Let $B(t)$ be a Brownian Motion. Show that the following processes are Brownian Motions on $[0,T].$












0












$begingroup$


1) $X(t)=-B(t).$



Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.



2) $X(t) = B(T-t)-B(T).$



I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
$$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
$$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
$$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
$$= t + s - 2min(T,s+t).$$
I don't know how to show that this is the same as $min(s,t).$



3) $X(t) = cB(t/c^2)$ where $Tleq infty.$



For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$



4) $X(t) = tB(1/t)$ and $X(0)=0.$



For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
$$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$






probability







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$endgroup$

















    0












    $begingroup$


    1) $X(t)=-B(t).$



    Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.



    2) $X(t) = B(T-t)-B(T).$



    I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
    $$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
    $$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
    $$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
    $$= t + s - 2min(T,s+t).$$
    I don't know how to show that this is the same as $min(s,t).$



    3) $X(t) = cB(t/c^2)$ where $Tleq infty.$



    For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
    $$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
    However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$



    4) $X(t) = tB(1/t)$ and $X(0)=0.$



    For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
    $$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
    However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$






    probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      1) $X(t)=-B(t).$



      Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.



      2) $X(t) = B(T-t)-B(T).$



      I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
      $$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
      $$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
      $$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
      $$= t + s - 2min(T,s+t).$$
      I don't know how to show that this is the same as $min(s,t).$



      3) $X(t) = cB(t/c^2)$ where $Tleq infty.$



      For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
      $$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
      However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$



      4) $X(t) = tB(1/t)$ and $X(0)=0.$



      For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
      $$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
      However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$






      probability







      share|cite|improve this question









      $endgroup$




      1) $X(t)=-B(t).$



      Since $-X(t)$ is a Gaussian Process with zero mean, $X(t)$ is also Gaussian Process with zero mean. We simply want to show that its covariance function is $min(t,s).$ This is trivial since $$gamma(s,t) = text{cov}(-B(t)times -B(s)) = text{cov}(B(t)B(s))=min(t,s).$$ Thus $X(t) = -B(t)$ is also a Brownian motion.



      2) $X(t) = B(T-t)-B(T).$



      I am not sure how to argue why $X(t)$ has zero mean. However for the covariance function we have:
      $$gamma(s,t) = text{cov}((B(T-s)-B(s))times (B(T-t)-B(t)))$$
      $$= E[(B(T-s)-B(s))times (B(T-t)-B(t))] $$
      $$= max(t,s)-min(T-s,t)-min(T-t,s)+min(s,t)$$
      $$= t + s - 2min(T,s+t).$$
      I don't know how to show that this is the same as $min(s,t).$



      3) $X(t) = cB(t/c^2)$ where $Tleq infty.$



      For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
      $$ = c^2E(B(s/c^2)B(t/c^2)) = c^2 min(s/c^2,t/c^2)=min(s,t).$$
      However, I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$



      4) $X(t) = tB(1/t)$ and $X(0)=0.$



      For this I can show that $gamma(s,t) = text{cov}(X(s),X(t))$
      $$ = tsE(B(1/s)B(1/t)) = ts min(1/t,1/s)=min(s,t).$$
      However, for this part too I am not sure how to show that $X(t)$ is a Gaussian process with mean $0.$






      probability




      probability






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      asked Dec 18 '18 at 21:12









      Hello_WorldHello_World

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          $begingroup$

          In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
          $$
          Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
          $$
          You should also mention that $X(0) = 0$, which is trivial of course.



          In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
          $$
          (X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
          $$
          The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.



          For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.



          In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            active

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            0












            $begingroup$

            In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
            $$
            Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
            $$
            You should also mention that $X(0) = 0$, which is trivial of course.



            In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
            $$
            (X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
            $$
            The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.



            For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.



            In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
              $$
              Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
              $$
              You should also mention that $X(0) = 0$, which is trivial of course.



              In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
              $$
              (X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
              $$
              The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.



              For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.



              In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
                $$
                Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
                $$
                You should also mention that $X(0) = 0$, which is trivial of course.



                In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
                $$
                (X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
                $$
                The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.



                For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.



                In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.






                share|cite|improve this answer











                $endgroup$



                In (2), you get directly $E[X(t)] = E[B(T-t)] - E[B(T)] = 0$ because $B$ is a centered Gaussian process. For the covariance, you made a small error. You want to calculate
                $$
                Cov(X(t),X(s)) = Cov(B(T-t)-B(T), B(T-s)-B(T)).
                $$
                You should also mention that $X(0) = 0$, which is trivial of course.



                In (3), you get similarly that $X(t)$ has mean $0$.Let's show that $X$ is a Gaussian process. To this end, let $0 leq t_1 < dots t_n leq T$. We have
                $$
                (X(t_1),dots,X(t_n)) = c(B(t_1/c^2),dots,B(t_n/c^2)).
                $$
                The right hand side has a multivariate normal distribution because $B$ is a Gaussian process.



                For (4), you need that $B$ is a Brownian Motion on $[0,infty)$ (not on a finite interval $[0,T]$). Using the same agumentation as above, you get that $X$ is Gaussion with mean $0$.



                In all cases, you also need to show that $X$ is continuous. (This is part of the definition of a Brownian Motion.) This is trivial for (1)-(3). In (4), however, the continuity in $0$ is not trivial.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 19 '18 at 14:26

























                answered Dec 19 '18 at 14:17









                Tki DenebTki Deneb

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