0
$\begingroup$

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 $$\lim_{x \rightarrow \infty} P_t(x,y)=0.$$

$\endgroup$
1
$\begingroup$

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) .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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