Describe discrete-time Langevin dynamics by its stationary distribution












1
















Let's consider stochastic dynamics with discrete time step $t$ and states $x_tinmathbb{R}^d$ that evolve as



$$
mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+f(x_t), Sigma).
$$



Moreover, we assume the existence of a unique stationary distribution $rho(x)$ that satisfies



$$
intmathrm{p}(x_t|x_{t-1})rho(x_{t-1})mathrm{d}x_{t-1}=rho(x_t).
$$



Can $f$ be written as a gradient descent on $lnrho$ transformed by a positive semi-definite matrix $R(x)$,



$$
mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+R(x_t)nabla_{x_t}lnrho(x_t),Sigma),?
$$



Do I additionally need to assume ergodicity or detailed balance for this to work?










probability stochastic-processes dynamical-systems markov-chains non-linear-dynamics







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    1
















    Let's consider stochastic dynamics with discrete time step $t$ and states $x_tinmathbb{R}^d$ that evolve as



    $$
    mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+f(x_t), Sigma).
    $$



    Moreover, we assume the existence of a unique stationary distribution $rho(x)$ that satisfies



    $$
    intmathrm{p}(x_t|x_{t-1})rho(x_{t-1})mathrm{d}x_{t-1}=rho(x_t).
    $$



    Can $f$ be written as a gradient descent on $lnrho$ transformed by a positive semi-definite matrix $R(x)$,



    $$
    mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+R(x_t)nabla_{x_t}lnrho(x_t),Sigma),?
    $$



    Do I additionally need to assume ergodicity or detailed balance for this to work?










    probability stochastic-processes dynamical-systems markov-chains non-linear-dynamics







    share|cite|improve this question



























      1












      1








      1









      Let's consider stochastic dynamics with discrete time step $t$ and states $x_tinmathbb{R}^d$ that evolve as



      $$
      mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+f(x_t), Sigma).
      $$



      Moreover, we assume the existence of a unique stationary distribution $rho(x)$ that satisfies



      $$
      intmathrm{p}(x_t|x_{t-1})rho(x_{t-1})mathrm{d}x_{t-1}=rho(x_t).
      $$



      Can $f$ be written as a gradient descent on $lnrho$ transformed by a positive semi-definite matrix $R(x)$,



      $$
      mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+R(x_t)nabla_{x_t}lnrho(x_t),Sigma),?
      $$



      Do I additionally need to assume ergodicity or detailed balance for this to work?










      probability stochastic-processes dynamical-systems markov-chains non-linear-dynamics







      share|cite|improve this question

















      Let's consider stochastic dynamics with discrete time step $t$ and states $x_tinmathbb{R}^d$ that evolve as



      $$
      mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+f(x_t), Sigma).
      $$



      Moreover, we assume the existence of a unique stationary distribution $rho(x)$ that satisfies



      $$
      intmathrm{p}(x_t|x_{t-1})rho(x_{t-1})mathrm{d}x_{t-1}=rho(x_t).
      $$



      Can $f$ be written as a gradient descent on $lnrho$ transformed by a positive semi-definite matrix $R(x)$,



      $$
      mathrm{p}(x_{t+1}|x_t)=mathcal{N}(x_t+R(x_t)nabla_{x_t}lnrho(x_t),Sigma),?
      $$



      Do I additionally need to assume ergodicity or detailed balance for this to work?










      probability stochastic-processes dynamical-systems markov-chains non-linear-dynamics




      probability stochastic-processes dynamical-systems markov-chains non-linear-dynamics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 23 at 18:02

























      asked Nov 20 at 14:47









      danijar

      220313




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          I don't think so. Counterexample:



          State space = ${0,1,2}$



          $p_1: x_{t+1}=x_t+1 $ mod $ 3$



          $p_2: x_{t+1}=x_t-1 $ mod $ 3$



          so $Sigma=0$.



          Then the same $rho(x_t)$ belongs to both $p_1$ and $p_2$.






          share|cite|improve this answer























          • Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
            – danijar
            Nov 22 at 20:12












          • They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
            – Ian
            Nov 22 at 23:06












          • @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
            – Smind
            Nov 23 at 11:17












          • Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
            – danijar
            Nov 23 at 15:18











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          I don't think so. Counterexample:



          State space = ${0,1,2}$



          $p_1: x_{t+1}=x_t+1 $ mod $ 3$



          $p_2: x_{t+1}=x_t-1 $ mod $ 3$



          so $Sigma=0$.



          Then the same $rho(x_t)$ belongs to both $p_1$ and $p_2$.






          share|cite|improve this answer























          • Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
            – danijar
            Nov 22 at 20:12












          • They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
            – Ian
            Nov 22 at 23:06












          • @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
            – Smind
            Nov 23 at 11:17












          • Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
            – danijar
            Nov 23 at 15:18
















          1














          I don't think so. Counterexample:



          State space = ${0,1,2}$



          $p_1: x_{t+1}=x_t+1 $ mod $ 3$



          $p_2: x_{t+1}=x_t-1 $ mod $ 3$



          so $Sigma=0$.



          Then the same $rho(x_t)$ belongs to both $p_1$ and $p_2$.






          share|cite|improve this answer























          • Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
            – danijar
            Nov 22 at 20:12












          • They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
            – Ian
            Nov 22 at 23:06












          • @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
            – Smind
            Nov 23 at 11:17












          • Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
            – danijar
            Nov 23 at 15:18














          1












          1








          1






          I don't think so. Counterexample:



          State space = ${0,1,2}$



          $p_1: x_{t+1}=x_t+1 $ mod $ 3$



          $p_2: x_{t+1}=x_t-1 $ mod $ 3$



          so $Sigma=0$.



          Then the same $rho(x_t)$ belongs to both $p_1$ and $p_2$.






          share|cite|improve this answer














          I don't think so. Counterexample:



          State space = ${0,1,2}$



          $p_1: x_{t+1}=x_t+1 $ mod $ 3$



          $p_2: x_{t+1}=x_t-1 $ mod $ 3$



          so $Sigma=0$.



          Then the same $rho(x_t)$ belongs to both $p_1$ and $p_2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 17:45

























          answered Nov 22 at 13:49









          Smind

          163




          163












          • Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
            – danijar
            Nov 22 at 20:12












          • They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
            – Ian
            Nov 22 at 23:06












          • @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
            – Smind
            Nov 23 at 11:17












          • Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
            – danijar
            Nov 23 at 15:18


















          • Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
            – danijar
            Nov 22 at 20:12












          • They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
            – Ian
            Nov 22 at 23:06












          • @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
            – Smind
            Nov 23 at 11:17












          • Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
            – danijar
            Nov 23 at 15:18
















          Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
          – danijar
          Nov 22 at 20:12






          Thanks! Do you mean the deterministic systems $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t+1 ,text{mod}, 3)$ and $p_1(x_{t+1}|x_t)=delta(x_{t+1}=x_t-1 ,text{mod}, 3)$? I think these systems have no stationary distribution $rho(x)$. Also, technically $x_tinmathbb{R}^d$, but an example with discrete states could already be helpful.
          – danijar
          Nov 22 at 20:12














          They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
          – Ian
          Nov 22 at 23:06






          They do have a stationary distribution, it's uniform. It isn't ergodic, but even with ergodicity you cannot do this. The canonical way to do it is called reversible dynamics.
          – Ian
          Nov 22 at 23:06














          @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
          – Smind
          Nov 23 at 11:17






          @danijar thanks, that's right. I agree with Ian that the stationary distribution is uniform. We can generalize this example to continuous $x_t$ by making $f_{1},f_2$ that go clockwise resp. counter clockwise around the same point. They'll have the same stationary distribution.
          – Smind
          Nov 23 at 11:17














          Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
          – danijar
          Nov 23 at 15:18




          Thank you. It makes sense that the stationary distribution is uniform, although it is not ergodic. @Ian, would you mind explaining how this can be done with reversible dynamics? What I had in mind was a discrete-time version of the equilibrium solution to Fokker-Planck in ergodic SDEs with Gaussians noise.
          – danijar
          Nov 23 at 15:18


















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