# Markov chain with Feller property

Does anybody know whether there is an analysis of when the monotone decreasing chain has the Feller property? The monotone decreasing is defined as a chain on $$\mathbb{N}$$ and the rate of going down $$n \mapsto n-1$$ in each step is $$q_n>0,$$ the one of staying at level $$n$$ is $$-q_n$$ and all others are zero. So your population only decreases or stays as it is. Of course, this chain then terminates at $$n=0.$$ I know that the chain is well-studied but I could not find an answer to my particular question.

This is why I started calculting the transition function by myself and ended up with

$$P_t(x,y) = \left(\prod_{k=y+1}^{x} q_k \right) \cdot \left( \sum_{l=y}^{x} \frac{e^{-q_l t}}{\prod_{p \in \{y,...,x\}\backslash \{l\}} (q_p-q_l)}\right)$$

for $$x \ge y$$ (and 0 otherwise) and $$q_p \neq q_l$$ for all possible combinations.

So to show that a Markov chain on a discrete space has the Feller property we need to see that $$\lim_{x \rightarrow \infty} P_t(x,y)=0$$ for any fixed $$y \in \mathbb{N}$$ and $$t \ge 0.$$

I suspected now that the answer is that this holds if and only if $$q_k \rightarrow 0$$ for $$k \rightarrow \infty,$$ but I don't see quite through this cumbersome expression for $$P_t.$$

Does anybody know if there is a treatment of this or how to get a suitable assumption on the $$q_k$$ such that $$\lim_{x \rightarrow \infty} P_t(x,y)=0.$$

Probably Feller unless $$\sum \frac 1 {q_i} < \infty$$. If the sum is finite, you reach 0 in bounded expected time starting from anywhere, and the Feller condition is not satisfied with the state 0 being a counterexample. If the sum is infinite, assume wlog that the $$q_i$$ are bounded below by $$1$$ for large $$i$$. If not, there are infinitely many less than $$1$$, and they will slow you up enough to keep you from reaching y. Let $$T_z$$ be the exponentially distributed time to make the transition from $$z$$ to $$z-1$$. The time to go from $$x$$ to $$y$$ is $$T_x + ... + T_{y+1}$$. $$P_t(x,y) \le P(T_x + ... + T_{x/2} < t)$$ and then calculate the mean and variance of $$T_x + ...+ T_{x/2}$$ to show that the latter probability is small. ( the simplifying assumption $$q_i > 1$$ makes the variance smaller than the mean) .