Mixing
Estimate for the distance from the initial value of a strong solution of an SDE
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous
$(X_t^x)_{tge0}$ be a continuous process on $(Omega,mathcal A,operatorname P)$ with $$X^x_t=x+int_0^tb(X^x_s):{rm d}s+int_0^tsigma(X^x_s):{rm d}W_s;;;text{for all }tge0text{ almost surely}tag1$$ for $xinmathbb R$
By Lipschitz continuity, $$|b(x)|^2+|sigma(x)|^2le c(1+|x|^2);;;text{for all }xinmathbb Rtag2$$ for some $cge0$. For simplicity of notation, write $|Y|_t^ast:=sup_{sin[0,:t]}|Y_s|$ for $tge0$ and any process $(Y_t)_{tge0}$.
Fix $tge0$ and $xinmathbb R$. By Hölder’s inequality and Fubini’s theorem$^1$, $$operatorname Eleft[{left|b(X^x)cdot[W]right|_t^ast}^2right]le ctleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag3$$ By the Burkholder-Davis-Gundy inequality and Fubini’s theorem, $$operatorname Eleft[{left|sigma(X^x)cdot Wright|_t^ast}^2right]le4cleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag4$$ Letting $c_1:=max(2,4c(t+4)t,4c(t+4))$, we obtain $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright)tag5$$ and hence $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1right)e^{c_1t}tag6$$ by Grönwall's inequality.
Question: How can we conclude $$operatorname Eleft[{left|X^x-xright|_t^ast}^2right]le c_2left(x^2+1right)ttag7$$ for some $c_2ge0$ from $(6)$?
$^1$ As usual, $(b(X^x)cdot[W])_t:=int_0^tb(X^x_s):{rm d}[W]_s$ and $(sigma(X^x)cdot W)_t:=int_0^tsigma(X^x_s):{rm d}W_s$.
probability-theory stochastic-processes stochastic-calculus stochastic-analysis sde
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
$endgroup$
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous
$(X_t^x)_{tge0}$ be a continuous process on $(Omega,mathcal A,operatorname P)$ with $$X^x_t=x+int_0^tb(X^x_s):{rm d}s+int_0^tsigma(X^x_s):{rm d}W_s;;;text{for all }tge0text{ almost surely}tag1$$ for $xinmathbb R$
By Lipschitz continuity, $$|b(x)|^2+|sigma(x)|^2le c(1+|x|^2);;;text{for all }xinmathbb Rtag2$$ for some $cge0$. For simplicity of notation, write $|Y|_t^ast:=sup_{sin[0,:t]}|Y_s|$ for $tge0$ and any process $(Y_t)_{tge0}$.
Fix $tge0$ and $xinmathbb R$. By Hölder’s inequality and Fubini’s theorem$^1$, $$operatorname Eleft[{left|b(X^x)cdot[W]right|_t^ast}^2right]le ctleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag3$$ By the Burkholder-Davis-Gundy inequality and Fubini’s theorem, $$operatorname Eleft[{left|sigma(X^x)cdot Wright|_t^ast}^2right]le4cleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag4$$ Letting $c_1:=max(2,4c(t+4)t,4c(t+4))$, we obtain $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright)tag5$$ and hence $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1right)e^{c_1t}tag6$$ by Grönwall's inequality.
Question: How can we conclude $$operatorname Eleft[{left|X^x-xright|_t^ast}^2right]le c_2left(x^2+1right)ttag7$$ for some $c_2ge0$ from $(6)$?
$^1$ As usual, $(b(X^x)cdot[W])_t:=int_0^tb(X^x_s):{rm d}[W]_s$ and $(sigma(X^x)cdot W)_t:=int_0^tsigma(X^x_s):{rm d}W_s$.
probability-theory stochastic-processes stochastic-calculus stochastic-analysis sde
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
$endgroup$
$begingroup$
Unless I'm missing something (7) fails to hold; just consider $b=0$, $sigma=1$.
$endgroup$
– saz
Jan 30 at 7:26
$begingroup$
@saz $(7)$ is claimed here in equation 9.18.
$endgroup$
– 0xbadf00d
Jan 30 at 9:47
$begingroup$
No, it's not. In (7) the power of t is wrong; in the book it's okay.
$endgroup$
– saz
Jan 30 at 9:57
$begingroup$
@saz That was a typo here.
$endgroup$
– 0xbadf00d
Jan 30 at 9:59
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous
$(X_t^x)_{tge0}$ be a continuous process on $(Omega,mathcal A,operatorname P)$ with $$X^x_t=x+int_0^tb(X^x_s):{rm d}s+int_0^tsigma(X^x_s):{rm d}W_s;;;text{for all }tge0text{ almost surely}tag1$$ for $xinmathbb R$
By Lipschitz continuity, $$|b(x)|^2+|sigma(x)|^2le c(1+|x|^2);;;text{for all }xinmathbb Rtag2$$ for some $cge0$. For simplicity of notation, write $|Y|_t^ast:=sup_{sin[0,:t]}|Y_s|$ for $tge0$ and any process $(Y_t)_{tge0}$.
Fix $tge0$ and $xinmathbb R$. By Hölder’s inequality and Fubini’s theorem$^1$, $$operatorname Eleft[{left|b(X^x)cdot[W]right|_t^ast}^2right]le ctleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag3$$ By the Burkholder-Davis-Gundy inequality and Fubini’s theorem, $$operatorname Eleft[{left|sigma(X^x)cdot Wright|_t^ast}^2right]le4cleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag4$$ Letting $c_1:=max(2,4c(t+4)t,4c(t+4))$, we obtain $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright)tag5$$ and hence $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1right)e^{c_1t}tag6$$ by Grönwall's inequality.
Question: How can we conclude $$operatorname Eleft[{left|X^x-xright|_t^ast}^2right]le c_2left(x^2+1right)ttag7$$ for some $c_2ge0$ from $(6)$?
$^1$ As usual, $(b(X^x)cdot[W])_t:=int_0^tb(X^x_s):{rm d}[W]_s$ and $(sigma(X^x)cdot W)_t:=int_0^tsigma(X^x_s):{rm d}W_s$.
probability-theory stochastic-processes stochastic-calculus stochastic-analysis sde
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
$endgroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous
$(X_t^x)_{tge0}$ be a continuous process on $(Omega,mathcal A,operatorname P)$ with $$X^x_t=x+int_0^tb(X^x_s):{rm d}s+int_0^tsigma(X^x_s):{rm d}W_s;;;text{for all }tge0text{ almost surely}tag1$$ for $xinmathbb R$
By Lipschitz continuity, $$|b(x)|^2+|sigma(x)|^2le c(1+|x|^2);;;text{for all }xinmathbb Rtag2$$ for some $cge0$. For simplicity of notation, write $|Y|_t^ast:=sup_{sin[0,:t]}|Y_s|$ for $tge0$ and any process $(Y_t)_{tge0}$.
Fix $tge0$ and $xinmathbb R$. By Hölder’s inequality and Fubini’s theorem$^1$, $$operatorname Eleft[{left|b(X^x)cdot[W]right|_t^ast}^2right]le ctleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag3$$ By the Burkholder-Davis-Gundy inequality and Fubini’s theorem, $$operatorname Eleft[{left|sigma(X^x)cdot Wright|_t^ast}^2right]le4cleft(t+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright).tag4$$ Letting $c_1:=max(2,4c(t+4)t,4c(t+4))$, we obtain $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1+int_0^toperatorname Eleft[{left|X^xright|_s^ast}^2right]{rm d}sright)tag5$$ and hence $$operatorname Eleft[{left|X^xright|_t^ast}^2right]le c_1left(x^2+1right)e^{c_1t}tag6$$ by Grönwall's inequality.
Question: How can we conclude $$operatorname Eleft[{left|X^x-xright|_t^ast}^2right]le c_2left(x^2+1right)ttag7$$ for some $c_2ge0$ from $(6)$?
$^1$ As usual, $(b(X^x)cdot[W])_t:=int_0^tb(X^x_s):{rm d}[W]_s$ and $(sigma(X^x)cdot W)_t:=int_0^tsigma(X^x_s):{rm d}W_s$.
probability-theory stochastic-processes stochastic-calculus stochastic-analysis sde
probability-theory stochastic-processes stochastic-calculus stochastic-analysis sde
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
edited Jan 30 at 9:59
0xbadf00d
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
asked Jan 29 at 23:24
0xbadf00d0xbadf00d
1,77641534
1,77641534
$begingroup$
Unless I'm missing something (7) fails to hold; just consider $b=0$, $sigma=1$.
$endgroup$
– saz
Jan 30 at 7:26
$begingroup$
@saz $(7)$ is claimed here in equation 9.18.
$endgroup$
– 0xbadf00d
Jan 30 at 9:47
$begingroup$
No, it's not. In (7) the power of t is wrong; in the book it's okay.
$endgroup$
– saz
Jan 30 at 9:57
$begingroup$
@saz That was a typo here.
$endgroup$
– 0xbadf00d
Jan 30 at 9:59
add a comment |
$begingroup$
Unless I'm missing something (7) fails to hold; just consider $b=0$, $sigma=1$.
$endgroup$
– saz
Jan 30 at 7:26
$begingroup$
@saz $(7)$ is claimed here in equation 9.18.
$endgroup$
– 0xbadf00d
Jan 30 at 9:47
$begingroup$
No, it's not. In (7) the power of t is wrong; in the book it's okay.
$endgroup$
– saz
Jan 30 at 9:57
$begingroup$
@saz That was a typo here.
$endgroup$
– 0xbadf00d
Jan 30 at 9:59
$begingroup$
Unless I'm missing something (7) fails to hold; just consider $b=0$, $sigma=1$.
$endgroup$
– saz
Jan 30 at 7:26
$begingroup$
Unless I'm missing something (7) fails to hold; just consider $b=0$, $sigma=1$.
$endgroup$
– saz
Jan 30 at 7:26
$begingroup$
@saz $(7)$ is claimed here in equation 9.18.
$endgroup$
– 0xbadf00d
Jan 30 at 9:47
$begingroup$
@saz $(7)$ is claimed here in equation 9.18.
$endgroup$
– 0xbadf00d
Jan 30 at 9:47
$begingroup$
No, it's not. In (7) the power of t is wrong; in the book it's okay.
$endgroup$
– saz
Jan 30 at 9:57
$begingroup$
No, it's not. In (7) the power of t is wrong; in the book it's okay.
$endgroup$
– saz
Jan 30 at 9:57
$begingroup$
@saz That was a typo here.
$endgroup$
– 0xbadf00d
Jan 30 at 9:59
$begingroup$
@saz That was a typo here.
$endgroup$
– 0xbadf00d
Jan 30 at 9:59
add a comment |
1 Answer
1
active
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$begingroup$
I doubt that it is possible to derive (7) from (6), but you can use (3) and (4) to get (7).
Since
$$|x-y|^2 leq 2x^2+2y^2, qquad x,y in mathbb{R},$$
it follows that
$$sup_{s leq t} |X_s|^2 = sup_{s leq t} |X_s-x+x|^2 leq 2 sup_{s leq t} |X_s-x|^2 + 2 x^2.$$
Using this estimate on the right-hand side of (3) and (4), respectively, we find that
$$mathbb{E} left( |b(X^x) bullet [W]|_t^{ast 2} + |sigma(X^x) bullet W|_t^{ast 2} right) leq C_1 left( t +t x^2 + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right),$$
for some constant $C_1>0$ and so
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 left( t (1+x^2) + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right).$$
Applying Gronwall's inequality we find that there exists a constant $C_3>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 t (x^2+1) e^{C_3 t}. $$
For $t in [0,T]$ this implies that there exists a constant $C=C(T)>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C t (x^2+1), qquad t in [0,T]. tag{8}$$
Note that we cannot expect to find a constant $C>0$ such that $(8)$ holds for all $t geq 0$ since the second moment may grow exponentially in $t$.
answered Jan 30 at 10:13
sazsaz
82k862131
$endgroup$
$begingroup$
I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
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– 0xbadf00d
Feb 19 at 19:47
add a comment |
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$begingroup$
I doubt that it is possible to derive (7) from (6), but you can use (3) and (4) to get (7).
Since
$$|x-y|^2 leq 2x^2+2y^2, qquad x,y in mathbb{R},$$
it follows that
$$sup_{s leq t} |X_s|^2 = sup_{s leq t} |X_s-x+x|^2 leq 2 sup_{s leq t} |X_s-x|^2 + 2 x^2.$$
Using this estimate on the right-hand side of (3) and (4), respectively, we find that
$$mathbb{E} left( |b(X^x) bullet [W]|_t^{ast 2} + |sigma(X^x) bullet W|_t^{ast 2} right) leq C_1 left( t +t x^2 + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right),$$
for some constant $C_1>0$ and so
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 left( t (1+x^2) + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right).$$
Applying Gronwall's inequality we find that there exists a constant $C_3>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 t (x^2+1) e^{C_3 t}. $$
For $t in [0,T]$ this implies that there exists a constant $C=C(T)>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C t (x^2+1), qquad t in [0,T]. tag{8}$$
Note that we cannot expect to find a constant $C>0$ such that $(8)$ holds for all $t geq 0$ since the second moment may grow exponentially in $t$.
answered Jan 30 at 10:13
sazsaz
82k862131
$endgroup$
$begingroup$
I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
$endgroup$
– 0xbadf00d
Feb 19 at 19:47
add a comment |
$begingroup$
I doubt that it is possible to derive (7) from (6), but you can use (3) and (4) to get (7).
Since
$$|x-y|^2 leq 2x^2+2y^2, qquad x,y in mathbb{R},$$
it follows that
$$sup_{s leq t} |X_s|^2 = sup_{s leq t} |X_s-x+x|^2 leq 2 sup_{s leq t} |X_s-x|^2 + 2 x^2.$$
Using this estimate on the right-hand side of (3) and (4), respectively, we find that
$$mathbb{E} left( |b(X^x) bullet [W]|_t^{ast 2} + |sigma(X^x) bullet W|_t^{ast 2} right) leq C_1 left( t +t x^2 + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right),$$
for some constant $C_1>0$ and so
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 left( t (1+x^2) + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right).$$
Applying Gronwall's inequality we find that there exists a constant $C_3>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 t (x^2+1) e^{C_3 t}. $$
For $t in [0,T]$ this implies that there exists a constant $C=C(T)>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C t (x^2+1), qquad t in [0,T]. tag{8}$$
Note that we cannot expect to find a constant $C>0$ such that $(8)$ holds for all $t geq 0$ since the second moment may grow exponentially in $t$.
answered Jan 30 at 10:13
sazsaz
82k862131
$endgroup$
$begingroup$
I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
$endgroup$
– 0xbadf00d
Feb 19 at 19:47
add a comment |
$begingroup$
I doubt that it is possible to derive (7) from (6), but you can use (3) and (4) to get (7).
Since
$$|x-y|^2 leq 2x^2+2y^2, qquad x,y in mathbb{R},$$
it follows that
$$sup_{s leq t} |X_s|^2 = sup_{s leq t} |X_s-x+x|^2 leq 2 sup_{s leq t} |X_s-x|^2 + 2 x^2.$$
Using this estimate on the right-hand side of (3) and (4), respectively, we find that
$$mathbb{E} left( |b(X^x) bullet [W]|_t^{ast 2} + |sigma(X^x) bullet W|_t^{ast 2} right) leq C_1 left( t +t x^2 + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right),$$
for some constant $C_1>0$ and so
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 left( t (1+x^2) + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right).$$
Applying Gronwall's inequality we find that there exists a constant $C_3>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 t (x^2+1) e^{C_3 t}. $$
For $t in [0,T]$ this implies that there exists a constant $C=C(T)>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C t (x^2+1), qquad t in [0,T]. tag{8}$$
Note that we cannot expect to find a constant $C>0$ such that $(8)$ holds for all $t geq 0$ since the second moment may grow exponentially in $t$.
answered Jan 30 at 10:13
sazsaz
82k862131
$endgroup$
I doubt that it is possible to derive (7) from (6), but you can use (3) and (4) to get (7).
Since
$$|x-y|^2 leq 2x^2+2y^2, qquad x,y in mathbb{R},$$
it follows that
$$sup_{s leq t} |X_s|^2 = sup_{s leq t} |X_s-x+x|^2 leq 2 sup_{s leq t} |X_s-x|^2 + 2 x^2.$$
Using this estimate on the right-hand side of (3) and (4), respectively, we find that
$$mathbb{E} left( |b(X^x) bullet [W]|_t^{ast 2} + |sigma(X^x) bullet W|_t^{ast 2} right) leq C_1 left( t +t x^2 + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right),$$
for some constant $C_1>0$ and so
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 left( t (1+x^2) + int_0^t mathbb{E}(|X^x-x|_s^{ast 2}) , ds right).$$
Applying Gronwall's inequality we find that there exists a constant $C_3>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C_2 t (x^2+1) e^{C_3 t}. $$
For $t in [0,T]$ this implies that there exists a constant $C=C(T)>0$ such that
$$mathbb{E}(|X^x-x|_t^{ast 2}) leq C t (x^2+1), qquad t in [0,T]. tag{8}$$
Note that we cannot expect to find a constant $C>0$ such that $(8)$ holds for all $t geq 0$ since the second moment may grow exponentially in $t$.
answered Jan 30 at 10:13
sazsaz
82k862131
answered Jan 30 at 10:13
sazsaz
82k862131
answered Jan 30 at 10:13
sazsaz
82k862131
answered Jan 30 at 10:13
sazsaz
82k862131
82k862131
$begingroup$
I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
$endgroup$
– 0xbadf00d
Feb 19 at 19:47
add a comment |
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I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
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– 0xbadf00d
Feb 19 at 19:47
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I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
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– 0xbadf00d
Feb 19 at 19:47
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I'm wondering how we can show that pathwise uniqueness implies uniqueness in law. Isn't that trivial in the scenario of this question? I'm asking in light of the Yamada-Watanabe result. Is the situation more complicated when one is considering weak solutions? I've asked a separate question for that, maybe you can take a look:math.stackexchange.com/questions/3118723/…
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– 0xbadf00d
Feb 19 at 19:47
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Unless I'm missing something (7) fails to hold; just consider $b=0$, $sigma=1$.
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– saz
Jan 30 at 7:26
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@saz $(7)$ is claimed here in equation 9.18.
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– 0xbadf00d
Jan 30 at 9:47
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No, it's not. In (7) the power of t is wrong; in the book it's okay.
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– saz
Jan 30 at 9:57
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@saz That was a typo here.
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– 0xbadf00d
Jan 30 at 9:59