Characteristic function of infinitely divisible measure has no zero.
1
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
78.5k758123
asked Nov 29 '18 at 16:00
user408858user408858
344110
add a comment |
1
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
78.5k758123
asked Nov 29 '18 at 16:00
user408858user408858
344110
You mean has no zeros?
– Richard Martin
Nov 29 '18 at 16:26
Yes, exactly. Sorry for writing root.
– user408858
Nov 29 '18 at 17:01
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.
– Richard Martin
Nov 29 '18 at 17:03
1
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
– saz
Nov 30 '18 at 6:52
add a comment |
1
1
1
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
78.5k758123
asked Nov 29 '18 at 16:00
user408858user408858
344110
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
78.5k758123
asked Nov 29 '18 at 16:00
user408858user408858
344110
edited Nov 29 '18 at 20:01
saz
78.5k758123
asked Nov 29 '18 at 16:00
user408858user408858
344110
edited Nov 29 '18 at 20:01
saz
78.5k758123
edited Nov 29 '18 at 20:01
saz
78.5k758123
edited Nov 29 '18 at 20:01
saz
78.5k758123
78.5k758123
asked Nov 29 '18 at 16:00
user408858user408858
344110
asked Nov 29 '18 at 16:00
user408858user408858
344110
asked Nov 29 '18 at 16:00
user408858user408858
344110
344110
You mean has no zeros?
– Richard Martin
Nov 29 '18 at 16:26
Yes, exactly. Sorry for writing root.
– user408858
Nov 29 '18 at 17:01
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.
– Richard Martin
Nov 29 '18 at 17:03
1
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
– saz
Nov 30 '18 at 6:52
add a comment |
You mean has no zeros?
– Richard Martin
Nov 29 '18 at 16:26
Yes, exactly. Sorry for writing root.
– user408858
Nov 29 '18 at 17:01
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.
– Richard Martin
Nov 29 '18 at 17:03
1
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
– saz
Nov 30 '18 at 6:52
You mean has no zeros?
– Richard Martin
Nov 29 '18 at 16:26
You mean has no zeros?
– Richard Martin
Nov 29 '18 at 16:26
Yes, exactly. Sorry for writing root.
– user408858
Nov 29 '18 at 17:01
Yes, exactly. Sorry for writing root.
– user408858
Nov 29 '18 at 17:01
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.
– Richard Martin
Nov 29 '18 at 17:03
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.
– Richard Martin
Nov 29 '18 at 17:03
1
1
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
– saz
Nov 30 '18 at 6:52
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
– saz
Nov 30 '18 at 6:52
add a comment |
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You mean has no zeros?
– Richard Martin
Nov 29 '18 at 16:26
Yes, exactly. Sorry for writing root.
– user408858
Nov 29 '18 at 17:01
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.
– Richard Martin
Nov 29 '18 at 17:03
1
This is a standard result which you can find for instance in Sato's book or in these lecture notes on p.14.
– saz
Nov 30 '18 at 6:52