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Let $Z$ be a Markov process on $\mathbb R$ given in the form $Z_{n+1} = f(Z_n,\xi_n)$ where $\xi_n$ is a sequence of iid real-valued random variables. The canonical space of $Z$ is the space of trajectories given by $ \Omega = \mathbb R^{\mathbb N_0} = \{\omega:\omega = (Z_0,Z_1,...,Z_n,...)\}. $ Let us suppose that for any z',z'',\xi\in \mathbb R such that z'\leq z'' it holds that f(z',\xi)\leq f(z'',\xi) and let $ g(z,n) = \mathsf P\left\{\left.\max\limits_{1\leq i\leq n}Z_i>1\right|Z_0 = z\right\}. $

How can I prove rigorously that g(z',n)\leq g(z'',n) for any z'\leq z'' and any fixed $n$?

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I would simply use coupling, i.e. construct another Markov process with a different starting point but with the same movements.

So let $y\geq z$, and let $Y$ be a Markov process on $\mathbb R$ defined by $Y_{n+1}=f(Y_n,\xi_n),$ where $\xi$ is the same sequence, with $Z_0=z$ and $Y_0=y$.

Surely (even better than almost surely!), $\forall n, Y_n\geq Z_n,$ so we have that \begin{align*} \max_{1\leq i\leq n} Y_i&\geq \max_{1\leq i\leq n} Z_i\\ \left\{\max_{1\leq i\leq n} Y_i > 1\right\} &\supset \left\{\max_{1\leq i\leq n} Z_i > 1\right\}\\ g(y,n)&\geq g(z,n) \end{align*} and indeed, $g$ is non-decreasing in its first argument.

Edit: the fact that $\forall n, Y_n\geq Z_n$ surely, should be obvious trajectory-wise.

If you really wish, you may use an induction on $n$: suppose that $\forall\omega\in\Omega$ (this is what I meant by "surely"), $Y_n(\omega)\geq Z_n(\omega).$ By definition of $f$, $f(Y_n(\omega),\xi_n(\omega))\geq f(Z_n(\omega),\xi_n(\omega)),$ so $Y_{n+1}(\omega)\geq Z_{n+1}(\omega).$

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    Thanks a lot, you resolve the confusion I've had2011-11-18