Mixing
Finding if $phi(t) = frac{cos(t)}{1 + t^4}$ is a characteristic function
$begingroup$
Let's consider a function:
$$phi(t) = frac{cos(t)}{1 + t^4} tag{1}.$$
How can I check whether $(1)$ is a characteristic function? I tried using Polya's criterion. Unfortunately it doesn't work here.
probability-theory characteristic-functions
edited Feb 1 at 17:53
saz
82.2k862131
asked Jan 31 at 13:51
HendrraHendrra
1,214516
$endgroup$
add a comment |
$begingroup$
Let's consider a function:
$$phi(t) = frac{cos(t)}{1 + t^4} tag{1}.$$
How can I check whether $(1)$ is a characteristic function? I tried using Polya's criterion. Unfortunately it doesn't work here.
probability-theory characteristic-functions
edited Feb 1 at 17:53
saz
82.2k862131
asked Jan 31 at 13:51
HendrraHendrra
1,214516
$endgroup$
add a comment |
$begingroup$
Let's consider a function:
$$phi(t) = frac{cos(t)}{1 + t^4} tag{1}.$$
How can I check whether $(1)$ is a characteristic function? I tried using Polya's criterion. Unfortunately it doesn't work here.
probability-theory characteristic-functions
edited Feb 1 at 17:53
saz
82.2k862131
asked Jan 31 at 13:51
HendrraHendrra
1,214516
$endgroup$
Let's consider a function:
$$phi(t) = frac{cos(t)}{1 + t^4} tag{1}.$$
How can I check whether $(1)$ is a characteristic function? I tried using Polya's criterion. Unfortunately it doesn't work here.
probability-theory characteristic-functions
probability-theory characteristic-functions
edited Feb 1 at 17:53
saz
82.2k862131
asked Jan 31 at 13:51
HendrraHendrra
1,214516
edited Feb 1 at 17:53
saz
82.2k862131
asked Jan 31 at 13:51
HendrraHendrra
1,214516
edited Feb 1 at 17:53
saz
82.2k862131
edited Feb 1 at 17:53
saz
82.2k862131
edited Feb 1 at 17:53
saz
82.2k862131
82.2k862131
asked Jan 31 at 13:51
HendrraHendrra
1,214516
asked Jan 31 at 13:51
HendrraHendrra
1,214516
asked Jan 31 at 13:51
HendrraHendrra
1,214516
1,214516
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$phi$ is clearly a smooth function. Calculating the fourth derivative shows
$$phi^{(4)}(0)=-23 <0.$$
By the following well-known statement, this implies that $phi$ is not a characteristic function.
Lemma: Let $f(t) = mathbb{E}e^{itX}$, $t in mathbb{R}$, be a characteristic function which is $4$ times differentiable. Then the $4$th derivative $f^{(4)}$ satisfies $$f^{(4)}(0) = mathbb{E}(X^4) geq 0.$$
answered Jan 31 at 15:47
sazsaz
82.2k862131
$endgroup$
add a comment |
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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
$begingroup$
$phi$ is clearly a smooth function. Calculating the fourth derivative shows
$$phi^{(4)}(0)=-23 <0.$$
By the following well-known statement, this implies that $phi$ is not a characteristic function.
Lemma: Let $f(t) = mathbb{E}e^{itX}$, $t in mathbb{R}$, be a characteristic function which is $4$ times differentiable. Then the $4$th derivative $f^{(4)}$ satisfies $$f^{(4)}(0) = mathbb{E}(X^4) geq 0.$$
answered Jan 31 at 15:47
sazsaz
82.2k862131
$endgroup$
add a comment |
$begingroup$
$phi$ is clearly a smooth function. Calculating the fourth derivative shows
$$phi^{(4)}(0)=-23 <0.$$
By the following well-known statement, this implies that $phi$ is not a characteristic function.
Lemma: Let $f(t) = mathbb{E}e^{itX}$, $t in mathbb{R}$, be a characteristic function which is $4$ times differentiable. Then the $4$th derivative $f^{(4)}$ satisfies $$f^{(4)}(0) = mathbb{E}(X^4) geq 0.$$
answered Jan 31 at 15:47
sazsaz
82.2k862131
$endgroup$
add a comment |
$begingroup$
$phi$ is clearly a smooth function. Calculating the fourth derivative shows
$$phi^{(4)}(0)=-23 <0.$$
By the following well-known statement, this implies that $phi$ is not a characteristic function.
Lemma: Let $f(t) = mathbb{E}e^{itX}$, $t in mathbb{R}$, be a characteristic function which is $4$ times differentiable. Then the $4$th derivative $f^{(4)}$ satisfies $$f^{(4)}(0) = mathbb{E}(X^4) geq 0.$$
answered Jan 31 at 15:47
sazsaz
82.2k862131
$endgroup$
$phi$ is clearly a smooth function. Calculating the fourth derivative shows
$$phi^{(4)}(0)=-23 <0.$$
By the following well-known statement, this implies that $phi$ is not a characteristic function.
Lemma: Let $f(t) = mathbb{E}e^{itX}$, $t in mathbb{R}$, be a characteristic function which is $4$ times differentiable. Then the $4$th derivative $f^{(4)}$ satisfies $$f^{(4)}(0) = mathbb{E}(X^4) geq 0.$$
answered Jan 31 at 15:47
sazsaz
82.2k862131
edited Jan 31 at 18:10
answered Jan 31 at 15:47
sazsaz
82.2k862131
answered Jan 31 at 15:47
sazsaz
82.2k862131
answered Jan 31 at 15:47
sazsaz
82.2k862131
82.2k862131
add a comment |
add a comment |
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