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
edited Jan 6 at 18:41
saz
82.3k862131
asked Jan 6 at 17:30
HendrraHendrra
1,214516
$endgroup$
add a comment |
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
edited Jan 6 at 18:41
saz
82.3k862131
asked Jan 6 at 17:30
HendrraHendrra
1,214516
$endgroup$
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
add a comment |
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
edited Jan 6 at 18:41
saz
82.3k862131
asked Jan 6 at 17:30
HendrraHendrra
1,214516
$endgroup$
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
edited Jan 6 at 18:41
saz
82.3k862131
asked Jan 6 at 17:30
HendrraHendrra
1,214516
edited Jan 6 at 18:41
saz
82.3k862131
asked Jan 6 at 17:30
HendrraHendrra
1,214516
edited Jan 6 at 18:41
saz
82.3k862131
edited Jan 6 at 18:41
saz
82.3k862131
edited Jan 6 at 18:41
saz
82.3k862131
82.3k862131
asked Jan 6 at 17:30
HendrraHendrra
1,214516
asked Jan 6 at 17:30
HendrraHendrra
1,214516
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
Hints:
- 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 $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$
- 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 $e^x = (e^{x/n})^n$.)
answered Jan 6 at 18:40
sazsaz
82.3k862131
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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3
$begingroup$
Hints:
- 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 $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$
- 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 $e^x = (e^{x/n})^n$.)
answered Jan 6 at 18:40
sazsaz
82.3k862131
$endgroup$
$begingroup$
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
add a comment |
3
$begingroup$
Hints:
- 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 $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$
- 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 $e^x = (e^{x/n})^n$.)
answered Jan 6 at 18:40
sazsaz
82.3k862131
$endgroup$
$begingroup$
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
add a comment |
3
3
3
$begingroup$
Hints:
- 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 $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$
- 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 $e^x = (e^{x/n})^n$.)
answered Jan 6 at 18:40
sazsaz
82.3k862131
$endgroup$
Hints:
- 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 $$psi(t) = (Phi(t))^n, qquad t in mathbb{R}^d.$$
- 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 $e^x = (e^{x/n})^n$.)
answered Jan 6 at 18:40
sazsaz
82.3k862131
edited Jan 7 at 15:42
answered Jan 6 at 18:40
sazsaz
82.3k862131
answered Jan 6 at 18:40
sazsaz
82.3k862131
answered Jan 6 at 18:40
sazsaz
82.3k862131
82.3k862131
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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