Multiple Martingale convergences: almost surely, but not in $L^{1}$ & almost surely and in $L^{2}$...












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Let $X_{i},iin mathbb{N}$ be a sequence of independent standard normal random variables, $mathscr{F}_{n}$ the associated natural filtration, and $S_{n}=sum_{i=1}^{n}X_{i}$. You are allowed to use that for a standard normal random variable we have $mathbb{E}(X^{2p})=(2p-1)!!=mathbb{Pi}^{p}_{i=1}(2i-1)$.



(a) Show that there exists a constant $C_{p}$ such that for all integers $pgeq1$
$$mathbb{E}((max_{1geq kgeq n}S_{k})^{2p})leq C_{p}n^{p}$$



This constant I found by using Doob's optimal inequality theorem and $C_{p}$ is equal to $(frac{2p}{2p-1})^{2p}(frac{2}{pi})^{p}$.



(b) Let $a_{n},ngeq0$ be a sequence of non-negative numbers such that $sum_{n}a_{n}^{2}<infty$. Show that infinite series



$$sum_{n=1}^{infty}a_{n}X_{n}$$



converges in $L^{2}$ and almost surely.



(c) Show that the martingale



$$M_{n}:=exp(S_{n}-frac{1}{2}n)$$



converges almost surely, but not in $L^{1}$.










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closed as off-topic by Davide Giraudo, Leucippus, Rebellos, KReiser, Did Nov 25 '18 at 9:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Leucippus, Rebellos, KReiser, Did

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Sorry, are you asking anything?
    $endgroup$
    – Will M.
    Nov 24 '18 at 22:35
















0












$begingroup$


Let $X_{i},iin mathbb{N}$ be a sequence of independent standard normal random variables, $mathscr{F}_{n}$ the associated natural filtration, and $S_{n}=sum_{i=1}^{n}X_{i}$. You are allowed to use that for a standard normal random variable we have $mathbb{E}(X^{2p})=(2p-1)!!=mathbb{Pi}^{p}_{i=1}(2i-1)$.



(a) Show that there exists a constant $C_{p}$ such that for all integers $pgeq1$
$$mathbb{E}((max_{1geq kgeq n}S_{k})^{2p})leq C_{p}n^{p}$$



This constant I found by using Doob's optimal inequality theorem and $C_{p}$ is equal to $(frac{2p}{2p-1})^{2p}(frac{2}{pi})^{p}$.



(b) Let $a_{n},ngeq0$ be a sequence of non-negative numbers such that $sum_{n}a_{n}^{2}<infty$. Show that infinite series



$$sum_{n=1}^{infty}a_{n}X_{n}$$



converges in $L^{2}$ and almost surely.



(c) Show that the martingale



$$M_{n}:=exp(S_{n}-frac{1}{2}n)$$



converges almost surely, but not in $L^{1}$.










share|cite|improve this question









$endgroup$



closed as off-topic by Davide Giraudo, Leucippus, Rebellos, KReiser, Did Nov 25 '18 at 9:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Leucippus, Rebellos, KReiser, Did

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Sorry, are you asking anything?
    $endgroup$
    – Will M.
    Nov 24 '18 at 22:35














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0








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1



$begingroup$


Let $X_{i},iin mathbb{N}$ be a sequence of independent standard normal random variables, $mathscr{F}_{n}$ the associated natural filtration, and $S_{n}=sum_{i=1}^{n}X_{i}$. You are allowed to use that for a standard normal random variable we have $mathbb{E}(X^{2p})=(2p-1)!!=mathbb{Pi}^{p}_{i=1}(2i-1)$.



(a) Show that there exists a constant $C_{p}$ such that for all integers $pgeq1$
$$mathbb{E}((max_{1geq kgeq n}S_{k})^{2p})leq C_{p}n^{p}$$



This constant I found by using Doob's optimal inequality theorem and $C_{p}$ is equal to $(frac{2p}{2p-1})^{2p}(frac{2}{pi})^{p}$.



(b) Let $a_{n},ngeq0$ be a sequence of non-negative numbers such that $sum_{n}a_{n}^{2}<infty$. Show that infinite series



$$sum_{n=1}^{infty}a_{n}X_{n}$$



converges in $L^{2}$ and almost surely.



(c) Show that the martingale



$$M_{n}:=exp(S_{n}-frac{1}{2}n)$$



converges almost surely, but not in $L^{1}$.










share|cite|improve this question









$endgroup$




Let $X_{i},iin mathbb{N}$ be a sequence of independent standard normal random variables, $mathscr{F}_{n}$ the associated natural filtration, and $S_{n}=sum_{i=1}^{n}X_{i}$. You are allowed to use that for a standard normal random variable we have $mathbb{E}(X^{2p})=(2p-1)!!=mathbb{Pi}^{p}_{i=1}(2i-1)$.



(a) Show that there exists a constant $C_{p}$ such that for all integers $pgeq1$
$$mathbb{E}((max_{1geq kgeq n}S_{k})^{2p})leq C_{p}n^{p}$$



This constant I found by using Doob's optimal inequality theorem and $C_{p}$ is equal to $(frac{2p}{2p-1})^{2p}(frac{2}{pi})^{p}$.



(b) Let $a_{n},ngeq0$ be a sequence of non-negative numbers such that $sum_{n}a_{n}^{2}<infty$. Show that infinite series



$$sum_{n=1}^{infty}a_{n}X_{n}$$



converges in $L^{2}$ and almost surely.



(c) Show that the martingale



$$M_{n}:=exp(S_{n}-frac{1}{2}n)$$



converges almost surely, but not in $L^{1}$.







probability-theory convergence martingales






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asked Nov 24 '18 at 21:09









R.SluijR.Sluij

236




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closed as off-topic by Davide Giraudo, Leucippus, Rebellos, KReiser, Did Nov 25 '18 at 9:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Leucippus, Rebellos, KReiser, Did

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Davide Giraudo, Leucippus, Rebellos, KReiser, Did Nov 25 '18 at 9:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Leucippus, Rebellos, KReiser, Did

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Sorry, are you asking anything?
    $endgroup$
    – Will M.
    Nov 24 '18 at 22:35


















  • $begingroup$
    Sorry, are you asking anything?
    $endgroup$
    – Will M.
    Nov 24 '18 at 22:35
















$begingroup$
Sorry, are you asking anything?
$endgroup$
– Will M.
Nov 24 '18 at 22:35




$begingroup$
Sorry, are you asking anything?
$endgroup$
– Will M.
Nov 24 '18 at 22:35










1 Answer
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b) is starightforward; just show that partial sums form a Cauchy sequnce in $L^{2}$. For c) use SLLN to conclude that $e^{(S_n-frac n 2)}=e^{n(frac {S_n} n-frac 1 2)}to 0$ almost surely. Since $Ee^{(S_n-frac n 2)}=(e^{frac 1 2 })^{n} e^{-frac n 2}=1$ for all $n$ we cannot have $L^{1}$ convergence.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    b) is starightforward; just show that partial sums form a Cauchy sequnce in $L^{2}$. For c) use SLLN to conclude that $e^{(S_n-frac n 2)}=e^{n(frac {S_n} n-frac 1 2)}to 0$ almost surely. Since $Ee^{(S_n-frac n 2)}=(e^{frac 1 2 })^{n} e^{-frac n 2}=1$ for all $n$ we cannot have $L^{1}$ convergence.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      b) is starightforward; just show that partial sums form a Cauchy sequnce in $L^{2}$. For c) use SLLN to conclude that $e^{(S_n-frac n 2)}=e^{n(frac {S_n} n-frac 1 2)}to 0$ almost surely. Since $Ee^{(S_n-frac n 2)}=(e^{frac 1 2 })^{n} e^{-frac n 2}=1$ for all $n$ we cannot have $L^{1}$ convergence.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        b) is starightforward; just show that partial sums form a Cauchy sequnce in $L^{2}$. For c) use SLLN to conclude that $e^{(S_n-frac n 2)}=e^{n(frac {S_n} n-frac 1 2)}to 0$ almost surely. Since $Ee^{(S_n-frac n 2)}=(e^{frac 1 2 })^{n} e^{-frac n 2}=1$ for all $n$ we cannot have $L^{1}$ convergence.






        share|cite|improve this answer









        $endgroup$



        b) is starightforward; just show that partial sums form a Cauchy sequnce in $L^{2}$. For c) use SLLN to conclude that $e^{(S_n-frac n 2)}=e^{n(frac {S_n} n-frac 1 2)}to 0$ almost surely. Since $Ee^{(S_n-frac n 2)}=(e^{frac 1 2 })^{n} e^{-frac n 2}=1$ for all $n$ we cannot have $L^{1}$ convergence.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 23:38









        Kavi Rama MurthyKavi Rama Murthy

        53.5k32055




        53.5k32055















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