What two probability distributions (other than the Gaussians) convolved together give a Gaussian pdf?
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 at 21:55
Monolite
1,5112925
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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 at 21:55
Monolite
1,5112925
add a comment |
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 at 21:55
Monolite
1,5112925
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 at 21:55
Monolite
1,5112925
asked Nov 26 at 21:55
Monolite
1,5112925
asked Nov 26 at 21:55
Monolite
1,5112925
asked Nov 26 at 21:55
Monolite
1,5112925
asked Nov 26 at 21:55
Monolite
1,5112925
1,5112925
add a comment |
add a comment |
1 Answer
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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 at 22:04
sds
3,5281129
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.
– Kavi Rama Murthy
Nov 26 at 23:39
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 22:04
sds
3,5281129
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.
– Kavi Rama Murthy
Nov 26 at 23:39
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
add a comment |
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 at 22:04
sds
3,5281129
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.
– Kavi Rama Murthy
Nov 26 at 23:39
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
add a comment |
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 at 22:04
sds
3,5281129
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 at 22:04
sds
3,5281129
edited Nov 26 at 23:46
answered Nov 26 at 22:04
sds
3,5281129
answered Nov 26 at 22:04
sds
3,5281129
answered Nov 26 at 22:04
sds
3,5281129
3,5281129
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.
– Kavi Rama Murthy
Nov 26 at 23:39
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
add a comment |
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.
– Kavi Rama Murthy
Nov 26 at 23:39
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
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.
– Kavi Rama Murthy
Nov 26 at 23:39
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.
– Kavi Rama Murthy
Nov 26 at 23:39
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
Also I can think to the Fourier transform pair of non-negative integrable functions $frac{2}{a^2+4pi^2 xi^2},e^{-a|x|} (a > 0)$
– reuns
Nov 26 at 23:56
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@reuns could you elaborate please? I am not sure I follow.
– Monolite
Nov 29 at 22:12
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@KaviRamaMurthy: I think I addressed your concern in an edit.
– sds
Nov 29 at 22:13
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
@reuns Is that a counterexample to my question?
– Monolite
Dec 1 at 17:28
add a comment |
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