Let $\{X_i\}_{i\ge 1}$ be a Markov chain with $X_i\ge 0, \forall i$ and define $Y_n:=\max_{k\le n} X_k$. Assuming we have access to $\Pr(Y_n<x)$ and $E Y_n$. Can we find an upper bound for $\Pr(X_n<x)$ or lower bound for $E X_n$?
While $P(X_n<x) \le P(Y_n<x)$ holds trivially, the inequality is not necessarily tight. I wonder if exists some function $f(n)$ such that $P(Y_n<x) < f(n) P(X_n<x)$, where desirably $f(n)$ grows logarithmically $f(n)\lesssim \log(n)^c$. While the CDF of maximum of independent random variables has a simple form (namely here), but in a Markov chain that independence is violated.
Assumption (optional): if the generic setting is not informative enough, one way assume an exponential-type bound for $\{Y_i\}_i$ such as $\Pr(Y_k<x)=1-e^{-k x}$
Implication: For many problems is much easier to bound the PDF or expectation of $Y_t$s. So if there are general approaches to convert that to a result for $X_t$s, paying a poly-logarithmic factor, it would be a very useful result.
