A Countable-Type Branching Process Model for the Tug-of-War Cancer Cell Dynamics

Abstract

We consider a time-continuous Markov branching process of proliferating cells with a countable collection of types. Among-type transitions are inspired by the Tug-of-War process introduced by McFarland et al. (Proc Natl Acad Sci 111(42):15138–15143, 2014) as a mathematical model for competition of advantageous driver mutations and deleterious passenger mutations in cancer cells. We introduce a version of the model in which a driver mutation pushes the type of the cell L-units up, while a passenger mutation pulls it 1-unit down. The distribution of time to divisions depends on the type (fitness) of cell, which is an integer. The extinction probability given any initial cell type is strictly less than 1, which allows us to investigate the transition between types (type transition) in an infinitely long cell lineage of cells. The analysis leads to the result that under driver dominance, the type transition process escapes to infinity, while under passenger dominance, it leads to a limit distribution. Implications in cancer cell dynamics and population genetics are discussed.

Appendices

Appendix

Model Construction

Assumptions: $s_{d}>s_{p}$, $\mu< \nu < d$.

The type of the branching process is determined by the number of driver and passenger mutations $(n_{d},n_{p})$. A type $(n_{d},n_{p})$ cell has fitness $b_{0}(1+s_{d})^{n_{d}}/(1+s_{p})^{n_{p}}$. We construct the model under the following driver-passenger relation.

$$\begin{aligned} \begin{aligned} \log _{(1+s_{p})}(1+s_{d})=L \in {\mathbb {N}}, L \ge 2. \end{aligned} \end{aligned}$$

(3)

Therefore, the fitness function can be rewritten as $b(i) = b_{0}(1+s_{p})^{i}$. Mutation rates for driver and passenger mutations are $\mu$ and $\nu$, respectively. The death rate is d and we define the total rate to be $\delta (i) = b(i)+\mu +\nu +d$.

Denote the continuous process as $({\textbf{Z}}_{t})_{t\in {\mathbb {R}}_{+}}$ and the jump chain as $({\textbf{J}}_{n})_{n\in {\mathbb {Z}}_{+}}$. In this construction, we assume that the parameters of the fitness function of drivers and passengers satisfy condition (3) for the reduced process. Therefore, driver mutation pushes the type of the cell L-units up, while a passenger mutation pulls it 1-unit down.

A naïve state space might be ${\mathbb {S}} \simeq \mathbb {Z_{+}}^{{\mathbb {Z}}}\simeq {\mathbb {R}}$ which is uncountable. However, we can define the state space in a more parsimonious way. We set $S_{m}=\{\cdots ,0,n_{-m},\cdots ,n_{m},0,\cdots \mid n_{k}\in {\mathbb {Z}}_{+}; -m\le k\le m\}$ and let ${\mathcal {S}}_{\infty }=\cup _{m\ge 0}S_{m}$. Since a countable union of countable sets is countable, the set $S_{\infty }$ is countable. Take an element in ${\textbf{s}} \in {\mathbb {S}}{\setminus } S_{\infty }$, then it has to have nonzero elements with indices arbitrarily small or arbitrarily large. Due to the proliferation mechanism and initial condition ${\textbf{J}}_{0}=\underline{e}_{0}$, ${\mathbb {P}}({\textbf{J}}_{n} ={\textbf{s}}) = 0$ for each $n\ge 0$. As long as the initial population is of finite size, we may discard the states with zero probabilities and the new state space $S_{\infty }$ is countable.

To construct the continuous-time process ${\textbf{Z}}_{t}$, we associate a holding time following a $Exp(\lambda _{{\textbf{s}}})$ distribution to state ${\textbf{s}} = (\cdots ,n_{-1},n_{0},n_{1},\cdots )$ where $\lambda _{{\textbf{s}}} = \sum _{k\in {\mathbb {Z}}}n_{k}\delta (k)$. Let j(n) be the time of the nth jump. The proliferation mechanism is as follows. After the nth jump, there are $({\textbf{Z}}_{j(n)})_{k}$ individuals of type k; then a single type i cell will be selected to proliferate, mutate, or die with probability

$$\begin{aligned} \frac{({\textbf{Z}}_{j(n)})_{i}\delta (i)}{\sum _{k\in {\mathbb {Z}}}({\textbf{Z}}_{j(n)})_{k}\delta (k)}. \end{aligned}$$

The probability is due to the competing exponential random variables argument. The cell will then proliferate, acquire a driver mutation, acquire a passenger mutation, or die with probabilities

$$\begin{aligned} \frac{b(i)}{\delta (i)}, \frac{\mu }{\delta (i)},\frac{\nu }{\delta (i)} \text {, or } \frac{d}{\delta (i)} \text {, respectively.} \end{aligned}$$

Finally, the probability space for the jump chain can be constructed according to Williams (1991) page 105 and the space can be extended to support the continuous process. In this space, the probabilities for the embedded branching process $({\textbf{E}}_{n})$ can be extracted.

The probability generating function for the embedded branching process is ${\textbf{G}}$ and its coordinates are defined by

$$\begin{aligned} G_{i}({\textbf{s}}) = \frac{d}{\delta (i)} + \frac{\nu }{\delta (i)}s_{i-1}+ \frac{b(i)}{\delta (i)}s_{i}^{2}+ \frac{\mu }{\delta (i)}s_{i+L}. \end{aligned}$$

The mean matrix is

$$\begin{aligned}&M_{i,i-1} = \frac{\nu }{\delta (i)}, M_{i,i} = 2\frac{b(i)}{\delta (i)}, M_{i,i+L} = \frac{\mu }{\delta (i)}; \forall i\in {\mathbb {Z}}. \end{aligned}$$

Propositions

Proposition 1

$\forall i,j \in {\mathbb {Z}},\lim _{n\rightarrow \infty }[(M^{n})_{i,j}]^{1/n}=2$, which implies ${\textbf{q}}\le \tilde{{\textbf{q}}}< \underline{1}$.

Proof

Recall that the mean matrix M is a matrix whose nonzero entries are

$$\begin{aligned}&M_{i,i-1} = \frac{\nu }{\delta (i)}, M_{i,i} = 2\frac{b(i)}{\delta (i)}, M_{i,i+L} = \frac{\mu }{\delta (i)}; \forall i\in {\mathbb {Z}}. \end{aligned}$$

The proof uses notation and results from Seneta (2006). Since all matrices there are indexed by $i,j \in {\mathbb {N}}$, we relabel our mean matrix M according to the relabeling function $\pi$ and denote it ${\widetilde{M}}$. The relabeling function $\pi$ is a mapping from ${\mathbb {Z}}$ to ${\mathbb {Z}}_{+}$ such that

$$\begin{aligned} \begin{aligned} \pi (0)&= 0,\pi (1)=1,\cdots , \pi (L) = L, \\ \pi (-1)&= L+1,\pi (-2) = L+2,\cdots , \pi (-L) = 2L, \\ \pi (L+1)&= 2L+1,\pi (L+2) = 2L+2,\cdots , \pi (2L) = 3L, \\&\quad \vdots \end{aligned} \end{aligned}$$

(4)

The nth truncation (the northwestern $n\times n$ submatrix) of ${\widetilde{M}}$ is irreducible for $n \ge L+1$. We will denote the nth truncation of ${\widetilde{M}}$ by $_{(n)}{\widetilde{M}}$. Define R to be the convergence parameter of ${\widetilde{M}}$ and $r = 1/R$. An explicit calculation of r does not seem feasible; however, the fact that $M_{i,i} \rightarrow 2$ as $i \rightarrow \infty$ simplifies the matter. For fixed $x\in (1,2)$, there exists n such that $M_{n,n} > x$. Since the matrix is non-negative,

$$\begin{aligned}&(M^{k})_{n,n} \ge (M_{n,n})^{k} \ge x^{k} \\&\quad \Rightarrow \lim _{k\rightarrow \infty }\left[ (M^{k})_{n,n}\right] ^{1/k} \ge x, \forall 1< x < 2\\&\quad \Rightarrow r\ge 2, R \le \frac{1}{2}. \end{aligned}$$

If the convergence parameter of the nth truncation of ${\widetilde{M}}$ is $_{(n)}R$ where $n \ge L+1$, then $1/{_{(n)}R}$ is the Perron-Frobenius eigenvalue of $_{(n)}{\widetilde{M}}$. By the Perron-Frobenius theorem, we have

$$\begin{aligned} \frac{1}{_{(n)}R} \le \max _{i}\sum _{j}\;_{(n)}{\widetilde{M}}_{i,j} \le 2. \end{aligned}$$

By Theorem 6.8 in Seneta (2006), $_{(n)}R \rightarrow R$ as $n\rightarrow \infty$. Therefore, $\frac{1}{R}\le 2$ and this implies $R = \frac{1}{2}$ or equivalently, $r = 2$.

Since the kth truncation of ${\widetilde{M}}$ is irreducible for $k\ge L+1$ and $r>1$, we invoke Proposition 4.1 in Hautphenne et al. (2013) and conclude $\tilde{{\textbf{q}}}<\underline{1}$, where $\underline{1}$ is the vector $(\cdots ,1,1,1,\cdots )^{T}$. $\square$

Remark

Although ${\widetilde{M}}$ is irreducible, it does not guarantee the PGF has at most 2 fixed points. In Bertacchi and Zucca (2014), authors show that in the case of non-strong local survival, it is possible to have ${\textbf{q}}<\tilde{{\textbf{q}}}<\underline{1}$ with an irreducible mean matrix. Corollary 4.4 of Bertacchi et al. (2020) states that $\sup _{i}{\tilde{q}}_{i}<1$ is sufficient for ${\textbf{q}}=\tilde{{\textbf{q}}}$. However, this condition is not satisfied in our model and we cannot conclude ${\textbf{q}}=\tilde{{\textbf{q}}}$.

Lemmas

Lemma 1

Under the assumption $(d+\nu ) > \mu L$, the continuous-time process ${\textbf{Z}}_{t}$ is non-explosive.

Proof

We compare the process ${\textbf{Z}}_{t}$ to a simpler process that is more likely to explode. Consider a new passenger mutations rate that incorporates the death rate $\xi = d+\nu$ and set the death rate of this new process to 0. For the new process to be non-explosive, it suffices to show

$$\begin{aligned} \sum _{n=0}^{\infty } \frac{1}{\lambda _{{\textbf{J}}_{n}}} = \infty \text { with probability } 1, \end{aligned}$$

where $({\textbf{J}}_{n})$ is the jump chain of the modified process with passenger mutation rate $\xi$. If the new process were a pure birth process, the above sum would be a multiple of the standard harmonic series and diverge to $\infty$ with probability 1. Under the assumption $\xi > \mu L$, a lineage is more likely to acquire L passenger mutations than acquire one driver mutation. To see this, let us focus on the transition between types along a lineage. We have

$$\begin{aligned} {\mathbb {P}}(\text {Consecutive } L \text { passenger mutations})&= \left( \frac{\xi }{\mu +\xi }\right) ^{L}\ge \left( \frac{L}{L+1}\right) ^{L} > \frac{1}{L+1} \\ {\mathbb {P}}(1 \text { driver mutation})&= \frac{\mu }{\mu +\xi } < \frac{1}{L+1}. \end{aligned}$$

Intuitively, this random sum of holding times should diverge to infinity almost surely since the fitness of cells on a lineage tends to decrease along generations. For a rigorous argument, since there are at most countably many lineages and we may enumerate them as $l_{1},l_{2},\cdots$. If the supremum of fitnesses of all lineages across generation is bounded by M from above, we have

$$\begin{aligned} \sum _{n=0}^{\infty } \frac{1}{\lambda _{{\textbf{J}}_{n}}} \ge \frac{1}{M+\mu +\xi } \sum _{n=0}^{\infty } \frac{1}{N_{n}} \ge \frac{1}{M+\mu +\xi } \sum _{k=1}^{\infty } \frac{1}{k} = \infty , \end{aligned}$$

where $N_{n}$ is the population size after the nth jump and it is non-decreasing with n. Hence, for an explosion to occur, we must have at least one lineage whose fitness is unbounded from above. We now show this event has probability zero. Select a lineage and model the transition of cell type as random walk $(R_{n})$ that increases by L with probability $\frac{\mu }{\mu +\xi }< \frac{1}{L+1}$ and decreases by 1 with probability $\frac{\xi }{\mu +\xi }> \frac{L}{L+1}$. According to Kolmogorov’s zero–one law, $\limsup _{n\rightarrow \infty } R_{n}$ and $\liminf _{n\rightarrow \infty } R_{n}$ are almost surely constants (see page 88 in Cinlar 2011). By Markov’s inequality, for some $\alpha > 0$ that will be specified later, we have

$$\begin{aligned} \begin{aligned} {\mathbb {P}}(R_{n} \ge K) = {\mathbb {P}}\left( e^{\alpha R_{n}} \ge e^{\alpha K}\right) \le \frac{\left( \frac{\xi }{\mu +\xi }e^{-\alpha }+\frac{\mu }{\mu +\xi }e^{L\alpha }\right) ^{n}}{e^{K\alpha }}. \end{aligned} \end{aligned}$$

(5)

If there exists $\alpha > 0$ such that $\frac{\xi }{\mu +\xi }e^{-\alpha }+\frac{\mu }{\mu +\xi }e^{L\alpha }< 1$, $\sum _{n=0}^{\infty }{\mathbb {P}}(R_{n} \ge K) < \infty$. Let $x = e^{\alpha }$, then the desired condition is equivalent to the existence of some $x > 1$ such that

$$\begin{aligned} \begin{aligned} \frac{\xi }{\mu +\xi }\frac{1}{x}+\frac{\mu }{\mu +\xi }x^{L}< 1 \iff \frac{\xi }{\mu +\xi }+\frac{\mu }{\mu +\xi }x^{L+1}- x<0. \end{aligned} \end{aligned}$$

(6)

Observe that the RHS of (6) is equal to 0 when $x=1$. The derivative of $\frac{\xi }{\mu +\xi }+\frac{\mu }{\mu +\xi }x^{L+1}- x$ is $\frac{\mu }{\mu +\xi }(L+1)x^{L}- 1$, and it is negative for x slightly greater than 1 since $\frac{\mu }{\mu +\xi } < \frac{1}{L+1}$. Therefore, there exists $\alpha >0$ such that $\frac{\xi }{\mu +\xi }e^{-\alpha }+\frac{\mu }{\mu +\xi }e^{L\alpha }< 1$. The inequality in (5) now proves ${\mathbb {P}}(R_{n}\ge K)$ is summable with respect to n. By Borel-Cantelli lemma, we have ${\textbf{1}}_{\{R_{n}\ge K\}} \rightarrow 0$ almost surely and this implies $\limsup _{n\rightarrow \infty }R_{n} < K$ for all K. Hence, $\lim _{n\rightarrow \infty } R_{n} = -\infty$ and finally

$$\begin{aligned} {\mathbb {P}}(\text {At least one lineage has unbounded fitness})&\!\le \! \sum _{i=1}^{\infty }{\mathbb {P}}(l_{i} \text { has unbounded fitness})=0. \end{aligned}$$

$\square$

Corollary

If $\nu + d > \mu L$, the continuous process $({\textbf{Z}}_{t})$ becomes extinct if and only if the embedded process $({\textbf{E}}_{n})$ becomes extinct.

Proof

To show the equivalence of extinctions in the continuous and embedded process under non-explosion, let $A_{n}$ denote the event that the embedded process becomes extinct at or before the nth generation. That is, $\omega \in A_{n}$ implies $||{\textbf{E}}_{n}(\omega )||_{\ell ^{1}} = 0$. Since the number of generations n is finite, the number of jumps is also finite. Therefore, $\lim _{t\rightarrow \infty }||{\textbf{Z}}_{t}(\omega )||_{\ell ^{1}} = 0$ for almost every $\omega \in A_{n}$.

On the other hand, let $B_{t}$ be the event that the continuous-time process becomes extinct at or before time $t \in {\mathbb {Q}}_{+}=\{q\in {\mathbb {Q}}\mid q \ge 0\}$. That is, $\omega \in B_{n}$ implies $||{\textbf{Z}}_{t}(\omega )||_{\ell ^{1}} = 0$. Since [0, t] is a finite interval and the process is non-explosive, the number of jumps in [0, t] is a.s. finite. Hence, the number of generations in [0, t] is also a.s. finite and $\lim _{n\rightarrow \infty }||{\textbf{E}}_{n}(\omega )||_{\ell ^{1}} = 0$ for almost every $\omega \in B_{t}$.

Finally, let $A = \cup _{n\ge 0}A_{n}$ and $B=\cup _{t\in {\mathbb {Q}}_{+}}B_{t}$, then

$$\begin{aligned} {\mathbb {P}}(A) = {\mathbb {P}}(\cup _{n> 0}A_{n}) = {\mathbb {P}}(\cup _{t\in {\mathbb {Q}}_{+}}B_{t}) = {\mathbb {P}}(B). \end{aligned}$$

$\square$

Lemma 2

Let $g(x) = \mu x^{L+1}-(\mu +\nu +d)x+\nu = 0$; then under condition of non-explosion ($\nu +d > \mu L$) and $d>\nu$, $\lim _{i\rightarrow -\infty }\frac{1-q_{i}}{1-q_{i-1}} = \alpha$ where $\alpha$ is the unique real solution to $g(x)=0$ such that $x \ge 1$. As a consequence, $\frac{\nu }{\mu L} < \alpha ^{L+1}$.

Proof

The difference equation defining extinction probabilities can be transformed into a perturbed linear system by the following manipulation. Recall that extinction probabilities satisfy

$$\begin{aligned} \begin{aligned} 1-s_{i-1} = \frac{\mu +\nu +d}{\nu }(1-s_{i}) - \frac{\mu }{\nu }(1-s_{i+L}) -\frac{b(i)}{\nu }s_{i}(1-s_{i}), \forall i\in {\mathbb {Z}}. \end{aligned} \end{aligned}$$

(7)

Define $y_{n} = 1-s_{-n}$, then we obtain the following difference equation.

$$\begin{aligned} \begin{aligned} y_{n+1} = \left( \frac{\mu +\nu +d}{\nu } - \frac{b(-n)}{\nu }(1-y_{n})\right) y_{n} - \frac{\mu }{\nu }y_{n-L}, \forall n\in {\mathbb {Z}}. \end{aligned} \end{aligned}$$

(8)

Since $y_{n} = 1-q_{-n}$ solves (8), replacing $1-y_{n}$ by $q_{-n}$ will result in a new system whose solution set contains $(1-q_{-n})_{n=-\infty }^{\infty }$. we obtain a perturbed linear system (9) with $(1-q_{-n})_{n\in {\mathbb {Z}}}$ being one of its solutions.

$$\begin{aligned} \begin{aligned} y_{n+1} = \left( \frac{\mu +\nu +d}{\nu } - \frac{b(-n)}{\nu }q_{-n}\right) y_{n} - \frac{\mu }{\nu }y_{n-L}, \forall n\in {\mathbb {Z}}. \end{aligned} \end{aligned}$$

(9)

Denote ${\textbf{y}}_{n} = (y_{n},\cdots ,y_{n-L})^{T}$. Then we can rewrite the above relation as

$$\begin{aligned} {\textbf{y}}_{n+1}&= (A+R(n)){\textbf{y}}_{n}, \\ A&= \begin{pmatrix} \frac{\mu +\nu +d}{\nu } &{}\underline{0}^{T} &{}-\frac{\mu }{\nu } \\ 1 &{}\underline{0}^{T} &{}0 \\ \underline{0} &{}I_{L-1} &{}\underline{0} \end{pmatrix}, \;\;R(n) = \text {diag}\left( -\frac{b(-n)}{\nu }q_{-n}, \cdots , -\frac{b(-n+L)}{\nu }q_{-n+L}\right) , \end{aligned}$$

where $\underline{0}$ is a $(L-1)-$column vector and $I_{L-1}$ is the $(L-1)\times (L-1)$ identity matrix. The characteristic polynomial of matrix A is a multiple of $f(x) = \nu x^{L+1} - (\mu +\nu +d)x^{L} + \mu$. Since $x=0$ is no a root of f, if we define $g(x) = x^{L+1}f(1/x) = \mu x^{L+1} - (\mu +\nu +d)x +\nu$ for $x\in {\mathbb {R}}$, there is a one-to-one correspondence between roots of f and roots of g. That is, $f(x)=0$ if and only if $g(1/x)=0$. Observe that $g(1) = -d < 0$ and $g''(x) > 0$ on $(0,\infty )$. Therefore, there is a unique positive real root of modulus greater than 1 for g and we denote it as $\alpha$. Note that $\frac{1}{\alpha }$ is the unique real root of f that has modulus less than 1. To show that roots of g are simple, consider the condition

$$\begin{aligned} g'(x) = 0 \Rightarrow x^{L} = \frac{\mu +\nu +d}{(L+1)\mu }. \end{aligned}$$

Substituting $x^{L}$ in g, we obtain

$$\begin{aligned} x\frac{\mu +\nu +d}{L+1} - (\mu +\nu +d)x +\nu = 0 \Rightarrow x = \frac{L+1}{L}\frac{\nu }{\mu +\nu +d}. \end{aligned}$$

Hence, $x^{L+1} = \frac{\nu }{\mu L}$ and if $x = \left( \frac{\nu }{\mu L}\right) ^{\frac{1}{L+1}}$ is not a root of g, all roots of g are simple and matrix A is diagonalizable. For contradiction, suppose $g\left( \left( \frac{\nu }{\mu L}\right) ^{\frac{1}{L+1}}\right) = 0$, which implies $\left( \frac{\nu }{\mu L}\right) ^{\frac{1}{L+1}} = \frac{L+1}{L}\frac{\nu }{\mu +\nu +d}$. Under the non-explosion condition $\nu +d > \mu L$, this yields

$$\begin{aligned} \mu +\nu +d> \mu (L+1) \Rightarrow \;&\frac{\nu }{\mu +\nu +d}< \frac{\nu }{\mu (L+1)} \\ \Rightarrow \;&\left( \frac{\nu }{\mu +\nu +d}\frac{L+1}{L}\right) ^{\frac{1}{L+1}}< \left( \frac{\nu }{\mu L}\right) ^{\frac{1}{L+1}}=\frac{L+1}{L}\frac{\nu }{\mu +\nu +d} \\ \Rightarrow \;&1< \frac{L+1}{L}\frac{\nu }{\mu +\nu +d} \\ \Rightarrow \;&L\mu + Ld < \nu , \text { which contradicts } d > \nu . \end{aligned}$$

We now express A as $A = T\Lambda T^{-1}$ with $\Lambda$ being a diagonal matrix containing distinct eigenvalues of A. Let us choose $N \in {\mathbb {N}}$, and investigate the asymptotic behavior of the perturbed linear system

$$\begin{aligned} {\textbf{z}}_{n+1} = T{\textbf{y}}_{n+1} = T(A+R(n))T^{-1}T{\textbf{y}}_{n} = \left( \Lambda +TR(n)T^{-1}\right) {\textbf{z}}_{n}, n\ge N. \end{aligned}$$

Since all eigenvalues of A are nonzero and simple and the operator norm of the perturbation $||TR(n)T^{-1}||$ satisfies

$$\begin{aligned}&||TR(n)T^{-1}|| \le ||T||\cdot ||R(n)||\cdot ||T^{-1}|| \le \frac{b(-n+L)}{\nu }||T||\cdot ||T^{-1}|| \\&\quad \Rightarrow \sum _{n\ge 0}||TR(n)T^{-1}|| < \infty . \end{aligned}$$

Hence, can invoke Theorem 3.4 in Bodine and Lutz (2015) (Discrete Version of Levinson’s Fundamental Theorem) with $n_{0}=0,K_{1}=K_{2}=1$. For convenience, the statement of the theorem is provided in Sect. E. The fundamental matrix for ${\textbf{z}}_{n}$ has the following form

$$\begin{aligned} (I + o(1)) \Lambda ^{n} \text { as } n \rightarrow \infty . \end{aligned}$$

Equivalently, the fundamental matrix for ${\textbf{y}}_{n}$ has the following form

$$\begin{aligned}&(I + o(1)) T^{-1}\Lambda ^{n} \text { as } n \rightarrow \infty , \text { where }\\&T^{-1} = \begin{pmatrix} \lambda _{1}^{L} &{}\cdots &{}\lambda _{L+1}^{L} \\ \lambda _{1}^{L-1} &{}\cdots &{}\lambda _{L+1}^{L-1} \\ \vdots &{} &{}\vdots \\ 1 &{}\cdots &{}1 \\ \end{pmatrix}. \end{aligned}$$

Since $(y_{n}) = (1-q_{-n})$ solves the perturbed system,

$$\begin{aligned} 1-q_{-n}&= (1+o(1))\sum _{k=1}^{L+1}c_{k}\lambda _{k}^{n}. \end{aligned}$$

The dominating eigenvalue is real. To prove this, suppose the dominating eigenvalue is a complex number $r\exp (\iota \theta )$ with $\theta \notin \{0,\pi \}$, then $r\exp (-\iota \theta )$ is also a dominating eigenvalue. Hence, as $n\rightarrow \infty$ and omit all n such that $\cos (n\theta ) = 0$,

$$\begin{aligned} \frac{1-q_{-n}}{r^{n}} \sim C \cdot \cos (n\theta ), C \ne 0 \text { is a constant}. \end{aligned}$$

This is impossible since LHS is always positive and the RHS takes negative values infinitely many times as $n\rightarrow \infty$. Similarly, we cannot have 2n dominating complex eigenvalues. Hence,

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1-q_{-n-1}}{1-q_{-n}} = \lambda _{k}\in {\mathbb {R}} \text { for some } 1\le k\le L+1. \end{aligned}$$

Recall that survival probabilities satisfy

$$\begin{aligned}&1-q_{-n} = \frac{\nu }{\delta (-n)}(1-q_{-n-1}) + \frac{b(-n)}{\delta (-n)}(1-q_{-n}^{2})+\frac{\mu }{\delta (-n)}(1-q_{-n+L}) \\&\quad \Rightarrow \frac{1-q_{-n}}{1-q_{-n-1}}= \frac{\nu }{\delta (-n)} + \frac{b(-n)}{\delta (-n)}\frac{1-q_{-n}^{2}}{(1-q_{-n-1})}+\frac{\mu }{\delta (-n)}\frac{1-q_{-n+L}}{1-q_{-n-1}} \\&\quad \Rightarrow \frac{1}{\lambda _{k}} = \frac{\nu }{\mu +\nu +d} + \frac{\mu }{\mu +\nu +d}\left( \frac{1}{\lambda _{k}}\right) ^{L+1}. \end{aligned}$$

By Theorem 1, monotonicity of extinction probabilities implies that $\lambda _{k}< 1$. Since $\lambda _{k}$ is the limit of ratios of real numbers, it must be a real number as well. Therefore, $\lambda _{k} = \frac{1}{\alpha }$ and $\alpha$ satisfies $g(\alpha ) = 0$. Finally, suppose $\frac{\nu }{\mu L} \ge \alpha ^{L+1}$, we arrive at a contradiction in the following way,

$$\begin{aligned} \frac{\nu }{\mu L} \ge \alpha ^{L+1} = \frac{(\mu +\nu +d)\alpha -\nu }{\mu } \Rightarrow \alpha \le \frac{L+1}{L}\frac{\nu }{\mu +\nu +d}< \frac{2}{1}\frac{\nu }{\nu +d} <1. \end{aligned}$$

$\square$

Corollary

The type transition process $(X_{t})$ is non-explosive.

Proof

According to Theorem 2.7.1 in Norris (1998), it suffices to show the supremum of rates is bounded.

$$\begin{aligned}&{\left\{ \begin{array}{ll} \lim _{i\rightarrow \infty }\delta (i)(1-T_{i,i}) = \lim _{i\rightarrow \infty }\mu \frac{1-q_{i+L}}{1-q_{i}}+\nu \frac{1-q_{i-1}}{1-q_{i}} = \mu +\nu \\ \lim _{i\rightarrow -\infty }\delta (i)(1-T_{i,i}) = \mu \alpha ^{L}+\nu \alpha ^{-1} = \mu +\nu +d \end{array}\right. } \\&\quad \Rightarrow \sup _{i\in {\mathbb {Z}}}\{\delta (i)(1-T_{i,i})\} <\infty . \end{aligned}$$

$\square$

Lemma 3

Under driver dominance $(\nu /\mu L\le 1)$, $\lim _{n\rightarrow \infty }{\mathbb {E}}(Y_{n}) = \infty$ and $\lim _{n\rightarrow \infty }{\mathbb {E}}(b(Y_{n}))\rightarrow \infty$. Similarly, $\lim _{t\rightarrow \infty }{\mathbb {E}}(X_{t}) = \infty$ and $\lim _{t\rightarrow \infty }{\mathbb {E}}(b(X_{t}))\rightarrow \infty$ due to non-explosion.

Proof

Notice that

$$\begin{aligned} \frac{J_{i,i+L}}{J_{i,i-1}} = \frac{\mu (1-q_{i+L})}{\nu (1-q_{i-1})} \ge \frac{1}{L}\frac{1-q_{i+L}}{1-q_{i-1}} > \frac{1}{L}. \end{aligned}$$

Therefore, $J_{i,i+L} > 1/(L+1)$ and $J_{i,i-1}<L/(L+1)$, which implies the expected increment of this random walk is positive for each state. In addition,

$$\begin{aligned} \lim _{i\rightarrow -\infty } J_{i,i+L} \ge \frac{\alpha ^{L+1}}{L+\alpha ^{L+1}} > \frac{1}{L+1}; \lim _{i\rightarrow -\infty } J_{i,i-1} \le \frac{L}{L+\alpha ^{L+1}}< \frac{L}{L+1}. \end{aligned}$$

This implies for all $S \in {\mathbb {Z}}$, the infimum of expected increments to the left of S is strictly positive, that is,

$$\begin{aligned} \inf _{i< S} \{LJ_{i,i+L} - J_{i,i-1}\} > 0. \end{aligned}$$

We prove $\lim _{n\rightarrow \infty }{\mathbb {E}}(Y_{n})=\infty$ by Fatou’s lemma. To show $\liminf _{n\rightarrow \infty }Y_{n} = \infty$, we impose an absorbing state $S>Y_{0}$ such that every state to the right of S collapses into state S. Denote the process with this absorbing barrier as $(Y^{(S)}_{n})$, then

$$\begin{aligned} \lim _{S\rightarrow \infty }\liminf _{n\rightarrow \infty }Y^{(S)}_{n}&= \sup _{S>0}\sup _{N \ge 1}\inf _{n\ge N}Y^{(S)}_{n} \\&= \sup _{N \ge 1}\sup _{S>0}\inf _{n\ge N}Y^{(S)}_{n} \\&\le \sup _{N \ge 1}\inf _{n\ge N}\sup _{S>0}Y^{(S)}_{n} \\&= \sup _{N \ge 1}\inf _{n\ge N}Y_{n} \\&= \liminf _{n\rightarrow \infty }Y_{n}. \end{aligned}$$

Since the truncated process $(Y^{(S)}_{n})$ is a random walk with an upper absorbing barrier whose expected increments are greater than a positive constant, $\liminf _{n\rightarrow \infty }Y^{(S)}_{n} = S$ almost surely. Consequently, $\liminf _{n\rightarrow \infty }Y_{n} = \infty$ and we conclude by Fatou’s lemma that ${\mathbb {E}}(Y_{n})\rightarrow \infty$ as $n\rightarrow \infty$. By convexity of $b(\cdot )$ and Jensen’s inequality, ${\mathbb {E}}(b(Y_{n})) \rightarrow \infty$ as $n\rightarrow \infty$.

Notice that $Y_{j(n)} = X_{n}$ where j(n) is the nth jump time of the continuous process. By non-explosion, $j(n) \rightarrow \infty$ as $n\rightarrow \infty$, which implies

$$\begin{aligned} \lim _{t\rightarrow \infty }{\mathbb {E}}(Y_{t}) = \infty \text { and } \lim _{t\rightarrow \infty }{\mathbb {E}}(b(Y_{t})) = \infty . \end{aligned}$$

$\square$

Theorems

Theorem 1

Under the condition of non-explosion in Lemma 1, $i> j$ implies $q_{i}\le q_{j}$, which further implies $\lim _{i\rightarrow \infty }q_{i} = 0$ and $\lim _{i\rightarrow -\infty }q_{i} = 1$.

Proof

Given the continuous-time process $({\textbf{Z}}_{t})$ is non-explosive, its extinction probability for the continuous-time process $({\textbf{Z}}_{t})$ is the same as that of the discrete embedded branching process $({\textbf{E}}_{n})$ by corollary of Lemma 1. We use a coupling argument to show the monotonicity of the extinction probabilities.

Let us fix $i > j$ and construct two continuous-time processes representing cancer populations with different initial types in the following way. Let superscript indicate the initial cell type, that is, ${\textbf{Z}}^{(j)}_0 = j$ and ${\textbf{Z}}^{(i)}_{0}=i$. The coupling of the process is described as follows.

At time 0, five exponentially distributed variables, B, U, V, D, and S are competing with each other with rates $b(j),\mu ,\nu ,d$ and $b(i)-b(j)$, respectively for $({\textbf{Z}}^{(i)}_{t})$ and the minimum of five random variables decides the actions of the initial type i cell. Similarly, there are four competing exponentially distributed random variables $B',U',V'$, and $D'$ with rates $b(j),\mu ,\nu$ and d for $({\textbf{Z}}^{(j)}_{t})$ at time 0. Actions of two initial cells are coupled by setting $B = B', U = U', V=V'$, and $D = D'$. That is,

• If the minimum is B, both initial type i cell and initial type j cell proliferate.

• If the minimum is U, both initial type i cell and initial type j cell acquire a driver mutation.

• If the minimum is V, both initial type i cell and initial type j cell acquire a passenger mutation.

• If the minimum is D, both initial type i cell and initial type j cell die.

• If the minimum is S, the initial type i cell proliferates and the initial type j cell takes no actions.

We further couple cells from two populations after the first action.

• If both initial cells proliferate, two new couples are formed, (i, j) and (i, j).

• If both initial cells acquire a driver mutation, the type $i+L$ cell is coupled with the type $j+L$ cell, forming a $(i+L,j+L)$ couple.

• If both initial cells acquire a passenger mutation, the type $i-1$ cell is coupled with the type $j-1$ cell, forming a $(i-1,j-1)$ couple.

• If both cells die, there will be no new couples.

• If the initial type i cell proliferates while the initial type j cell takes no action, one type i cell is coupled with the initial type j cell, forming a (i, j) couple.

Notice that in the last scenario, the uncoupled type i cell will evolve (proliferate, mutate, or die) freely, independent of the $\left( {\textbf{Z}}^{(j)}_{t}\right)$ population. Therefore, if we continue this construction, we have $||{\textbf{Z}}^{(i)}_{t}||_{\ell ^{1}} \ge ||{\textbf{Z}}^{(j)}_{t}||_{\ell ^{1}}$ for all $t \ge 0$. This implies $q_{i} \le q_{j}$ and it then follows that $\lim _{i\rightarrow \infty }q_{i}=q_{\infty }$ as well as $\lim _{i\rightarrow -\infty }q_{i}=q_{-\infty }$ exist and

$$\begin{aligned} q_{\infty }&= q_{\infty }^{2} \Rightarrow q_{\infty } \in \{0,1\} \\ q_{-\infty }&= \frac{d}{\mu +\nu +d}+\frac{\nu }{\mu +\nu +d}q_{-\infty }+\frac{\mu }{\mu +\nu +d}q_{-\infty } \Rightarrow q_{-\infty } = 1. \end{aligned}$$

If $q_{\infty } = 1$, we have $q_{i} = 0$ for all $i\in {\mathbb {Z}}$ and this contradicts the fact that all extinction probabilities are strictly less than one. Therefore, $\lim _{i\rightarrow \infty }q_{i} = 0$. $\square$

Theorem 2

When $\frac{\nu }{\mu L} \le 1$, $(Y_{n})$ and $(X_{t})$ are transient.

Proof

It suffices to show that there exists a $1-$subinvariant measure that is not invariant by Theorem 5.4 in Seneta (2006). We claim that ${\textbf{x}} = (\delta (n)(1-q_{n})(1-T_{n,n}))_{n\in {\mathbb {Z}}}$ is the desired measure.

$$\begin{aligned} {\textbf{x}}^{T}J \lneqq {\textbf{x}}^{T} \iff&\mu (1-q_{i})+\nu (1-q_{i}) \le \delta (i)(1-q_{i})(1-T_{i,i}) \\ \iff&\frac{\mu }{\delta (i)}+\frac{b(i)}{\delta (i)}\frac{1-q_{i}^{2}}{1-q_{i}}+\frac{\nu }{\delta (i)} \le 1 \\ \iff&b(i)q_{i} \le d \\ \iff&q_{i} \le \frac{d}{b(i)}. \end{aligned}$$

Equality cannot hold for all i since $\frac{d}{b(i)} > 1$ for small i. Define ${\textbf{q}}'= (\min (1,\frac{d}{b(i)}))_{i=-\infty }^{\infty }$. To show $q_{i} \le \frac{d}{b(i)}$, is suffices to show ${\textbf{q}}' \ge {\textbf{G}}({\textbf{q}}')$. To see this, notice that

$$\begin{aligned}&{\textbf{q}}' \ge {\textbf{G}}({\textbf{q}}') \Rightarrow {\textbf{G}}({\textbf{q}}') \ge {\textbf{G}}^{\circ 2}({\textbf{q}}') \Rightarrow \cdots \Rightarrow {\textbf{G}}^{\circ n}({\textbf{q}}') \ge {\textbf{G}}^{\circ (n+1)}({\textbf{q}}') \\&\quad \Rightarrow \;{\textbf{q}}' \ge \lim _{n\rightarrow \infty }{\textbf{G}}^{\circ n}({\textbf{q}}') \;\ge \; {\textbf{q}}, \text { since } {\textbf{q}} \text { is the minimal fixed point,} \end{aligned}$$

and ${\textbf{G}}^{\circ n}$ is the n-fold composition of ${\textbf{G}}$ with itself. Now, we verify ${\textbf{q}}' \ge {\textbf{G}}({\textbf{q}}')$. When all three quantities $\left( q'_{i-1},q'_{i},q'_{i+L}\right)$ are less than or equal to 1,

$$\begin{aligned} {\textbf{q}}' \ge {\textbf{G}}({\textbf{q}}') \iff&\frac{d}{b(i)} \ge \frac{d}{\delta (i)}+\frac{\nu }{\delta (i)}\frac{d}{b(i-1)}+\frac{b(i)}{\delta (i)}\left( \frac{d}{b(i)}\right) ^{2}+\frac{\mu }{\delta (i)}\frac{d}{b(i+L)} \\ \iff&d\delta (i) \ge db(i) + \nu d(1+s_{p}) + d + \mu d(1+s_{p})^{-L} \\ \iff&\mu \left( 1-(1+s_{p})^{-L}\right) \ge \nu s_{p}. \end{aligned}$$

Since $(1+s_{p})^{-L} \ge 1-Ls_{p}$, we have

$$\begin{aligned} \mu L&\ge \nu \Rightarrow \mu Ls_{p} \ge \nu s_{p} \\&\Rightarrow \mu Ls_{p} \ge \nu s_{p} \\&\Rightarrow \mu \left( 1-(1+s_{p})^{-L}\right) \ge \nu s_{p}. \end{aligned}$$

If $\frac{d}{b(i-1)} > 1$ and $\frac{d}{b(i)} \le 1$, by the same computation, we have

$$\begin{aligned} \frac{d}{b(i)}&\ge \frac{d}{\delta (i)}+\frac{\nu }{\delta (i)}\frac{d}{b(i-1)}+\frac{b(i)}{\delta (i)} \left( \frac{d}{b(i)}\right) ^{2}+\frac{\mu }{\delta (i)}\frac{d}{b(i+L)} \\&\ge \frac{d}{\delta (i)} +\frac{\nu }{\delta (i)}+\frac{b(i)}{\delta (i)}\left( \frac{d}{b(i)}\right) ^{2}+\frac{\mu }{\delta (i)}\frac{d}{b(i+L)} \\&=\frac{d}{\delta (i)} +\frac{\nu }{\delta (i)}q_{i-1}'+\frac{b(i)}{\delta (i)}q_{i}'^{2}+\frac{\mu }{\delta (i)}q_{i+L}'. \end{aligned}$$

If $\frac{d}{b(i-1)} > 1$ and $\frac{d}{b(i)} > 1$, $q'_{i} = q'_{i-1} = 1$ and the inequality holds trivially. Since the process is non-explosive, $(Y_{n})$ is transient implies $(X_{t})$ is transient as well (Theorem 3.4.1 in Norris 1998). $\square$

Theorem 3

When $\frac{\nu }{\mu L} > 1$, the jump chain $(Y_{n})$ is positive recurrent. Consequently, $(X_{t})$ admits a limiting distribution.

Proof

It is clear that the jump chain is irreducible with period $L+1$. To show it is positive recurrent, we prove by contradiction. Suppose the jump chain is transient or null recurrent, then by Theorem 1.8.5 in Norris (1998), for any $x \in {\mathbb {Z}}_{+}$,

$$\begin{aligned} \lim _{n\rightarrow \infty } {\mathbb {P}}(|Y_{n}|>x)=1-\lim _{n\rightarrow \infty } {\mathbb {P}}(|Y_{n}|\le x)= 1-\lim _{n\rightarrow \infty }\sum _{k=-x}^{x}{\mathbb {P}}(|Y_{n}|= k) = 1. \end{aligned}$$

To show it is not the case, we bound the right and left tail probabilities $(i\rightarrow \infty \text { and } i\rightarrow -\infty )$.

Observe that as $i\rightarrow \infty$, the jump chain tends to a random walk with

$$\begin{aligned} p_{i,i-1} = \frac{\nu }{\mu +\nu }>\frac{L}{L+1}, p_{i,i+L} = \frac{\mu }{\mu +\nu }<\frac{1}{L+1}. \end{aligned}$$

Therefore, there exists $I_{r}>0$ such that for all $i \ge I_{r}$, $p_{i,i-1} > q \in \left( \frac{L}{L+1},\frac{\nu }{\mu +\nu }\right)$ and $p_{i,i+L} < p \in \left( \frac{\mu }{\mu +\nu },\frac{1}{L+1}\right)$ and $p+q = 1$. Let $(R_{n})$ be a random walk with a retaining barrier starting from $R_{0}=I_{r}$ and for $i\ge I_{r}$,

$$\begin{aligned} p_{i,\max \{I_{r},i-1\}} = q, p_{i,i+L} = p. \end{aligned}$$

This random walk is more likely to increase than the original process for $i \ge I_{r}$. Since this random walk is more likely to decrease by L than increase by L, $(R_{n})$ is positive recurrent. To see this,

$$\begin{aligned} q^{L} \ge \left( \frac{L}{L+1}\right) ^{L} > \frac{1}{L+1} \ge p. \end{aligned}$$

As already stated, by Lemma 2, as $i \rightarrow -\infty$, the jump chain approaches a random walk with

$$\begin{aligned} p_{i, i-1} = \frac{\nu }{\mu +\nu +d}\alpha ^{-1} < \frac{L}{L+1}, p_{i,i+L} = \frac{\mu }{\mu +\nu +d}\alpha ^{L} > \frac{1}{L+1}. \end{aligned}$$

Again, there exists $I_{l}<0$ such that for all $j\le I_{l}$, $p_{j, j-1} < q \in (\frac{\nu }{\mu +\nu +d}\alpha ^{-1},\frac{L}{L+1})$ and $p_{j,j+L} > p \in (\frac{1}{L+1},\frac{\mu }{\mu +\nu +d}\alpha ^{L})$ and $p+q = 1$. Let $(L_{n})$ be a random walk starting from $I_{l}$ with a retaining barrier on the right, and take $q \in (\frac{\nu }{\mu +\nu +d}\alpha ^{-1},\frac{L}{L+1}), p \in (\frac{1}{L+1},\frac{\mu }{\mu +\nu +d}\alpha ^{L})$ such that for all $j \le I_{l}$,

$$\begin{aligned} p_{j,j-1} = q, p_{j,\min \{I_{l},j+L\}} = p. \end{aligned}$$

This is a left-continuous random walk with a retaining barrier whose expected increment $Lp-q$ is greater than 0. Therefore, by Spitzer (2001) Case (i) on page 188 and P1 on page 191, the process is positive-recurrent.

Fix $\epsilon \in (0,1)$; then there exist $u>I_{r}>0$ and $0>I_{l}>v$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }{\mathbb {P}}(R_{n}>u)< \frac{\epsilon }{2} \text { and } \lim _{n\rightarrow \infty }{\mathbb {P}}(L_{n}<v) < \frac{\epsilon }{2}. \end{aligned}$$

Since $(R_{n})$ (resp. $(L_{n})$) has a heavier right (resp. left) tail than the jump chain,

$$\begin{aligned} \limsup _{n\rightarrow \infty } {\mathbb {P}}(Y_{n}>u)< \frac{\epsilon }{2} \text { and } \limsup _{n\rightarrow \infty } {\mathbb {P}}(Y_{n}<v) < \frac{\epsilon }{2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \limsup _{n\rightarrow \infty } {\mathbb {P}}(|Y_{n}|>\max \{u,|v|\}) \le \limsup _{n\rightarrow \infty } {\mathbb {P}}(R_{n}>u) + \limsup _{n\rightarrow \infty } {\mathbb {P}}(L_{n}<v)< \epsilon <1. \end{aligned}$$

This is a contradiction, so the jump chain must be positive recurrent. Consequently, it admits a unique invariant distribution ${\textbf{y}}$. That is, for all $i\in {\mathbb {Z}}$,

$$\begin{aligned} y_{i-L}J_{i-L,i} + y_{i+1}J_{i+1,i} = y_{i} \text { and } \sum _{i\in {\mathbb {Z}}} y_{i} = 1. \end{aligned}$$

Recall that the holding time for state i follows an exponential distribution with $\delta (i)(1-T_{i,i})$, and we have

$$\begin{aligned}&{\left\{ \begin{array}{ll} \lim _{i\rightarrow \infty }\delta (i)(1-T_{i,i}) = \lim _{i\rightarrow \infty }\mu \frac{1-q_{i+L}}{1-q_{i}}+\nu \frac{1-q_{i-1}}{1-q_{i}} = \mu +\nu \\ \lim _{i\rightarrow -\infty }\delta (i)(1-T_{i,i}) = \mu \alpha ^{L}+\nu \alpha ^{-1} = \mu +\nu +d \end{array}\right. } \\&\quad \Rightarrow \inf _{i\in {\mathbb {Z}}}\{\delta (i)(1-T_{i,i})\} > 0. \end{aligned}$$

Since the infimum of the rates is strictly greater than 0, $y_{i}/(\delta (i)(1-T_{i,i}))$ a summable invariant measure for the generator Q, which concludes that $(X_{t})$ admits a limiting distribution (Theorem 3.6.2 of Norris 1998). $\square$

Discrete Version of Levinson’s Fundamental Theorem

Let $y(n)\in {\mathbb {C}}^{p}$ and $y(n+1) = [\Lambda (n)+R(n)]y(n)$, where $\Lambda (n) = \text {diag}(\lambda _{1}(n),\cdots ,\lambda _{p}(n))$. Suppose $\Lambda (n)$ is invertible for $n\ge n_{0}$ and following conditions are satisfied. Since any two matrix norms are equivalent, we use the operator norm, denoted $||\cdot ||$.

$$\begin{aligned}&\exists K_{1},K_{2} \text { s.t. } \forall i\ne j \text { either } {\left\{ \begin{array}{ll} &{}[\prod _{k=n_{0}}^{n}|\frac{\lambda _{j}(k)}{\lambda _{i}(k)}| \rightarrow 0 \text { as } n\rightarrow \infty \text { and } \\ &{}\qquad \prod _{k=n_{n_{1}}}^{n_{2}}|\frac{\lambda _{j}(k)}{\lambda _{i}(k)}| \le K_{1}, \forall n_{0}\le n_{1}\le n_{2}] \\ &{}\text {or } \prod _{k=n_{n_{1}}}^{n_{2}}|\frac{\lambda _{j}(k)}{\lambda _{i}(k)}| \ge K_{2}, \forall n_{0}\le n_{1}\le n_{2}, \end{array}\right. } \\&\text {and }\sup _{1\le i\le p}\sum _{n=n_{0}}^{\infty }\frac{||R(n)||}{|\lambda _{i}(n)|} < \infty . \end{aligned}$$

Then, the perturbed linear system $y(n+1) = [\Lambda (n)+R(n)]y(n)$ has a fundamental matrix satisfying

$$\begin{aligned} Y(n) = (I + o(1))\prod _{k=n_{0}}^{n-1}\Lambda (k) \text { as } n\rightarrow \infty . \end{aligned}$$

Columns of Y(n) are linearly independent solution vectors such that $y(n) = Y(n)y(0)$.