What two probability distributions (other than the Gaussians) convolved together give a Gaussian pdf?
$begingroup$
It is well know that the convolution of two Gaussian probability distribution gives a Gaussian, are there any other choices of initial pdfs that still give a Gaussian after having performed the convultion?
probability probability-theory probability-distributions normal-distribution
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
$endgroup$
add a comment |
$begingroup$
It is well know that the convolution of two Gaussian probability distribution gives a Gaussian, are there any other choices of initial pdfs that still give a Gaussian after having performed the convultion?
probability probability-theory probability-distributions normal-distribution
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
$endgroup$
add a comment |
$begingroup$
It is well know that the convolution of two Gaussian probability distribution gives a Gaussian, are there any other choices of initial pdfs that still give a Gaussian after having performed the convultion?
probability probability-theory probability-distributions normal-distribution
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
$endgroup$
It is well know that the convolution of two Gaussian probability distribution gives a Gaussian, are there any other choices of initial pdfs that still give a Gaussian after having performed the convultion?
probability probability-theory probability-distributions normal-distribution
probability probability-theory probability-distributions normal-distribution
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
asked Nov 26 '18 at 21:55
MonoliteMonolite
1,5412925
1,5412925
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given that Fourier transform turns convolution into product, the answer is either trivially no (the only PDF which convolved with itself gives a Gaussian is Gaussian itself) or trivially yes (for any PDF with a nowhere vanishing Fourier transform there is another PDF which convolved with it gives a Gaussian).
answered Nov 26 '18 at 22:04
sdssds
3,5281129
$endgroup$
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
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: "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%2f3014992%2fwhat-two-probability-distributions-other-than-the-gaussians-convolved-together%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
$begingroup$
Given that Fourier transform turns convolution into product, the answer is either trivially no (the only PDF which convolved with itself gives a Gaussian is Gaussian itself) or trivially yes (for any PDF with a nowhere vanishing Fourier transform there is another PDF which convolved with it gives a Gaussian).
answered Nov 26 '18 at 22:04
sdssds
3,5281129
$endgroup$
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
add a comment |
$begingroup$
Given that Fourier transform turns convolution into product, the answer is either trivially no (the only PDF which convolved with itself gives a Gaussian is Gaussian itself) or trivially yes (for any PDF with a nowhere vanishing Fourier transform there is another PDF which convolved with it gives a Gaussian).
answered Nov 26 '18 at 22:04
sdssds
3,5281129
$endgroup$
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
add a comment |
$begingroup$
Given that Fourier transform turns convolution into product, the answer is either trivially no (the only PDF which convolved with itself gives a Gaussian is Gaussian itself) or trivially yes (for any PDF with a nowhere vanishing Fourier transform there is another PDF which convolved with it gives a Gaussian).
answered Nov 26 '18 at 22:04
sdssds
3,5281129
$endgroup$
Given that Fourier transform turns convolution into product, the answer is either trivially no (the only PDF which convolved with itself gives a Gaussian is Gaussian itself) or trivially yes (for any PDF with a nowhere vanishing Fourier transform there is another PDF which convolved with it gives a Gaussian).
answered Nov 26 '18 at 22:04
sdssds
3,5281129
edited Nov 26 '18 at 23:46
answered Nov 26 '18 at 22:04
sdssds
3,5281129
answered Nov 26 '18 at 22:04
sdssds
3,5281129
answered Nov 26 '18 at 22:04
sdssds
3,5281129
3,5281129
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
add a comment |
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
The second part is wrong. If the characteristic function of the given PDF has a zero then you cannot get Gaussian characteristic function by multiplying it with another characteristic function.
$endgroup$
– Kavi Rama Murthy
Nov 26 '18 at 23:39
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
$endgroup$
– reuns
Nov 26 '18 at 23:56
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@reuns could you elaborate please? I am not sure I follow.
$endgroup$
– Monolite
Nov 29 '18 at 22:12
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@KaviRamaMurthy: I think I addressed your concern in an edit.
$endgroup$
– sds
Nov 29 '18 at 22:13
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
$begingroup$
@reuns Is that a counterexample to my question?
$endgroup$
– Monolite
Dec 1 '18 at 17:28
add a comment |
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.
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%2f3014992%2fwhat-two-probability-distributions-other-than-the-gaussians-convolved-together%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
