Hermite polynomial with brownian motion is martingale












3












$begingroup$


Let $(B_t)_{tge 0}$ be a standard brownian motion. I want to show that $(H_n(B_t,t))_{tge 0}$ is a martingale, where



$$H_n(x,t)=frac{d^n}{du^n}e^{ux-frac{u^2}{2}t}Big|_{u=0},$$



such that the Taylor-formula is



$$e^{ux-frac{u^2}{2}t}=sum_{nge 0}H_n(x,t)frac{u^n}{n!}$$



How do I calculate $E[|H_n(B_t,t)|]$? Is it possible to interchange integration and differentiation? I tried to calculate $H_n(x,t)$ first to find a formula, but I do not get a nice solution. Can someone give me a hint on this? Thanks in advance!






probability-theory stochastic-processes stochastic-calculus







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    3












    $begingroup$


    Let $(B_t)_{tge 0}$ be a standard brownian motion. I want to show that $(H_n(B_t,t))_{tge 0}$ is a martingale, where



    $$H_n(x,t)=frac{d^n}{du^n}e^{ux-frac{u^2}{2}t}Big|_{u=0},$$



    such that the Taylor-formula is



    $$e^{ux-frac{u^2}{2}t}=sum_{nge 0}H_n(x,t)frac{u^n}{n!}$$



    How do I calculate $E[|H_n(B_t,t)|]$? Is it possible to interchange integration and differentiation? I tried to calculate $H_n(x,t)$ first to find a formula, but I do not get a nice solution. Can someone give me a hint on this? Thanks in advance!






    probability-theory stochastic-processes stochastic-calculus







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Let $(B_t)_{tge 0}$ be a standard brownian motion. I want to show that $(H_n(B_t,t))_{tge 0}$ is a martingale, where



      $$H_n(x,t)=frac{d^n}{du^n}e^{ux-frac{u^2}{2}t}Big|_{u=0},$$



      such that the Taylor-formula is



      $$e^{ux-frac{u^2}{2}t}=sum_{nge 0}H_n(x,t)frac{u^n}{n!}$$



      How do I calculate $E[|H_n(B_t,t)|]$? Is it possible to interchange integration and differentiation? I tried to calculate $H_n(x,t)$ first to find a formula, but I do not get a nice solution. Can someone give me a hint on this? Thanks in advance!






      probability-theory stochastic-processes stochastic-calculus







      share|cite|improve this question









      $endgroup$




      Let $(B_t)_{tge 0}$ be a standard brownian motion. I want to show that $(H_n(B_t,t))_{tge 0}$ is a martingale, where



      $$H_n(x,t)=frac{d^n}{du^n}e^{ux-frac{u^2}{2}t}Big|_{u=0},$$



      such that the Taylor-formula is



      $$e^{ux-frac{u^2}{2}t}=sum_{nge 0}H_n(x,t)frac{u^n}{n!}$$



      How do I calculate $E[|H_n(B_t,t)|]$? Is it possible to interchange integration and differentiation? I tried to calculate $H_n(x,t)$ first to find a formula, but I do not get a nice solution. Can someone give me a hint on this? Thanks in advance!






      probability-theory stochastic-processes stochastic-calculus




      probability-theory stochastic-processes stochastic-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 9 at 22:41









      user628255user628255

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          $begingroup$

          You may do it inductively.




          Lemma 1. We have for $nge 1$,
          $$
          frac{{rm d}^n}{{rm d}u^n}left(uf(u)right)=uf^{(n)}(u)+nf^{(n-1)}(u).
          $$




          This Lemma 1 can be easily shown by mathematical induction. Hence I skip its proof here.




          Lemma 2. We have for $nge 2$,
          $$
          H_{n+1}(x,t)=xH_n(x,t)-ntH_{n-1}(x,t).
          $$




          Proof. For $nge 2$, we have
          begin{align}
          H_{n+1}(x,t)&=frac{{rm d}^{n+1}}{{rm d}u^{n+1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}\
          &=frac{{rm d}^n}{{rm d}u^n}left(frac{rm d}{{rm d}u}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
          &=frac{{rm d}^n}{{rm d}u^n}left(e^{ux-frac{t}{2}u^2}left(x-turight)right)bigg|_{u=0}\
          &=xfrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}bigg|_{u=0}-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
          &=xH_n(x,t)-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
          end{align}
          Now, apply Lemma 1 to the last term, and we obtain
          $$
          frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)=ufrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}+nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}.
          $$
          Thus by taking $u=0$, we have
          $$
          frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}=nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}=nH_{n-1}(x,t).
          $$
          Thanks to this result, Lemma 2 follows immediately.#



          Note that the last equation also proves




          Lemma 3. We have for $nge 2$,
          $$
          frac{partial H_n}{partial x}(x,t)=nH_{n-1}(x,t),
          $$




          because
          begin{align}
          frac{partial H_n}{partial x}(x,t)&=frac{partial}{partial x}left(frac{partial^n}{partial u^n}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
          &=frac{partial^n}{partial u^n}left(frac{partial}{partial x}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
          &=frac{partial^n}{partial u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
          end{align}



          Now, your question can be figured out inductively.



          Initially, note that
          begin{align}
          H_1(B_t,t)&=B_t\
          H_2(B_t,t)&=B_t^2-t
          end{align}
          are both martingales.



          Inductively, suppose $H_n(B_t,t)$ and $H_{n-1}(B_t,t)$ are martingales. We need to show that $H_{n+1}(B_t,t)$ is a martingale. By Lemma 2,
          begin{align}
          {rm d}H_{n+1}(B_t,t)&={rm d}left(B_tH_n(B_t,t)-ntH_{n-1}(B_t,t)right)\
          &=H_n(B_t,t){rm d}B_t+B_t{rm d}H_n(B_t,t)+{rm d}left<B_t,H_n(B_t,t)right>-nH_{n-1}(B_t,t){rm d}t-nt{rm d}H_{n-1}(B_t,t),
          end{align}
          where the last line uses Ito's formula. Note that by Ito's formula again,
          begin{align}
          {rm d}H_n(B_t,t)&=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}left<B_tright>\
          &=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}t\
          &=left(frac{partial H_n}{partial t}(B_t,t)+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t)right){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t\
          &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t,
          end{align}
          where the last line is due to the inductive assumption that $H_n(B_t,t)$ is a martingale, for which the coefficients in front of ${rm d}t$ must vanish. Similarly, we have
          $$
          {rm d}H_{n-1}(B_t,t)=frac{partial H_{n-1}}{partial x}(B_t,t){rm d}B_t.
          $$
          In addition, we also have
          begin{align}
          {rm d}left<B_t,H_n(B_t,t)right>&={rm d}B_t{rm d}H_n(B_t,t)\
          &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t{rm d}B_t\
          &=frac{partial H_n}{partial x}(B_t,t){rm d}t.
          end{align}



          Thanks to all these results, we eventually have
          begin{align}
          {rm d}H_{n+1}(B_t,t)&=left(frac{partial H_n}{partial x}(B_t,t)-nH_{n-1}(B_t,t)right){rm d}t\
          &+left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t\
          &=left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t,
          end{align}
          where the last line results from Lemma 3. This result shows that $H_{n+1}(B_t,t)$ is a stochastic integral of $B_t$, which obviously indicates that it is a martingale.



          To sum up, we may conclude that $H_n(B_t,t)$ is a martingale for all $nge 1$.






          share|cite|improve this answer









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            $begingroup$

            You may do it inductively.




            Lemma 1. We have for $nge 1$,
            $$
            frac{{rm d}^n}{{rm d}u^n}left(uf(u)right)=uf^{(n)}(u)+nf^{(n-1)}(u).
            $$




            This Lemma 1 can be easily shown by mathematical induction. Hence I skip its proof here.




            Lemma 2. We have for $nge 2$,
            $$
            H_{n+1}(x,t)=xH_n(x,t)-ntH_{n-1}(x,t).
            $$




            Proof. For $nge 2$, we have
            begin{align}
            H_{n+1}(x,t)&=frac{{rm d}^{n+1}}{{rm d}u^{n+1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}\
            &=frac{{rm d}^n}{{rm d}u^n}left(frac{rm d}{{rm d}u}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
            &=frac{{rm d}^n}{{rm d}u^n}left(e^{ux-frac{t}{2}u^2}left(x-turight)right)bigg|_{u=0}\
            &=xfrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}bigg|_{u=0}-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
            &=xH_n(x,t)-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
            end{align}
            Now, apply Lemma 1 to the last term, and we obtain
            $$
            frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)=ufrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}+nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}.
            $$
            Thus by taking $u=0$, we have
            $$
            frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}=nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}=nH_{n-1}(x,t).
            $$
            Thanks to this result, Lemma 2 follows immediately.#



            Note that the last equation also proves




            Lemma 3. We have for $nge 2$,
            $$
            frac{partial H_n}{partial x}(x,t)=nH_{n-1}(x,t),
            $$




            because
            begin{align}
            frac{partial H_n}{partial x}(x,t)&=frac{partial}{partial x}left(frac{partial^n}{partial u^n}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
            &=frac{partial^n}{partial u^n}left(frac{partial}{partial x}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
            &=frac{partial^n}{partial u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
            end{align}



            Now, your question can be figured out inductively.



            Initially, note that
            begin{align}
            H_1(B_t,t)&=B_t\
            H_2(B_t,t)&=B_t^2-t
            end{align}
            are both martingales.



            Inductively, suppose $H_n(B_t,t)$ and $H_{n-1}(B_t,t)$ are martingales. We need to show that $H_{n+1}(B_t,t)$ is a martingale. By Lemma 2,
            begin{align}
            {rm d}H_{n+1}(B_t,t)&={rm d}left(B_tH_n(B_t,t)-ntH_{n-1}(B_t,t)right)\
            &=H_n(B_t,t){rm d}B_t+B_t{rm d}H_n(B_t,t)+{rm d}left<B_t,H_n(B_t,t)right>-nH_{n-1}(B_t,t){rm d}t-nt{rm d}H_{n-1}(B_t,t),
            end{align}
            where the last line uses Ito's formula. Note that by Ito's formula again,
            begin{align}
            {rm d}H_n(B_t,t)&=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}left<B_tright>\
            &=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}t\
            &=left(frac{partial H_n}{partial t}(B_t,t)+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t)right){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t\
            &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t,
            end{align}
            where the last line is due to the inductive assumption that $H_n(B_t,t)$ is a martingale, for which the coefficients in front of ${rm d}t$ must vanish. Similarly, we have
            $$
            {rm d}H_{n-1}(B_t,t)=frac{partial H_{n-1}}{partial x}(B_t,t){rm d}B_t.
            $$
            In addition, we also have
            begin{align}
            {rm d}left<B_t,H_n(B_t,t)right>&={rm d}B_t{rm d}H_n(B_t,t)\
            &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t{rm d}B_t\
            &=frac{partial H_n}{partial x}(B_t,t){rm d}t.
            end{align}



            Thanks to all these results, we eventually have
            begin{align}
            {rm d}H_{n+1}(B_t,t)&=left(frac{partial H_n}{partial x}(B_t,t)-nH_{n-1}(B_t,t)right){rm d}t\
            &+left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t\
            &=left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t,
            end{align}
            where the last line results from Lemma 3. This result shows that $H_{n+1}(B_t,t)$ is a stochastic integral of $B_t$, which obviously indicates that it is a martingale.



            To sum up, we may conclude that $H_n(B_t,t)$ is a martingale for all $nge 1$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              You may do it inductively.




              Lemma 1. We have for $nge 1$,
              $$
              frac{{rm d}^n}{{rm d}u^n}left(uf(u)right)=uf^{(n)}(u)+nf^{(n-1)}(u).
              $$




              This Lemma 1 can be easily shown by mathematical induction. Hence I skip its proof here.




              Lemma 2. We have for $nge 2$,
              $$
              H_{n+1}(x,t)=xH_n(x,t)-ntH_{n-1}(x,t).
              $$




              Proof. For $nge 2$, we have
              begin{align}
              H_{n+1}(x,t)&=frac{{rm d}^{n+1}}{{rm d}u^{n+1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}\
              &=frac{{rm d}^n}{{rm d}u^n}left(frac{rm d}{{rm d}u}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
              &=frac{{rm d}^n}{{rm d}u^n}left(e^{ux-frac{t}{2}u^2}left(x-turight)right)bigg|_{u=0}\
              &=xfrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}bigg|_{u=0}-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
              &=xH_n(x,t)-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
              end{align}
              Now, apply Lemma 1 to the last term, and we obtain
              $$
              frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)=ufrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}+nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}.
              $$
              Thus by taking $u=0$, we have
              $$
              frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}=nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}=nH_{n-1}(x,t).
              $$
              Thanks to this result, Lemma 2 follows immediately.#



              Note that the last equation also proves




              Lemma 3. We have for $nge 2$,
              $$
              frac{partial H_n}{partial x}(x,t)=nH_{n-1}(x,t),
              $$




              because
              begin{align}
              frac{partial H_n}{partial x}(x,t)&=frac{partial}{partial x}left(frac{partial^n}{partial u^n}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
              &=frac{partial^n}{partial u^n}left(frac{partial}{partial x}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
              &=frac{partial^n}{partial u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
              end{align}



              Now, your question can be figured out inductively.



              Initially, note that
              begin{align}
              H_1(B_t,t)&=B_t\
              H_2(B_t,t)&=B_t^2-t
              end{align}
              are both martingales.



              Inductively, suppose $H_n(B_t,t)$ and $H_{n-1}(B_t,t)$ are martingales. We need to show that $H_{n+1}(B_t,t)$ is a martingale. By Lemma 2,
              begin{align}
              {rm d}H_{n+1}(B_t,t)&={rm d}left(B_tH_n(B_t,t)-ntH_{n-1}(B_t,t)right)\
              &=H_n(B_t,t){rm d}B_t+B_t{rm d}H_n(B_t,t)+{rm d}left<B_t,H_n(B_t,t)right>-nH_{n-1}(B_t,t){rm d}t-nt{rm d}H_{n-1}(B_t,t),
              end{align}
              where the last line uses Ito's formula. Note that by Ito's formula again,
              begin{align}
              {rm d}H_n(B_t,t)&=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}left<B_tright>\
              &=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}t\
              &=left(frac{partial H_n}{partial t}(B_t,t)+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t)right){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t\
              &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t,
              end{align}
              where the last line is due to the inductive assumption that $H_n(B_t,t)$ is a martingale, for which the coefficients in front of ${rm d}t$ must vanish. Similarly, we have
              $$
              {rm d}H_{n-1}(B_t,t)=frac{partial H_{n-1}}{partial x}(B_t,t){rm d}B_t.
              $$
              In addition, we also have
              begin{align}
              {rm d}left<B_t,H_n(B_t,t)right>&={rm d}B_t{rm d}H_n(B_t,t)\
              &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t{rm d}B_t\
              &=frac{partial H_n}{partial x}(B_t,t){rm d}t.
              end{align}



              Thanks to all these results, we eventually have
              begin{align}
              {rm d}H_{n+1}(B_t,t)&=left(frac{partial H_n}{partial x}(B_t,t)-nH_{n-1}(B_t,t)right){rm d}t\
              &+left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t\
              &=left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t,
              end{align}
              where the last line results from Lemma 3. This result shows that $H_{n+1}(B_t,t)$ is a stochastic integral of $B_t$, which obviously indicates that it is a martingale.



              To sum up, we may conclude that $H_n(B_t,t)$ is a martingale for all $nge 1$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                You may do it inductively.




                Lemma 1. We have for $nge 1$,
                $$
                frac{{rm d}^n}{{rm d}u^n}left(uf(u)right)=uf^{(n)}(u)+nf^{(n-1)}(u).
                $$




                This Lemma 1 can be easily shown by mathematical induction. Hence I skip its proof here.




                Lemma 2. We have for $nge 2$,
                $$
                H_{n+1}(x,t)=xH_n(x,t)-ntH_{n-1}(x,t).
                $$




                Proof. For $nge 2$, we have
                begin{align}
                H_{n+1}(x,t)&=frac{{rm d}^{n+1}}{{rm d}u^{n+1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}\
                &=frac{{rm d}^n}{{rm d}u^n}left(frac{rm d}{{rm d}u}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=frac{{rm d}^n}{{rm d}u^n}left(e^{ux-frac{t}{2}u^2}left(x-turight)right)bigg|_{u=0}\
                &=xfrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}bigg|_{u=0}-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=xH_n(x,t)-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
                end{align}
                Now, apply Lemma 1 to the last term, and we obtain
                $$
                frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)=ufrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}+nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}.
                $$
                Thus by taking $u=0$, we have
                $$
                frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}=nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}=nH_{n-1}(x,t).
                $$
                Thanks to this result, Lemma 2 follows immediately.#



                Note that the last equation also proves




                Lemma 3. We have for $nge 2$,
                $$
                frac{partial H_n}{partial x}(x,t)=nH_{n-1}(x,t),
                $$




                because
                begin{align}
                frac{partial H_n}{partial x}(x,t)&=frac{partial}{partial x}left(frac{partial^n}{partial u^n}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=frac{partial^n}{partial u^n}left(frac{partial}{partial x}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=frac{partial^n}{partial u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
                end{align}



                Now, your question can be figured out inductively.



                Initially, note that
                begin{align}
                H_1(B_t,t)&=B_t\
                H_2(B_t,t)&=B_t^2-t
                end{align}
                are both martingales.



                Inductively, suppose $H_n(B_t,t)$ and $H_{n-1}(B_t,t)$ are martingales. We need to show that $H_{n+1}(B_t,t)$ is a martingale. By Lemma 2,
                begin{align}
                {rm d}H_{n+1}(B_t,t)&={rm d}left(B_tH_n(B_t,t)-ntH_{n-1}(B_t,t)right)\
                &=H_n(B_t,t){rm d}B_t+B_t{rm d}H_n(B_t,t)+{rm d}left<B_t,H_n(B_t,t)right>-nH_{n-1}(B_t,t){rm d}t-nt{rm d}H_{n-1}(B_t,t),
                end{align}
                where the last line uses Ito's formula. Note that by Ito's formula again,
                begin{align}
                {rm d}H_n(B_t,t)&=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}left<B_tright>\
                &=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}t\
                &=left(frac{partial H_n}{partial t}(B_t,t)+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t)right){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t\
                &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t,
                end{align}
                where the last line is due to the inductive assumption that $H_n(B_t,t)$ is a martingale, for which the coefficients in front of ${rm d}t$ must vanish. Similarly, we have
                $$
                {rm d}H_{n-1}(B_t,t)=frac{partial H_{n-1}}{partial x}(B_t,t){rm d}B_t.
                $$
                In addition, we also have
                begin{align}
                {rm d}left<B_t,H_n(B_t,t)right>&={rm d}B_t{rm d}H_n(B_t,t)\
                &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t{rm d}B_t\
                &=frac{partial H_n}{partial x}(B_t,t){rm d}t.
                end{align}



                Thanks to all these results, we eventually have
                begin{align}
                {rm d}H_{n+1}(B_t,t)&=left(frac{partial H_n}{partial x}(B_t,t)-nH_{n-1}(B_t,t)right){rm d}t\
                &+left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t\
                &=left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t,
                end{align}
                where the last line results from Lemma 3. This result shows that $H_{n+1}(B_t,t)$ is a stochastic integral of $B_t$, which obviously indicates that it is a martingale.



                To sum up, we may conclude that $H_n(B_t,t)$ is a martingale for all $nge 1$.






                share|cite|improve this answer









                $endgroup$



                You may do it inductively.




                Lemma 1. We have for $nge 1$,
                $$
                frac{{rm d}^n}{{rm d}u^n}left(uf(u)right)=uf^{(n)}(u)+nf^{(n-1)}(u).
                $$




                This Lemma 1 can be easily shown by mathematical induction. Hence I skip its proof here.




                Lemma 2. We have for $nge 2$,
                $$
                H_{n+1}(x,t)=xH_n(x,t)-ntH_{n-1}(x,t).
                $$




                Proof. For $nge 2$, we have
                begin{align}
                H_{n+1}(x,t)&=frac{{rm d}^{n+1}}{{rm d}u^{n+1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}\
                &=frac{{rm d}^n}{{rm d}u^n}left(frac{rm d}{{rm d}u}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=frac{{rm d}^n}{{rm d}u^n}left(e^{ux-frac{t}{2}u^2}left(x-turight)right)bigg|_{u=0}\
                &=xfrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}bigg|_{u=0}-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=xH_n(x,t)-tfrac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
                end{align}
                Now, apply Lemma 1 to the last term, and we obtain
                $$
                frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)=ufrac{{rm d}^n}{{rm d}u^n}e^{ux-frac{t}{2}u^2}+nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}.
                $$
                Thus by taking $u=0$, we have
                $$
                frac{{rm d}^n}{{rm d}u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}=nfrac{{rm d}^{n-1}}{{rm d}u^{n-1}}e^{ux-frac{t}{2}u^2}bigg|_{u=0}=nH_{n-1}(x,t).
                $$
                Thanks to this result, Lemma 2 follows immediately.#



                Note that the last equation also proves




                Lemma 3. We have for $nge 2$,
                $$
                frac{partial H_n}{partial x}(x,t)=nH_{n-1}(x,t),
                $$




                because
                begin{align}
                frac{partial H_n}{partial x}(x,t)&=frac{partial}{partial x}left(frac{partial^n}{partial u^n}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=frac{partial^n}{partial u^n}left(frac{partial}{partial x}e^{ux-frac{t}{2}u^2}right)bigg|_{u=0}\
                &=frac{partial^n}{partial u^n}left(ue^{ux-frac{t}{2}u^2}right)bigg|_{u=0}.
                end{align}



                Now, your question can be figured out inductively.



                Initially, note that
                begin{align}
                H_1(B_t,t)&=B_t\
                H_2(B_t,t)&=B_t^2-t
                end{align}
                are both martingales.



                Inductively, suppose $H_n(B_t,t)$ and $H_{n-1}(B_t,t)$ are martingales. We need to show that $H_{n+1}(B_t,t)$ is a martingale. By Lemma 2,
                begin{align}
                {rm d}H_{n+1}(B_t,t)&={rm d}left(B_tH_n(B_t,t)-ntH_{n-1}(B_t,t)right)\
                &=H_n(B_t,t){rm d}B_t+B_t{rm d}H_n(B_t,t)+{rm d}left<B_t,H_n(B_t,t)right>-nH_{n-1}(B_t,t){rm d}t-nt{rm d}H_{n-1}(B_t,t),
                end{align}
                where the last line uses Ito's formula. Note that by Ito's formula again,
                begin{align}
                {rm d}H_n(B_t,t)&=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}left<B_tright>\
                &=frac{partial H_n}{partial t}(B_t,t){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t){rm d}t\
                &=left(frac{partial H_n}{partial t}(B_t,t)+frac{1}{2}frac{partial^2H_n}{partial x^2}(B_t,t)right){rm d}t+frac{partial H_n}{partial x}(B_t,t){rm d}B_t\
                &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t,
                end{align}
                where the last line is due to the inductive assumption that $H_n(B_t,t)$ is a martingale, for which the coefficients in front of ${rm d}t$ must vanish. Similarly, we have
                $$
                {rm d}H_{n-1}(B_t,t)=frac{partial H_{n-1}}{partial x}(B_t,t){rm d}B_t.
                $$
                In addition, we also have
                begin{align}
                {rm d}left<B_t,H_n(B_t,t)right>&={rm d}B_t{rm d}H_n(B_t,t)\
                &=frac{partial H_n}{partial x}(B_t,t){rm d}B_t{rm d}B_t\
                &=frac{partial H_n}{partial x}(B_t,t){rm d}t.
                end{align}



                Thanks to all these results, we eventually have
                begin{align}
                {rm d}H_{n+1}(B_t,t)&=left(frac{partial H_n}{partial x}(B_t,t)-nH_{n-1}(B_t,t)right){rm d}t\
                &+left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t\
                &=left(H_n(B_t,t)+B_tfrac{partial H_n}{partial x}(B_t,t)-ntfrac{partial H_{n-1}}{partial x}(B_t,t)right){rm d}B_t,
                end{align}
                where the last line results from Lemma 3. This result shows that $H_{n+1}(B_t,t)$ is a stochastic integral of $B_t$, which obviously indicates that it is a martingale.



                To sum up, we may conclude that $H_n(B_t,t)$ is a martingale for all $nge 1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 4:06









                hypernovahypernova

                4,694314




                4,694314






























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