Show that $sum_{k=1}^{infty} sigma^2_k <infty$ implies $|sum_{k=1}^{infty} X_k|<infty $ almost surely.












3












$begingroup$


Suppose that $X_1, X_2, X_3,ldots$ are sequence of independent random variables such that $mu_k= 0$ and $ sigma^2_k =operatorname{Var}(X_k)< infty$ for all $k$. Then
show that $sum_{k=1}^{infty} sigma^2_k <infty$ implies $|sum_{k=1}^{infty} X_k|<infty $ almost surely.



I was wondering how could I use the facts: 1) if a martingale $M$ is bounded in $L^2 $ then $lim M_n$ exists almost surely.
2) Orthogonality of increments of $M$ to prove the above statement. I would like to see the solution in explained way.






sequences-and-series probability-theory random-variables martingales variance







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $P(|Z| ge b) ge a$ then $Var(Z) ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) to 0$ show that the sequence of random variables $Y_n = sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely.
    $endgroup$
    – reuns
    Apr 7 '17 at 0:15












  • $begingroup$
    Then where are the orthogonality applied?
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 11:56










  • $begingroup$
    $Var(Y_n) = sum_{k=1}^n Var(X_k)$
    $endgroup$
    – reuns
    Apr 7 '17 at 11:59










  • $begingroup$
    Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/…
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 12:11










  • $begingroup$
    Did you complete this question ? Write your own answer then.
    $endgroup$
    – reuns
    Apr 7 '17 at 12:12
















3












$begingroup$


Suppose that $X_1, X_2, X_3,ldots$ are sequence of independent random variables such that $mu_k= 0$ and $ sigma^2_k =operatorname{Var}(X_k)< infty$ for all $k$. Then
show that $sum_{k=1}^{infty} sigma^2_k <infty$ implies $|sum_{k=1}^{infty} X_k|<infty $ almost surely.



I was wondering how could I use the facts: 1) if a martingale $M$ is bounded in $L^2 $ then $lim M_n$ exists almost surely.
2) Orthogonality of increments of $M$ to prove the above statement. I would like to see the solution in explained way.






sequences-and-series probability-theory random-variables martingales variance







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $P(|Z| ge b) ge a$ then $Var(Z) ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) to 0$ show that the sequence of random variables $Y_n = sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely.
    $endgroup$
    – reuns
    Apr 7 '17 at 0:15












  • $begingroup$
    Then where are the orthogonality applied?
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 11:56










  • $begingroup$
    $Var(Y_n) = sum_{k=1}^n Var(X_k)$
    $endgroup$
    – reuns
    Apr 7 '17 at 11:59










  • $begingroup$
    Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/…
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 12:11










  • $begingroup$
    Did you complete this question ? Write your own answer then.
    $endgroup$
    – reuns
    Apr 7 '17 at 12:12














3












3








3





$begingroup$


Suppose that $X_1, X_2, X_3,ldots$ are sequence of independent random variables such that $mu_k= 0$ and $ sigma^2_k =operatorname{Var}(X_k)< infty$ for all $k$. Then
show that $sum_{k=1}^{infty} sigma^2_k <infty$ implies $|sum_{k=1}^{infty} X_k|<infty $ almost surely.



I was wondering how could I use the facts: 1) if a martingale $M$ is bounded in $L^2 $ then $lim M_n$ exists almost surely.
2) Orthogonality of increments of $M$ to prove the above statement. I would like to see the solution in explained way.






sequences-and-series probability-theory random-variables martingales variance







share|cite|improve this question











$endgroup$




Suppose that $X_1, X_2, X_3,ldots$ are sequence of independent random variables such that $mu_k= 0$ and $ sigma^2_k =operatorname{Var}(X_k)< infty$ for all $k$. Then
show that $sum_{k=1}^{infty} sigma^2_k <infty$ implies $|sum_{k=1}^{infty} X_k|<infty $ almost surely.



I was wondering how could I use the facts: 1) if a martingale $M$ is bounded in $L^2 $ then $lim M_n$ exists almost surely.
2) Orthogonality of increments of $M$ to prove the above statement. I would like to see the solution in explained way.






sequences-and-series probability-theory random-variables martingales variance




sequences-and-series probability-theory random-variables martingales variance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 22:23









Davide Giraudo

125k16150261




125k16150261










asked Apr 6 '17 at 23:51









00012 suxn00012 suxn

39919




39919












  • $begingroup$
    If $P(|Z| ge b) ge a$ then $Var(Z) ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) to 0$ show that the sequence of random variables $Y_n = sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely.
    $endgroup$
    – reuns
    Apr 7 '17 at 0:15












  • $begingroup$
    Then where are the orthogonality applied?
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 11:56










  • $begingroup$
    $Var(Y_n) = sum_{k=1}^n Var(X_k)$
    $endgroup$
    – reuns
    Apr 7 '17 at 11:59










  • $begingroup$
    Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/…
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 12:11










  • $begingroup$
    Did you complete this question ? Write your own answer then.
    $endgroup$
    – reuns
    Apr 7 '17 at 12:12


















  • $begingroup$
    If $P(|Z| ge b) ge a$ then $Var(Z) ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) to 0$ show that the sequence of random variables $Y_n = sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely.
    $endgroup$
    – reuns
    Apr 7 '17 at 0:15












  • $begingroup$
    Then where are the orthogonality applied?
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 11:56










  • $begingroup$
    $Var(Y_n) = sum_{k=1}^n Var(X_k)$
    $endgroup$
    – reuns
    Apr 7 '17 at 11:59










  • $begingroup$
    Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/…
    $endgroup$
    – 00012 suxn
    Apr 7 '17 at 12:11










  • $begingroup$
    Did you complete this question ? Write your own answer then.
    $endgroup$
    – reuns
    Apr 7 '17 at 12:12
















$begingroup$
If $P(|Z| ge b) ge a$ then $Var(Z) ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) to 0$ show that the sequence of random variables $Y_n = sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely.
$endgroup$
– reuns
Apr 7 '17 at 0:15






$begingroup$
If $P(|Z| ge b) ge a$ then $Var(Z) ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) to 0$ show that the sequence of random variables $Y_n = sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely.
$endgroup$
– reuns
Apr 7 '17 at 0:15














$begingroup$
Then where are the orthogonality applied?
$endgroup$
– 00012 suxn
Apr 7 '17 at 11:56




$begingroup$
Then where are the orthogonality applied?
$endgroup$
– 00012 suxn
Apr 7 '17 at 11:56












$begingroup$
$Var(Y_n) = sum_{k=1}^n Var(X_k)$
$endgroup$
– reuns
Apr 7 '17 at 11:59




$begingroup$
$Var(Y_n) = sum_{k=1}^n Var(X_k)$
$endgroup$
– reuns
Apr 7 '17 at 11:59












$begingroup$
Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/…
$endgroup$
– 00012 suxn
Apr 7 '17 at 12:11




$begingroup$
Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/…
$endgroup$
– 00012 suxn
Apr 7 '17 at 12:11












$begingroup$
Did you complete this question ? Write your own answer then.
$endgroup$
– reuns
Apr 7 '17 at 12:12




$begingroup$
Did you complete this question ? Write your own answer then.
$endgroup$
– reuns
Apr 7 '17 at 12:12










1 Answer
1






active

oldest

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0












$begingroup$

Let $M_n:=sum_{i=1}^nX_i$. Defining $mathcal F_n$ as the $sigma$-algebra generated by $X_i$, $1leqslant ileqslant n$. Then $(S_n,mathcal F_n)$ is a martingale and
using orthogonality of increments,
$$
mathbb Eleft[M_n^2right]=sum_{k=1}^nmathbb Eleft[X_k^2right]=sum_{k=1}^nsigma_k^2
$$
hence
$$
sup_nmathbb Eleft[M_n^2right]leqslant sum_{k=1}^{+infty}sigma_k^2.
$$






share|cite|improve this answer









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    0












    $begingroup$

    Let $M_n:=sum_{i=1}^nX_i$. Defining $mathcal F_n$ as the $sigma$-algebra generated by $X_i$, $1leqslant ileqslant n$. Then $(S_n,mathcal F_n)$ is a martingale and
    using orthogonality of increments,
    $$
    mathbb Eleft[M_n^2right]=sum_{k=1}^nmathbb Eleft[X_k^2right]=sum_{k=1}^nsigma_k^2
    $$
    hence
    $$
    sup_nmathbb Eleft[M_n^2right]leqslant sum_{k=1}^{+infty}sigma_k^2.
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $M_n:=sum_{i=1}^nX_i$. Defining $mathcal F_n$ as the $sigma$-algebra generated by $X_i$, $1leqslant ileqslant n$. Then $(S_n,mathcal F_n)$ is a martingale and
      using orthogonality of increments,
      $$
      mathbb Eleft[M_n^2right]=sum_{k=1}^nmathbb Eleft[X_k^2right]=sum_{k=1}^nsigma_k^2
      $$
      hence
      $$
      sup_nmathbb Eleft[M_n^2right]leqslant sum_{k=1}^{+infty}sigma_k^2.
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $M_n:=sum_{i=1}^nX_i$. Defining $mathcal F_n$ as the $sigma$-algebra generated by $X_i$, $1leqslant ileqslant n$. Then $(S_n,mathcal F_n)$ is a martingale and
        using orthogonality of increments,
        $$
        mathbb Eleft[M_n^2right]=sum_{k=1}^nmathbb Eleft[X_k^2right]=sum_{k=1}^nsigma_k^2
        $$
        hence
        $$
        sup_nmathbb Eleft[M_n^2right]leqslant sum_{k=1}^{+infty}sigma_k^2.
        $$






        share|cite|improve this answer









        $endgroup$



        Let $M_n:=sum_{i=1}^nX_i$. Defining $mathcal F_n$ as the $sigma$-algebra generated by $X_i$, $1leqslant ileqslant n$. Then $(S_n,mathcal F_n)$ is a martingale and
        using orthogonality of increments,
        $$
        mathbb Eleft[M_n^2right]=sum_{k=1}^nmathbb Eleft[X_k^2right]=sum_{k=1}^nsigma_k^2
        $$
        hence
        $$
        sup_nmathbb Eleft[M_n^2right]leqslant sum_{k=1}^{+infty}sigma_k^2.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 22:14









        Davide GiraudoDavide Giraudo

        125k16150261




        125k16150261






























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