Law of Iterated Logarithm when the mean does not exist
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Let $X_1,X_2,ldots$ be an i.i.d. sequence of random variables such that $X_1geq 0$ a.s. and $mathbb P[X_1>x]sim x^{-alpha}$, where $alpha<1$. This implies that $X_1$ does not have finite mean. Does the following law of iterated logarithm hold?
$$
limsup_{ntoinfty} frac{X_1+cdots+X_n}{n^{alpha}(loglog n)^{1-alpha}}=c,quad text{a.s.},
$$
where $c$ is a constant depending on the distribution of $X_1$.
A continuous-time version holds (see here). Also, a special case of the claim exists here (Theorem 3) regarding zeros of the simple random walk, in which $alpha=frac 12$.
probability-theory probability-limit-theorems
asked Nov 14 at 5:54
Ali Khezeli
40328
add a comment |
up vote
5
down vote
favorite
Let $X_1,X_2,ldots$ be an i.i.d. sequence of random variables such that $X_1geq 0$ a.s. and $mathbb P[X_1>x]sim x^{-alpha}$, where $alpha<1$. This implies that $X_1$ does not have finite mean. Does the following law of iterated logarithm hold?
$$
limsup_{ntoinfty} frac{X_1+cdots+X_n}{n^{alpha}(loglog n)^{1-alpha}}=c,quad text{a.s.},
$$
where $c$ is a constant depending on the distribution of $X_1$.
A continuous-time version holds (see here). Also, a special case of the claim exists here (Theorem 3) regarding zeros of the simple random walk, in which $alpha=frac 12$.
probability-theory probability-limit-theorems
asked Nov 14 at 5:54
Ali Khezeli
40328
When $alpha=3/2$ in $mathbb P[X_1>x]sim x^{-alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $infty$, so the mean should lead to the order $xcdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or?
– Diger
Nov 23 at 23:41
@Diger you are right, the condition should be $alpha <1$ not $alpha <2$. I will correct it.
– Ali Khezeli
Nov 24 at 3:13
$alpha leq 1$ but fair enough ;)
– Diger
Nov 24 at 13:16
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $X_1,X_2,ldots$ be an i.i.d. sequence of random variables such that $X_1geq 0$ a.s. and $mathbb P[X_1>x]sim x^{-alpha}$, where $alpha<1$. This implies that $X_1$ does not have finite mean. Does the following law of iterated logarithm hold?
$$
limsup_{ntoinfty} frac{X_1+cdots+X_n}{n^{alpha}(loglog n)^{1-alpha}}=c,quad text{a.s.},
$$
where $c$ is a constant depending on the distribution of $X_1$.
A continuous-time version holds (see here). Also, a special case of the claim exists here (Theorem 3) regarding zeros of the simple random walk, in which $alpha=frac 12$.
probability-theory probability-limit-theorems
asked Nov 14 at 5:54
Ali Khezeli
40328
Let $X_1,X_2,ldots$ be an i.i.d. sequence of random variables such that $X_1geq 0$ a.s. and $mathbb P[X_1>x]sim x^{-alpha}$, where $alpha<1$. This implies that $X_1$ does not have finite mean. Does the following law of iterated logarithm hold?
$$
limsup_{ntoinfty} frac{X_1+cdots+X_n}{n^{alpha}(loglog n)^{1-alpha}}=c,quad text{a.s.},
$$
where $c$ is a constant depending on the distribution of $X_1$.
A continuous-time version holds (see here). Also, a special case of the claim exists here (Theorem 3) regarding zeros of the simple random walk, in which $alpha=frac 12$.
probability-theory probability-limit-theorems
probability-theory probability-limit-theorems
asked Nov 14 at 5:54
Ali Khezeli
40328
asked Nov 14 at 5:54
Ali Khezeli
40328
edited Nov 24 at 3:14
asked Nov 14 at 5:54
Ali Khezeli
40328
asked Nov 14 at 5:54
Ali Khezeli
40328
asked Nov 14 at 5:54
Ali Khezeli
40328
40328
When $alpha=3/2$ in $mathbb P[X_1>x]sim x^{-alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $infty$, so the mean should lead to the order $xcdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or?
– Diger
Nov 23 at 23:41
@Diger you are right, the condition should be $alpha <1$ not $alpha <2$. I will correct it.
– Ali Khezeli
Nov 24 at 3:13
$alpha leq 1$ but fair enough ;)
– Diger
Nov 24 at 13:16
add a comment |
When $alpha=3/2$ in $mathbb P[X_1>x]sim x^{-alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $infty$, so the mean should lead to the order $xcdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or?
– Diger
Nov 23 at 23:41
@Diger you are right, the condition should be $alpha <1$ not $alpha <2$. I will correct it.
– Ali Khezeli
Nov 24 at 3:13
$alpha leq 1$ but fair enough ;)
– Diger
Nov 24 at 13:16
When $alpha=3/2$ in $mathbb P[X_1>x]sim x^{-alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $infty$, so the mean should lead to the order $xcdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or?
– Diger
Nov 23 at 23:41
When $alpha=3/2$ in $mathbb P[X_1>x]sim x^{-alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $infty$, so the mean should lead to the order $xcdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or?
– Diger
Nov 23 at 23:41
@Diger you are right, the condition should be $alpha <1$ not $alpha <2$. I will correct it.
– Ali Khezeli
Nov 24 at 3:13
@Diger you are right, the condition should be $alpha <1$ not $alpha <2$. I will correct it.
– Ali Khezeli
Nov 24 at 3:13
$alpha leq 1$ but fair enough ;)
– Diger
Nov 24 at 13:16
$alpha leq 1$ but fair enough ;)
– Diger
Nov 24 at 13:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes, it holds. Thanks to Jean Bertoin for telling me the answer, the problem is solved in Theorem 4 of this paper.
answered Nov 22 at 11:49
Ali Khezeli
40328
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, it holds. Thanks to Jean Bertoin for telling me the answer, the problem is solved in Theorem 4 of this paper.
answered Nov 22 at 11:49
Ali Khezeli
40328
add a comment |
up vote
2
down vote
accepted
Yes, it holds. Thanks to Jean Bertoin for telling me the answer, the problem is solved in Theorem 4 of this paper.
answered Nov 22 at 11:49
Ali Khezeli
40328
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, it holds. Thanks to Jean Bertoin for telling me the answer, the problem is solved in Theorem 4 of this paper.
answered Nov 22 at 11:49
Ali Khezeli
40328
Yes, it holds. Thanks to Jean Bertoin for telling me the answer, the problem is solved in Theorem 4 of this paper.
answered Nov 22 at 11:49
Ali Khezeli
40328
answered Nov 22 at 11:49
Ali Khezeli
40328
answered Nov 22 at 11:49
Ali Khezeli
40328
answered Nov 22 at 11:49
Ali Khezeli
40328
40328
add a comment |
add a comment |
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When $alpha=3/2$ in $mathbb P[X_1>x]sim x^{-alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $infty$, so the mean should lead to the order $xcdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or?
– Diger
Nov 23 at 23:41
@Diger you are right, the condition should be $alpha <1$ not $alpha <2$. I will correct it.
– Ali Khezeli
Nov 24 at 3:13
$alpha leq 1$ but fair enough ;)
– Diger
Nov 24 at 13:16