continuous local martingale brownian motion












2












$begingroup$


$B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
$X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



$f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
I would be thankful for any help.






probability-theory finance local-martingales







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    $B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
    $X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



    $f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
    I would be thankful for any help.






    probability-theory finance local-martingales







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      $B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
      $X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



      $f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
      I would be thankful for any help.






      probability-theory finance local-martingales







      share|cite|improve this question











      $endgroup$




      $B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
      $X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



      $f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
      I would be thankful for any help.






      probability-theory finance local-martingales




      probability-theory finance local-martingales






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 17 '16 at 21:57







      Mathfreak

















      asked Apr 17 '16 at 12:56









      MathfreakMathfreak

      9210




      9210






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
          $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
          and according to Ito's lemma:
          $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
          Computing the partial derivatives:
          begin{align}
          frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
          frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
          frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
          end{align}
          It follows that
          $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
          Hence, we obtain
          $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
          which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






          share|cite|improve this answer











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1746366%2fcontinuous-local-martingale-brownian-motion%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
            $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
            and according to Ito's lemma:
            $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
            Computing the partial derivatives:
            begin{align}
            frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
            frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
            frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
            end{align}
            It follows that
            $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
            Hence, we obtain
            $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
            which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
              $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
              and according to Ito's lemma:
              $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
              Computing the partial derivatives:
              begin{align}
              frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
              frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
              frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
              end{align}
              It follows that
              $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
              Hence, we obtain
              $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
              which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
                $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
                and according to Ito's lemma:
                $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
                Computing the partial derivatives:
                begin{align}
                frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
                frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
                frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
                end{align}
                It follows that
                $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
                Hence, we obtain
                $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
                which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






                share|cite|improve this answer











                $endgroup$



                Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
                $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
                and according to Ito's lemma:
                $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
                Computing the partial derivatives:
                begin{align}
                frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
                frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
                frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
                end{align}
                It follows that
                $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
                Hence, we obtain
                $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
                which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 17 '18 at 22:24

























                answered Apr 17 '16 at 22:32









                SironSiron

                1,205614




                1,205614






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1746366%2fcontinuous-local-martingale-brownian-motion%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                    Read more

                    Puebla de Zaragoza

                    Image
                    Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                    Read more

                    Musa

                    Image
                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                    Read more