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) .


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