Let $\lambda\_k,\mu\_k\in\mathbb R\_{\ge0}$ $(k\ge1)$ be nonnegative real numbers, let $S=\mathbb Z\_{\ge0}$ be the nonnegative integers, let $T=\mathbb R\_{\ge0}$ be the nonnegative real numbers and consider the continuous-time Markov chain $X=(X\_t)\_{t\in T}$ on $S$ with rates
$$Q(n,n+k)=(n+1)\lambda\_k\quad(k\ge1),\qquad Q(n,n-k)=(n+1-k)\mu\_k\quad(1\le k\le n).$$
(This Markov chain appears in biology as a model of the length of an evolving DNA sequence (Miklós et. al. 2004). I have also examined some properties of this process in a [Math.StackExchange post](https://math.stackexchange.com/q/3706336/279330).)


For example, if $0=\lambda\_k=u\_k$ for all integers $k\ge2,$ then we recover the *linear birth-death process with immigration* with birth rate $\lambda\_1,$ death rate $\mu\_1$ and immigration rate $\lambda\_1,$ whose nonzero rates are
$$Q(n,n+1)=(n+1)\lambda\_1\quad(n\ge0,k\ge1),\qquad Q(n,n-1)=n\mu\_1\quad(n\ge1).$$
Or, for example, given parameters $\mu\in\mathbb R\_{>0},\gamma,r\in(0,1),$ we can let $\mu\_k=\mu(1-r)^2r^{k-1}$ and $\lambda\_k=\mu(1-r)^2\gamma^kr^{k-1}$ for all $k\ge1.$ Both these examples have been used in, and are of interest in, computational biology.


Now, **for which parameters $\lambda\_k,\mu\_k$ does such a Markov chain exist?** I have heard that the Hille-Yoshida theorem may be helpful and we need a ``dissipative" condition on the growth rate of the terms $Q(n,m).$ However, I don't know how to apply such a theorem here.


In addition, **for which parameters $\lambda\_k,\mu\_k$ does there exist such a Markov chain with all the regularity properties that are important for applications?** (E.g., conservative, standard (maybe?), maybe more....)


I have leafed through parts of Anderson's *Continuous-Time Markov Chains*, Karlin & Taylor's *Second Course* and Ethier & Kurtz's *Markov Processes*, but none of these books contains anything directly helpful.




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Miklós, I., Lunter, G. A., & Holmes, I. (2004). A “long indel” model for evolutionary sequence alignment. Molecular Biology and Evolution, 21(3), 529-540.



Existence of Markov chain on nonnegative integers with specified rates
Define first the modified rates
$$ \tilde Q(n,m) = \frac{Q(n,m)}{n + 1} \, . $$
Clearly, $\tilde Q(n, n+k) = \lambda\_k$, and $\tilde Q(n, n-k) \leqslant \mu\_k$. Assuming that $\lambda\_k$ is summable (otherwise the problem is clearly ill-posed), $\tilde Q$ corresponds to a unique conservative continuous-time Markov chain $\tilde X\_t$.


Now $Q$ corresponds to a time-change of $\tilde X\_t$: the corresponding Markov chain $X\_t$ follows the same path as $\tilde X\_t$, but the holding times at $n$ are $(n+1)$ times shorter.


The only thing that can go wrong with $X\_t$ is a finite-time explosion: if $\tilde X\_t$ goes to infinity too fast, then $X\_t$ may diverge to infinity in finite time. More precisely, the life-time of $X\_t$ is
$$ \tau = \int\_0^\infty \frac{1}{\tilde X\_t + 1} \, dt . $$
Thus, your question can be phrased equivalently: when is $\tau$ infinite almost surely?


* If $k \lambda\_k$ is a summable sequence, then it is not very difficult to show that $\limsup (\tilde X\_t / t) < \infty$, and consequently $\tau = \infty$. This requires a pointwise comparison of $\tilde X\_t$ with a continuous-time random walk that only has positive jumps with rates $\lambda\_k$, plus the strong law of large numbers.
* If, on the other hand, $\mu\_k = 0$ for $k$ large enough (or at least $\mu\_k$ decays sufficiently fast) and $\lambda\_k \asymp k^{-1-\alpha}$ for some $\alpha \in (0, 1)$, then it can be proved that $\tilde X\_t$ is of the order $t^{1/\alpha}$, and consequently $\tau$ is infinite.
* However, if $\mu\_k$ decays sufficiently slowly (or perhaps grows sufficiently fast), it can well compensate the slow decay of $\lambda\_k$. Here, I suppose, things get complicated (and interesting!).




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*Edited: a note on the construction of $\tilde X\_t$.*


Suppose that $\lambda\_k$ is a summable sequence. Then the overall transition rates from state $n$: $$Q(n) = \sum\_{m \ne n} \tilde Q(n, m) = \sum\_{k = 1}^\infty \lambda\_k + \sum\_{k = 1}^n \frac{n + 1 - k}{n + 1} \mu\_k$$ are finite.


Consider the usual construction of a Markov chain: let $E\_n$ be a sequence of standard exponentially distributed random variables, let $Z\_n$ be a discrete-time Markov chain with transition probabilities $(Q(n))^{-1} \tilde Q(n, m)$ (or zero if $n = m$), define
$$ T\_n = \sum\_{j = 0}^{n - 1} \frac{E\_j}{Q(Z\_j)} \, , $$
and
$$ \tilde X\_t = Z\_n \qquad \text{for $t \in [T\_n, T\_{n+1})$.}$$
In other words, $\tilde X\_t$ follows the path of $Z\_n$, with state-dependent holding times given by $E\_n / Q(Z\_n)$.


If $T\_n$ go to infinity as $n \to \infty$, then it is a standard exercise to verify that $\tilde X\_t$ is a continuous-time Markov chain (and this is exactly how these are introduced in some textbooks; I do not have a reference off the top of my head, though). Thus, we need to verify that indeed $T\_n \to \infty$ as $n \to \infty$.


This is true in greater generality: if the overall transition rate of *positive* jumps, $$\sum\_{m > n} \tilde Q(n, m),$$ is bounded as $n \to \infty$. Perhaps there is a neat, single-line argument for that. A somewhat involved proof goes roughly as follows.


Let $n\_1 < n\_2 < \ldots$ be the enumeration of all positive jumps of $Z\_n$, that is, all $n$ such that $Z\_n > Z\_{n-1}$. Then it is a nice (but rather technical) exercise to see that $T\_{n\_{j+1}} - T\_{n\_j}$ (the waiting time for a positive jump of $\tilde X\_t$) is exponentially distributed, with mean $(\sum\_{k = 1}^\infty \lambda\_k)^{-1}$. (Note that in the more general situation described above, $T\_{n\_{j+1}} - T\_{n\_j}$ is no longer exponentially distributed, but it is bounded from below by some exponentially distributed random variable with a fixed mean). Therefore,
$$ \lim\_{n \to \infty} T\_n = \sum\_{j = 0}^\infty (T\_{n\_{j+1}} - T\_{n\_j}) = \infty $$
almost surely, as desired.


