Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n$?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $$Y_{n+1}(\omega)\geq Z_{n+1}(\omega).$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.