What's the distribution of $int tW(t)dt$ for Brownian motion $W(t)$












2












$begingroup$


Let's say a standard Brownian motion is written as $B(t)$. Then, define $W(t) = sigma B(t)$.
I know that $int_0^1 W(t)dt $ has the distribution Normal(0,$sigma^2/3$).
But, what about $int_0^1 tW(t)dt$?



Also, it seems that the two distributions $int_0^1 W(t)dt $ and $int_0^1 tW(t)dt$ are correlated. Is there a way to calculate their covariance?



Thanks!






probability stochastic-calculus brownian-motion







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let's say a standard Brownian motion is written as $B(t)$. Then, define $W(t) = sigma B(t)$.
    I know that $int_0^1 W(t)dt $ has the distribution Normal(0,$sigma^2/3$).
    But, what about $int_0^1 tW(t)dt$?



    Also, it seems that the two distributions $int_0^1 W(t)dt $ and $int_0^1 tW(t)dt$ are correlated. Is there a way to calculate their covariance?



    Thanks!






    probability stochastic-calculus brownian-motion







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let's say a standard Brownian motion is written as $B(t)$. Then, define $W(t) = sigma B(t)$.
      I know that $int_0^1 W(t)dt $ has the distribution Normal(0,$sigma^2/3$).
      But, what about $int_0^1 tW(t)dt$?



      Also, it seems that the two distributions $int_0^1 W(t)dt $ and $int_0^1 tW(t)dt$ are correlated. Is there a way to calculate their covariance?



      Thanks!






      probability stochastic-calculus brownian-motion







      share|cite|improve this question









      $endgroup$




      Let's say a standard Brownian motion is written as $B(t)$. Then, define $W(t) = sigma B(t)$.
      I know that $int_0^1 W(t)dt $ has the distribution Normal(0,$sigma^2/3$).
      But, what about $int_0^1 tW(t)dt$?



      Also, it seems that the two distributions $int_0^1 W(t)dt $ and $int_0^1 tW(t)dt$ are correlated. Is there a way to calculate their covariance?



      Thanks!






      probability stochastic-calculus brownian-motion




      probability stochastic-calculus brownian-motion






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 1:25









      JackieJackie

      423




      423






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          $newcommand{E}{mathbb{E}}$
          Yes, indeed it is normally distributed, and covariance can be calculated. Without loss of generality, let $sigma^2 = 1$, as we can simply multiply by $sigma$ at the end to recover. To see that it's normally distributed, note first $t W(t)$ is almost-surely continuous, and thus is integrable on $[0,1]$. Since $W$ is a Gaussian process, every Riemann sum is normally distributed, and the integral $int_0^1 t W(t) ,dt$ can be written as the almost-sure limit of a sequence of Gaussians. Recalling that the weak limit of a sequence of Gaussians (when it exists), is Gaussian shows that the integral is normal. All that remains to do is compute the covariance.



          We compute begin{align*}
          Eleft[int_0^1W(t), dt cdot int_{0}^1 sW(s),ds right] &= Eleft[ iint s W(s) W(t), ds, dtright] \
          &= int_{0}^1 int_0^s sE[W(s)W(t)],dt,ds + int_{0}^1 int_0^t sE[W(s)W(t)],ds,dt \
          &=int_0^1int_0^s ts ,dt,ds + int_0^1 int_0^t s^2 ,ds,dt\
          &= frac{1}{8} + frac{1}{12} \
          &=frac{5}{24},.
          end{align*}






          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%2f3038810%2fwhats-the-distribution-of-int-twtdt-for-brownian-motion-wt%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









            1












            $begingroup$

            $newcommand{E}{mathbb{E}}$
            Yes, indeed it is normally distributed, and covariance can be calculated. Without loss of generality, let $sigma^2 = 1$, as we can simply multiply by $sigma$ at the end to recover. To see that it's normally distributed, note first $t W(t)$ is almost-surely continuous, and thus is integrable on $[0,1]$. Since $W$ is a Gaussian process, every Riemann sum is normally distributed, and the integral $int_0^1 t W(t) ,dt$ can be written as the almost-sure limit of a sequence of Gaussians. Recalling that the weak limit of a sequence of Gaussians (when it exists), is Gaussian shows that the integral is normal. All that remains to do is compute the covariance.



            We compute begin{align*}
            Eleft[int_0^1W(t), dt cdot int_{0}^1 sW(s),ds right] &= Eleft[ iint s W(s) W(t), ds, dtright] \
            &= int_{0}^1 int_0^s sE[W(s)W(t)],dt,ds + int_{0}^1 int_0^t sE[W(s)W(t)],ds,dt \
            &=int_0^1int_0^s ts ,dt,ds + int_0^1 int_0^t s^2 ,ds,dt\
            &= frac{1}{8} + frac{1}{12} \
            &=frac{5}{24},.
            end{align*}






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $newcommand{E}{mathbb{E}}$
              Yes, indeed it is normally distributed, and covariance can be calculated. Without loss of generality, let $sigma^2 = 1$, as we can simply multiply by $sigma$ at the end to recover. To see that it's normally distributed, note first $t W(t)$ is almost-surely continuous, and thus is integrable on $[0,1]$. Since $W$ is a Gaussian process, every Riemann sum is normally distributed, and the integral $int_0^1 t W(t) ,dt$ can be written as the almost-sure limit of a sequence of Gaussians. Recalling that the weak limit of a sequence of Gaussians (when it exists), is Gaussian shows that the integral is normal. All that remains to do is compute the covariance.



              We compute begin{align*}
              Eleft[int_0^1W(t), dt cdot int_{0}^1 sW(s),ds right] &= Eleft[ iint s W(s) W(t), ds, dtright] \
              &= int_{0}^1 int_0^s sE[W(s)W(t)],dt,ds + int_{0}^1 int_0^t sE[W(s)W(t)],ds,dt \
              &=int_0^1int_0^s ts ,dt,ds + int_0^1 int_0^t s^2 ,ds,dt\
              &= frac{1}{8} + frac{1}{12} \
              &=frac{5}{24},.
              end{align*}






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $newcommand{E}{mathbb{E}}$
                Yes, indeed it is normally distributed, and covariance can be calculated. Without loss of generality, let $sigma^2 = 1$, as we can simply multiply by $sigma$ at the end to recover. To see that it's normally distributed, note first $t W(t)$ is almost-surely continuous, and thus is integrable on $[0,1]$. Since $W$ is a Gaussian process, every Riemann sum is normally distributed, and the integral $int_0^1 t W(t) ,dt$ can be written as the almost-sure limit of a sequence of Gaussians. Recalling that the weak limit of a sequence of Gaussians (when it exists), is Gaussian shows that the integral is normal. All that remains to do is compute the covariance.



                We compute begin{align*}
                Eleft[int_0^1W(t), dt cdot int_{0}^1 sW(s),ds right] &= Eleft[ iint s W(s) W(t), ds, dtright] \
                &= int_{0}^1 int_0^s sE[W(s)W(t)],dt,ds + int_{0}^1 int_0^t sE[W(s)W(t)],ds,dt \
                &=int_0^1int_0^s ts ,dt,ds + int_0^1 int_0^t s^2 ,ds,dt\
                &= frac{1}{8} + frac{1}{12} \
                &=frac{5}{24},.
                end{align*}






                share|cite|improve this answer









                $endgroup$



                $newcommand{E}{mathbb{E}}$
                Yes, indeed it is normally distributed, and covariance can be calculated. Without loss of generality, let $sigma^2 = 1$, as we can simply multiply by $sigma$ at the end to recover. To see that it's normally distributed, note first $t W(t)$ is almost-surely continuous, and thus is integrable on $[0,1]$. Since $W$ is a Gaussian process, every Riemann sum is normally distributed, and the integral $int_0^1 t W(t) ,dt$ can be written as the almost-sure limit of a sequence of Gaussians. Recalling that the weak limit of a sequence of Gaussians (when it exists), is Gaussian shows that the integral is normal. All that remains to do is compute the covariance.



                We compute begin{align*}
                Eleft[int_0^1W(t), dt cdot int_{0}^1 sW(s),ds right] &= Eleft[ iint s W(s) W(t), ds, dtright] \
                &= int_{0}^1 int_0^s sE[W(s)W(t)],dt,ds + int_{0}^1 int_0^t sE[W(s)W(t)],ds,dt \
                &=int_0^1int_0^s ts ,dt,ds + int_0^1 int_0^t s^2 ,ds,dt\
                &= frac{1}{8} + frac{1}{12} \
                &=frac{5}{24},.
                end{align*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 14 '18 at 3:24









                Marcus MMarcus M

                8,7931947




                8,7931947






























                    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%2f3038810%2fwhats-the-distribution-of-int-twtdt-for-brownian-motion-wt%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

                    Wiesbaden

                    Image
                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                    Read more

                    Marschland

                    Image
                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more