idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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48,501 | How to do regression when there is a mix of numerical and non-numerical predictor variables? | First off, it depends what your dependent variable (Y) is. If it is numerical then most multiple regression models would be sufficient. If it (Y) is categorical then you need a logistic regression or a similar categorical regression model.
As for how to handle independent variables, the numerical ones will fit neatly i... | How to do regression when there is a mix of numerical and non-numerical predictor variables? | First off, it depends what your dependent variable (Y) is. If it is numerical then most multiple regression models would be sufficient. If it (Y) is categorical then you need a logistic regression or | How to do regression when there is a mix of numerical and non-numerical predictor variables?
First off, it depends what your dependent variable (Y) is. If it is numerical then most multiple regression models would be sufficient. If it (Y) is categorical then you need a logistic regression or a similar categorical regre... | How to do regression when there is a mix of numerical and non-numerical predictor variables?
First off, it depends what your dependent variable (Y) is. If it is numerical then most multiple regression models would be sufficient. If it (Y) is categorical then you need a logistic regression or |
48,502 | How to do regression when there is a mix of numerical and non-numerical predictor variables? | Almost all the regression algorithms take care of both numerical and categorical variables. For categorical variables, there are different "coding" can be used.
The Simple example is binary coding. For example, for gender, you can use $0$ to represent male and $1$ to represent female. If the variables has more then $2$... | How to do regression when there is a mix of numerical and non-numerical predictor variables? | Almost all the regression algorithms take care of both numerical and categorical variables. For categorical variables, there are different "coding" can be used.
The Simple example is binary coding. Fo | How to do regression when there is a mix of numerical and non-numerical predictor variables?
Almost all the regression algorithms take care of both numerical and categorical variables. For categorical variables, there are different "coding" can be used.
The Simple example is binary coding. For example, for gender, you ... | How to do regression when there is a mix of numerical and non-numerical predictor variables?
Almost all the regression algorithms take care of both numerical and categorical variables. For categorical variables, there are different "coding" can be used.
The Simple example is binary coding. Fo |
48,503 | Training performance jumps up after epoch, Dev performance jumps down | Yes, this reduction is normal
When training any kind of machine learning algorithm, if you continue training, your algorithm overfits the training set and starts learning the details of the noise in the training set instead of utilizing the generalizable information. When it does this, your algorithm often loses it's g... | Training performance jumps up after epoch, Dev performance jumps down | Yes, this reduction is normal
When training any kind of machine learning algorithm, if you continue training, your algorithm overfits the training set and starts learning the details of the noise in t | Training performance jumps up after epoch, Dev performance jumps down
Yes, this reduction is normal
When training any kind of machine learning algorithm, if you continue training, your algorithm overfits the training set and starts learning the details of the noise in the training set instead of utilizing the generaliz... | Training performance jumps up after epoch, Dev performance jumps down
Yes, this reduction is normal
When training any kind of machine learning algorithm, if you continue training, your algorithm overfits the training set and starts learning the details of the noise in t |
48,504 | Training performance jumps up after epoch, Dev performance jumps down | I believe this can be an artefact on how you calculate your metrics. If you are using an moving average to calculate a metric and reseting the moving average at epoch boundaries than it is reasonable to expect a jump in epoch boundaries.
For example if you accuracies are going from 0 to 50 in the first epoch you movin... | Training performance jumps up after epoch, Dev performance jumps down | I believe this can be an artefact on how you calculate your metrics. If you are using an moving average to calculate a metric and reseting the moving average at epoch boundaries than it is reasonable | Training performance jumps up after epoch, Dev performance jumps down
I believe this can be an artefact on how you calculate your metrics. If you are using an moving average to calculate a metric and reseting the moving average at epoch boundaries than it is reasonable to expect a jump in epoch boundaries.
For example... | Training performance jumps up after epoch, Dev performance jumps down
I believe this can be an artefact on how you calculate your metrics. If you are using an moving average to calculate a metric and reseting the moving average at epoch boundaries than it is reasonable |
48,505 | Number of Bernoulli trials to first success, with changing $p$ | We all come in contact with a special case of this distribution when going Bernoulli $\rightarrow$ Geometric distribution.
The Bernoulli distribution is the realization of an event $X_k=1$ with $\Pr(X_k=1)=\theta$. The Geometric distribution $Y$ is the realization of $y-1$ non-events and $1$ event, all iid trials i.... | Number of Bernoulli trials to first success, with changing $p$ | We all come in contact with a special case of this distribution when going Bernoulli $\rightarrow$ Geometric distribution.
The Bernoulli distribution is the realization of an event $X_k=1$ with $\Pr | Number of Bernoulli trials to first success, with changing $p$
We all come in contact with a special case of this distribution when going Bernoulli $\rightarrow$ Geometric distribution.
The Bernoulli distribution is the realization of an event $X_k=1$ with $\Pr(X_k=1)=\theta$. The Geometric distribution $Y$ is the re... | Number of Bernoulli trials to first success, with changing $p$
We all come in contact with a special case of this distribution when going Bernoulli $\rightarrow$ Geometric distribution.
The Bernoulli distribution is the realization of an event $X_k=1$ with $\Pr |
48,506 | A sequence of random variables, how to understand it in the convergence theory? | As I understand this. You should have some Randome Variables $X_n$ which depends on $n$.
For example, if you take a look at this post:
Convergence in distribution, probability, and 2nd mean
You'll find that if $n \rightarrow \infty$ then $X_n$ converges in probability. Depeding on RVs you have different types of conver... | A sequence of random variables, how to understand it in the convergence theory? | As I understand this. You should have some Randome Variables $X_n$ which depends on $n$.
For example, if you take a look at this post:
Convergence in distribution, probability, and 2nd mean
You'll fin | A sequence of random variables, how to understand it in the convergence theory?
As I understand this. You should have some Randome Variables $X_n$ which depends on $n$.
For example, if you take a look at this post:
Convergence in distribution, probability, and 2nd mean
You'll find that if $n \rightarrow \infty$ then $X... | A sequence of random variables, how to understand it in the convergence theory?
As I understand this. You should have some Randome Variables $X_n$ which depends on $n$.
For example, if you take a look at this post:
Convergence in distribution, probability, and 2nd mean
You'll fin |
48,507 | A sequence of random variables, how to understand it in the convergence theory? | The following contents are just copy-paste from: Sequence of Random Variables.
Here, we would like to discuss what we precisely mean by a sequence of random variables. Remember that, in any probability model, we have a sample space $S$ and a probability measure $P$. For simplicity, suppose that our sample space consist... | A sequence of random variables, how to understand it in the convergence theory? | The following contents are just copy-paste from: Sequence of Random Variables.
Here, we would like to discuss what we precisely mean by a sequence of random variables. Remember that, in any probabilit | A sequence of random variables, how to understand it in the convergence theory?
The following contents are just copy-paste from: Sequence of Random Variables.
Here, we would like to discuss what we precisely mean by a sequence of random variables. Remember that, in any probability model, we have a sample space $S$ and ... | A sequence of random variables, how to understand it in the convergence theory?
The following contents are just copy-paste from: Sequence of Random Variables.
Here, we would like to discuss what we precisely mean by a sequence of random variables. Remember that, in any probabilit |
48,508 | How did Likert calculate sigma values in his original 1932 paper? | Thorndike's Table 22 displays the expected value of a doubly-truncated normal distribution, which can be seen as a conditional expectation given that the variate is in an interval specified by quantiles:
$$\mathbb{E}(Z \mid z_p<Z<z_{p+q}) = \frac{\phi(z_p)-\phi(z_{p+q})}{q}$$
where $z_p$ is the lower $p$th quantile of ... | How did Likert calculate sigma values in his original 1932 paper? | Thorndike's Table 22 displays the expected value of a doubly-truncated normal distribution, which can be seen as a conditional expectation given that the variate is in an interval specified by quantil | How did Likert calculate sigma values in his original 1932 paper?
Thorndike's Table 22 displays the expected value of a doubly-truncated normal distribution, which can be seen as a conditional expectation given that the variate is in an interval specified by quantiles:
$$\mathbb{E}(Z \mid z_p<Z<z_{p+q}) = \frac{\phi(z_... | How did Likert calculate sigma values in his original 1932 paper?
Thorndike's Table 22 displays the expected value of a doubly-truncated normal distribution, which can be seen as a conditional expectation given that the variate is in an interval specified by quantil |
48,509 | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix? | Instruction how you can compute sums of squares SSt, SSb, SSw out of matrix of distances (euclidean) between cases (data points) without having at hand the cases x variables dataset. You don't need to know the centroids' coordinates (the group means) - they pass invisibly "on the background": euclidean geometry laws al... | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix? | Instruction how you can compute sums of squares SSt, SSb, SSw out of matrix of distances (euclidean) between cases (data points) without having at hand the cases x variables dataset. You don't need to | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix?
Instruction how you can compute sums of squares SSt, SSb, SSw out of matrix of distances (euclidean) between cases (data points) without having at hand the cases x variables dataset. You don't need to know the centroids' coordinates (... | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix?
Instruction how you can compute sums of squares SSt, SSb, SSw out of matrix of distances (euclidean) between cases (data points) without having at hand the cases x variables dataset. You don't need to |
48,510 | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix? | Sum of squares is closely tied to Euclidean distance. Hamming (on bits) is a special case, as it is the same as Euclidean (on bits), but you cannot conclude from an arbitrary distance matrix what the SSQ etc. are.
Recall how the sum-of-squares are usually defined, compared to Euclidean distance:
$$
SSQ(A,B) = \sum_{a\i... | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix? | Sum of squares is closely tied to Euclidean distance. Hamming (on bits) is a special case, as it is the same as Euclidean (on bits), but you cannot conclude from an arbitrary distance matrix what the | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix?
Sum of squares is closely tied to Euclidean distance. Hamming (on bits) is a special case, as it is the same as Euclidean (on bits), but you cannot conclude from an arbitrary distance matrix what the SSQ etc. are.
Recall how the sum-o... | Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix?
Sum of squares is closely tied to Euclidean distance. Hamming (on bits) is a special case, as it is the same as Euclidean (on bits), but you cannot conclude from an arbitrary distance matrix what the |
48,511 | Negative Binomial MGF converges to Poisson MGF | You make a mistake by ignoring $p^r$: If you consider your MGF $$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}\,,$$
then $$\log\{M_X(t)\} = r\log(p)-r\log\{1-e^t(1-p)\}$$ and using the asymptotic equivalences
\begin{align*}
r\log(p)-r\log\{1-e^t(1-p)\}
&= r\log(1-[1-p])-r\log\{1-e^t(1-p)\}\\
&\approx -r[1-p]+re^t(1-p)\\
&\approx... | Negative Binomial MGF converges to Poisson MGF | You make a mistake by ignoring $p^r$: If you consider your MGF $$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}\,,$$
then $$\log\{M_X(t)\} = r\log(p)-r\log\{1-e^t(1-p)\}$$ and using the asymptotic equivalences
\ | Negative Binomial MGF converges to Poisson MGF
You make a mistake by ignoring $p^r$: If you consider your MGF $$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}\,,$$
then $$\log\{M_X(t)\} = r\log(p)-r\log\{1-e^t(1-p)\}$$ and using the asymptotic equivalences
\begin{align*}
r\log(p)-r\log\{1-e^t(1-p)\}
&= r\log(1-[1-p])-r\log\{1-e^t... | Negative Binomial MGF converges to Poisson MGF
You make a mistake by ignoring $p^r$: If you consider your MGF $$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}\,,$$
then $$\log\{M_X(t)\} = r\log(p)-r\log\{1-e^t(1-p)\}$$ and using the asymptotic equivalences
\ |
48,512 | Negative Binomial MGF converges to Poisson MGF | $$\eqalign{ M_{x}(t)&=\left[\frac{p}{1-\frac{e^{t}(1-p)}{p}}\right]^r \\
&=\left[\frac{1-(1-p)}{1-\frac{e^{t}(1-p)}{p}}\right]\\
&= \left[\frac{1-\frac{r(1-p)}{r}}{1-\frac{e^{t}r(1-p)}{rp}}\right]\\
&=\frac{e^{-r(1-p)}}{e^{{-e^{t}}r(1-p)}}\\
&=e^{-\lambda}e^{e^{t}\... | Negative Binomial MGF converges to Poisson MGF | $$\eqalign{ M_{x}(t)&=\left[\frac{p}{1-\frac{e^{t}(1-p)}{p}}\right]^r \\
&=\left[\frac{1-(1-p)}{1-\frac{e^{t}(1-p)}{p}}\right]\\
&= \left[\frac{1-\frac{r(1-p)}{r}}{1-\frac | Negative Binomial MGF converges to Poisson MGF
$$\eqalign{ M_{x}(t)&=\left[\frac{p}{1-\frac{e^{t}(1-p)}{p}}\right]^r \\
&=\left[\frac{1-(1-p)}{1-\frac{e^{t}(1-p)}{p}}\right]\\
&= \left[\frac{1-\frac{r(1-p)}{r}}{1-\frac{e^{t}r(1-p)}{rp}}\right]\\
&=\frac{e^{-r(1-p)}}{e^{{-e^{t}}... | Negative Binomial MGF converges to Poisson MGF
$$\eqalign{ M_{x}(t)&=\left[\frac{p}{1-\frac{e^{t}(1-p)}{p}}\right]^r \\
&=\left[\frac{1-(1-p)}{1-\frac{e^{t}(1-p)}{p}}\right]\\
&= \left[\frac{1-\frac{r(1-p)}{r}}{1-\frac |
48,513 | Negative Binomial MGF converges to Poisson MGF | The above answer was correct that you ignored the numerator $p^r$, but I found the illustration is a little confusing, so here we go for a standard way to solve the problem
$\begin{aligned} \lim_{r\to \infty} M_{NB}(t) & = \lim_{r\to \infty} (\frac{p}{1-(1-p)e^t})^r \\ &= \lim_{r\to \infty} (\frac{1-(1-p)}{1-(1-p)e^t})... | Negative Binomial MGF converges to Poisson MGF | The above answer was correct that you ignored the numerator $p^r$, but I found the illustration is a little confusing, so here we go for a standard way to solve the problem
$\begin{aligned} \lim_{r\to | Negative Binomial MGF converges to Poisson MGF
The above answer was correct that you ignored the numerator $p^r$, but I found the illustration is a little confusing, so here we go for a standard way to solve the problem
$\begin{aligned} \lim_{r\to \infty} M_{NB}(t) & = \lim_{r\to \infty} (\frac{p}{1-(1-p)e^t})^r \\ &= ... | Negative Binomial MGF converges to Poisson MGF
The above answer was correct that you ignored the numerator $p^r$, but I found the illustration is a little confusing, so here we go for a standard way to solve the problem
$\begin{aligned} \lim_{r\to |
48,514 | Convergence of the Independent Metropolis-Hastings algorithm | A more relevant paper about the convergence of Metropolis-Hastings algorithms is the one by Mengersen and Tweedie (1996) since it is both quite readable and general. In this paper, two major results can be singled out:
The independent Metropolis-Hastings algorithm with target $p$ and proposal $q$ leads to a uniformly ... | Convergence of the Independent Metropolis-Hastings algorithm | A more relevant paper about the convergence of Metropolis-Hastings algorithms is the one by Mengersen and Tweedie (1996) since it is both quite readable and general. In this paper, two major results c | Convergence of the Independent Metropolis-Hastings algorithm
A more relevant paper about the convergence of Metropolis-Hastings algorithms is the one by Mengersen and Tweedie (1996) since it is both quite readable and general. In this paper, two major results can be singled out:
The independent Metropolis-Hastings alg... | Convergence of the Independent Metropolis-Hastings algorithm
A more relevant paper about the convergence of Metropolis-Hastings algorithms is the one by Mengersen and Tweedie (1996) since it is both quite readable and general. In this paper, two major results c |
48,515 | "Error: no valid set of coefficients has been found: please supply starting values" when trying to get confidence intervals in R | I can't solve this problem (yet), but I can diagnose some of what's going wrong.
This is a summary of your data:
with(dd,table(Treatment,Count))
Count
Treatment 0 1 2 4
1 13 0 0 0
2 4 8 1 0
3 13 0 0 0
4 5 6 1 1
You can see that treatments 1 and 3 have all of ... | "Error: no valid set of coefficients has been found: please supply starting values" when trying to g | I can't solve this problem (yet), but I can diagnose some of what's going wrong.
This is a summary of your data:
with(dd,table(Treatment,Count))
Count
Treatment 0 1 2 4
1 13 0 | "Error: no valid set of coefficients has been found: please supply starting values" when trying to get confidence intervals in R
I can't solve this problem (yet), but I can diagnose some of what's going wrong.
This is a summary of your data:
with(dd,table(Treatment,Count))
Count
Treatment 0 1 2 4
... | "Error: no valid set of coefficients has been found: please supply starting values" when trying to g
I can't solve this problem (yet), but I can diagnose some of what's going wrong.
This is a summary of your data:
with(dd,table(Treatment,Count))
Count
Treatment 0 1 2 4
1 13 0 |
48,516 | How can I make the correct time-series analysis for my data? | You would likely be looking to calculate the residual autocorrelation function (RACF), also called residual cross-correlation function, which is the autocorrelation function on the residuals of fitted models for two different time series (fitted using ARMA models, auto-regressive moving average models). The calculation... | How can I make the correct time-series analysis for my data? | You would likely be looking to calculate the residual autocorrelation function (RACF), also called residual cross-correlation function, which is the autocorrelation function on the residuals of fitted | How can I make the correct time-series analysis for my data?
You would likely be looking to calculate the residual autocorrelation function (RACF), also called residual cross-correlation function, which is the autocorrelation function on the residuals of fitted models for two different time series (fitted using ARMA mo... | How can I make the correct time-series analysis for my data?
You would likely be looking to calculate the residual autocorrelation function (RACF), also called residual cross-correlation function, which is the autocorrelation function on the residuals of fitted |
48,517 | Correctly expressing improvement in AUC? | You can equivalently use the following expressions:
50% relative improvement
25% absolute improvement, or a 25 percentage point improvement.
Speaking of correct terms, AUROC (if that is your metric) should be used instead of AUC, as the latter term is ambiguous. | Correctly expressing improvement in AUC? | You can equivalently use the following expressions:
50% relative improvement
25% absolute improvement, or a 25 percentage point improvement.
Speaking of correct terms, AUROC (if that is your metric) | Correctly expressing improvement in AUC?
You can equivalently use the following expressions:
50% relative improvement
25% absolute improvement, or a 25 percentage point improvement.
Speaking of correct terms, AUROC (if that is your metric) should be used instead of AUC, as the latter term is ambiguous. | Correctly expressing improvement in AUC?
You can equivalently use the following expressions:
50% relative improvement
25% absolute improvement, or a 25 percentage point improvement.
Speaking of correct terms, AUROC (if that is your metric) |
48,518 | Correctly expressing improvement in AUC? | I just posted a somewhat related question here: Compare and quantify relative improvement in ROC AUC scores? that has this as a component to it.
I'm not sure, though given the ROC AUC represents the probability that a classifier scores a positive case higher than a negative case, perhaps comparing the magnitude of the... | Correctly expressing improvement in AUC? | I just posted a somewhat related question here: Compare and quantify relative improvement in ROC AUC scores? that has this as a component to it.
I'm not sure, though given the ROC AUC represents the | Correctly expressing improvement in AUC?
I just posted a somewhat related question here: Compare and quantify relative improvement in ROC AUC scores? that has this as a component to it.
I'm not sure, though given the ROC AUC represents the probability that a classifier scores a positive case higher than a negative cas... | Correctly expressing improvement in AUC?
I just posted a somewhat related question here: Compare and quantify relative improvement in ROC AUC scores? that has this as a component to it.
I'm not sure, though given the ROC AUC represents the |
48,519 | Correctly expressing improvement in AUC? | To calculate your relative improvement assumed to be $x$, you should use $x=(0.75-0.5)/0.5$.
I believe the relative improvement is more intuitive in the AUC case. Also, AUC is commomly used in literature, so both AUC and AUROC arr okay | Correctly expressing improvement in AUC? | To calculate your relative improvement assumed to be $x$, you should use $x=(0.75-0.5)/0.5$.
I believe the relative improvement is more intuitive in the AUC case. Also, AUC is commomly used in litera | Correctly expressing improvement in AUC?
To calculate your relative improvement assumed to be $x$, you should use $x=(0.75-0.5)/0.5$.
I believe the relative improvement is more intuitive in the AUC case. Also, AUC is commomly used in literature, so both AUC and AUROC arr okay | Correctly expressing improvement in AUC?
To calculate your relative improvement assumed to be $x$, you should use $x=(0.75-0.5)/0.5$.
I believe the relative improvement is more intuitive in the AUC case. Also, AUC is commomly used in litera |
48,520 | Basic question on link function in GLM | All of the basic regression-type models that people use are for the mean. With OLS regression, where the response is assumed conditionally normal, your predicted values, $\hat y_i$, are the conditional means (cf., here). So in a GLiM context more broadly, where the response is distributed as something else like a Ber... | Basic question on link function in GLM | All of the basic regression-type models that people use are for the mean. With OLS regression, where the response is assumed conditionally normal, your predicted values, $\hat y_i$, are the condition | Basic question on link function in GLM
All of the basic regression-type models that people use are for the mean. With OLS regression, where the response is assumed conditionally normal, your predicted values, $\hat y_i$, are the conditional means (cf., here). So in a GLiM context more broadly, where the response is d... | Basic question on link function in GLM
All of the basic regression-type models that people use are for the mean. With OLS regression, where the response is assumed conditionally normal, your predicted values, $\hat y_i$, are the condition |
48,521 | How can we convert values proportional to probabilities to Bernoulli probabilities? | Since $p(1)=p$ and $p(0)=1-p$ are both proportional to a known expression* (the unscaled probabilities, $u(i)=c.p(i)$, with the same unknown constant of proportionality, $c$) and you know the $p(i)$ values must add to $1$, then $u(0)+u(1)=c$.
Which is to say $p(i) = \frac{u(i)}{u(0)+u(1)},\: i=0,1$.
(This notion is wi... | How can we convert values proportional to probabilities to Bernoulli probabilities? | Since $p(1)=p$ and $p(0)=1-p$ are both proportional to a known expression* (the unscaled probabilities, $u(i)=c.p(i)$, with the same unknown constant of proportionality, $c$) and you know the $p(i)$ v | How can we convert values proportional to probabilities to Bernoulli probabilities?
Since $p(1)=p$ and $p(0)=1-p$ are both proportional to a known expression* (the unscaled probabilities, $u(i)=c.p(i)$, with the same unknown constant of proportionality, $c$) and you know the $p(i)$ values must add to $1$, then $u(0)+u(... | How can we convert values proportional to probabilities to Bernoulli probabilities?
Since $p(1)=p$ and $p(0)=1-p$ are both proportional to a known expression* (the unscaled probabilities, $u(i)=c.p(i)$, with the same unknown constant of proportionality, $c$) and you know the $p(i)$ v |
48,522 | Why is regularization interpreted as a gaussian prior on my weights? | Since we're using MAP we are trying to maximize the probability of the parameters given the data.
$$
P(W|x,y) = \frac {P(x,y|W) P(W)} {P(x,y)}
$$
$P(x,y)$ can be ignored since it's fixed for our data. So we are trying to maximize $log(P(x,y|W)) + log(P(W))$. Let's look at $P(W)$.
If each element of $W$ is drawn ind... | Why is regularization interpreted as a gaussian prior on my weights? | Since we're using MAP we are trying to maximize the probability of the parameters given the data.
$$
P(W|x,y) = \frac {P(x,y|W) P(W)} {P(x,y)}
$$
$P(x,y)$ can be ignored since it's fixed for our da | Why is regularization interpreted as a gaussian prior on my weights?
Since we're using MAP we are trying to maximize the probability of the parameters given the data.
$$
P(W|x,y) = \frac {P(x,y|W) P(W)} {P(x,y)}
$$
$P(x,y)$ can be ignored since it's fixed for our data. So we are trying to maximize $log(P(x,y|W)) + l... | Why is regularization interpreted as a gaussian prior on my weights?
Since we're using MAP we are trying to maximize the probability of the parameters given the data.
$$
P(W|x,y) = \frac {P(x,y|W) P(W)} {P(x,y)}
$$
$P(x,y)$ can be ignored since it's fixed for our da |
48,523 | Improving Chebyshev-type bound for discrete uniform distribution | If you know that the the random variable is bounded then you know a lot about the random variable and the Chebyshev and Markov inequality will not be tight in general.
A better inequality is the Hoeffding's inequality (consider the general case and not the binomial version). This takes into account implicitly that you... | Improving Chebyshev-type bound for discrete uniform distribution | If you know that the the random variable is bounded then you know a lot about the random variable and the Chebyshev and Markov inequality will not be tight in general.
A better inequality is the Hoeff | Improving Chebyshev-type bound for discrete uniform distribution
If you know that the the random variable is bounded then you know a lot about the random variable and the Chebyshev and Markov inequality will not be tight in general.
A better inequality is the Hoeffding's inequality (consider the general case and not th... | Improving Chebyshev-type bound for discrete uniform distribution
If you know that the the random variable is bounded then you know a lot about the random variable and the Chebyshev and Markov inequality will not be tight in general.
A better inequality is the Hoeff |
48,524 | Expectation of gradients | $$ E_Q\left[\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)}\right] = \int\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)} Q_\phi(h|x) = \int{\nabla_\phi Q_\phi(h|x)}$$
Assuming that you can exchange the integral and the gradient operators (deep waters)
$$ = \nabla_\phi\int{ Q_\phi(h|x)} = \nabla_\phi E_Q[1] = \nabla_\phi 1 =0$... | Expectation of gradients | $$ E_Q\left[\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)}\right] = \int\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)} Q_\phi(h|x) = \int{\nabla_\phi Q_\phi(h|x)}$$
Assuming that you can exchange the integr | Expectation of gradients
$$ E_Q\left[\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)}\right] = \int\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)} Q_\phi(h|x) = \int{\nabla_\phi Q_\phi(h|x)}$$
Assuming that you can exchange the integral and the gradient operators (deep waters)
$$ = \nabla_\phi\int{ Q_\phi(h|x)} = \nabla_\phi E... | Expectation of gradients
$$ E_Q\left[\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)}\right] = \int\frac{\nabla_\phi Q_\phi(h|x)}{Q_\phi(h|x)} Q_\phi(h|x) = \int{\nabla_\phi Q_\phi(h|x)}$$
Assuming that you can exchange the integr |
48,525 | What is the proper way to use rfImpute? (Imputation by Random Forest in R) | I'm not entirely sure if this is an answer to your question, but maybe you'll find it useful.
Maybe the author of the randomForest package would disagree with me, but I feel like the rfImpute() function is mostly used or called upon other imputation packages in their algorithms to impute many variables. If you only hav... | What is the proper way to use rfImpute? (Imputation by Random Forest in R) | I'm not entirely sure if this is an answer to your question, but maybe you'll find it useful.
Maybe the author of the randomForest package would disagree with me, but I feel like the rfImpute() functi | What is the proper way to use rfImpute? (Imputation by Random Forest in R)
I'm not entirely sure if this is an answer to your question, but maybe you'll find it useful.
Maybe the author of the randomForest package would disagree with me, but I feel like the rfImpute() function is mostly used or called upon other imputa... | What is the proper way to use rfImpute? (Imputation by Random Forest in R)
I'm not entirely sure if this is an answer to your question, but maybe you'll find it useful.
Maybe the author of the randomForest package would disagree with me, but I feel like the rfImpute() functi |
48,526 | Combining ARIMA model with regression | Yes. You can either use an ARIMAX model, or a regression with ARIMA errors. Rob Hyndman explains the difference in his blog post "The ARIMAX model muddle". In R, you can use the forecast package to fit regressions with ARIMA errors, or the TSA package to fit an ARIMAX model. | Combining ARIMA model with regression | Yes. You can either use an ARIMAX model, or a regression with ARIMA errors. Rob Hyndman explains the difference in his blog post "The ARIMAX model muddle". In R, you can use the forecast package to fi | Combining ARIMA model with regression
Yes. You can either use an ARIMAX model, or a regression with ARIMA errors. Rob Hyndman explains the difference in his blog post "The ARIMAX model muddle". In R, you can use the forecast package to fit regressions with ARIMA errors, or the TSA package to fit an ARIMAX model. | Combining ARIMA model with regression
Yes. You can either use an ARIMAX model, or a regression with ARIMA errors. Rob Hyndman explains the difference in his blog post "The ARIMAX model muddle". In R, you can use the forecast package to fi |
48,527 | Finding pdf of transformed variable for uniform distribution | One way to check whether you are right, or the webpage is right is to see if the pdfs integrate to 1. $X$ is defined between 0 and 1, so $Y = X^3$ is also defined between 0 and 1.
$$\int_0^1 \dfrac{1}{3} \dfrac{1}{y^2}dy = \dfrac{1}{3}\left[-\dfrac{1}{y} \right]^1_0 = \text{does not converge}. $$
So clearly the webpag... | Finding pdf of transformed variable for uniform distribution | One way to check whether you are right, or the webpage is right is to see if the pdfs integrate to 1. $X$ is defined between 0 and 1, so $Y = X^3$ is also defined between 0 and 1.
$$\int_0^1 \dfrac{1} | Finding pdf of transformed variable for uniform distribution
One way to check whether you are right, or the webpage is right is to see if the pdfs integrate to 1. $X$ is defined between 0 and 1, so $Y = X^3$ is also defined between 0 and 1.
$$\int_0^1 \dfrac{1}{3} \dfrac{1}{y^2}dy = \dfrac{1}{3}\left[-\dfrac{1}{y} \ri... | Finding pdf of transformed variable for uniform distribution
One way to check whether you are right, or the webpage is right is to see if the pdfs integrate to 1. $X$ is defined between 0 and 1, so $Y = X^3$ is also defined between 0 and 1.
$$\int_0^1 \dfrac{1} |
48,528 | Stacking: Do more base classifiers always improve accuracy? | As with any classifier, adding new input features can improve classification accuracy when when the new features contain new information about the labels. This performance improvement isn't guaranteed because classifiers are imperfect and may not be able to exploit the information. If the new features share information... | Stacking: Do more base classifiers always improve accuracy? | As with any classifier, adding new input features can improve classification accuracy when when the new features contain new information about the labels. This performance improvement isn't guaranteed | Stacking: Do more base classifiers always improve accuracy?
As with any classifier, adding new input features can improve classification accuracy when when the new features contain new information about the labels. This performance improvement isn't guaranteed because classifiers are imperfect and may not be able to ex... | Stacking: Do more base classifiers always improve accuracy?
As with any classifier, adding new input features can improve classification accuracy when when the new features contain new information about the labels. This performance improvement isn't guaranteed |
48,529 | Generalised Boosted Models (GBM) Assumptions | After contacting the author of the paper directly, I can answer the question myself. Assumptions:
1) Independence of observations
2) Assumptions related to the interaction depth. If set to $1$, strictly additive model is assumed. As we increase the interaction depth, this assumption is relaxed. | Generalised Boosted Models (GBM) Assumptions | After contacting the author of the paper directly, I can answer the question myself. Assumptions:
1) Independence of observations
2) Assumptions related to the interaction depth. If set to $1$, stric | Generalised Boosted Models (GBM) Assumptions
After contacting the author of the paper directly, I can answer the question myself. Assumptions:
1) Independence of observations
2) Assumptions related to the interaction depth. If set to $1$, strictly additive model is assumed. As we increase the interaction depth, this a... | Generalised Boosted Models (GBM) Assumptions
After contacting the author of the paper directly, I can answer the question myself. Assumptions:
1) Independence of observations
2) Assumptions related to the interaction depth. If set to $1$, stric |
48,530 | Why do normal-pseudo residuals measure the deviation from the median? | Background
In this paper, $Y$ is a random variable with continuous distribution function $F(y)=\Pr(Y \le y)$. One way to measure how extreme a small value of $Y$ may be is to report the "probability of observing an equal or more extremely (small) value under the model [$F$]": in other words, when $F(y)$ is close to $0... | Why do normal-pseudo residuals measure the deviation from the median? | Background
In this paper, $Y$ is a random variable with continuous distribution function $F(y)=\Pr(Y \le y)$. One way to measure how extreme a small value of $Y$ may be is to report the "probability | Why do normal-pseudo residuals measure the deviation from the median?
Background
In this paper, $Y$ is a random variable with continuous distribution function $F(y)=\Pr(Y \le y)$. One way to measure how extreme a small value of $Y$ may be is to report the "probability of observing an equal or more extremely (small) va... | Why do normal-pseudo residuals measure the deviation from the median?
Background
In this paper, $Y$ is a random variable with continuous distribution function $F(y)=\Pr(Y \le y)$. One way to measure how extreme a small value of $Y$ may be is to report the "probability |
48,531 | Why must kernel functions be scalar products [duplicate] | AFAIK the kernel trick is only applied when the data only appear in the form of scalar products like $x_1'x_2$. For many problems we need the dual representation in such forms so that the kernel trick can be applied.
If the kernel can be written as scalar products in some feature space $k(x_1, x_2)=\phi(x_1)'\phi(x_2)$... | Why must kernel functions be scalar products [duplicate] | AFAIK the kernel trick is only applied when the data only appear in the form of scalar products like $x_1'x_2$. For many problems we need the dual representation in such forms so that the kernel trick | Why must kernel functions be scalar products [duplicate]
AFAIK the kernel trick is only applied when the data only appear in the form of scalar products like $x_1'x_2$. For many problems we need the dual representation in such forms so that the kernel trick can be applied.
If the kernel can be written as scalar product... | Why must kernel functions be scalar products [duplicate]
AFAIK the kernel trick is only applied when the data only appear in the form of scalar products like $x_1'x_2$. For many problems we need the dual representation in such forms so that the kernel trick |
48,532 | What are some of the best approaches for variable selection in Poisson regression? | You can use the lasso or elastic net regularisation. Both are available in glmnet, if you are an R user, with a poisson dependent variable, using the family=poisson option. Hopefully you have sufficient observations to be able to split the dataset and do cross validation.
Stepwise selection methods are generally best a... | What are some of the best approaches for variable selection in Poisson regression? | You can use the lasso or elastic net regularisation. Both are available in glmnet, if you are an R user, with a poisson dependent variable, using the family=poisson option. Hopefully you have sufficie | What are some of the best approaches for variable selection in Poisson regression?
You can use the lasso or elastic net regularisation. Both are available in glmnet, if you are an R user, with a poisson dependent variable, using the family=poisson option. Hopefully you have sufficient observations to be able to split t... | What are some of the best approaches for variable selection in Poisson regression?
You can use the lasso or elastic net regularisation. Both are available in glmnet, if you are an R user, with a poisson dependent variable, using the family=poisson option. Hopefully you have sufficie |
48,533 | Distribution of Quotient of 2 dependent random variables | Although the brute force method I mentioned in the comments may work, there is an easier way which does not rely on Basu's theorem, and it also avoids integration of the joint density of transformed random variables. I leave some details out because the question is of homework-style.
Write
$$
Z_n = \frac{\frac{1}{n}U... | Distribution of Quotient of 2 dependent random variables | Although the brute force method I mentioned in the comments may work, there is an easier way which does not rely on Basu's theorem, and it also avoids integration of the joint density of transformed r | Distribution of Quotient of 2 dependent random variables
Although the brute force method I mentioned in the comments may work, there is an easier way which does not rely on Basu's theorem, and it also avoids integration of the joint density of transformed random variables. I leave some details out because the question ... | Distribution of Quotient of 2 dependent random variables
Although the brute force method I mentioned in the comments may work, there is an easier way which does not rely on Basu's theorem, and it also avoids integration of the joint density of transformed r |
48,534 | What happens if I flip targets and predictions in cross-entropy? | I'll rewrite the cross entropy in terms of distributions $q$ and $z$, so we can save your notation $t$ and $p$ for explicitly talking about classifiers. The cross entropy is:
$$H(q, z) = -\sum_{x} q(x) \log z(x)$$
The information theoretic meaning of the cross entropy is this: Say $q$ is a distribution that generates s... | What happens if I flip targets and predictions in cross-entropy? | I'll rewrite the cross entropy in terms of distributions $q$ and $z$, so we can save your notation $t$ and $p$ for explicitly talking about classifiers. The cross entropy is:
$$H(q, z) = -\sum_{x} q(x | What happens if I flip targets and predictions in cross-entropy?
I'll rewrite the cross entropy in terms of distributions $q$ and $z$, so we can save your notation $t$ and $p$ for explicitly talking about classifiers. The cross entropy is:
$$H(q, z) = -\sum_{x} q(x) \log z(x)$$
The information theoretic meaning of the ... | What happens if I flip targets and predictions in cross-entropy?
I'll rewrite the cross entropy in terms of distributions $q$ and $z$, so we can save your notation $t$ and $p$ for explicitly talking about classifiers. The cross entropy is:
$$H(q, z) = -\sum_{x} q(x |
48,535 | Expectation with indicator function | The original random variable $X_{t+1}$ is normally distributed. Call its distribution $P_{X_{t+1}}$
Define a function
$$g(v) = v \cdot 1_{\{v > z_t\}}$$
where $1_{\{\cdot\}}$ is the indicator function. This can also be written as:
$$g(v) = \left \{
\begin{array}{cl}
v & v > z_t \\
0 & \text{Otherwi... | Expectation with indicator function | The original random variable $X_{t+1}$ is normally distributed. Call its distribution $P_{X_{t+1}}$
Define a function
$$g(v) = v \cdot 1_{\{v > z_t\}}$$
where $1_{\{\cdot\}}$ is the indicator functio | Expectation with indicator function
The original random variable $X_{t+1}$ is normally distributed. Call its distribution $P_{X_{t+1}}$
Define a function
$$g(v) = v \cdot 1_{\{v > z_t\}}$$
where $1_{\{\cdot\}}$ is the indicator function. This can also be written as:
$$g(v) = \left \{
\begin{array}{cl}
v & ... | Expectation with indicator function
The original random variable $X_{t+1}$ is normally distributed. Call its distribution $P_{X_{t+1}}$
Define a function
$$g(v) = v \cdot 1_{\{v > z_t\}}$$
where $1_{\{\cdot\}}$ is the indicator functio |
48,536 | Scikit-learn's SGDClassifier code question | As you said, with $L_2$ regularization, the update is:
\begin{equation}
w \leftarrow w - \eta_t\lambda w - \eta_t y_t x_t 1_{\{...\}}
\end{equation}
The first two terms corresponds to a simple scaling. It is updated with the line:
w.scale(max(0, 1.0 - ((1.0 - l1_ratio) * eta * alpha)))
where the weight w are not really... | Scikit-learn's SGDClassifier code question | As you said, with $L_2$ regularization, the update is:
\begin{equation}
w \leftarrow w - \eta_t\lambda w - \eta_t y_t x_t 1_{\{...\}}
\end{equation}
The first two terms corresponds to a simple scaling | Scikit-learn's SGDClassifier code question
As you said, with $L_2$ regularization, the update is:
\begin{equation}
w \leftarrow w - \eta_t\lambda w - \eta_t y_t x_t 1_{\{...\}}
\end{equation}
The first two terms corresponds to a simple scaling. It is updated with the line:
w.scale(max(0, 1.0 - ((1.0 - l1_ratio) * eta *... | Scikit-learn's SGDClassifier code question
As you said, with $L_2$ regularization, the update is:
\begin{equation}
w \leftarrow w - \eta_t\lambda w - \eta_t y_t x_t 1_{\{...\}}
\end{equation}
The first two terms corresponds to a simple scaling |
48,537 | Scikit-learn's SGDClassifier code question | While i can't give you a complete analysis, i'm pretty sure, that these differences are due to the fact, that the implementation is targeting sparsity in the data.
Have a look at the Bottou paper you linked, especially part 5.1! There is a special treatment mentioned for sparse input and the update rule is splitted int... | Scikit-learn's SGDClassifier code question | While i can't give you a complete analysis, i'm pretty sure, that these differences are due to the fact, that the implementation is targeting sparsity in the data.
Have a look at the Bottou paper you | Scikit-learn's SGDClassifier code question
While i can't give you a complete analysis, i'm pretty sure, that these differences are due to the fact, that the implementation is targeting sparsity in the data.
Have a look at the Bottou paper you linked, especially part 5.1! There is a special treatment mentioned for spars... | Scikit-learn's SGDClassifier code question
While i can't give you a complete analysis, i'm pretty sure, that these differences are due to the fact, that the implementation is targeting sparsity in the data.
Have a look at the Bottou paper you |
48,538 | Correlation of order statistics from Uniform parent | You appear to understand the steps involved. I am not sure if this is an assignment or exercise, so it may not be appropriate to show workings anyway, but am happy to sketch out the approach ...
Given: $X \sim \text{Uniform}(0,1)$ with pdf $f(x)$:
Then, the joint pdf of the $r^{\text{th}}$ and $s^{\text{th}}$ order st... | Correlation of order statistics from Uniform parent | You appear to understand the steps involved. I am not sure if this is an assignment or exercise, so it may not be appropriate to show workings anyway, but am happy to sketch out the approach ...
Given | Correlation of order statistics from Uniform parent
You appear to understand the steps involved. I am not sure if this is an assignment or exercise, so it may not be appropriate to show workings anyway, but am happy to sketch out the approach ...
Given: $X \sim \text{Uniform}(0,1)$ with pdf $f(x)$:
Then, the joint pdf... | Correlation of order statistics from Uniform parent
You appear to understand the steps involved. I am not sure if this is an assignment or exercise, so it may not be appropriate to show workings anyway, but am happy to sketch out the approach ...
Given |
48,539 | Testing for publication bias in meta-analysis when effect is raw mean | The "standard" Egger regression test uses $\sqrt{v_i} = SE_i$ (i.e., the standard error of the outcomes) as the predictor. This is sometimes not advisable, especially when the standard error is a function of the outcome measure itself. For example, suppose you are not meta-analyzing means, but raw (Pearson product-mome... | Testing for publication bias in meta-analysis when effect is raw mean | The "standard" Egger regression test uses $\sqrt{v_i} = SE_i$ (i.e., the standard error of the outcomes) as the predictor. This is sometimes not advisable, especially when the standard error is a func | Testing for publication bias in meta-analysis when effect is raw mean
The "standard" Egger regression test uses $\sqrt{v_i} = SE_i$ (i.e., the standard error of the outcomes) as the predictor. This is sometimes not advisable, especially when the standard error is a function of the outcome measure itself. For example, s... | Testing for publication bias in meta-analysis when effect is raw mean
The "standard" Egger regression test uses $\sqrt{v_i} = SE_i$ (i.e., the standard error of the outcomes) as the predictor. This is sometimes not advisable, especially when the standard error is a func |
48,540 | monte carlo simulation using exponential distributions | The whole point of simulation is to show you that such variation is realistic. In fact, there appears to be nothing wrong with your results--except that you haven't yet done enough simulations to appreciate what they are telling you. In this case, the results are especially erratic because the starting population is ... | monte carlo simulation using exponential distributions | The whole point of simulation is to show you that such variation is realistic. In fact, there appears to be nothing wrong with your results--except that you haven't yet done enough simulations to app | monte carlo simulation using exponential distributions
The whole point of simulation is to show you that such variation is realistic. In fact, there appears to be nothing wrong with your results--except that you haven't yet done enough simulations to appreciate what they are telling you. In this case, the results are... | monte carlo simulation using exponential distributions
The whole point of simulation is to show you that such variation is realistic. In fact, there appears to be nothing wrong with your results--except that you haven't yet done enough simulations to app |
48,541 | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log odds ratios from between-subject designs? | I'll focus in my answer purely on the question on how to compute a (log) OR based on a within-subjects design that is comparable to that from a between-subjects design.
Suppose you have a within-subjects design with these data:
condition2
decision1 decision2 ... | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log | I'll focus in my answer purely on the question on how to compute a (log) OR based on a within-subjects design that is comparable to that from a between-subjects design.
Suppose you have a within-subje | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log odds ratios from between-subject designs?
I'll focus in my answer purely on the question on how to compute a (log) OR based on a within-subjects design that is comparable to that from a between-subjects design.
Suppose y... | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log
I'll focus in my answer purely on the question on how to compute a (log) OR based on a within-subjects design that is comparable to that from a between-subjects design.
Suppose you have a within-subje |
48,542 | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log odds ratios from between-subject designs? | You have the individual participant data so you have two choices (a) a one step approach which since you use R can be implemented in lme4 or (b) a two-step approach where you reduce each study to a single summary statistic and then meta-analyse them using, as you suggest metafor. If you do the single step approach you ... | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log | You have the individual participant data so you have two choices (a) a one step approach which since you use R can be implemented in lme4 or (b) a two-step approach where you reduce each study to a si | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log odds ratios from between-subject designs?
You have the individual participant data so you have two choices (a) a one step approach which since you use R can be implemented in lme4 or (b) a two-step approach where you red... | How can I compute a log odds ratio for a within-subject design that could be meta-analyzed with log
You have the individual participant data so you have two choices (a) a one step approach which since you use R can be implemented in lme4 or (b) a two-step approach where you reduce each study to a si |
48,543 | Linear Regression for a discrete count dependent variable? | By "GLM", I assume you mean the General Linear Model, which generalizes multiple regression and ANOVA. For what it's worth, many people (and I) just call that 'multiple regression'; I typically reserve "GLM" for generalized linear model. You are right that the general linear model requires normality. For the sake of... | Linear Regression for a discrete count dependent variable? | By "GLM", I assume you mean the General Linear Model, which generalizes multiple regression and ANOVA. For what it's worth, many people (and I) just call that 'multiple regression'; I typically reser | Linear Regression for a discrete count dependent variable?
By "GLM", I assume you mean the General Linear Model, which generalizes multiple regression and ANOVA. For what it's worth, many people (and I) just call that 'multiple regression'; I typically reserve "GLM" for generalized linear model. You are right that th... | Linear Regression for a discrete count dependent variable?
By "GLM", I assume you mean the General Linear Model, which generalizes multiple regression and ANOVA. For what it's worth, many people (and I) just call that 'multiple regression'; I typically reser |
48,544 | Understanding Kernel Functions for SVMs | By solving the optimization problem of SVM in its dual form, it turns out that the dependency of the problem on the training data $\{x_i\}_{i=1}^n$ is only through their inner products. That is, you only need $\{x_i^\top x_j\}_{i, j=1}^n$ i.e., inner products of all pairs of points you have. So to train an SVM, you onl... | Understanding Kernel Functions for SVMs | By solving the optimization problem of SVM in its dual form, it turns out that the dependency of the problem on the training data $\{x_i\}_{i=1}^n$ is only through their inner products. That is, you o | Understanding Kernel Functions for SVMs
By solving the optimization problem of SVM in its dual form, it turns out that the dependency of the problem on the training data $\{x_i\}_{i=1}^n$ is only through their inner products. That is, you only need $\{x_i^\top x_j\}_{i, j=1}^n$ i.e., inner products of all pairs of poin... | Understanding Kernel Functions for SVMs
By solving the optimization problem of SVM in its dual form, it turns out that the dependency of the problem on the training data $\{x_i\}_{i=1}^n$ is only through their inner products. That is, you o |
48,545 | Understanding Kernel Functions for SVMs | First, the radial basis function (RBF) is given by $k\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$, with
$$
k(x,y)=\exp(-\gamma\lVert x-y \rVert^2),
$$
where $\gamma$ is a positive parameter.
What is really useful in SVMs is the so-called "kernel trick". Briefly, you don't need to explicitly know the mapping from the o... | Understanding Kernel Functions for SVMs | First, the radial basis function (RBF) is given by $k\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$, with
$$
k(x,y)=\exp(-\gamma\lVert x-y \rVert^2),
$$
where $\gamma$ is a positive parameter.
What is | Understanding Kernel Functions for SVMs
First, the radial basis function (RBF) is given by $k\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$, with
$$
k(x,y)=\exp(-\gamma\lVert x-y \rVert^2),
$$
where $\gamma$ is a positive parameter.
What is really useful in SVMs is the so-called "kernel trick". Briefly, you don't need t... | Understanding Kernel Functions for SVMs
First, the radial basis function (RBF) is given by $k\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$, with
$$
k(x,y)=\exp(-\gamma\lVert x-y \rVert^2),
$$
where $\gamma$ is a positive parameter.
What is |
48,546 | Why is it that xgb.cv performs well but xgb.train does not | I just lost a couple of days on perhaps the same issue. TL;DR: are you sure your watchlist has the same number and order of columns as your ce.dmatrix?
In the current implementation of xgb.cv, any watchlist argument that gets passed in is going to get ignored. xgb.cv ends up calling xgb.cv.mknfold, which forcibly set... | Why is it that xgb.cv performs well but xgb.train does not | I just lost a couple of days on perhaps the same issue. TL;DR: are you sure your watchlist has the same number and order of columns as your ce.dmatrix?
In the current implementation of xgb.cv, any wa | Why is it that xgb.cv performs well but xgb.train does not
I just lost a couple of days on perhaps the same issue. TL;DR: are you sure your watchlist has the same number and order of columns as your ce.dmatrix?
In the current implementation of xgb.cv, any watchlist argument that gets passed in is going to get ignored.... | Why is it that xgb.cv performs well but xgb.train does not
I just lost a couple of days on perhaps the same issue. TL;DR: are you sure your watchlist has the same number and order of columns as your ce.dmatrix?
In the current implementation of xgb.cv, any wa |
48,547 | Why is it that xgb.cv performs well but xgb.train does not | Documentation is bit nebulous to me, but whole point of cross validation is to select the best hyper parameters to avoid overfitting. So xgb.cv is using cross validation to tune the parameters before testing, therefore avoiding overfitting. | Why is it that xgb.cv performs well but xgb.train does not | Documentation is bit nebulous to me, but whole point of cross validation is to select the best hyper parameters to avoid overfitting. So xgb.cv is using cross validation to tune the parameters before | Why is it that xgb.cv performs well but xgb.train does not
Documentation is bit nebulous to me, but whole point of cross validation is to select the best hyper parameters to avoid overfitting. So xgb.cv is using cross validation to tune the parameters before testing, therefore avoiding overfitting. | Why is it that xgb.cv performs well but xgb.train does not
Documentation is bit nebulous to me, but whole point of cross validation is to select the best hyper parameters to avoid overfitting. So xgb.cv is using cross validation to tune the parameters before |
48,548 | Why is it that xgb.cv performs well but xgb.train does not | Why would you need a watchlist in the CV method? The respective CV folds ARE the watchlist! I don't know the R command, but in Python verbose_eval=True returns the proper output that you are looking for. My guess is that since CV is only used for hyperparameter tuning and doesn't return a model by itself, the usage of ... | Why is it that xgb.cv performs well but xgb.train does not | Why would you need a watchlist in the CV method? The respective CV folds ARE the watchlist! I don't know the R command, but in Python verbose_eval=True returns the proper output that you are looking f | Why is it that xgb.cv performs well but xgb.train does not
Why would you need a watchlist in the CV method? The respective CV folds ARE the watchlist! I don't know the R command, but in Python verbose_eval=True returns the proper output that you are looking for. My guess is that since CV is only used for hyperparameter... | Why is it that xgb.cv performs well but xgb.train does not
Why would you need a watchlist in the CV method? The respective CV folds ARE the watchlist! I don't know the R command, but in Python verbose_eval=True returns the proper output that you are looking f |
48,549 | Nesting terminology in mixed models | I would describe the final model as simply having
a random intercept for every observed combination of A and B.
Although it looks like an interaction because you're just using R's interaction syntax, you're really just redefine a grouping variable in a slightly less structured way than in the previous specifications,... | Nesting terminology in mixed models | I would describe the final model as simply having
a random intercept for every observed combination of A and B.
Although it looks like an interaction because you're just using R's interaction syntax | Nesting terminology in mixed models
I would describe the final model as simply having
a random intercept for every observed combination of A and B.
Although it looks like an interaction because you're just using R's interaction syntax, you're really just redefine a grouping variable in a slightly less structured way ... | Nesting terminology in mixed models
I would describe the final model as simply having
a random intercept for every observed combination of A and B.
Although it looks like an interaction because you're just using R's interaction syntax |
48,550 | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms | Coding the targets as $\{0.1, 0.9\}$ is an example of label smoothing. It's one of the tricks highlighted in this review and it appears to originate in "Rethinking the Inception Architecture for Computer Vision" by Christian Szegedy et al. as a certain kind of regularization. It's used as a regularization strategy to d... | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms | Coding the targets as $\{0.1, 0.9\}$ is an example of label smoothing. It's one of the tricks highlighted in this review and it appears to originate in "Rethinking the Inception Architecture for Compu | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms
Coding the targets as $\{0.1, 0.9\}$ is an example of label smoothing. It's one of the tricks highlighted in this review and it appears to originate in "Rethinking the Inception Architecture for Computer Vision" by Christian Szeged... | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms
Coding the targets as $\{0.1, 0.9\}$ is an example of label smoothing. It's one of the tricks highlighted in this review and it appears to originate in "Rethinking the Inception Architecture for Compu |
48,551 | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms | I suspect the reasons they suggested this approach was that the magnitudes of the gradient descent steps in back-propagation are proportional to the derivative of the activation function. If you use a logistic activation function then the derivative is numerically zero before the output reaches 1 or 0. As a result the... | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms | I suspect the reasons they suggested this approach was that the magnitudes of the gradient descent steps in back-propagation are proportional to the derivative of the activation function. If you use a | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms
I suspect the reasons they suggested this approach was that the magnitudes of the gradient descent steps in back-propagation are proportional to the derivative of the activation function. If you use a logistic activation function t... | Targets of 0.1/0.9 instead of 0/1 in neural networks and other classification algorithms
I suspect the reasons they suggested this approach was that the magnitudes of the gradient descent steps in back-propagation are proportional to the derivative of the activation function. If you use a |
48,552 | Minimum "recommended" sample size for boxplots? Boxplots for different sample sizes | [I thought I had written an answer to the first question but I can't locate one.]
With 5 or fewer observations, you might as well just plot the actual points.
It doesn't matter when comparing across samples that the samples aren't the same size, but if you have much larger samples in some groups you should see more poi... | Minimum "recommended" sample size for boxplots? Boxplots for different sample sizes | [I thought I had written an answer to the first question but I can't locate one.]
With 5 or fewer observations, you might as well just plot the actual points.
It doesn't matter when comparing across s | Minimum "recommended" sample size for boxplots? Boxplots for different sample sizes
[I thought I had written an answer to the first question but I can't locate one.]
With 5 or fewer observations, you might as well just plot the actual points.
It doesn't matter when comparing across samples that the samples aren't the s... | Minimum "recommended" sample size for boxplots? Boxplots for different sample sizes
[I thought I had written an answer to the first question but I can't locate one.]
With 5 or fewer observations, you might as well just plot the actual points.
It doesn't matter when comparing across s |
48,553 | Covariance of two time series driven by a restricted VAR(1) model | One direction to go in might be something like:
$$
X_n = \rho_x X_{n-1} + \epsilon_n
$$
and
$$
Y_n = \rho_y Y_{n-1} + \rho_{xy}X_n +v_n
$$
Substitute for $X_n$ in the second equation:
$$Y_n = \rho_y Y_{n-1} + \rho_{xy} \rho_x X_{n-1} + \rho_{xy} \epsilon_n +v_n$$
Let $Z_n = \left[ \begin{array}{c} X_n \\ Y_n \end{arra... | Covariance of two time series driven by a restricted VAR(1) model | One direction to go in might be something like:
$$
X_n = \rho_x X_{n-1} + \epsilon_n
$$
and
$$
Y_n = \rho_y Y_{n-1} + \rho_{xy}X_n +v_n
$$
Substitute for $X_n$ in the second equation:
$$Y_n = \rho_y Y | Covariance of two time series driven by a restricted VAR(1) model
One direction to go in might be something like:
$$
X_n = \rho_x X_{n-1} + \epsilon_n
$$
and
$$
Y_n = \rho_y Y_{n-1} + \rho_{xy}X_n +v_n
$$
Substitute for $X_n$ in the second equation:
$$Y_n = \rho_y Y_{n-1} + \rho_{xy} \rho_x X_{n-1} + \rho_{xy} \epsilon... | Covariance of two time series driven by a restricted VAR(1) model
One direction to go in might be something like:
$$
X_n = \rho_x X_{n-1} + \epsilon_n
$$
and
$$
Y_n = \rho_y Y_{n-1} + \rho_{xy}X_n +v_n
$$
Substitute for $X_n$ in the second equation:
$$Y_n = \rho_y Y |
48,554 | Covariance of two time series driven by a restricted VAR(1) model | After taking Richard's hint into account, you can appeal to the general result of an autocovariance function of a stable $VAR(1)$
$$y_t=c+\Phi y_{t-1}+\epsilon_t,$$
with $\Omega$ the variance-covariance matrix of $\epsilon_t$:
$$
\Gamma_j=\sum_{i=0}^\infty\Phi^{j+i}\Omega(\Phi^{i})^\top
$$
This follows from wri... | Covariance of two time series driven by a restricted VAR(1) model | After taking Richard's hint into account, you can appeal to the general result of an autocovariance function of a stable $VAR(1)$
$$y_t=c+\Phi y_{t-1}+\epsilon_t,$$
with $\Omega$ the variance-covari | Covariance of two time series driven by a restricted VAR(1) model
After taking Richard's hint into account, you can appeal to the general result of an autocovariance function of a stable $VAR(1)$
$$y_t=c+\Phi y_{t-1}+\epsilon_t,$$
with $\Omega$ the variance-covariance matrix of $\epsilon_t$:
$$
\Gamma_j=\sum_{i=... | Covariance of two time series driven by a restricted VAR(1) model
After taking Richard's hint into account, you can appeal to the general result of an autocovariance function of a stable $VAR(1)$
$$y_t=c+\Phi y_{t-1}+\epsilon_t,$$
with $\Omega$ the variance-covari |
48,555 | Bias of Tibshirani's Lasso estimator | Suppose that we have a rank deficient least squares problem
$\min \| X\beta -y \|_{2}$
where $\alpha$ is a nonzero vector in the null space of $X$. That is,
$X\alpha=0$.
The lasso estimator can be formulated either as a constrained least squares problem or as unconstrained problem with an objective that includes t... | Bias of Tibshirani's Lasso estimator | Suppose that we have a rank deficient least squares problem
$\min \| X\beta -y \|_{2}$
where $\alpha$ is a nonzero vector in the null space of $X$. That is,
$X\alpha=0$.
The lasso estimator can b | Bias of Tibshirani's Lasso estimator
Suppose that we have a rank deficient least squares problem
$\min \| X\beta -y \|_{2}$
where $\alpha$ is a nonzero vector in the null space of $X$. That is,
$X\alpha=0$.
The lasso estimator can be formulated either as a constrained least squares problem or as unconstrained prob... | Bias of Tibshirani's Lasso estimator
Suppose that we have a rank deficient least squares problem
$\min \| X\beta -y \|_{2}$
where $\alpha$ is a nonzero vector in the null space of $X$. That is,
$X\alpha=0$.
The lasso estimator can b |
48,556 | Standard reference for K-means [duplicate] | According to wikipedia, the term k-means was first introduced in the reference you refer to. The usual reference in the computer vision community for the algorithm, which solves the k-means problem, is:
Lloyd, Stuart P. "Least squares quantization in PCM." Information Theory, IEEE Transactions on 28.2 (1982): 129-137... | Standard reference for K-means [duplicate] | According to wikipedia, the term k-means was first introduced in the reference you refer to. The usual reference in the computer vision community for the algorithm, which solves the k-means problem, i | Standard reference for K-means [duplicate]
According to wikipedia, the term k-means was first introduced in the reference you refer to. The usual reference in the computer vision community for the algorithm, which solves the k-means problem, is:
Lloyd, Stuart P. "Least squares quantization in PCM." Information Theory... | Standard reference for K-means [duplicate]
According to wikipedia, the term k-means was first introduced in the reference you refer to. The usual reference in the computer vision community for the algorithm, which solves the k-means problem, i |
48,557 | Distribution of differences in beta-distribution | If you look at the joint density of $(X_{(n-1)},X_{(n)})$ the two largest observation, their joint density is given by
$$g_n(x,y)=\dfrac{n!}{(n-2)!0!0!}F(x)^{n-2}[F(y)-F(x)]^0[1-F(y)]^0f(x)f(y)\mathbb{I}_{x\le y}$$
(where the $0$'s are considering the special case $j=n-1$, $k=n$ in the generic formula).
From there the... | Distribution of differences in beta-distribution | If you look at the joint density of $(X_{(n-1)},X_{(n)})$ the two largest observation, their joint density is given by
$$g_n(x,y)=\dfrac{n!}{(n-2)!0!0!}F(x)^{n-2}[F(y)-F(x)]^0[1-F(y)]^0f(x)f(y)\mathb | Distribution of differences in beta-distribution
If you look at the joint density of $(X_{(n-1)},X_{(n)})$ the two largest observation, their joint density is given by
$$g_n(x,y)=\dfrac{n!}{(n-2)!0!0!}F(x)^{n-2}[F(y)-F(x)]^0[1-F(y)]^0f(x)f(y)\mathbb{I}_{x\le y}$$
(where the $0$'s are considering the special case $j=n-... | Distribution of differences in beta-distribution
If you look at the joint density of $(X_{(n-1)},X_{(n)})$ the two largest observation, their joint density is given by
$$g_n(x,y)=\dfrac{n!}{(n-2)!0!0!}F(x)^{n-2}[F(y)-F(x)]^0[1-F(y)]^0f(x)f(y)\mathb |
48,558 | Conditional Expectation of sum of uniform random variables? | It is perhaps easier to see what is going on with a diagram
So your calculation will be $$\dfrac{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} x \,dy \, dx}{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} 1 \,dy \, dx} = \dfrac{23}{30} \approx 0.76666667$$ | Conditional Expectation of sum of uniform random variables? | It is perhaps easier to see what is going on with a diagram
So your calculation will be $$\dfrac{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} x \,dy \, dx}{\displaystyle \int_{x=0.3}^{1}\int_{y= | Conditional Expectation of sum of uniform random variables?
It is perhaps easier to see what is going on with a diagram
So your calculation will be $$\dfrac{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} x \,dy \, dx}{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} 1 \,dy \, dx} = \dfrac{23}{30} \approx 0.76666667... | Conditional Expectation of sum of uniform random variables?
It is perhaps easier to see what is going on with a diagram
So your calculation will be $$\dfrac{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} x \,dy \, dx}{\displaystyle \int_{x=0.3}^{1}\int_{y= |
48,559 | Choosing IRLS over gradient descent in logistic regression | Yes, IRLS could be faster, as I said in my answer to your previous question. For example, if the log-likelihood is nearly quadratic (which it will usually be if you are able to start fairly close to the maximum and the sample size isn't very small), then it may converge in only a couple of steps. Note, in fact that on ... | Choosing IRLS over gradient descent in logistic regression | Yes, IRLS could be faster, as I said in my answer to your previous question. For example, if the log-likelihood is nearly quadratic (which it will usually be if you are able to start fairly close to t | Choosing IRLS over gradient descent in logistic regression
Yes, IRLS could be faster, as I said in my answer to your previous question. For example, if the log-likelihood is nearly quadratic (which it will usually be if you are able to start fairly close to the maximum and the sample size isn't very small), then it may... | Choosing IRLS over gradient descent in logistic regression
Yes, IRLS could be faster, as I said in my answer to your previous question. For example, if the log-likelihood is nearly quadratic (which it will usually be if you are able to start fairly close to t |
48,560 | Does Random Forest ever compare the splitting of one node to the slitting of a **different** node? | Random forests are based on greedy induction of decision trees, which means that the best attribute to split on and best cut-off is computed independently for each internal node in the tree. Thus, nodes are not directly compared.
The article you cited does not specify that a split might be induced on a smaller number o... | Does Random Forest ever compare the splitting of one node to the slitting of a **different** node? | Random forests are based on greedy induction of decision trees, which means that the best attribute to split on and best cut-off is computed independently for each internal node in the tree. Thus, nod | Does Random Forest ever compare the splitting of one node to the slitting of a **different** node?
Random forests are based on greedy induction of decision trees, which means that the best attribute to split on and best cut-off is computed independently for each internal node in the tree. Thus, nodes are not directly c... | Does Random Forest ever compare the splitting of one node to the slitting of a **different** node?
Random forests are based on greedy induction of decision trees, which means that the best attribute to split on and best cut-off is computed independently for each internal node in the tree. Thus, nod |
48,561 | The distribution of $\bf{x}$ given an underdetermined system $A{\bf x}={\bf b}\sim N(0,\sigma^2 I)$ | Consider a simple case where $A = \left[\begin{array}{cc}1& 1\end{array} \right] $ and $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
Let $x_1$, $x_2$, and $b$ be random variables. We know from $A {\bf x} = {b}$ that $x_1 + x_2 = b$. We also know that $b$ is a normally distributed random variable.
Must $x_1... | The distribution of $\bf{x}$ given an underdetermined system $A{\bf x}={\bf b}\sim N(0,\sigma^2 I)$ | Consider a simple case where $A = \left[\begin{array}{cc}1& 1\end{array} \right] $ and $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
Let $x_1$, $x_2$, and $b$ be random variables. We know f | The distribution of $\bf{x}$ given an underdetermined system $A{\bf x}={\bf b}\sim N(0,\sigma^2 I)$
Consider a simple case where $A = \left[\begin{array}{cc}1& 1\end{array} \right] $ and $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
Let $x_1$, $x_2$, and $b$ be random variables. We know from $A {\bf x} = {b}... | The distribution of $\bf{x}$ given an underdetermined system $A{\bf x}={\bf b}\sim N(0,\sigma^2 I)$
Consider a simple case where $A = \left[\begin{array}{cc}1& 1\end{array} \right] $ and $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
Let $x_1$, $x_2$, and $b$ be random variables. We know f |
48,562 | What types of functions can be implemented in a layer of a Neural Networks? | There are plenty of functions that are hard to optimise. In fact when you look at the successful implementation, they are mainly image/signal processing using sigmoid or ReLU activation functions. (maybe you want to clarify what your target function is. I.e. do you mean the function you want to approximate or are you i... | What types of functions can be implemented in a layer of a Neural Networks? | There are plenty of functions that are hard to optimise. In fact when you look at the successful implementation, they are mainly image/signal processing using sigmoid or ReLU activation functions. (ma | What types of functions can be implemented in a layer of a Neural Networks?
There are plenty of functions that are hard to optimise. In fact when you look at the successful implementation, they are mainly image/signal processing using sigmoid or ReLU activation functions. (maybe you want to clarify what your target fun... | What types of functions can be implemented in a layer of a Neural Networks?
There are plenty of functions that are hard to optimise. In fact when you look at the successful implementation, they are mainly image/signal processing using sigmoid or ReLU activation functions. (ma |
48,563 | What types of functions can be implemented in a layer of a Neural Networks? | My question is: are there specific types of functions/layers that are
believed to be hard/impossible to train in neural nets and using back
propagation? Is it likely to get much better results using neural nets
if we were to use more sophisticated optimization methods?
With backpropagation alone you can't comput... | What types of functions can be implemented in a layer of a Neural Networks? | My question is: are there specific types of functions/layers that are
believed to be hard/impossible to train in neural nets and using back
propagation? Is it likely to get much better results usi | What types of functions can be implemented in a layer of a Neural Networks?
My question is: are there specific types of functions/layers that are
believed to be hard/impossible to train in neural nets and using back
propagation? Is it likely to get much better results using neural nets
if we were to use more soph... | What types of functions can be implemented in a layer of a Neural Networks?
My question is: are there specific types of functions/layers that are
believed to be hard/impossible to train in neural nets and using back
propagation? Is it likely to get much better results usi |
48,564 | How to compute partial log-likelihood function in Cox proportional hazards model? | This is technically a programming question with an easy programming answer. If you simply want the partial likelihood, why not fool R into giving it to you? Simply initialize beta and allow no iterations, then extract the loglik value from the coxph object. (see ?coxph.object).
For example:
## artificial data
library(s... | How to compute partial log-likelihood function in Cox proportional hazards model? | This is technically a programming question with an easy programming answer. If you simply want the partial likelihood, why not fool R into giving it to you? Simply initialize beta and allow no iterati | How to compute partial log-likelihood function in Cox proportional hazards model?
This is technically a programming question with an easy programming answer. If you simply want the partial likelihood, why not fool R into giving it to you? Simply initialize beta and allow no iterations, then extract the loglik value fro... | How to compute partial log-likelihood function in Cox proportional hazards model?
This is technically a programming question with an easy programming answer. If you simply want the partial likelihood, why not fool R into giving it to you? Simply initialize beta and allow no iterati |
48,565 | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image classification tasks | For image classification task, how can a stacked auto-encoder help an traditional Convolutional Neural Network?
As mentioned in the paper, we can use the pre-trained weights to initialize CNN layers, although that essentially doesn't add anything to the CNN, it normally helps setting a good starting point for training... | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image clas | For image classification task, how can a stacked auto-encoder help an traditional Convolutional Neural Network?
As mentioned in the paper, we can use the pre-trained weights to initialize CNN layers, | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image classification tasks
For image classification task, how can a stacked auto-encoder help an traditional Convolutional Neural Network?
As mentioned in the paper, we can use the pre-trained weights to initialize CNN layers, al... | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image clas
For image classification task, how can a stacked auto-encoder help an traditional Convolutional Neural Network?
As mentioned in the paper, we can use the pre-trained weights to initialize CNN layers, |
48,566 | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image classification tasks | As dontloo mentions, an autoencoder can be used to initialize weights for a CNN and thus act as an additional layer before the CNN. In particular, the the hierarchical nature of a stacked autoencoder allows us to encode different types of (progressively more complex) features in each hidden layer, similarly to CNNs.
St... | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image clas | As dontloo mentions, an autoencoder can be used to initialize weights for a CNN and thus act as an additional layer before the CNN. In particular, the the hierarchical nature of a stacked autoencoder | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image classification tasks
As dontloo mentions, an autoencoder can be used to initialize weights for a CNN and thus act as an additional layer before the CNN. In particular, the the hierarchical nature of a stacked autoencoder all... | How does a Stacked AutoEncoder increases performance of a Convolutional Neural Network in image clas
As dontloo mentions, an autoencoder can be used to initialize weights for a CNN and thus act as an additional layer before the CNN. In particular, the the hierarchical nature of a stacked autoencoder |
48,567 | Non-normality of residuals in a negative binomial GLMM | Reginald, you have probably moved on by now, but for people that stumble across this post I would like to note that the DHARMa package (available from CRAN, see here) that I have created solves this problem, i.e. it will allow you to test if the residuals are compatible with the assumptions of nb (or any other distribu... | Non-normality of residuals in a negative binomial GLMM | Reginald, you have probably moved on by now, but for people that stumble across this post I would like to note that the DHARMa package (available from CRAN, see here) that I have created solves this p | Non-normality of residuals in a negative binomial GLMM
Reginald, you have probably moved on by now, but for people that stumble across this post I would like to note that the DHARMa package (available from CRAN, see here) that I have created solves this problem, i.e. it will allow you to test if the residuals are compa... | Non-normality of residuals in a negative binomial GLMM
Reginald, you have probably moved on by now, but for people that stumble across this post I would like to note that the DHARMa package (available from CRAN, see here) that I have created solves this p |
48,568 | Expectation of a chi-squared distribution | The PDFs are
$$f_U(u) = C(1)u^{-1/2} e^{-u/2}$$
and
$$f_V(v) = C(n)v^{n/2-1}e^{-v/2}$$
where
$$C(k) = \frac{1}{2^{k/2}\Gamma(\frac{k}{2})}$$
are the normalizing constants. Use polar-like coordinates $u=(r\cos(\theta))^2$ and $v=(r\sin(\theta))^2$ to evaluate the expectation, after first computing
$$\eqalign{du\wedge d... | Expectation of a chi-squared distribution | The PDFs are
$$f_U(u) = C(1)u^{-1/2} e^{-u/2}$$
and
$$f_V(v) = C(n)v^{n/2-1}e^{-v/2}$$
where
$$C(k) = \frac{1}{2^{k/2}\Gamma(\frac{k}{2})}$$
are the normalizing constants. Use polar-like coordinates | Expectation of a chi-squared distribution
The PDFs are
$$f_U(u) = C(1)u^{-1/2} e^{-u/2}$$
and
$$f_V(v) = C(n)v^{n/2-1}e^{-v/2}$$
where
$$C(k) = \frac{1}{2^{k/2}\Gamma(\frac{k}{2})}$$
are the normalizing constants. Use polar-like coordinates $u=(r\cos(\theta))^2$ and $v=(r\sin(\theta))^2$ to evaluate the expectation, a... | Expectation of a chi-squared distribution
The PDFs are
$$f_U(u) = C(1)u^{-1/2} e^{-u/2}$$
and
$$f_V(v) = C(n)v^{n/2-1}e^{-v/2}$$
where
$$C(k) = \frac{1}{2^{k/2}\Gamma(\frac{k}{2})}$$
are the normalizing constants. Use polar-like coordinates |
48,569 | How to show mean and standard deviation of Normal distribution? | Since only the terms involving $\mu$ are relevant, I will be dropping multiplicative terms not involving it without warning.
\begin{align*}
[\mu | x_1,\ldots,x_n,\sigma^2] &\propto [x_1,\ldots,x_n|\mu,\sigma^2] \times [\mu|\sigma^2]\\
&\propto
\prod_i \exp(-\frac{(x_i-\mu)^2}{2\sigma^2}) \times \exp(-\frac{(\mu-\beta... | How to show mean and standard deviation of Normal distribution? | Since only the terms involving $\mu$ are relevant, I will be dropping multiplicative terms not involving it without warning.
\begin{align*}
[\mu | x_1,\ldots,x_n,\sigma^2] &\propto [x_1,\ldots,x_n|\m | How to show mean and standard deviation of Normal distribution?
Since only the terms involving $\mu$ are relevant, I will be dropping multiplicative terms not involving it without warning.
\begin{align*}
[\mu | x_1,\ldots,x_n,\sigma^2] &\propto [x_1,\ldots,x_n|\mu,\sigma^2] \times [\mu|\sigma^2]\\
&\propto
\prod_i \e... | How to show mean and standard deviation of Normal distribution?
Since only the terms involving $\mu$ are relevant, I will be dropping multiplicative terms not involving it without warning.
\begin{align*}
[\mu | x_1,\ldots,x_n,\sigma^2] &\propto [x_1,\ldots,x_n|\m |
48,570 | How to decide on Theil-Sen outliers? | I assume that you are familiar with the notion of breakdown point of an estimator. A similar concept exists for outlier identification rules (see [3]).
(a) The breakdown point of the Theil-Sen estimator at 2D data is $1-\frac{1}{\sqrt{2}}$.
(b) Furthermore, the 2D Theil-Sen estimator is residual admissible (meaning ... | How to decide on Theil-Sen outliers? | I assume that you are familiar with the notion of breakdown point of an estimator. A similar concept exists for outlier identification rules (see [3]).
(a) The breakdown point of the Theil-Sen estim | How to decide on Theil-Sen outliers?
I assume that you are familiar with the notion of breakdown point of an estimator. A similar concept exists for outlier identification rules (see [3]).
(a) The breakdown point of the Theil-Sen estimator at 2D data is $1-\frac{1}{\sqrt{2}}$.
(b) Furthermore, the 2D Theil-Sen estim... | How to decide on Theil-Sen outliers?
I assume that you are familiar with the notion of breakdown point of an estimator. A similar concept exists for outlier identification rules (see [3]).
(a) The breakdown point of the Theil-Sen estim |
48,571 | Quantreg : Unbalanced residuals | You are conducting a median regression (tau=0.5) on asymmetrically distributed data. Here is a simpler example to show what is going on.
Suppose your asymmetric data are lognormal:
set.seed(1)
xx <- rlnorm(100,0,1)
Then what you are doing amounts to finding the median of your data.
median(xx)
[1] 1.121518
Now, the me... | Quantreg : Unbalanced residuals | You are conducting a median regression (tau=0.5) on asymmetrically distributed data. Here is a simpler example to show what is going on.
Suppose your asymmetric data are lognormal:
set.seed(1)
xx <- r | Quantreg : Unbalanced residuals
You are conducting a median regression (tau=0.5) on asymmetrically distributed data. Here is a simpler example to show what is going on.
Suppose your asymmetric data are lognormal:
set.seed(1)
xx <- rlnorm(100,0,1)
Then what you are doing amounts to finding the median of your data.
medi... | Quantreg : Unbalanced residuals
You are conducting a median regression (tau=0.5) on asymmetrically distributed data. Here is a simpler example to show what is going on.
Suppose your asymmetric data are lognormal:
set.seed(1)
xx <- r |
48,572 | ICC in a multi level model with two random effects | There are several ways to calculate and interpret the ICC for a mixed model. A useful thread is here. To calculate, it is the amount of variance from certain factor(s) divided by the total variance. That would be akin to your B and D calculations. This can be interpreted as the correlation of two randomly drawn uni... | ICC in a multi level model with two random effects | There are several ways to calculate and interpret the ICC for a mixed model. A useful thread is here. To calculate, it is the amount of variance from certain factor(s) divided by the total variance. | ICC in a multi level model with two random effects
There are several ways to calculate and interpret the ICC for a mixed model. A useful thread is here. To calculate, it is the amount of variance from certain factor(s) divided by the total variance. That would be akin to your B and D calculations. This can be inter... | ICC in a multi level model with two random effects
There are several ways to calculate and interpret the ICC for a mixed model. A useful thread is here. To calculate, it is the amount of variance from certain factor(s) divided by the total variance. |
48,573 | interpreting confidence intervals in t.test | .2532 is always going to be in the confidence interval since this is how the interval was constructed. The formula is
$(\bar{x_1} - \bar{x_2}) \pm t^*_{df} \times SE(\bar{x_1} - \bar{x_2})$
Further more $\bar{x_1} - \bar{x_2}$ is called your observed result (or sample difference in means) and as you can see this is w... | interpreting confidence intervals in t.test | .2532 is always going to be in the confidence interval since this is how the interval was constructed. The formula is
$(\bar{x_1} - \bar{x_2}) \pm t^*_{df} \times SE(\bar{x_1} - \bar{x_2})$
Further | interpreting confidence intervals in t.test
.2532 is always going to be in the confidence interval since this is how the interval was constructed. The formula is
$(\bar{x_1} - \bar{x_2}) \pm t^*_{df} \times SE(\bar{x_1} - \bar{x_2})$
Further more $\bar{x_1} - \bar{x_2}$ is called your observed result (or sample diffe... | interpreting confidence intervals in t.test
.2532 is always going to be in the confidence interval since this is how the interval was constructed. The formula is
$(\bar{x_1} - \bar{x_2}) \pm t^*_{df} \times SE(\bar{x_1} - \bar{x_2})$
Further |
48,574 | interpreting confidence intervals in t.test | Why do smart people have such trouble explaining simple things.
In laymen terms, p-value = 0.03236 means there is a 97% chance sample A > sample B.
You're 95% confident the difference between sample A and B is between 0.02 and 0.48. This is a weird result given the difference between A and B is only 0.25. This indicat... | interpreting confidence intervals in t.test | Why do smart people have such trouble explaining simple things.
In laymen terms, p-value = 0.03236 means there is a 97% chance sample A > sample B.
You're 95% confident the difference between sample | interpreting confidence intervals in t.test
Why do smart people have such trouble explaining simple things.
In laymen terms, p-value = 0.03236 means there is a 97% chance sample A > sample B.
You're 95% confident the difference between sample A and B is between 0.02 and 0.48. This is a weird result given the differenc... | interpreting confidence intervals in t.test
Why do smart people have such trouble explaining simple things.
In laymen terms, p-value = 0.03236 means there is a 97% chance sample A > sample B.
You're 95% confident the difference between sample |
48,575 | Books for statistical computing course? | Numerical Analysis for Statisticians, by Kenneth Lange, is a wonderful book for this purpose. It provides most of the necessary background in calculus and some algebra to conduct rigorous numerical analyses of statistical problems. This includes expansions, eigen-analysis, optimisation, integration, approximation theor... | Books for statistical computing course? | Numerical Analysis for Statisticians, by Kenneth Lange, is a wonderful book for this purpose. It provides most of the necessary background in calculus and some algebra to conduct rigorous numerical an | Books for statistical computing course?
Numerical Analysis for Statisticians, by Kenneth Lange, is a wonderful book for this purpose. It provides most of the necessary background in calculus and some algebra to conduct rigorous numerical analyses of statistical problems. This includes expansions, eigen-analysis, optimi... | Books for statistical computing course?
Numerical Analysis for Statisticians, by Kenneth Lange, is a wonderful book for this purpose. It provides most of the necessary background in calculus and some algebra to conduct rigorous numerical an |
48,576 | What is the difference between sample space and population? | The population is the set of all units a random process can pick. The sample space S is the set of all possible outcome of a random variable.
For example, the population can be the complete population of the US. Then your random process picks a person, John Smith.
If your random variable asks the color of hair of a p... | What is the difference between sample space and population? | The population is the set of all units a random process can pick. The sample space S is the set of all possible outcome of a random variable.
For example, the population can be the complete populatio | What is the difference between sample space and population?
The population is the set of all units a random process can pick. The sample space S is the set of all possible outcome of a random variable.
For example, the population can be the complete population of the US. Then your random process picks a person, John S... | What is the difference between sample space and population?
The population is the set of all units a random process can pick. The sample space S is the set of all possible outcome of a random variable.
For example, the population can be the complete populatio |
48,577 | What is the difference between sample space and population? | In both Probability and Statistics, we refer to the (probability-theoretic) experiment of drawing a SAMPLE,
size n, from a (statistical) POPULATION. The SAMPLE SPACE for this experiment is therefore the set of all n-element samples. If n = 1, then the sample space is the same as the population.
Now, I understand ... | What is the difference between sample space and population? | In both Probability and Statistics, we refer to the (probability-theoretic) experiment of drawing a SAMPLE,
size n, from a (statistical) POPULATION. The SAMPLE SPACE for this experiment is therefo | What is the difference between sample space and population?
In both Probability and Statistics, we refer to the (probability-theoretic) experiment of drawing a SAMPLE,
size n, from a (statistical) POPULATION. The SAMPLE SPACE for this experiment is therefore the set of all n-element samples. If n = 1, then the sam... | What is the difference between sample space and population?
In both Probability and Statistics, we refer to the (probability-theoretic) experiment of drawing a SAMPLE,
size n, from a (statistical) POPULATION. The SAMPLE SPACE for this experiment is therefo |
48,578 | Variance of the Poisson Binomial Distribution | Think of the case where $n=2$. If $p_1 = p_2 = 0.5$, this maximizes the variance of X. If $p_1 = 0$ and $p_2 = 1$, then $X=1$ and there is no variance. | Variance of the Poisson Binomial Distribution | Think of the case where $n=2$. If $p_1 = p_2 = 0.5$, this maximizes the variance of X. If $p_1 = 0$ and $p_2 = 1$, then $X=1$ and there is no variance. | Variance of the Poisson Binomial Distribution
Think of the case where $n=2$. If $p_1 = p_2 = 0.5$, this maximizes the variance of X. If $p_1 = 0$ and $p_2 = 1$, then $X=1$ and there is no variance. | Variance of the Poisson Binomial Distribution
Think of the case where $n=2$. If $p_1 = p_2 = 0.5$, this maximizes the variance of X. If $p_1 = 0$ and $p_2 = 1$, then $X=1$ and there is no variance. |
48,579 | Full rank assumption in the linear regression model explanation: | There is an error. It should be $\beta_4'=\beta_4-a$. If we substitute these new $\beta$s into the regression equation we get:
\begin{align}
C &= \beta_1 + \beta_2'X_{non-labor-income}+\beta_3'X_{non-labor-income}+\beta_4'X_{total-income}+\varepsilon \\
& = \beta_1 + (\beta_2 +a)X_{non-labor-income}+(\beta_3+a)X_{salar... | Full rank assumption in the linear regression model explanation: | There is an error. It should be $\beta_4'=\beta_4-a$. If we substitute these new $\beta$s into the regression equation we get:
\begin{align}
C &= \beta_1 + \beta_2'X_{non-labor-income}+\beta_3'X_{non- | Full rank assumption in the linear regression model explanation:
There is an error. It should be $\beta_4'=\beta_4-a$. If we substitute these new $\beta$s into the regression equation we get:
\begin{align}
C &= \beta_1 + \beta_2'X_{non-labor-income}+\beta_3'X_{non-labor-income}+\beta_4'X_{total-income}+\varepsilon \\
&... | Full rank assumption in the linear regression model explanation:
There is an error. It should be $\beta_4'=\beta_4-a$. If we substitute these new $\beta$s into the regression equation we get:
\begin{align}
C &= \beta_1 + \beta_2'X_{non-labor-income}+\beta_3'X_{non- |
48,580 | Full rank assumption in the linear regression model explanation: | Old post but since I've been wondering about the same question here's my take on it. I came up a with the following numerical example to help me understand why the design matrix $\textbf{X}$ must be full rank.
First, with a linear model we want to solve (leaving out the error term $\boldsymbol{\epsilon}$):
$$\textbf{y}... | Full rank assumption in the linear regression model explanation: | Old post but since I've been wondering about the same question here's my take on it. I came up a with the following numerical example to help me understand why the design matrix $\textbf{X}$ must be f | Full rank assumption in the linear regression model explanation:
Old post but since I've been wondering about the same question here's my take on it. I came up a with the following numerical example to help me understand why the design matrix $\textbf{X}$ must be full rank.
First, with a linear model we want to solve (... | Full rank assumption in the linear regression model explanation:
Old post but since I've been wondering about the same question here's my take on it. I came up a with the following numerical example to help me understand why the design matrix $\textbf{X}$ must be f |
48,581 | Does the sample size influence the number of PCs needed to explain a fixed percentage of variance? | You are probably in the $n\ll p$ situation. As @Glen_b wrote in the comment above, such a behaviour is expected.
Illustration and intuition
Let $p=1000$. Let the population be a multivariate normal with mean zero and some arbitrary covariance matrix $\boldsymbol\Sigma$. Let us sample $n$ points from this population and... | Does the sample size influence the number of PCs needed to explain a fixed percentage of variance? | You are probably in the $n\ll p$ situation. As @Glen_b wrote in the comment above, such a behaviour is expected.
Illustration and intuition
Let $p=1000$. Let the population be a multivariate normal wi | Does the sample size influence the number of PCs needed to explain a fixed percentage of variance?
You are probably in the $n\ll p$ situation. As @Glen_b wrote in the comment above, such a behaviour is expected.
Illustration and intuition
Let $p=1000$. Let the population be a multivariate normal with mean zero and some... | Does the sample size influence the number of PCs needed to explain a fixed percentage of variance?
You are probably in the $n\ll p$ situation. As @Glen_b wrote in the comment above, such a behaviour is expected.
Illustration and intuition
Let $p=1000$. Let the population be a multivariate normal wi |
48,582 | What is the meaning of the correlation in glm | You are actually mentioning the answer to your question in your question's body. The coefficients you see are actually estimated. This means coefficients themselves are actually random variables that follow a distribution. What you see is one value of the random variable. The calculated correlation is the correlation b... | What is the meaning of the correlation in glm | You are actually mentioning the answer to your question in your question's body. The coefficients you see are actually estimated. This means coefficients themselves are actually random variables that | What is the meaning of the correlation in glm
You are actually mentioning the answer to your question in your question's body. The coefficients you see are actually estimated. This means coefficients themselves are actually random variables that follow a distribution. What you see is one value of the random variable. T... | What is the meaning of the correlation in glm
You are actually mentioning the answer to your question in your question's body. The coefficients you see are actually estimated. This means coefficients themselves are actually random variables that |
48,583 | What is the meaning of the correlation in glm | The correlation of estimates is a scaling parameter so that the standard deviation of the estimated probabilities are constant regardless of linear transformations of the predictor variable. They have no substantive meaning. They only set the metric of the covariance. | What is the meaning of the correlation in glm | The correlation of estimates is a scaling parameter so that the standard deviation of the estimated probabilities are constant regardless of linear transformations of the predictor variable. They hav | What is the meaning of the correlation in glm
The correlation of estimates is a scaling parameter so that the standard deviation of the estimated probabilities are constant regardless of linear transformations of the predictor variable. They have no substantive meaning. They only set the metric of the covariance. | What is the meaning of the correlation in glm
The correlation of estimates is a scaling parameter so that the standard deviation of the estimated probabilities are constant regardless of linear transformations of the predictor variable. They hav |
48,584 | Method for a hypothesis testing non normal distribution number of retweets | You could use Mann-Withney U-test
In statistics, the Mann–Whitney U test (also called the
Mann–Whitney–Wilcoxon (MWW), Wilcoxon rank-sum test (WRS), or
Wilcoxon–Mann–Whitney test) is a nonparametric test of the null
hypothesis that two samples come from the same population against an
alternative hypothesis, es... | Method for a hypothesis testing non normal distribution number of retweets | You could use Mann-Withney U-test
In statistics, the Mann–Whitney U test (also called the
Mann–Whitney–Wilcoxon (MWW), Wilcoxon rank-sum test (WRS), or
Wilcoxon–Mann–Whitney test) is a nonparamet | Method for a hypothesis testing non normal distribution number of retweets
You could use Mann-Withney U-test
In statistics, the Mann–Whitney U test (also called the
Mann–Whitney–Wilcoxon (MWW), Wilcoxon rank-sum test (WRS), or
Wilcoxon–Mann–Whitney test) is a nonparametric test of the null
hypothesis that two sa... | Method for a hypothesis testing non normal distribution number of retweets
You could use Mann-Withney U-test
In statistics, the Mann–Whitney U test (also called the
Mann–Whitney–Wilcoxon (MWW), Wilcoxon rank-sum test (WRS), or
Wilcoxon–Mann–Whitney test) is a nonparamet |
48,585 | Nonrandomized better performance | The importance of randomized experiments is that we can get causal inference from them: by randomly assigning treatments to subjects, we know that the outcome should be approximately independent of any other covariates, other than treatment. Thus, the only systematic difference between treatment and control groups shou... | Nonrandomized better performance | The importance of randomized experiments is that we can get causal inference from them: by randomly assigning treatments to subjects, we know that the outcome should be approximately independent of an | Nonrandomized better performance
The importance of randomized experiments is that we can get causal inference from them: by randomly assigning treatments to subjects, we know that the outcome should be approximately independent of any other covariates, other than treatment. Thus, the only systematic difference between ... | Nonrandomized better performance
The importance of randomized experiments is that we can get causal inference from them: by randomly assigning treatments to subjects, we know that the outcome should be approximately independent of an |
48,586 | Nonrandomized better performance | I think (this interpretation of) Jaynes' statement is false.
In theory of computer science, for example there is lots of effort put into derandomizing randomized algorithms. One (not very CS-y) example is the Miller Rabin algorithm. If you'd like to dig deeper into this, computer science theory stackexchange is the p... | Nonrandomized better performance | I think (this interpretation of) Jaynes' statement is false.
In theory of computer science, for example there is lots of effort put into derandomizing randomized algorithms. One (not very CS-y) exam | Nonrandomized better performance
I think (this interpretation of) Jaynes' statement is false.
In theory of computer science, for example there is lots of effort put into derandomizing randomized algorithms. One (not very CS-y) example is the Miller Rabin algorithm. If you'd like to dig deeper into this, computer scie... | Nonrandomized better performance
I think (this interpretation of) Jaynes' statement is false.
In theory of computer science, for example there is lots of effort put into derandomizing randomized algorithms. One (not very CS-y) exam |
48,587 | Nonrandomized better performance | An example that I'm familiar with: Sometimes experimenters want to design an experiment to minimize the maximum response variance over the design space (G-optimal designs). Early work used genetic algorithms to find such designs. Later work used meta-models and a more traditional optimization approach to find design... | Nonrandomized better performance | An example that I'm familiar with: Sometimes experimenters want to design an experiment to minimize the maximum response variance over the design space (G-optimal designs). Early work used genetic a | Nonrandomized better performance
An example that I'm familiar with: Sometimes experimenters want to design an experiment to minimize the maximum response variance over the design space (G-optimal designs). Early work used genetic algorithms to find such designs. Later work used meta-models and a more traditional opt... | Nonrandomized better performance
An example that I'm familiar with: Sometimes experimenters want to design an experiment to minimize the maximum response variance over the design space (G-optimal designs). Early work used genetic a |
48,588 | How can I estimate the sliding window standard deviation of a stream? | You might be able to adapt a technique dating from the dark ages when people calculated standard deviations with hand-operated calculators, so they kept running tallies of both the sums of the observations and of the sums of the squares of the observations. Quoting from the Wikipedia page on mean squared error
the "co... | How can I estimate the sliding window standard deviation of a stream? | You might be able to adapt a technique dating from the dark ages when people calculated standard deviations with hand-operated calculators, so they kept running tallies of both the sums of the observa | How can I estimate the sliding window standard deviation of a stream?
You might be able to adapt a technique dating from the dark ages when people calculated standard deviations with hand-operated calculators, so they kept running tallies of both the sums of the observations and of the sums of the squares of the observ... | How can I estimate the sliding window standard deviation of a stream?
You might be able to adapt a technique dating from the dark ages when people calculated standard deviations with hand-operated calculators, so they kept running tallies of both the sums of the observa |
48,589 | How can I estimate the sliding window standard deviation of a stream? | I implemented the frugal streaming algorithm and made a small enhancement that improved the convergence and reduced the error. I am well satisfied with the results for quantiles: less than 5% error, 95% of the time. Using this to compute the first and third quartiles and other code that estimates the max and min, I the... | How can I estimate the sliding window standard deviation of a stream? | I implemented the frugal streaming algorithm and made a small enhancement that improved the convergence and reduced the error. I am well satisfied with the results for quantiles: less than 5% error, 9 | How can I estimate the sliding window standard deviation of a stream?
I implemented the frugal streaming algorithm and made a small enhancement that improved the convergence and reduced the error. I am well satisfied with the results for quantiles: less than 5% error, 95% of the time. Using this to compute the first an... | How can I estimate the sliding window standard deviation of a stream?
I implemented the frugal streaming algorithm and made a small enhancement that improved the convergence and reduced the error. I am well satisfied with the results for quantiles: less than 5% error, 9 |
48,590 | Which iterative algorithm lmer uses for REML estimation? | Neither of the above. The paper describing the algorithms used in lmer [Bates et al. J. Statistical Software (2015) 1-48, available via vignette("lmer",package="lme4")] specifies that the likelihood conditional on top-level variance-covariance parameters is computed by penalized least squares (section 3.6); the algori... | Which iterative algorithm lmer uses for REML estimation? | Neither of the above. The paper describing the algorithms used in lmer [Bates et al. J. Statistical Software (2015) 1-48, available via vignette("lmer",package="lme4")] specifies that the likelihood | Which iterative algorithm lmer uses for REML estimation?
Neither of the above. The paper describing the algorithms used in lmer [Bates et al. J. Statistical Software (2015) 1-48, available via vignette("lmer",package="lme4")] specifies that the likelihood conditional on top-level variance-covariance parameters is comp... | Which iterative algorithm lmer uses for REML estimation?
Neither of the above. The paper describing the algorithms used in lmer [Bates et al. J. Statistical Software (2015) 1-48, available via vignette("lmer",package="lme4")] specifies that the likelihood |
48,591 | Problems generating a sample from a custom distribution with log | We can generate random variates from this distribution by inverting it.
This means solving $1 - F(x) = q$ (which lies between $0$ and $1$, implying $a\gt 0$) for $x$. Notice that $\log^b(x)$ will be undefined or negative (which won't work) unless $x \gt 1$. Leave aside the case $b=0$ for the moment. The solutions are... | Problems generating a sample from a custom distribution with log | We can generate random variates from this distribution by inverting it.
This means solving $1 - F(x) = q$ (which lies between $0$ and $1$, implying $a\gt 0$) for $x$. Notice that $\log^b(x)$ will be | Problems generating a sample from a custom distribution with log
We can generate random variates from this distribution by inverting it.
This means solving $1 - F(x) = q$ (which lies between $0$ and $1$, implying $a\gt 0$) for $x$. Notice that $\log^b(x)$ will be undefined or negative (which won't work) unless $x \gt ... | Problems generating a sample from a custom distribution with log
We can generate random variates from this distribution by inverting it.
This means solving $1 - F(x) = q$ (which lies between $0$ and $1$, implying $a\gt 0$) for $x$. Notice that $\log^b(x)$ will be |
48,592 | Limitation of LDA (latent dirichlet allocation) | Common LDA limitations:
Fixed K (the number of topics is fixed and must be known ahead of time)
Uncorrelated topics (Dirichlet topic distribution cannot capture correlations)
Non-hierarchical (in data-limited regimes hierarchical models allow sharing of data)
Static (no evolution of topics over time)
Bag of words (as... | Limitation of LDA (latent dirichlet allocation) | Common LDA limitations:
Fixed K (the number of topics is fixed and must be known ahead of time)
Uncorrelated topics (Dirichlet topic distribution cannot capture correlations)
Non-hierarchical (in dat | Limitation of LDA (latent dirichlet allocation)
Common LDA limitations:
Fixed K (the number of topics is fixed and must be known ahead of time)
Uncorrelated topics (Dirichlet topic distribution cannot capture correlations)
Non-hierarchical (in data-limited regimes hierarchical models allow sharing of data)
Static (no ... | Limitation of LDA (latent dirichlet allocation)
Common LDA limitations:
Fixed K (the number of topics is fixed and must be known ahead of time)
Uncorrelated topics (Dirichlet topic distribution cannot capture correlations)
Non-hierarchical (in dat |
48,593 | Is the minimum of the Doksum ratio always unbounded? | It is geometrically obvious that shifting $G$ changes nothing and that rescaling will rescale the denominator by the same factor--it's merely a matter of keeping track of units of measurement. The following is a rigorous algebraic demonstration. Your conclusion follows immediately.
Let $a\gt 0$ and $b$ be real numbe... | Is the minimum of the Doksum ratio always unbounded? | It is geometrically obvious that shifting $G$ changes nothing and that rescaling will rescale the denominator by the same factor--it's merely a matter of keeping track of units of measurement. The fo | Is the minimum of the Doksum ratio always unbounded?
It is geometrically obvious that shifting $G$ changes nothing and that rescaling will rescale the denominator by the same factor--it's merely a matter of keeping track of units of measurement. The following is a rigorous algebraic demonstration. Your conclusion fol... | Is the minimum of the Doksum ratio always unbounded?
It is geometrically obvious that shifting $G$ changes nothing and that rescaling will rescale the denominator by the same factor--it's merely a matter of keeping track of units of measurement. The fo |
48,594 | Sum of squared Negative Binomial probability masses | Take two independent Poisson random variables $X$ and $X'$, with means $\lambda$ and $\lambda'$.
The formula answering your question in the Poisson case is a particular case of the identity $$\Pr(X = X' \mid \lambda, \lambda') = \exp(-\lambda)\exp(-\lambda')I_0(2\sqrt{\lambda}{\sqrt{\lambda'}}).$$
> lambda <- 1; lam... | Sum of squared Negative Binomial probability masses | Take two independent Poisson random variables $X$ and $X'$, with means $\lambda$ and $\lambda'$.
The formula answering your question in the Poisson case is a particular case of the identity $$\Pr(X | Sum of squared Negative Binomial probability masses
Take two independent Poisson random variables $X$ and $X'$, with means $\lambda$ and $\lambda'$.
The formula answering your question in the Poisson case is a particular case of the identity $$\Pr(X = X' \mid \lambda, \lambda') = \exp(-\lambda)\exp(-\lambda')I_0(2\s... | Sum of squared Negative Binomial probability masses
Take two independent Poisson random variables $X$ and $X'$, with means $\lambda$ and $\lambda'$.
The formula answering your question in the Poisson case is a particular case of the identity $$\Pr(X |
48,595 | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as opposed to just $\theta^Tx$ | You need power to get rid of negative values. When you raise positive number to the power - you will always get positive value. For negative power - the result is just small and for positive - it's big and grows exponentially.
By using softmax you will never get negative probability nor probability higher then 1 and wi... | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as oppose | You need power to get rid of negative values. When you raise positive number to the power - you will always get positive value. For negative power - the result is just small and for positive - it's bi | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as opposed to just $\theta^Tx$
You need power to get rid of negative values. When you raise positive number to the power - you will always get positive value. For negative power - the result is just small and for positive - it's ... | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as oppose
You need power to get rid of negative values. When you raise positive number to the power - you will always get positive value. For negative power - the result is just small and for positive - it's bi |
48,596 | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as opposed to just $\theta^Tx$ | There is an intuitive definition. I tried to explain the softmax in this answer. To put it simply, you are interpreting the unbounded $\theta_i^Tx$ as log-odds, and the softmax converts them to probabilities in $[0,1]$. Your formula has no such interpretation. | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as oppose | There is an intuitive definition. I tried to explain the softmax in this answer. To put it simply, you are interpreting the unbounded $\theta_i^Tx$ as log-odds, and the softmax converts them to prob | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as opposed to just $\theta^Tx$
There is an intuitive definition. I tried to explain the softmax in this answer. To put it simply, you are interpreting the unbounded $\theta_i^Tx$ as log-odds, and the softmax converts them to pr... | Softmax regression: Intuition about why distribution of $y$ is in terms of $e^{\theta^Tx}$ as oppose
There is an intuitive definition. I tried to explain the softmax in this answer. To put it simply, you are interpreting the unbounded $\theta_i^Tx$ as log-odds, and the softmax converts them to prob |
48,597 | Understanding a characterization of minimal sufficient statistics | The symbols $\boldsymbol{x,y}$ refer to data. Associated with each possible value $v$ of the statistic $S$ is a collection of possible data values $S^{-1}(v)$ for which $S$ has the value $v$. Since $T$ has the same value (call it $w$) on every such dataset, we may define $H(v) = w$.
The case $S(\boldsymbol{x}) \ne S(... | Understanding a characterization of minimal sufficient statistics | The symbols $\boldsymbol{x,y}$ refer to data. Associated with each possible value $v$ of the statistic $S$ is a collection of possible data values $S^{-1}(v)$ for which $S$ has the value $v$. Since | Understanding a characterization of minimal sufficient statistics
The symbols $\boldsymbol{x,y}$ refer to data. Associated with each possible value $v$ of the statistic $S$ is a collection of possible data values $S^{-1}(v)$ for which $S$ has the value $v$. Since $T$ has the same value (call it $w$) on every such dat... | Understanding a characterization of minimal sufficient statistics
The symbols $\boldsymbol{x,y}$ refer to data. Associated with each possible value $v$ of the statistic $S$ is a collection of possible data values $S^{-1}(v)$ for which $S$ has the value $v$. Since |
48,598 | Bayesian neural networks: very multimodal posterior? | Regarding the question how the non-identifiability can be addressed, I can recommend to have a look at Improving the Identifiability of Neural Networks for Bayesian Inference, which "eliminates" the (discrete) combinatorial non-identifiability problem through ordering of nodes (as one of the comments suspected). The pa... | Bayesian neural networks: very multimodal posterior? | Regarding the question how the non-identifiability can be addressed, I can recommend to have a look at Improving the Identifiability of Neural Networks for Bayesian Inference, which "eliminates" the ( | Bayesian neural networks: very multimodal posterior?
Regarding the question how the non-identifiability can be addressed, I can recommend to have a look at Improving the Identifiability of Neural Networks for Bayesian Inference, which "eliminates" the (discrete) combinatorial non-identifiability problem through orderin... | Bayesian neural networks: very multimodal posterior?
Regarding the question how the non-identifiability can be addressed, I can recommend to have a look at Improving the Identifiability of Neural Networks for Bayesian Inference, which "eliminates" the ( |
48,599 | Bayesian neural networks: very multimodal posterior? | In Keras the posterior distribution is specified explicitly. If multimodality of posteriors is expected, it has to be specified.
Here is one example http://ezcodesample.com/Multimodal.html, showing how to change unimodal posterior into multimodal.
That means Keras or tensorflow provide you best parameters for the distr... | Bayesian neural networks: very multimodal posterior? | In Keras the posterior distribution is specified explicitly. If multimodality of posteriors is expected, it has to be specified.
Here is one example http://ezcodesample.com/Multimodal.html, showing ho | Bayesian neural networks: very multimodal posterior?
In Keras the posterior distribution is specified explicitly. If multimodality of posteriors is expected, it has to be specified.
Here is one example http://ezcodesample.com/Multimodal.html, showing how to change unimodal posterior into multimodal.
That means Keras or... | Bayesian neural networks: very multimodal posterior?
In Keras the posterior distribution is specified explicitly. If multimodality of posteriors is expected, it has to be specified.
Here is one example http://ezcodesample.com/Multimodal.html, showing ho |
48,600 | Is it possible for test error to be lower than training error | Totally possible, though it probably means that you aren't training quite as much as you could be. Typically when you look at test/train accuracies over time you get a graph like this:
The test/train stages can be (very broadly) categorized as follows:
first you start training and the test/train accuracy is noisier, ... | Is it possible for test error to be lower than training error | Totally possible, though it probably means that you aren't training quite as much as you could be. Typically when you look at test/train accuracies over time you get a graph like this:
The test/train | Is it possible for test error to be lower than training error
Totally possible, though it probably means that you aren't training quite as much as you could be. Typically when you look at test/train accuracies over time you get a graph like this:
The test/train stages can be (very broadly) categorized as follows:
fir... | Is it possible for test error to be lower than training error
Totally possible, though it probably means that you aren't training quite as much as you could be. Typically when you look at test/train accuracies over time you get a graph like this:
The test/train |
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