Rate of convergence of sum of random variables

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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







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  • Markov's inequality for part (a)?
    – 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
    – Mike Earnest
    Aug 7 at 21:58










  • This is going to be extremely useful. Thanks a ton!
    – iYOA
    Aug 7 at 22:07










  • Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
    – Mike Earnest
    Aug 7 at 22:08














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







share|cite|improve this question





















  • Markov's inequality for part (a)?
    – 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
    – Mike Earnest
    Aug 7 at 21:58










  • This is going to be extremely useful. Thanks a ton!
    – iYOA
    Aug 7 at 22:07










  • Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
    – Mike Earnest
    Aug 7 at 22:08












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







share|cite|improve this question













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









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








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)?
    – 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
    – Mike Earnest
    Aug 7 at 21:58










  • This is going to be extremely useful. Thanks a ton!
    – iYOA
    Aug 7 at 22:07










  • Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
    – Mike Earnest
    Aug 7 at 22:08
















  • Markov's inequality for part (a)?
    – 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
    – Mike Earnest
    Aug 7 at 21:58










  • This is going to be extremely useful. Thanks a ton!
    – iYOA
    Aug 7 at 22:07










  • Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
    – Mike Earnest
    Aug 7 at 22:08















Markov's inequality for part (a)?
– Brian Tung
Aug 7 at 21:29




Markov's inequality for part (a)?
– 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
– 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
– Mike Earnest
Aug 7 at 21:58












This is going to be extremely useful. Thanks a ton!
– iYOA
Aug 7 at 22:07




This is going to be extremely useful. Thanks a ton!
– iYOA
Aug 7 at 22:07












Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
– Mike Earnest
Aug 7 at 22:08




Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly.
– Mike Earnest
Aug 7 at 22:08










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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:



  1. 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 $1/alpha$.


  2. 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 $n/|S_n|to 1/|EX_1|$.






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    1 Answer
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    active

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    active

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    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:



    1. 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 $1/alpha$.


    2. 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 $n/|S_n|to 1/|EX_1|$.






    share|cite|improve this answer

























      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:



      1. 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 $1/alpha$.


      2. 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 $n/|S_n|to 1/|EX_1|$.






      share|cite|improve this answer























        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:



        1. 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 $1/alpha$.


        2. 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 $n/|S_n|to 1/|EX_1|$.






        share|cite|improve this answer













        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:



        1. 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 $1/alpha$.


        2. 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 $n/|S_n|to 1/|EX_1|$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 7 at 21:33









        Mike Earnest

        15.7k11645




        15.7k11645






















             

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