Reversal of an Autoregressive Cauchy Markov Chain
Let $mu_0 (dx)$ be the standard one-dimensional Cauchy distribution, i.e.
begin{align}
mu_0 (dx) = frac{1}{pi} frac{1}{1+x^2} dx.
end{align}
Suppose I fix $rho in [0, 1]$, and form a Markov chain ${X_n}_{n geqslant 0}$ as follows:
- At step $n$, I sample $Y_n sim mu_0$.
- I then set $X_{n+1} = rho X_n + (1 - rho) Y_n$
It is not so hard to show that this chain admits $mu_0$ as a stationary measure, as this essentially comes from the fact that Cauchy distribution is a stable distribution.
What I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel $q (x to dy)$, I want to understand the Markov kernel $r$ such that
begin{align}
mu_0 (dx) q (x to dy) = mu_0 (dy) r (y to dx).
end{align}
Fortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what $r (y to dx)$ is:
begin{align}
r (y to dx) = frac{1 + y^2}{pi} frac{1}{1 + x^2} frac{1-rho}{ (1 - rho)^2 + (y - rho x)^2}.
end{align}
My question is then: is there a simple, elegant way to draw exact samples from $r$?
I would highlight that this is not a purely algorithmic question; I'd really like to understand what this reversal kernel $r$ is doing. A nice byproduct of that would then be that I could simulate from it easily.
For completeness, some of the `purely algorithmic' solutions I had considered were the following.
- I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.
- I could try something like the inverse CDF method, but it seems to me that the CDF of $r$ is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.
probability markov-chains
asked Nov 26 at 13:49
πr8
9,8483924
add a comment |
Let $mu_0 (dx)$ be the standard one-dimensional Cauchy distribution, i.e.
begin{align}
mu_0 (dx) = frac{1}{pi} frac{1}{1+x^2} dx.
end{align}
Suppose I fix $rho in [0, 1]$, and form a Markov chain ${X_n}_{n geqslant 0}$ as follows:
- At step $n$, I sample $Y_n sim mu_0$.
- I then set $X_{n+1} = rho X_n + (1 - rho) Y_n$
It is not so hard to show that this chain admits $mu_0$ as a stationary measure, as this essentially comes from the fact that Cauchy distribution is a stable distribution.
What I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel $q (x to dy)$, I want to understand the Markov kernel $r$ such that
begin{align}
mu_0 (dx) q (x to dy) = mu_0 (dy) r (y to dx).
end{align}
Fortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what $r (y to dx)$ is:
begin{align}
r (y to dx) = frac{1 + y^2}{pi} frac{1}{1 + x^2} frac{1-rho}{ (1 - rho)^2 + (y - rho x)^2}.
end{align}
My question is then: is there a simple, elegant way to draw exact samples from $r$?
I would highlight that this is not a purely algorithmic question; I'd really like to understand what this reversal kernel $r$ is doing. A nice byproduct of that would then be that I could simulate from it easily.
For completeness, some of the `purely algorithmic' solutions I had considered were the following.
- I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.
- I could try something like the inverse CDF method, but it seems to me that the CDF of $r$ is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.
probability markov-chains
asked Nov 26 at 13:49
πr8
9,8483924
add a comment |
Let $mu_0 (dx)$ be the standard one-dimensional Cauchy distribution, i.e.
begin{align}
mu_0 (dx) = frac{1}{pi} frac{1}{1+x^2} dx.
end{align}
Suppose I fix $rho in [0, 1]$, and form a Markov chain ${X_n}_{n geqslant 0}$ as follows:
- At step $n$, I sample $Y_n sim mu_0$.
- I then set $X_{n+1} = rho X_n + (1 - rho) Y_n$
It is not so hard to show that this chain admits $mu_0$ as a stationary measure, as this essentially comes from the fact that Cauchy distribution is a stable distribution.
What I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel $q (x to dy)$, I want to understand the Markov kernel $r$ such that
begin{align}
mu_0 (dx) q (x to dy) = mu_0 (dy) r (y to dx).
end{align}
Fortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what $r (y to dx)$ is:
begin{align}
r (y to dx) = frac{1 + y^2}{pi} frac{1}{1 + x^2} frac{1-rho}{ (1 - rho)^2 + (y - rho x)^2}.
end{align}
My question is then: is there a simple, elegant way to draw exact samples from $r$?
I would highlight that this is not a purely algorithmic question; I'd really like to understand what this reversal kernel $r$ is doing. A nice byproduct of that would then be that I could simulate from it easily.
For completeness, some of the `purely algorithmic' solutions I had considered were the following.
- I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.
- I could try something like the inverse CDF method, but it seems to me that the CDF of $r$ is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.
probability markov-chains
asked Nov 26 at 13:49
πr8
9,8483924
Let $mu_0 (dx)$ be the standard one-dimensional Cauchy distribution, i.e.
begin{align}
mu_0 (dx) = frac{1}{pi} frac{1}{1+x^2} dx.
end{align}
Suppose I fix $rho in [0, 1]$, and form a Markov chain ${X_n}_{n geqslant 0}$ as follows:
- At step $n$, I sample $Y_n sim mu_0$.
- I then set $X_{n+1} = rho X_n + (1 - rho) Y_n$
It is not so hard to show that this chain admits $mu_0$ as a stationary measure, as this essentially comes from the fact that Cauchy distribution is a stable distribution.
What I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel $q (x to dy)$, I want to understand the Markov kernel $r$ such that
begin{align}
mu_0 (dx) q (x to dy) = mu_0 (dy) r (y to dx).
end{align}
Fortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what $r (y to dx)$ is:
begin{align}
r (y to dx) = frac{1 + y^2}{pi} frac{1}{1 + x^2} frac{1-rho}{ (1 - rho)^2 + (y - rho x)^2}.
end{align}
My question is then: is there a simple, elegant way to draw exact samples from $r$?
I would highlight that this is not a purely algorithmic question; I'd really like to understand what this reversal kernel $r$ is doing. A nice byproduct of that would then be that I could simulate from it easily.
For completeness, some of the `purely algorithmic' solutions I had considered were the following.
- I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.
- I could try something like the inverse CDF method, but it seems to me that the CDF of $r$ is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.
probability markov-chains
probability markov-chains
asked Nov 26 at 13:49
πr8
9,8483924
asked Nov 26 at 13:49
πr8
9,8483924
edited Nov 26 at 13:55
asked Nov 26 at 13:49
πr8
9,8483924
asked Nov 26 at 13:49
πr8
9,8483924
asked Nov 26 at 13:49
πr8
9,8483924
9,8483924
add a comment |
add a comment |
active
oldest
votes
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3014357%2freversal-of-an-autoregressive-cauchy-markov-chain%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f3014357%2freversal-of-an-autoregressive-cauchy-markov-chain%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
