Show that $psi(t)=e^{lambda (varphi(t)-1)}$ is infinitely divisble for any characteristic function $varphi$












3












$begingroup$


I am given a function
$$e^{lambda(varphi(t) -1)} tag{1},$$
where $varphi(t)$ is a characteristic function. I managed to show that $(1)$ is a characteristic function too.

Now I am to show that $(1)$ is an infinitely divisible function. What does it mean?



I know that a distribution is infinitely divisible if it can be expressed as the probability distribution of the sum of an arbitrary number of independent and identically distributed random variables.



Do I have to find the distribution of my characteristic function and then show that it is infinitely divisible?






probability-theory characteristic-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Since $psi:=e^{lambda(phi-1)}$ is a characteristic function, we can associate a probability measure, say $mu$, with $psi$ through the relation $$psi(t) = int e^{it x} , mu(dx).$$ You are supposed to show that $mu$ is infinitely divisible.
    $endgroup$
    – saz
    Jan 6 at 17:37












  • $begingroup$
    @saz thanks for your comment. Why are we considering only $varphi(t)$? Not the full expression that contains it?
    $endgroup$
    – Hendrra
    Jan 6 at 17:40






  • 1




    $begingroup$
    Ah, sorry, my mistake... please see my edited comment.
    $endgroup$
    – saz
    Jan 6 at 17:42










  • $begingroup$
    Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $mu$ or is there a method of dealing with such things?
    $endgroup$
    – Hendrra
    Jan 6 at 17:59
















3












$begingroup$


I am given a function
$$e^{lambda(varphi(t) -1)} tag{1},$$
where $varphi(t)$ is a characteristic function. I managed to show that $(1)$ is a characteristic function too.

Now I am to show that $(1)$ is an infinitely divisible function. What does it mean?



I know that a distribution is infinitely divisible if it can be expressed as the probability distribution of the sum of an arbitrary number of independent and identically distributed random variables.



Do I have to find the distribution of my characteristic function and then show that it is infinitely divisible?






probability-theory characteristic-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Since $psi:=e^{lambda(phi-1)}$ is a characteristic function, we can associate a probability measure, say $mu$, with $psi$ through the relation $$psi(t) = int e^{it x} , mu(dx).$$ You are supposed to show that $mu$ is infinitely divisible.
    $endgroup$
    – saz
    Jan 6 at 17:37












  • $begingroup$
    @saz thanks for your comment. Why are we considering only $varphi(t)$? Not the full expression that contains it?
    $endgroup$
    – Hendrra
    Jan 6 at 17:40






  • 1




    $begingroup$
    Ah, sorry, my mistake... please see my edited comment.
    $endgroup$
    – saz
    Jan 6 at 17:42










  • $begingroup$
    Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $mu$ or is there a method of dealing with such things?
    $endgroup$
    – Hendrra
    Jan 6 at 17:59














3












3








3





$begingroup$


I am given a function
$$e^{lambda(varphi(t) -1)} tag{1},$$
where $varphi(t)$ is a characteristic function. I managed to show that $(1)$ is a characteristic function too.

Now I am to show that $(1)$ is an infinitely divisible function. What does it mean?



I know that a distribution is infinitely divisible if it can be expressed as the probability distribution of the sum of an arbitrary number of independent and identically distributed random variables.



Do I have to find the distribution of my characteristic function and then show that it is infinitely divisible?






probability-theory characteristic-functions







share|cite|improve this question











$endgroup$




I am given a function
$$e^{lambda(varphi(t) -1)} tag{1},$$
where $varphi(t)$ is a characteristic function. I managed to show that $(1)$ is a characteristic function too.

Now I am to show that $(1)$ is an infinitely divisible function. What does it mean?



I know that a distribution is infinitely divisible if it can be expressed as the probability distribution of the sum of an arbitrary number of independent and identically distributed random variables.



Do I have to find the distribution of my characteristic function and then show that it is infinitely divisible?






probability-theory characteristic-functions




probability-theory characteristic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 18:41









saz

82.4k862131




82.4k862131










asked Jan 6 at 17:30









HendrraHendrra

1,214516




1,214516








  • 2




    $begingroup$
    Since $psi:=e^{lambda(phi-1)}$ is a characteristic function, we can associate a probability measure, say $mu$, with $psi$ through the relation $$psi(t) = int e^{it x} , mu(dx).$$ You are supposed to show that $mu$ is infinitely divisible.
    $endgroup$
    – saz
    Jan 6 at 17:37












  • $begingroup$
    @saz thanks for your comment. Why are we considering only $varphi(t)$? Not the full expression that contains it?
    $endgroup$
    – Hendrra
    Jan 6 at 17:40






  • 1




    $begingroup$
    Ah, sorry, my mistake... please see my edited comment.
    $endgroup$
    – saz
    Jan 6 at 17:42










  • $begingroup$
    Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $mu$ or is there a method of dealing with such things?
    $endgroup$
    – Hendrra
    Jan 6 at 17:59














  • 2




    $begingroup$
    Since $psi:=e^{lambda(phi-1)}$ is a characteristic function, we can associate a probability measure, say $mu$, with $psi$ through the relation $$psi(t) = int e^{it x} , mu(dx).$$ You are supposed to show that $mu$ is infinitely divisible.
    $endgroup$
    – saz
    Jan 6 at 17:37












  • $begingroup$
    @saz thanks for your comment. Why are we considering only $varphi(t)$? Not the full expression that contains it?
    $endgroup$
    – Hendrra
    Jan 6 at 17:40






  • 1




    $begingroup$
    Ah, sorry, my mistake... please see my edited comment.
    $endgroup$
    – saz
    Jan 6 at 17:42










  • $begingroup$
    Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $mu$ or is there a method of dealing with such things?
    $endgroup$
    – Hendrra
    Jan 6 at 17:59








2




2




$begingroup$
Since $psi:=e^{lambda(phi-1)}$ is a characteristic function, we can associate a probability measure, say $mu$, with $psi$ through the relation $$psi(t) = int e^{it x} , mu(dx).$$ You are supposed to show that $mu$ is infinitely divisible.
$endgroup$
– saz
Jan 6 at 17:37






$begingroup$
Since $psi:=e^{lambda(phi-1)}$ is a characteristic function, we can associate a probability measure, say $mu$, with $psi$ through the relation $$psi(t) = int e^{it x} , mu(dx).$$ You are supposed to show that $mu$ is infinitely divisible.
$endgroup$
– saz
Jan 6 at 17:37














$begingroup$
@saz thanks for your comment. Why are we considering only $varphi(t)$? Not the full expression that contains it?
$endgroup$
– Hendrra
Jan 6 at 17:40




$begingroup$
@saz thanks for your comment. Why are we considering only $varphi(t)$? Not the full expression that contains it?
$endgroup$
– Hendrra
Jan 6 at 17:40




1




1




$begingroup$
Ah, sorry, my mistake... please see my edited comment.
$endgroup$
– saz
Jan 6 at 17:42




$begingroup$
Ah, sorry, my mistake... please see my edited comment.
$endgroup$
– saz
Jan 6 at 17:42












$begingroup$
Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $mu$ or is there a method of dealing with such things?
$endgroup$
– Hendrra
Jan 6 at 17:59




$begingroup$
Thanks. Now I understand! We don't know if there is a density function, do we? Am I to find $mu$ or is there a method of dealing with such things?
$endgroup$
– Hendrra
Jan 6 at 17:59










1 Answer
1






active

oldest

votes


















3












$begingroup$

Hints:




  1. Let $mu$ be a probability distribution with characteristic function $psi$. Show that $mu$ is infinitely divisible if for any $n in mathbb{N}$ there exists a characteristic function $Phi$ such that $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$

  2. Use Step 1 for $psi(t) = e^{lambda (varphi(t)-1)}$. Try to find a suitable characteristic function $Phi$. (Hint: No need for complicated calculations. Use $e^x = (e^{x/n})^n$.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
    $endgroup$
    – Hendrra
    Jan 6 at 19:03






  • 1




    $begingroup$
    @Hendrra Yes, that's it.
    $endgroup$
    – saz
    Jan 6 at 19:47












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Hints:




  1. Let $mu$ be a probability distribution with characteristic function $psi$. Show that $mu$ is infinitely divisible if for any $n in mathbb{N}$ there exists a characteristic function $Phi$ such that $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$

  2. Use Step 1 for $psi(t) = e^{lambda (varphi(t)-1)}$. Try to find a suitable characteristic function $Phi$. (Hint: No need for complicated calculations. Use $e^x = (e^{x/n})^n$.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
    $endgroup$
    – Hendrra
    Jan 6 at 19:03






  • 1




    $begingroup$
    @Hendrra Yes, that's it.
    $endgroup$
    – saz
    Jan 6 at 19:47
















3












$begingroup$

Hints:




  1. Let $mu$ be a probability distribution with characteristic function $psi$. Show that $mu$ is infinitely divisible if for any $n in mathbb{N}$ there exists a characteristic function $Phi$ such that $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$

  2. Use Step 1 for $psi(t) = e^{lambda (varphi(t)-1)}$. Try to find a suitable characteristic function $Phi$. (Hint: No need for complicated calculations. Use $e^x = (e^{x/n})^n$.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
    $endgroup$
    – Hendrra
    Jan 6 at 19:03






  • 1




    $begingroup$
    @Hendrra Yes, that's it.
    $endgroup$
    – saz
    Jan 6 at 19:47














3












3








3





$begingroup$

Hints:




  1. Let $mu$ be a probability distribution with characteristic function $psi$. Show that $mu$ is infinitely divisible if for any $n in mathbb{N}$ there exists a characteristic function $Phi$ such that $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$

  2. Use Step 1 for $psi(t) = e^{lambda (varphi(t)-1)}$. Try to find a suitable characteristic function $Phi$. (Hint: No need for complicated calculations. Use $e^x = (e^{x/n})^n$.)






share|cite|improve this answer











$endgroup$



Hints:




  1. Let $mu$ be a probability distribution with characteristic function $psi$. Show that $mu$ is infinitely divisible if for any $n in mathbb{N}$ there exists a characteristic function $Phi$ such that $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$

  2. Use Step 1 for $psi(t) = e^{lambda (varphi(t)-1)}$. Try to find a suitable characteristic function $Phi$. (Hint: No need for complicated calculations. Use $e^x = (e^{x/n})^n$.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 15:42

























answered Jan 6 at 18:40









sazsaz

82.4k862131




82.4k862131












  • $begingroup$
    Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
    $endgroup$
    – Hendrra
    Jan 6 at 19:03






  • 1




    $begingroup$
    @Hendrra Yes, that's it.
    $endgroup$
    – saz
    Jan 6 at 19:47


















  • $begingroup$
    Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
    $endgroup$
    – Hendrra
    Jan 6 at 19:03






  • 1




    $begingroup$
    @Hendrra Yes, that's it.
    $endgroup$
    – saz
    Jan 6 at 19:47
















$begingroup$
Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
$endgroup$
– Hendrra
Jan 6 at 19:03




$begingroup$
Thanks! That was easier than I thought. So to sum up you showed that there exist a function $Phi(t) = e^{lambda/n(varphi(t) - 1)}$ which is a characteristic function thus $mu$ is infinitely divisible and $psi$ by definition is an infinitely divisible function?
$endgroup$
– Hendrra
Jan 6 at 19:03




1




1




$begingroup$
@Hendrra Yes, that's it.
$endgroup$
– saz
Jan 6 at 19:47




$begingroup$
@Hendrra Yes, that's it.
$endgroup$
– saz
Jan 6 at 19:47


















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