# Rigorous proof of a simple fact dealing with event space in probability theory

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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could you please explain why $Y_n\geq Z_n$ in a formal way? that fact I was willing to make rigor in my considerations –  Ilya Nov 15 '11 at 15:00
Thanks a lot, you resolve the confusion I've had –  Ilya Nov 18 '11 at 21:27