If $X$ is an Ito process, is $mathbb E(int X mathrm d X)$ convex?












5












$begingroup$


Consider the functional $F$, which is defined for each Ito process



$$X(t) = int_0^t mu(s) mathrm d s + int_0^t sigma(s) mathrm d W(s)$$



as



$$F(X) := mathbb Ebigg(int_0^T X(s) mathrm dX(s)bigg)$$



Now I would like to prove that $F$ is convex, which seems to be intuitive because it is true for absolutely continuous processes.



Due to the Ito formula / product rule,



$$
int_0^T X(s) mathrm dX(s)
= frac 1 2 X(T)^2 - int_0^T sigma(s)^2 mathrm ds
$$



If we restrict ourselves on Ito processes for which we have $sigma = 0$, the proof is thus very easy. However, the quadratic variation term for $sigma ne 0$ seems to mess everything up. On the other hand, the $sigma$ is also included in the $X(T)^2$ term, so we don't immediately get a counterexample.



Is there another proof for the claim? Or is the claim wrong and there is a counterexample?






functional-analysis stochastic-processes stochastic-calculus stochastic-analysis







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Consider the functional $F$, which is defined for each Ito process



    $$X(t) = int_0^t mu(s) mathrm d s + int_0^t sigma(s) mathrm d W(s)$$



    as



    $$F(X) := mathbb Ebigg(int_0^T X(s) mathrm dX(s)bigg)$$



    Now I would like to prove that $F$ is convex, which seems to be intuitive because it is true for absolutely continuous processes.



    Due to the Ito formula / product rule,



    $$
    int_0^T X(s) mathrm dX(s)
    = frac 1 2 X(T)^2 - int_0^T sigma(s)^2 mathrm ds
    $$



    If we restrict ourselves on Ito processes for which we have $sigma = 0$, the proof is thus very easy. However, the quadratic variation term for $sigma ne 0$ seems to mess everything up. On the other hand, the $sigma$ is also included in the $X(T)^2$ term, so we don't immediately get a counterexample.



    Is there another proof for the claim? Or is the claim wrong and there is a counterexample?






    functional-analysis stochastic-processes stochastic-calculus stochastic-analysis







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      3



      $begingroup$


      Consider the functional $F$, which is defined for each Ito process



      $$X(t) = int_0^t mu(s) mathrm d s + int_0^t sigma(s) mathrm d W(s)$$



      as



      $$F(X) := mathbb Ebigg(int_0^T X(s) mathrm dX(s)bigg)$$



      Now I would like to prove that $F$ is convex, which seems to be intuitive because it is true for absolutely continuous processes.



      Due to the Ito formula / product rule,



      $$
      int_0^T X(s) mathrm dX(s)
      = frac 1 2 X(T)^2 - int_0^T sigma(s)^2 mathrm ds
      $$



      If we restrict ourselves on Ito processes for which we have $sigma = 0$, the proof is thus very easy. However, the quadratic variation term for $sigma ne 0$ seems to mess everything up. On the other hand, the $sigma$ is also included in the $X(T)^2$ term, so we don't immediately get a counterexample.



      Is there another proof for the claim? Or is the claim wrong and there is a counterexample?






      functional-analysis stochastic-processes stochastic-calculus stochastic-analysis







      share|cite|improve this question











      $endgroup$




      Consider the functional $F$, which is defined for each Ito process



      $$X(t) = int_0^t mu(s) mathrm d s + int_0^t sigma(s) mathrm d W(s)$$



      as



      $$F(X) := mathbb Ebigg(int_0^T X(s) mathrm dX(s)bigg)$$



      Now I would like to prove that $F$ is convex, which seems to be intuitive because it is true for absolutely continuous processes.



      Due to the Ito formula / product rule,



      $$
      int_0^T X(s) mathrm dX(s)
      = frac 1 2 X(T)^2 - int_0^T sigma(s)^2 mathrm ds
      $$



      If we restrict ourselves on Ito processes for which we have $sigma = 0$, the proof is thus very easy. However, the quadratic variation term for $sigma ne 0$ seems to mess everything up. On the other hand, the $sigma$ is also included in the $X(T)^2$ term, so we don't immediately get a counterexample.



      Is there another proof for the claim? Or is the claim wrong and there is a counterexample?






      functional-analysis stochastic-processes stochastic-calculus stochastic-analysis




      functional-analysis stochastic-processes stochastic-calculus stochastic-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 15:47







      Kolodez

















      asked Jan 12 at 14:47









      KolodezKolodez

      776




      776






















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          $begingroup$

          I believe that $F$ fails to be convex.



          First of all, if $sigma$ is a "nice" function, then the stochastic integral $M_t := int_0^{t} sigma(s) , dW_s$ is a martingale which implies that $$mathbb{E} left( int_0^T X_s , dM_s right)=0,$$ i.e.



          $$F(X) = mathbb{E} left( int_0^T X_s mu(s) , ds right). tag{1}$$



          Now consider $$X_t := int_0^t W_s , ds. $$ It follows from the very definition of $F$ that $$F(W) = mathbb{E} left( int_0^T W_s , dW_s right)=0,$$ and $(1)$ yields $$begin{align*} F(X) = mathbb{E} left( int_0^T X_s W_s , ds right) &= mathbb{E} left( int_0^T int_0^s W_s W_r , dr , ds right) \ &= int_0^T int_0^s underbrace{mathbb{E}(W_s W_r)}_{=r} , dr , ds \ &= frac{T^3}{6}. end{align*}$$For any $lambda in [0,1]$ we have by (1) that



          $$begin{align*} F(lambda X + (1-lambda) W) &= mathbb{E} bigg( int_0^T (lambda X_s + (1-lambda) W_s) (lambda W_s+0) , ds bigg) \ &=lambda^2 underbrace{mathbb{E}left( int_0^T X_s W_s , ds right)}_{stackrel{(2)}{=} T^3/6} + lambda (1-lambda) mathbb{E} left( int_0^T W_s^2 , ds right) \ &= lambda^2 frac{T^3}{6} + lambda (1-lambda) frac{T^2}{2}. end{align*}$$



          If we choose $lambda=1/2$ and $T$ small enough (e.g. $T=1$), then



          $$F(tfrac{1}{2} X + tfrac{1}{2} W) = frac{T^3}{24} + frac{T^2}{8}$$



          is strictly larger than



          $$frac{1}{2} F(X) + frac{1}{2} F(W) = frac{T^3}{12}.$$



          This means that $F$ is not convex.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            I believe that $F$ fails to be convex.



            First of all, if $sigma$ is a "nice" function, then the stochastic integral $M_t := int_0^{t} sigma(s) , dW_s$ is a martingale which implies that $$mathbb{E} left( int_0^T X_s , dM_s right)=0,$$ i.e.



            $$F(X) = mathbb{E} left( int_0^T X_s mu(s) , ds right). tag{1}$$



            Now consider $$X_t := int_0^t W_s , ds. $$ It follows from the very definition of $F$ that $$F(W) = mathbb{E} left( int_0^T W_s , dW_s right)=0,$$ and $(1)$ yields $$begin{align*} F(X) = mathbb{E} left( int_0^T X_s W_s , ds right) &= mathbb{E} left( int_0^T int_0^s W_s W_r , dr , ds right) \ &= int_0^T int_0^s underbrace{mathbb{E}(W_s W_r)}_{=r} , dr , ds \ &= frac{T^3}{6}. end{align*}$$For any $lambda in [0,1]$ we have by (1) that



            $$begin{align*} F(lambda X + (1-lambda) W) &= mathbb{E} bigg( int_0^T (lambda X_s + (1-lambda) W_s) (lambda W_s+0) , ds bigg) \ &=lambda^2 underbrace{mathbb{E}left( int_0^T X_s W_s , ds right)}_{stackrel{(2)}{=} T^3/6} + lambda (1-lambda) mathbb{E} left( int_0^T W_s^2 , ds right) \ &= lambda^2 frac{T^3}{6} + lambda (1-lambda) frac{T^2}{2}. end{align*}$$



            If we choose $lambda=1/2$ and $T$ small enough (e.g. $T=1$), then



            $$F(tfrac{1}{2} X + tfrac{1}{2} W) = frac{T^3}{24} + frac{T^2}{8}$$



            is strictly larger than



            $$frac{1}{2} F(X) + frac{1}{2} F(W) = frac{T^3}{12}.$$



            This means that $F$ is not convex.






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              I believe that $F$ fails to be convex.



              First of all, if $sigma$ is a "nice" function, then the stochastic integral $M_t := int_0^{t} sigma(s) , dW_s$ is a martingale which implies that $$mathbb{E} left( int_0^T X_s , dM_s right)=0,$$ i.e.



              $$F(X) = mathbb{E} left( int_0^T X_s mu(s) , ds right). tag{1}$$



              Now consider $$X_t := int_0^t W_s , ds. $$ It follows from the very definition of $F$ that $$F(W) = mathbb{E} left( int_0^T W_s , dW_s right)=0,$$ and $(1)$ yields $$begin{align*} F(X) = mathbb{E} left( int_0^T X_s W_s , ds right) &= mathbb{E} left( int_0^T int_0^s W_s W_r , dr , ds right) \ &= int_0^T int_0^s underbrace{mathbb{E}(W_s W_r)}_{=r} , dr , ds \ &= frac{T^3}{6}. end{align*}$$For any $lambda in [0,1]$ we have by (1) that



              $$begin{align*} F(lambda X + (1-lambda) W) &= mathbb{E} bigg( int_0^T (lambda X_s + (1-lambda) W_s) (lambda W_s+0) , ds bigg) \ &=lambda^2 underbrace{mathbb{E}left( int_0^T X_s W_s , ds right)}_{stackrel{(2)}{=} T^3/6} + lambda (1-lambda) mathbb{E} left( int_0^T W_s^2 , ds right) \ &= lambda^2 frac{T^3}{6} + lambda (1-lambda) frac{T^2}{2}. end{align*}$$



              If we choose $lambda=1/2$ and $T$ small enough (e.g. $T=1$), then



              $$F(tfrac{1}{2} X + tfrac{1}{2} W) = frac{T^3}{24} + frac{T^2}{8}$$



              is strictly larger than



              $$frac{1}{2} F(X) + frac{1}{2} F(W) = frac{T^3}{12}.$$



              This means that $F$ is not convex.






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                I believe that $F$ fails to be convex.



                First of all, if $sigma$ is a "nice" function, then the stochastic integral $M_t := int_0^{t} sigma(s) , dW_s$ is a martingale which implies that $$mathbb{E} left( int_0^T X_s , dM_s right)=0,$$ i.e.



                $$F(X) = mathbb{E} left( int_0^T X_s mu(s) , ds right). tag{1}$$



                Now consider $$X_t := int_0^t W_s , ds. $$ It follows from the very definition of $F$ that $$F(W) = mathbb{E} left( int_0^T W_s , dW_s right)=0,$$ and $(1)$ yields $$begin{align*} F(X) = mathbb{E} left( int_0^T X_s W_s , ds right) &= mathbb{E} left( int_0^T int_0^s W_s W_r , dr , ds right) \ &= int_0^T int_0^s underbrace{mathbb{E}(W_s W_r)}_{=r} , dr , ds \ &= frac{T^3}{6}. end{align*}$$For any $lambda in [0,1]$ we have by (1) that



                $$begin{align*} F(lambda X + (1-lambda) W) &= mathbb{E} bigg( int_0^T (lambda X_s + (1-lambda) W_s) (lambda W_s+0) , ds bigg) \ &=lambda^2 underbrace{mathbb{E}left( int_0^T X_s W_s , ds right)}_{stackrel{(2)}{=} T^3/6} + lambda (1-lambda) mathbb{E} left( int_0^T W_s^2 , ds right) \ &= lambda^2 frac{T^3}{6} + lambda (1-lambda) frac{T^2}{2}. end{align*}$$



                If we choose $lambda=1/2$ and $T$ small enough (e.g. $T=1$), then



                $$F(tfrac{1}{2} X + tfrac{1}{2} W) = frac{T^3}{24} + frac{T^2}{8}$$



                is strictly larger than



                $$frac{1}{2} F(X) + frac{1}{2} F(W) = frac{T^3}{12}.$$



                This means that $F$ is not convex.






                share|cite|improve this answer











                $endgroup$



                I believe that $F$ fails to be convex.



                First of all, if $sigma$ is a "nice" function, then the stochastic integral $M_t := int_0^{t} sigma(s) , dW_s$ is a martingale which implies that $$mathbb{E} left( int_0^T X_s , dM_s right)=0,$$ i.e.



                $$F(X) = mathbb{E} left( int_0^T X_s mu(s) , ds right). tag{1}$$



                Now consider $$X_t := int_0^t W_s , ds. $$ It follows from the very definition of $F$ that $$F(W) = mathbb{E} left( int_0^T W_s , dW_s right)=0,$$ and $(1)$ yields $$begin{align*} F(X) = mathbb{E} left( int_0^T X_s W_s , ds right) &= mathbb{E} left( int_0^T int_0^s W_s W_r , dr , ds right) \ &= int_0^T int_0^s underbrace{mathbb{E}(W_s W_r)}_{=r} , dr , ds \ &= frac{T^3}{6}. end{align*}$$For any $lambda in [0,1]$ we have by (1) that



                $$begin{align*} F(lambda X + (1-lambda) W) &= mathbb{E} bigg( int_0^T (lambda X_s + (1-lambda) W_s) (lambda W_s+0) , ds bigg) \ &=lambda^2 underbrace{mathbb{E}left( int_0^T X_s W_s , ds right)}_{stackrel{(2)}{=} T^3/6} + lambda (1-lambda) mathbb{E} left( int_0^T W_s^2 , ds right) \ &= lambda^2 frac{T^3}{6} + lambda (1-lambda) frac{T^2}{2}. end{align*}$$



                If we choose $lambda=1/2$ and $T$ small enough (e.g. $T=1$), then



                $$F(tfrac{1}{2} X + tfrac{1}{2} W) = frac{T^3}{24} + frac{T^2}{8}$$



                is strictly larger than



                $$frac{1}{2} F(X) + frac{1}{2} F(W) = frac{T^3}{12}.$$



                This means that $F$ is not convex.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 12 at 20:44

























                answered Jan 12 at 18:23









                sazsaz

                81.7k861128




                81.7k861128






























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