Ito diffusion: Connection between backward Kolmogorov equation and stationary distribution
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Suppose we have an Ito diffusion
$$ dX_t = b(X_t)dt + sigma(X_t) dB_t, tag{1}$$
where $dB_t$ is Brownian motion. Also assume we know that this diffusion process converges to a stationary distribution $pi$. I am interested in the quantity
$$u_{infty}(x) = lim_{t rightarrow infty} mathbb E[psi(X_t)] = mathbb E_pi[psi(X)],$$
for some function $psi : mathbb R rightarrow mathbb R$; assuming the expectation exists.
I am wondering if there is a way to connect this quantity with the backward Kolmogorov equation, i.e. define $u(x,t) = mathbb E^x[psi(X_t)]$; then the backward Kolmogorov equation says that $u$ solves
begin{align}
frac{partial u}{partial t} &= mathcal A u, tag{2}\
u(0,x) &= psi(x)
end{align}
where $mathcal A$ is the infitesimal generator for (1).
To me it seems like $u_infty(x)$ should be the stationary solution of (2), so we might be able to set $frac{partial u}{partial t} = 0$ to reduce things to an ODE. But I can't wrap my mind around the boundary conditions that should be in place for this to work, or if I've missed an obvious reason for why this wouldn't work.
pde stochastic-processes stationary-processes
asked Nov 14 at 13:00
Jack
83
add a comment |
up vote
0
down vote
favorite
Suppose we have an Ito diffusion
$$ dX_t = b(X_t)dt + sigma(X_t) dB_t, tag{1}$$
where $dB_t$ is Brownian motion. Also assume we know that this diffusion process converges to a stationary distribution $pi$. I am interested in the quantity
$$u_{infty}(x) = lim_{t rightarrow infty} mathbb E[psi(X_t)] = mathbb E_pi[psi(X)],$$
for some function $psi : mathbb R rightarrow mathbb R$; assuming the expectation exists.
I am wondering if there is a way to connect this quantity with the backward Kolmogorov equation, i.e. define $u(x,t) = mathbb E^x[psi(X_t)]$; then the backward Kolmogorov equation says that $u$ solves
begin{align}
frac{partial u}{partial t} &= mathcal A u, tag{2}\
u(0,x) &= psi(x)
end{align}
where $mathcal A$ is the infitesimal generator for (1).
To me it seems like $u_infty(x)$ should be the stationary solution of (2), so we might be able to set $frac{partial u}{partial t} = 0$ to reduce things to an ODE. But I can't wrap my mind around the boundary conditions that should be in place for this to work, or if I've missed an obvious reason for why this wouldn't work.
pde stochastic-processes stationary-processes
asked Nov 14 at 13:00
Jack
83
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have an Ito diffusion
$$ dX_t = b(X_t)dt + sigma(X_t) dB_t, tag{1}$$
where $dB_t$ is Brownian motion. Also assume we know that this diffusion process converges to a stationary distribution $pi$. I am interested in the quantity
$$u_{infty}(x) = lim_{t rightarrow infty} mathbb E[psi(X_t)] = mathbb E_pi[psi(X)],$$
for some function $psi : mathbb R rightarrow mathbb R$; assuming the expectation exists.
I am wondering if there is a way to connect this quantity with the backward Kolmogorov equation, i.e. define $u(x,t) = mathbb E^x[psi(X_t)]$; then the backward Kolmogorov equation says that $u$ solves
begin{align}
frac{partial u}{partial t} &= mathcal A u, tag{2}\
u(0,x) &= psi(x)
end{align}
where $mathcal A$ is the infitesimal generator for (1).
To me it seems like $u_infty(x)$ should be the stationary solution of (2), so we might be able to set $frac{partial u}{partial t} = 0$ to reduce things to an ODE. But I can't wrap my mind around the boundary conditions that should be in place for this to work, or if I've missed an obvious reason for why this wouldn't work.
pde stochastic-processes stationary-processes
asked Nov 14 at 13:00
Jack
83
Suppose we have an Ito diffusion
$$ dX_t = b(X_t)dt + sigma(X_t) dB_t, tag{1}$$
where $dB_t$ is Brownian motion. Also assume we know that this diffusion process converges to a stationary distribution $pi$. I am interested in the quantity
$$u_{infty}(x) = lim_{t rightarrow infty} mathbb E[psi(X_t)] = mathbb E_pi[psi(X)],$$
for some function $psi : mathbb R rightarrow mathbb R$; assuming the expectation exists.
I am wondering if there is a way to connect this quantity with the backward Kolmogorov equation, i.e. define $u(x,t) = mathbb E^x[psi(X_t)]$; then the backward Kolmogorov equation says that $u$ solves
begin{align}
frac{partial u}{partial t} &= mathcal A u, tag{2}\
u(0,x) &= psi(x)
end{align}
where $mathcal A$ is the infitesimal generator for (1).
To me it seems like $u_infty(x)$ should be the stationary solution of (2), so we might be able to set $frac{partial u}{partial t} = 0$ to reduce things to an ODE. But I can't wrap my mind around the boundary conditions that should be in place for this to work, or if I've missed an obvious reason for why this wouldn't work.
pde stochastic-processes stationary-processes
pde stochastic-processes stationary-processes
asked Nov 14 at 13:00
Jack
83
asked Nov 14 at 13:00
Jack
83
asked Nov 14 at 13:00
Jack
83
asked Nov 14 at 13:00
Jack
83
asked Nov 14 at 13:00
Jack
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $b(x)/sigma^2(x)$).
answered Nov 14 at 13:07
Richard Martin
1,5468
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $b(x)/sigma^2(x)$).
answered Nov 14 at 13:07
Richard Martin
1,5468
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
add a comment |
up vote
0
down vote
The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $b(x)/sigma^2(x)$).
answered Nov 14 at 13:07
Richard Martin
1,5468
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
add a comment |
up vote
0
down vote
up vote
0
down vote
The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $b(x)/sigma^2(x)$).
answered Nov 14 at 13:07
Richard Martin
1,5468
The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $b(x)/sigma^2(x)$).
answered Nov 14 at 13:07
Richard Martin
1,5468
answered Nov 14 at 13:07
Richard Martin
1,5468
answered Nov 14 at 13:07
Richard Martin
1,5468
answered Nov 14 at 13:07
Richard Martin
1,5468
1,5468
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
add a comment |
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself.
– Jack
Nov 14 at 13:31
add a comment |
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