A problem in probability theory about expected value and Fubini's theorem











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Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,



i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$



ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.




Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,



$$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$



I think in order to prove the claim, I can combine the above statement with the following result



$$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$



My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.



I'd appreciate any help.










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    Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,



    i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
    Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$



    ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.




    Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,



    $$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$



    I think in order to prove the claim, I can combine the above statement with the following result



    $$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$



    My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.



    I'd appreciate any help.










    share|cite|improve this question
























      up vote
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      favorite









      up vote
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      down vote

      favorite












      Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,



      i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
      Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$



      ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.




      Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,



      $$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$



      I think in order to prove the claim, I can combine the above statement with the following result



      $$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$



      My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.



      I'd appreciate any help.










      share|cite|improve this question














      Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,



      i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
      Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$



      ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.




      Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,



      $$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$



      I think in order to prove the claim, I can combine the above statement with the following result



      $$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$



      My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.



      I'd appreciate any help.







      probability-theory convergence expected-value






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      asked Nov 15 at 16:27









      Weak Nullstellensatz

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          Observe that
          $$
          mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
          =int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
          $$

          Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
          $$
          tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
          $$






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            Observe that
            $$
            mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
            =int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
            $$

            Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
            $$
            tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
            $$






            share|cite|improve this answer

























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              0
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              Observe that
              $$
              mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
              =int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
              $$

              Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
              $$
              tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
              $$






              share|cite|improve this answer























                up vote
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                up vote
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                Observe that
                $$
                mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
                =int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
                $$

                Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
                $$
                tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
                $$






                share|cite|improve this answer












                Observe that
                $$
                mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
                =int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
                $$

                Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
                $$
                tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 16:12









                Davide Giraudo

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