Mixing
Large Deviations rate function and Cramérs Theorem
$begingroup$
Given $X_1,...$ of iid random variables. We know that if the moment generating function $M(theta) < infty, forall theta $ from Cramérs Theorem we get:
$lim_{nto infty} frac{1}{n}log mathbb{P}(S_n ge na) = -I(a)$ where
$I(a) = sup_theta(atheta - log M(theta))$.
Question: What happens if $M(theta)$ isn't finite for all values of $theta$? Namely, is there another version of Cramérs theorem to help when calculating the rate function $I$ when this situation comes up.
large-deviation-theory
asked Jan 22 at 20:24
all.overall.over
537
$endgroup$
add a comment |
$begingroup$
Given $X_1,...$ of iid random variables. We know that if the moment generating function $M(theta) < infty, forall theta $ from Cramérs Theorem we get:
$lim_{nto infty} frac{1}{n}log mathbb{P}(S_n ge na) = -I(a)$ where
$I(a) = sup_theta(atheta - log M(theta))$.
Question: What happens if $M(theta)$ isn't finite for all values of $theta$? Namely, is there another version of Cramérs theorem to help when calculating the rate function $I$ when this situation comes up.
large-deviation-theory
asked Jan 22 at 20:24
all.overall.over
537
$endgroup$
add a comment |
$begingroup$
Given $X_1,...$ of iid random variables. We know that if the moment generating function $M(theta) < infty, forall theta $ from Cramérs Theorem we get:
$lim_{nto infty} frac{1}{n}log mathbb{P}(S_n ge na) = -I(a)$ where
$I(a) = sup_theta(atheta - log M(theta))$.
Question: What happens if $M(theta)$ isn't finite for all values of $theta$? Namely, is there another version of Cramérs theorem to help when calculating the rate function $I$ when this situation comes up.
large-deviation-theory
asked Jan 22 at 20:24
all.overall.over
537
$endgroup$
Given $X_1,...$ of iid random variables. We know that if the moment generating function $M(theta) < infty, forall theta $ from Cramérs Theorem we get:
$lim_{nto infty} frac{1}{n}log mathbb{P}(S_n ge na) = -I(a)$ where
$I(a) = sup_theta(atheta - log M(theta))$.
Question: What happens if $M(theta)$ isn't finite for all values of $theta$? Namely, is there another version of Cramérs theorem to help when calculating the rate function $I$ when this situation comes up.
large-deviation-theory
large-deviation-theory
asked Jan 22 at 20:24
all.overall.over
537
asked Jan 22 at 20:24
all.overall.over
537
edited Jan 22 at 20:47
all.over
asked Jan 22 at 20:24
all.overall.over
537
asked Jan 22 at 20:24
all.overall.over
537
asked Jan 22 at 20:24
all.overall.over
537
537
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes. Let $Theta$ be the set of $theta$ such that $M(theta)<infty$. It is easy to check that $Theta$ is convex, that is, an interval. As long as $theta$ is in the interior of $Theta$, and solves $M'(theta)/M(theta)=a$, that is, the maximization problem defining $I(a)$ is solved the calculus way, the conclusion holds for that value of $a$. The condition that you cite, that $Theta=mathbb R$, is introduced for convenience of exposition.
I'm away from my books at the moment, so I can't give a reference to this version.
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
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1 Answer
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$begingroup$
Yes. Let $Theta$ be the set of $theta$ such that $M(theta)<infty$. It is easy to check that $Theta$ is convex, that is, an interval. As long as $theta$ is in the interior of $Theta$, and solves $M'(theta)/M(theta)=a$, that is, the maximization problem defining $I(a)$ is solved the calculus way, the conclusion holds for that value of $a$. The condition that you cite, that $Theta=mathbb R$, is introduced for convenience of exposition.
I'm away from my books at the moment, so I can't give a reference to this version.
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
$endgroup$
add a comment |
$begingroup$
Yes. Let $Theta$ be the set of $theta$ such that $M(theta)<infty$. It is easy to check that $Theta$ is convex, that is, an interval. As long as $theta$ is in the interior of $Theta$, and solves $M'(theta)/M(theta)=a$, that is, the maximization problem defining $I(a)$ is solved the calculus way, the conclusion holds for that value of $a$. The condition that you cite, that $Theta=mathbb R$, is introduced for convenience of exposition.
I'm away from my books at the moment, so I can't give a reference to this version.
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
$endgroup$
add a comment |
$begingroup$
Yes. Let $Theta$ be the set of $theta$ such that $M(theta)<infty$. It is easy to check that $Theta$ is convex, that is, an interval. As long as $theta$ is in the interior of $Theta$, and solves $M'(theta)/M(theta)=a$, that is, the maximization problem defining $I(a)$ is solved the calculus way, the conclusion holds for that value of $a$. The condition that you cite, that $Theta=mathbb R$, is introduced for convenience of exposition.
I'm away from my books at the moment, so I can't give a reference to this version.
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
$endgroup$
Yes. Let $Theta$ be the set of $theta$ such that $M(theta)<infty$. It is easy to check that $Theta$ is convex, that is, an interval. As long as $theta$ is in the interior of $Theta$, and solves $M'(theta)/M(theta)=a$, that is, the maximization problem defining $I(a)$ is solved the calculus way, the conclusion holds for that value of $a$. The condition that you cite, that $Theta=mathbb R$, is introduced for convenience of exposition.
I'm away from my books at the moment, so I can't give a reference to this version.
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
answered Jan 23 at 23:50
kimchi loverkimchi lover
11.1k31229
11.1k31229
add a comment |
add a comment |
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