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.










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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.










    share|cite|improve this question
























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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.










      share|cite|improve this question













      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






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      asked Nov 14 at 13:00









      Jack

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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 $b(x)/sigma^2(x)$).






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          • 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











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          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 $b(x)/sigma^2(x)$).






          share|cite|improve this answer





















          • 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















          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 $b(x)/sigma^2(x)$).






          share|cite|improve this answer





















          • 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













          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 $b(x)/sigma^2(x)$).






          share|cite|improve this answer












          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 $b(x)/sigma^2(x)$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















           

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