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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. This question is also presented here.

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up vote 1 down vote accepted

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.

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