# Stationary distribution of a Markov chain on the nonnegative integers

Let $$\lambda_k,\mu_k\in\mathbb R_{\ge0}$$ $$(k\ge1)$$ be nonnegative real numbers such that $$\sum_{k=1}^\infty k\lambda_k<\infty,$$ 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). A MathOverflow post contains more information about this process.) 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 $$Q(n,n+1)=(n+1)\lambda_1\quad(n\ge0,k\ge1),\qquad Q(n,n-1)=n\mu_1\quad(n\ge1).$$

Now, assume this chain is reversible and irreducible, and let $$\nu$$ be the stationary measure. Then detailed balance implies $$\forall n\in\mathbb Z_{\ge0}~\forall k\in\mathbb Z_{\ge1}\quad \nu_n\cdot(n+1)\lambda_k=\nu_{n+k}\cdot(n+1)\mu_k,$$ so that $$\nu_n\lambda_k=\nu_{n+k}\mu_k.$$ From reversibility and irreducibility it follows that $$\nu_i>0$$ for all $$i\in S,$$ so that $$\forall k\in\mathbb Z_{\ge1}\quad\lambda_k=0\leftrightarrow\mu_k=0.$$

Now, let $$A=\{k\in\mathbb Z_{\ge1}:\mu_k\ne0\}~(=\{k\in\mathbb Z_{\ge1}:\lambda_k\ne0\}),$$ and let $$d=\gcd A.$$ If $$d\ne1,$$ then $$d>1,$$ and so considering things mod $$d$$ we see that $$X$$ is not irreducible, contradiction; thus $$\gcd A=d=1.$$ And if $$1\in A,$$ then for all $$n\in S$$ we have $$\nu_n\lambda_1=\nu_{n+1}\mu_1,$$ and since $$\mu_1\ne0$$ we can say $$\nu_{n+1}=\nu_n\lambda_1/\mu_1,$$ so that $$\forall k\in S\quad\nu_k=\nu_0(\lambda_1/\mu_1)^k.$$ Normalizing, we have $$\forall k\in S\quad\nu_k=(1-\lambda_1/\mu_1)(\lambda_1/\mu_1)^k.$$ Note that detailed balance also yields $$\forall k\in A\quad\lambda_k/\mu_k=(\lambda_1/\mu_1)^k.$$

Now, my question is: What is the stationary distribution, if $$1\not\in A?$$ I tried the following example: Take $$A=\{2,3\}.$$ In addition, let ($$r$$ for ratio) $$r_2=\lambda_2/\mu_2,$$ $$r_3=\lambda_3/\mu_3.$$ From detailed balance, we have for all $$n\in S$$ that $$\nu_nr_3=\nu_3=\nu_{n+1}r_2,$$ so that $$\nu_{n+1}=\nu_nr_3/r_2.$$ It follows that $$\forall n\in S\quad\nu_n=(1-r_3/r_2)(r_3/r_2)^n.$$ (Note that $$r_3/r_2=\lambda_1/\mu_1$$ if $$\{1,2,3\}\subseteq A.$$) However, I cannot think of how to generalize this example to arbitrary sets $$A.$$

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

The reversibility assumption (and detailed balance) are doing a lot of heavy lifting here. If it holds, then we have $$\nu_{n+k} = r_k \nu_n$$ for all $$n \in \mathbb Z_{\ge 0}$$ and all $$k \in A$$, which essentially tells us $$\nu$$, we just have to massage it into the right form.
Since $$\gcd A = 1$$, there is some integer linear combination $$\sum_{k \in A} c_k \cdot k = 1$$, and it follows from $$\nu_{n+k} = r_k \nu_n$$ that $$\nu_{n+1} = \nu_n \prod_{k \in A} r_k^{c_k}.$$ So we have $$\nu_n = (1-\rho) \rho^n$$, where $$\rho = \prod_{k \in A} r_k^{c_k}$$. In your example, we can take $$c_2 = -1$$ and $$c_3 = 1$$, so $$\rho = \frac{r_3}{r_2}$$, and we get back your formula. Of course, we can also take $$c_2 = 2$$ and $$c_3 = -1$$, and get a formula with $$\rho = \frac{r_2^2}{r_3}$$. Detailed balance only holds if all of these expressions for $$\rho$$ are equal; for any $$j,k \in A$$ we must have $$r_j^k = r_k^j$$. In that case, we actually just have $$r_k = \rho^k$$ for all $$k \in A$$, and the process is not meaningfully different in the end from the $$A = \{1\}$$ case.
(In other words, assuming detailed balance holds, we can just pick any $$k \in A$$ and say that $$\nu_n = (1-r_k^{1/k}) r_k^{n/k}$$, ignoring all other elements of $$A$$.)