Quantifying dependence of Cauchy random variables
Given two Cauchy random variables $theta_1 sim mathrm{Cauchy}(x_0^{(1)}, gamma^{(1)})$ and $theta_2 sim mathrm{Cauchy}(x_0^{(2)}, gamma^{(2)})$. That are not independent. The dependence structure of random variables can often be quantified with their covariance or correlation coefficient. However, these Cauchy random variables have no moments. Thus, covariance and correlation do not exist.
Are there other ways of representing the dependence of the random variables? Is it possible to estimate those with Monte Carlo?
covariance independence copula heavy-tailed
add a comment |
Given two Cauchy random variables $theta_1 sim mathrm{Cauchy}(x_0^{(1)}, gamma^{(1)})$ and $theta_2 sim mathrm{Cauchy}(x_0^{(2)}, gamma^{(2)})$. That are not independent. The dependence structure of random variables can often be quantified with their covariance or correlation coefficient. However, these Cauchy random variables have no moments. Thus, covariance and correlation do not exist.
Are there other ways of representing the dependence of the random variables? Is it possible to estimate those with Monte Carlo?
covariance independence copula heavy-tailed
2
May consider general dependence metrics such as mutual information: en.wikipedia.org/wiki/Mutual_information
– John Madden
2 hours ago
add a comment |
Given two Cauchy random variables $theta_1 sim mathrm{Cauchy}(x_0^{(1)}, gamma^{(1)})$ and $theta_2 sim mathrm{Cauchy}(x_0^{(2)}, gamma^{(2)})$. That are not independent. The dependence structure of random variables can often be quantified with their covariance or correlation coefficient. However, these Cauchy random variables have no moments. Thus, covariance and correlation do not exist.
Are there other ways of representing the dependence of the random variables? Is it possible to estimate those with Monte Carlo?
covariance independence copula heavy-tailed
Given two Cauchy random variables $theta_1 sim mathrm{Cauchy}(x_0^{(1)}, gamma^{(1)})$ and $theta_2 sim mathrm{Cauchy}(x_0^{(2)}, gamma^{(2)})$. That are not independent. The dependence structure of random variables can often be quantified with their covariance or correlation coefficient. However, these Cauchy random variables have no moments. Thus, covariance and correlation do not exist.
Are there other ways of representing the dependence of the random variables? Is it possible to estimate those with Monte Carlo?
covariance independence copula heavy-tailed
covariance independence copula heavy-tailed
asked 2 hours ago
JonasJonas
46510
46510
2
May consider general dependence metrics such as mutual information: en.wikipedia.org/wiki/Mutual_information
– John Madden
2 hours ago
add a comment |
2
May consider general dependence metrics such as mutual information: en.wikipedia.org/wiki/Mutual_information
– John Madden
2 hours ago
2
2
May consider general dependence metrics such as mutual information: en.wikipedia.org/wiki/Mutual_information
– John Madden
2 hours ago
May consider general dependence metrics such as mutual information: en.wikipedia.org/wiki/Mutual_information
– John Madden
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
Just because they don't have a covariance doesn't mean that the basic $x^tSigma^{-1} x$ structure usually associated with covariances can't be used. In fact, the multivariate ($k$-dimensional) Cauchy can be written as:
$$f({mathbf x}; {mathbfmu},{mathbfSigma}, k)= frac{Gammaleft(frac{1+k}{2}right)}{Gamma(frac{1}{2})pi^{frac{k}{2}}left|{mathbfSigma}right|^{frac{1}{2}}left[1+({mathbf x}-{mathbfmu})^T{mathbfSigma}^{-1}({mathbf x}-{mathbfmu})right]^{frac{1+k}{2}}} $$
which I have lifted from the Wikipedia page. This is just a multivariate Student-$t$ distribution with one degree of freedom.
For the purposes of developing intuition, I would just use the normalized off-diagonal elements of $Sigma$ as if they were correlations, even though they are not. They reflect the strength of the linear relationship between the variables in a way very similar to that of a correlation; $Sigma$ has to be positive definite symmetric; if $Sigma$ is diagonal, the variates are independent, etc.
Maximum likelihood estimation of the parameters can be done using the E-M algorithm, which in this case is easily implemented. The log of the likelihood function is:
$$mathcal{L}(mu, Sigma) = -{nover 2}|Sigma| - {k+1 over 2}sum_{i=1}^nlog(1+s_i)$$
where $s_i = (x_i-mu)^TSigma^{-1}(x_i-mu)$. Differentiating leads to the following simple expressions:
$$mu = sum w_ix_i/sum w_i$$
$$Sigma = {1 over n}sum w_i(x-mu)(x-mu)^T$$
$$w_i = (1+k)/(1+s_i)$$
The E-M algorithm just iterates over these three expressions, substituting the most recent estimates of all the parameters at each step.
For more on this, see Estimation Methods for the Multivariate t Distribution, Nadarajah and Kotz, 2008.
add a comment |
While $text{cov}(X,Y)$ does not exist, for a pair of variates with Cauchy marginals, $text{cov}(Phi(X),Phi(Y))$ does exist for bounded functions $Phi(cdot)$. Borrowing from the concept of copulas, one can turn $X$ and $Y$ into Uniform $(0,1)$ variates, by using their marginal cdfs, $Phi_X(X)$ and $Phi_Y(Y)$, and look at the covariance or correlation of the resulting variates.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f386036%2fquantifying-dependence-of-cauchy-random-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Just because they don't have a covariance doesn't mean that the basic $x^tSigma^{-1} x$ structure usually associated with covariances can't be used. In fact, the multivariate ($k$-dimensional) Cauchy can be written as:
$$f({mathbf x}; {mathbfmu},{mathbfSigma}, k)= frac{Gammaleft(frac{1+k}{2}right)}{Gamma(frac{1}{2})pi^{frac{k}{2}}left|{mathbfSigma}right|^{frac{1}{2}}left[1+({mathbf x}-{mathbfmu})^T{mathbfSigma}^{-1}({mathbf x}-{mathbfmu})right]^{frac{1+k}{2}}} $$
which I have lifted from the Wikipedia page. This is just a multivariate Student-$t$ distribution with one degree of freedom.
For the purposes of developing intuition, I would just use the normalized off-diagonal elements of $Sigma$ as if they were correlations, even though they are not. They reflect the strength of the linear relationship between the variables in a way very similar to that of a correlation; $Sigma$ has to be positive definite symmetric; if $Sigma$ is diagonal, the variates are independent, etc.
Maximum likelihood estimation of the parameters can be done using the E-M algorithm, which in this case is easily implemented. The log of the likelihood function is:
$$mathcal{L}(mu, Sigma) = -{nover 2}|Sigma| - {k+1 over 2}sum_{i=1}^nlog(1+s_i)$$
where $s_i = (x_i-mu)^TSigma^{-1}(x_i-mu)$. Differentiating leads to the following simple expressions:
$$mu = sum w_ix_i/sum w_i$$
$$Sigma = {1 over n}sum w_i(x-mu)(x-mu)^T$$
$$w_i = (1+k)/(1+s_i)$$
The E-M algorithm just iterates over these three expressions, substituting the most recent estimates of all the parameters at each step.
For more on this, see Estimation Methods for the Multivariate t Distribution, Nadarajah and Kotz, 2008.
add a comment |
Just because they don't have a covariance doesn't mean that the basic $x^tSigma^{-1} x$ structure usually associated with covariances can't be used. In fact, the multivariate ($k$-dimensional) Cauchy can be written as:
$$f({mathbf x}; {mathbfmu},{mathbfSigma}, k)= frac{Gammaleft(frac{1+k}{2}right)}{Gamma(frac{1}{2})pi^{frac{k}{2}}left|{mathbfSigma}right|^{frac{1}{2}}left[1+({mathbf x}-{mathbfmu})^T{mathbfSigma}^{-1}({mathbf x}-{mathbfmu})right]^{frac{1+k}{2}}} $$
which I have lifted from the Wikipedia page. This is just a multivariate Student-$t$ distribution with one degree of freedom.
For the purposes of developing intuition, I would just use the normalized off-diagonal elements of $Sigma$ as if they were correlations, even though they are not. They reflect the strength of the linear relationship between the variables in a way very similar to that of a correlation; $Sigma$ has to be positive definite symmetric; if $Sigma$ is diagonal, the variates are independent, etc.
Maximum likelihood estimation of the parameters can be done using the E-M algorithm, which in this case is easily implemented. The log of the likelihood function is:
$$mathcal{L}(mu, Sigma) = -{nover 2}|Sigma| - {k+1 over 2}sum_{i=1}^nlog(1+s_i)$$
where $s_i = (x_i-mu)^TSigma^{-1}(x_i-mu)$. Differentiating leads to the following simple expressions:
$$mu = sum w_ix_i/sum w_i$$
$$Sigma = {1 over n}sum w_i(x-mu)(x-mu)^T$$
$$w_i = (1+k)/(1+s_i)$$
The E-M algorithm just iterates over these three expressions, substituting the most recent estimates of all the parameters at each step.
For more on this, see Estimation Methods for the Multivariate t Distribution, Nadarajah and Kotz, 2008.
add a comment |
Just because they don't have a covariance doesn't mean that the basic $x^tSigma^{-1} x$ structure usually associated with covariances can't be used. In fact, the multivariate ($k$-dimensional) Cauchy can be written as:
$$f({mathbf x}; {mathbfmu},{mathbfSigma}, k)= frac{Gammaleft(frac{1+k}{2}right)}{Gamma(frac{1}{2})pi^{frac{k}{2}}left|{mathbfSigma}right|^{frac{1}{2}}left[1+({mathbf x}-{mathbfmu})^T{mathbfSigma}^{-1}({mathbf x}-{mathbfmu})right]^{frac{1+k}{2}}} $$
which I have lifted from the Wikipedia page. This is just a multivariate Student-$t$ distribution with one degree of freedom.
For the purposes of developing intuition, I would just use the normalized off-diagonal elements of $Sigma$ as if they were correlations, even though they are not. They reflect the strength of the linear relationship between the variables in a way very similar to that of a correlation; $Sigma$ has to be positive definite symmetric; if $Sigma$ is diagonal, the variates are independent, etc.
Maximum likelihood estimation of the parameters can be done using the E-M algorithm, which in this case is easily implemented. The log of the likelihood function is:
$$mathcal{L}(mu, Sigma) = -{nover 2}|Sigma| - {k+1 over 2}sum_{i=1}^nlog(1+s_i)$$
where $s_i = (x_i-mu)^TSigma^{-1}(x_i-mu)$. Differentiating leads to the following simple expressions:
$$mu = sum w_ix_i/sum w_i$$
$$Sigma = {1 over n}sum w_i(x-mu)(x-mu)^T$$
$$w_i = (1+k)/(1+s_i)$$
The E-M algorithm just iterates over these three expressions, substituting the most recent estimates of all the parameters at each step.
For more on this, see Estimation Methods for the Multivariate t Distribution, Nadarajah and Kotz, 2008.
Just because they don't have a covariance doesn't mean that the basic $x^tSigma^{-1} x$ structure usually associated with covariances can't be used. In fact, the multivariate ($k$-dimensional) Cauchy can be written as:
$$f({mathbf x}; {mathbfmu},{mathbfSigma}, k)= frac{Gammaleft(frac{1+k}{2}right)}{Gamma(frac{1}{2})pi^{frac{k}{2}}left|{mathbfSigma}right|^{frac{1}{2}}left[1+({mathbf x}-{mathbfmu})^T{mathbfSigma}^{-1}({mathbf x}-{mathbfmu})right]^{frac{1+k}{2}}} $$
which I have lifted from the Wikipedia page. This is just a multivariate Student-$t$ distribution with one degree of freedom.
For the purposes of developing intuition, I would just use the normalized off-diagonal elements of $Sigma$ as if they were correlations, even though they are not. They reflect the strength of the linear relationship between the variables in a way very similar to that of a correlation; $Sigma$ has to be positive definite symmetric; if $Sigma$ is diagonal, the variates are independent, etc.
Maximum likelihood estimation of the parameters can be done using the E-M algorithm, which in this case is easily implemented. The log of the likelihood function is:
$$mathcal{L}(mu, Sigma) = -{nover 2}|Sigma| - {k+1 over 2}sum_{i=1}^nlog(1+s_i)$$
where $s_i = (x_i-mu)^TSigma^{-1}(x_i-mu)$. Differentiating leads to the following simple expressions:
$$mu = sum w_ix_i/sum w_i$$
$$Sigma = {1 over n}sum w_i(x-mu)(x-mu)^T$$
$$w_i = (1+k)/(1+s_i)$$
The E-M algorithm just iterates over these three expressions, substituting the most recent estimates of all the parameters at each step.
For more on this, see Estimation Methods for the Multivariate t Distribution, Nadarajah and Kotz, 2008.
edited 2 hours ago
answered 2 hours ago
jbowmanjbowman
23.7k34278
23.7k34278
add a comment |
add a comment |
While $text{cov}(X,Y)$ does not exist, for a pair of variates with Cauchy marginals, $text{cov}(Phi(X),Phi(Y))$ does exist for bounded functions $Phi(cdot)$. Borrowing from the concept of copulas, one can turn $X$ and $Y$ into Uniform $(0,1)$ variates, by using their marginal cdfs, $Phi_X(X)$ and $Phi_Y(Y)$, and look at the covariance or correlation of the resulting variates.
add a comment |
While $text{cov}(X,Y)$ does not exist, for a pair of variates with Cauchy marginals, $text{cov}(Phi(X),Phi(Y))$ does exist for bounded functions $Phi(cdot)$. Borrowing from the concept of copulas, one can turn $X$ and $Y$ into Uniform $(0,1)$ variates, by using their marginal cdfs, $Phi_X(X)$ and $Phi_Y(Y)$, and look at the covariance or correlation of the resulting variates.
add a comment |
While $text{cov}(X,Y)$ does not exist, for a pair of variates with Cauchy marginals, $text{cov}(Phi(X),Phi(Y))$ does exist for bounded functions $Phi(cdot)$. Borrowing from the concept of copulas, one can turn $X$ and $Y$ into Uniform $(0,1)$ variates, by using their marginal cdfs, $Phi_X(X)$ and $Phi_Y(Y)$, and look at the covariance or correlation of the resulting variates.
While $text{cov}(X,Y)$ does not exist, for a pair of variates with Cauchy marginals, $text{cov}(Phi(X),Phi(Y))$ does exist for bounded functions $Phi(cdot)$. Borrowing from the concept of copulas, one can turn $X$ and $Y$ into Uniform $(0,1)$ variates, by using their marginal cdfs, $Phi_X(X)$ and $Phi_Y(Y)$, and look at the covariance or correlation of the resulting variates.
answered 2 hours ago
Xi'anXi'an
54k690348
54k690348
add a comment |
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f386036%2fquantifying-dependence-of-cauchy-random-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
May consider general dependence metrics such as mutual information: en.wikipedia.org/wiki/Mutual_information
– John Madden
2 hours ago