Considering an iterated system described by $$ X_n =\gamma_nX_{n-1} , $$ where $\gamma_i$ are non-negative i.i.d. variables. It is easy to show that the expectation will grow unbounded for $\mathbb E\gamma_i > 1$, but the trajectory that the state follows will (with high probability) be given by $$ X_n\sim e^{n\lambda+o(n)} $$ where $\lambda$ is the Lyapunov exponent given by $\mathbb E\log\gamma$. Since $\log\mathbb E\gamma\geq\mathbb E\log \gamma$, divergence of the average process does not imply divergence of the instantaneous trajectory.
Now, consider the altered system $$ X_n =\min\{\gamma_nX_{n-1},P\} $$ where $P$ is deterministic. It is clear now that the expectation will no longer grow unbounded. I am running simulations which seem to suggest that (unlike for the unbounded system), when the Lyapunov exponent of the unbounded system is negative, $\mathbb E X_n$ decays asymptotically for the bounded system.
My Question is: Are there any theorems that state that, for a bounded Markovian system, if its unbounded counterpart has a negative Lyapunov exponents then the average of the bounded system will decay? Or something similar?
Let $X$ denote the "unbounded" system $$X\to\gamma X,$$ with $E\gamma\gt1$ and $E\log\gamma\lt0$, then $X_n\to0$ almost surely and $E(X_n)\to\infty$. Let $X^P$ denote the "bounded" system $$X^P\to\min\{\gamma X^P,P\},$$ then one can couple $X$ and $X^P$ in such a way that, almost surely and for every $n$, $$X^P_n\leqslant\min\{X_n,P\}.$$ Then $X_n^P\to0$ almost surely because $X_n$ does, and $X_n^P\leqslant P$ almost surely and for every $n$, hence $$E(X_n^P)\to0.$$ To determine whether this convergence in the mean to $0$ is exponential or not, requires other arguments.