Mixing
Proving that if $varphi(t)$ is an infinitely divisible characteristic function then $|varphi(t)|$ as well
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Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.
My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
$$varphi(t) = big(varphi_1(t) big)^n.$$
I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
$$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
$$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
I got stuck here. What can I do next? Is my attempt a correct one?
probability characteristic-functions
asked Jan 8 at 15:51
HendrraHendrra
1,168516
$endgroup$
add a comment |
$begingroup$
Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.
My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
$$varphi(t) = big(varphi_1(t) big)^n.$$
I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
$$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
$$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
I got stuck here. What can I do next? Is my attempt a correct one?
probability characteristic-functions
asked Jan 8 at 15:51
HendrraHendrra
1,168516
$endgroup$
add a comment |
$begingroup$
Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.
My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
$$varphi(t) = big(varphi_1(t) big)^n.$$
I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
$$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
$$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
I got stuck here. What can I do next? Is my attempt a correct one?
probability characteristic-functions
asked Jan 8 at 15:51
HendrraHendrra
1,168516
$endgroup$
Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.
My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
$$varphi(t) = big(varphi_1(t) big)^n.$$
I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
$$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
$$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
I got stuck here. What can I do next? Is my attempt a correct one?
probability characteristic-functions
probability characteristic-functions
asked Jan 8 at 15:51
HendrraHendrra
1,168516
asked Jan 8 at 15:51
HendrraHendrra
1,168516
asked Jan 8 at 15:51
HendrraHendrra
1,168516
asked Jan 8 at 15:51
HendrraHendrra
1,168516
asked Jan 8 at 15:51
HendrraHendrra
1,168516
1,168516
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $varphi_j(t)$ be the characteristic function such that
$$
varphi(t) = [varphi_j(t)]^j
$$ holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
[phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
$$ and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.
answered Jan 8 at 17:05
SongSong
11.2k628
$endgroup$
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Let $varphi_j(t)$ be the characteristic function such that
$$
varphi(t) = [varphi_j(t)]^j
$$ holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
[phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
$$ and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.
answered Jan 8 at 17:05
SongSong
11.2k628
$endgroup$
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
add a comment |
$begingroup$
Let $varphi_j(t)$ be the characteristic function such that
$$
varphi(t) = [varphi_j(t)]^j
$$ holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
[phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
$$ and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.
answered Jan 8 at 17:05
SongSong
11.2k628
$endgroup$
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
add a comment |
$begingroup$
Let $varphi_j(t)$ be the characteristic function such that
$$
varphi(t) = [varphi_j(t)]^j
$$ holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
[phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
$$ and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.
answered Jan 8 at 17:05
SongSong
11.2k628
$endgroup$
Let $varphi_j(t)$ be the characteristic function such that
$$
varphi(t) = [varphi_j(t)]^j
$$ holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
[phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
$$ and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.
answered Jan 8 at 17:05
SongSong
11.2k628
answered Jan 8 at 17:05
SongSong
11.2k628
answered Jan 8 at 17:05
SongSong
11.2k628
answered Jan 8 at 17:05
SongSong
11.2k628
11.2k628
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
add a comment |
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
$begingroup$
Thank you. Very smart solution :)
$endgroup$
– Hendrra
Jan 9 at 14:17
add a comment |
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