Why is $limsup S_n $ a constant in $[-infty,infty]$ according to Hewitt-Savage $0-1$ law?
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This question is based on a statement in Durrett's book, 4th edition, page 181, in the proof the theorem in that page.
$S_n = X_1+...+X_n$, with $X_i$ iid.
I'm supposed to notice that $limsup_n S_n =c$ is a permutable event, and hence, by the Hewitt-Savage $0-1$ law, we have $exists_c in [-infty,infty] P(limsup_n S_n =c)in 0,1$.
But why should it be $1$ and not $0$?
probability-theory
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
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up vote
0
down vote
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This question is based on a statement in Durrett's book, 4th edition, page 181, in the proof the theorem in that page.
$S_n = X_1+...+X_n$, with $X_i$ iid.
I'm supposed to notice that $limsup_n S_n =c$ is a permutable event, and hence, by the Hewitt-Savage $0-1$ law, we have $exists_c in [-infty,infty] P(limsup_n S_n =c)in 0,1$.
But why should it be $1$ and not $0$?
probability-theory
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
Well hang on, $limsup S_n(omega)$ exists and is in $[-infty, infty]$ for any $omega$, you don't need Hewitt-Savage to prove that. Are you sure that's what the text says?
– Jack M
Aug 26 at 10:12
@JackM First line of the proof on page 181: «Theorem 4.1.1 [Hewitt-Savage law] implies limsup $S_n$ is a constant $c in [-infty, infty]$.
– An old man in the sea.
Aug 26 at 10:16
Okay, then the formula after "Hence, by..." in your question is not quite right. You are trying to prove $exists c P(limsup S_n=c)in0,1$.
– Jack M
Aug 26 at 11:11
To clarify: the event $liminf S_nin[-infty, infty]$ is indeed a permutable event... because it's equal to the entire probability space!
– Jack M
Aug 26 at 11:18
I've edited the question
– An old man in the sea.
Aug 27 at 8:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is based on a statement in Durrett's book, 4th edition, page 181, in the proof the theorem in that page.
$S_n = X_1+...+X_n$, with $X_i$ iid.
I'm supposed to notice that $limsup_n S_n =c$ is a permutable event, and hence, by the Hewitt-Savage $0-1$ law, we have $exists_c in [-infty,infty] P(limsup_n S_n =c)in 0,1$.
But why should it be $1$ and not $0$?
probability-theory
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
This question is based on a statement in Durrett's book, 4th edition, page 181, in the proof the theorem in that page.
$S_n = X_1+...+X_n$, with $X_i$ iid.
I'm supposed to notice that $limsup_n S_n =c$ is a permutable event, and hence, by the Hewitt-Savage $0-1$ law, we have $exists_c in [-infty,infty] P(limsup_n S_n =c)in 0,1$.
But why should it be $1$ and not $0$?
probability-theory
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
edited Aug 26 at 11:33
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
asked Aug 26 at 10:07
An old man in the sea.
1,2901929
1,2901929
Well hang on, $limsup S_n(omega)$ exists and is in $[-infty, infty]$ for any $omega$, you don't need Hewitt-Savage to prove that. Are you sure that's what the text says?
– Jack M
Aug 26 at 10:12
@JackM First line of the proof on page 181: «Theorem 4.1.1 [Hewitt-Savage law] implies limsup $S_n$ is a constant $c in [-infty, infty]$.
– An old man in the sea.
Aug 26 at 10:16
Okay, then the formula after "Hence, by..." in your question is not quite right. You are trying to prove $exists c P(limsup S_n=c)in0,1$.
– Jack M
Aug 26 at 11:11
To clarify: the event $liminf S_nin[-infty, infty]$ is indeed a permutable event... because it's equal to the entire probability space!
– Jack M
Aug 26 at 11:18
I've edited the question
– An old man in the sea.
Aug 27 at 8:56
add a comment |Â
Well hang on, $limsup S_n(omega)$ exists and is in $[-infty, infty]$ for any $omega$, you don't need Hewitt-Savage to prove that. Are you sure that's what the text says?
– Jack M
Aug 26 at 10:12
@JackM First line of the proof on page 181: «Theorem 4.1.1 [Hewitt-Savage law] implies limsup $S_n$ is a constant $c in [-infty, infty]$.
– An old man in the sea.
Aug 26 at 10:16
Okay, then the formula after "Hence, by..." in your question is not quite right. You are trying to prove $exists c P(limsup S_n=c)in0,1$.
– Jack M
Aug 26 at 11:11
To clarify: the event $liminf S_nin[-infty, infty]$ is indeed a permutable event... because it's equal to the entire probability space!
– Jack M
Aug 26 at 11:18
I've edited the question
– An old man in the sea.
Aug 27 at 8:56
Well hang on, $limsup S_n(omega)$ exists and is in $[-infty, infty]$ for any $omega$, you don't need Hewitt-Savage to prove that. Are you sure that's what the text says?
– Jack M
Aug 26 at 10:12
Well hang on, $limsup S_n(omega)$ exists and is in $[-infty, infty]$ for any $omega$, you don't need Hewitt-Savage to prove that. Are you sure that's what the text says?
– Jack M
Aug 26 at 10:12
@JackM First line of the proof on page 181: «Theorem 4.1.1 [Hewitt-Savage law] implies limsup $S_n$ is a constant $c in [-infty, infty]$.
– An old man in the sea.
Aug 26 at 10:16
@JackM First line of the proof on page 181: «Theorem 4.1.1 [Hewitt-Savage law] implies limsup $S_n$ is a constant $c in [-infty, infty]$.
– An old man in the sea.
Aug 26 at 10:16
Okay, then the formula after "Hence, by..." in your question is not quite right. You are trying to prove $exists c P(limsup S_n=c)in0,1$.
– Jack M
Aug 26 at 11:11
Okay, then the formula after "Hence, by..." in your question is not quite right. You are trying to prove $exists c P(limsup S_n=c)in0,1$.
– Jack M
Aug 26 at 11:11
To clarify: the event $liminf S_nin[-infty, infty]$ is indeed a permutable event... because it's equal to the entire probability space!
– Jack M
Aug 26 at 11:18
To clarify: the event $liminf S_nin[-infty, infty]$ is indeed a permutable event... because it's equal to the entire probability space!
– Jack M
Aug 26 at 11:18
I've edited the question
– An old man in the sea.
Aug 27 at 8:56
I've edited the question
– An old man in the sea.
Aug 27 at 8:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The actual argument is slightly subtle. Note that $limsup S_n$ is a well-defined random variable, so let's examine its distribution. We would like to prove that this distribution is actually a Dirac distribution centered on some point. For any particular constant $c$, the event
$$limsup S_nleq c$$
is a permutable event, and thus occurs with probability either $0$ or $1$. So the cumulative distribution function of $limsup S_n$ only takes on the values $0$ or $1$, which implies the distribution is indeed Dirac.
Maybe a slightly more intuitive way of doing this is to use, instead of a constant $c$, an arbitrary interval $I$. So we have that for any interval, $limsup S_nin I$ occurs with probability $0$ or $1$, and so the distribution is Dirac.
answered Aug 26 at 11:17
Jack M
17.4k33473
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The actual argument is slightly subtle. Note that $limsup S_n$ is a well-defined random variable, so let's examine its distribution. We would like to prove that this distribution is actually a Dirac distribution centered on some point. For any particular constant $c$, the event
$$limsup S_nleq c$$
is a permutable event, and thus occurs with probability either $0$ or $1$. So the cumulative distribution function of $limsup S_n$ only takes on the values $0$ or $1$, which implies the distribution is indeed Dirac.
Maybe a slightly more intuitive way of doing this is to use, instead of a constant $c$, an arbitrary interval $I$. So we have that for any interval, $limsup S_nin I$ occurs with probability $0$ or $1$, and so the distribution is Dirac.
answered Aug 26 at 11:17
Jack M
17.4k33473
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
add a comment |Â
up vote
1
down vote
accepted
The actual argument is slightly subtle. Note that $limsup S_n$ is a well-defined random variable, so let's examine its distribution. We would like to prove that this distribution is actually a Dirac distribution centered on some point. For any particular constant $c$, the event
$$limsup S_nleq c$$
is a permutable event, and thus occurs with probability either $0$ or $1$. So the cumulative distribution function of $limsup S_n$ only takes on the values $0$ or $1$, which implies the distribution is indeed Dirac.
Maybe a slightly more intuitive way of doing this is to use, instead of a constant $c$, an arbitrary interval $I$. So we have that for any interval, $limsup S_nin I$ occurs with probability $0$ or $1$, and so the distribution is Dirac.
answered Aug 26 at 11:17
Jack M
17.4k33473
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The actual argument is slightly subtle. Note that $limsup S_n$ is a well-defined random variable, so let's examine its distribution. We would like to prove that this distribution is actually a Dirac distribution centered on some point. For any particular constant $c$, the event
$$limsup S_nleq c$$
is a permutable event, and thus occurs with probability either $0$ or $1$. So the cumulative distribution function of $limsup S_n$ only takes on the values $0$ or $1$, which implies the distribution is indeed Dirac.
Maybe a slightly more intuitive way of doing this is to use, instead of a constant $c$, an arbitrary interval $I$. So we have that for any interval, $limsup S_nin I$ occurs with probability $0$ or $1$, and so the distribution is Dirac.
answered Aug 26 at 11:17
Jack M
17.4k33473
The actual argument is slightly subtle. Note that $limsup S_n$ is a well-defined random variable, so let's examine its distribution. We would like to prove that this distribution is actually a Dirac distribution centered on some point. For any particular constant $c$, the event
$$limsup S_nleq c$$
is a permutable event, and thus occurs with probability either $0$ or $1$. So the cumulative distribution function of $limsup S_n$ only takes on the values $0$ or $1$, which implies the distribution is indeed Dirac.
Maybe a slightly more intuitive way of doing this is to use, instead of a constant $c$, an arbitrary interval $I$. So we have that for any interval, $limsup S_nin I$ occurs with probability $0$ or $1$, and so the distribution is Dirac.
answered Aug 26 at 11:17
Jack M
17.4k33473
edited Aug 26 at 11:28
answered Aug 26 at 11:17
Jack M
17.4k33473
answered Aug 26 at 11:17
Jack M
17.4k33473
answered Aug 26 at 11:17
Jack M
17.4k33473
17.4k33473
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
add a comment |Â
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
Hi Jack, many thanks ;) +1
– An old man in the sea.
Aug 27 at 9:01
add a comment |Â
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Well hang on, $limsup S_n(omega)$ exists and is in $[-infty, infty]$ for any $omega$, you don't need Hewitt-Savage to prove that. Are you sure that's what the text says?
– Jack M
Aug 26 at 10:12
@JackM First line of the proof on page 181: «Theorem 4.1.1 [Hewitt-Savage law] implies limsup $S_n$ is a constant $c in [-infty, infty]$.
– An old man in the sea.
Aug 26 at 10:16
Okay, then the formula after "Hence, by..." in your question is not quite right. You are trying to prove $exists c P(limsup S_n=c)in0,1$.
– Jack M
Aug 26 at 11:11
To clarify: the event $liminf S_nin[-infty, infty]$ is indeed a permutable event... because it's equal to the entire probability space!
– Jack M
Aug 26 at 11:18
I've edited the question
– An old man in the sea.
Aug 27 at 8:56