0
$\begingroup$

Consider a real-valued Markov process $X$ with a transition density $f(x,y)$, i.e. $ \mathsf P\{X\in A|X_0 = x\} = \int\limits_A f(x,y)\,dy. $ For this process I want to find $ u(x) = \mathsf P\{X_n < 0\text{ for some }n\geq 0|X_0 = x\}. $

Since for this problem does not matter the distribution of this process for $X_0 = x<0$ I would like to modify it in the following way. I define a modification $Y$ taking values in $\mathbb R^+\cup\{a\}$ with $a<0$ such that:

  1. for any $A\subset \mathbb R^+$ holds $\mathsf P\{Y_1\in A|Y_0 = x\} = \mathsf P\{X_1\in A|X_0 = x\}$;

  2. for $x\geq0$ we pur $\mathsf P\{Y_1 = a|Y_0 = x\} = \mathsf P\{X_1\leq 0|X_0 = x\}$;

  3. $\mathsf P\{Y_1 = a|Y_0 = a\} = 1$.

I hope that for this process $ v(x) = \mathsf P\{Y_n < 0\text{ for some }n\geq 0|Y_0 = x\} $ coincides with $u(x)$ but I don't know how to prove this fact. I also asked it here: https://mathoverflow.net/questions/68377/modification-of-a-markov-process-on-the-real-line

1 Answers 1

1

Here is a realization of the process $(Y_n)$, which shows that $u(x)=v(x)$ for every $x\ge0$. Fix $x\ge0$, consider $(X_n)$ starting from $X_0=x$, introduce the stopping time $\tau=\inf\{n\ge0;X_n<0\}$ and the function $\varphi$ such that $\varphi(y)=y$ for every $y\ge0$ and $\varphi(y)=a$ for every $y<0$, and finally define $ Y_n=\varphi(X_{n\wedge\tau}), $ for every $n\ge0$. You are done.

  • 0
    Double question, double answer. One more method in my collection. Will it be a Markov process?2011-06-21