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Consider the stochastic iterative updates \begin{align} \theta_{t+1} \leftarrow \theta_t + \alpha_t \cdot \left [ h(\theta_t) + M_t \right ], \end{align} where $\theta_t \in \mathrm{R}^d$, $h \colon \mathrm{R} ^d \rightarrow \mathrm{R}^d$ is Lipschitz continuous, $M_t$ is a martingale difference. Here the stepsize $\alpha_t $ satisfies \begin{align} \sum_{t\geq 0} \alpha_t = \infty, ~~~~\sum_{t \geq 0} \alpha_t^2 < \infty. \end{align} Standard stochastic approximation results suggest that $\{ \theta_t \}_{t\geq 0}$ converges to the limit point of an ODE \begin{align} \dot \theta(t) = h( \theta(t) ). \end{align} Moreover, under the assumption that the ODE has an unique global equilibrium $\theta^*$, it can be shown that $\theta_t \rightarrow \theta^*$ as $t$ goes to infinity.

However, one interesting question is whether $\{ \theta_t\}_{t\geq 0}$ converges if $h$ has multiple zeros. That is, there exists multiple $\bar \theta$'s such that $h( \bar \theta) = 0$. Under what condition can we show that the algorithm converges to any one of these $\bar \theta$'s (equilibria)? Moreover, under what conditions can we show that the algorithm only converges to a locally stable equilibrium?

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  • $\begingroup$ Do you mean to compare to $\theta' = -h(\theta(t))$ instead of $\theta' = h(\theta(t))$? $\endgroup$
    – Michael
    Commented Nov 6, 2017 at 22:44
  • $\begingroup$ Anyway $\theta_t$ will not "converge" even under the conditions you state. For example, $\{M_t\}_{t=0}^{\infty}$ can be iid uniform over $[-1,1]$. Then the noise $M_t$ will always make $\theta_t$ jitter without converging to anything. $\endgroup$
    – Michael
    Commented Nov 6, 2017 at 23:04
  • $\begingroup$ @Michael Thanks for your comment, I've edited the question regarding the minus sign. For the second comment, the stepsizes $\alpha_t$ is decaying, which satisfy $\sum_{t \geq 0} \alpha_t =\infty$ and $\sum_{t \geq 0} \alpha^2 _t < \infty$. In this case, I think your example will still converge by the martingale convergence theorem. $\endgroup$
    – Steve
    Commented Nov 7, 2017 at 2:52
  • $\begingroup$ That stepsize does not appear to be multiplying your noise $M_t$ and so certainly $\theta_t$ cannot converge to anything given the $\{M_t\}_{t=0}^{\infty}$ iid uniform $[-1,1]$. It means that $$P[\theta_{t+1}\geq \theta_t + \alpha_t h(\theta_t) + 1/2] = P[\theta_{t+1} \leq \theta_t + \alpha_t h(\theta_t) - 1/2]=1/4$$ It cannot converge to a constant when it has such a wide range of values every time step, and it cannot converge to a random variable since the noise is iid over time steps, and it keeps wildly swinging its values every step. $\endgroup$
    – Michael
    Commented Nov 7, 2017 at 6:20
  • $\begingroup$ @Michael Sorry, the stepsize $\alpha_t$ should be multiplied to $M_t$. This setting is the same as standard stochastic approximation setting. $\endgroup$
    – Steve
    Commented Nov 7, 2017 at 10:55

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