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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52,001 | Nonlinear Statistics? | I'd add that the distinction you sense is less than it seems. "Nonlinear statistics" is not really a coherent or standard label, not least because it is defined negatively rather than positively.
On linear: the big positive is that many nonlinearities are coped with by models that are linear in the parameters. You do... | Nonlinear Statistics? | I'd add that the distinction you sense is less than it seems. "Nonlinear statistics" is not really a coherent or standard label, not least because it is defined negatively rather than positively.
On | Nonlinear Statistics?
I'd add that the distinction you sense is less than it seems. "Nonlinear statistics" is not really a coherent or standard label, not least because it is defined negatively rather than positively.
On linear: the big positive is that many nonlinearities are coped with by models that are linear in ... | Nonlinear Statistics?
I'd add that the distinction you sense is less than it seems. "Nonlinear statistics" is not really a coherent or standard label, not least because it is defined negatively rather than positively.
On |
52,002 | Nonlinear Statistics? | The best answer to this may depend on your discipline, but I like Bruce Hansen's Econometrics textbook (free, online): http://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf
See Section 9.1 for nonlinear least squares -- these results are really not very different from the linear case.
You might also be interes... | Nonlinear Statistics? | The best answer to this may depend on your discipline, but I like Bruce Hansen's Econometrics textbook (free, online): http://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf
See Section 9.1 fo | Nonlinear Statistics?
The best answer to this may depend on your discipline, but I like Bruce Hansen's Econometrics textbook (free, online): http://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf
See Section 9.1 for nonlinear least squares -- these results are really not very different from the linear case.
You... | Nonlinear Statistics?
The best answer to this may depend on your discipline, but I like Bruce Hansen's Econometrics textbook (free, online): http://www.ssc.wisc.edu/~bhansen/econometrics/Econometrics.pdf
See Section 9.1 fo |
52,003 | Probability Distribution - which to use Normal or Hypergeometric | The random variables $\{X_i\}$, which are defined as the amount of money family $i$ spends in a given month is known to be uniformly distributed, that is, $X_i \sim \text{U}(500, 4500)$.
For a given sample of size 10, you are asked to compute the probability that at least 2 of them will spend more that \$3000. This is ... | Probability Distribution - which to use Normal or Hypergeometric | The random variables $\{X_i\}$, which are defined as the amount of money family $i$ spends in a given month is known to be uniformly distributed, that is, $X_i \sim \text{U}(500, 4500)$.
For a given s | Probability Distribution - which to use Normal or Hypergeometric
The random variables $\{X_i\}$, which are defined as the amount of money family $i$ spends in a given month is known to be uniformly distributed, that is, $X_i \sim \text{U}(500, 4500)$.
For a given sample of size 10, you are asked to compute the probabil... | Probability Distribution - which to use Normal or Hypergeometric
The random variables $\{X_i\}$, which are defined as the amount of money family $i$ spends in a given month is known to be uniformly distributed, that is, $X_i \sim \text{U}(500, 4500)$.
For a given s |
52,004 | Probability Distribution - which to use Normal or Hypergeometric | Neither the normal distribution nor the hypergeometric distribution apply to this question. There is no underlying assumption of normality here: the probability distribution for the expenses for a single family selected at random is given to be uniform between 500 and 4500. Furthermore, the hypergeometric distributi... | Probability Distribution - which to use Normal or Hypergeometric | Neither the normal distribution nor the hypergeometric distribution apply to this question. There is no underlying assumption of normality here: the probability distribution for the expenses for a s | Probability Distribution - which to use Normal or Hypergeometric
Neither the normal distribution nor the hypergeometric distribution apply to this question. There is no underlying assumption of normality here: the probability distribution for the expenses for a single family selected at random is given to be uniform ... | Probability Distribution - which to use Normal or Hypergeometric
Neither the normal distribution nor the hypergeometric distribution apply to this question. There is no underlying assumption of normality here: the probability distribution for the expenses for a s |
52,005 | Probability Distribution - which to use Normal or Hypergeometric | Thanks everyone. I've attached my working with the corrected numbers. Now I understand the rationale behind the workings above. :) | Probability Distribution - which to use Normal or Hypergeometric | Thanks everyone. I've attached my working with the corrected numbers. Now I understand the rationale behind the workings above. :) | Probability Distribution - which to use Normal or Hypergeometric
Thanks everyone. I've attached my working with the corrected numbers. Now I understand the rationale behind the workings above. :) | Probability Distribution - which to use Normal or Hypergeometric
Thanks everyone. I've attached my working with the corrected numbers. Now I understand the rationale behind the workings above. :) |
52,006 | Probability Distribution - which to use Normal or Hypergeometric | Although the correct answer is given, I'll give you some hints and a thought process here that may help you understand why the answer turns out the way it does.
The most important thing to arrive at the correct answer is to realize that you should be using the Binomial Distribution. How to see this?
The binomial distri... | Probability Distribution - which to use Normal or Hypergeometric | Although the correct answer is given, I'll give you some hints and a thought process here that may help you understand why the answer turns out the way it does.
The most important thing to arrive at t | Probability Distribution - which to use Normal or Hypergeometric
Although the correct answer is given, I'll give you some hints and a thought process here that may help you understand why the answer turns out the way it does.
The most important thing to arrive at the correct answer is to realize that you should be usin... | Probability Distribution - which to use Normal or Hypergeometric
Although the correct answer is given, I'll give you some hints and a thought process here that may help you understand why the answer turns out the way it does.
The most important thing to arrive at t |
52,007 | What tools do Machine Learning experts use in the real world? | MATLAB was primarily developed for optimization and mathematical simulations in engineering problems. But yes it has performance issues when it comes to machine learning, optimization, etc., in terms of customizing ability.
Over time, most statistical analysis / machine learning has shifted to R and Python, because of ... | What tools do Machine Learning experts use in the real world? | MATLAB was primarily developed for optimization and mathematical simulations in engineering problems. But yes it has performance issues when it comes to machine learning, optimization, etc., in terms | What tools do Machine Learning experts use in the real world?
MATLAB was primarily developed for optimization and mathematical simulations in engineering problems. But yes it has performance issues when it comes to machine learning, optimization, etc., in terms of customizing ability.
Over time, most statistical analys... | What tools do Machine Learning experts use in the real world?
MATLAB was primarily developed for optimization and mathematical simulations in engineering problems. But yes it has performance issues when it comes to machine learning, optimization, etc., in terms |
52,008 | What tools do Machine Learning experts use in the real world? | If scale is not an issue then any solution you probably already know is fine. It is more a matter of personal and company choice (cost/legacy issues). So it doesn't really matter if it's going be R or MATLAB, Python or Java, Weka or Rapidminer, open source tools or propriety code. However, big players like those you me... | What tools do Machine Learning experts use in the real world? | If scale is not an issue then any solution you probably already know is fine. It is more a matter of personal and company choice (cost/legacy issues). So it doesn't really matter if it's going be R or | What tools do Machine Learning experts use in the real world?
If scale is not an issue then any solution you probably already know is fine. It is more a matter of personal and company choice (cost/legacy issues). So it doesn't really matter if it's going be R or MATLAB, Python or Java, Weka or Rapidminer, open source t... | What tools do Machine Learning experts use in the real world?
If scale is not an issue then any solution you probably already know is fine. It is more a matter of personal and company choice (cost/legacy issues). So it doesn't really matter if it's going be R or |
52,009 | What tools do Machine Learning experts use in the real world? | MATLAB is a great tool. However for Machine Learning simulation there is an increasing interest in R. R is emerging as a great platform for Machine Learning, Data Mining, Statistical Modelling (etc.) tasks. R has got a rich range of packages for statistical modelling.
With the emergence of Big Data R has got a positive... | What tools do Machine Learning experts use in the real world? | MATLAB is a great tool. However for Machine Learning simulation there is an increasing interest in R. R is emerging as a great platform for Machine Learning, Data Mining, Statistical Modelling (etc.) | What tools do Machine Learning experts use in the real world?
MATLAB is a great tool. However for Machine Learning simulation there is an increasing interest in R. R is emerging as a great platform for Machine Learning, Data Mining, Statistical Modelling (etc.) tasks. R has got a rich range of packages for statistical ... | What tools do Machine Learning experts use in the real world?
MATLAB is a great tool. However for Machine Learning simulation there is an increasing interest in R. R is emerging as a great platform for Machine Learning, Data Mining, Statistical Modelling (etc.) |
52,010 | Why do they call it "sampling distribution of the sample mean"? | Within a particular setting where the type of distribution is known or implied, "distribution of the sample mean" works just fine. But in general would the "distribution of the sample mean" be its sampling distribution, a bootstrap distribution, a permutation distribution, or perhaps something else?
The existence of ... | Why do they call it "sampling distribution of the sample mean"? | Within a particular setting where the type of distribution is known or implied, "distribution of the sample mean" works just fine. But in general would the "distribution of the sample mean" be its sa | Why do they call it "sampling distribution of the sample mean"?
Within a particular setting where the type of distribution is known or implied, "distribution of the sample mean" works just fine. But in general would the "distribution of the sample mean" be its sampling distribution, a bootstrap distribution, a permuta... | Why do they call it "sampling distribution of the sample mean"?
Within a particular setting where the type of distribution is known or implied, "distribution of the sample mean" works just fine. But in general would the "distribution of the sample mean" be its sa |
52,011 | Why do they call it "sampling distribution of the sample mean"? | For a given data set, the sample mean provides a single estimate of the population mean. This estimate is a constant and thus its distribution is rather boring.
In contrast, the sampling distribution of the mean refers to the frequentist approach of considering the distribution of the sample means between many hypothe... | Why do they call it "sampling distribution of the sample mean"? | For a given data set, the sample mean provides a single estimate of the population mean. This estimate is a constant and thus its distribution is rather boring.
In contrast, the sampling distribution | Why do they call it "sampling distribution of the sample mean"?
For a given data set, the sample mean provides a single estimate of the population mean. This estimate is a constant and thus its distribution is rather boring.
In contrast, the sampling distribution of the mean refers to the frequentist approach of consi... | Why do they call it "sampling distribution of the sample mean"?
For a given data set, the sample mean provides a single estimate of the population mean. This estimate is a constant and thus its distribution is rather boring.
In contrast, the sampling distribution |
52,012 | Why do they call it "sampling distribution of the sample mean"? | You can have a sampling distribution of other statistics than the mean, such as the estimated median, or estimated variance.
Sometimes "sampling distribution" might be a loose term referring to the estimated mean and estimated variance of the sample taken together (with the unspoken assumption that the distribution of ... | Why do they call it "sampling distribution of the sample mean"? | You can have a sampling distribution of other statistics than the mean, such as the estimated median, or estimated variance.
Sometimes "sampling distribution" might be a loose term referring to the es | Why do they call it "sampling distribution of the sample mean"?
You can have a sampling distribution of other statistics than the mean, such as the estimated median, or estimated variance.
Sometimes "sampling distribution" might be a loose term referring to the estimated mean and estimated variance of the sample taken ... | Why do they call it "sampling distribution of the sample mean"?
You can have a sampling distribution of other statistics than the mean, such as the estimated median, or estimated variance.
Sometimes "sampling distribution" might be a loose term referring to the es |
52,013 | What is the test statistic in Kolmogorov–Smirnov test? | Note that the Kolmogorov-Smirnov test statistic is very clearly defined in the immediately previous section:
$$D_n=\sup_x|F_n(x)−F(x)|\,.$$
The reason they discuss $\sqrt{n}D_n$ in the next section is that the standard deviation of the distribution of $D_n$ goes down as $1/\sqrt n$, while $\sqrt{n}D_n$ converges in di... | What is the test statistic in Kolmogorov–Smirnov test? | Note that the Kolmogorov-Smirnov test statistic is very clearly defined in the immediately previous section:
$$D_n=\sup_x|F_n(x)−F(x)|\,.$$
The reason they discuss $\sqrt{n}D_n$ in the next section i | What is the test statistic in Kolmogorov–Smirnov test?
Note that the Kolmogorov-Smirnov test statistic is very clearly defined in the immediately previous section:
$$D_n=\sup_x|F_n(x)−F(x)|\,.$$
The reason they discuss $\sqrt{n}D_n$ in the next section is that the standard deviation of the distribution of $D_n$ goes d... | What is the test statistic in Kolmogorov–Smirnov test?
Note that the Kolmogorov-Smirnov test statistic is very clearly defined in the immediately previous section:
$$D_n=\sup_x|F_n(x)−F(x)|\,.$$
The reason they discuss $\sqrt{n}D_n$ in the next section i |
52,014 | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | It is difficult to tell without the data to reproduce things. You didn't list the caret code so I'm not sure what was done there.
I think that the bottom line is that you have 44 samples and 10-fold CV, known to have high variance, is not going to give you repeatable results. I would suggest using several repeats of 10... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | It is difficult to tell without the data to reproduce things. You didn't list the caret code so I'm not sure what was done there.
I think that the bottom line is that you have 44 samples and 10-fold C | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
It is difficult to tell without the data to reproduce things. You didn't list the caret code so I'm not sure what was done there.
I think that the bottom line is that you have 44 samples and 10-fold CV, known to have high variance,... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
It is difficult to tell without the data to reproduce things. You didn't list the caret code so I'm not sure what was done there.
I think that the bottom line is that you have 44 samples and 10-fold C |
52,015 | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | The reason why you're getting different lambda values is because every time you call
cv.glmnet(xMatrix, y, alpha=0.5, nfolds=10), you're essentially creating different cross validation folds.To retain the same lambda value, you need to make sure you're using the same cross validation folds every time, thus you might wa... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | The reason why you're getting different lambda values is because every time you call
cv.glmnet(xMatrix, y, alpha=0.5, nfolds=10), you're essentially creating different cross validation folds.To retain | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
The reason why you're getting different lambda values is because every time you call
cv.glmnet(xMatrix, y, alpha=0.5, nfolds=10), you're essentially creating different cross validation folds.To retain the same lambda value, you nee... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
The reason why you're getting different lambda values is because every time you call
cv.glmnet(xMatrix, y, alpha=0.5, nfolds=10), you're essentially creating different cross validation folds.To retain |
52,016 | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | If you want to get repeatable optimal lambda & alpha, you can use leave-one-out CV (which doesn't support AUC though).
cv <- cv.glmnet(x,y,alpha=1,nfolds=nrow(x))
It's pretty normal for an n-fold CV to bring you very huge variance, since the partitions of sample are generated randomly. | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | If you want to get repeatable optimal lambda & alpha, you can use leave-one-out CV (which doesn't support AUC though).
cv <- cv.glmnet(x,y,alpha=1,nfolds=nrow(x))
It's pretty normal for an n-fold CV | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
If you want to get repeatable optimal lambda & alpha, you can use leave-one-out CV (which doesn't support AUC though).
cv <- cv.glmnet(x,y,alpha=1,nfolds=nrow(x))
It's pretty normal for an n-fold CV to bring you very huge varianc... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
If you want to get repeatable optimal lambda & alpha, you can use leave-one-out CV (which doesn't support AUC though).
cv <- cv.glmnet(x,y,alpha=1,nfolds=nrow(x))
It's pretty normal for an n-fold CV |
52,017 | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | Your methodology is not great for reproducibility: you set.seed(1) once then run cv.glmnet() 100 times. Each of those calls to cv.glmnet() is itself calling sample() N times. So if the length of your data ever changes, the reproducibility changes.
Better to explicitly set.seed() right before each run. Or else keep the ... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test? | Your methodology is not great for reproducibility: you set.seed(1) once then run cv.glmnet() 100 times. Each of those calls to cv.glmnet() is itself calling sample() N times. So if the length of your | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
Your methodology is not great for reproducibility: you set.seed(1) once then run cv.glmnet() 100 times. Each of those calls to cv.glmnet() is itself calling sample() N times. So if the length of your data ever changes, the reproduc... | Why does lambda.min value in glmnet tuning cross-validation change, when repeating test?
Your methodology is not great for reproducibility: you set.seed(1) once then run cv.glmnet() 100 times. Each of those calls to cv.glmnet() is itself calling sample() N times. So if the length of your |
52,018 | What is lambda in an elastic net model (penalized regression)? | You're confused; $\alpha$ and $\lambda$ are totally different.
$\alpha$ sets the degree of mixing between ridge regression and lasso: when $\alpha = 0$, the elastic net does the former, and when $\alpha = 1$, it does the latter. Values of $\alpha$ between those extremes will give a result that is a blend of the two.
Me... | What is lambda in an elastic net model (penalized regression)? | You're confused; $\alpha$ and $\lambda$ are totally different.
$\alpha$ sets the degree of mixing between ridge regression and lasso: when $\alpha = 0$, the elastic net does the former, and when $\alp | What is lambda in an elastic net model (penalized regression)?
You're confused; $\alpha$ and $\lambda$ are totally different.
$\alpha$ sets the degree of mixing between ridge regression and lasso: when $\alpha = 0$, the elastic net does the former, and when $\alpha = 1$, it does the latter. Values of $\alpha$ between t... | What is lambda in an elastic net model (penalized regression)?
You're confused; $\alpha$ and $\lambda$ are totally different.
$\alpha$ sets the degree of mixing between ridge regression and lasso: when $\alpha = 0$, the elastic net does the former, and when $\alp |
52,019 | Repeated measures within factors settings for G*Power power calculation | GPower is assuming you have your data set up so that a row is a case (often a person), and a column is a measure.
For example, if we measured Y on three occasions, we'd have Y1, Y2, Y3, and we'd have three measures.
The groups are when you have a between case predictor - for example gender or experimental group. So whe... | Repeated measures within factors settings for G*Power power calculation | GPower is assuming you have your data set up so that a row is a case (often a person), and a column is a measure.
For example, if we measured Y on three occasions, we'd have Y1, Y2, Y3, and we'd have | Repeated measures within factors settings for G*Power power calculation
GPower is assuming you have your data set up so that a row is a case (often a person), and a column is a measure.
For example, if we measured Y on three occasions, we'd have Y1, Y2, Y3, and we'd have three measures.
The groups are when you have a b... | Repeated measures within factors settings for G*Power power calculation
GPower is assuming you have your data set up so that a row is a case (often a person), and a column is a measure.
For example, if we measured Y on three occasions, we'd have Y1, Y2, Y3, and we'd have |
52,020 | Repeated measures within factors settings for G*Power power calculation | I had the same question, so I sent an e-mail to the G*Power team. They informed me that the current version of G*Power (3.1.9.2) cannot conveniently do power analyses for repeated measures designs with more than one within-subject or between-subject factor. It is possible using the "Generic F test" option, but this is ... | Repeated measures within factors settings for G*Power power calculation | I had the same question, so I sent an e-mail to the G*Power team. They informed me that the current version of G*Power (3.1.9.2) cannot conveniently do power analyses for repeated measures designs wit | Repeated measures within factors settings for G*Power power calculation
I had the same question, so I sent an e-mail to the G*Power team. They informed me that the current version of G*Power (3.1.9.2) cannot conveniently do power analyses for repeated measures designs with more than one within-subject or between-subjec... | Repeated measures within factors settings for G*Power power calculation
I had the same question, so I sent an e-mail to the G*Power team. They informed me that the current version of G*Power (3.1.9.2) cannot conveniently do power analyses for repeated measures designs wit |
52,021 | Repeated measures within factors settings for G*Power power calculation | Jumping in a bit late, but I figured I'd build on @Jeremy's response and add some clarifying examples (as the 2x2 RM design is a bit ambiguous to me).
Assuming the "ANOVA: RM, within factors" option is where we're at, I believe the "number of groups" refers to the number of between-subjects LEVELS (not factors) that yo... | Repeated measures within factors settings for G*Power power calculation | Jumping in a bit late, but I figured I'd build on @Jeremy's response and add some clarifying examples (as the 2x2 RM design is a bit ambiguous to me).
Assuming the "ANOVA: RM, within factors" option i | Repeated measures within factors settings for G*Power power calculation
Jumping in a bit late, but I figured I'd build on @Jeremy's response and add some clarifying examples (as the 2x2 RM design is a bit ambiguous to me).
Assuming the "ANOVA: RM, within factors" option is where we're at, I believe the "number of group... | Repeated measures within factors settings for G*Power power calculation
Jumping in a bit late, but I figured I'd build on @Jeremy's response and add some clarifying examples (as the 2x2 RM design is a bit ambiguous to me).
Assuming the "ANOVA: RM, within factors" option i |
52,022 | Repeated measures within factors settings for G*Power power calculation | The problem is that you cannot enter "1" in the "number of groups" window.
Therefore, seems it is not possible to use G*Power for RM ANOVAs with no between-subject factors. | Repeated measures within factors settings for G*Power power calculation | The problem is that you cannot enter "1" in the "number of groups" window.
Therefore, seems it is not possible to use G*Power for RM ANOVAs with no between-subject factors. | Repeated measures within factors settings for G*Power power calculation
The problem is that you cannot enter "1" in the "number of groups" window.
Therefore, seems it is not possible to use G*Power for RM ANOVAs with no between-subject factors. | Repeated measures within factors settings for G*Power power calculation
The problem is that you cannot enter "1" in the "number of groups" window.
Therefore, seems it is not possible to use G*Power for RM ANOVAs with no between-subject factors. |
52,023 | Omit 0 lag order in ACF plot | Another possible solution is as follows:
# Create an "acf" object called z
z <- acf(dummy)
# Check class of the object
class(z)
# View attributes of the "acf" object
attributes(z)
# Use "acf" attribute to view the first 13 elements (1 = lag at 0)
z$acf[1:13]
# Get rid of the first element (i.e. lag 0)
z$acf[2:13]
# Plo... | Omit 0 lag order in ACF plot | Another possible solution is as follows:
# Create an "acf" object called z
z <- acf(dummy)
# Check class of the object
class(z)
# View attributes of the "acf" object
attributes(z)
# Use "acf" attribut | Omit 0 lag order in ACF plot
Another possible solution is as follows:
# Create an "acf" object called z
z <- acf(dummy)
# Check class of the object
class(z)
# View attributes of the "acf" object
attributes(z)
# Use "acf" attribute to view the first 13 elements (1 = lag at 0)
z$acf[1:13]
# Get rid of the first element (... | Omit 0 lag order in ACF plot
Another possible solution is as follows:
# Create an "acf" object called z
z <- acf(dummy)
# Check class of the object
class(z)
# View attributes of the "acf" object
attributes(z)
# Use "acf" attribut |
52,024 | Omit 0 lag order in ACF plot | Use the Acf function from the forecast package. | Omit 0 lag order in ACF plot | Use the Acf function from the forecast package. | Omit 0 lag order in ACF plot
Use the Acf function from the forecast package. | Omit 0 lag order in ACF plot
Use the Acf function from the forecast package. |
52,025 | Omit 0 lag order in ACF plot | Use this code:
suppose;
x = rnorm(100) ## A typical white noise process
plot(acf(x,plot=F)[1:20]) | Omit 0 lag order in ACF plot | Use this code:
suppose;
x = rnorm(100) ## A typical white noise process
plot(acf(x,plot=F)[1:20]) | Omit 0 lag order in ACF plot
Use this code:
suppose;
x = rnorm(100) ## A typical white noise process
plot(acf(x,plot=F)[1:20]) | Omit 0 lag order in ACF plot
Use this code:
suppose;
x = rnorm(100) ## A typical white noise process
plot(acf(x,plot=F)[1:20]) |
52,026 | Omit 0 lag order in ACF plot | Set the xlim and ylim. For example:
acf(x,20,xlim(1,20),ylim(-0.2,0.5)) | Omit 0 lag order in ACF plot | Set the xlim and ylim. For example:
acf(x,20,xlim(1,20),ylim(-0.2,0.5)) | Omit 0 lag order in ACF plot
Set the xlim and ylim. For example:
acf(x,20,xlim(1,20),ylim(-0.2,0.5)) | Omit 0 lag order in ACF plot
Set the xlim and ylim. For example:
acf(x,20,xlim(1,20),ylim(-0.2,0.5)) |
52,027 | Multiple curves when plotting a random forest [closed] | When there is no test result (ytest was empty for training), plot shows:
for classification, black solid line for overall OOB error and a bunch of colour lines, one for each class' error (i.e. 1-this class recall).
for regression, one black solid line for OOB MSE error.
When test is present, documentation (?plot.rand... | Multiple curves when plotting a random forest [closed] | When there is no test result (ytest was empty for training), plot shows:
for classification, black solid line for overall OOB error and a bunch of colour lines, one for each class' error (i.e. 1-this | Multiple curves when plotting a random forest [closed]
When there is no test result (ytest was empty for training), plot shows:
for classification, black solid line for overall OOB error and a bunch of colour lines, one for each class' error (i.e. 1-this class recall).
for regression, one black solid line for OOB MSE ... | Multiple curves when plotting a random forest [closed]
When there is no test result (ytest was empty for training), plot shows:
for classification, black solid line for overall OOB error and a bunch of colour lines, one for each class' error (i.e. 1-this |
52,028 | Multiple curves when plotting a random forest [closed] | I have come across with the same issue and I found a link where it shows the graph you get when plotting the random forest model: http://statweb.stanford.edu/~jtaylo/courses/stats202/ensemble.html
If you scroll down, there is the plot(model) graph and the graph with four curves: one for the oob error in black and three... | Multiple curves when plotting a random forest [closed] | I have come across with the same issue and I found a link where it shows the graph you get when plotting the random forest model: http://statweb.stanford.edu/~jtaylo/courses/stats202/ensemble.html
If | Multiple curves when plotting a random forest [closed]
I have come across with the same issue and I found a link where it shows the graph you get when plotting the random forest model: http://statweb.stanford.edu/~jtaylo/courses/stats202/ensemble.html
If you scroll down, there is the plot(model) graph and the graph wit... | Multiple curves when plotting a random forest [closed]
I have come across with the same issue and I found a link where it shows the graph you get when plotting the random forest model: http://statweb.stanford.edu/~jtaylo/courses/stats202/ensemble.html
If |
52,029 | Multiple curves when plotting a random forest [closed] | I know this post is a little bit old, but I just came accross the same problem and solved it like this:
rndF1 <- randomForest(train.X, train.Y, test.X, test.Y)
plot(rndF1)
rndF1.legend <- if (is.null(rndF1$test$err.rate)) {colnames(rndF1$err.rate)}
else {colnames(rndF1$test$err.rate)}
legend("top", cex... | Multiple curves when plotting a random forest [closed] | I know this post is a little bit old, but I just came accross the same problem and solved it like this:
rndF1 <- randomForest(train.X, train.Y, test.X, test.Y)
plot(rndF1)
rndF1.legend <- if (is.null( | Multiple curves when plotting a random forest [closed]
I know this post is a little bit old, but I just came accross the same problem and solved it like this:
rndF1 <- randomForest(train.X, train.Y, test.X, test.Y)
plot(rndF1)
rndF1.legend <- if (is.null(rndF1$test$err.rate)) {colnames(rndF1$err.rate)}
... | Multiple curves when plotting a random forest [closed]
I know this post is a little bit old, but I just came accross the same problem and solved it like this:
rndF1 <- randomForest(train.X, train.Y, test.X, test.Y)
plot(rndF1)
rndF1.legend <- if (is.null( |
52,030 | Multiple curves when plotting a random forest [closed] | If the lines are very close, I think it is not necessary to distinct them. Otherwise, you can output the value by $err.rate. and compare with the line. Or you can use only several trees in the function randomForest(Species~., iris,importance=TRUE,ntree=24) and then plot it, then you it would be easier to tell and find ... | Multiple curves when plotting a random forest [closed] | If the lines are very close, I think it is not necessary to distinct them. Otherwise, you can output the value by $err.rate. and compare with the line. Or you can use only several trees in the functio | Multiple curves when plotting a random forest [closed]
If the lines are very close, I think it is not necessary to distinct them. Otherwise, you can output the value by $err.rate. and compare with the line. Or you can use only several trees in the function randomForest(Species~., iris,importance=TRUE,ntree=24) and then... | Multiple curves when plotting a random forest [closed]
If the lines are very close, I think it is not necessary to distinct them. Otherwise, you can output the value by $err.rate. and compare with the line. Or you can use only several trees in the functio |
52,031 | Multiple curves when plotting a random forest [closed] | The plot() function can be very useful here. In fact, you can do something like this to change the labels that appear on the plot. The plot of the RandomForest will show error rates, and are very useful for analyzing the performance of the algorithm, for the different number of trees.
I recently used this:
cat(paste0(... | Multiple curves when plotting a random forest [closed] | The plot() function can be very useful here. In fact, you can do something like this to change the labels that appear on the plot. The plot of the RandomForest will show error rates, and are very use | Multiple curves when plotting a random forest [closed]
The plot() function can be very useful here. In fact, you can do something like this to change the labels that appear on the plot. The plot of the RandomForest will show error rates, and are very useful for analyzing the performance of the algorithm, for the diffe... | Multiple curves when plotting a random forest [closed]
The plot() function can be very useful here. In fact, you can do something like this to change the labels that appear on the plot. The plot of the RandomForest will show error rates, and are very use |
52,032 | How to determine if a variable is categorical? | Tax rates are not categorical, they are continuous. A tax rate can vary - e.g. the sales tax in New York City is, I believe, 8.825%.
It appears that the data you have only has certain tax rates. But that is a feature of your data, not an underlying characteristic of the variable. Categorical variables CANNOT take valu... | How to determine if a variable is categorical? | Tax rates are not categorical, they are continuous. A tax rate can vary - e.g. the sales tax in New York City is, I believe, 8.825%.
It appears that the data you have only has certain tax rates. But t | How to determine if a variable is categorical?
Tax rates are not categorical, they are continuous. A tax rate can vary - e.g. the sales tax in New York City is, I believe, 8.825%.
It appears that the data you have only has certain tax rates. But that is a feature of your data, not an underlying characteristic of the va... | How to determine if a variable is categorical?
Tax rates are not categorical, they are continuous. A tax rate can vary - e.g. the sales tax in New York City is, I believe, 8.825%.
It appears that the data you have only has certain tax rates. But t |
52,033 | How to determine if a variable is categorical? | It certainly looks as if the variable plotted along the X axis can only take certain discrete values.
However ... a categorical variable is one that takes values in a sample space where neither magnitude nor order have any meaning. Example: a medical study might record the gender of the patient (male/female), which is... | How to determine if a variable is categorical? | It certainly looks as if the variable plotted along the X axis can only take certain discrete values.
However ... a categorical variable is one that takes values in a sample space where neither magni | How to determine if a variable is categorical?
It certainly looks as if the variable plotted along the X axis can only take certain discrete values.
However ... a categorical variable is one that takes values in a sample space where neither magnitude nor order have any meaning. Example: a medical study might record th... | How to determine if a variable is categorical?
It certainly looks as if the variable plotted along the X axis can only take certain discrete values.
However ... a categorical variable is one that takes values in a sample space where neither magni |
52,034 | How robust is ANOVA to violations of normality? | Don't look at it as a binary thing: "either I can trust the results or I can't." Look at it as a spectrum. With all assumptions perfectly satisfied (including the in most cases crucial one of random sampling), statistics such as F- and p-values will allow you to make accurate sample-to-population inferences. The fa... | How robust is ANOVA to violations of normality? | Don't look at it as a binary thing: "either I can trust the results or I can't." Look at it as a spectrum. With all assumptions perfectly satisfied (including the in most cases crucial one of rando | How robust is ANOVA to violations of normality?
Don't look at it as a binary thing: "either I can trust the results or I can't." Look at it as a spectrum. With all assumptions perfectly satisfied (including the in most cases crucial one of random sampling), statistics such as F- and p-values will allow you to make a... | How robust is ANOVA to violations of normality?
Don't look at it as a binary thing: "either I can trust the results or I can't." Look at it as a spectrum. With all assumptions perfectly satisfied (including the in most cases crucial one of rando |
52,035 | How robust is ANOVA to violations of normality? | Let me state a couple of things. First, I think it's best to understand repeated measures ANOVA as actually a multi-level model in disguise, and that may create additional complexities here. I should let one of CV's contributors who are more expert on multi-level models address that issue.
However, in general, it's... | How robust is ANOVA to violations of normality? | Let me state a couple of things. First, I think it's best to understand repeated measures ANOVA as actually a multi-level model in disguise, and that may create additional complexities here. I shoul | How robust is ANOVA to violations of normality?
Let me state a couple of things. First, I think it's best to understand repeated measures ANOVA as actually a multi-level model in disguise, and that may create additional complexities here. I should let one of CV's contributors who are more expert on multi-level models... | How robust is ANOVA to violations of normality?
Let me state a couple of things. First, I think it's best to understand repeated measures ANOVA as actually a multi-level model in disguise, and that may create additional complexities here. I shoul |
52,036 | How robust is ANOVA to violations of normality? | My understanding is that ANOVA including repeated measures is robust to violations to normality of errors assumptions. However there is indications that the errors should be equal in their variation across different factor levels.
Can I trust ANOVA results for a non-normally distributed DV? | How robust is ANOVA to violations of normality? | My understanding is that ANOVA including repeated measures is robust to violations to normality of errors assumptions. However there is indications that the errors should be equal in their variation | How robust is ANOVA to violations of normality?
My understanding is that ANOVA including repeated measures is robust to violations to normality of errors assumptions. However there is indications that the errors should be equal in their variation across different factor levels.
Can I trust ANOVA results for a non-norm... | How robust is ANOVA to violations of normality?
My understanding is that ANOVA including repeated measures is robust to violations to normality of errors assumptions. However there is indications that the errors should be equal in their variation |
52,037 | How to do a pretty scatter plot in R? [closed] | You can to do this in the new version of ggplot2 (0.9).
You can try it out:
library(ggplot2) #make sure the newest is installed
df <- data.frame(v1 = runif(1000), v2 = runif(1000))
bin.plot<-qplot(data=df,
x=v1,
y=v2,
z=v2)
bin.plot+stat_summary_hex(fun=function(z)len... | How to do a pretty scatter plot in R? [closed] | You can to do this in the new version of ggplot2 (0.9).
You can try it out:
library(ggplot2) #make sure the newest is installed
df <- data.frame(v1 = runif(1000), v2 = runif(1000))
bin.plot<-qplot(d | How to do a pretty scatter plot in R? [closed]
You can to do this in the new version of ggplot2 (0.9).
You can try it out:
library(ggplot2) #make sure the newest is installed
df <- data.frame(v1 = runif(1000), v2 = runif(1000))
bin.plot<-qplot(data=df,
x=v1,
y=v2,
z=v2)... | How to do a pretty scatter plot in R? [closed]
You can to do this in the new version of ggplot2 (0.9).
You can try it out:
library(ggplot2) #make sure the newest is installed
df <- data.frame(v1 = runif(1000), v2 = runif(1000))
bin.plot<-qplot(d |
52,038 | How to do a pretty scatter plot in R? [closed] | You may want to look at these two entries from 'SAS and R':
http://sas-and-r.blogspot.com/2011/07/example-91-scatterplots-with-binning.html
http://sas-and-r.blogspot.com/2011/07/example-92-transparency-and-bivariate.html
They cover the use of binning, transparency and bivariate kernel density estimators for scatter pl... | How to do a pretty scatter plot in R? [closed] | You may want to look at these two entries from 'SAS and R':
http://sas-and-r.blogspot.com/2011/07/example-91-scatterplots-with-binning.html
http://sas-and-r.blogspot.com/2011/07/example-92-transparen | How to do a pretty scatter plot in R? [closed]
You may want to look at these two entries from 'SAS and R':
http://sas-and-r.blogspot.com/2011/07/example-91-scatterplots-with-binning.html
http://sas-and-r.blogspot.com/2011/07/example-92-transparency-and-bivariate.html
They cover the use of binning, transparency and biv... | How to do a pretty scatter plot in R? [closed]
You may want to look at these two entries from 'SAS and R':
http://sas-and-r.blogspot.com/2011/07/example-91-scatterplots-with-binning.html
http://sas-and-r.blogspot.com/2011/07/example-92-transparen |
52,039 | How to do a pretty scatter plot in R? [closed] | It's not really an answer to your question about binning one easy solution in ggplot2 to deal with large amount of data in scatterplots is to use the alpha parameter to set some transparency
> df <- data.frame(v1 = rnorm(100000), v2 = rnorm(100000))
> ggplot(df, aes(x=v1, y=v2)) + geom_point(alpha = .01) + theme_bw() | How to do a pretty scatter plot in R? [closed] | It's not really an answer to your question about binning one easy solution in ggplot2 to deal with large amount of data in scatterplots is to use the alpha parameter to set some transparency
> df <- d | How to do a pretty scatter plot in R? [closed]
It's not really an answer to your question about binning one easy solution in ggplot2 to deal with large amount of data in scatterplots is to use the alpha parameter to set some transparency
> df <- data.frame(v1 = rnorm(100000), v2 = rnorm(100000))
> ggplot(df, aes(x=v1, ... | How to do a pretty scatter plot in R? [closed]
It's not really an answer to your question about binning one easy solution in ggplot2 to deal with large amount of data in scatterplots is to use the alpha parameter to set some transparency
> df <- d |
52,040 | Orthogonalized regression reference? | I think you misremember the end of the process. In R, it would go like this:
# generating random x1 x2 x3 in (0,1) (10 values each)
> x1 <- runif(10)
> x2 <- runif(10)
> x3 <- runif(10)
# generating y
> y <- x1 + 2*x2 + 3*x3 + rnorm(10)
# classical regression
> lm(y ~ x1 + x2 + x3)
Call:
lm(formula = y ~ x1 + x2 + x... | Orthogonalized regression reference? | I think you misremember the end of the process. In R, it would go like this:
# generating random x1 x2 x3 in (0,1) (10 values each)
> x1 <- runif(10)
> x2 <- runif(10)
> x3 <- runif(10)
# generating | Orthogonalized regression reference?
I think you misremember the end of the process. In R, it would go like this:
# generating random x1 x2 x3 in (0,1) (10 values each)
> x1 <- runif(10)
> x2 <- runif(10)
> x3 <- runif(10)
# generating y
> y <- x1 + 2*x2 + 3*x3 + rnorm(10)
# classical regression
> lm(y ~ x1 + x2 + x3... | Orthogonalized regression reference?
I think you misremember the end of the process. In R, it would go like this:
# generating random x1 x2 x3 in (0,1) (10 values each)
> x1 <- runif(10)
> x2 <- runif(10)
> x3 <- runif(10)
# generating |
52,041 | Orthogonalized regression reference? | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
This is the Frisch Waugh Lovell theorem in action | Orthogonalized regression reference? | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| Orthogonalized regression reference?
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
This is the Frisc... | Orthogonalized regression reference?
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
|
52,042 | Orthogonalized regression reference? | Ruud's An Introduction to Classical Econometric Theory rides that FWL pony about as far as possible. It's a really interesting geometric take on regression. | Orthogonalized regression reference? | Ruud's An Introduction to Classical Econometric Theory rides that FWL pony about as far as possible. It's a really interesting geometric take on regression. | Orthogonalized regression reference?
Ruud's An Introduction to Classical Econometric Theory rides that FWL pony about as far as possible. It's a really interesting geometric take on regression. | Orthogonalized regression reference?
Ruud's An Introduction to Classical Econometric Theory rides that FWL pony about as far as possible. It's a really interesting geometric take on regression. |
52,043 | Orthogonalized regression reference? | Model can be reparametrized in such a way that two new likelihood equations emerge, each with just one unknown parameter. This will facilitate solving the likelihood equations and also help the general interpretation and use of regression models. (7.2.2 in [hendry2007econometric])
Suppose you want to reparametrize the... | Orthogonalized regression reference? | Model can be reparametrized in such a way that two new likelihood equations emerge, each with just one unknown parameter. This will facilitate solving the likelihood equations and also help the genera | Orthogonalized regression reference?
Model can be reparametrized in such a way that two new likelihood equations emerge, each with just one unknown parameter. This will facilitate solving the likelihood equations and also help the general interpretation and use of regression models. (7.2.2 in [hendry2007econometric])
... | Orthogonalized regression reference?
Model can be reparametrized in such a way that two new likelihood equations emerge, each with just one unknown parameter. This will facilitate solving the likelihood equations and also help the genera |
52,044 | Using control variates & antithetic method with Monte Carlo | There is no one way to implement either control or antithetic variates, however, a couple of examples may help.
Antithetic variables: Imagine instead of the random number generator you actually used, you generated a $U(0,1)$ variate, call it $u$, and ran it through the inverse CDF of the Weibull(1,5) distribution, th... | Using control variates & antithetic method with Monte Carlo | There is no one way to implement either control or antithetic variates, however, a couple of examples may help.
Antithetic variables: Imagine instead of the random number generator you actually used | Using control variates & antithetic method with Monte Carlo
There is no one way to implement either control or antithetic variates, however, a couple of examples may help.
Antithetic variables: Imagine instead of the random number generator you actually used, you generated a $U(0,1)$ variate, call it $u$, and ran it ... | Using control variates & antithetic method with Monte Carlo
There is no one way to implement either control or antithetic variates, however, a couple of examples may help.
Antithetic variables: Imagine instead of the random number generator you actually used |
52,045 | Using control variates & antithetic method with Monte Carlo | As correctly explained by jbowman, you have to create a negative correlation between your simulations to implement antithetic variables: with a generic wblrndfunction this is not possible. You thus have to get back to the definition of a Weibull $\mathcal{W}(\lambda,k)$, which is a scale transform of the $k$th power of... | Using control variates & antithetic method with Monte Carlo | As correctly explained by jbowman, you have to create a negative correlation between your simulations to implement antithetic variables: with a generic wblrndfunction this is not possible. You thus ha | Using control variates & antithetic method with Monte Carlo
As correctly explained by jbowman, you have to create a negative correlation between your simulations to implement antithetic variables: with a generic wblrndfunction this is not possible. You thus have to get back to the definition of a Weibull $\mathcal{W}(\... | Using control variates & antithetic method with Monte Carlo
As correctly explained by jbowman, you have to create a negative correlation between your simulations to implement antithetic variables: with a generic wblrndfunction this is not possible. You thus ha |
52,046 | Splitting a numeric column for a dataframe | df_split<- strsplit(as.character(df$position), split=":")
df <- transform(df, seq_name= sapply(df_split, "[[", 1),pos2= sapply(df_split, "[[", 2))
>
> df
name position pos seq_name pos2
1 HLA 1:1-15 1:1-15 1 1-15
2 HLA 1:2-16 1:2-16 1 2-16
3 HLA 1:3-17 1:3-17 1 3-17 | Splitting a numeric column for a dataframe | df_split<- strsplit(as.character(df$position), split=":")
df <- transform(df, seq_name= sapply(df_split, "[[", 1),pos2= sapply(df_split, "[[", 2))
>
> df
name position pos seq_name pos2
1 HLA | Splitting a numeric column for a dataframe
df_split<- strsplit(as.character(df$position), split=":")
df <- transform(df, seq_name= sapply(df_split, "[[", 1),pos2= sapply(df_split, "[[", 2))
>
> df
name position pos seq_name pos2
1 HLA 1:1-15 1:1-15 1 1-15
2 HLA 1:2-16 1:2-16 1 2-16
3 HLA 1:... | Splitting a numeric column for a dataframe
df_split<- strsplit(as.character(df$position), split=":")
df <- transform(df, seq_name= sapply(df_split, "[[", 1),pos2= sapply(df_split, "[[", 2))
>
> df
name position pos seq_name pos2
1 HLA |
52,047 | Splitting a numeric column for a dataframe | Here is a one line method using tidyr.separate():
library(tidyr)
df <- separate(df, position, into = c("seq","position"), sep = ":", extra = "merge") | Splitting a numeric column for a dataframe | Here is a one line method using tidyr.separate():
library(tidyr)
df <- separate(df, position, into = c("seq","position"), sep = ":", extra = "merge") | Splitting a numeric column for a dataframe
Here is a one line method using tidyr.separate():
library(tidyr)
df <- separate(df, position, into = c("seq","position"), sep = ":", extra = "merge") | Splitting a numeric column for a dataframe
Here is a one line method using tidyr.separate():
library(tidyr)
df <- separate(df, position, into = c("seq","position"), sep = ":", extra = "merge") |
52,048 | Splitting a numeric column for a dataframe | The "trick" is to use do.call.
> a <- data.frame(x = c("1:1-15", "1:2-16", "1:3-17"))
> a
x
1 1:1-15
2 1:2-16
3 1:3-17
> a$x <- as.character(a$x)
> a.split <- strsplit(a$x, split = ":")
> tmp <-do.call(rbind, a.split)
> data.frame(a, tmp)
x X1 X2
1 1:1-15 1 1-15
2 1:2-16 1 2-16
3 1:3-17 1 3-17 | Splitting a numeric column for a dataframe | The "trick" is to use do.call.
> a <- data.frame(x = c("1:1-15", "1:2-16", "1:3-17"))
> a
x
1 1:1-15
2 1:2-16
3 1:3-17
> a$x <- as.character(a$x)
> a.split <- strsplit(a$x, split = ":")
> tmp < | Splitting a numeric column for a dataframe
The "trick" is to use do.call.
> a <- data.frame(x = c("1:1-15", "1:2-16", "1:3-17"))
> a
x
1 1:1-15
2 1:2-16
3 1:3-17
> a$x <- as.character(a$x)
> a.split <- strsplit(a$x, split = ":")
> tmp <-do.call(rbind, a.split)
> data.frame(a, tmp)
x X1 X2
1 1:1-15 1 1-... | Splitting a numeric column for a dataframe
The "trick" is to use do.call.
> a <- data.frame(x = c("1:1-15", "1:2-16", "1:3-17"))
> a
x
1 1:1-15
2 1:2-16
3 1:3-17
> a$x <- as.character(a$x)
> a.split <- strsplit(a$x, split = ":")
> tmp < |
52,049 | Increasing Exam Expected Mark | First a couple of assumptions:
1. All marks are equally likely.
1. If you guess your mark to be 95 and you get 95, your return mark is 100 not
105.
1. Similarly, if your exam mark is 1 and you guess 50 (say), then your return
mark is 0 not -4.
1. I'm only considering discrete marks, that is, values 0, ..., 100.
S... | Increasing Exam Expected Mark | First a couple of assumptions:
1. All marks are equally likely.
1. If you guess your mark to be 95 and you get 95, your return mark is 100 not
105.
1. Similarly, if your exam mark is 1 and you gues | Increasing Exam Expected Mark
First a couple of assumptions:
1. All marks are equally likely.
1. If you guess your mark to be 95 and you get 95, your return mark is 100 not
105.
1. Similarly, if your exam mark is 1 and you guess 50 (say), then your return
mark is 0 not -4.
1. I'm only considering discrete marks, ... | Increasing Exam Expected Mark
First a couple of assumptions:
1. All marks are equally likely.
1. If you guess your mark to be 95 and you get 95, your return mark is 100 not
105.
1. Similarly, if your exam mark is 1 and you gues |
52,050 | Increasing Exam Expected Mark | I'm not sure if this would be a funny game or your professor is mildly sadistic. It would be torturous for students who are right on the edge of passing (which we may expect them to be the worst guessers!) Sorry not an answer but I couldn't help myself. | Increasing Exam Expected Mark | I'm not sure if this would be a funny game or your professor is mildly sadistic. It would be torturous for students who are right on the edge of passing (which we may expect them to be the worst guess | Increasing Exam Expected Mark
I'm not sure if this would be a funny game or your professor is mildly sadistic. It would be torturous for students who are right on the edge of passing (which we may expect them to be the worst guessers!) Sorry not an answer but I couldn't help myself. | Increasing Exam Expected Mark
I'm not sure if this would be a funny game or your professor is mildly sadistic. It would be torturous for students who are right on the edge of passing (which we may expect them to be the worst guess |
52,051 | Increasing Exam Expected Mark | Use the bootstrap! Take lots of practice exams and estimate what your score will be on the real exam. If it does not improve your estimate, it will probably be good preparation! | Increasing Exam Expected Mark | Use the bootstrap! Take lots of practice exams and estimate what your score will be on the real exam. If it does not improve your estimate, it will probably be good preparation! | Increasing Exam Expected Mark
Use the bootstrap! Take lots of practice exams and estimate what your score will be on the real exam. If it does not improve your estimate, it will probably be good preparation! | Increasing Exam Expected Mark
Use the bootstrap! Take lots of practice exams and estimate what your score will be on the real exam. If it does not improve your estimate, it will probably be good preparation! |
52,052 | R-squared result in linear regression and "unexplained variance" | $R^2$ is the squared correlation of the OLS prediction $\hat{Y}$ and the DV $Y$. In a multiple regression with three predictors $X_{1}, X_{2}, X_{3}$:
# generate some data
> N <- 100
> X1 <- rnorm(N, 175, 7) # predictor 1
> X2 <- rnorm(N, 30, 8) # predic... | R-squared result in linear regression and "unexplained variance" | $R^2$ is the squared correlation of the OLS prediction $\hat{Y}$ and the DV $Y$. In a multiple regression with three predictors $X_{1}, X_{2}, X_{3}$:
# generate some data
> N <- 100
> X1 <- rnorm(N, | R-squared result in linear regression and "unexplained variance"
$R^2$ is the squared correlation of the OLS prediction $\hat{Y}$ and the DV $Y$. In a multiple regression with three predictors $X_{1}, X_{2}, X_{3}$:
# generate some data
> N <- 100
> X1 <- rnorm(N, 175, 7) # predictor 1
... | R-squared result in linear regression and "unexplained variance"
$R^2$ is the squared correlation of the OLS prediction $\hat{Y}$ and the DV $Y$. In a multiple regression with three predictors $X_{1}, X_{2}, X_{3}$:
# generate some data
> N <- 100
> X1 <- rnorm(N, |
52,053 | R-squared result in linear regression and "unexplained variance" | R^2 is the percent of variance in the DV accounted for by the whole model. That is, your intercept and your IVS combined account for that much of the variance, using the linear regression model.
In your case, you got an R^2 of 0.85, indicating that intercept, plus cdd plus pmax combined account for 85% of the variance... | R-squared result in linear regression and "unexplained variance" | R^2 is the percent of variance in the DV accounted for by the whole model. That is, your intercept and your IVS combined account for that much of the variance, using the linear regression model.
In y | R-squared result in linear regression and "unexplained variance"
R^2 is the percent of variance in the DV accounted for by the whole model. That is, your intercept and your IVS combined account for that much of the variance, using the linear regression model.
In your case, you got an R^2 of 0.85, indicating that inter... | R-squared result in linear regression and "unexplained variance"
R^2 is the percent of variance in the DV accounted for by the whole model. That is, your intercept and your IVS combined account for that much of the variance, using the linear regression model.
In y |
52,054 | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function? | I mistrust all but the lowest-level functions in Excel, and for good reason: many procedures that go beyond simple arithmetic operations have flaws or errors and most of them are poorly documented. This includes all the probability distribution functions.
Numerical flaws are inevitable due to limitations in floating p... | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function? | I mistrust all but the lowest-level functions in Excel, and for good reason: many procedures that go beyond simple arithmetic operations have flaws or errors and most of them are poorly documented. T | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function?
I mistrust all but the lowest-level functions in Excel, and for good reason: many procedures that go beyond simple arithmetic operations have flaws or errors and most of them are poorly documented. This includes all the probability d... | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function?
I mistrust all but the lowest-level functions in Excel, and for good reason: many procedures that go beyond simple arithmetic operations have flaws or errors and most of them are poorly documented. T |
52,055 | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function? | On Microsoft Office for Mac 2008 LOGNORMDIST(5;1;1) gives 0,728882893 in Excel. On R plnorm(5,1,1) gives 0.7288829. In R you need to supply mean and standard deviation on log scale, so it seems that in Excel you need to do the same. | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function? | On Microsoft Office for Mac 2008 LOGNORMDIST(5;1;1) gives 0,728882893 in Excel. On R plnorm(5,1,1) gives 0.7288829. In R you need to supply mean and standard deviation on log scale, so it seems that i | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function?
On Microsoft Office for Mac 2008 LOGNORMDIST(5;1;1) gives 0,728882893 in Excel. On R plnorm(5,1,1) gives 0.7288829. In R you need to supply mean and standard deviation on log scale, so it seems that in Excel you need to do the same. | How to use the LOGNORMALDIST function to generate a Cumulative Distribution Function?
On Microsoft Office for Mac 2008 LOGNORMDIST(5;1;1) gives 0,728882893 in Excel. On R plnorm(5,1,1) gives 0.7288829. In R you need to supply mean and standard deviation on log scale, so it seems that i |
52,056 | Calculate median without access to raw data | The question can be construed as requesting a nonparametric estimator of the median of a sample in the form f(min, mean, max, sd). In this circumstance, by contemplating extreme (two-point) distributions, we can trivially establish that
$$ 2\ \text{mean} - \text{max} \le \text{median} \le 2\ \text{mean} - \text{min}.$... | Calculate median without access to raw data | The question can be construed as requesting a nonparametric estimator of the median of a sample in the form f(min, mean, max, sd). In this circumstance, by contemplating extreme (two-point) distribut | Calculate median without access to raw data
The question can be construed as requesting a nonparametric estimator of the median of a sample in the form f(min, mean, max, sd). In this circumstance, by contemplating extreme (two-point) distributions, we can trivially establish that
$$ 2\ \text{mean} - \text{max} \le \te... | Calculate median without access to raw data
The question can be construed as requesting a nonparametric estimator of the median of a sample in the form f(min, mean, max, sd). In this circumstance, by contemplating extreme (two-point) distribut |
52,057 | Calculate median without access to raw data | If you know underlying distribution of the data, you can.
For example, for normal distributed data, the mean and median are same (median=mode=mean).
Or for exponential distribution with mean $\lambda^{-1}$ the median is $\lambda^{-1} ln(2)$.
it is impossible to obtain median without having raw data or knowing the actu... | Calculate median without access to raw data | If you know underlying distribution of the data, you can.
For example, for normal distributed data, the mean and median are same (median=mode=mean).
Or for exponential distribution with mean $\lambda | Calculate median without access to raw data
If you know underlying distribution of the data, you can.
For example, for normal distributed data, the mean and median are same (median=mode=mean).
Or for exponential distribution with mean $\lambda^{-1}$ the median is $\lambda^{-1} ln(2)$.
it is impossible to obtain median... | Calculate median without access to raw data
If you know underlying distribution of the data, you can.
For example, for normal distributed data, the mean and median are same (median=mode=mean).
Or for exponential distribution with mean $\lambda |
52,058 | Difference between Excel's RAND(), RAND()*RAND(), etc | Standardization is good, but it's not the right standardization for this situation. It helps to see that multiplying values of RAND() is the same as adding their logarithms (followed by a subsequent exponentiation). Because the different calls to RAND() are supposed to be independent, those logarithms are still indep... | Difference between Excel's RAND(), RAND()*RAND(), etc | Standardization is good, but it's not the right standardization for this situation. It helps to see that multiplying values of RAND() is the same as adding their logarithms (followed by a subsequent | Difference between Excel's RAND(), RAND()*RAND(), etc
Standardization is good, but it's not the right standardization for this situation. It helps to see that multiplying values of RAND() is the same as adding their logarithms (followed by a subsequent exponentiation). Because the different calls to RAND() are suppo... | Difference between Excel's RAND(), RAND()*RAND(), etc
Standardization is good, but it's not the right standardization for this situation. It helps to see that multiplying values of RAND() is the same as adding their logarithms (followed by a subsequent |
52,059 | Difference between Excel's RAND(), RAND()*RAND(), etc | "In Excel, the Rand function returns a random number that is greater than or equal to 0 and less than 1. The Rand function returns a new random number each time your spreadsheet recalculates." -http://www.techonthenet.com/excel/formulas/rand.php
Because RAND() is always less than one and greater than zero, multiplying ... | Difference between Excel's RAND(), RAND()*RAND(), etc | "In Excel, the Rand function returns a random number that is greater than or equal to 0 and less than 1. The Rand function returns a new random number each time your spreadsheet recalculates." -http:/ | Difference between Excel's RAND(), RAND()*RAND(), etc
"In Excel, the Rand function returns a random number that is greater than or equal to 0 and less than 1. The Rand function returns a new random number each time your spreadsheet recalculates." -http://www.techonthenet.com/excel/formulas/rand.php
Because RAND() is a... | Difference between Excel's RAND(), RAND()*RAND(), etc
"In Excel, the Rand function returns a random number that is greater than or equal to 0 and less than 1. The Rand function returns a new random number each time your spreadsheet recalculates." -http:/ |
52,060 | Difference between Excel's RAND(), RAND()*RAND(), etc | I am not sure why your graph has values from -2 to 4 but for what it is worth here is the answer to the text of your question:
Suppose that $U \sim U[0,1]$. Then the cdf of $U$ is given by $F(u) = u$ for $u \in (0,1)$ and 1 otherwise.
When you multiply different iid realizations of the random draws you are essentially ... | Difference between Excel's RAND(), RAND()*RAND(), etc | I am not sure why your graph has values from -2 to 4 but for what it is worth here is the answer to the text of your question:
Suppose that $U \sim U[0,1]$. Then the cdf of $U$ is given by $F(u) = u$ | Difference between Excel's RAND(), RAND()*RAND(), etc
I am not sure why your graph has values from -2 to 4 but for what it is worth here is the answer to the text of your question:
Suppose that $U \sim U[0,1]$. Then the cdf of $U$ is given by $F(u) = u$ for $u \in (0,1)$ and 1 otherwise.
When you multiply different ii... | Difference between Excel's RAND(), RAND()*RAND(), etc
I am not sure why your graph has values from -2 to 4 but for what it is worth here is the answer to the text of your question:
Suppose that $U \sim U[0,1]$. Then the cdf of $U$ is given by $F(u) = u$ |
52,061 | Difference between Excel's RAND(), RAND()*RAND(), etc | There is no mysterious reason. If you multiply a bunch of numbers between 0 an 1, the result will forcibly be close to 0. The average result for RAND()*RAND()*RAND()*RAND()*RAND()*RAND() should be something close to (0.5^6), that is, 0.015625.
Be careful using Excel's RAND() function, though. It's not the best random n... | Difference between Excel's RAND(), RAND()*RAND(), etc | There is no mysterious reason. If you multiply a bunch of numbers between 0 an 1, the result will forcibly be close to 0. The average result for RAND()*RAND()*RAND()*RAND()*RAND()*RAND() should be som | Difference between Excel's RAND(), RAND()*RAND(), etc
There is no mysterious reason. If you multiply a bunch of numbers between 0 an 1, the result will forcibly be close to 0. The average result for RAND()*RAND()*RAND()*RAND()*RAND()*RAND() should be something close to (0.5^6), that is, 0.015625.
Be careful using Exce... | Difference between Excel's RAND(), RAND()*RAND(), etc
There is no mysterious reason. If you multiply a bunch of numbers between 0 an 1, the result will forcibly be close to 0. The average result for RAND()*RAND()*RAND()*RAND()*RAND()*RAND() should be som |
52,062 | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gumbel distribution? | The NIST page on Gumbel distributions shows the method of moments estimators for the parameters of both the maximum and minimum extreme-value distributions. Those are easily calculated and should provide reliable initial estimates.
The parameterization of the density function for the minimum extreme value distribution ... | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gum | The NIST page on Gumbel distributions shows the method of moments estimators for the parameters of both the maximum and minimum extreme-value distributions. Those are easily calculated and should prov | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gumbel distribution?
The NIST page on Gumbel distributions shows the method of moments estimators for the parameters of both the maximum and minimum extreme-value distributions. Those are easily calculated and should provid... | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gum
The NIST page on Gumbel distributions shows the method of moments estimators for the parameters of both the maximum and minimum extreme-value distributions. Those are easily calculated and should prov |
52,063 | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gumbel distribution? | Putting EdM's answer into code, which seems to work well and is very concise:
library(evd)
library(fitdistrplus)
library(survival)
time <- seq(0, 1000, by = 1)
deathTime <- lung$time[lung$status == 2]
scale_est <- (sd(deathTime)*sqrt(6))/pi
loc_est <- mean(deathTime) + 0.5772157*scale_est
fitGum <- fitdist(deathTime... | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gum | Putting EdM's answer into code, which seems to work well and is very concise:
library(evd)
library(fitdistrplus)
library(survival)
time <- seq(0, 1000, by = 1)
deathTime <- lung$time[lung$status == 2 | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gumbel distribution?
Putting EdM's answer into code, which seems to work well and is very concise:
library(evd)
library(fitdistrplus)
library(survival)
time <- seq(0, 1000, by = 1)
deathTime <- lung$time[lung$status == 2]
... | Is there a clean way to derive the start parameters for running the `fitdist()` function for the Gum
Putting EdM's answer into code, which seems to work well and is very concise:
library(evd)
library(fitdistrplus)
library(survival)
time <- seq(0, 1000, by = 1)
deathTime <- lung$time[lung$status == 2 |
52,064 | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance? | The result that we obtain from linear regression is a function of random variables, so the parameters are random variables. You can calculate variance for any random variable. The variance tells us how uncertain the estimates are (in absolutely noiseless data, or with an overfitting model, the variances would be zeros)... | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance? | The result that we obtain from linear regression is a function of random variables, so the parameters are random variables. You can calculate variance for any random variable. The variance tells us ho | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance?
The result that we obtain from linear regression is a function of random variables, so the parameters are random variables. You can calculate variance for any random variable. The variance tells us how uncertain the estimates are (in absolutely n... | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance?
The result that we obtain from linear regression is a function of random variables, so the parameters are random variables. You can calculate variance for any random variable. The variance tells us ho |
52,065 | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance? | The data you are feeding into your OLS model are random draws from some underlying population. You could either be drawing from the joint distribution of the predictors and the outcome, or from the distribution of the outcome conditional on the predictors (so you consider the predictors fixed - this is the assumption i... | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance? | The data you are feeding into your OLS model are random draws from some underlying population. You could either be drawing from the joint distribution of the predictors and the outcome, or from the di | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance?
The data you are feeding into your OLS model are random draws from some underlying population. You could either be drawing from the joint distribution of the predictors and the outcome, or from the distribution of the outcome conditional on the p... | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance?
The data you are feeding into your OLS model are random draws from some underlying population. You could either be drawing from the joint distribution of the predictors and the outcome, or from the di |
52,066 | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance? | The other answers are correct, but I think it might be helpful to simulate what is happening.
library(ggplot2)
set.seed(2023)
# Define sample size
#
N <- 100
# Define number of times to repeat the simulation
#
R <- 1000
# Fix values of x
#
x <- seq(0, 1, 1/(N - 1))
# Define conditional expected values of y as E[y|x... | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance? | The other answers are correct, but I think it might be helpful to simulate what is happening.
library(ggplot2)
set.seed(2023)
# Define sample size
#
N <- 100
# Define number of times to repeat the s | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance?
The other answers are correct, but I think it might be helpful to simulate what is happening.
library(ggplot2)
set.seed(2023)
# Define sample size
#
N <- 100
# Define number of times to repeat the simulation
#
R <- 1000
# Fix values of x
#
x <... | What does it mean for $\hat\beta_1$ and $\hat\beta_0$ to have a variance?
The other answers are correct, but I think it might be helpful to simulate what is happening.
library(ggplot2)
set.seed(2023)
# Define sample size
#
N <- 100
# Define number of times to repeat the s |
52,067 | Is a data size in a binomial distribution random variable? | One can well think of situations in which $N$ is random, even with a frequentist interpretation (there may be a sequence of experiments with different $N$ that can be interpreted as varying randomly). For example, a service center may be interested in what percentage of calls they receive is about a certain product. Th... | Is a data size in a binomial distribution random variable? | One can well think of situations in which $N$ is random, even with a frequentist interpretation (there may be a sequence of experiments with different $N$ that can be interpreted as varying randomly). | Is a data size in a binomial distribution random variable?
One can well think of situations in which $N$ is random, even with a frequentist interpretation (there may be a sequence of experiments with different $N$ that can be interpreted as varying randomly). For example, a service center may be interested in what perc... | Is a data size in a binomial distribution random variable?
One can well think of situations in which $N$ is random, even with a frequentist interpretation (there may be a sequence of experiments with different $N$ that can be interpreted as varying randomly). |
52,068 | Is a data size in a binomial distribution random variable? | Binomial distribution has two parameters: sample size and probability of success. The parameters can either be fixed or be random variables, can be known or need to be estimated. If they are to be estimated, then depending on your preferred approach they can be treated as random variables (Bayesian school of thought) o... | Is a data size in a binomial distribution random variable? | Binomial distribution has two parameters: sample size and probability of success. The parameters can either be fixed or be random variables, can be known or need to be estimated. If they are to be est | Is a data size in a binomial distribution random variable?
Binomial distribution has two parameters: sample size and probability of success. The parameters can either be fixed or be random variables, can be known or need to be estimated. If they are to be estimated, then depending on your preferred approach they can be... | Is a data size in a binomial distribution random variable?
Binomial distribution has two parameters: sample size and probability of success. The parameters can either be fixed or be random variables, can be known or need to be estimated. If they are to be est |
52,069 | What does the I operator stand for in the context of time series modeling? | It is the identity operator, $IX_t=X_t$, and is typically used in ARIMA type formulas where you also have the backshift operator $B$ (sometimes people use $\nabla$ for the backshift), or polynomials in $B$. Essentially, it's a way to make notation more compact.
For instance, if you have a time series $(X_t)$, then
$$(I... | What does the I operator stand for in the context of time series modeling? | It is the identity operator, $IX_t=X_t$, and is typically used in ARIMA type formulas where you also have the backshift operator $B$ (sometimes people use $\nabla$ for the backshift), or polynomials i | What does the I operator stand for in the context of time series modeling?
It is the identity operator, $IX_t=X_t$, and is typically used in ARIMA type formulas where you also have the backshift operator $B$ (sometimes people use $\nabla$ for the backshift), or polynomials in $B$. Essentially, it's a way to make notati... | What does the I operator stand for in the context of time series modeling?
It is the identity operator, $IX_t=X_t$, and is typically used in ARIMA type formulas where you also have the backshift operator $B$ (sometimes people use $\nabla$ for the backshift), or polynomials i |
52,070 | What does the I operator stand for in the context of time series modeling? | The main answer by Stephan is correct, but it is odd to use the identity operator (with the symbol $I$) in a scalar context instead of just using the number one. It would be simpler here if they just said that:
$$(1-B)X_t = X_t-X_{t-1}.$$ | What does the I operator stand for in the context of time series modeling? | The main answer by Stephan is correct, but it is odd to use the identity operator (with the symbol $I$) in a scalar context instead of just using the number one. It would be simpler here if they just | What does the I operator stand for in the context of time series modeling?
The main answer by Stephan is correct, but it is odd to use the identity operator (with the symbol $I$) in a scalar context instead of just using the number one. It would be simpler here if they just said that:
$$(1-B)X_t = X_t-X_{t-1}.$$ | What does the I operator stand for in the context of time series modeling?
The main answer by Stephan is correct, but it is odd to use the identity operator (with the symbol $I$) in a scalar context instead of just using the number one. It would be simpler here if they just |
52,071 | Why is error of OLS not zero? | The major issue in your proof is $ (X^TX)^{-1}X^T = X^T(XX^T)^{-1} $
One obvious proof of the statement does not hold is by choosing a random matrix and then compute value on both sides. You would find they are different.
(I try X= np.array([[1,2,3],[4,5,6]])).
The reason of your proof goes wrong is because when you pe... | Why is error of OLS not zero? | The major issue in your proof is $ (X^TX)^{-1}X^T = X^T(XX^T)^{-1} $
One obvious proof of the statement does not hold is by choosing a random matrix and then compute value on both sides. You would fin | Why is error of OLS not zero?
The major issue in your proof is $ (X^TX)^{-1}X^T = X^T(XX^T)^{-1} $
One obvious proof of the statement does not hold is by choosing a random matrix and then compute value on both sides. You would find they are different.
(I try X= np.array([[1,2,3],[4,5,6]])).
The reason of your proof goe... | Why is error of OLS not zero?
The major issue in your proof is $ (X^TX)^{-1}X^T = X^T(XX^T)^{-1} $
One obvious proof of the statement does not hold is by choosing a random matrix and then compute value on both sides. You would fin |
52,072 | Why is error of OLS not zero? | How do you go from 3b to 3c?
You seem to be using the following $$(V\Sigma\Sigma V^T)^{-1}V\Sigma = V\Sigma^{-1}$$ But, that is only allowed when $\Sigma$ is square (ie when $n=p$). In that case $$(V\Sigma\Sigma V^T)^{-1} = (V^T)^{-1}\Sigma^{-1}\Sigma^{-1} V^{-1}$$
For $p=n$ it is correct that your cost function is zer... | Why is error of OLS not zero? | How do you go from 3b to 3c?
You seem to be using the following $$(V\Sigma\Sigma V^T)^{-1}V\Sigma = V\Sigma^{-1}$$ But, that is only allowed when $\Sigma$ is square (ie when $n=p$). In that case $$(V\ | Why is error of OLS not zero?
How do you go from 3b to 3c?
You seem to be using the following $$(V\Sigma\Sigma V^T)^{-1}V\Sigma = V\Sigma^{-1}$$ But, that is only allowed when $\Sigma$ is square (ie when $n=p$). In that case $$(V\Sigma\Sigma V^T)^{-1} = (V^T)^{-1}\Sigma^{-1}\Sigma^{-1} V^{-1}$$
For $p=n$ it is correct ... | Why is error of OLS not zero?
How do you go from 3b to 3c?
You seem to be using the following $$(V\Sigma\Sigma V^T)^{-1}V\Sigma = V\Sigma^{-1}$$ But, that is only allowed when $\Sigma$ is square (ie when $n=p$). In that case $$(V\ |
52,073 | Is the probability of a continuous variable obtained via integrating over an interval of the probability density curve *cumulative* probability? | For easier reading, I have combined three extensive Comments (now deleted) into
an Answer:
You don't have the true PDF $f(x)$ from density in R. From the code, we know $X$ is standard normal, so the exact value of $p=P(0<X<1)$ could be found from in R as $0.3413447.$
diff(pnorm(c(0,1)))
[1] 0.3413447
However, I suppos... | Is the probability of a continuous variable obtained via integrating over an interval of the probabi | For easier reading, I have combined three extensive Comments (now deleted) into
an Answer:
You don't have the true PDF $f(x)$ from density in R. From the code, we know $X$ is standard normal, so the e | Is the probability of a continuous variable obtained via integrating over an interval of the probability density curve *cumulative* probability?
For easier reading, I have combined three extensive Comments (now deleted) into
an Answer:
You don't have the true PDF $f(x)$ from density in R. From the code, we know $X$ is ... | Is the probability of a continuous variable obtained via integrating over an interval of the probabi
For easier reading, I have combined three extensive Comments (now deleted) into
an Answer:
You don't have the true PDF $f(x)$ from density in R. From the code, we know $X$ is standard normal, so the e |
52,074 | Is the probability of a continuous variable obtained via integrating over an interval of the probability density curve *cumulative* probability? | The cumulative density function (cdf, $F(x)$) and the probability density function (pdf, $f(x)$) are related by the equation:
$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(x)dx$
By the fundamental theorem of calculus, we have that $\frac{\partial}{\partial x}F(x) = f(x)$
(We can replace $-\infty$ with the lower limit of t... | Is the probability of a continuous variable obtained via integrating over an interval of the probabi | The cumulative density function (cdf, $F(x)$) and the probability density function (pdf, $f(x)$) are related by the equation:
$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(x)dx$
By the fundamental theore | Is the probability of a continuous variable obtained via integrating over an interval of the probability density curve *cumulative* probability?
The cumulative density function (cdf, $F(x)$) and the probability density function (pdf, $f(x)$) are related by the equation:
$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(x)dx$
... | Is the probability of a continuous variable obtained via integrating over an interval of the probabi
The cumulative density function (cdf, $F(x)$) and the probability density function (pdf, $f(x)$) are related by the equation:
$F(x) = P(X \leq x) = \int_{-\infty}^{x} f(x)dx$
By the fundamental theore |
52,075 | Which statistical tests can I conduct to analyse the trend of series data? | The plot itself is perhaps the best way to present the tendency.
Consider supplementing it with a robust visual indication of trend, such as a lightly colored line or curve. Building on psychometric principles (lightly and with some diffidence), I would favor an exponential curve determined by, say, the median value... | Which statistical tests can I conduct to analyse the trend of series data? | The plot itself is perhaps the best way to present the tendency.
Consider supplementing it with a robust visual indication of trend, such as a lightly colored line or curve. Building on psychometri | Which statistical tests can I conduct to analyse the trend of series data?
The plot itself is perhaps the best way to present the tendency.
Consider supplementing it with a robust visual indication of trend, such as a lightly colored line or curve. Building on psychometric principles (lightly and with some diffidenc... | Which statistical tests can I conduct to analyse the trend of series data?
The plot itself is perhaps the best way to present the tendency.
Consider supplementing it with a robust visual indication of trend, such as a lightly colored line or curve. Building on psychometri |
52,076 | Which statistical tests can I conduct to analyse the trend of series data? | While your question variables are categorical, they could also be treated as ordinal, since they are done in sequence, so there is a natural ordering of the questions. In that case something like Spearman's rho correlation coefficient would be... ok. | Which statistical tests can I conduct to analyse the trend of series data? | While your question variables are categorical, they could also be treated as ordinal, since they are done in sequence, so there is a natural ordering of the questions. In that case something like Spea | Which statistical tests can I conduct to analyse the trend of series data?
While your question variables are categorical, they could also be treated as ordinal, since they are done in sequence, so there is a natural ordering of the questions. In that case something like Spearman's rho correlation coefficient would be..... | Which statistical tests can I conduct to analyse the trend of series data?
While your question variables are categorical, they could also be treated as ordinal, since they are done in sequence, so there is a natural ordering of the questions. In that case something like Spea |
52,077 | Which statistical tests can I conduct to analyse the trend of series data? | Caveat The OP presents an interesting experiment that produced (up to) 200x20 = 4000 measurements. It's best to analyze the data at the student level, not the 20 averages per question, using for example spline regression as the averages don't follow a simple trend and the variances don't look constant either. That bein... | Which statistical tests can I conduct to analyse the trend of series data? | Caveat The OP presents an interesting experiment that produced (up to) 200x20 = 4000 measurements. It's best to analyze the data at the student level, not the 20 averages per question, using for examp | Which statistical tests can I conduct to analyse the trend of series data?
Caveat The OP presents an interesting experiment that produced (up to) 200x20 = 4000 measurements. It's best to analyze the data at the student level, not the 20 averages per question, using for example spline regression as the averages don't fo... | Which statistical tests can I conduct to analyse the trend of series data?
Caveat The OP presents an interesting experiment that produced (up to) 200x20 = 4000 measurements. It's best to analyze the data at the student level, not the 20 averages per question, using for examp |
52,078 | Which statistical tests can I conduct to analyse the trend of series data? | I've written a second answer inspired by a newer question which asks how fit a nonlinear model for tree growth: nls() singular gradient matrix at initial parameter estimates error.
On the surface the study of how trees grow doesn't have much in common with the study of how students learn. However, if we flip a growth c... | Which statistical tests can I conduct to analyse the trend of series data? | I've written a second answer inspired by a newer question which asks how fit a nonlinear model for tree growth: nls() singular gradient matrix at initial parameter estimates error.
On the surface the | Which statistical tests can I conduct to analyse the trend of series data?
I've written a second answer inspired by a newer question which asks how fit a nonlinear model for tree growth: nls() singular gradient matrix at initial parameter estimates error.
On the surface the study of how trees grow doesn't have much in ... | Which statistical tests can I conduct to analyse the trend of series data?
I've written a second answer inspired by a newer question which asks how fit a nonlinear model for tree growth: nls() singular gradient matrix at initial parameter estimates error.
On the surface the |
52,079 | How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth? | Let's do the calculations to see what the answers are.
By changing the units of measurement of $x$ to the origin $x_0$ we may assume $x_0=0$ (to simplify the work and the notation) and--therefore--the middle of the curve is at $x=0.$ Thus
$$\frac{\mathrm{d}}{\mathrm{d}x} \log(f(x)) = \frac{\mathrm{d}}{\mathrm{d}x}\lef... | How does logistic growth rate coincide with the slope of the line in the exponential phase of the gr | Let's do the calculations to see what the answers are.
By changing the units of measurement of $x$ to the origin $x_0$ we may assume $x_0=0$ (to simplify the work and the notation) and--therefore--the | How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth?
Let's do the calculations to see what the answers are.
By changing the units of measurement of $x$ to the origin $x_0$ we may assume $x_0=0$ (to simplify the work and the notation) and--therefore--the middle of the... | How does logistic growth rate coincide with the slope of the line in the exponential phase of the gr
Let's do the calculations to see what the answers are.
By changing the units of measurement of $x$ to the origin $x_0$ we may assume $x_0=0$ (to simplify the work and the notation) and--therefore--the |
52,080 | How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth? | Here are some exponential curves:
$e^x$ in cyan
$e^{2x}$ in pink
$1-e^{-x}$ in green
$1-e^{-2x}$ in orange
In a sense on the left the pink curve has twice the exponential growth rate of the cyan curve, and symmetrically on the right the orange curve has twice the exponential growth rate of the green curve though in ... | How does logistic growth rate coincide with the slope of the line in the exponential phase of the gr | Here are some exponential curves:
$e^x$ in cyan
$e^{2x}$ in pink
$1-e^{-x}$ in green
$1-e^{-2x}$ in orange
In a sense on the left the pink curve has twice the exponential growth rate of the cyan cu | How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth?
Here are some exponential curves:
$e^x$ in cyan
$e^{2x}$ in pink
$1-e^{-x}$ in green
$1-e^{-2x}$ in orange
In a sense on the left the pink curve has twice the exponential growth rate of the cyan curve, and symme... | How does logistic growth rate coincide with the slope of the line in the exponential phase of the gr
Here are some exponential curves:
$e^x$ in cyan
$e^{2x}$ in pink
$1-e^{-x}$ in green
$1-e^{-2x}$ in orange
In a sense on the left the pink curve has twice the exponential growth rate of the cyan cu |
52,081 | How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth? | If L = 1 and one transforms to the logit (log odds) scale $p(x)$ then the slope is exactly k. The right hand side of the first equals is by definition of log odds and the second is by algebraic simplification.
$p(x) = log(\frac{f(x)}{1-f(x)}) = k(x - x_0)$
Now
$dp(x)/dx = k$
This is the usual interpretation of k used ... | How does logistic growth rate coincide with the slope of the line in the exponential phase of the gr | If L = 1 and one transforms to the logit (log odds) scale $p(x)$ then the slope is exactly k. The right hand side of the first equals is by definition of log odds and the second is by algebraic simpl | How does logistic growth rate coincide with the slope of the line in the exponential phase of the growth?
If L = 1 and one transforms to the logit (log odds) scale $p(x)$ then the slope is exactly k. The right hand side of the first equals is by definition of log odds and the second is by algebraic simplification.
$p(... | How does logistic growth rate coincide with the slope of the line in the exponential phase of the gr
If L = 1 and one transforms to the logit (log odds) scale $p(x)$ then the slope is exactly k. The right hand side of the first equals is by definition of log odds and the second is by algebraic simpl |
52,082 | Is classification only a machine learning problem? | You are exactly right. Machine/statistical learning is one approach to classification, but not the only one. Simple rules created by humans are probably more common in computer programs than ones created by ML. | Is classification only a machine learning problem? | You are exactly right. Machine/statistical learning is one approach to classification, but not the only one. Simple rules created by humans are probably more common in computer programs than ones crea | Is classification only a machine learning problem?
You are exactly right. Machine/statistical learning is one approach to classification, but not the only one. Simple rules created by humans are probably more common in computer programs than ones created by ML. | Is classification only a machine learning problem?
You are exactly right. Machine/statistical learning is one approach to classification, but not the only one. Simple rules created by humans are probably more common in computer programs than ones crea |
52,083 | Is classification only a machine learning problem? | Actually, classification methodology has been around in classical probability and statistics for the better part of a century, well before "machine learning" was a deal. See, e.g., the classic multivariate analysis text by Johnson and Wichern, Applied Multivariate Statistical Analysis, 6th Edition, Section IV. CLASSI... | Is classification only a machine learning problem? | Actually, classification methodology has been around in classical probability and statistics for the better part of a century, well before "machine learning" was a deal. See, e.g., the classic multiv | Is classification only a machine learning problem?
Actually, classification methodology has been around in classical probability and statistics for the better part of a century, well before "machine learning" was a deal. See, e.g., the classic multivariate analysis text by Johnson and Wichern, Applied Multivariate Sta... | Is classification only a machine learning problem?
Actually, classification methodology has been around in classical probability and statistics for the better part of a century, well before "machine learning" was a deal. See, e.g., the classic multiv |
52,084 | Expectation of conditional uniform variates [duplicate] | No, the event $X_1>X_2$ provides some information. Consider a more general case where you have independent $X_1,..,X_n$ and the event $\bigcap_{i=2}^n X_1>X_i$, surely you have the right to suspect that $X_1$ is closer to $1$ more probably than $0$.
For your question, there are various ways to calculate it, normalise t... | Expectation of conditional uniform variates [duplicate] | No, the event $X_1>X_2$ provides some information. Consider a more general case where you have independent $X_1,..,X_n$ and the event $\bigcap_{i=2}^n X_1>X_i$, surely you have the right to suspect th | Expectation of conditional uniform variates [duplicate]
No, the event $X_1>X_2$ provides some information. Consider a more general case where you have independent $X_1,..,X_n$ and the event $\bigcap_{i=2}^n X_1>X_i$, surely you have the right to suspect that $X_1$ is closer to $1$ more probably than $0$.
For your quest... | Expectation of conditional uniform variates [duplicate]
No, the event $X_1>X_2$ provides some information. Consider a more general case where you have independent $X_1,..,X_n$ and the event $\bigcap_{i=2}^n X_1>X_i$, surely you have the right to suspect th |
52,085 | Expectation of conditional uniform variates [duplicate] | Comment. illustrating @gunes (+1) argument via simulation in R.
set.seed(2021)
X1 = runif(10^6); X2 = runif(10^6)
mean(X1[X1>X2])
[1] 0.6668622 # aprx 2/3
In the figure below, $E(X_1|X_1 >X_2)= 2/3$ is the
the average horizontal value of the blue points.
#smaller samples for clearer figure
x1 = X1[1:30000]; x2 = X2... | Expectation of conditional uniform variates [duplicate] | Comment. illustrating @gunes (+1) argument via simulation in R.
set.seed(2021)
X1 = runif(10^6); X2 = runif(10^6)
mean(X1[X1>X2])
[1] 0.6668622 # aprx 2/3
In the figure below, $E(X_1|X_1 >X_2)= 2/3 | Expectation of conditional uniform variates [duplicate]
Comment. illustrating @gunes (+1) argument via simulation in R.
set.seed(2021)
X1 = runif(10^6); X2 = runif(10^6)
mean(X1[X1>X2])
[1] 0.6668622 # aprx 2/3
In the figure below, $E(X_1|X_1 >X_2)= 2/3$ is the
the average horizontal value of the blue points.
#small... | Expectation of conditional uniform variates [duplicate]
Comment. illustrating @gunes (+1) argument via simulation in R.
set.seed(2021)
X1 = runif(10^6); X2 = runif(10^6)
mean(X1[X1>X2])
[1] 0.6668622 # aprx 2/3
In the figure below, $E(X_1|X_1 >X_2)= 2/3 |
52,086 | One class classifier vs binary classifier | Suppose you are trying to perform two-class classification on faulty and non-faulty machinery data, where each example in the dataset is represented using the feature vector $\mathbf{x} = [x_1 \ x_2]^T$. This could be done as shown below:
Here, the faulty machinery data is represented by the orange area, and the non-f... | One class classifier vs binary classifier | Suppose you are trying to perform two-class classification on faulty and non-faulty machinery data, where each example in the dataset is represented using the feature vector $\mathbf{x} = [x_1 \ x_2]^ | One class classifier vs binary classifier
Suppose you are trying to perform two-class classification on faulty and non-faulty machinery data, where each example in the dataset is represented using the feature vector $\mathbf{x} = [x_1 \ x_2]^T$. This could be done as shown below:
Here, the faulty machinery data is rep... | One class classifier vs binary classifier
Suppose you are trying to perform two-class classification on faulty and non-faulty machinery data, where each example in the dataset is represented using the feature vector $\mathbf{x} = [x_1 \ x_2]^ |
52,087 | One class classifier vs binary classifier | You could, but why would you? There is much more to gain by training a model on both classes, so that your model can learn from both classes and better distinguish between them. Imagine trying to learn to distinguish between cats and dogs, would it help you to see only images of cats, and then try to guess which image ... | One class classifier vs binary classifier | You could, but why would you? There is much more to gain by training a model on both classes, so that your model can learn from both classes and better distinguish between them. Imagine trying to lear | One class classifier vs binary classifier
You could, but why would you? There is much more to gain by training a model on both classes, so that your model can learn from both classes and better distinguish between them. Imagine trying to learn to distinguish between cats and dogs, would it help you to see only images o... | One class classifier vs binary classifier
You could, but why would you? There is much more to gain by training a model on both classes, so that your model can learn from both classes and better distinguish between them. Imagine trying to lear |
52,088 | Probability Mass Function making the Truncated Normal Discrete | The other answer here uses the normal density values at the exact points. Another similar method would be to take the normal probabilities across intervals centred on those points. In the latter case, taking the support to be $X = 1,...,m$ you get:
$$p_X(x) = \frac{\Phi \Big( \frac{x - \mu + 1/2}{\sigma} \Big) - \Phi... | Probability Mass Function making the Truncated Normal Discrete | The other answer here uses the normal density values at the exact points. Another similar method would be to take the normal probabilities across intervals centred on those points. In the latter cas | Probability Mass Function making the Truncated Normal Discrete
The other answer here uses the normal density values at the exact points. Another similar method would be to take the normal probabilities across intervals centred on those points. In the latter case, taking the support to be $X = 1,...,m$ you get:
$$p_X(... | Probability Mass Function making the Truncated Normal Discrete
The other answer here uses the normal density values at the exact points. Another similar method would be to take the normal probabilities across intervals centred on those points. In the latter cas |
52,089 | Probability Mass Function making the Truncated Normal Discrete | When you use the values of the normal density then you get automatically the first four conditions satisfied
$$P(X = x\vert \mu,\sigma,a,b,c) = \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sum_{y \in \Omega} e^{-\frac{(y-\mu)^2}{2\sigma^2}}}$$
where $\Omega = \lbrace a,a+c,a+2c,\dots b-c,c \rbrace$ all the values in the su... | Probability Mass Function making the Truncated Normal Discrete | When you use the values of the normal density then you get automatically the first four conditions satisfied
$$P(X = x\vert \mu,\sigma,a,b,c) = \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sum_{y \in \Ome | Probability Mass Function making the Truncated Normal Discrete
When you use the values of the normal density then you get automatically the first four conditions satisfied
$$P(X = x\vert \mu,\sigma,a,b,c) = \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sum_{y \in \Omega} e^{-\frac{(y-\mu)^2}{2\sigma^2}}}$$
where $\Omega = \... | Probability Mass Function making the Truncated Normal Discrete
When you use the values of the normal density then you get automatically the first four conditions satisfied
$$P(X = x\vert \mu,\sigma,a,b,c) = \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sum_{y \in \Ome |
52,090 | Probability Mass Function making the Truncated Normal Discrete | The most natural way would be as follows:
Let $X$ have support $0,1,...T$, where $T \in \{2,3,...,\}$ (i.e T can be taken as the limit to $\infty$, since the infinite sum can be bounded by the integral over the kernel of a gaussian pdf, but it will be intractable to work with in practice).
$$f_X(X=x;\mu, \sigma) = \fra... | Probability Mass Function making the Truncated Normal Discrete | The most natural way would be as follows:
Let $X$ have support $0,1,...T$, where $T \in \{2,3,...,\}$ (i.e T can be taken as the limit to $\infty$, since the infinite sum can be bounded by the integra | Probability Mass Function making the Truncated Normal Discrete
The most natural way would be as follows:
Let $X$ have support $0,1,...T$, where $T \in \{2,3,...,\}$ (i.e T can be taken as the limit to $\infty$, since the infinite sum can be bounded by the integral over the kernel of a gaussian pdf, but it will be intra... | Probability Mass Function making the Truncated Normal Discrete
The most natural way would be as follows:
Let $X$ have support $0,1,...T$, where $T \in \{2,3,...,\}$ (i.e T can be taken as the limit to $\infty$, since the infinite sum can be bounded by the integra |
52,091 | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | One potential option is to add a tiny bit of random noise to each observation. In that way fewer points will overlap.
You can either add it directly and use R's basic plotting capabilities or look into the jitter type layer that comes with the GGplot package that adds the noise automatically. | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | One potential option is to add a tiny bit of random noise to each observation. In that way fewer points will overlap.
You can either add it directly and use R's basic plotting capabilities or look int | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
One potential option is to add a tiny bit of random noise to each observation. In that way fewer points will overlap.
You can either add it directly and use R's basic plotting capabilities or look into the jitter type layer that comes w... | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
One potential option is to add a tiny bit of random noise to each observation. In that way fewer points will overlap.
You can either add it directly and use R's basic plotting capabilities or look int |
52,092 | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | Mosaic plots are a good way of doing this.
https://cran.r-project.org/web/packages/ggmosaic/vignettes/ggmosaic.html | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | Mosaic plots are a good way of doing this.
https://cran.r-project.org/web/packages/ggmosaic/vignettes/ggmosaic.html | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
Mosaic plots are a good way of doing this.
https://cran.r-project.org/web/packages/ggmosaic/vignettes/ggmosaic.html | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
Mosaic plots are a good way of doing this.
https://cran.r-project.org/web/packages/ggmosaic/vignettes/ggmosaic.html |
52,093 | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | The ggplot2 library should handle something like this. There are example of the specific code out on the internet. I’ll just address the idea, since this is CV.SE, not SO.
I would represent the points in a data frame with three columns. One column would have the x-coordinate, one column would have the y-coordinate, and... | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | The ggplot2 library should handle something like this. There are example of the specific code out on the internet. I’ll just address the idea, since this is CV.SE, not SO.
I would represent the points | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
The ggplot2 library should handle something like this. There are example of the specific code out on the internet. I’ll just address the idea, since this is CV.SE, not SO.
I would represent the points in a data frame with three columns.... | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
The ggplot2 library should handle something like this. There are example of the specific code out on the internet. I’ll just address the idea, since this is CV.SE, not SO.
I would represent the points |
52,094 | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | Similar to what Dave proposes, but in base R: visualize table counts using grayscale, with darker grays for cells with higher counts.
set.seed(1)
nn <- 1e6
aa <- sample(1:10,nn,prob=(1:10)^2-5*(1:10)+20,replace=TRUE)
bb <- sample(1:10,nn,prob=20-(1:10),replace=TRUE)
data_table <- table(aa,bb)
grayscale <- function (... | The best way to plot high amount of discrete data with 2 variables in R [duplicate] | Similar to what Dave proposes, but in base R: visualize table counts using grayscale, with darker grays for cells with higher counts.
set.seed(1)
nn <- 1e6
aa <- sample(1:10,nn,prob=(1:10)^2-5*(1:10) | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
Similar to what Dave proposes, but in base R: visualize table counts using grayscale, with darker grays for cells with higher counts.
set.seed(1)
nn <- 1e6
aa <- sample(1:10,nn,prob=(1:10)^2-5*(1:10)+20,replace=TRUE)
bb <- sample(1:10,... | The best way to plot high amount of discrete data with 2 variables in R [duplicate]
Similar to what Dave proposes, but in base R: visualize table counts using grayscale, with darker grays for cells with higher counts.
set.seed(1)
nn <- 1e6
aa <- sample(1:10,nn,prob=(1:10)^2-5*(1:10) |
52,095 | Why is the correlation between independent variables/regressor and residuals zero for OLS? | In any model with an intercept, the residuals are uncorrelated with the predictors $X$ by construction; this is true whether or not the linear model is a good fit and it has nothing to do with assumptions.
It's important here to distinguish between the residuals and the unobserved things often called the errors.
The co... | Why is the correlation between independent variables/regressor and residuals zero for OLS? | In any model with an intercept, the residuals are uncorrelated with the predictors $X$ by construction; this is true whether or not the linear model is a good fit and it has nothing to do with assumpt | Why is the correlation between independent variables/regressor and residuals zero for OLS?
In any model with an intercept, the residuals are uncorrelated with the predictors $X$ by construction; this is true whether or not the linear model is a good fit and it has nothing to do with assumptions.
It's important here to ... | Why is the correlation between independent variables/regressor and residuals zero for OLS?
In any model with an intercept, the residuals are uncorrelated with the predictors $X$ by construction; this is true whether or not the linear model is a good fit and it has nothing to do with assumpt |
52,096 | Why is the correlation between independent variables/regressor and residuals zero for OLS? | Consider the model
$$Y_i = 3 + 4x_i + e_i,$$
where $e_i \stackrel{iid}{\sim} \mathsf{Norm}(0, \sigma=1).$
A version of this is simulated in R as follows:
set.seed(625)
x = runif(20, 1, 23)
y = 3 + 4*x + rnorm(20, 0, 1)
Of course, one anticipates a linear association between $x_i$ and $Y_i,$
otherwise there is not much... | Why is the correlation between independent variables/regressor and residuals zero for OLS? | Consider the model
$$Y_i = 3 + 4x_i + e_i,$$
where $e_i \stackrel{iid}{\sim} \mathsf{Norm}(0, \sigma=1).$
A version of this is simulated in R as follows:
set.seed(625)
x = runif(20, 1, 23)
y = 3 + 4*x | Why is the correlation between independent variables/regressor and residuals zero for OLS?
Consider the model
$$Y_i = 3 + 4x_i + e_i,$$
where $e_i \stackrel{iid}{\sim} \mathsf{Norm}(0, \sigma=1).$
A version of this is simulated in R as follows:
set.seed(625)
x = runif(20, 1, 23)
y = 3 + 4*x + rnorm(20, 0, 1)
Of course... | Why is the correlation between independent variables/regressor and residuals zero for OLS?
Consider the model
$$Y_i = 3 + 4x_i + e_i,$$
where $e_i \stackrel{iid}{\sim} \mathsf{Norm}(0, \sigma=1).$
A version of this is simulated in R as follows:
set.seed(625)
x = runif(20, 1, 23)
y = 3 + 4*x |
52,097 | Defining continuous random variables via uncountable sets | The problem with both characterizations is that they ignore the underlying probabilities.
Recall that a random variable $X$ is a function that assigns real numbers to elements of the sample space. If a considerable part of the domain of $X$ has no probability, then the range of $X$ may have virtually any property what... | Defining continuous random variables via uncountable sets | The problem with both characterizations is that they ignore the underlying probabilities.
Recall that a random variable $X$ is a function that assigns real numbers to elements of the sample space. If | Defining continuous random variables via uncountable sets
The problem with both characterizations is that they ignore the underlying probabilities.
Recall that a random variable $X$ is a function that assigns real numbers to elements of the sample space. If a considerable part of the domain of $X$ has no probability, ... | Defining continuous random variables via uncountable sets
The problem with both characterizations is that they ignore the underlying probabilities.
Recall that a random variable $X$ is a function that assigns real numbers to elements of the sample space. If |
52,098 | Defining continuous random variables via uncountable sets | Well, even if the range (or support set) of the random variable $X$ is uncountable, $X$ do not necessarily have a density. The answer by @Sebastian mentions measure, and specifically counting measure. But counting measure on an uncountable set isn't very useful, for instance, it is not $\sigma$-finite. So not very usef... | Defining continuous random variables via uncountable sets | Well, even if the range (or support set) of the random variable $X$ is uncountable, $X$ do not necessarily have a density. The answer by @Sebastian mentions measure, and specifically counting measure. | Defining continuous random variables via uncountable sets
Well, even if the range (or support set) of the random variable $X$ is uncountable, $X$ do not necessarily have a density. The answer by @Sebastian mentions measure, and specifically counting measure. But counting measure on an uncountable set isn't very useful,... | Defining continuous random variables via uncountable sets
Well, even if the range (or support set) of the random variable $X$ is uncountable, $X$ do not necessarily have a density. The answer by @Sebastian mentions measure, and specifically counting measure. |
52,099 | Defining continuous random variables via uncountable sets | Consider the example where your sample space $\Omega$ = $\mathbb{R}$. This is uncountably infinite. However whether the RV is continuous depends on the used measure. If you would use $\mu = \#$ (i.e. the counting-measure) you could still easily defined a density with respect to $\mu$ that would induce a discrete distri... | Defining continuous random variables via uncountable sets | Consider the example where your sample space $\Omega$ = $\mathbb{R}$. This is uncountably infinite. However whether the RV is continuous depends on the used measure. If you would use $\mu = \#$ (i.e. | Defining continuous random variables via uncountable sets
Consider the example where your sample space $\Omega$ = $\mathbb{R}$. This is uncountably infinite. However whether the RV is continuous depends on the used measure. If you would use $\mu = \#$ (i.e. the counting-measure) you could still easily defined a density... | Defining continuous random variables via uncountable sets
Consider the example where your sample space $\Omega$ = $\mathbb{R}$. This is uncountably infinite. However whether the RV is continuous depends on the used measure. If you would use $\mu = \#$ (i.e. |
52,100 | Does Length Scale of the Kernel in Gaussian Process directly Relates to Correlation Length? | The Gaussian RBF kernel, also known as the squared exponential or exponentiated quadratic kernel, is
$$
k(x, y) = \exp\left( - \frac{\lVert x - y \rVert^2}{2 \ell^2} \right)
,$$
where $\ell$ is often called the lengthscale. Remember that for $f \sim \mathcal{GP}(0, k)$, the correlation between $f(x)$ and $f(y)$ is exac... | Does Length Scale of the Kernel in Gaussian Process directly Relates to Correlation Length? | The Gaussian RBF kernel, also known as the squared exponential or exponentiated quadratic kernel, is
$$
k(x, y) = \exp\left( - \frac{\lVert x - y \rVert^2}{2 \ell^2} \right)
,$$
where $\ell$ is often | Does Length Scale of the Kernel in Gaussian Process directly Relates to Correlation Length?
The Gaussian RBF kernel, also known as the squared exponential or exponentiated quadratic kernel, is
$$
k(x, y) = \exp\left( - \frac{\lVert x - y \rVert^2}{2 \ell^2} \right)
,$$
where $\ell$ is often called the lengthscale. Reme... | Does Length Scale of the Kernel in Gaussian Process directly Relates to Correlation Length?
The Gaussian RBF kernel, also known as the squared exponential or exponentiated quadratic kernel, is
$$
k(x, y) = \exp\left( - \frac{\lVert x - y \rVert^2}{2 \ell^2} \right)
,$$
where $\ell$ is often |
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