Integral representation of fractional Brownian motion and covariance function
$begingroup$
It is well known that a gaussian process is identified by its covariance function and its mean function. Given the covariance function:
$$
R left( t , s right) = E left[B_t^H B_s^H right] = frac{1}{2} left ( t^{2H} + s^{2H} - left | t- s right |^{2H} right)
$$
fractional Brownian motion (fBm) is the centered gaussian process with covariance function $R(t,s)$.
Moreover in https://projecteuclid.org/download/pdf_1/euclid.aop/1008956692 is given an integral representation of the fBm as
$$
B^H_t = int_0^t K(t,s) dW_s
$$
where $W$ is the wiener process and $K(t,s)$ is the Kernel:
$$
K(t,s) = c_H ( t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}} F_1left ( frac{t}{s} right )
$$
where
$$
F_1(z) = c_H left (frac{1}{2} - H right ) int_0^{z-1} theta^{H - frac{3}{2}} left ( 1 - left (theta + 1right )^{H - frac{1}{2}} right ) dtheta
$$
and $c_H$ is a constant that depends on the Hurst parameter $H$.
Given this representation, the covariance function of such process is given by
$$
int_0^{t wedge s} K(t, r) K(s, r) dr.
$$
For instance note that if $H = frac{1}{2}$, setting $c_H = 1$, we obtain the standard Wiener process as expected.
Consider the case $H < frac{1}{2}$ I want to prove that
$$
R(t,s) = int_0^{t wedge s} K(t, r) K(s, r) dr
$$
where $R(t,s)$ is the covariance function of the fBm. However I am having problems in handling such integral and I was wondering if there are other approaches than mere computations.
Thanks in advance to whoever has read the whole thing I wrote.
probability-theory stochastic-processes stochastic-calculus brownian-motion
asked Dec 14 '18 at 11:42
JCFJCF
339112
$endgroup$
add a comment |
$begingroup$
It is well known that a gaussian process is identified by its covariance function and its mean function. Given the covariance function:
$$
R left( t , s right) = E left[B_t^H B_s^H right] = frac{1}{2} left ( t^{2H} + s^{2H} - left | t- s right |^{2H} right)
$$
fractional Brownian motion (fBm) is the centered gaussian process with covariance function $R(t,s)$.
Moreover in https://projecteuclid.org/download/pdf_1/euclid.aop/1008956692 is given an integral representation of the fBm as
$$
B^H_t = int_0^t K(t,s) dW_s
$$
where $W$ is the wiener process and $K(t,s)$ is the Kernel:
$$
K(t,s) = c_H ( t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}} F_1left ( frac{t}{s} right )
$$
where
$$
F_1(z) = c_H left (frac{1}{2} - H right ) int_0^{z-1} theta^{H - frac{3}{2}} left ( 1 - left (theta + 1right )^{H - frac{1}{2}} right ) dtheta
$$
and $c_H$ is a constant that depends on the Hurst parameter $H$.
Given this representation, the covariance function of such process is given by
$$
int_0^{t wedge s} K(t, r) K(s, r) dr.
$$
For instance note that if $H = frac{1}{2}$, setting $c_H = 1$, we obtain the standard Wiener process as expected.
Consider the case $H < frac{1}{2}$ I want to prove that
$$
R(t,s) = int_0^{t wedge s} K(t, r) K(s, r) dr
$$
where $R(t,s)$ is the covariance function of the fBm. However I am having problems in handling such integral and I was wondering if there are other approaches than mere computations.
Thanks in advance to whoever has read the whole thing I wrote.
probability-theory stochastic-processes stochastic-calculus brownian-motion
asked Dec 14 '18 at 11:42
JCFJCF
339112
$endgroup$
add a comment |
$begingroup$
It is well known that a gaussian process is identified by its covariance function and its mean function. Given the covariance function:
$$
R left( t , s right) = E left[B_t^H B_s^H right] = frac{1}{2} left ( t^{2H} + s^{2H} - left | t- s right |^{2H} right)
$$
fractional Brownian motion (fBm) is the centered gaussian process with covariance function $R(t,s)$.
Moreover in https://projecteuclid.org/download/pdf_1/euclid.aop/1008956692 is given an integral representation of the fBm as
$$
B^H_t = int_0^t K(t,s) dW_s
$$
where $W$ is the wiener process and $K(t,s)$ is the Kernel:
$$
K(t,s) = c_H ( t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}} F_1left ( frac{t}{s} right )
$$
where
$$
F_1(z) = c_H left (frac{1}{2} - H right ) int_0^{z-1} theta^{H - frac{3}{2}} left ( 1 - left (theta + 1right )^{H - frac{1}{2}} right ) dtheta
$$
and $c_H$ is a constant that depends on the Hurst parameter $H$.
Given this representation, the covariance function of such process is given by
$$
int_0^{t wedge s} K(t, r) K(s, r) dr.
$$
For instance note that if $H = frac{1}{2}$, setting $c_H = 1$, we obtain the standard Wiener process as expected.
Consider the case $H < frac{1}{2}$ I want to prove that
$$
R(t,s) = int_0^{t wedge s} K(t, r) K(s, r) dr
$$
where $R(t,s)$ is the covariance function of the fBm. However I am having problems in handling such integral and I was wondering if there are other approaches than mere computations.
Thanks in advance to whoever has read the whole thing I wrote.
probability-theory stochastic-processes stochastic-calculus brownian-motion
asked Dec 14 '18 at 11:42
JCFJCF
339112
$endgroup$
It is well known that a gaussian process is identified by its covariance function and its mean function. Given the covariance function:
$$
R left( t , s right) = E left[B_t^H B_s^H right] = frac{1}{2} left ( t^{2H} + s^{2H} - left | t- s right |^{2H} right)
$$
fractional Brownian motion (fBm) is the centered gaussian process with covariance function $R(t,s)$.
Moreover in https://projecteuclid.org/download/pdf_1/euclid.aop/1008956692 is given an integral representation of the fBm as
$$
B^H_t = int_0^t K(t,s) dW_s
$$
where $W$ is the wiener process and $K(t,s)$ is the Kernel:
$$
K(t,s) = c_H ( t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}} F_1left ( frac{t}{s} right )
$$
where
$$
F_1(z) = c_H left (frac{1}{2} - H right ) int_0^{z-1} theta^{H - frac{3}{2}} left ( 1 - left (theta + 1right )^{H - frac{1}{2}} right ) dtheta
$$
and $c_H$ is a constant that depends on the Hurst parameter $H$.
Given this representation, the covariance function of such process is given by
$$
int_0^{t wedge s} K(t, r) K(s, r) dr.
$$
For instance note that if $H = frac{1}{2}$, setting $c_H = 1$, we obtain the standard Wiener process as expected.
Consider the case $H < frac{1}{2}$ I want to prove that
$$
R(t,s) = int_0^{t wedge s} K(t, r) K(s, r) dr
$$
where $R(t,s)$ is the covariance function of the fBm. However I am having problems in handling such integral and I was wondering if there are other approaches than mere computations.
Thanks in advance to whoever has read the whole thing I wrote.
probability-theory stochastic-processes stochastic-calculus brownian-motion
probability-theory stochastic-processes stochastic-calculus brownian-motion
asked Dec 14 '18 at 11:42
JCFJCF
339112
asked Dec 14 '18 at 11:42
JCFJCF
339112
edited Dec 14 '18 at 13:51
JCF
asked Dec 14 '18 at 11:42
JCFJCF
339112
asked Dec 14 '18 at 11:42
JCFJCF
339112
asked Dec 14 '18 at 11:42
JCFJCF
339112
339112
add a comment |
add a comment |
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