How is Grönwall's inequality applied here?
$begingroup$
Let $hin C^1(mathbb R)$ such that $h'$ is Lipschitz continuous and $$Lvarphi:=-h'varphi'+varphi'';;;text{for }varphiin C^2(mathbb R).$$ Now, let $(X_t)_{tge0}$ be the unique strong solution of $${rm d}X_t=-h'(X_t){rm d}t+sqrt2{rm d}W_ttag1,$$ where $(W_t)_{tge0}$ is a Brownian motion.
I've read (in this paper, below Assumption 2.4) that if $V:mathbb Rto[0,infty)$ with $$V(x)xrightarrow{xtoinfty}inftytag2$$ and $a,d>0$ with $$LVle-aV+dtag3,$$ then $$operatorname Eleft[V(X_t)midmathcal F_s^Xright]le e^{-a(t-s)}V(X_s)+frac da(1-e^{-a(t-s)})tag4.$$ Why does $(4)$ hold?
Obviously, $(4)$ is an application of (the Itō formula and) a Gronwall-type lemma. Actually, it's precisely Theorem 6 here. However, in order to apply that theorem, we should need that the process $left(operatorname Eleft[V(X_t)midmathcal F_s^Xright]right)_{tge0}$ is continuous. This shouldn't hold, unless $V$ is (continuous and) bounded (which would allow an application of Lebesgue's dominated convergence theorem). But $V$ is clearly assumed to be unbounded by $(2)$. So, what am I missing?
(Clearly, the authors of the paper are missing assumptions on $V$ anyway. In order for $(3)$ to make sense, $V$ needs to be twice differentiable (at least in some weak sense).
stochastic-processes conditional-expectation martingales sde lyapunov-functions
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
$endgroup$
add a comment |
$begingroup$
Let $hin C^1(mathbb R)$ such that $h'$ is Lipschitz continuous and $$Lvarphi:=-h'varphi'+varphi'';;;text{for }varphiin C^2(mathbb R).$$ Now, let $(X_t)_{tge0}$ be the unique strong solution of $${rm d}X_t=-h'(X_t){rm d}t+sqrt2{rm d}W_ttag1,$$ where $(W_t)_{tge0}$ is a Brownian motion.
I've read (in this paper, below Assumption 2.4) that if $V:mathbb Rto[0,infty)$ with $$V(x)xrightarrow{xtoinfty}inftytag2$$ and $a,d>0$ with $$LVle-aV+dtag3,$$ then $$operatorname Eleft[V(X_t)midmathcal F_s^Xright]le e^{-a(t-s)}V(X_s)+frac da(1-e^{-a(t-s)})tag4.$$ Why does $(4)$ hold?
Obviously, $(4)$ is an application of (the Itō formula and) a Gronwall-type lemma. Actually, it's precisely Theorem 6 here. However, in order to apply that theorem, we should need that the process $left(operatorname Eleft[V(X_t)midmathcal F_s^Xright]right)_{tge0}$ is continuous. This shouldn't hold, unless $V$ is (continuous and) bounded (which would allow an application of Lebesgue's dominated convergence theorem). But $V$ is clearly assumed to be unbounded by $(2)$. So, what am I missing?
(Clearly, the authors of the paper are missing assumptions on $V$ anyway. In order for $(3)$ to make sense, $V$ needs to be twice differentiable (at least in some weak sense).
stochastic-processes conditional-expectation martingales sde lyapunov-functions
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
$endgroup$
add a comment |
$begingroup$
Let $hin C^1(mathbb R)$ such that $h'$ is Lipschitz continuous and $$Lvarphi:=-h'varphi'+varphi'';;;text{for }varphiin C^2(mathbb R).$$ Now, let $(X_t)_{tge0}$ be the unique strong solution of $${rm d}X_t=-h'(X_t){rm d}t+sqrt2{rm d}W_ttag1,$$ where $(W_t)_{tge0}$ is a Brownian motion.
I've read (in this paper, below Assumption 2.4) that if $V:mathbb Rto[0,infty)$ with $$V(x)xrightarrow{xtoinfty}inftytag2$$ and $a,d>0$ with $$LVle-aV+dtag3,$$ then $$operatorname Eleft[V(X_t)midmathcal F_s^Xright]le e^{-a(t-s)}V(X_s)+frac da(1-e^{-a(t-s)})tag4.$$ Why does $(4)$ hold?
Obviously, $(4)$ is an application of (the Itō formula and) a Gronwall-type lemma. Actually, it's precisely Theorem 6 here. However, in order to apply that theorem, we should need that the process $left(operatorname Eleft[V(X_t)midmathcal F_s^Xright]right)_{tge0}$ is continuous. This shouldn't hold, unless $V$ is (continuous and) bounded (which would allow an application of Lebesgue's dominated convergence theorem). But $V$ is clearly assumed to be unbounded by $(2)$. So, what am I missing?
(Clearly, the authors of the paper are missing assumptions on $V$ anyway. In order for $(3)$ to make sense, $V$ needs to be twice differentiable (at least in some weak sense).
stochastic-processes conditional-expectation martingales sde lyapunov-functions
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
$endgroup$
Let $hin C^1(mathbb R)$ such that $h'$ is Lipschitz continuous and $$Lvarphi:=-h'varphi'+varphi'';;;text{for }varphiin C^2(mathbb R).$$ Now, let $(X_t)_{tge0}$ be the unique strong solution of $${rm d}X_t=-h'(X_t){rm d}t+sqrt2{rm d}W_ttag1,$$ where $(W_t)_{tge0}$ is a Brownian motion.
I've read (in this paper, below Assumption 2.4) that if $V:mathbb Rto[0,infty)$ with $$V(x)xrightarrow{xtoinfty}inftytag2$$ and $a,d>0$ with $$LVle-aV+dtag3,$$ then $$operatorname Eleft[V(X_t)midmathcal F_s^Xright]le e^{-a(t-s)}V(X_s)+frac da(1-e^{-a(t-s)})tag4.$$ Why does $(4)$ hold?
Obviously, $(4)$ is an application of (the Itō formula and) a Gronwall-type lemma. Actually, it's precisely Theorem 6 here. However, in order to apply that theorem, we should need that the process $left(operatorname Eleft[V(X_t)midmathcal F_s^Xright]right)_{tge0}$ is continuous. This shouldn't hold, unless $V$ is (continuous and) bounded (which would allow an application of Lebesgue's dominated convergence theorem). But $V$ is clearly assumed to be unbounded by $(2)$. So, what am I missing?
(Clearly, the authors of the paper are missing assumptions on $V$ anyway. In order for $(3)$ to make sense, $V$ needs to be twice differentiable (at least in some weak sense).
stochastic-processes conditional-expectation martingales sde lyapunov-functions
stochastic-processes conditional-expectation martingales sde lyapunov-functions
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
edited Jan 18 at 23:47
0xbadf00d
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
asked Jan 18 at 21:56
0xbadf00d0xbadf00d
1,62141534
1,62141534
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It appears to be implicit that $V$ is $C^2$, hence locally bounded. By Ito,
$$
V(X_t) = e^{-a(t-s)}V(X_s)+{dover a}(1-e^{-a(t-s)})+int_s^t K_u,du +M_t-M_s,
$$
where $M$ is a martingale and $K_u=LV(X_u)=aV(X_u)-dle 0$. Now take he conditional expectation.
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078802%2fhow-is-gr%25c3%25b6nwalls-inequality-applied-here%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It appears to be implicit that $V$ is $C^2$, hence locally bounded. By Ito,
$$
V(X_t) = e^{-a(t-s)}V(X_s)+{dover a}(1-e^{-a(t-s)})+int_s^t K_u,du +M_t-M_s,
$$
where $M$ is a martingale and $K_u=LV(X_u)=aV(X_u)-dle 0$. Now take he conditional expectation.
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
add a comment |
$begingroup$
It appears to be implicit that $V$ is $C^2$, hence locally bounded. By Ito,
$$
V(X_t) = e^{-a(t-s)}V(X_s)+{dover a}(1-e^{-a(t-s)})+int_s^t K_u,du +M_t-M_s,
$$
where $M$ is a martingale and $K_u=LV(X_u)=aV(X_u)-dle 0$. Now take he conditional expectation.
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
add a comment |
$begingroup$
It appears to be implicit that $V$ is $C^2$, hence locally bounded. By Ito,
$$
V(X_t) = e^{-a(t-s)}V(X_s)+{dover a}(1-e^{-a(t-s)})+int_s^t K_u,du +M_t-M_s,
$$
where $M$ is a martingale and $K_u=LV(X_u)=aV(X_u)-dle 0$. Now take he conditional expectation.
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
It appears to be implicit that $V$ is $C^2$, hence locally bounded. By Ito,
$$
V(X_t) = e^{-a(t-s)}V(X_s)+{dover a}(1-e^{-a(t-s)})+int_s^t K_u,du +M_t-M_s,
$$
where $M$ is a martingale and $K_u=LV(X_u)=aV(X_u)-dle 0$. Now take he conditional expectation.
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
answered Jan 19 at 18:25
John DawkinsJohn Dawkins
13.4k11017
13.4k11017
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
add a comment |
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Did you mean $K_u=LV(X_u)+aV_u-d$? And isn't a factor $e^{au}$ missing in the integrand?
$endgroup$
– LutzL
Jan 19 at 21:53
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
$begingroup$
Yes. That factor is missing.
$endgroup$
– John Dawkins
Jan 20 at 16:44
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078802%2fhow-is-gr%25c3%25b6nwalls-inequality-applied-here%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
