Variance of the Wiener Increment












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I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.



The Wiener increment $Delta W$ is defined as:



$Delta W(t) = W(t + Delta t) - W(t) $



which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.



I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:



$(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $



$langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$



$langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$



I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!






statistics normal-distribution random-walk statistical-mechanics







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    0












    $begingroup$


    I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.



    The Wiener increment $Delta W$ is defined as:



    $Delta W(t) = W(t + Delta t) - W(t) $



    which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.



    I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:



    $(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $



    $langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$



    $langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$



    I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!






    statistics normal-distribution random-walk statistical-mechanics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.



      The Wiener increment $Delta W$ is defined as:



      $Delta W(t) = W(t + Delta t) - W(t) $



      which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.



      I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:



      $(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $



      $langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$



      $langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$



      I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!






      statistics normal-distribution random-walk statistical-mechanics







      share|cite|improve this question











      $endgroup$




      I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.



      The Wiener increment $Delta W$ is defined as:



      $Delta W(t) = W(t + Delta t) - W(t) $



      which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.



      I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:



      $(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $



      $langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$



      $langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$



      I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!






      statistics normal-distribution random-walk statistical-mechanics




      statistics normal-distribution random-walk statistical-mechanics






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      share|cite|improve this question













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      edited Dec 7 '18 at 17:28









      user10354138

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      asked Dec 7 '18 at 17:08









      asphasph

      32




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          $begingroup$

          So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.



          But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
            $endgroup$
            – asph
            Dec 9 '18 at 15:42













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          $begingroup$

          So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.



          But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
            $endgroup$
            – asph
            Dec 9 '18 at 15:42


















          0












          $begingroup$

          So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.



          But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
            $endgroup$
            – asph
            Dec 9 '18 at 15:42
















          0












          0








          0





          $begingroup$

          So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.



          But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.






          share|cite|improve this answer









          $endgroup$



          So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.



          But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 17:18









          IanIan

          67.6k25387




          67.6k25387












          • $begingroup$
            Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
            $endgroup$
            – asph
            Dec 9 '18 at 15:42




















          • $begingroup$
            Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
            $endgroup$
            – asph
            Dec 9 '18 at 15:42


















          $begingroup$
          Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
          $endgroup$
          – asph
          Dec 9 '18 at 15:42






          $begingroup$
          Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
          $endgroup$
          – asph
          Dec 9 '18 at 15:42




















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