Rate of convergence of sum of random variables
Clash Royale CLAN TAG#URR8PPP
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I'm working on the following problem and I'm a little stuck
Suppose $X_1,X_2,dots$ are iid.
(a) If $E|X_1|^alpha$ is finite for some $alpha>0$, show that $max_1le kle n |X_k|/n^1/alphato 0$ a.s.
(b) If $EX_1$ is finite and nonzero, show that $max_1le k le n|X_k|/|S_n|to 0$ a.s.
For part a) I'm thinking of using Borel-Cantelli by considering $sum P(max|X_k|^alpha
> epsilon n)$ but I'm having trouble showing it's finite.
For part b) I think you would just use part a) and split up the product into $$fracmaxn^1/alphafracn^1/alpha$$ Then it's just a matter of showing $$fracn^1/alpha$$ converges to something finite, which seems like an SLLN type problem, but not quite. Any advice would be appreciated.
source: Spring 1997
probability probability-theory law-of-large-numbers borel-cantelli-lemmas
add a comment |Â
up vote
2
down vote
favorite
I'm working on the following problem and I'm a little stuck
Suppose $X_1,X_2,dots$ are iid.
(a) If $E|X_1|^alpha$ is finite for some $alpha>0$, show that $max_1le kle n |X_k|/n^1/alphato 0$ a.s.
(b) If $EX_1$ is finite and nonzero, show that $max_1le k le n|X_k|/|S_n|to 0$ a.s.
For part a) I'm thinking of using Borel-Cantelli by considering $sum P(max|X_k|^alpha
> epsilon n)$ but I'm having trouble showing it's finite.
For part b) I think you would just use part a) and split up the product into $$fracmaxn^1/alphafracn^1/alpha$$ Then it's just a matter of showing $$fracn^1/alpha$$ converges to something finite, which seems like an SLLN type problem, but not quite. Any advice would be appreciated.
source: Spring 1997
probability probability-theory law-of-large-numbers borel-cantelli-lemmas
Markov's inequality for part (a)?
â Brian Tung
Aug 7 at 21:29
1
You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view
â Mike Earnest
Aug 7 at 21:58
This is going to be extremely useful. Thanks a ton!
â iYOA
Aug 7 at 22:07
Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
â Mike Earnest
Aug 7 at 22:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm working on the following problem and I'm a little stuck
Suppose $X_1,X_2,dots$ are iid.
(a) If $E|X_1|^alpha$ is finite for some $alpha>0$, show that $max_1le kle n |X_k|/n^1/alphato 0$ a.s.
(b) If $EX_1$ is finite and nonzero, show that $max_1le k le n|X_k|/|S_n|to 0$ a.s.
For part a) I'm thinking of using Borel-Cantelli by considering $sum P(max|X_k|^alpha
> epsilon n)$ but I'm having trouble showing it's finite.
For part b) I think you would just use part a) and split up the product into $$fracmaxn^1/alphafracn^1/alpha$$ Then it's just a matter of showing $$fracn^1/alpha$$ converges to something finite, which seems like an SLLN type problem, but not quite. Any advice would be appreciated.
source: Spring 1997
probability probability-theory law-of-large-numbers borel-cantelli-lemmas
I'm working on the following problem and I'm a little stuck
Suppose $X_1,X_2,dots$ are iid.
(a) If $E|X_1|^alpha$ is finite for some $alpha>0$, show that $max_1le kle n |X_k|/n^1/alphato 0$ a.s.
(b) If $EX_1$ is finite and nonzero, show that $max_1le k le n|X_k|/|S_n|to 0$ a.s.
For part a) I'm thinking of using Borel-Cantelli by considering $sum P(max|X_k|^alpha
> epsilon n)$ but I'm having trouble showing it's finite.
For part b) I think you would just use part a) and split up the product into $$fracmaxn^1/alphafracn^1/alpha$$ Then it's just a matter of showing $$fracn^1/alpha$$ converges to something finite, which seems like an SLLN type problem, but not quite. Any advice would be appreciated.
source: Spring 1997
probability probability-theory law-of-large-numbers borel-cantelli-lemmas
edited Aug 8 at 1:16
Mike Earnest
15.7k11645
15.7k11645
asked Aug 7 at 20:38
iYOA
60549
60549
Markov's inequality for part (a)?
â Brian Tung
Aug 7 at 21:29
1
You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view
â Mike Earnest
Aug 7 at 21:58
This is going to be extremely useful. Thanks a ton!
â iYOA
Aug 7 at 22:07
Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
â Mike Earnest
Aug 7 at 22:08
add a comment |Â
Markov's inequality for part (a)?
â Brian Tung
Aug 7 at 21:29
1
You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view
â Mike Earnest
Aug 7 at 21:58
This is going to be extremely useful. Thanks a ton!
â iYOA
Aug 7 at 22:07
Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
â Mike Earnest
Aug 7 at 22:08
Markov's inequality for part (a)?
â Brian Tung
Aug 7 at 21:29
Markov's inequality for part (a)?
â Brian Tung
Aug 7 at 21:29
1
1
You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view
â Mike Earnest
Aug 7 at 21:58
You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view
â Mike Earnest
Aug 7 at 21:58
This is going to be extremely useful. Thanks a ton!
â iYOA
Aug 7 at 22:07
This is going to be extremely useful. Thanks a ton!
â iYOA
Aug 7 at 22:07
Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
â Mike Earnest
Aug 7 at 22:08
Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
â Mike Earnest
Aug 7 at 22:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
I will use the following two Lemmas, try to prove them on your own:
If $X_i$ is an iid sequence with $mathbb E|X_1|<infty$, then $X_i/nto 0$ a.s.
If $a_n$ is a sequence of real numbers with $a_n/nto 0$, then $frac1n max_1le ile na_ito 0.$
Onto your problems:
Since $E|X_1|^alpha<infty$, the first Lemma implies $|X_1|^alpha/nto 0$ a.s, then the second lemma implies $(max_1le kle n |X_k|^alpha)/nto 0$ a.s. Conclude by raising each term to the $1/alpha$.
You should split it up like $$fracX_kncdot fracn.$$The first fraction converges to $0$ by part (a). Since $n/S_nto 1/EX_1$, you have $n/|S_n|to 1/|EX_1|$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I will use the following two Lemmas, try to prove them on your own:
If $X_i$ is an iid sequence with $mathbb E|X_1|<infty$, then $X_i/nto 0$ a.s.
If $a_n$ is a sequence of real numbers with $a_n/nto 0$, then $frac1n max_1le ile na_ito 0.$
Onto your problems:
Since $E|X_1|^alpha<infty$, the first Lemma implies $|X_1|^alpha/nto 0$ a.s, then the second lemma implies $(max_1le kle n |X_k|^alpha)/nto 0$ a.s. Conclude by raising each term to the $1/alpha$.
You should split it up like $$fracX_kncdot fracn.$$The first fraction converges to $0$ by part (a). Since $n/S_nto 1/EX_1$, you have $n/|S_n|to 1/|EX_1|$.
add a comment |Â
up vote
3
down vote
accepted
I will use the following two Lemmas, try to prove them on your own:
If $X_i$ is an iid sequence with $mathbb E|X_1|<infty$, then $X_i/nto 0$ a.s.
If $a_n$ is a sequence of real numbers with $a_n/nto 0$, then $frac1n max_1le ile na_ito 0.$
Onto your problems:
Since $E|X_1|^alpha<infty$, the first Lemma implies $|X_1|^alpha/nto 0$ a.s, then the second lemma implies $(max_1le kle n |X_k|^alpha)/nto 0$ a.s. Conclude by raising each term to the $1/alpha$.
You should split it up like $$fracX_kncdot fracn.$$The first fraction converges to $0$ by part (a). Since $n/S_nto 1/EX_1$, you have $n/|S_n|to 1/|EX_1|$.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I will use the following two Lemmas, try to prove them on your own:
If $X_i$ is an iid sequence with $mathbb E|X_1|<infty$, then $X_i/nto 0$ a.s.
If $a_n$ is a sequence of real numbers with $a_n/nto 0$, then $frac1n max_1le ile na_ito 0.$
Onto your problems:
Since $E|X_1|^alpha<infty$, the first Lemma implies $|X_1|^alpha/nto 0$ a.s, then the second lemma implies $(max_1le kle n |X_k|^alpha)/nto 0$ a.s. Conclude by raising each term to the $1/alpha$.
You should split it up like $$fracX_kncdot fracn.$$The first fraction converges to $0$ by part (a). Since $n/S_nto 1/EX_1$, you have $n/|S_n|to 1/|EX_1|$.
I will use the following two Lemmas, try to prove them on your own:
If $X_i$ is an iid sequence with $mathbb E|X_1|<infty$, then $X_i/nto 0$ a.s.
If $a_n$ is a sequence of real numbers with $a_n/nto 0$, then $frac1n max_1le ile na_ito 0.$
Onto your problems:
Since $E|X_1|^alpha<infty$, the first Lemma implies $|X_1|^alpha/nto 0$ a.s, then the second lemma implies $(max_1le kle n |X_k|^alpha)/nto 0$ a.s. Conclude by raising each term to the $1/alpha$.
You should split it up like $$fracX_kncdot fracn.$$The first fraction converges to $0$ by part (a). Since $n/S_nto 1/EX_1$, you have $n/|S_n|to 1/|EX_1|$.
answered Aug 7 at 21:33
Mike Earnest
15.7k11645
15.7k11645
add a comment |Â
add a comment |Â
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Markov's inequality for part (a)?
â Brian Tung
Aug 7 at 21:29
1
You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view
â Mike Earnest
Aug 7 at 21:58
This is going to be extremely useful. Thanks a ton!
â iYOA
Aug 7 at 22:07
Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
â Mike Earnest
Aug 7 at 22:08