Mixing
Coefficients of Mandelbrot - van Ness integral representation of fractional Brownian motion
$begingroup$
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?
$$
frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
$$
brownian-motion
asked Mar 4 '18 at 17:56
user166233user166233
111
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add a comment |
$begingroup$
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?
$$
frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
$$
brownian-motion
asked Mar 4 '18 at 17:56
user166233user166233
111
$endgroup$
add a comment |
$begingroup$
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?
$$
frac{1}{Gamma(1/2+H)} quad text{ .vs. } quad sqrt{frac{2Htimes Gamma(3/2- H)}{Gamma(1/2 + H)times Gamma(2-2H)}}quad ???
$$
brownian-motion
asked Mar 4 '18 at 17:56
user166233user166233
111
$endgroup$
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
brownian-motion
asked Mar 4 '18 at 17:56
user166233user166233
111
asked Mar 4 '18 at 17:56
user166233user166233
111
asked Mar 4 '18 at 17:56
user166233user166233
111
asked Mar 4 '18 at 17:56
user166233user166233
111
asked Mar 4 '18 at 17:56
user166233user166233
111
111
add a comment |
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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), $B_{0}^{H}=b_0$.
answered Mar 28 '18 at 8:59
stevensteven
297
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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.
answered Jan 9 at 8:37
Paul HPaul H
567
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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), $B_{0}^{H}=b_0$.
answered Mar 28 '18 at 8:59
stevensteven
297
$endgroup$
add a comment |
$begingroup$
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), $B_{0}^{H}=b_0$.
answered Mar 28 '18 at 8:59
stevensteven
297
$endgroup$
add a comment |
$begingroup$
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), $B_{0}^{H}=b_0$.
answered Mar 28 '18 at 8:59
stevensteven
297
$endgroup$
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), $B_{0}^{H}=b_0$.
answered Mar 28 '18 at 8:59
stevensteven
297
answered Mar 28 '18 at 8:59
stevensteven
297
answered Mar 28 '18 at 8:59
stevensteven
297
answered Mar 28 '18 at 8:59
stevensteven
297
297
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add a comment |
$begingroup$
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.
answered Jan 9 at 8:37
Paul HPaul H
567
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 8:37
Paul HPaul H
567
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 8:37
Paul HPaul H
567
$endgroup$
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.
answered Jan 9 at 8:37
Paul HPaul H
567
answered Jan 9 at 8:37
Paul HPaul H
567
answered Jan 9 at 8:37
Paul HPaul H
567
answered Jan 9 at 8:37
Paul HPaul H
567
567
add a comment |
add a comment |
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