distributed with expectation:
$E[\frac{\partial f}{\partial x}] = \frac{\partial}{\partial x}E[f]$
And Covarian | Calculating the expression for the derivative of a Gaussian process
For simplicity I assume x has one dimension.
$\frac{\partial f}{\partial x}$ is normally distributed with expectation:
$E[\frac{\partial f}{\partial x}] = \frac{\partial}{\partial x}E[f]$
And Covariance:
$\text{Cov}(\frac{\partial f_1}{\partial x_1},\f... | Calculating the expression for the derivative of a Gaussian process
For simplicity I assume x has one dimension.
$\frac{\partial f}{\partial x}$ is normally distributed with expectation:
$E[\frac{\partial f}{\partial x}] = \frac{\partial}{\partial x}E[f]$
And Covarian |
54,335 | Calculating the expression for the derivative of a Gaussian process | Since your domain for $f$ is $n$-dimensional you will actually have $n$ derivative processes $\frac{\partial f}{\partial x_j}$ with $j=1,\ldots n$. You need to calculate the mean and correlation function of $\frac{\partial f}{\partial x_j}$. From the linked answer you know that those are the derivative of $f$'s mean fu... | Calculating the expression for the derivative of a Gaussian process | Since your domain for $f$ is $n$-dimensional you will actually have $n$ derivative processes $\frac{\partial f}{\partial x_j}$ with $j=1,\ldots n$. You need to calculate the mean and correlation funct | Calculating the expression for the derivative of a Gaussian process
Since your domain for $f$ is $n$-dimensional you will actually have $n$ derivative processes $\frac{\partial f}{\partial x_j}$ with $j=1,\ldots n$. You need to calculate the mean and correlation function of $\frac{\partial f}{\partial x_j}$. From the l... | Calculating the expression for the derivative of a Gaussian process
Since your domain for $f$ is $n$-dimensional you will actually have $n$ derivative processes $\frac{\partial f}{\partial x_j}$ with $j=1,\ldots n$. You need to calculate the mean and correlation funct |
54,336 | Calculating the expression for the derivative of a Gaussian process | You can use sympy in Python, it will calculate any derivatives including integral defined one.
diffn(ff,x0,kk) :
dffk= Derivative(ff(x),x,kk)
dffk1= simplify( dffk.doit())
dffx0= simplify(Subs(dffk1, (x), (x0)).doit())
return dffx0 | Calculating the expression for the derivative of a Gaussian process | You can use sympy in Python, it will calculate any derivatives including integral defined one.
diffn(ff,x0,kk) :
dffk= Derivative(ff(x),x,kk)
dffk1= simplify( dffk.doit())
dffx0= simplify(Sub | Calculating the expression for the derivative of a Gaussian process
You can use sympy in Python, it will calculate any derivatives including integral defined one.
diffn(ff,x0,kk) :
dffk= Derivative(ff(x),x,kk)
dffk1= simplify( dffk.doit())
dffx0= simplify(Subs(dffk1, (x), (x0)).doit())
return dffx0 | Calculating the expression for the derivative of a Gaussian process
You can use sympy in Python, it will calculate any derivatives including integral defined one.
diffn(ff,x0,kk) :
dffk= Derivative(ff(x),x,kk)
dffk1= simplify( dffk.doit())
dffx0= simplify(Sub |
54,337 | t test for intercept? | It is also $H_0:\beta_{0,0}=0$ in your example. You can infer that from the general formulation of a t-ratio
$$
t=\frac{\hat\beta_j-\beta_{j,0}}{std.error(\hat\beta_j)},
$$
where $\beta_{j,0}$ is the hypothesis formulated on $\beta_j$ and $\beta_0$ is the coefficient on the intercept.
In your case, we have
$$
.837=\fr... | t test for intercept? | It is also $H_0:\beta_{0,0}=0$ in your example. You can infer that from the general formulation of a t-ratio
$$
t=\frac{\hat\beta_j-\beta_{j,0}}{std.error(\hat\beta_j)},
$$
where $\beta_{j,0}$ is the | t test for intercept?
It is also $H_0:\beta_{0,0}=0$ in your example. You can infer that from the general formulation of a t-ratio
$$
t=\frac{\hat\beta_j-\beta_{j,0}}{std.error(\hat\beta_j)},
$$
where $\beta_{j,0}$ is the hypothesis formulated on $\beta_j$ and $\beta_0$ is the coefficient on the intercept.
In your case... | t test for intercept?
It is also $H_0:\beta_{0,0}=0$ in your example. You can infer that from the general formulation of a t-ratio
$$
t=\frac{\hat\beta_j-\beta_{j,0}}{std.error(\hat\beta_j)},
$$
where $\beta_{j,0}$ is the |
54,338 | t test for intercept? | Hypothesis testing is like mathematical ''proof by contradiction''; if you want to prove something, then you assume that the opposite is true and using this ''opposite is true'' assumption you try to find a contradiction. As contradictions are impossible, the assumption ''opposite is true'' must be false.
In hypothe... | t test for intercept? | Hypothesis testing is like mathematical ''proof by contradiction''; if you want to prove something, then you assume that the opposite is true and using this ''opposite is true'' assumption you try to | t test for intercept?
Hypothesis testing is like mathematical ''proof by contradiction''; if you want to prove something, then you assume that the opposite is true and using this ''opposite is true'' assumption you try to find a contradiction. As contradictions are impossible, the assumption ''opposite is true'' must ... | t test for intercept?
Hypothesis testing is like mathematical ''proof by contradiction''; if you want to prove something, then you assume that the opposite is true and using this ''opposite is true'' assumption you try to |
54,339 | Relationship between R2 and correlation coefficient [duplicate] | The usual way of interpreting the coefficient of determination R^{2} is to see it as the percentage of the variation of the dependent variable y (Var(y)) can be explained by our model.
For the proof we have to know the following (taken from OLS theory and general statistics):
I hope this answer clears your doubt. | Relationship between R2 and correlation coefficient [duplicate] | The usual way of interpreting the coefficient of determination R^{2} is to see it as the percentage of the variation of the dependent variable y (Var(y)) can be explained by our model.
For the proof w | Relationship between R2 and correlation coefficient [duplicate]
The usual way of interpreting the coefficient of determination R^{2} is to see it as the percentage of the variation of the dependent variable y (Var(y)) can be explained by our model.
For the proof we have to know the following (taken from OLS theory and ... | Relationship between R2 and correlation coefficient [duplicate]
The usual way of interpreting the coefficient of determination R^{2} is to see it as the percentage of the variation of the dependent variable y (Var(y)) can be explained by our model.
For the proof w |
54,340 | How to update weights in a neural network using gradient descent with mini-batches? | Mini-batch is implemented basically as you describe in 2.
Epoch starts. We sample and feedforward a minibatch, get the error and backprop it, i.e. update the weights. We repeat this until we have sampled the full data set. Epoch over.
Assuming that the network is minimizing the following objective function:
$$
\fra... | How to update weights in a neural network using gradient descent with mini-batches? | Mini-batch is implemented basically as you describe in 2.
Epoch starts. We sample and feedforward a minibatch, get the error and backprop it, i.e. update the weights. We repeat this until we have sa | How to update weights in a neural network using gradient descent with mini-batches?
Mini-batch is implemented basically as you describe in 2.
Epoch starts. We sample and feedforward a minibatch, get the error and backprop it, i.e. update the weights. We repeat this until we have sampled the full data set. Epoch over.... | How to update weights in a neural network using gradient descent with mini-batches?
Mini-batch is implemented basically as you describe in 2.
Epoch starts. We sample and feedforward a minibatch, get the error and backprop it, i.e. update the weights. We repeat this until we have sa |
54,341 | Why is ridge regression giving different results in Matlab and Python? | MATLAB always uses the centred and scaled variables for the computations within ridge. It just back-transforms them before returning them. As you have a really small matrix this probably makes a noticeable difference. You can reproduce the Python results in MATLAB easily:
X = [1 1 2 ; 3 4 2 ; 6 5 2 ; 5 5 3];
Y = [1 0 0... | Why is ridge regression giving different results in Matlab and Python? | MATLAB always uses the centred and scaled variables for the computations within ridge. It just back-transforms them before returning them. As you have a really small matrix this probably makes a notic | Why is ridge regression giving different results in Matlab and Python?
MATLAB always uses the centred and scaled variables for the computations within ridge. It just back-transforms them before returning them. As you have a really small matrix this probably makes a noticeable difference. You can reproduce the Python re... | Why is ridge regression giving different results in Matlab and Python?
MATLAB always uses the centred and scaled variables for the computations within ridge. It just back-transforms them before returning them. As you have a really small matrix this probably makes a notic |
54,342 | Probability of product of two random variables | Since the distribution of $XY$ is characterised by its cdf, you want to compute $\mathbb{P}(XY<z)$ for an arbitrary value $z$. Let us assume first that $Y$ is always positive. Then
$$\eqalign{\mathbb{P}(XY<z)&=\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]\\&=\mathbb{E}[\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]|Y]]\\&=\mathb... | Probability of product of two random variables | Since the distribution of $XY$ is characterised by its cdf, you want to compute $\mathbb{P}(XY<z)$ for an arbitrary value $z$. Let us assume first that $Y$ is always positive. Then
$$\eqalign{\mathbb | Probability of product of two random variables
Since the distribution of $XY$ is characterised by its cdf, you want to compute $\mathbb{P}(XY<z)$ for an arbitrary value $z$. Let us assume first that $Y$ is always positive. Then
$$\eqalign{\mathbb{P}(XY<z)&=\mathbb{E}[\mathbb{I}_{(-\infty,z)}(XY)]\\&=\mathbb{E}[\mathbb... | Probability of product of two random variables
Since the distribution of $XY$ is characterised by its cdf, you want to compute $\mathbb{P}(XY<z)$ for an arbitrary value $z$. Let us assume first that $Y$ is always positive. Then
$$\eqalign{\mathbb |
54,343 | Probability of product of two random variables | $ P(XY=k)= \sum_{t}P(X=t,Y=k/t)=\sum_{t}P(Y=k/t|X=t)P(X=t)$
Now,if $X$ and $Y$ are independent,then $P(Y=k/t|X=t)=P(Y=k/t)$ | Probability of product of two random variables | $ P(XY=k)= \sum_{t}P(X=t,Y=k/t)=\sum_{t}P(Y=k/t|X=t)P(X=t)$
Now,if $X$ and $Y$ are independent,then $P(Y=k/t|X=t)=P(Y=k/t)$ | Probability of product of two random variables
$ P(XY=k)= \sum_{t}P(X=t,Y=k/t)=\sum_{t}P(Y=k/t|X=t)P(X=t)$
Now,if $X$ and $Y$ are independent,then $P(Y=k/t|X=t)=P(Y=k/t)$ | Probability of product of two random variables
$ P(XY=k)= \sum_{t}P(X=t,Y=k/t)=\sum_{t}P(Y=k/t|X=t)P(X=t)$
Now,if $X$ and $Y$ are independent,then $P(Y=k/t|X=t)=P(Y=k/t)$ |
54,344 | Multilevel model with nested repeated measures design | The only random part here is the individual. Both Time and Treatment are fixed parts. As I understand it, you want global (ie. fixed) estimates of the effect of
Time
Each level of Treatment (except for the reference level)
The interaction between each level of Treatment (except for the
reference level) and Time.
T... | Multilevel model with nested repeated measures design | The only random part here is the individual. Both Time and Treatment are fixed parts. As I understand it, you want global (ie. fixed) estimates of the effect of
Time
Each level of Treatment (excep | Multilevel model with nested repeated measures design
The only random part here is the individual. Both Time and Treatment are fixed parts. As I understand it, you want global (ie. fixed) estimates of the effect of
Time
Each level of Treatment (except for the reference level)
The interaction between each level of T... | Multilevel model with nested repeated measures design
The only random part here is the individual. Both Time and Treatment are fixed parts. As I understand it, you want global (ie. fixed) estimates of the effect of
Time
Each level of Treatment (excep |
54,345 | gam smoother vs parametric term (concurvity difference) | The concurvity moves from the stated smooth terms to the parametric terms, which concurvity groups in total under the para column of the matrix or matrices returned.
Here's a modified example from ?concurvity
library("mgcv")
## simulate data with concurvity...
set.seed(8)
n<- 200
f2 <- function(x) 0.2 * x^11 * (10 * (1... | gam smoother vs parametric term (concurvity difference) | The concurvity moves from the stated smooth terms to the parametric terms, which concurvity groups in total under the para column of the matrix or matrices returned.
Here's a modified example from ?co | gam smoother vs parametric term (concurvity difference)
The concurvity moves from the stated smooth terms to the parametric terms, which concurvity groups in total under the para column of the matrix or matrices returned.
Here's a modified example from ?concurvity
library("mgcv")
## simulate data with concurvity...
set... | gam smoother vs parametric term (concurvity difference)
The concurvity moves from the stated smooth terms to the parametric terms, which concurvity groups in total under the para column of the matrix or matrices returned.
Here's a modified example from ?co |
54,346 | Empirical verification of the probability integral transform | I believe your code just does not do what you want it to do. Here's what you want:
set.seed(154)
x <- rnorm(10000)
hist(pnorm(x))
This histogram looks uniform.
I believe
plot(ppoints(1000), pnorm(ppoints(1000)))
results in a plot of a portion of the graph of the normal cdf.
Here's a quick verification
plot((1:100 - ... | Empirical verification of the probability integral transform | I believe your code just does not do what you want it to do. Here's what you want:
set.seed(154)
x <- rnorm(10000)
hist(pnorm(x))
This histogram looks uniform.
I believe
plot(ppoints(1000), pnorm(pp | Empirical verification of the probability integral transform
I believe your code just does not do what you want it to do. Here's what you want:
set.seed(154)
x <- rnorm(10000)
hist(pnorm(x))
This histogram looks uniform.
I believe
plot(ppoints(1000), pnorm(ppoints(1000)))
results in a plot of a portion of the graph ... | Empirical verification of the probability integral transform
I believe your code just does not do what you want it to do. Here's what you want:
set.seed(154)
x <- rnorm(10000)
hist(pnorm(x))
This histogram looks uniform.
I believe
plot(ppoints(1000), pnorm(pp |
54,347 | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of the random vector also gaussian? | A random vector $X$ have the multinormal distribution if all linear combinations are normally distributed. Just take the linear combination with coefficient vector $e_i = (0,0,\dots, 1,0,\dots,0)$ where the one is in place $i$. Then $e_i^T X=X_i$, the $i$th component, hence that is normally distributed.
For the additio... | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of th | A random vector $X$ have the multinormal distribution if all linear combinations are normally distributed. Just take the linear combination with coefficient vector $e_i = (0,0,\dots, 1,0,\dots,0)$ whe | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of the random vector also gaussian?
A random vector $X$ have the multinormal distribution if all linear combinations are normally distributed. Just take the linear combination with coefficient vector $e_i = (0,0,\dots, 1,0,\d... | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of th
A random vector $X$ have the multinormal distribution if all linear combinations are normally distributed. Just take the linear combination with coefficient vector $e_i = (0,0,\dots, 1,0,\dots,0)$ whe |
54,348 | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of the random vector also gaussian? | kjetil's answer (+1) addresses the more general case of linear combinations of components of your multivariate normal distribution.
Your specific case concerns the so-called marginals of the multivariate normal distribution. And yes, the marginals of a multivariate normal distribution are again normal (multivariate nor... | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of th | kjetil's answer (+1) addresses the more general case of linear combinations of components of your multivariate normal distribution.
Your specific case concerns the so-called marginals of the multivari | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of the random vector also gaussian?
kjetil's answer (+1) addresses the more general case of linear combinations of components of your multivariate normal distribution.
Your specific case concerns the so-called marginals of th... | If I have a k-dimensional random vector distributed as multivariate gaussian, are the elements of th
kjetil's answer (+1) addresses the more general case of linear combinations of components of your multivariate normal distribution.
Your specific case concerns the so-called marginals of the multivari |
54,349 | Is there a book on applied linear algebra | My favourite book on linear algebra is this one. And it's quite inexpensive only 9.99 the kindle version. So related to your question this book teaches linear algebra from practical applications using python working on challenging problems, so some python experience is required though some people have learned python th... | Is there a book on applied linear algebra | My favourite book on linear algebra is this one. And it's quite inexpensive only 9.99 the kindle version. So related to your question this book teaches linear algebra from practical applications using | Is there a book on applied linear algebra
My favourite book on linear algebra is this one. And it's quite inexpensive only 9.99 the kindle version. So related to your question this book teaches linear algebra from practical applications using python working on challenging problems, so some python experience is required... | Is there a book on applied linear algebra
My favourite book on linear algebra is this one. And it's quite inexpensive only 9.99 the kindle version. So related to your question this book teaches linear algebra from practical applications using |
54,350 | Is there a book on applied linear algebra | Gilbert Strang's linear algebra is good.
If you want to learn linear algebra from scratch I think this one is pretty good too. It is even simpler than Strang's book.
http://www.amazon.com/Linear-Algebra-Applications-8th-Edition/dp/0136009298 | Is there a book on applied linear algebra | Gilbert Strang's linear algebra is good.
If you want to learn linear algebra from scratch I think this one is pretty good too. It is even simpler than Strang's book.
http://www.amazon.com/Linear-Alge | Is there a book on applied linear algebra
Gilbert Strang's linear algebra is good.
If you want to learn linear algebra from scratch I think this one is pretty good too. It is even simpler than Strang's book.
http://www.amazon.com/Linear-Algebra-Applications-8th-Edition/dp/0136009298 | Is there a book on applied linear algebra
Gilbert Strang's linear algebra is good.
If you want to learn linear algebra from scratch I think this one is pretty good too. It is even simpler than Strang's book.
http://www.amazon.com/Linear-Alge |
54,351 | Recommendations for books regarding statistical consulting | Statistical Sleuth does not describe the consulting process, but teaches methods using case studies.
To quote:
The Sleuth was written to train graduate students in disciplines other than Statistics to correctly draw and communicate statistical conclusions for their Master's and Doctoral theses, and for their eventual ... | Recommendations for books regarding statistical consulting | Statistical Sleuth does not describe the consulting process, but teaches methods using case studies.
To quote:
The Sleuth was written to train graduate students in disciplines other than Statistics t | Recommendations for books regarding statistical consulting
Statistical Sleuth does not describe the consulting process, but teaches methods using case studies.
To quote:
The Sleuth was written to train graduate students in disciplines other than Statistics to correctly draw and communicate statistical conclusions for ... | Recommendations for books regarding statistical consulting
Statistical Sleuth does not describe the consulting process, but teaches methods using case studies.
To quote:
The Sleuth was written to train graduate students in disciplines other than Statistics t |
54,352 | Recommendations for books regarding statistical consulting | There is some books dedicated to that topic of training statistical consultants, I do not have personal experience with this books, but here are some:
"Statistical Consulting" by Javier Cabrera and Andrew McDougall (had a positive review in the American Statistician)
"Statistical Consulting: A Guide to Effective Co... | Recommendations for books regarding statistical consulting | There is some books dedicated to that topic of training statistical consultants, I do not have personal experience with this books, but here are some:
"Statistical Consulting" by Javier Cabrera and | Recommendations for books regarding statistical consulting
There is some books dedicated to that topic of training statistical consultants, I do not have personal experience with this books, but here are some:
"Statistical Consulting" by Javier Cabrera and Andrew McDougall (had a positive review in the American Sta... | Recommendations for books regarding statistical consulting
There is some books dedicated to that topic of training statistical consultants, I do not have personal experience with this books, but here are some:
"Statistical Consulting" by Javier Cabrera and |
54,353 | Are products of independent random variables independent? | You are making this problem a lot harder than it needs to be because the
random variables in question are two-valued, and the problem can be
treated as one of independence of events rather than independence of
random variables. In what follows, I will treat the independence of
events even though the events will be sta... | Are products of independent random variables independent? | You are making this problem a lot harder than it needs to be because the
random variables in question are two-valued, and the problem can be
treated as one of independence of events rather than indepe | Are products of independent random variables independent?
You are making this problem a lot harder than it needs to be because the
random variables in question are two-valued, and the problem can be
treated as one of independence of events rather than independence of
random variables. In what follows, I will treat the... | Are products of independent random variables independent?
You are making this problem a lot harder than it needs to be because the
random variables in question are two-valued, and the problem can be
treated as one of independence of events rather than indepe |
54,354 | Are products of independent random variables independent? | ... How do I state this precisely, if it is right? $\forall i \leq n, \sigma(X_i) \subseteq \sigma(X_n)$ ?
Your have the right idea, but I would recommend using the definition of the Markov property to state this, namely that we have $P(X_n\mid X_0,\dots,X_{n-1})=P(X_n \mid X_{n-1})$. There is nothing imprecise about... | Are products of independent random variables independent? | ... How do I state this precisely, if it is right? $\forall i \leq n, \sigma(X_i) \subseteq \sigma(X_n)$ ?
Your have the right idea, but I would recommend using the definition of the Markov property | Are products of independent random variables independent?
... How do I state this precisely, if it is right? $\forall i \leq n, \sigma(X_i) \subseteq \sigma(X_n)$ ?
Your have the right idea, but I would recommend using the definition of the Markov property to state this, namely that we have $P(X_n\mid X_0,\dots,X_{n-... | Are products of independent random variables independent?
... How do I state this precisely, if it is right? $\forall i \leq n, \sigma(X_i) \subseteq \sigma(X_n)$ ?
Your have the right idea, but I would recommend using the definition of the Markov property |
54,355 | Comparing the within-subject variance between two groups of subjects | You can test this by fitting a linear mixed model. A linear mixed model is like a multiple regression model but you can have random effects. The random effects part is needed because you have multiple tests per observation. You will then model score as a function of sex and test, and the subjects are your random effect... | Comparing the within-subject variance between two groups of subjects | You can test this by fitting a linear mixed model. A linear mixed model is like a multiple regression model but you can have random effects. The random effects part is needed because you have multiple | Comparing the within-subject variance between two groups of subjects
You can test this by fitting a linear mixed model. A linear mixed model is like a multiple regression model but you can have random effects. The random effects part is needed because you have multiple tests per observation. You will then model score a... | Comparing the within-subject variance between two groups of subjects
You can test this by fitting a linear mixed model. A linear mixed model is like a multiple regression model but you can have random effects. The random effects part is needed because you have multiple |
54,356 | Comparing the within-subject variance between two groups of subjects | The average of variances is probably not the best variable to use.
Variance is a squared result. In many cases, the standard deviation may be the better result to use, because it is on the same scale as your original data. You can then decide on whether use the arithmetic mean, or e.g. root-mean-square again.
Avoid bli... | Comparing the within-subject variance between two groups of subjects | The average of variances is probably not the best variable to use.
Variance is a squared result. In many cases, the standard deviation may be the better result to use, because it is on the same scale | Comparing the within-subject variance between two groups of subjects
The average of variances is probably not the best variable to use.
Variance is a squared result. In many cases, the standard deviation may be the better result to use, because it is on the same scale as your original data. You can then decide on wheth... | Comparing the within-subject variance between two groups of subjects
The average of variances is probably not the best variable to use.
Variance is a squared result. In many cases, the standard deviation may be the better result to use, because it is on the same scale |
54,357 | Comparing the within-subject variance between two groups of subjects | The fact the same subject does the test multiple times introduces correlation among the observations. The goal is to get an estimate of the var-covar matrix of your data. We will assume that the subjects are independent, but that the score for the same subject on the test are dependent. This means that that var-cova... | Comparing the within-subject variance between two groups of subjects | The fact the same subject does the test multiple times introduces correlation among the observations. The goal is to get an estimate of the var-covar matrix of your data. We will assume that the sub | Comparing the within-subject variance between two groups of subjects
The fact the same subject does the test multiple times introduces correlation among the observations. The goal is to get an estimate of the var-covar matrix of your data. We will assume that the subjects are independent, but that the score for the s... | Comparing the within-subject variance between two groups of subjects
The fact the same subject does the test multiple times introduces correlation among the observations. The goal is to get an estimate of the var-covar matrix of your data. We will assume that the sub |
54,358 | Comparing the within-subject variance between two groups of subjects | I'm not sure if you can assume normality of the performance variances, but I'm still looking into it. What you're saying makes sense.
Since you are using a within-subjects design (prone to carryover effects) I think it would be most interesting to look at how the variances are changing over time. Are your subjects pe... | Comparing the within-subject variance between two groups of subjects | I'm not sure if you can assume normality of the performance variances, but I'm still looking into it. What you're saying makes sense.
Since you are using a within-subjects design (prone to carryover | Comparing the within-subject variance between two groups of subjects
I'm not sure if you can assume normality of the performance variances, but I'm still looking into it. What you're saying makes sense.
Since you are using a within-subjects design (prone to carryover effects) I think it would be most interesting to lo... | Comparing the within-subject variance between two groups of subjects
I'm not sure if you can assume normality of the performance variances, but I'm still looking into it. What you're saying makes sense.
Since you are using a within-subjects design (prone to carryover |
54,359 | Comparing the within-subject variance between two groups of subjects | It is not very well known but as an alternative to linear mixed model suggested by Jonas Berge you can perform a F test of equality of variances
https://en.wikipedia.org/wiki/F-test_of_equality_of_variances | Comparing the within-subject variance between two groups of subjects | It is not very well known but as an alternative to linear mixed model suggested by Jonas Berge you can perform a F test of equality of variances
https://en.wikipedia.org/wiki/F-test_of_equality_of_var | Comparing the within-subject variance between two groups of subjects
It is not very well known but as an alternative to linear mixed model suggested by Jonas Berge you can perform a F test of equality of variances
https://en.wikipedia.org/wiki/F-test_of_equality_of_variances | Comparing the within-subject variance between two groups of subjects
It is not very well known but as an alternative to linear mixed model suggested by Jonas Berge you can perform a F test of equality of variances
https://en.wikipedia.org/wiki/F-test_of_equality_of_var |
54,360 | Bias inputs in an RNN | This is basically correct.
The bias is an "offset" added to each unit in a neural network layer that's independent of the input to the layer. The bias permits a layer to model a data space that's centered around some point other than the origin.
Mathematically, a feedforward neural network layer without bias is written... | Bias inputs in an RNN | This is basically correct.
The bias is an "offset" added to each unit in a neural network layer that's independent of the input to the layer. The bias permits a layer to model a data space that's cent | Bias inputs in an RNN
This is basically correct.
The bias is an "offset" added to each unit in a neural network layer that's independent of the input to the layer. The bias permits a layer to model a data space that's centered around some point other than the origin.
Mathematically, a feedforward neural network layer w... | Bias inputs in an RNN
This is basically correct.
The bias is an "offset" added to each unit in a neural network layer that's independent of the input to the layer. The bias permits a layer to model a data space that's cent |
54,361 | Central Limit Theorem for Normal Distribution of Negative Binomial | You also can use CLT directly,one form of CLT states:
$\frac{\sum_{i=1}^nX_i-n\mu}{\sigma\sqrt{n}}\sim N(0,1)=\Rightarrow\sum_{i=1}^nX_i\sim N(n\mu,n\sigma^2)$
Above equations invovle two theorems:
The first one is one form CLT
The second related to multivariate normal distribution, but it also apply to 1-dimensional... | Central Limit Theorem for Normal Distribution of Negative Binomial | You also can use CLT directly,one form of CLT states:
$\frac{\sum_{i=1}^nX_i-n\mu}{\sigma\sqrt{n}}\sim N(0,1)=\Rightarrow\sum_{i=1}^nX_i\sim N(n\mu,n\sigma^2)$
Above equations invovle two theorems:
T | Central Limit Theorem for Normal Distribution of Negative Binomial
You also can use CLT directly,one form of CLT states:
$\frac{\sum_{i=1}^nX_i-n\mu}{\sigma\sqrt{n}}\sim N(0,1)=\Rightarrow\sum_{i=1}^nX_i\sim N(n\mu,n\sigma^2)$
Above equations invovle two theorems:
The first one is one form CLT
The second related to m... | Central Limit Theorem for Normal Distribution of Negative Binomial
You also can use CLT directly,one form of CLT states:
$\frac{\sum_{i=1}^nX_i-n\mu}{\sigma\sqrt{n}}\sim N(0,1)=\Rightarrow\sum_{i=1}^nX_i\sim N(n\mu,n\sigma^2)$
Above equations invovle two theorems:
T |
54,362 | Central Limit Theorem for Normal Distribution of Negative Binomial | 1.6(c) From the Central Limit Theorem we know that as the number of samples from any distribution increases, it becomes better approximated by a normal distribution.
This is not what the central limit theorem says. The CLT does not hold for every distribution, and in its standard form it concerns properly scaled and ... | Central Limit Theorem for Normal Distribution of Negative Binomial | 1.6(c) From the Central Limit Theorem we know that as the number of samples from any distribution increases, it becomes better approximated by a normal distribution.
This is not what the central lim | Central Limit Theorem for Normal Distribution of Negative Binomial
1.6(c) From the Central Limit Theorem we know that as the number of samples from any distribution increases, it becomes better approximated by a normal distribution.
This is not what the central limit theorem says. The CLT does not hold for every dist... | Central Limit Theorem for Normal Distribution of Negative Binomial
1.6(c) From the Central Limit Theorem we know that as the number of samples from any distribution increases, it becomes better approximated by a normal distribution.
This is not what the central lim |
54,363 | Central Limit Theorem for Normal Distribution of Negative Binomial | The Central Limit Theorem makes a limiting-distribution statement for sums of random variables from which sum we have subtracted the sum's expected value, and which we have divided by its standard deviation. Denoting $\sum_{i=1}^kY_i \equiv S_k$ the CLT can be written as
$$\frac {S_k - E(S_k)}{\sqrt {{\rm Var}(S_k)}} \... | Central Limit Theorem for Normal Distribution of Negative Binomial | The Central Limit Theorem makes a limiting-distribution statement for sums of random variables from which sum we have subtracted the sum's expected value, and which we have divided by its standard dev | Central Limit Theorem for Normal Distribution of Negative Binomial
The Central Limit Theorem makes a limiting-distribution statement for sums of random variables from which sum we have subtracted the sum's expected value, and which we have divided by its standard deviation. Denoting $\sum_{i=1}^kY_i \equiv S_k$ the CLT... | Central Limit Theorem for Normal Distribution of Negative Binomial
The Central Limit Theorem makes a limiting-distribution statement for sums of random variables from which sum we have subtracted the sum's expected value, and which we have divided by its standard dev |
54,364 | Multidimensional dynamic time warping | There are two ways to do it. The way you describe is DTWI, but other way, DTWD can be better, because it pools the information before warping.
There is an explanation of the differences, and an empirical study here.
http://www.cs.ucr.edu/~eamonn/Multi-Dimensional_DTW_Journal.pdf | Multidimensional dynamic time warping | There are two ways to do it. The way you describe is DTWI, but other way, DTWD can be better, because it pools the information before warping.
There is an explanation of the differences, and an empiri | Multidimensional dynamic time warping
There are two ways to do it. The way you describe is DTWI, but other way, DTWD can be better, because it pools the information before warping.
There is an explanation of the differences, and an empirical study here.
http://www.cs.ucr.edu/~eamonn/Multi-Dimensional_DTW_Journal.pdf | Multidimensional dynamic time warping
There are two ways to do it. The way you describe is DTWI, but other way, DTWD can be better, because it pools the information before warping.
There is an explanation of the differences, and an empiri |
54,365 | Name of single sample multinomial distribution | It's called the categorical distribution (among other things).
http://en.wikipedia.org/wiki/Categorical_distribution
This article mentions the following names: categorical, multinoulli, generalized Bernoulli | Name of single sample multinomial distribution | It's called the categorical distribution (among other things).
http://en.wikipedia.org/wiki/Categorical_distribution
This article mentions the following names: categorical, multinoulli, generalized Be | Name of single sample multinomial distribution
It's called the categorical distribution (among other things).
http://en.wikipedia.org/wiki/Categorical_distribution
This article mentions the following names: categorical, multinoulli, generalized Bernoulli | Name of single sample multinomial distribution
It's called the categorical distribution (among other things).
http://en.wikipedia.org/wiki/Categorical_distribution
This article mentions the following names: categorical, multinoulli, generalized Be |
54,366 | Naming PCA factors: is it a minor art? | You focus on "naming", but I would say that the real problem is understanding what principal components mean. You are right: this is an art. It often turns out that they are very difficult to interpret, hence all the attempts (especially in factor analysis literature and practice) to rotate the components/factors in o... | Naming PCA factors: is it a minor art? | You focus on "naming", but I would say that the real problem is understanding what principal components mean. You are right: this is an art. It often turns out that they are very difficult to interpre | Naming PCA factors: is it a minor art?
You focus on "naming", but I would say that the real problem is understanding what principal components mean. You are right: this is an art. It often turns out that they are very difficult to interpret, hence all the attempts (especially in factor analysis literature and practice... | Naming PCA factors: is it a minor art?
You focus on "naming", but I would say that the real problem is understanding what principal components mean. You are right: this is an art. It often turns out that they are very difficult to interpre |
54,367 | Is there a two-way Friedman's test? | Your data are ordinal ratings, so you need some form of ordinal logistic regression. But I also gather that your data are not independent ("... 15 measures per patient..."), so that needs to be taken into account as well. Thus, the appropriate method here is a mixed effects ordinal logistic regression. In R, mixed e... | Is there a two-way Friedman's test? | Your data are ordinal ratings, so you need some form of ordinal logistic regression. But I also gather that your data are not independent ("... 15 measures per patient..."), so that needs to be taken | Is there a two-way Friedman's test?
Your data are ordinal ratings, so you need some form of ordinal logistic regression. But I also gather that your data are not independent ("... 15 measures per patient..."), so that needs to be taken into account as well. Thus, the appropriate method here is a mixed effects ordinal... | Is there a two-way Friedman's test?
Your data are ordinal ratings, so you need some form of ordinal logistic regression. But I also gather that your data are not independent ("... 15 measures per patient..."), so that needs to be taken |
54,368 | Is there a two-way Friedman's test? | You can use simple linear regression with interactions to determine relations of proteins, locations (and their interaction) with concentations. Following is the output from the data that you have provided:
> summary(lm(Concentration~Protein*Location, data=prtdf))
Call:
lm(formula = Concentration ~ Protein * Location,... | Is there a two-way Friedman's test? | You can use simple linear regression with interactions to determine relations of proteins, locations (and their interaction) with concentations. Following is the output from the data that you have pro | Is there a two-way Friedman's test?
You can use simple linear regression with interactions to determine relations of proteins, locations (and their interaction) with concentations. Following is the output from the data that you have provided:
> summary(lm(Concentration~Protein*Location, data=prtdf))
Call:
lm(formula =... | Is there a two-way Friedman's test?
You can use simple linear regression with interactions to determine relations of proteins, locations (and their interaction) with concentations. Following is the output from the data that you have pro |
54,369 | Is there a two-way Friedman's test? | After reading a little bit more, I think I got the answer to my own question.
Since the response variable is ordinal and the other factors are categorical, I shouldn't use simple linear regression, so I've used Logistic Regression Model.
There are two packages in R that can handle ordinal variables: lrm {rms} and polr ... | Is there a two-way Friedman's test? | After reading a little bit more, I think I got the answer to my own question.
Since the response variable is ordinal and the other factors are categorical, I shouldn't use simple linear regression, so | Is there a two-way Friedman's test?
After reading a little bit more, I think I got the answer to my own question.
Since the response variable is ordinal and the other factors are categorical, I shouldn't use simple linear regression, so I've used Logistic Regression Model.
There are two packages in R that can handle or... | Is there a two-way Friedman's test?
After reading a little bit more, I think I got the answer to my own question.
Since the response variable is ordinal and the other factors are categorical, I shouldn't use simple linear regression, so |
54,370 | How do I calculate a t-score from a p-value (gain scores and N also available) | I'm guessing (hoping) this is a one-sided one-sample t test, where the 'gain' for
the $i$th subject is a difference $d_i$ and the test statistic is
$t = \bar d \sqrt{n}/S_d,$ in which $\bar d$ and $S_d$ are the mean
and standard deviation, respectively, of the $n$ differences.
Let $\delta$ be the population mean of $d_... | How do I calculate a t-score from a p-value (gain scores and N also available) | I'm guessing (hoping) this is a one-sided one-sample t test, where the 'gain' for
the $i$th subject is a difference $d_i$ and the test statistic is
$t = \bar d \sqrt{n}/S_d,$ in which $\bar d$ and $S_ | How do I calculate a t-score from a p-value (gain scores and N also available)
I'm guessing (hoping) this is a one-sided one-sample t test, where the 'gain' for
the $i$th subject is a difference $d_i$ and the test statistic is
$t = \bar d \sqrt{n}/S_d,$ in which $\bar d$ and $S_d$ are the mean
and standard deviation, r... | How do I calculate a t-score from a p-value (gain scores and N also available)
I'm guessing (hoping) this is a one-sided one-sample t test, where the 'gain' for
the $i$th subject is a difference $d_i$ and the test statistic is
$t = \bar d \sqrt{n}/S_d,$ in which $\bar d$ and $S_ |
54,371 | How do I calculate a t-score from a p-value (gain scores and N also available) | Why not just look it up a T Table or just punch the numbers in excel? I get the elaborate explanation above and good job at it but feels a bit overkill. In excel you can use T.INV. Just that you need degrees of freedom (which for the t distribution is n-1) for the second argument in T.INV(). And then just adjust based ... | How do I calculate a t-score from a p-value (gain scores and N also available) | Why not just look it up a T Table or just punch the numbers in excel? I get the elaborate explanation above and good job at it but feels a bit overkill. In excel you can use T.INV. Just that you need | How do I calculate a t-score from a p-value (gain scores and N also available)
Why not just look it up a T Table or just punch the numbers in excel? I get the elaborate explanation above and good job at it but feels a bit overkill. In excel you can use T.INV. Just that you need degrees of freedom (which for the t distr... | How do I calculate a t-score from a p-value (gain scores and N also available)
Why not just look it up a T Table or just punch the numbers in excel? I get the elaborate explanation above and good job at it but feels a bit overkill. In excel you can use T.INV. Just that you need |
54,372 | Can I Interpret the impact of variables like positive or negative on the model by Random Forest, as I can do by Logistic Regression | The short answer is No.
The long answer follows, for which I fit a random forest to demonstrate variable importance (a.k.a variable ranking):
if(!require('randomForest')) { install.packages("randomForest"); require("randomForest") }
# Observe iris data
pairs(iris)
# Train & Test split
train = sample (1: nrow(iris ... | Can I Interpret the impact of variables like positive or negative on the model by Random Forest, as | The short answer is No.
The long answer follows, for which I fit a random forest to demonstrate variable importance (a.k.a variable ranking):
if(!require('randomForest')) { install.packages("randomFo | Can I Interpret the impact of variables like positive or negative on the model by Random Forest, as I can do by Logistic Regression
The short answer is No.
The long answer follows, for which I fit a random forest to demonstrate variable importance (a.k.a variable ranking):
if(!require('randomForest')) { install.packag... | Can I Interpret the impact of variables like positive or negative on the model by Random Forest, as
The short answer is No.
The long answer follows, for which I fit a random forest to demonstrate variable importance (a.k.a variable ranking):
if(!require('randomForest')) { install.packages("randomFo |
54,373 | Selecting the most similar subset from an alternative dataset | This problem seem to be the same as selecting control groups for case-control studies. In the famous Doll and Hill study of smoking and lung cancer, the authors identified patients with lung cancer (cases) and then examined a control group "deliberately selected to be closely comparable in age and sex with the carcinom... | Selecting the most similar subset from an alternative dataset | This problem seem to be the same as selecting control groups for case-control studies. In the famous Doll and Hill study of smoking and lung cancer, the authors identified patients with lung cancer (c | Selecting the most similar subset from an alternative dataset
This problem seem to be the same as selecting control groups for case-control studies. In the famous Doll and Hill study of smoking and lung cancer, the authors identified patients with lung cancer (cases) and then examined a control group "deliberately sele... | Selecting the most similar subset from an alternative dataset
This problem seem to be the same as selecting control groups for case-control studies. In the famous Doll and Hill study of smoking and lung cancer, the authors identified patients with lung cancer (c |
54,374 | Selecting the most similar subset from an alternative dataset | EdM's answer seems like a good match, and you should absolutely look into that literature. Here's a possible alternative approach to explore.
One reasonable way of measuring distances between distributions is known as the maximum mean discrepancy (MMD); a thorough overview is given by Gretton, Borgwardt, Rasch, Schölko... | Selecting the most similar subset from an alternative dataset | EdM's answer seems like a good match, and you should absolutely look into that literature. Here's a possible alternative approach to explore.
One reasonable way of measuring distances between distribu | Selecting the most similar subset from an alternative dataset
EdM's answer seems like a good match, and you should absolutely look into that literature. Here's a possible alternative approach to explore.
One reasonable way of measuring distances between distributions is known as the maximum mean discrepancy (MMD); a th... | Selecting the most similar subset from an alternative dataset
EdM's answer seems like a good match, and you should absolutely look into that literature. Here's a possible alternative approach to explore.
One reasonable way of measuring distances between distribu |
54,375 | Selecting the most similar subset from an alternative dataset | Thanks to EdM, this problem is called matching. And I found an excellent package in R called Matching for this purpose. I found an excellent resource in this related document Genetic Matching for Estimating Causal Effects!. | Selecting the most similar subset from an alternative dataset | Thanks to EdM, this problem is called matching. And I found an excellent package in R called Matching for this purpose. I found an excellent resource in this related document Genetic Matching for Esti | Selecting the most similar subset from an alternative dataset
Thanks to EdM, this problem is called matching. And I found an excellent package in R called Matching for this purpose. I found an excellent resource in this related document Genetic Matching for Estimating Causal Effects!. | Selecting the most similar subset from an alternative dataset
Thanks to EdM, this problem is called matching. And I found an excellent package in R called Matching for this purpose. I found an excellent resource in this related document Genetic Matching for Esti |
54,376 | Why does the linear test statistic of GLM follow F-distribution? | Why does the linear test static of GLM follow F-distribution?
It doesn't.
Then, the test statistic will follow an $F$-distribution [...] does this hold for all generalized linear models?
There's no result that establishes it in the general case, and indeed we can show (e.g. by simulation in particular instances) tha... | Why does the linear test statistic of GLM follow F-distribution? | Why does the linear test static of GLM follow F-distribution?
It doesn't.
Then, the test statistic will follow an $F$-distribution [...] does this hold for all generalized linear models?
There's no | Why does the linear test statistic of GLM follow F-distribution?
Why does the linear test static of GLM follow F-distribution?
It doesn't.
Then, the test statistic will follow an $F$-distribution [...] does this hold for all generalized linear models?
There's no result that establishes it in the general case, and in... | Why does the linear test statistic of GLM follow F-distribution?
Why does the linear test static of GLM follow F-distribution?
It doesn't.
Then, the test statistic will follow an $F$-distribution [...] does this hold for all generalized linear models?
There's no |
54,377 | Shifted log-normal distribution and moments | We have
$Y^n=(aX+b)^n=\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k}$
so
$\mathbb{E}Y^n=\mathbb{E}(\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k})=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k \mathbb{E}X^k$.
The rest remains on what is $X$ (i.e. what are its $\mathbb{E}X^k$ moments). For log-normal distribution we have
$\mathbb{E}X^k=e^{k\... | Shifted log-normal distribution and moments | We have
$Y^n=(aX+b)^n=\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k}$
so
$\mathbb{E}Y^n=\mathbb{E}(\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k})=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k \mathbb{E}X^k$.
The rest rema | Shifted log-normal distribution and moments
We have
$Y^n=(aX+b)^n=\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k}$
so
$\mathbb{E}Y^n=\mathbb{E}(\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k})=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k \mathbb{E}X^k$.
The rest remains on what is $X$ (i.e. what are its $\mathbb{E}X^k$ moments). For log-norm... | Shifted log-normal distribution and moments
We have
$Y^n=(aX+b)^n=\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k}$
so
$\mathbb{E}Y^n=\mathbb{E}(\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k})=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k \mathbb{E}X^k$.
The rest rema |
54,378 | Shifted log-normal distribution and moments | You have $Y=aX+b$, but the multiplication by $a$ still leaves you with a lognormal, not really changing anything. If $X\sim \text{logN}(\mu,\sigma^2)$ then $aX\sim \text{logN}(\mu+\log(a),\sigma^2)$, so if you know how to compute moments for a lognormal you can do it for $aX$ as easily as for $X$.
So for $X\sim \text{l... | Shifted log-normal distribution and moments | You have $Y=aX+b$, but the multiplication by $a$ still leaves you with a lognormal, not really changing anything. If $X\sim \text{logN}(\mu,\sigma^2)$ then $aX\sim \text{logN}(\mu+\log(a),\sigma^2)$, | Shifted log-normal distribution and moments
You have $Y=aX+b$, but the multiplication by $a$ still leaves you with a lognormal, not really changing anything. If $X\sim \text{logN}(\mu,\sigma^2)$ then $aX\sim \text{logN}(\mu+\log(a),\sigma^2)$, so if you know how to compute moments for a lognormal you can do it for $aX$... | Shifted log-normal distribution and moments
You have $Y=aX+b$, but the multiplication by $a$ still leaves you with a lognormal, not really changing anything. If $X\sim \text{logN}(\mu,\sigma^2)$ then $aX\sim \text{logN}(\mu+\log(a),\sigma^2)$, |
54,379 | Number of Gaussian mixture components needed to approximate any distribution | I am afraid this is an absurd question: there is no magical number and no upper bound on the number of components in a Gaussian mixture for approximating (in which sense?) any distribution. Just think of the Gaussian mixture with 19 components... | Number of Gaussian mixture components needed to approximate any distribution | I am afraid this is an absurd question: there is no magical number and no upper bound on the number of components in a Gaussian mixture for approximating (in which sense?) any distribution. Just think | Number of Gaussian mixture components needed to approximate any distribution
I am afraid this is an absurd question: there is no magical number and no upper bound on the number of components in a Gaussian mixture for approximating (in which sense?) any distribution. Just think of the Gaussian mixture with 19 components... | Number of Gaussian mixture components needed to approximate any distribution
I am afraid this is an absurd question: there is no magical number and no upper bound on the number of components in a Gaussian mixture for approximating (in which sense?) any distribution. Just think |
54,380 | Why discrepancy between lasso and randomForest? | This could be because you're measuring two different things. The lasso coefficients are essentially effect sizes, and shrinkage helps distinguish "zero" effects from "nonzero" effects. Importance of a variable in the random forest model measures the improvement in predictive accuracy due to including that variable.
So ... | Why discrepancy between lasso and randomForest? | This could be because you're measuring two different things. The lasso coefficients are essentially effect sizes, and shrinkage helps distinguish "zero" effects from "nonzero" effects. Importance of a | Why discrepancy between lasso and randomForest?
This could be because you're measuring two different things. The lasso coefficients are essentially effect sizes, and shrinkage helps distinguish "zero" effects from "nonzero" effects. Importance of a variable in the random forest model measures the improvement in predict... | Why discrepancy between lasso and randomForest?
This could be because you're measuring two different things. The lasso coefficients are essentially effect sizes, and shrinkage helps distinguish "zero" effects from "nonzero" effects. Importance of a |
54,381 | What is the shape of the decision surface of a Gaussian Process classifier? | As you can see in the example I crafted below, the probability surface in the case of the squared exponential (Gaussian) Kernel as covariance function for Gaussian processes looks like a smooth density.
A good read about covariance functions in the context of Gaussian processes is Chapter 4 - Covariance Functions [1]
... | What is the shape of the decision surface of a Gaussian Process classifier? | As you can see in the example I crafted below, the probability surface in the case of the squared exponential (Gaussian) Kernel as covariance function for Gaussian processes looks like a smooth densit | What is the shape of the decision surface of a Gaussian Process classifier?
As you can see in the example I crafted below, the probability surface in the case of the squared exponential (Gaussian) Kernel as covariance function for Gaussian processes looks like a smooth density.
A good read about covariance functions in... | What is the shape of the decision surface of a Gaussian Process classifier?
As you can see in the example I crafted below, the probability surface in the case of the squared exponential (Gaussian) Kernel as covariance function for Gaussian processes looks like a smooth densit |
54,382 | Bayesian analysis: Estimate whether a parameter is 0 or not | If you are interested in TESTING B=0, then the standard Bayesian solution (i.e. the most accepted), to that problem is the Bayes Factor (BF). Suppose you want to test model 1 against model 2. Let
$\pi_1(\theta_1|y)= \frac{L(\theta_1)\pi_1(\theta_1)}{p_1(y)}$, $p_1(y)=\int L(\theta_1)\pi_1(\theta_1)d\theta_1$
$\pi_2(\t... | Bayesian analysis: Estimate whether a parameter is 0 or not | If you are interested in TESTING B=0, then the standard Bayesian solution (i.e. the most accepted), to that problem is the Bayes Factor (BF). Suppose you want to test model 1 against model 2. Let
$\p | Bayesian analysis: Estimate whether a parameter is 0 or not
If you are interested in TESTING B=0, then the standard Bayesian solution (i.e. the most accepted), to that problem is the Bayes Factor (BF). Suppose you want to test model 1 against model 2. Let
$\pi_1(\theta_1|y)= \frac{L(\theta_1)\pi_1(\theta_1)}{p_1(y)}$,... | Bayesian analysis: Estimate whether a parameter is 0 or not
If you are interested in TESTING B=0, then the standard Bayesian solution (i.e. the most accepted), to that problem is the Bayes Factor (BF). Suppose you want to test model 1 against model 2. Let
$\p |
54,383 | Bayesian analysis: Estimate whether a parameter is 0 or not | The model you describe is simple univariate linear regression and what you want to do is to test wether your intercept is greater then zero. You can use MCMC obtaining posterior distribution of such model and then you can check what is the proportion of cases where $B > 0$:
$$ \Pr(B > 0) = \frac{1}N \sum^N_{i=1} \mathb... | Bayesian analysis: Estimate whether a parameter is 0 or not | The model you describe is simple univariate linear regression and what you want to do is to test wether your intercept is greater then zero. You can use MCMC obtaining posterior distribution of such m | Bayesian analysis: Estimate whether a parameter is 0 or not
The model you describe is simple univariate linear regression and what you want to do is to test wether your intercept is greater then zero. You can use MCMC obtaining posterior distribution of such model and then you can check what is the proportion of cases ... | Bayesian analysis: Estimate whether a parameter is 0 or not
The model you describe is simple univariate linear regression and what you want to do is to test wether your intercept is greater then zero. You can use MCMC obtaining posterior distribution of such m |
54,384 | Extremly poor polynomial fitting with SVR in sklearn | In short, you need to tune your parameters. Here's the sklearn docs:
The free parameters in the model are C and epsilon.
and their descriptions:
C : float, optional (default=1.0)
Penalty parameter C of the error term.
epsilon : float, optional (default=0.1)
Epsilon in the epsilon-SVR model. It specifies the epsilon... | Extremly poor polynomial fitting with SVR in sklearn | In short, you need to tune your parameters. Here's the sklearn docs:
The free parameters in the model are C and epsilon.
and their descriptions:
C : float, optional (default=1.0)
Penalty parameter | Extremly poor polynomial fitting with SVR in sklearn
In short, you need to tune your parameters. Here's the sklearn docs:
The free parameters in the model are C and epsilon.
and their descriptions:
C : float, optional (default=1.0)
Penalty parameter C of the error term.
epsilon : float, optional (default=0.1)
Epsil... | Extremly poor polynomial fitting with SVR in sklearn
In short, you need to tune your parameters. Here's the sklearn docs:
The free parameters in the model are C and epsilon.
and their descriptions:
C : float, optional (default=1.0)
Penalty parameter |
54,385 | Extremly poor polynomial fitting with SVR in sklearn | For the polynomial kernel, specify kernel='poly' and also try also rescaling your data, as well as tuning your parameters C and epsilon as Matthew described.
>>> from sklearn import preprocessing
>>> X = [[ 1., -1., 2.],
... [ 2., 0., 0.],
... [ 0., 1., -1.]]
>>> X_scaled = preprocessing.scale(X)
>>> X_s... | Extremly poor polynomial fitting with SVR in sklearn | For the polynomial kernel, specify kernel='poly' and also try also rescaling your data, as well as tuning your parameters C and epsilon as Matthew described.
>>> from sklearn import preprocessing
>>> | Extremly poor polynomial fitting with SVR in sklearn
For the polynomial kernel, specify kernel='poly' and also try also rescaling your data, as well as tuning your parameters C and epsilon as Matthew described.
>>> from sklearn import preprocessing
>>> X = [[ 1., -1., 2.],
... [ 2., 0., 0.],
... [ 0., 1.,... | Extremly poor polynomial fitting with SVR in sklearn
For the polynomial kernel, specify kernel='poly' and also try also rescaling your data, as well as tuning your parameters C and epsilon as Matthew described.
>>> from sklearn import preprocessing
>>> |
54,386 | Is it feasible to transform each variable differently while doing multiple regression | Yes. Sure. The key is to understand that in the expression "linear regression" the word "linear" means "linear with respect to the coefficients in front of variables". So, you can not only transform each variable differently, but, for example, make two different transformations of each variable and include both in regr... | Is it feasible to transform each variable differently while doing multiple regression | Yes. Sure. The key is to understand that in the expression "linear regression" the word "linear" means "linear with respect to the coefficients in front of variables". So, you can not only transform e | Is it feasible to transform each variable differently while doing multiple regression
Yes. Sure. The key is to understand that in the expression "linear regression" the word "linear" means "linear with respect to the coefficients in front of variables". So, you can not only transform each variable differently, but, for... | Is it feasible to transform each variable differently while doing multiple regression
Yes. Sure. The key is to understand that in the expression "linear regression" the word "linear" means "linear with respect to the coefficients in front of variables". So, you can not only transform e |
54,387 | In a model with several parameters, which one should be tuned via cross validation first? | As these hyperparameters interact with each other, it is best to tune them together. Generally, a hyperparameter response surface is very complex, which means that tuning parameters separately usually leads to poor results.
The standard approach to tuning is grid search, e.g. test predetermined tuples of hyperparamete... | In a model with several parameters, which one should be tuned via cross validation first? | As these hyperparameters interact with each other, it is best to tune them together. Generally, a hyperparameter response surface is very complex, which means that tuning parameters separately usually | In a model with several parameters, which one should be tuned via cross validation first?
As these hyperparameters interact with each other, it is best to tune them together. Generally, a hyperparameter response surface is very complex, which means that tuning parameters separately usually leads to poor results.
The s... | In a model with several parameters, which one should be tuned via cross validation first?
As these hyperparameters interact with each other, it is best to tune them together. Generally, a hyperparameter response surface is very complex, which means that tuning parameters separately usually |
54,388 | Autocorrelation and Partial Correlation plots in ARMA models | The blue shaded part joins the boundaries of an approximate 95% interval for the individual correlations assuming the series is independent. So if your data were white noise, about 5% of those autocorrelations would be expected to lie outside those bounds.
The PACF is basically the lagged correlations adjusted for the ... | Autocorrelation and Partial Correlation plots in ARMA models | The blue shaded part joins the boundaries of an approximate 95% interval for the individual correlations assuming the series is independent. So if your data were white noise, about 5% of those autocor | Autocorrelation and Partial Correlation plots in ARMA models
The blue shaded part joins the boundaries of an approximate 95% interval for the individual correlations assuming the series is independent. So if your data were white noise, about 5% of those autocorrelations would be expected to lie outside those bounds.
Th... | Autocorrelation and Partial Correlation plots in ARMA models
The blue shaded part joins the boundaries of an approximate 95% interval for the individual correlations assuming the series is independent. So if your data were white noise, about 5% of those autocor |
54,389 | Autocorrelation and Partial Correlation plots in ARMA models | The blue shaded areas are used to test the statistical significance of the autocorrelation and partial autocorrelation coefficients. In the ACF, these bands are sometimes based on Bartlett's standard errors, which go back to a paper published in 1946. They are calculated using the formula below, where $r_{k}$ denotes t... | Autocorrelation and Partial Correlation plots in ARMA models | The blue shaded areas are used to test the statistical significance of the autocorrelation and partial autocorrelation coefficients. In the ACF, these bands are sometimes based on Bartlett's standard | Autocorrelation and Partial Correlation plots in ARMA models
The blue shaded areas are used to test the statistical significance of the autocorrelation and partial autocorrelation coefficients. In the ACF, these bands are sometimes based on Bartlett's standard errors, which go back to a paper published in 1946. They ar... | Autocorrelation and Partial Correlation plots in ARMA models
The blue shaded areas are used to test the statistical significance of the autocorrelation and partial autocorrelation coefficients. In the ACF, these bands are sometimes based on Bartlett's standard |
54,390 | Marginal, joint, and conditional distributions of a multivariate normal | Alrighty, y'all. I have an answer. Sorry it took me so long to get it posted here. School was absolutely hectic this week. Spring break is here, though, and I can type up my answer.
First we need to find the joint distribution of $(Y_1, Y_3)$. Since $Y\sim MVN( \mu, \Sigma)$ we know that any subset of the components ... | Marginal, joint, and conditional distributions of a multivariate normal | Alrighty, y'all. I have an answer. Sorry it took me so long to get it posted here. School was absolutely hectic this week. Spring break is here, though, and I can type up my answer.
First we need to | Marginal, joint, and conditional distributions of a multivariate normal
Alrighty, y'all. I have an answer. Sorry it took me so long to get it posted here. School was absolutely hectic this week. Spring break is here, though, and I can type up my answer.
First we need to find the joint distribution of $(Y_1, Y_3)$. Si... | Marginal, joint, and conditional distributions of a multivariate normal
Alrighty, y'all. I have an answer. Sorry it took me so long to get it posted here. School was absolutely hectic this week. Spring break is here, though, and I can type up my answer.
First we need to |
54,391 | Intuitive interpretation of Bayes risk $R(\delta, \lambda) = \int_{\Omega}R(\theta, \delta) \lambda(\theta) d\theta$ | The Bayes risk is the frequentist risk averaged over the parameter space against to the prior distribution $\lambda$. The notion turns a function of $\theta$, $R(\theta,\delta)$, into a positive number, $R(\delta,\lambda)$, and hence allows for a total ordering of estimators $\delta$, hence for the definition of the Ba... | Intuitive interpretation of Bayes risk $R(\delta, \lambda) = \int_{\Omega}R(\theta, \delta) \lambda( | The Bayes risk is the frequentist risk averaged over the parameter space against to the prior distribution $\lambda$. The notion turns a function of $\theta$, $R(\theta,\delta)$, into a positive numbe | Intuitive interpretation of Bayes risk $R(\delta, \lambda) = \int_{\Omega}R(\theta, \delta) \lambda(\theta) d\theta$
The Bayes risk is the frequentist risk averaged over the parameter space against to the prior distribution $\lambda$. The notion turns a function of $\theta$, $R(\theta,\delta)$, into a positive number, ... | Intuitive interpretation of Bayes risk $R(\delta, \lambda) = \int_{\Omega}R(\theta, \delta) \lambda(
The Bayes risk is the frequentist risk averaged over the parameter space against to the prior distribution $\lambda$. The notion turns a function of $\theta$, $R(\theta,\delta)$, into a positive numbe |
54,392 | Using an asymmetric distance matrix for clustering | Full metric properties are rarely required if you don't need strong theoretical results. In particular, $d(x,y)=0\Rightarrow x=y$ is not realistic on natural data. Just because two observed values are identical does not imply they are the same observation (this property can easily be restored by working on equivalence ... | Using an asymmetric distance matrix for clustering | Full metric properties are rarely required if you don't need strong theoretical results. In particular, $d(x,y)=0\Rightarrow x=y$ is not realistic on natural data. Just because two observed values are | Using an asymmetric distance matrix for clustering
Full metric properties are rarely required if you don't need strong theoretical results. In particular, $d(x,y)=0\Rightarrow x=y$ is not realistic on natural data. Just because two observed values are identical does not imply they are the same observation (this propert... | Using an asymmetric distance matrix for clustering
Full metric properties are rarely required if you don't need strong theoretical results. In particular, $d(x,y)=0\Rightarrow x=y$ is not realistic on natural data. Just because two observed values are |
54,393 | How to compute the sum of a mixture distribution with another distribution? | Let's start with notation. When you need to refer to different things, you have to give them different names. To that end let $f$ refer to the first PDF and $g$ the second.
Second, let's focus on the form of the PDFs and strip away unnecessary details:
$f$ is a positive linear combination of $\phi_b$ and $\phi_l$: t... | How to compute the sum of a mixture distribution with another distribution? | Let's start with notation. When you need to refer to different things, you have to give them different names. To that end let $f$ refer to the first PDF and $g$ the second.
Second, let's focus on th | How to compute the sum of a mixture distribution with another distribution?
Let's start with notation. When you need to refer to different things, you have to give them different names. To that end let $f$ refer to the first PDF and $g$ the second.
Second, let's focus on the form of the PDFs and strip away unnecessar... | How to compute the sum of a mixture distribution with another distribution?
Let's start with notation. When you need to refer to different things, you have to give them different names. To that end let $f$ refer to the first PDF and $g$ the second.
Second, let's focus on th |
54,394 | Generalized linear models and central limit theorem | In short: CLT alone isn't sufficient; $n$ isn't always near enough to infinity; and the shape of the distribution of the sample mean isn't the only consideration
If you're using a hypothesis test in your comparison of treatment means in ANOVA, you rely on the distribution of a ratio of two quantities having an F-distr... | Generalized linear models and central limit theorem | In short: CLT alone isn't sufficient; $n$ isn't always near enough to infinity; and the shape of the distribution of the sample mean isn't the only consideration
If you're using a hypothesis test in | Generalized linear models and central limit theorem
In short: CLT alone isn't sufficient; $n$ isn't always near enough to infinity; and the shape of the distribution of the sample mean isn't the only consideration
If you're using a hypothesis test in your comparison of treatment means in ANOVA, you rely on the distrib... | Generalized linear models and central limit theorem
In short: CLT alone isn't sufficient; $n$ isn't always near enough to infinity; and the shape of the distribution of the sample mean isn't the only consideration
If you're using a hypothesis test in |
54,395 | Proving Bayesian Network must be acyclic | You're not going to find any contradictions by creating a cyclic graph. It's not that Bayes nets (or as I've heard them called, conditional independence networks since they really don't have anything to do with Bayesianism besides conditional independence rules) "have to be acyclic". We assume them to be acyclic to get... | Proving Bayesian Network must be acyclic | You're not going to find any contradictions by creating a cyclic graph. It's not that Bayes nets (or as I've heard them called, conditional independence networks since they really don't have anything | Proving Bayesian Network must be acyclic
You're not going to find any contradictions by creating a cyclic graph. It's not that Bayes nets (or as I've heard them called, conditional independence networks since they really don't have anything to do with Bayesianism besides conditional independence rules) "have to be acyc... | Proving Bayesian Network must be acyclic
You're not going to find any contradictions by creating a cyclic graph. It's not that Bayes nets (or as I've heard them called, conditional independence networks since they really don't have anything |
54,396 | Proving Bayesian Network must be acyclic | A Bayesian Network can be viewed as a data structure that provides the skeleton for representing a joint distribution compactly in a factorized way. For any valid joint distribution two restrictions should be satisfied:
1) All probabilities in the distribution should be non negative;
2) All the probabilities should s... | Proving Bayesian Network must be acyclic | A Bayesian Network can be viewed as a data structure that provides the skeleton for representing a joint distribution compactly in a factorized way. For any valid joint distribution two restrictions s | Proving Bayesian Network must be acyclic
A Bayesian Network can be viewed as a data structure that provides the skeleton for representing a joint distribution compactly in a factorized way. For any valid joint distribution two restrictions should be satisfied:
1) All probabilities in the distribution should be non ne... | Proving Bayesian Network must be acyclic
A Bayesian Network can be viewed as a data structure that provides the skeleton for representing a joint distribution compactly in a factorized way. For any valid joint distribution two restrictions s |
54,397 | Proving Bayesian Network must be acyclic | I am afraid that Nick's answer might be incomplete.
BNs must be acyclic in order to guarantee that their underlying probability distribution is normalized to 1. It is quite easy to prove that this is the case, by starting at a vertex with no parents (which must exist, otherwise the graph would contain a cycle) and marg... | Proving Bayesian Network must be acyclic | I am afraid that Nick's answer might be incomplete.
BNs must be acyclic in order to guarantee that their underlying probability distribution is normalized to 1. It is quite easy to prove that this is | Proving Bayesian Network must be acyclic
I am afraid that Nick's answer might be incomplete.
BNs must be acyclic in order to guarantee that their underlying probability distribution is normalized to 1. It is quite easy to prove that this is the case, by starting at a vertex with no parents (which must exist, otherwise ... | Proving Bayesian Network must be acyclic
I am afraid that Nick's answer might be incomplete.
BNs must be acyclic in order to guarantee that their underlying probability distribution is normalized to 1. It is quite easy to prove that this is |
54,398 | Proving Bayesian Network must be acyclic | Bayesian Network is defined to be a DAG (Directed Acyclic Graph), you cannot prove a definition. Take a look at this explanation. | Proving Bayesian Network must be acyclic | Bayesian Network is defined to be a DAG (Directed Acyclic Graph), you cannot prove a definition. Take a look at this explanation. | Proving Bayesian Network must be acyclic
Bayesian Network is defined to be a DAG (Directed Acyclic Graph), you cannot prove a definition. Take a look at this explanation. | Proving Bayesian Network must be acyclic
Bayesian Network is defined to be a DAG (Directed Acyclic Graph), you cannot prove a definition. Take a look at this explanation. |
54,399 | Proving Bayesian Network must be acyclic | Cycles are redundant, that's all. And it is pretty easy to show if you think about the transitivity.
A -> B -> C -> A implies A -> B and B -> A
So the directionality is immediately lost (for every pair in the loop).
A -> B -> C -> A is also deterministic, the multiplier is one.
For example:
Say B = (1/3)A + e1 and ... | Proving Bayesian Network must be acyclic | Cycles are redundant, that's all. And it is pretty easy to show if you think about the transitivity.
A -> B -> C -> A implies A -> B and B -> A
So the directionality is immediately lost (for every p | Proving Bayesian Network must be acyclic
Cycles are redundant, that's all. And it is pretty easy to show if you think about the transitivity.
A -> B -> C -> A implies A -> B and B -> A
So the directionality is immediately lost (for every pair in the loop).
A -> B -> C -> A is also deterministic, the multiplier is on... | Proving Bayesian Network must be acyclic
Cycles are redundant, that's all. And it is pretty easy to show if you think about the transitivity.
A -> B -> C -> A implies A -> B and B -> A
So the directionality is immediately lost (for every p |
54,400 | Does triple interaction need to include all main effect variables? | It's awkward to try to give direct, literal answers to "should I?" and "do I have to?" questions on this site. It's preferable to talk about what the consequences of a certain decision are likely to be.
If you include a second-order, AxBxD interaction term without the first-order BxD term, you are liable to mistake ... | Does triple interaction need to include all main effect variables? | It's awkward to try to give direct, literal answers to "should I?" and "do I have to?" questions on this site. It's preferable to talk about what the consequences of a certain decision are likely to | Does triple interaction need to include all main effect variables?
It's awkward to try to give direct, literal answers to "should I?" and "do I have to?" questions on this site. It's preferable to talk about what the consequences of a certain decision are likely to be.
If you include a second-order, AxBxD interactio... | Does triple interaction need to include all main effect variables?
It's awkward to try to give direct, literal answers to "should I?" and "do I have to?" questions on this site. It's preferable to talk about what the consequences of a certain decision are likely to |
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