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







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$endgroup$

















    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




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          $begingroup$

          $newcommand{E}{mathbb{E}}$
          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









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            1 Answer
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            active

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            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






























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