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?










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$endgroup$












  • $begingroup$
    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
















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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 20:01









saz

78.9k858123




78.9k858123










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


















  • $begingroup$
    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










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