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Suppose that we have a Markov process $\{Z_t\}_{t=0}^\infty$, where $Z_t \geq 0$ for any $t$. Assume that, conditioning on $Z_t = z_t$, we have $ \mathbb{E}\{Z_{t+1}|Z_t = z_t\} \leq \kappa z_t^2 $. Here $\kappa > 0$ is a constant.

Question: Conditioning on that the realization of $Z_0$ is sufficiently small, can we prove that $\mathbb{E}\{Z_{t}\} \leq c\exp(-t^2)$, where $c$ is an constant, or something like $\mathbb{E}\{Z_{t}^2\} \leq c\exp(-t^2)$? If not, what additional conditions on $\kappa$ or the value of $Z_0$ do we need? Or is there any counter example for this claim?

If we can further assume the boundedness of $Z_t$ for all $t$, (how) can we prove this claim?

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  • $\begingroup$ $Z_t$ are bounded in what sense? You need to clarify this definition. $\endgroup$ – Henry.L Jun 3 '17 at 2:36
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It is not true in general.

If $Z_t$ is not bounded, the expectation can diverge to infinity. For example, define: $$ Z_t = \left\{\begin{array}{lc}2^{2^{t}} & \text{with probability $\epsilon/2$}\\0 &\text{otherwise}\end{array}\right.$$

Then $E[Z_0]=\epsilon$ and $Z_{t+1} = (Z_t)^2$ but $E[Z_t]$ diverges to infinity.

If $Z_t$ is bounded, the convergence might be simply exponential. For example take $Z_1=c$ and \begin{align*} Z_{t+1} = \left\{ \begin{array}{ll} c &\text{ with probability $\kappa Z_t^2$}\\ 0 &\text{ otherwise} \end{array} \right. \end{align*} We have $E[Z_{t+1}|Z_t] = \kappa Z_t^2 $ and $E[Z_t]=c (\kappa c^2)^t$.

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  • $\begingroup$ If we further assume the boundedness of $Z_t$ for all $t$, then can we prove this claim? $\endgroup$ – Minkov May 27 '17 at 16:14
  • $\begingroup$ If I am not mistaken, it still does not work even if $Z_t$ is bounded. I updated my answer. $\endgroup$ – N. Gast Jun 10 '17 at 7:56

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