Mixing
Fokker-Planck equation for a Markov semigroup with densities
$begingroup$
Let
$(E,mathcal E)$ be a measurable space
$(kappa_t)_{tge0}$ be a Markov semigroup on $(E,mathcal E)$ and $$kappa_tf:=intkappa_t(;cdot;,{rm d}y)f(y)$$ for $$fin F_0:=left{f:Etomathbb Rmid ftext{ is }mathcal Etext{-measurable and }left(kappa_tleft|fright|right)(x)<inftytext{ for all }xin Etext{ and }tge0right}$$
$Fsubseteq F_0$ be a $mathbb R$-Banach space containing the elementary $mathcal E$-measurable functions as a dense subset
Assume $$T(t)f:=kappa_tf;;;text{for }fin Ftext{ and }tge0$$ is a $C^0$-semigroup on $F$. Let $$operatorname{orb}f:[0,infty)to F;,;;;tmapsto T(t)f$$ for $fin F$ and $(mathcal D(A),A)$ denote the infinitesimal generator of $(T(t))_{tge0}$. We know that $$left(operatorname{orb}fright)'(t)=AT(t)f=T(t)Af;;;text{for all }finmathcal D(A)text{ and }tge0.tag1$$
Now, assume that there is a $mathcal Eotimesmathcal E$-measurable $p_t:Etimes Eto[0,infty)$ with $$kappa_t(x,B)=int_Bp(x,y)lambdaleft({rm d}yright);;;text{for all }(x,B)in Etimesmathcal Etag2$$ for some measure $lambda$ on $(E,mathcal E)$ for all $tge0$. It's easy to see that $$p_{s+t}(x,;cdot;)=int p_s(x,y)p_t(y,;cdot;)lambdaleft({rm d}yright);;;lambdatext{-almost everywhere for all }xin Etag3$$ for all $s,tge0$.
Does $(p_t)_{tge0}$ satisfy an identity like $(1)$?
I guess the answer is yes and the corresponding identity is called the Fokker-PLanck equation. However, I couldn't find this equation in this general setting. Clearly, we need regularity assumptions on $(p_t)_{tge0}$, but which assumptions do we really need to impose?
I could imagine that we're able to show something like $$fracpartial{partial t}p_t(x,y)=(Ap_t(x,;cdot;)(y).$$
pde markov-process stochastic-analysis sde stochastic-pde
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
$endgroup$
|
show 13 more comments
$begingroup$
Let
$(E,mathcal E)$ be a measurable space
$(kappa_t)_{tge0}$ be a Markov semigroup on $(E,mathcal E)$ and $$kappa_tf:=intkappa_t(;cdot;,{rm d}y)f(y)$$ for $$fin F_0:=left{f:Etomathbb Rmid ftext{ is }mathcal Etext{-measurable and }left(kappa_tleft|fright|right)(x)<inftytext{ for all }xin Etext{ and }tge0right}$$
$Fsubseteq F_0$ be a $mathbb R$-Banach space containing the elementary $mathcal E$-measurable functions as a dense subset
Assume $$T(t)f:=kappa_tf;;;text{for }fin Ftext{ and }tge0$$ is a $C^0$-semigroup on $F$. Let $$operatorname{orb}f:[0,infty)to F;,;;;tmapsto T(t)f$$ for $fin F$ and $(mathcal D(A),A)$ denote the infinitesimal generator of $(T(t))_{tge0}$. We know that $$left(operatorname{orb}fright)'(t)=AT(t)f=T(t)Af;;;text{for all }finmathcal D(A)text{ and }tge0.tag1$$
Now, assume that there is a $mathcal Eotimesmathcal E$-measurable $p_t:Etimes Eto[0,infty)$ with $$kappa_t(x,B)=int_Bp(x,y)lambdaleft({rm d}yright);;;text{for all }(x,B)in Etimesmathcal Etag2$$ for some measure $lambda$ on $(E,mathcal E)$ for all $tge0$. It's easy to see that $$p_{s+t}(x,;cdot;)=int p_s(x,y)p_t(y,;cdot;)lambdaleft({rm d}yright);;;lambdatext{-almost everywhere for all }xin Etag3$$ for all $s,tge0$.
Does $(p_t)_{tge0}$ satisfy an identity like $(1)$?
I guess the answer is yes and the corresponding identity is called the Fokker-PLanck equation. However, I couldn't find this equation in this general setting. Clearly, we need regularity assumptions on $(p_t)_{tge0}$, but which assumptions do we really need to impose?
I could imagine that we're able to show something like $$fracpartial{partial t}p_t(x,y)=(Ap_t(x,;cdot;)(y).$$
pde markov-process stochastic-analysis sde stochastic-pde
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
$endgroup$
$begingroup$
The equation at the very end of your question is known as Kolmogorov backward equation. I'm not aware of general conditions for the equation to hold, apart from the diffusion case. It already starts with the problem that it is not clear that $p$ is regular enough to be in the domain of $A$ and to admit a time derivative...
$endgroup$
– saz
Jan 20 at 7:40
$begingroup$
@saz Do you you have a good reference for the diffusion case?
$endgroup$
– 0xbadf00d
Jan 20 at 9:45
$begingroup$
@saz Some thoughts: Let $f∈mathcal D(A)$ and $$u(t,x):=(T(t)f)(x);;;text{for }(t,x)∈[0,∞)×E.$$ For fixed $t≥0$, let $$v(x):=u(t,x);;;text{for }x∈E.$$ Since $f∈mathcal D(A)$, $v∈mathcal D(A)$ and $$frac{u(t+h,;⋅;)-u(t,;⋅;)}h=frac{T(h)v-v}hxrightarrow{h→0}Av.$$
$endgroup$
– 0xbadf00d
Jan 20 at 11:51
1
$begingroup$
Ah, I see, sorry, my fault. So, yes, then the topology should be finer than the topology of pointwise convergence.
$endgroup$
– saz
Jan 20 at 12:12
1
$begingroup$
In general, the time derivative of the transition density of a Markov process does not need to be locally bounded (in the sense which you stated in your previous comment). If it was locally bounded, then indeed everything would work out quite nicely.
$endgroup$
– saz
Jan 24 at 16:54
|
show 13 more comments
$begingroup$
Let
$(E,mathcal E)$ be a measurable space
$(kappa_t)_{tge0}$ be a Markov semigroup on $(E,mathcal E)$ and $$kappa_tf:=intkappa_t(;cdot;,{rm d}y)f(y)$$ for $$fin F_0:=left{f:Etomathbb Rmid ftext{ is }mathcal Etext{-measurable and }left(kappa_tleft|fright|right)(x)<inftytext{ for all }xin Etext{ and }tge0right}$$
$Fsubseteq F_0$ be a $mathbb R$-Banach space containing the elementary $mathcal E$-measurable functions as a dense subset
Assume $$T(t)f:=kappa_tf;;;text{for }fin Ftext{ and }tge0$$ is a $C^0$-semigroup on $F$. Let $$operatorname{orb}f:[0,infty)to F;,;;;tmapsto T(t)f$$ for $fin F$ and $(mathcal D(A),A)$ denote the infinitesimal generator of $(T(t))_{tge0}$. We know that $$left(operatorname{orb}fright)'(t)=AT(t)f=T(t)Af;;;text{for all }finmathcal D(A)text{ and }tge0.tag1$$
Now, assume that there is a $mathcal Eotimesmathcal E$-measurable $p_t:Etimes Eto[0,infty)$ with $$kappa_t(x,B)=int_Bp(x,y)lambdaleft({rm d}yright);;;text{for all }(x,B)in Etimesmathcal Etag2$$ for some measure $lambda$ on $(E,mathcal E)$ for all $tge0$. It's easy to see that $$p_{s+t}(x,;cdot;)=int p_s(x,y)p_t(y,;cdot;)lambdaleft({rm d}yright);;;lambdatext{-almost everywhere for all }xin Etag3$$ for all $s,tge0$.
Does $(p_t)_{tge0}$ satisfy an identity like $(1)$?
I guess the answer is yes and the corresponding identity is called the Fokker-PLanck equation. However, I couldn't find this equation in this general setting. Clearly, we need regularity assumptions on $(p_t)_{tge0}$, but which assumptions do we really need to impose?
I could imagine that we're able to show something like $$fracpartial{partial t}p_t(x,y)=(Ap_t(x,;cdot;)(y).$$
pde markov-process stochastic-analysis sde stochastic-pde
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
$endgroup$
Let
$(E,mathcal E)$ be a measurable space
$(kappa_t)_{tge0}$ be a Markov semigroup on $(E,mathcal E)$ and $$kappa_tf:=intkappa_t(;cdot;,{rm d}y)f(y)$$ for $$fin F_0:=left{f:Etomathbb Rmid ftext{ is }mathcal Etext{-measurable and }left(kappa_tleft|fright|right)(x)<inftytext{ for all }xin Etext{ and }tge0right}$$
$Fsubseteq F_0$ be a $mathbb R$-Banach space containing the elementary $mathcal E$-measurable functions as a dense subset
Assume $$T(t)f:=kappa_tf;;;text{for }fin Ftext{ and }tge0$$ is a $C^0$-semigroup on $F$. Let $$operatorname{orb}f:[0,infty)to F;,;;;tmapsto T(t)f$$ for $fin F$ and $(mathcal D(A),A)$ denote the infinitesimal generator of $(T(t))_{tge0}$. We know that $$left(operatorname{orb}fright)'(t)=AT(t)f=T(t)Af;;;text{for all }finmathcal D(A)text{ and }tge0.tag1$$
Now, assume that there is a $mathcal Eotimesmathcal E$-measurable $p_t:Etimes Eto[0,infty)$ with $$kappa_t(x,B)=int_Bp(x,y)lambdaleft({rm d}yright);;;text{for all }(x,B)in Etimesmathcal Etag2$$ for some measure $lambda$ on $(E,mathcal E)$ for all $tge0$. It's easy to see that $$p_{s+t}(x,;cdot;)=int p_s(x,y)p_t(y,;cdot;)lambdaleft({rm d}yright);;;lambdatext{-almost everywhere for all }xin Etag3$$ for all $s,tge0$.
Does $(p_t)_{tge0}$ satisfy an identity like $(1)$?
I guess the answer is yes and the corresponding identity is called the Fokker-PLanck equation. However, I couldn't find this equation in this general setting. Clearly, we need regularity assumptions on $(p_t)_{tge0}$, but which assumptions do we really need to impose?
I could imagine that we're able to show something like $$fracpartial{partial t}p_t(x,y)=(Ap_t(x,;cdot;)(y).$$
pde markov-process stochastic-analysis sde stochastic-pde
pde markov-process stochastic-analysis sde stochastic-pde
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
asked Jan 19 at 23:08
0xbadf00d0xbadf00d
1,94341532
1,94341532
$begingroup$
The equation at the very end of your question is known as Kolmogorov backward equation. I'm not aware of general conditions for the equation to hold, apart from the diffusion case. It already starts with the problem that it is not clear that $p$ is regular enough to be in the domain of $A$ and to admit a time derivative...
$endgroup$
– saz
Jan 20 at 7:40
$begingroup$
@saz Do you you have a good reference for the diffusion case?
$endgroup$
– 0xbadf00d
Jan 20 at 9:45
$begingroup$
@saz Some thoughts: Let $f∈mathcal D(A)$ and $$u(t,x):=(T(t)f)(x);;;text{for }(t,x)∈[0,∞)×E.$$ For fixed $t≥0$, let $$v(x):=u(t,x);;;text{for }x∈E.$$ Since $f∈mathcal D(A)$, $v∈mathcal D(A)$ and $$frac{u(t+h,;⋅;)-u(t,;⋅;)}h=frac{T(h)v-v}hxrightarrow{h→0}Av.$$
$endgroup$
– 0xbadf00d
Jan 20 at 11:51
1
$begingroup$
Ah, I see, sorry, my fault. So, yes, then the topology should be finer than the topology of pointwise convergence.
$endgroup$
– saz
Jan 20 at 12:12
1
$begingroup$
In general, the time derivative of the transition density of a Markov process does not need to be locally bounded (in the sense which you stated in your previous comment). If it was locally bounded, then indeed everything would work out quite nicely.
$endgroup$
– saz
Jan 24 at 16:54
|
show 13 more comments
$begingroup$
The equation at the very end of your question is known as Kolmogorov backward equation. I'm not aware of general conditions for the equation to hold, apart from the diffusion case. It already starts with the problem that it is not clear that $p$ is regular enough to be in the domain of $A$ and to admit a time derivative...
$endgroup$
– saz
Jan 20 at 7:40
$begingroup$
@saz Do you you have a good reference for the diffusion case?
$endgroup$
– 0xbadf00d
Jan 20 at 9:45
$begingroup$
@saz Some thoughts: Let $f∈mathcal D(A)$ and $$u(t,x):=(T(t)f)(x);;;text{for }(t,x)∈[0,∞)×E.$$ For fixed $t≥0$, let $$v(x):=u(t,x);;;text{for }x∈E.$$ Since $f∈mathcal D(A)$, $v∈mathcal D(A)$ and $$frac{u(t+h,;⋅;)-u(t,;⋅;)}h=frac{T(h)v-v}hxrightarrow{h→0}Av.$$
$endgroup$
– 0xbadf00d
Jan 20 at 11:51
1
$begingroup$
Ah, I see, sorry, my fault. So, yes, then the topology should be finer than the topology of pointwise convergence.
$endgroup$
– saz
Jan 20 at 12:12
1
$begingroup$
In general, the time derivative of the transition density of a Markov process does not need to be locally bounded (in the sense which you stated in your previous comment). If it was locally bounded, then indeed everything would work out quite nicely.
$endgroup$
– saz
Jan 24 at 16:54
$begingroup$
The equation at the very end of your question is known as Kolmogorov backward equation. I'm not aware of general conditions for the equation to hold, apart from the diffusion case. It already starts with the problem that it is not clear that $p$ is regular enough to be in the domain of $A$ and to admit a time derivative...
$endgroup$
– saz
Jan 20 at 7:40
$begingroup$
The equation at the very end of your question is known as Kolmogorov backward equation. I'm not aware of general conditions for the equation to hold, apart from the diffusion case. It already starts with the problem that it is not clear that $p$ is regular enough to be in the domain of $A$ and to admit a time derivative...
$endgroup$
– saz
Jan 20 at 7:40
$begingroup$
@saz Do you you have a good reference for the diffusion case?
$endgroup$
– 0xbadf00d
Jan 20 at 9:45
$begingroup$
@saz Do you you have a good reference for the diffusion case?
$endgroup$
– 0xbadf00d
Jan 20 at 9:45
$begingroup$
@saz Some thoughts: Let $f∈mathcal D(A)$ and $$u(t,x):=(T(t)f)(x);;;text{for }(t,x)∈[0,∞)×E.$$ For fixed $t≥0$, let $$v(x):=u(t,x);;;text{for }x∈E.$$ Since $f∈mathcal D(A)$, $v∈mathcal D(A)$ and $$frac{u(t+h,;⋅;)-u(t,;⋅;)}h=frac{T(h)v-v}hxrightarrow{h→0}Av.$$
$endgroup$
– 0xbadf00d
Jan 20 at 11:51
$begingroup$
@saz Some thoughts: Let $f∈mathcal D(A)$ and $$u(t,x):=(T(t)f)(x);;;text{for }(t,x)∈[0,∞)×E.$$ For fixed $t≥0$, let $$v(x):=u(t,x);;;text{for }x∈E.$$ Since $f∈mathcal D(A)$, $v∈mathcal D(A)$ and $$frac{u(t+h,;⋅;)-u(t,;⋅;)}h=frac{T(h)v-v}hxrightarrow{h→0}Av.$$
$endgroup$
– 0xbadf00d
Jan 20 at 11:51
1
1
$begingroup$
Ah, I see, sorry, my fault. So, yes, then the topology should be finer than the topology of pointwise convergence.
$endgroup$
– saz
Jan 20 at 12:12
$begingroup$
Ah, I see, sorry, my fault. So, yes, then the topology should be finer than the topology of pointwise convergence.
$endgroup$
– saz
Jan 20 at 12:12
1
1
$begingroup$
In general, the time derivative of the transition density of a Markov process does not need to be locally bounded (in the sense which you stated in your previous comment). If it was locally bounded, then indeed everything would work out quite nicely.
$endgroup$
– saz
Jan 24 at 16:54
$begingroup$
In general, the time derivative of the transition density of a Markov process does not need to be locally bounded (in the sense which you stated in your previous comment). If it was locally bounded, then indeed everything would work out quite nicely.
$endgroup$
– saz
Jan 24 at 16:54
|
show 13 more comments
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$begingroup$
The equation at the very end of your question is known as Kolmogorov backward equation. I'm not aware of general conditions for the equation to hold, apart from the diffusion case. It already starts with the problem that it is not clear that $p$ is regular enough to be in the domain of $A$ and to admit a time derivative...
$endgroup$
– saz
Jan 20 at 7:40
$begingroup$
@saz Do you you have a good reference for the diffusion case?
$endgroup$
– 0xbadf00d
Jan 20 at 9:45
$begingroup$
@saz Some thoughts: Let $f∈mathcal D(A)$ and $$u(t,x):=(T(t)f)(x);;;text{for }(t,x)∈[0,∞)×E.$$ For fixed $t≥0$, let $$v(x):=u(t,x);;;text{for }x∈E.$$ Since $f∈mathcal D(A)$, $v∈mathcal D(A)$ and $$frac{u(t+h,;⋅;)-u(t,;⋅;)}h=frac{T(h)v-v}hxrightarrow{h→0}Av.$$
$endgroup$
– 0xbadf00d
Jan 20 at 11:51
1
$begingroup$
Ah, I see, sorry, my fault. So, yes, then the topology should be finer than the topology of pointwise convergence.
$endgroup$
– saz
Jan 20 at 12:12
1
$begingroup$
In general, the time derivative of the transition density of a Markov process does not need to be locally bounded (in the sense which you stated in your previous comment). If it was locally bounded, then indeed everything would work out quite nicely.
$endgroup$
– saz
Jan 24 at 16:54