Coefficients of Mandelbrot - van Ness integral representation of fractional Brownian motion












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There are several integral representations of fractional Brownian motion (Hurst parameter $H$) with respect to standard Brownian motion. One of the most commonly used one is Mandelbrot-van Ness. However, I've seen two slightly different versions of Mandelbrot-van Ness and am not sure how they are consistent with each other:



(1) Mandelbrot and van Ness, 1968: $t>0$,
$$
W_t^{H} = frac{1}{Gamma(1/2+H)}left{ int_{-infty}^0 left[ frac{1}{(t-s)^{1/2-H}} - frac{1}{(-s)^{1/2-H}} right] dW_s + int_0^t frac{dW_s}{(t-s)^{1/2-H}} right}.
$$



(2) Mandelbrot - van Ness, alternative form (see Jost 2008): $tin mathbb{R}$,
$$
W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right},
$$
where $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



I noticed that the domain of $t$ are slightly different in the above two versions, one for $t>0$ and one for $tinmathbb{R}$. Also, the terms are arranged slightly differently in the above two versions. Supposedly, both versions should satisfy the same correlation function:
$$
mathbb{E}left[ W_{t'}^{H} W_t^{H} right] = frac{1}{2}left{ t^{2H} + t'^{2H} - |t-t'|^{2H} right}.
$$
My question is that why are the normalizing coefficients in the two versions so different?
$$
frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
$$










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    There are several integral representations of fractional Brownian motion (Hurst parameter $H$) with respect to standard Brownian motion. One of the most commonly used one is Mandelbrot-van Ness. However, I've seen two slightly different versions of Mandelbrot-van Ness and am not sure how they are consistent with each other:



    (1) Mandelbrot and van Ness, 1968: $t>0$,
    $$
    W_t^{H} = frac{1}{Gamma(1/2+H)}left{ int_{-infty}^0 left[ frac{1}{(t-s)^{1/2-H}} - frac{1}{(-s)^{1/2-H}} right] dW_s + int_0^t frac{dW_s}{(t-s)^{1/2-H}} right}.
    $$



    (2) Mandelbrot - van Ness, alternative form (see Jost 2008): $tin mathbb{R}$,
    $$
    W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right},
    $$
    where $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



    I noticed that the domain of $t$ are slightly different in the above two versions, one for $t>0$ and one for $tinmathbb{R}$. Also, the terms are arranged slightly differently in the above two versions. Supposedly, both versions should satisfy the same correlation function:
    $$
    mathbb{E}left[ W_{t'}^{H} W_t^{H} right] = frac{1}{2}left{ t^{2H} + t'^{2H} - |t-t'|^{2H} right}.
    $$
    My question is that why are the normalizing coefficients in the two versions so different?
    $$
    frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
    $$










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      There are several integral representations of fractional Brownian motion (Hurst parameter $H$) with respect to standard Brownian motion. One of the most commonly used one is Mandelbrot-van Ness. However, I've seen two slightly different versions of Mandelbrot-van Ness and am not sure how they are consistent with each other:



      (1) Mandelbrot and van Ness, 1968: $t>0$,
      $$
      W_t^{H} = frac{1}{Gamma(1/2+H)}left{ int_{-infty}^0 left[ frac{1}{(t-s)^{1/2-H}} - frac{1}{(-s)^{1/2-H}} right] dW_s + int_0^t frac{dW_s}{(t-s)^{1/2-H}} right}.
      $$



      (2) Mandelbrot - van Ness, alternative form (see Jost 2008): $tin mathbb{R}$,
      $$
      W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right},
      $$
      where $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



      I noticed that the domain of $t$ are slightly different in the above two versions, one for $t>0$ and one for $tinmathbb{R}$. Also, the terms are arranged slightly differently in the above two versions. Supposedly, both versions should satisfy the same correlation function:
      $$
      mathbb{E}left[ W_{t'}^{H} W_t^{H} right] = frac{1}{2}left{ t^{2H} + t'^{2H} - |t-t'|^{2H} right}.
      $$
      My question is that why are the normalizing coefficients in the two versions so different?
      $$
      frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
      $$










      share|cite|improve this question









      $endgroup$




      There are several integral representations of fractional Brownian motion (Hurst parameter $H$) with respect to standard Brownian motion. One of the most commonly used one is Mandelbrot-van Ness. However, I've seen two slightly different versions of Mandelbrot-van Ness and am not sure how they are consistent with each other:



      (1) Mandelbrot and van Ness, 1968: $t>0$,
      $$
      W_t^{H} = frac{1}{Gamma(1/2+H)}left{ int_{-infty}^0 left[ frac{1}{(t-s)^{1/2-H}} - frac{1}{(-s)^{1/2-H}} right] dW_s + int_0^t frac{dW_s}{(t-s)^{1/2-H}} right}.
      $$



      (2) Mandelbrot - van Ness, alternative form (see Jost 2008): $tin mathbb{R}$,
      $$
      W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right},
      $$
      where $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



      I noticed that the domain of $t$ are slightly different in the above two versions, one for $t>0$ and one for $tinmathbb{R}$. Also, the terms are arranged slightly differently in the above two versions. Supposedly, both versions should satisfy the same correlation function:
      $$
      mathbb{E}left[ W_{t'}^{H} W_t^{H} right] = frac{1}{2}left{ t^{2H} + t'^{2H} - |t-t'|^{2H} right}.
      $$
      My question is that why are the normalizing coefficients in the two versions so different?
      $$
      frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
      $$







      brownian-motion






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      asked Mar 4 '18 at 17:56









      user166233user166233

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          $begingroup$

          If you let $B_{0}^{H}=0$, then $W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right}$ with $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



          In Definition 2.1. of Mandelbrot and Van Ness (1968), $B_{0}^{H}=b_0$.






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            $begingroup$

            The fractional Brownian motion as defined by the Mandelbrot Van Ness Representation actually defines a processes $W^H$ which has the correlation structure $$mathbb{E}W^H_s W^H_t = frac{V_H}{2}{|t|^{2H} + |s|^{2H} - |t-s|^{2H}},$$
            with $$V_H = left(frac{1}{Gamma(H+frac{1}{2})}right)^2left{int_{0}^{infty}left((1+s)^{H-frac{1}{2}} - s^{H-frac{1}{2}}right)^2ds + frac{1}{2H}right}.$$ See Corollary 3.4. in [Mandelbrot and Van Ness 1968].



            Now if we assume that one can show that $C_H^{-2} = Gamma(H+frac{1}{2})^{2} V_H$ then the to representations are consistent.



            To the question: Why did Mandelbrot and Van Ness chose $Gamma(H+frac{1}{2})^{-1}$ as a scaling factor? I can only suspect that they were motivated to take the same scaling factor as for the Riemann-Liouville Brownian motion, which in turn comes from the scaling of the fractional integration operator.






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              2 Answers
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              2 Answers
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              $begingroup$

              If you let $B_{0}^{H}=0$, then $W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right}$ with $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



              In Definition 2.1. of Mandelbrot and Van Ness (1968), $B_{0}^{H}=b_0$.






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                $begingroup$

                If you let $B_{0}^{H}=0$, then $W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right}$ with $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



                In Definition 2.1. of Mandelbrot and Van Ness (1968), $B_{0}^{H}=b_0$.






                share|cite|improve this answer









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                  0












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                  $begingroup$

                  If you let $B_{0}^{H}=0$, then $W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right}$ with $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



                  In Definition 2.1. of Mandelbrot and Van Ness (1968), $B_{0}^{H}=b_0$.






                  share|cite|improve this answer









                  $endgroup$



                  If you let $B_{0}^{H}=0$, then $W_t^{H} = C_H left{ int_{-infty}^t frac{dW_s}{(t-s)^{1/2-H}} - int_{-infty}^0 frac{dW_s}{(-s)^{1/2-H}}right}$ with $C_H = sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}$.



                  In Definition 2.1. of Mandelbrot and Van Ness (1968), $B_{0}^{H}=b_0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 28 '18 at 8:59









                  stevensteven

                  297




                  297























                      0












                      $begingroup$

                      The fractional Brownian motion as defined by the Mandelbrot Van Ness Representation actually defines a processes $W^H$ which has the correlation structure $$mathbb{E}W^H_s W^H_t = frac{V_H}{2}{|t|^{2H} + |s|^{2H} - |t-s|^{2H}},$$
                      with $$V_H = left(frac{1}{Gamma(H+frac{1}{2})}right)^2left{int_{0}^{infty}left((1+s)^{H-frac{1}{2}} - s^{H-frac{1}{2}}right)^2ds + frac{1}{2H}right}.$$ See Corollary 3.4. in [Mandelbrot and Van Ness 1968].



                      Now if we assume that one can show that $C_H^{-2} = Gamma(H+frac{1}{2})^{2} V_H$ then the to representations are consistent.



                      To the question: Why did Mandelbrot and Van Ness chose $Gamma(H+frac{1}{2})^{-1}$ as a scaling factor? I can only suspect that they were motivated to take the same scaling factor as for the Riemann-Liouville Brownian motion, which in turn comes from the scaling of the fractional integration operator.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The fractional Brownian motion as defined by the Mandelbrot Van Ness Representation actually defines a processes $W^H$ which has the correlation structure $$mathbb{E}W^H_s W^H_t = frac{V_H}{2}{|t|^{2H} + |s|^{2H} - |t-s|^{2H}},$$
                        with $$V_H = left(frac{1}{Gamma(H+frac{1}{2})}right)^2left{int_{0}^{infty}left((1+s)^{H-frac{1}{2}} - s^{H-frac{1}{2}}right)^2ds + frac{1}{2H}right}.$$ See Corollary 3.4. in [Mandelbrot and Van Ness 1968].



                        Now if we assume that one can show that $C_H^{-2} = Gamma(H+frac{1}{2})^{2} V_H$ then the to representations are consistent.



                        To the question: Why did Mandelbrot and Van Ness chose $Gamma(H+frac{1}{2})^{-1}$ as a scaling factor? I can only suspect that they were motivated to take the same scaling factor as for the Riemann-Liouville Brownian motion, which in turn comes from the scaling of the fractional integration operator.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The fractional Brownian motion as defined by the Mandelbrot Van Ness Representation actually defines a processes $W^H$ which has the correlation structure $$mathbb{E}W^H_s W^H_t = frac{V_H}{2}{|t|^{2H} + |s|^{2H} - |t-s|^{2H}},$$
                          with $$V_H = left(frac{1}{Gamma(H+frac{1}{2})}right)^2left{int_{0}^{infty}left((1+s)^{H-frac{1}{2}} - s^{H-frac{1}{2}}right)^2ds + frac{1}{2H}right}.$$ See Corollary 3.4. in [Mandelbrot and Van Ness 1968].



                          Now if we assume that one can show that $C_H^{-2} = Gamma(H+frac{1}{2})^{2} V_H$ then the to representations are consistent.



                          To the question: Why did Mandelbrot and Van Ness chose $Gamma(H+frac{1}{2})^{-1}$ as a scaling factor? I can only suspect that they were motivated to take the same scaling factor as for the Riemann-Liouville Brownian motion, which in turn comes from the scaling of the fractional integration operator.






                          share|cite|improve this answer









                          $endgroup$



                          The fractional Brownian motion as defined by the Mandelbrot Van Ness Representation actually defines a processes $W^H$ which has the correlation structure $$mathbb{E}W^H_s W^H_t = frac{V_H}{2}{|t|^{2H} + |s|^{2H} - |t-s|^{2H}},$$
                          with $$V_H = left(frac{1}{Gamma(H+frac{1}{2})}right)^2left{int_{0}^{infty}left((1+s)^{H-frac{1}{2}} - s^{H-frac{1}{2}}right)^2ds + frac{1}{2H}right}.$$ See Corollary 3.4. in [Mandelbrot and Van Ness 1968].



                          Now if we assume that one can show that $C_H^{-2} = Gamma(H+frac{1}{2})^{2} V_H$ then the to representations are consistent.



                          To the question: Why did Mandelbrot and Van Ness chose $Gamma(H+frac{1}{2})^{-1}$ as a scaling factor? I can only suspect that they were motivated to take the same scaling factor as for the Riemann-Liouville Brownian motion, which in turn comes from the scaling of the fractional integration operator.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 9 at 8:37









                          Paul HPaul H

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