Characteristic function of infinitely divisible measure has no zero.
1
$begingroup$
I want to show that the characteristic function of an infinitely divisible measure $P$ has no zero, directly by using the fact that for all $ninmathbb{N}$, there is a measure $P_n$, such that:
$$varphi_P(t)=Big(varphi_{P_n}(t)Big)^n$$
Can someone give me a hint?
probability-theory probability-distributions characteristic-functions
edited Nov 29 '18 at 20:01
saz
78.9k858123
asked Nov 29 '18 at 16:00
user408858user408858
470110
$endgroup$
add a comment |
1
$begingroup$
I want to show that the characteristic function of an infinitely divisible measure $P$ has no zero, directly by using the fact that for all $ninmathbb{N}$, there is a measure $P_n$, such that:
$$varphi_P(t)=Big(varphi_{P_n}(t)Big)^n$$
Can someone give me a hint?
probability-theory probability-distributions characteristic-functions
edited Nov 29 '18 at 20:01
saz
78.9k858123
asked Nov 29 '18 at 16:00
user408858user408858
470110
$endgroup$
$begingroup$
You mean has no zeros?
$endgroup$
– Richard Martin
Nov 29 '18 at 16:26
$begingroup$
Yes, exactly. Sorry for writing root.
$endgroup$
– user408858
Nov 29 '18 at 17:01
$begingroup$
It's because in a neighbourhood of that point you'd be taking an $n$th root, so $varphi$ wouldn't be analytic. usual argument about branches of log function etc.
$endgroup$
– Richard Martin
Nov 29 '18 at 17:03
1
$begingroup$
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
$endgroup$
– saz
Nov 30 '18 at 6:52
add a comment |
1
1
1
$begingroup$
I want to show that the characteristic function of an infinitely divisible measure $P$ has no zero, directly by using the fact that for all $ninmathbb{N}$, there is a measure $P_n$, such that:
$$varphi_P(t)=Big(varphi_{P_n}(t)Big)^n$$
Can someone give me a hint?
probability-theory probability-distributions characteristic-functions
edited Nov 29 '18 at 20:01
saz
78.9k858123
asked Nov 29 '18 at 16:00
user408858user408858
470110
$endgroup$
I want to show that the characteristic function of an infinitely divisible measure $P$ has no zero, directly by using the fact that for all $ninmathbb{N}$, there is a measure $P_n$, such that:
$$varphi_P(t)=Big(varphi_{P_n}(t)Big)^n$$
Can someone give me a hint?
probability-theory probability-distributions characteristic-functions
probability-theory probability-distributions characteristic-functions
edited Nov 29 '18 at 20:01
saz
78.9k858123
asked Nov 29 '18 at 16:00
user408858user408858
470110
edited Nov 29 '18 at 20:01
saz
78.9k858123
asked Nov 29 '18 at 16:00
user408858user408858
470110
edited Nov 29 '18 at 20:01
saz
78.9k858123
edited Nov 29 '18 at 20:01
saz
78.9k858123
edited Nov 29 '18 at 20:01
saz
78.9k858123
78.9k858123
asked Nov 29 '18 at 16:00
user408858user408858
470110
asked Nov 29 '18 at 16:00
user408858user408858
470110
asked Nov 29 '18 at 16:00
user408858user408858
470110
470110
$begingroup$
You mean has no zeros?
$endgroup$
– Richard Martin
Nov 29 '18 at 16:26
$begingroup$
Yes, exactly. Sorry for writing root.
$endgroup$
– user408858
Nov 29 '18 at 17:01
$begingroup$
It's because in a neighbourhood of that point you'd be taking an $n$th root, so $varphi$ wouldn't be analytic. usual argument about branches of log function etc.
$endgroup$
– Richard Martin
Nov 29 '18 at 17:03
1
$begingroup$
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
$endgroup$
– saz
Nov 30 '18 at 6:52
add a comment |
$begingroup$
You mean has no zeros?
$endgroup$
– Richard Martin
Nov 29 '18 at 16:26
$begingroup$
Yes, exactly. Sorry for writing root.
$endgroup$
– user408858
Nov 29 '18 at 17:01
$begingroup$
It's because in a neighbourhood of that point you'd be taking an $n$th root, so $varphi$ wouldn't be analytic. usual argument about branches of log function etc.
$endgroup$
– Richard Martin
Nov 29 '18 at 17:03
1
$begingroup$
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
$endgroup$
– saz
Nov 30 '18 at 6:52
$begingroup$
You mean has no zeros?
$endgroup$
– Richard Martin
Nov 29 '18 at 16:26
$begingroup$
You mean has no zeros?
$endgroup$
– Richard Martin
Nov 29 '18 at 16:26
$begingroup$
Yes, exactly. Sorry for writing root.
$endgroup$
– user408858
Nov 29 '18 at 17:01
$begingroup$
Yes, exactly. Sorry for writing root.
$endgroup$
– user408858
Nov 29 '18 at 17:01
$begingroup$
It's because in a neighbourhood of that point you'd be taking an $n$th root, so $varphi$ wouldn't be analytic. usual argument about branches of log function etc.
$endgroup$
– Richard Martin
Nov 29 '18 at 17:03
$begingroup$
It's because in a neighbourhood of that point you'd be taking an $n$th root, so $varphi$ wouldn't be analytic. usual argument about branches of log function etc.
$endgroup$
– Richard Martin
Nov 29 '18 at 17:03
1
1
$begingroup$
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
$endgroup$
– saz
Nov 30 '18 at 6:52
$begingroup$
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
$endgroup$
– saz
Nov 30 '18 at 6:52
add a comment |
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$begingroup$
You mean has no zeros?
$endgroup$
– Richard Martin
Nov 29 '18 at 16:26
$begingroup$
Yes, exactly. Sorry for writing root.
$endgroup$
– user408858
Nov 29 '18 at 17:01
$begingroup$
It's because in a neighbourhood of that point you'd be taking an $n$th root, so $varphi$ wouldn't be analytic. usual argument about branches of log function etc.
$endgroup$
– Richard Martin
Nov 29 '18 at 17:03
1
$begingroup$
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
$endgroup$
– saz
Nov 30 '18 at 6:52