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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.)

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.


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

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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!).


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.

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  • $\begingroup$ Why is the problem ill-posed if $\lambda_k$ is not summable? I don't see how. $\endgroup$ – xFioraMstr18 Jun 5 '20 at 14:08
  • $\begingroup$ And could you elaborate on why $\tilde Q$ corresponds to a unique conservative continuous-time Markov chain if $\lambda_k$ is summable? $\endgroup$ – xFioraMstr18 Jun 5 '20 at 14:22
  • $\begingroup$ @xFioraMstr18: The holding time at $n$ is exponential with mean $(\sum_{m\ne n} Q(n,m))^{-1}$. Regarding the other question: if $\lambda_k$ and $\mu_k$ are summable, then the holding time of $\tilde{X}_t$ at any state $n$ has mean greater than a constant, and so the usual construction of a continuous-time Markov chain can be applied. If $\mu_k$ are not summable, the construction is more involved: essentially one shows that the exit times from $\{1, 2, \ldots, N\}$ diverge to infinity as $N \to \infty$. Unfortunately I do not have enough time now to elaborate. $\endgroup$ – Mateusz Kwaśnicki Jun 5 '20 at 14:29
  • $\begingroup$ Okay, thanks! And if in the future you could return to elaborate further (if you are compelled), I would greatly appreciate it. $\endgroup$ – xFioraMstr18 Jun 5 '20 at 14:47

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