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? 



Superlinear Convergence of a Markov Chain
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 $E[Z\_t]=c (\kappa
c^2)^t$. 


