Maximal inequality for the Itō integral












3












$begingroup$


Let $W$ be a Brownian motion and $X$ be a predictable process with $$operatorname Eleft[int_0^t|X_s|^2:{rm d}sright]<infty;;;text{for all }tge0.$$ Now, let $pge2$.




How can we show that $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le Coperatorname Eleft[left|int_0^tleft|X_sright|^2:{rm d}sright|^{frac p2}right]tag1$$ for some $Cge0$.




Let $$q:=frac p{p-1}.$$ Clearly, we somehow need to apply Doob's inequality and the Itō formula. Doob's inequality yields $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le q^poperatorname Eleft[left(left|int_0^tX_s:{rm d}W_sright|^2right)^{frac p2}right]tag2.$$ How do we need to proceed from here?






probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis







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$endgroup$

















    3












    $begingroup$


    Let $W$ be a Brownian motion and $X$ be a predictable process with $$operatorname Eleft[int_0^t|X_s|^2:{rm d}sright]<infty;;;text{for all }tge0.$$ Now, let $pge2$.




    How can we show that $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le Coperatorname Eleft[left|int_0^tleft|X_sright|^2:{rm d}sright|^{frac p2}right]tag1$$ for some $Cge0$.




    Let $$q:=frac p{p-1}.$$ Clearly, we somehow need to apply Doob's inequality and the Itō formula. Doob's inequality yields $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le q^poperatorname Eleft[left(left|int_0^tX_s:{rm d}W_sright|^2right)^{frac p2}right]tag2.$$ How do we need to proceed from here?






    probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Let $W$ be a Brownian motion and $X$ be a predictable process with $$operatorname Eleft[int_0^t|X_s|^2:{rm d}sright]<infty;;;text{for all }tge0.$$ Now, let $pge2$.




      How can we show that $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le Coperatorname Eleft[left|int_0^tleft|X_sright|^2:{rm d}sright|^{frac p2}right]tag1$$ for some $Cge0$.




      Let $$q:=frac p{p-1}.$$ Clearly, we somehow need to apply Doob's inequality and the Itō formula. Doob's inequality yields $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le q^poperatorname Eleft[left(left|int_0^tX_s:{rm d}W_sright|^2right)^{frac p2}right]tag2.$$ How do we need to proceed from here?






      probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis







      share|cite|improve this question









      $endgroup$




      Let $W$ be a Brownian motion and $X$ be a predictable process with $$operatorname Eleft[int_0^t|X_s|^2:{rm d}sright]<infty;;;text{for all }tge0.$$ Now, let $pge2$.




      How can we show that $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le Coperatorname Eleft[left|int_0^tleft|X_sright|^2:{rm d}sright|^{frac p2}right]tag1$$ for some $Cge0$.




      Let $$q:=frac p{p-1}.$$ Clearly, we somehow need to apply Doob's inequality and the Itō formula. Doob's inequality yields $$operatorname Eleft[sup_{sin[0,:t]}left|int_0^sX_r:{rm d}W_rright|^pright]le q^poperatorname Eleft[left(left|int_0^tX_s:{rm d}W_sright|^2right)^{frac p2}right]tag2.$$ How do we need to proceed from here?






      probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis




      probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis






      share|cite|improve this question













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      asked Dec 22 '18 at 0:29









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          $begingroup$

          If $p=2$ then the assertion is a direct consequence of Itô's isometry. From now on I will assume that $p>2$. Moreover, by using a standard stopping technique, we may assume without loss of generality that $$M_t := int_0^t X_s , dW_s$$ and its quadratic variation $$langle M rangle_t = int_0^t X_s^2 , ds$$ are bounded processes. As $p > 2$ the mapping $x mapsto |x|^p$ is twice continuously differentiable, and therefore an application of Itô's formula gives



          $$|M_t|^p = p int_0^t |M_s|^{p-1} dM_s + frac{p(p-1)}{2} int_0^t |M_s|^{p-2} , dlangle M rangle_s.$$



          Since $M$ is bounded, the first term on the right-hand side is a martingale; thus



          $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p mathbb{E}(|M_t|^p) = q^p frac{p (p-1)}{2} mathbb{E} int_0^t |M_s|^{p-2} , dlangle M rangle_s$$



          and so



          $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2} mathbb{E} left( langle M rangle_t sup_{s leq t} |M_s|^{p-2} right).$$



          Now an application of Hölder's inequality yields



          $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2}left( mathbb{E} left[ sup_{s leq t} |M_s|^p right] right)^{1-2/p} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}.$$



          Hence,



          $$left(mathbb{E}left[ sup_{s leq t} |M_s|^p right] right)^{2/p}leq q^p frac{p(p-1)}{2} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}$$



          and this proves the assertion.



          Remark: The Burkholder-Davis-Gundy inequality states that $$mathbb{E} left( sup_{s leq t} left| int_0^s X_s , dW_s right|^p right)$$ is comparable with $$mathbb{E} left( left| int_0^t X_s^2 , ds right|^{p/2} right),$$



          see e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch for a proof.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            If $p=2$ then the assertion is a direct consequence of Itô's isometry. From now on I will assume that $p>2$. Moreover, by using a standard stopping technique, we may assume without loss of generality that $$M_t := int_0^t X_s , dW_s$$ and its quadratic variation $$langle M rangle_t = int_0^t X_s^2 , ds$$ are bounded processes. As $p > 2$ the mapping $x mapsto |x|^p$ is twice continuously differentiable, and therefore an application of Itô's formula gives



            $$|M_t|^p = p int_0^t |M_s|^{p-1} dM_s + frac{p(p-1)}{2} int_0^t |M_s|^{p-2} , dlangle M rangle_s.$$



            Since $M$ is bounded, the first term on the right-hand side is a martingale; thus



            $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p mathbb{E}(|M_t|^p) = q^p frac{p (p-1)}{2} mathbb{E} int_0^t |M_s|^{p-2} , dlangle M rangle_s$$



            and so



            $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2} mathbb{E} left( langle M rangle_t sup_{s leq t} |M_s|^{p-2} right).$$



            Now an application of Hölder's inequality yields



            $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2}left( mathbb{E} left[ sup_{s leq t} |M_s|^p right] right)^{1-2/p} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}.$$



            Hence,



            $$left(mathbb{E}left[ sup_{s leq t} |M_s|^p right] right)^{2/p}leq q^p frac{p(p-1)}{2} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}$$



            and this proves the assertion.



            Remark: The Burkholder-Davis-Gundy inequality states that $$mathbb{E} left( sup_{s leq t} left| int_0^s X_s , dW_s right|^p right)$$ is comparable with $$mathbb{E} left( left| int_0^t X_s^2 , ds right|^{p/2} right),$$



            see e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch for a proof.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              If $p=2$ then the assertion is a direct consequence of Itô's isometry. From now on I will assume that $p>2$. Moreover, by using a standard stopping technique, we may assume without loss of generality that $$M_t := int_0^t X_s , dW_s$$ and its quadratic variation $$langle M rangle_t = int_0^t X_s^2 , ds$$ are bounded processes. As $p > 2$ the mapping $x mapsto |x|^p$ is twice continuously differentiable, and therefore an application of Itô's formula gives



              $$|M_t|^p = p int_0^t |M_s|^{p-1} dM_s + frac{p(p-1)}{2} int_0^t |M_s|^{p-2} , dlangle M rangle_s.$$



              Since $M$ is bounded, the first term on the right-hand side is a martingale; thus



              $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p mathbb{E}(|M_t|^p) = q^p frac{p (p-1)}{2} mathbb{E} int_0^t |M_s|^{p-2} , dlangle M rangle_s$$



              and so



              $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2} mathbb{E} left( langle M rangle_t sup_{s leq t} |M_s|^{p-2} right).$$



              Now an application of Hölder's inequality yields



              $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2}left( mathbb{E} left[ sup_{s leq t} |M_s|^p right] right)^{1-2/p} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}.$$



              Hence,



              $$left(mathbb{E}left[ sup_{s leq t} |M_s|^p right] right)^{2/p}leq q^p frac{p(p-1)}{2} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}$$



              and this proves the assertion.



              Remark: The Burkholder-Davis-Gundy inequality states that $$mathbb{E} left( sup_{s leq t} left| int_0^s X_s , dW_s right|^p right)$$ is comparable with $$mathbb{E} left( left| int_0^t X_s^2 , ds right|^{p/2} right),$$



              see e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch for a proof.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                If $p=2$ then the assertion is a direct consequence of Itô's isometry. From now on I will assume that $p>2$. Moreover, by using a standard stopping technique, we may assume without loss of generality that $$M_t := int_0^t X_s , dW_s$$ and its quadratic variation $$langle M rangle_t = int_0^t X_s^2 , ds$$ are bounded processes. As $p > 2$ the mapping $x mapsto |x|^p$ is twice continuously differentiable, and therefore an application of Itô's formula gives



                $$|M_t|^p = p int_0^t |M_s|^{p-1} dM_s + frac{p(p-1)}{2} int_0^t |M_s|^{p-2} , dlangle M rangle_s.$$



                Since $M$ is bounded, the first term on the right-hand side is a martingale; thus



                $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p mathbb{E}(|M_t|^p) = q^p frac{p (p-1)}{2} mathbb{E} int_0^t |M_s|^{p-2} , dlangle M rangle_s$$



                and so



                $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2} mathbb{E} left( langle M rangle_t sup_{s leq t} |M_s|^{p-2} right).$$



                Now an application of Hölder's inequality yields



                $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2}left( mathbb{E} left[ sup_{s leq t} |M_s|^p right] right)^{1-2/p} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}.$$



                Hence,



                $$left(mathbb{E}left[ sup_{s leq t} |M_s|^p right] right)^{2/p}leq q^p frac{p(p-1)}{2} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}$$



                and this proves the assertion.



                Remark: The Burkholder-Davis-Gundy inequality states that $$mathbb{E} left( sup_{s leq t} left| int_0^s X_s , dW_s right|^p right)$$ is comparable with $$mathbb{E} left( left| int_0^t X_s^2 , ds right|^{p/2} right),$$



                see e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch for a proof.






                share|cite|improve this answer











                $endgroup$



                If $p=2$ then the assertion is a direct consequence of Itô's isometry. From now on I will assume that $p>2$. Moreover, by using a standard stopping technique, we may assume without loss of generality that $$M_t := int_0^t X_s , dW_s$$ and its quadratic variation $$langle M rangle_t = int_0^t X_s^2 , ds$$ are bounded processes. As $p > 2$ the mapping $x mapsto |x|^p$ is twice continuously differentiable, and therefore an application of Itô's formula gives



                $$|M_t|^p = p int_0^t |M_s|^{p-1} dM_s + frac{p(p-1)}{2} int_0^t |M_s|^{p-2} , dlangle M rangle_s.$$



                Since $M$ is bounded, the first term on the right-hand side is a martingale; thus



                $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p mathbb{E}(|M_t|^p) = q^p frac{p (p-1)}{2} mathbb{E} int_0^t |M_s|^{p-2} , dlangle M rangle_s$$



                and so



                $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2} mathbb{E} left( langle M rangle_t sup_{s leq t} |M_s|^{p-2} right).$$



                Now an application of Hölder's inequality yields



                $$mathbb{E}left( sup_{s leq t} |M_s|^p right) leq q^p frac{p(p-1)}{2}left( mathbb{E} left[ sup_{s leq t} |M_s|^p right] right)^{1-2/p} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}.$$



                Hence,



                $$left(mathbb{E}left[ sup_{s leq t} |M_s|^p right] right)^{2/p}leq q^p frac{p(p-1)}{2} (mathbb{E}[langle M rangle_t^{p/2}]^{2/p}$$



                and this proves the assertion.



                Remark: The Burkholder-Davis-Gundy inequality states that $$mathbb{E} left( sup_{s leq t} left| int_0^s X_s , dW_s right|^p right)$$ is comparable with $$mathbb{E} left( left| int_0^t X_s^2 , ds right|^{p/2} right),$$



                see e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch for a proof.







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                edited Dec 30 '18 at 11:14

























                answered Dec 22 '18 at 8:22









                sazsaz

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