Convergence of the Matérn covariance function to the squared exponential
$begingroup$
The Matérn covariance function converges to the squared exponential
covariance function.
Many sources, amongst them the GPML book and Wikipedia,
state this result. None of them provide details.
I am looking for references that provide details and a proof. In particular, I wonder about the kind of convergence.
mathematical-statistics covariance convergence gaussian-process rbf-kernel
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The Matérn covariance function converges to the squared exponential
covariance function.
Many sources, amongst them the GPML book and Wikipedia,
state this result. None of them provide details.
I am looking for references that provide details and a proof. In particular, I wonder about the kind of convergence.
mathematical-statistics covariance convergence gaussian-process rbf-kernel
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The Matérn covariance function converges to the squared exponential
covariance function.
Many sources, amongst them the GPML book and Wikipedia,
state this result. None of them provide details.
I am looking for references that provide details and a proof. In particular, I wonder about the kind of convergence.
mathematical-statistics covariance convergence gaussian-process rbf-kernel
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The Matérn covariance function converges to the squared exponential
covariance function.
Many sources, amongst them the GPML book and Wikipedia,
state this result. None of them provide details.
I am looking for references that provide details and a proof. In particular, I wonder about the kind of convergence.
mathematical-statistics covariance convergence gaussian-process rbf-kernel
mathematical-statistics covariance convergence gaussian-process rbf-kernel
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 14 at 13:14
parisparis
361
361
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The Matérn function can be written in terms of
$$f_{nu}(x) = C_nu |x|^{nu} K_{nu}left(|x|right)tag{*}$$
where $C_nu$ is a normalizing constant (to make the value of $f_nu(0)$ equal to $1$) and $x = sqrt{2nu}, d/rho.$
(This agrees with the Wikipedia notation where $x$ represents $sqrt{2nu} d/rho.$)
As shown at Moment generating function of the inner product of two gaussian random vectors (using elementary techniques),
the Matern function is proportional to the density function for the distribution of the dot product of two random vectors where each has $2nu+1$ components and all components are independently distributed as standard Normal variables.
Such an inner product is the sum of the $2nu+1$ independent and identically distributed products of corresponding components of the vectors. Each of those is the product of two independent standard Normal variables $X$ and $Y$ and therefore has mean $0$ and variance
$$operatorname{Var}(XY) = E[(XY)^2] = E[X^2]E[Y^2] = (1)(1) = 1.$$
Consequently the inner product has mean $(2nu+1)(0) = 0$ and variance $(2nu+1)(1)=2nu+1.$
The Central Limit Theorem asserts that the normalized versions of these inner products therefore approach a standard Normal distribution almost surely. The effect of normalization is to replace $x$ by the square root of its variance, $xsqrt{2nu+1},$ which changes the probability element $f_{nu}(x)mathrm{d}x$ by
$$f_{nu}(xsqrt{2nu+1})mathrm{d}(xsqrt{2nu+1}) = sqrt{2nu+1} f_{nu}(xsqrt{2nu+1})mathrm{d}x.$$
This differs from $(*)$ (where we may take $rho=1$ without any loss of generality, because it merely establishes the distance unit of measurement) only insofar as $x$ is multiplied by $sqrt{2nu+1}$ instead of $sqrt{2nu}.$ Since the ratio of these terms approaches unity, in the limit it makes no difference which one is used. Consequently, the convergence is almost sure.
One tiny nicety is that because $f_nu$ is normalized to have a peak height of $1,$ which is $sqrt{2pi}$ times the peak height of the standard Normal density, the convergence is to $sqrt{2pi}$ times the standard Normal density rather than the density itself. Re-introducing the scale factor $rho$, we have deduced--using purely statistical thinking!--that
$$lim_{nutoinfty} f_nu(d) = expleft(-frac{d^2}{2rho^2}right)$$ almost surely.
This agrees with what Wikipedia asserts.

This plot shows graphs of $f_2$ (blue), $f_5$ (red), and the limiting Gaussian (gold). The convergence occurs by pulling the tail in to fill out the peak.
$endgroup$
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
});
}
});
paris is a new contributor. Be nice, and check out our Code of Conduct.
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%2f387113%2fconvergence-of-the-mat%25c3%25a9rn-covariance-function-to-the-squared-exponential%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Matérn function can be written in terms of
$$f_{nu}(x) = C_nu |x|^{nu} K_{nu}left(|x|right)tag{*}$$
where $C_nu$ is a normalizing constant (to make the value of $f_nu(0)$ equal to $1$) and $x = sqrt{2nu}, d/rho.$
(This agrees with the Wikipedia notation where $x$ represents $sqrt{2nu} d/rho.$)
As shown at Moment generating function of the inner product of two gaussian random vectors (using elementary techniques),
the Matern function is proportional to the density function for the distribution of the dot product of two random vectors where each has $2nu+1$ components and all components are independently distributed as standard Normal variables.
Such an inner product is the sum of the $2nu+1$ independent and identically distributed products of corresponding components of the vectors. Each of those is the product of two independent standard Normal variables $X$ and $Y$ and therefore has mean $0$ and variance
$$operatorname{Var}(XY) = E[(XY)^2] = E[X^2]E[Y^2] = (1)(1) = 1.$$
Consequently the inner product has mean $(2nu+1)(0) = 0$ and variance $(2nu+1)(1)=2nu+1.$
The Central Limit Theorem asserts that the normalized versions of these inner products therefore approach a standard Normal distribution almost surely. The effect of normalization is to replace $x$ by the square root of its variance, $xsqrt{2nu+1},$ which changes the probability element $f_{nu}(x)mathrm{d}x$ by
$$f_{nu}(xsqrt{2nu+1})mathrm{d}(xsqrt{2nu+1}) = sqrt{2nu+1} f_{nu}(xsqrt{2nu+1})mathrm{d}x.$$
This differs from $(*)$ (where we may take $rho=1$ without any loss of generality, because it merely establishes the distance unit of measurement) only insofar as $x$ is multiplied by $sqrt{2nu+1}$ instead of $sqrt{2nu}.$ Since the ratio of these terms approaches unity, in the limit it makes no difference which one is used. Consequently, the convergence is almost sure.
One tiny nicety is that because $f_nu$ is normalized to have a peak height of $1,$ which is $sqrt{2pi}$ times the peak height of the standard Normal density, the convergence is to $sqrt{2pi}$ times the standard Normal density rather than the density itself. Re-introducing the scale factor $rho$, we have deduced--using purely statistical thinking!--that
$$lim_{nutoinfty} f_nu(d) = expleft(-frac{d^2}{2rho^2}right)$$ almost surely.
This agrees with what Wikipedia asserts.

This plot shows graphs of $f_2$ (blue), $f_5$ (red), and the limiting Gaussian (gold). The convergence occurs by pulling the tail in to fill out the peak.
$endgroup$
add a comment |
$begingroup$
The Matérn function can be written in terms of
$$f_{nu}(x) = C_nu |x|^{nu} K_{nu}left(|x|right)tag{*}$$
where $C_nu$ is a normalizing constant (to make the value of $f_nu(0)$ equal to $1$) and $x = sqrt{2nu}, d/rho.$
(This agrees with the Wikipedia notation where $x$ represents $sqrt{2nu} d/rho.$)
As shown at Moment generating function of the inner product of two gaussian random vectors (using elementary techniques),
the Matern function is proportional to the density function for the distribution of the dot product of two random vectors where each has $2nu+1$ components and all components are independently distributed as standard Normal variables.
Such an inner product is the sum of the $2nu+1$ independent and identically distributed products of corresponding components of the vectors. Each of those is the product of two independent standard Normal variables $X$ and $Y$ and therefore has mean $0$ and variance
$$operatorname{Var}(XY) = E[(XY)^2] = E[X^2]E[Y^2] = (1)(1) = 1.$$
Consequently the inner product has mean $(2nu+1)(0) = 0$ and variance $(2nu+1)(1)=2nu+1.$
The Central Limit Theorem asserts that the normalized versions of these inner products therefore approach a standard Normal distribution almost surely. The effect of normalization is to replace $x$ by the square root of its variance, $xsqrt{2nu+1},$ which changes the probability element $f_{nu}(x)mathrm{d}x$ by
$$f_{nu}(xsqrt{2nu+1})mathrm{d}(xsqrt{2nu+1}) = sqrt{2nu+1} f_{nu}(xsqrt{2nu+1})mathrm{d}x.$$
This differs from $(*)$ (where we may take $rho=1$ without any loss of generality, because it merely establishes the distance unit of measurement) only insofar as $x$ is multiplied by $sqrt{2nu+1}$ instead of $sqrt{2nu}.$ Since the ratio of these terms approaches unity, in the limit it makes no difference which one is used. Consequently, the convergence is almost sure.
One tiny nicety is that because $f_nu$ is normalized to have a peak height of $1,$ which is $sqrt{2pi}$ times the peak height of the standard Normal density, the convergence is to $sqrt{2pi}$ times the standard Normal density rather than the density itself. Re-introducing the scale factor $rho$, we have deduced--using purely statistical thinking!--that
$$lim_{nutoinfty} f_nu(d) = expleft(-frac{d^2}{2rho^2}right)$$ almost surely.
This agrees with what Wikipedia asserts.

This plot shows graphs of $f_2$ (blue), $f_5$ (red), and the limiting Gaussian (gold). The convergence occurs by pulling the tail in to fill out the peak.
$endgroup$
add a comment |
$begingroup$
The Matérn function can be written in terms of
$$f_{nu}(x) = C_nu |x|^{nu} K_{nu}left(|x|right)tag{*}$$
where $C_nu$ is a normalizing constant (to make the value of $f_nu(0)$ equal to $1$) and $x = sqrt{2nu}, d/rho.$
(This agrees with the Wikipedia notation where $x$ represents $sqrt{2nu} d/rho.$)
As shown at Moment generating function of the inner product of two gaussian random vectors (using elementary techniques),
the Matern function is proportional to the density function for the distribution of the dot product of two random vectors where each has $2nu+1$ components and all components are independently distributed as standard Normal variables.
Such an inner product is the sum of the $2nu+1$ independent and identically distributed products of corresponding components of the vectors. Each of those is the product of two independent standard Normal variables $X$ and $Y$ and therefore has mean $0$ and variance
$$operatorname{Var}(XY) = E[(XY)^2] = E[X^2]E[Y^2] = (1)(1) = 1.$$
Consequently the inner product has mean $(2nu+1)(0) = 0$ and variance $(2nu+1)(1)=2nu+1.$
The Central Limit Theorem asserts that the normalized versions of these inner products therefore approach a standard Normal distribution almost surely. The effect of normalization is to replace $x$ by the square root of its variance, $xsqrt{2nu+1},$ which changes the probability element $f_{nu}(x)mathrm{d}x$ by
$$f_{nu}(xsqrt{2nu+1})mathrm{d}(xsqrt{2nu+1}) = sqrt{2nu+1} f_{nu}(xsqrt{2nu+1})mathrm{d}x.$$
This differs from $(*)$ (where we may take $rho=1$ without any loss of generality, because it merely establishes the distance unit of measurement) only insofar as $x$ is multiplied by $sqrt{2nu+1}$ instead of $sqrt{2nu}.$ Since the ratio of these terms approaches unity, in the limit it makes no difference which one is used. Consequently, the convergence is almost sure.
One tiny nicety is that because $f_nu$ is normalized to have a peak height of $1,$ which is $sqrt{2pi}$ times the peak height of the standard Normal density, the convergence is to $sqrt{2pi}$ times the standard Normal density rather than the density itself. Re-introducing the scale factor $rho$, we have deduced--using purely statistical thinking!--that
$$lim_{nutoinfty} f_nu(d) = expleft(-frac{d^2}{2rho^2}right)$$ almost surely.
This agrees with what Wikipedia asserts.

This plot shows graphs of $f_2$ (blue), $f_5$ (red), and the limiting Gaussian (gold). The convergence occurs by pulling the tail in to fill out the peak.
$endgroup$
The Matérn function can be written in terms of
$$f_{nu}(x) = C_nu |x|^{nu} K_{nu}left(|x|right)tag{*}$$
where $C_nu$ is a normalizing constant (to make the value of $f_nu(0)$ equal to $1$) and $x = sqrt{2nu}, d/rho.$
(This agrees with the Wikipedia notation where $x$ represents $sqrt{2nu} d/rho.$)
As shown at Moment generating function of the inner product of two gaussian random vectors (using elementary techniques),
the Matern function is proportional to the density function for the distribution of the dot product of two random vectors where each has $2nu+1$ components and all components are independently distributed as standard Normal variables.
Such an inner product is the sum of the $2nu+1$ independent and identically distributed products of corresponding components of the vectors. Each of those is the product of two independent standard Normal variables $X$ and $Y$ and therefore has mean $0$ and variance
$$operatorname{Var}(XY) = E[(XY)^2] = E[X^2]E[Y^2] = (1)(1) = 1.$$
Consequently the inner product has mean $(2nu+1)(0) = 0$ and variance $(2nu+1)(1)=2nu+1.$
The Central Limit Theorem asserts that the normalized versions of these inner products therefore approach a standard Normal distribution almost surely. The effect of normalization is to replace $x$ by the square root of its variance, $xsqrt{2nu+1},$ which changes the probability element $f_{nu}(x)mathrm{d}x$ by
$$f_{nu}(xsqrt{2nu+1})mathrm{d}(xsqrt{2nu+1}) = sqrt{2nu+1} f_{nu}(xsqrt{2nu+1})mathrm{d}x.$$
This differs from $(*)$ (where we may take $rho=1$ without any loss of generality, because it merely establishes the distance unit of measurement) only insofar as $x$ is multiplied by $sqrt{2nu+1}$ instead of $sqrt{2nu}.$ Since the ratio of these terms approaches unity, in the limit it makes no difference which one is used. Consequently, the convergence is almost sure.
One tiny nicety is that because $f_nu$ is normalized to have a peak height of $1,$ which is $sqrt{2pi}$ times the peak height of the standard Normal density, the convergence is to $sqrt{2pi}$ times the standard Normal density rather than the density itself. Re-introducing the scale factor $rho$, we have deduced--using purely statistical thinking!--that
$$lim_{nutoinfty} f_nu(d) = expleft(-frac{d^2}{2rho^2}right)$$ almost surely.
This agrees with what Wikipedia asserts.

This plot shows graphs of $f_2$ (blue), $f_5$ (red), and the limiting Gaussian (gold). The convergence occurs by pulling the tail in to fill out the peak.
answered Jan 14 at 15:41
whuber♦whuber
202k33440808
202k33440808
add a comment |
add a comment |
paris is a new contributor. Be nice, and check out our Code of Conduct.
paris is a new contributor. Be nice, and check out our Code of Conduct.
paris is a new contributor. Be nice, and check out our Code of Conduct.
paris is a new contributor. Be nice, and check out our Code of Conduct.
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.
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%2f387113%2fconvergence-of-the-mat%25c3%25a9rn-covariance-function-to-the-squared-exponential%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