Integrability in Dynkin's formula
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$W$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$
$b,sigma:mathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$
$(X_t)_{tge0}$ be a strong solution of $${rm d}X_t=b(t,X_t){rm d}t+sigma(t,X_t){rm d}W_ttag3$$
Moreover, let $$(L_tf)(x):=b(t,x)f'(x)+frac12sigma^2(t,x)f''(x);;;text{for }xinmathbb Rtext{ and }fin C^2(mathbb R)$$ for $tge0$. Now, let $fin C^2(mathbb R)$. By the Itō formula, $$f(t,X_t)=f(0,X_0)+int_0^t(L_sf)(X_s):{rm d}s+underbrace{int_0^tsigma f'(X_s):{rm d}W_s}_{=::M_t};;;text{for all }tge0tag4.$$ Assuming $f'$ is bounded, we obtain that $M$ is a martingale.
Are we able to show that $$operatorname Eleft[int_0^tleft|(L_sf)(X_s)right|:{rm d}sright]<inftytag5$$ for all $tge0$?
I've read here (in Corollary 6.5) that this would be the case. However, I don't see how we can bound $left|f''(X_t)right|$.
probability-theory stochastic-processes stochastic-calculus martingales stochastic-analysis
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
$endgroup$
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$W$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$
$b,sigma:mathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$
$(X_t)_{tge0}$ be a strong solution of $${rm d}X_t=b(t,X_t){rm d}t+sigma(t,X_t){rm d}W_ttag3$$
Moreover, let $$(L_tf)(x):=b(t,x)f'(x)+frac12sigma^2(t,x)f''(x);;;text{for }xinmathbb Rtext{ and }fin C^2(mathbb R)$$ for $tge0$. Now, let $fin C^2(mathbb R)$. By the Itō formula, $$f(t,X_t)=f(0,X_0)+int_0^t(L_sf)(X_s):{rm d}s+underbrace{int_0^tsigma f'(X_s):{rm d}W_s}_{=::M_t};;;text{for all }tge0tag4.$$ Assuming $f'$ is bounded, we obtain that $M$ is a martingale.
Are we able to show that $$operatorname Eleft[int_0^tleft|(L_sf)(X_s)right|:{rm d}sright]<inftytag5$$ for all $tge0$?
I've read here (in Corollary 6.5) that this would be the case. However, I don't see how we can bound $left|f''(X_t)right|$.
probability-theory stochastic-processes stochastic-calculus martingales stochastic-analysis
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
$endgroup$
$begingroup$
What are the "conditions of either of the above theorems" which are mentioned in the statement of the result?
$endgroup$
– saz
Dec 25 '18 at 15:15
$begingroup$
@saz The first one is $f'$ being bounded (as assumed here) and the second one is that all partial derivatives have at most exponential growth.
$endgroup$
– 0xbadf00d
Dec 27 '18 at 10:14
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$W$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$
$b,sigma:mathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$
$(X_t)_{tge0}$ be a strong solution of $${rm d}X_t=b(t,X_t){rm d}t+sigma(t,X_t){rm d}W_ttag3$$
Moreover, let $$(L_tf)(x):=b(t,x)f'(x)+frac12sigma^2(t,x)f''(x);;;text{for }xinmathbb Rtext{ and }fin C^2(mathbb R)$$ for $tge0$. Now, let $fin C^2(mathbb R)$. By the Itō formula, $$f(t,X_t)=f(0,X_0)+int_0^t(L_sf)(X_s):{rm d}s+underbrace{int_0^tsigma f'(X_s):{rm d}W_s}_{=::M_t};;;text{for all }tge0tag4.$$ Assuming $f'$ is bounded, we obtain that $M$ is a martingale.
Are we able to show that $$operatorname Eleft[int_0^tleft|(L_sf)(X_s)right|:{rm d}sright]<inftytag5$$ for all $tge0$?
I've read here (in Corollary 6.5) that this would be the case. However, I don't see how we can bound $left|f''(X_t)right|$.
probability-theory stochastic-processes stochastic-calculus martingales stochastic-analysis
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
$endgroup$
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$W$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$
$b,sigma:mathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$
$(X_t)_{tge0}$ be a strong solution of $${rm d}X_t=b(t,X_t){rm d}t+sigma(t,X_t){rm d}W_ttag3$$
Moreover, let $$(L_tf)(x):=b(t,x)f'(x)+frac12sigma^2(t,x)f''(x);;;text{for }xinmathbb Rtext{ and }fin C^2(mathbb R)$$ for $tge0$. Now, let $fin C^2(mathbb R)$. By the Itō formula, $$f(t,X_t)=f(0,X_0)+int_0^t(L_sf)(X_s):{rm d}s+underbrace{int_0^tsigma f'(X_s):{rm d}W_s}_{=::M_t};;;text{for all }tge0tag4.$$ Assuming $f'$ is bounded, we obtain that $M$ is a martingale.
Are we able to show that $$operatorname Eleft[int_0^tleft|(L_sf)(X_s)right|:{rm d}sright]<inftytag5$$ for all $tge0$?
I've read here (in Corollary 6.5) that this would be the case. However, I don't see how we can bound $left|f''(X_t)right|$.
probability-theory stochastic-processes stochastic-calculus martingales stochastic-analysis
probability-theory stochastic-processes stochastic-calculus martingales stochastic-analysis
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
asked Dec 25 '18 at 12:47
0xbadf00d0xbadf00d
1,87441532
1,87441532
$begingroup$
What are the "conditions of either of the above theorems" which are mentioned in the statement of the result?
$endgroup$
– saz
Dec 25 '18 at 15:15
$begingroup$
@saz The first one is $f'$ being bounded (as assumed here) and the second one is that all partial derivatives have at most exponential growth.
$endgroup$
– 0xbadf00d
Dec 27 '18 at 10:14
add a comment |
$begingroup$
What are the "conditions of either of the above theorems" which are mentioned in the statement of the result?
$endgroup$
– saz
Dec 25 '18 at 15:15
$begingroup$
@saz The first one is $f'$ being bounded (as assumed here) and the second one is that all partial derivatives have at most exponential growth.
$endgroup$
– 0xbadf00d
Dec 27 '18 at 10:14
$begingroup$
What are the "conditions of either of the above theorems" which are mentioned in the statement of the result?
$endgroup$
– saz
Dec 25 '18 at 15:15
$begingroup$
What are the "conditions of either of the above theorems" which are mentioned in the statement of the result?
$endgroup$
– saz
Dec 25 '18 at 15:15
$begingroup$
@saz The first one is $f'$ being bounded (as assumed here) and the second one is that all partial derivatives have at most exponential growth.
$endgroup$
– 0xbadf00d
Dec 27 '18 at 10:14
$begingroup$
@saz The first one is $f'$ being bounded (as assumed here) and the second one is that all partial derivatives have at most exponential growth.
$endgroup$
– 0xbadf00d
Dec 27 '18 at 10:14
add a comment |
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$begingroup$
What are the "conditions of either of the above theorems" which are mentioned in the statement of the result?
$endgroup$
– saz
Dec 25 '18 at 15:15
$begingroup$
@saz The first one is $f'$ being bounded (as assumed here) and the second one is that all partial derivatives have at most exponential growth.
$endgroup$
– 0xbadf00d
Dec 27 '18 at 10:14