A construction of a Stratonovich type integral for fractional Brownian motion












1












$begingroup$


I'm studying this article https://projecteuclid.org/download/pdf_1/euclid.twjm/1500574954 and I'm having problems understanding the proof of lemma 3.



Let me recall some of the criminals involved. Let $(u_t)_{t in [0,T]}$ be a simple bounded process of the form



$$
u = sum_{j = 0}^{n-1} F_j 1_{(t_s, t_{j+1}]},
$$
let us define
$$
u^epsilon_t = frac{1}{2 epsilon}int_{t - epsilon}^{t+epsilon} u_r dr.
$$
Let us define the seminorm
$$
lvert| phi |rvert_K^2 = int_0^T phi(s)^2 K(T,s)^2 ds + int_0^T left(int_s^T left | phi(t) - phi(s) right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
where $K$ satisfies
$$ left | K(t,s) right | leq c((t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}}) $$ and
$$left | frac{partial K}{partial t} (t,s) right | leq c (t-s)^{H - frac{3}{2}}$$
where $H < frac{1}{2}$. Let $mathcal{H}_K$ be the completion of the set of step functions with respect to the norm defined above. Then at page 614 it is stated that $u_t^epsilon$ converges to $u$ in $mathbb{D}^{1,2}(mathcal{H}_K)$. The key step that I'm missing is the integral estimate done at page 615, in fact I obtain and estimate that also involves the integral:
$$
int_{t_i + 2 epsilon}^{t_{i+1} - 2 epsilon} left( int_{s}^{t_i + 2 epsilon} left| u_t^epsilon - u_s^epsilon right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
and I am not able to state that this integral converges to zero. Any suggestions?






probability-theory measure-theory stochastic-calculus stochastic-analysis malliavin-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Regarding the last integral, don't we have $|u_t^epsilon-u_s^epsilon|=0$ for $s,tin(t_i+2epsilon,t_{i+1}-2epsilon)$?
    $endgroup$
    – AddSup
    Dec 9 '18 at 12:11












  • $begingroup$
    Yes, that is true. Anyway in the integral $int_s^{t_i + 2 epsilon} dots$, $s$ belongs to $[0, t_i + 2 epsilon]$ and that shall not be null.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:29












  • $begingroup$
    I meant $t$ not $s$.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:40










  • $begingroup$
    ? In the inner integral, $t$ lies between $s$ and $t_i+2epsilon$ while $s$ lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$, so $t$, too, lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$. Where does $0$ in $[0,t_i+2epsilon]$ come from?
    $endgroup$
    – AddSup
    Dec 9 '18 at 16:54










  • $begingroup$
    You're right, stupidly I didn't notice it.
    $endgroup$
    – JCF
    Dec 9 '18 at 17:48
















1












$begingroup$


I'm studying this article https://projecteuclid.org/download/pdf_1/euclid.twjm/1500574954 and I'm having problems understanding the proof of lemma 3.



Let me recall some of the criminals involved. Let $(u_t)_{t in [0,T]}$ be a simple bounded process of the form



$$
u = sum_{j = 0}^{n-1} F_j 1_{(t_s, t_{j+1}]},
$$
let us define
$$
u^epsilon_t = frac{1}{2 epsilon}int_{t - epsilon}^{t+epsilon} u_r dr.
$$
Let us define the seminorm
$$
lvert| phi |rvert_K^2 = int_0^T phi(s)^2 K(T,s)^2 ds + int_0^T left(int_s^T left | phi(t) - phi(s) right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
where $K$ satisfies
$$ left | K(t,s) right | leq c((t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}}) $$ and
$$left | frac{partial K}{partial t} (t,s) right | leq c (t-s)^{H - frac{3}{2}}$$
where $H < frac{1}{2}$. Let $mathcal{H}_K$ be the completion of the set of step functions with respect to the norm defined above. Then at page 614 it is stated that $u_t^epsilon$ converges to $u$ in $mathbb{D}^{1,2}(mathcal{H}_K)$. The key step that I'm missing is the integral estimate done at page 615, in fact I obtain and estimate that also involves the integral:
$$
int_{t_i + 2 epsilon}^{t_{i+1} - 2 epsilon} left( int_{s}^{t_i + 2 epsilon} left| u_t^epsilon - u_s^epsilon right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
and I am not able to state that this integral converges to zero. Any suggestions?






probability-theory measure-theory stochastic-calculus stochastic-analysis malliavin-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Regarding the last integral, don't we have $|u_t^epsilon-u_s^epsilon|=0$ for $s,tin(t_i+2epsilon,t_{i+1}-2epsilon)$?
    $endgroup$
    – AddSup
    Dec 9 '18 at 12:11












  • $begingroup$
    Yes, that is true. Anyway in the integral $int_s^{t_i + 2 epsilon} dots$, $s$ belongs to $[0, t_i + 2 epsilon]$ and that shall not be null.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:29












  • $begingroup$
    I meant $t$ not $s$.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:40










  • $begingroup$
    ? In the inner integral, $t$ lies between $s$ and $t_i+2epsilon$ while $s$ lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$, so $t$, too, lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$. Where does $0$ in $[0,t_i+2epsilon]$ come from?
    $endgroup$
    – AddSup
    Dec 9 '18 at 16:54










  • $begingroup$
    You're right, stupidly I didn't notice it.
    $endgroup$
    – JCF
    Dec 9 '18 at 17:48














1












1








1


1



$begingroup$


I'm studying this article https://projecteuclid.org/download/pdf_1/euclid.twjm/1500574954 and I'm having problems understanding the proof of lemma 3.



Let me recall some of the criminals involved. Let $(u_t)_{t in [0,T]}$ be a simple bounded process of the form



$$
u = sum_{j = 0}^{n-1} F_j 1_{(t_s, t_{j+1}]},
$$
let us define
$$
u^epsilon_t = frac{1}{2 epsilon}int_{t - epsilon}^{t+epsilon} u_r dr.
$$
Let us define the seminorm
$$
lvert| phi |rvert_K^2 = int_0^T phi(s)^2 K(T,s)^2 ds + int_0^T left(int_s^T left | phi(t) - phi(s) right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
where $K$ satisfies
$$ left | K(t,s) right | leq c((t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}}) $$ and
$$left | frac{partial K}{partial t} (t,s) right | leq c (t-s)^{H - frac{3}{2}}$$
where $H < frac{1}{2}$. Let $mathcal{H}_K$ be the completion of the set of step functions with respect to the norm defined above. Then at page 614 it is stated that $u_t^epsilon$ converges to $u$ in $mathbb{D}^{1,2}(mathcal{H}_K)$. The key step that I'm missing is the integral estimate done at page 615, in fact I obtain and estimate that also involves the integral:
$$
int_{t_i + 2 epsilon}^{t_{i+1} - 2 epsilon} left( int_{s}^{t_i + 2 epsilon} left| u_t^epsilon - u_s^epsilon right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
and I am not able to state that this integral converges to zero. Any suggestions?






probability-theory measure-theory stochastic-calculus stochastic-analysis malliavin-calculus







share|cite|improve this question











$endgroup$




I'm studying this article https://projecteuclid.org/download/pdf_1/euclid.twjm/1500574954 and I'm having problems understanding the proof of lemma 3.



Let me recall some of the criminals involved. Let $(u_t)_{t in [0,T]}$ be a simple bounded process of the form



$$
u = sum_{j = 0}^{n-1} F_j 1_{(t_s, t_{j+1}]},
$$
let us define
$$
u^epsilon_t = frac{1}{2 epsilon}int_{t - epsilon}^{t+epsilon} u_r dr.
$$
Let us define the seminorm
$$
lvert| phi |rvert_K^2 = int_0^T phi(s)^2 K(T,s)^2 ds + int_0^T left(int_s^T left | phi(t) - phi(s) right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
where $K$ satisfies
$$ left | K(t,s) right | leq c((t-s)^{H - frac{1}{2}} + s^{H - frac{1}{2}}) $$ and
$$left | frac{partial K}{partial t} (t,s) right | leq c (t-s)^{H - frac{3}{2}}$$
where $H < frac{1}{2}$. Let $mathcal{H}_K$ be the completion of the set of step functions with respect to the norm defined above. Then at page 614 it is stated that $u_t^epsilon$ converges to $u$ in $mathbb{D}^{1,2}(mathcal{H}_K)$. The key step that I'm missing is the integral estimate done at page 615, in fact I obtain and estimate that also involves the integral:
$$
int_{t_i + 2 epsilon}^{t_{i+1} - 2 epsilon} left( int_{s}^{t_i + 2 epsilon} left| u_t^epsilon - u_s^epsilon right | (t-s)^{H - frac{3}{2}} dt right)^2 ds
$$
and I am not able to state that this integral converges to zero. Any suggestions?






probability-theory measure-theory stochastic-calculus stochastic-analysis malliavin-calculus




probability-theory measure-theory stochastic-calculus stochastic-analysis malliavin-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 17:23







JCF

















asked Dec 8 '18 at 17:09









JCFJCF

349112




349112












  • $begingroup$
    Regarding the last integral, don't we have $|u_t^epsilon-u_s^epsilon|=0$ for $s,tin(t_i+2epsilon,t_{i+1}-2epsilon)$?
    $endgroup$
    – AddSup
    Dec 9 '18 at 12:11












  • $begingroup$
    Yes, that is true. Anyway in the integral $int_s^{t_i + 2 epsilon} dots$, $s$ belongs to $[0, t_i + 2 epsilon]$ and that shall not be null.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:29












  • $begingroup$
    I meant $t$ not $s$.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:40










  • $begingroup$
    ? In the inner integral, $t$ lies between $s$ and $t_i+2epsilon$ while $s$ lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$, so $t$, too, lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$. Where does $0$ in $[0,t_i+2epsilon]$ come from?
    $endgroup$
    – AddSup
    Dec 9 '18 at 16:54










  • $begingroup$
    You're right, stupidly I didn't notice it.
    $endgroup$
    – JCF
    Dec 9 '18 at 17:48


















  • $begingroup$
    Regarding the last integral, don't we have $|u_t^epsilon-u_s^epsilon|=0$ for $s,tin(t_i+2epsilon,t_{i+1}-2epsilon)$?
    $endgroup$
    – AddSup
    Dec 9 '18 at 12:11












  • $begingroup$
    Yes, that is true. Anyway in the integral $int_s^{t_i + 2 epsilon} dots$, $s$ belongs to $[0, t_i + 2 epsilon]$ and that shall not be null.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:29












  • $begingroup$
    I meant $t$ not $s$.
    $endgroup$
    – JCF
    Dec 9 '18 at 15:40










  • $begingroup$
    ? In the inner integral, $t$ lies between $s$ and $t_i+2epsilon$ while $s$ lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$, so $t$, too, lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$. Where does $0$ in $[0,t_i+2epsilon]$ come from?
    $endgroup$
    – AddSup
    Dec 9 '18 at 16:54










  • $begingroup$
    You're right, stupidly I didn't notice it.
    $endgroup$
    – JCF
    Dec 9 '18 at 17:48
















$begingroup$
Regarding the last integral, don't we have $|u_t^epsilon-u_s^epsilon|=0$ for $s,tin(t_i+2epsilon,t_{i+1}-2epsilon)$?
$endgroup$
– AddSup
Dec 9 '18 at 12:11






$begingroup$
Regarding the last integral, don't we have $|u_t^epsilon-u_s^epsilon|=0$ for $s,tin(t_i+2epsilon,t_{i+1}-2epsilon)$?
$endgroup$
– AddSup
Dec 9 '18 at 12:11














$begingroup$
Yes, that is true. Anyway in the integral $int_s^{t_i + 2 epsilon} dots$, $s$ belongs to $[0, t_i + 2 epsilon]$ and that shall not be null.
$endgroup$
– JCF
Dec 9 '18 at 15:29






$begingroup$
Yes, that is true. Anyway in the integral $int_s^{t_i + 2 epsilon} dots$, $s$ belongs to $[0, t_i + 2 epsilon]$ and that shall not be null.
$endgroup$
– JCF
Dec 9 '18 at 15:29














$begingroup$
I meant $t$ not $s$.
$endgroup$
– JCF
Dec 9 '18 at 15:40




$begingroup$
I meant $t$ not $s$.
$endgroup$
– JCF
Dec 9 '18 at 15:40












$begingroup$
? In the inner integral, $t$ lies between $s$ and $t_i+2epsilon$ while $s$ lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$, so $t$, too, lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$. Where does $0$ in $[0,t_i+2epsilon]$ come from?
$endgroup$
– AddSup
Dec 9 '18 at 16:54




$begingroup$
? In the inner integral, $t$ lies between $s$ and $t_i+2epsilon$ while $s$ lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$, so $t$, too, lies between $t_i+2epsilon$ and $t_{i+1}-2epsilon$. Where does $0$ in $[0,t_i+2epsilon]$ come from?
$endgroup$
– AddSup
Dec 9 '18 at 16:54












$begingroup$
You're right, stupidly I didn't notice it.
$endgroup$
– JCF
Dec 9 '18 at 17:48




$begingroup$
You're right, stupidly I didn't notice it.
$endgroup$
– JCF
Dec 9 '18 at 17:48










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