Show that $g_1(mu_1,sigma_1)ast g_2(mu_2,sigma_2)=gleft(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2}right)$
The definition (convention) I have been using for the Fourier transform is
$$mathscr{F}[f(t)]=g(omega)=frac{1}{sqrt{2 pi}}int_{t=-infty}^{infty}f(t)e^{iomega t}dttag{1}$$
and the inverse as
$$mathscr{F}^{-1}[g(omega)]=f(t)=frac{1}{sqrt{2 pi}}int_{omega=-infty}^{infty}g(omega)e^{-iomega t}domega$$
Let $g_1(t; mu_1,sigma_1)$ and $g_1(t;mu_2,sigma_2)$ be two Gaussian functions of $t$ with mean $mu$ and width
$sigma$ as indicated.
Show that $g_1(mu_1,sigma_1)ast g_2(mu_2,sigma_2)=gleft(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2}right)$
By the convolution theorem
$$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}mathscr{F}(g_1)mathscr{F}(g_2)tag{2}$$ and insertion of $(1)$ into $(2)$
$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}left[frac{1}{sqrt{2 pi}}operatorname{Largeint}_{t_1=-infty}^{infty}expleft(-frac{(t_1-mu_1)^2}{2sigma_1^2}right)e^{iomega t_1}dt_1right]left[frac{1}{sqrt{2 pi}}{Largeint}_{t_2=-infty}^{infty}expleft(-frac{(t_2-mu_2)^2}{2sigma_2^2}right)e^{iomega t_2}dt_2right]$
it is at this point for which I am completely stuck and don't know how to proceed to complete the proof.
The answer given by the author is:
I don't understand the authors' solution for 3 reasons; firstly, I don't understand why there aren't any integral signs on the RHS of equation $(20)$ & $(21)$.
Secondly, $t$ and $omega$ are the Fourier pairs in this transformation so why is the author using $mu_1$ and $mu_2$ in the complex exponentials?
Lastly, the author mentions the translation property for Fourier transforms:
$$mathscr{F}(t-t_0)=e^{i omega t_0} g(omega)$$
but I fail to see how this is being utilized in the answer.
Is there anyone that could please help me understand the authors' solution or give me any hints or tips about how to complete the proof I began?
integration fourier-analysis fourier-transform convolution
add a comment |
The definition (convention) I have been using for the Fourier transform is
$$mathscr{F}[f(t)]=g(omega)=frac{1}{sqrt{2 pi}}int_{t=-infty}^{infty}f(t)e^{iomega t}dttag{1}$$
and the inverse as
$$mathscr{F}^{-1}[g(omega)]=f(t)=frac{1}{sqrt{2 pi}}int_{omega=-infty}^{infty}g(omega)e^{-iomega t}domega$$
Let $g_1(t; mu_1,sigma_1)$ and $g_1(t;mu_2,sigma_2)$ be two Gaussian functions of $t$ with mean $mu$ and width
$sigma$ as indicated.
Show that $g_1(mu_1,sigma_1)ast g_2(mu_2,sigma_2)=gleft(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2}right)$
By the convolution theorem
$$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}mathscr{F}(g_1)mathscr{F}(g_2)tag{2}$$ and insertion of $(1)$ into $(2)$
$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}left[frac{1}{sqrt{2 pi}}operatorname{Largeint}_{t_1=-infty}^{infty}expleft(-frac{(t_1-mu_1)^2}{2sigma_1^2}right)e^{iomega t_1}dt_1right]left[frac{1}{sqrt{2 pi}}{Largeint}_{t_2=-infty}^{infty}expleft(-frac{(t_2-mu_2)^2}{2sigma_2^2}right)e^{iomega t_2}dt_2right]$
it is at this point for which I am completely stuck and don't know how to proceed to complete the proof.
The answer given by the author is:
I don't understand the authors' solution for 3 reasons; firstly, I don't understand why there aren't any integral signs on the RHS of equation $(20)$ & $(21)$.
Secondly, $t$ and $omega$ are the Fourier pairs in this transformation so why is the author using $mu_1$ and $mu_2$ in the complex exponentials?
Lastly, the author mentions the translation property for Fourier transforms:
$$mathscr{F}(t-t_0)=e^{i omega t_0} g(omega)$$
but I fail to see how this is being utilized in the answer.
Is there anyone that could please help me understand the authors' solution or give me any hints or tips about how to complete the proof I began?
integration fourier-analysis fourier-transform convolution
The author is using a rule like cse.yorku.ca/~kosta/CompVis_Notes/… to transform, and using the translate to account for the mu shift in the original Gaussian. The mu is being used for $t_0$.
– Gaffney
Jan 4 at 1:13
The Fourier transform of a Gaussian is another Gaussian
– Dylan
Jan 4 at 8:14
add a comment |
The definition (convention) I have been using for the Fourier transform is
$$mathscr{F}[f(t)]=g(omega)=frac{1}{sqrt{2 pi}}int_{t=-infty}^{infty}f(t)e^{iomega t}dttag{1}$$
and the inverse as
$$mathscr{F}^{-1}[g(omega)]=f(t)=frac{1}{sqrt{2 pi}}int_{omega=-infty}^{infty}g(omega)e^{-iomega t}domega$$
Let $g_1(t; mu_1,sigma_1)$ and $g_1(t;mu_2,sigma_2)$ be two Gaussian functions of $t$ with mean $mu$ and width
$sigma$ as indicated.
Show that $g_1(mu_1,sigma_1)ast g_2(mu_2,sigma_2)=gleft(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2}right)$
By the convolution theorem
$$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}mathscr{F}(g_1)mathscr{F}(g_2)tag{2}$$ and insertion of $(1)$ into $(2)$
$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}left[frac{1}{sqrt{2 pi}}operatorname{Largeint}_{t_1=-infty}^{infty}expleft(-frac{(t_1-mu_1)^2}{2sigma_1^2}right)e^{iomega t_1}dt_1right]left[frac{1}{sqrt{2 pi}}{Largeint}_{t_2=-infty}^{infty}expleft(-frac{(t_2-mu_2)^2}{2sigma_2^2}right)e^{iomega t_2}dt_2right]$
it is at this point for which I am completely stuck and don't know how to proceed to complete the proof.
The answer given by the author is:
I don't understand the authors' solution for 3 reasons; firstly, I don't understand why there aren't any integral signs on the RHS of equation $(20)$ & $(21)$.
Secondly, $t$ and $omega$ are the Fourier pairs in this transformation so why is the author using $mu_1$ and $mu_2$ in the complex exponentials?
Lastly, the author mentions the translation property for Fourier transforms:
$$mathscr{F}(t-t_0)=e^{i omega t_0} g(omega)$$
but I fail to see how this is being utilized in the answer.
Is there anyone that could please help me understand the authors' solution or give me any hints or tips about how to complete the proof I began?
integration fourier-analysis fourier-transform convolution
The definition (convention) I have been using for the Fourier transform is
$$mathscr{F}[f(t)]=g(omega)=frac{1}{sqrt{2 pi}}int_{t=-infty}^{infty}f(t)e^{iomega t}dttag{1}$$
and the inverse as
$$mathscr{F}^{-1}[g(omega)]=f(t)=frac{1}{sqrt{2 pi}}int_{omega=-infty}^{infty}g(omega)e^{-iomega t}domega$$
Let $g_1(t; mu_1,sigma_1)$ and $g_1(t;mu_2,sigma_2)$ be two Gaussian functions of $t$ with mean $mu$ and width
$sigma$ as indicated.
Show that $g_1(mu_1,sigma_1)ast g_2(mu_2,sigma_2)=gleft(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2}right)$
By the convolution theorem
$$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}mathscr{F}(g_1)mathscr{F}(g_2)tag{2}$$ and insertion of $(1)$ into $(2)$
$mathscr{F}(g_1 ast g_ 2)=sqrt{2pi}left[frac{1}{sqrt{2 pi}}operatorname{Largeint}_{t_1=-infty}^{infty}expleft(-frac{(t_1-mu_1)^2}{2sigma_1^2}right)e^{iomega t_1}dt_1right]left[frac{1}{sqrt{2 pi}}{Largeint}_{t_2=-infty}^{infty}expleft(-frac{(t_2-mu_2)^2}{2sigma_2^2}right)e^{iomega t_2}dt_2right]$
it is at this point for which I am completely stuck and don't know how to proceed to complete the proof.
The answer given by the author is:
I don't understand the authors' solution for 3 reasons; firstly, I don't understand why there aren't any integral signs on the RHS of equation $(20)$ & $(21)$.
Secondly, $t$ and $omega$ are the Fourier pairs in this transformation so why is the author using $mu_1$ and $mu_2$ in the complex exponentials?
Lastly, the author mentions the translation property for Fourier transforms:
$$mathscr{F}(t-t_0)=e^{i omega t_0} g(omega)$$
but I fail to see how this is being utilized in the answer.
Is there anyone that could please help me understand the authors' solution or give me any hints or tips about how to complete the proof I began?
integration fourier-analysis fourier-transform convolution
integration fourier-analysis fourier-transform convolution
asked Jan 4 at 0:22
BLAZE
6,071112754
6,071112754
The author is using a rule like cse.yorku.ca/~kosta/CompVis_Notes/… to transform, and using the translate to account for the mu shift in the original Gaussian. The mu is being used for $t_0$.
– Gaffney
Jan 4 at 1:13
The Fourier transform of a Gaussian is another Gaussian
– Dylan
Jan 4 at 8:14
add a comment |
The author is using a rule like cse.yorku.ca/~kosta/CompVis_Notes/… to transform, and using the translate to account for the mu shift in the original Gaussian. The mu is being used for $t_0$.
– Gaffney
Jan 4 at 1:13
The Fourier transform of a Gaussian is another Gaussian
– Dylan
Jan 4 at 8:14
The author is using a rule like cse.yorku.ca/~kosta/CompVis_Notes/… to transform, and using the translate to account for the mu shift in the original Gaussian. The mu is being used for $t_0$.
– Gaffney
Jan 4 at 1:13
The author is using a rule like cse.yorku.ca/~kosta/CompVis_Notes/… to transform, and using the translate to account for the mu shift in the original Gaussian. The mu is being used for $t_0$.
– Gaffney
Jan 4 at 1:13
The Fourier transform of a Gaussian is another Gaussian
– Dylan
Jan 4 at 8:14
The Fourier transform of a Gaussian is another Gaussian
– Dylan
Jan 4 at 8:14
add a comment |
1 Answer
1
active
oldest
votes
The trick is to simplify everything be completing the square and recognizing that any normal distribution will integrate to 1.
For example
$$mathscr{F}(g_1) = frac{1}{sqrt{2pi}} frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-mu_1)^2}{2sigma_1^2} + iomega t_1)dt_1$$
by completing the square
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{t_1^2-2t_1(mu_1+isigma_1^2omega)+(mu_1+isigma_1^2omega)^2 - 2mu_1 isigma_1^2omega + 4sigma_1^4omega^2}{2sigma_1^2})dt_1
$$
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-(mu_1+isigma_1^2omega))^2}{2sigma_1^2})
exp(mu_1 iomega - 2sigma_1^2omega^2)
dt_1
$$
Integrating out the normal distribution with mean $mu_1+isigma_1^2omega$ and variance $sigma_1^2$ makes this
$$=frac{1}{sqrt{2pi}}exp(mu_1 iomega - 2sigma_1^2omega^2)$$
Similarly
$$mathscr{F}(g_2) = frac{1}{sqrt{2pi}} exp(mu_2 iomega - 2sigma_2^2omega^2)$$
So
$$mathscr{F}(g_1 * g_2) = frac{1}{sqrt{2pi}}
exp((mu_1+mu_2) iomega - 2(sigma_1^2+sigma_2^2)omega^2)$$
Then the inverse Fourier transform of this is
$$frac{1}{2pi}int_{-infty}^{infty}
exp((mu_1+mu_2-t) iomega - 2(sigma_1^2+sigma_2^2)omega^2)
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{omega^2-iomegafrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}
-(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2
+(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{(omega^2-ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
exp((frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2)
domega
$$
Integrating out the normal distribution with mean $ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}$ and variance $frac{1}{sigma_1^2+sigma_2^2}$ makes this
$$=frac{1}{sqrt{2pi(sigma_1^2+sigma_2^2)}}
exp((frac{t-(mu_1+mu_2)}{2(sigma_1^2+sigma_2^2)})^2)$$
which is $g(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2})$ as desired.
New contributor
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%2f3061182%2fshow-that-g-1-mu-1-sigma-1-ast-g-2-mu-2-sigma-2-g-left-mu-1-mu-2-sqrt%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
The trick is to simplify everything be completing the square and recognizing that any normal distribution will integrate to 1.
For example
$$mathscr{F}(g_1) = frac{1}{sqrt{2pi}} frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-mu_1)^2}{2sigma_1^2} + iomega t_1)dt_1$$
by completing the square
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{t_1^2-2t_1(mu_1+isigma_1^2omega)+(mu_1+isigma_1^2omega)^2 - 2mu_1 isigma_1^2omega + 4sigma_1^4omega^2}{2sigma_1^2})dt_1
$$
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-(mu_1+isigma_1^2omega))^2}{2sigma_1^2})
exp(mu_1 iomega - 2sigma_1^2omega^2)
dt_1
$$
Integrating out the normal distribution with mean $mu_1+isigma_1^2omega$ and variance $sigma_1^2$ makes this
$$=frac{1}{sqrt{2pi}}exp(mu_1 iomega - 2sigma_1^2omega^2)$$
Similarly
$$mathscr{F}(g_2) = frac{1}{sqrt{2pi}} exp(mu_2 iomega - 2sigma_2^2omega^2)$$
So
$$mathscr{F}(g_1 * g_2) = frac{1}{sqrt{2pi}}
exp((mu_1+mu_2) iomega - 2(sigma_1^2+sigma_2^2)omega^2)$$
Then the inverse Fourier transform of this is
$$frac{1}{2pi}int_{-infty}^{infty}
exp((mu_1+mu_2-t) iomega - 2(sigma_1^2+sigma_2^2)omega^2)
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{omega^2-iomegafrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}
-(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2
+(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{(omega^2-ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
exp((frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2)
domega
$$
Integrating out the normal distribution with mean $ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}$ and variance $frac{1}{sigma_1^2+sigma_2^2}$ makes this
$$=frac{1}{sqrt{2pi(sigma_1^2+sigma_2^2)}}
exp((frac{t-(mu_1+mu_2)}{2(sigma_1^2+sigma_2^2)})^2)$$
which is $g(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2})$ as desired.
New contributor
add a comment |
The trick is to simplify everything be completing the square and recognizing that any normal distribution will integrate to 1.
For example
$$mathscr{F}(g_1) = frac{1}{sqrt{2pi}} frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-mu_1)^2}{2sigma_1^2} + iomega t_1)dt_1$$
by completing the square
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{t_1^2-2t_1(mu_1+isigma_1^2omega)+(mu_1+isigma_1^2omega)^2 - 2mu_1 isigma_1^2omega + 4sigma_1^4omega^2}{2sigma_1^2})dt_1
$$
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-(mu_1+isigma_1^2omega))^2}{2sigma_1^2})
exp(mu_1 iomega - 2sigma_1^2omega^2)
dt_1
$$
Integrating out the normal distribution with mean $mu_1+isigma_1^2omega$ and variance $sigma_1^2$ makes this
$$=frac{1}{sqrt{2pi}}exp(mu_1 iomega - 2sigma_1^2omega^2)$$
Similarly
$$mathscr{F}(g_2) = frac{1}{sqrt{2pi}} exp(mu_2 iomega - 2sigma_2^2omega^2)$$
So
$$mathscr{F}(g_1 * g_2) = frac{1}{sqrt{2pi}}
exp((mu_1+mu_2) iomega - 2(sigma_1^2+sigma_2^2)omega^2)$$
Then the inverse Fourier transform of this is
$$frac{1}{2pi}int_{-infty}^{infty}
exp((mu_1+mu_2-t) iomega - 2(sigma_1^2+sigma_2^2)omega^2)
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{omega^2-iomegafrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}
-(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2
+(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{(omega^2-ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
exp((frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2)
domega
$$
Integrating out the normal distribution with mean $ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}$ and variance $frac{1}{sigma_1^2+sigma_2^2}$ makes this
$$=frac{1}{sqrt{2pi(sigma_1^2+sigma_2^2)}}
exp((frac{t-(mu_1+mu_2)}{2(sigma_1^2+sigma_2^2)})^2)$$
which is $g(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2})$ as desired.
New contributor
add a comment |
The trick is to simplify everything be completing the square and recognizing that any normal distribution will integrate to 1.
For example
$$mathscr{F}(g_1) = frac{1}{sqrt{2pi}} frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-mu_1)^2}{2sigma_1^2} + iomega t_1)dt_1$$
by completing the square
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{t_1^2-2t_1(mu_1+isigma_1^2omega)+(mu_1+isigma_1^2omega)^2 - 2mu_1 isigma_1^2omega + 4sigma_1^4omega^2}{2sigma_1^2})dt_1
$$
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-(mu_1+isigma_1^2omega))^2}{2sigma_1^2})
exp(mu_1 iomega - 2sigma_1^2omega^2)
dt_1
$$
Integrating out the normal distribution with mean $mu_1+isigma_1^2omega$ and variance $sigma_1^2$ makes this
$$=frac{1}{sqrt{2pi}}exp(mu_1 iomega - 2sigma_1^2omega^2)$$
Similarly
$$mathscr{F}(g_2) = frac{1}{sqrt{2pi}} exp(mu_2 iomega - 2sigma_2^2omega^2)$$
So
$$mathscr{F}(g_1 * g_2) = frac{1}{sqrt{2pi}}
exp((mu_1+mu_2) iomega - 2(sigma_1^2+sigma_2^2)omega^2)$$
Then the inverse Fourier transform of this is
$$frac{1}{2pi}int_{-infty}^{infty}
exp((mu_1+mu_2-t) iomega - 2(sigma_1^2+sigma_2^2)omega^2)
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{omega^2-iomegafrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}
-(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2
+(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{(omega^2-ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
exp((frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2)
domega
$$
Integrating out the normal distribution with mean $ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}$ and variance $frac{1}{sigma_1^2+sigma_2^2}$ makes this
$$=frac{1}{sqrt{2pi(sigma_1^2+sigma_2^2)}}
exp((frac{t-(mu_1+mu_2)}{2(sigma_1^2+sigma_2^2)})^2)$$
which is $g(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2})$ as desired.
New contributor
The trick is to simplify everything be completing the square and recognizing that any normal distribution will integrate to 1.
For example
$$mathscr{F}(g_1) = frac{1}{sqrt{2pi}} frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-mu_1)^2}{2sigma_1^2} + iomega t_1)dt_1$$
by completing the square
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{t_1^2-2t_1(mu_1+isigma_1^2omega)+(mu_1+isigma_1^2omega)^2 - 2mu_1 isigma_1^2omega + 4sigma_1^4omega^2}{2sigma_1^2})dt_1
$$
$$=frac{1}{sqrt{2pi}}frac{1}{sqrt{2pisigma_1^2}}
int_{-infty}^{infty}exp(-frac{(t_1-(mu_1+isigma_1^2omega))^2}{2sigma_1^2})
exp(mu_1 iomega - 2sigma_1^2omega^2)
dt_1
$$
Integrating out the normal distribution with mean $mu_1+isigma_1^2omega$ and variance $sigma_1^2$ makes this
$$=frac{1}{sqrt{2pi}}exp(mu_1 iomega - 2sigma_1^2omega^2)$$
Similarly
$$mathscr{F}(g_2) = frac{1}{sqrt{2pi}} exp(mu_2 iomega - 2sigma_2^2omega^2)$$
So
$$mathscr{F}(g_1 * g_2) = frac{1}{sqrt{2pi}}
exp((mu_1+mu_2) iomega - 2(sigma_1^2+sigma_2^2)omega^2)$$
Then the inverse Fourier transform of this is
$$frac{1}{2pi}int_{-infty}^{infty}
exp((mu_1+mu_2-t) iomega - 2(sigma_1^2+sigma_2^2)omega^2)
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{omega^2-iomegafrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}
-(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2
+(frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
domega
$$
$$=frac{1}{2pi}int_{-infty}^{infty}
exp(-frac{(omega^2-ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2}
{frac{1}{2(sigma_1^2+sigma_2^2)}})
exp((frac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)})^2)
domega
$$
Integrating out the normal distribution with mean $ifrac{mu_1+mu_2-t}{2(sigma_1^2+sigma_2^2)}$ and variance $frac{1}{sigma_1^2+sigma_2^2}$ makes this
$$=frac{1}{sqrt{2pi(sigma_1^2+sigma_2^2)}}
exp((frac{t-(mu_1+mu_2)}{2(sigma_1^2+sigma_2^2)})^2)$$
which is $g(mu_1+mu_2,sqrt{sigma_1^2+sigma_2^2})$ as desired.
New contributor
New contributor
answered Jan 4 at 3:50
Erik Parkinson
9159
9159
New contributor
New contributor
add a comment |
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.
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%2f3061182%2fshow-that-g-1-mu-1-sigma-1-ast-g-2-mu-2-sigma-2-g-left-mu-1-mu-2-sqrt%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
The author is using a rule like cse.yorku.ca/~kosta/CompVis_Notes/… to transform, and using the translate to account for the mu shift in the original Gaussian. The mu is being used for $t_0$.
– Gaffney
Jan 4 at 1:13
The Fourier transform of a Gaussian is another Gaussian
– Dylan
Jan 4 at 8:14