Could anyone tell me about the convergence rate of the following
$$ X_{n+1}= f_{\omega_n}(X_n)$$ where $f_1,\dots, f_s$ are finite number of bounded Lipschitz functions (with Lipschitz constants $L_i$ and $\sum_{k=1}^{s} p_kL_k <1$ )on $\mathcal X$, a metric space. $\omega_0,\omega_1,\dots, \omega_n,\dots$ are i.i.d discrete random variable taking values in $\{1,2,\dots,s\}$. $p_k=\mathbb P(\omega_i=k)$. It is given that a unique invariant probability measure exists for such a chain. Thanks for any help!
Since $f_k, k=1,2,\cdots,s$ are Lipschitz, $ \left\vert X_{n+1} \right\vert = \left\vert f_{\omega_n}(X_n) \right\vert\le L_{\omega_n}\left\vert X_n \right\vert \le \cdots \le \prod_{k=0}^nL_{\omega_k}|X_0|$. Hence the ratio $X_n/X_0$ is bounded by $$ R_{n} = \prod_{k=0}^{n-1}L_{\omega_k} = \exp\left( \sum_{k=0}^{n-1}\ln \left(L_{\omega_k}\right)\right). $$
The ratio bound $R_n$ can be considered as the convergence ratio since the bounds can be achieved by simply letting $f_k(x) \equiv L_k \cdot x$.
Note $\ln \left(L_{\omega_k}\right)$ is i.i.d. with probabilities being $p_k, k=1,2,\cdots,s$. By law of large numbers, $\frac{\sum_{k=0}^{n-1}\ln \left(L_{\omega_k}\right)}{n}\rightarrow E\left(L_{\omega_k}\right)=\sum_{i=1}^sp_i\ln \left(L_{i}\right)\equiv R$ a.s.
Recall that $\ln(\cdot)$ is concave function. By Jensen's inequality, we have $$ R\equiv \sum_{i=1}^sp_i\ln \left(L_{i}\right) \le \ln \left( \sum_{i=1}^sp_iL_i \right) < 0. $$ We further have $e^R=\prod_{k=1}^nL_i^{p_i} < 1$. Hence the convergence rate for large $n$ is $$ R_n \approx exp(nR) = {\left( e^R \right)}^n={\left(\prod_{k=1}^sL_i^{p_i}\right)}^n. $$