Spatial Moran models, II: cancer initiation in spatially structured tissue

Abstract

We study the accumulation and spread of advantageous mutations in a spatial stochastic model of cancer initiation on a lattice. The parameters of this general model can be tuned to study a variety of cancer types and genetic progression pathways. This investigation contributes to an understanding of how the selective advantage of cancer cells together with the rates of mutations driving cancer, impact the process and timing of carcinogenesis. These results can be used to give insights into tumor heterogeneity and the “cancer field effect,” the observation that a malignancy is often surrounded by cells that have undergone premalignant transformation.

Appendices

Appendix A: Construction and duality

To make the paper self-contained and to recall some facts that may not be widely known, we will now construct the two type biased voter model and explain its duality with coalescing branching random walk.

To construct the biased voter model, we follow the approach in Griffeath (1978). Associated with each order pair (x, y) of nearest neighbors, we have two Poisson processes, $T_n^{x,y,v}$, $n \ge 1$ and $T_n^{x,y,b}$, $n \ge 1$ with rates 1 / 2d and $(\lambda -1)/2d$; v is for voter and b is for birth. Here, all of the Poisson processes are independent, and together constitute the graphical representation. At each time $t=T^{x,y,v}$ we draw an arrow $(y,t) \rightarrow (x,t)$ and put a $\delta$ at (x, t), while at each time $t=T^{x,y,b}$ we draw an arrow $(y,t) \rightarrow (x,t)$. We think of arrows as little tubes that allow fluid to flow in the direction indicated, while the $\delta$’s are dams that stop the passage of the fluid. The $\delta$’s occur just before the arrows so they don’t block the fluid that flows through them.

Given an initial set A of sites that are occupied by 1’s, the set of sites that are occupied by 1’s at time t is the set $\xi ^A_t$ of points that are wet if fluid is injected at points of A at time 0. By checking cases, one can see that the effect of an arrow-$\delta$ from y to x is as follows:

Before	After
$x \quad \, y$	$x \quad \, y$
0    0	0    0
0    1	1    1
1    0	0    0
1    1	1    1

The first case should be clear. In the second the arrow spreads the fluid from y to x. In the third the $\delta$ at x stops the fluid, but there is nothing from y to replace it, while in the fourth case there is. Thus the arrow-$\delta$ produces a voter model step: x imitates y.

If we remove the $\delta$, then only the third line changes and the overall result is a birth from y to x, with the two particles coalescing to one if x is already occupied:

Before	After
$x \quad \, y$	$x \quad \, y$
0    0	0    0
0    1	1    1
1    0	1    0
1    1	1    1

The graphical representation has the useful property that it constructs the biased voter model for all initial conditions on the same probability space. As Harris (1976) noted, this implies that the constructed processes are additive:

$$\begin{aligned} \xi ^{A\cup B}_t = \xi ^A_t \cup \xi ^B_t. \end{aligned}$$

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since a space-time point can be reached from $A\cup B$ at time 0 if and only if it can be reached from A or from B. A consequence of additivity is that $A \rightarrow \xi ^A_t$ is increasing, a property that is called “attractive.”

An important reason for constructing a process from a graphical representation is that it allows us to construct a dual process. Let $\zeta ^{x,t}_r$ be the set of points at time $t-r$ that can be reached by a path starting from x at time t that goes down the graphical representation and crosses the arrows in the direction OPPOSITE their orientation. If we recall that the $\delta$’s occur just before the arrows on the way up then we see that the effect of an arrow-$\delta$ from y to x on the dual process is

Before	After
$x \quad \, y$	$x \quad \, y$
0    0	0    0
0    1	0    1
1    0	0    1
1    1	0    1

In words this is a coalescing random walk. If there is a particle at x (corresponding to a 1) it jumps to y. If there is also a particle at y the two coalesce to 1. It is easy to see that an arrow without a $\delta$ has the same effect in the dual as it did in the forward process except that now the birth is from x to y.

Given a set of sites B, let $\zeta ^{B,t}_s = \cup _{x\in B} \zeta ^{x,t}_s$. It is immediate from the definitions that

$$\begin{aligned} \{ \xi ^A_t \cap B \ne \emptyset \} = \{ \zeta ^{B,t}_t \cap A \ne \emptyset \}, \end{aligned}$$

i.e., the two events are equal. To get rid of the superscript t from the dual process, we note that if $t < t'$ then the distribution of $\zeta ^{B,t'}_r$ for $r\le t$ is the same as $\zeta ^{B,t}_r$ for $r\le t$. Invoking Kolmogorov’s extension theorem there is process $\zeta ^B_r$ defined for all time $r\ge 0$ that has the same distribution as $\zeta ^{B,t}_r$ for $r\le t$. This process satisfies

$$\begin{aligned} P( \xi ^A_t \cap B \ne \emptyset ) = P( \zeta ^B_t \cap A \ne \emptyset ). \end{aligned}$$

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In what follows, we will be interested in the biased voter model with mutation $0 \rightarrow 1$ at rate $u_1$. Mutation can be incorporated into the graphical representation by adding independent Poisson processes $T^{x,\mu }_n$, $n\ge 1$ with rate $u_1$. If we let $\hat{\xi }^A_t$ be the biased voter model with mutation starting with A occupied at time t, and suppose that there are mutations at $x_i$ at times $t_1 < t_2 < \cdots t_k < t$ then

$$\begin{aligned} \hat{\xi }_t^A = \xi _t^A \cup \xi ^{x_1,t_1}_t \cdots \cup \xi ^{x_k,t_k}_t \end{aligned}$$

where $\xi ^{x_i,t_i}_t$ is the biased voter model without mutation starting with $x_i$ occupied at time $t_i$.

In our proofs, it will be useful to be able to quantify the notion that two processes, $\xi _t^A$ and $\xi _t^B$ or $\xi ^{x_1,t_1}_t$ and $\xi ^{x_2,t_2}_t$ are independent when they don’t hit each other. To do this we use a coupling due to Griffeath (1978). We define the first process on a graphical representation, and the second on an independent graphical representation with the caveat that events in the second process that involve an edge (x, y) where x or y is occupied in the first process must use the first graphical representation, so that the pair of processes has the same joint distribution as if they were both defined on the same graphical representation.

Appendix B: Proof of Lemma 1

In this section we find a space-time box that the sub-critical biased voter model does not escape with high probability. To prove this in $d\ge 2$ we will use a result of Merle (2008), who studied the likelihood of the voter model to wander far from its starting point. The result extends easily to the subcritical biased voter model because it is dominated by the voter model.

Merle’s (2008) results are based on the work Bramson et al. (2001), who studied the ordinary voter model with kernel p(x, y). That is, voter at x changes opinions at rate 1, and imitates the one at y with probability p(x, y) where $p(x,y) = p(0,y-x)$ is irreducible and symmetric with $p(0,0)=0$ and $\sum _x p(0,x) x_i x_j = \sigma ^2 \delta (i,j)$. Here $\delta (i,j)=1$ if $i=j$ and 0 otherwise. To get a limit, they scale space so that the voters live on ${\mathbb {Z}}^d/\sqrt{n}$, run time at rate n and denote the resulting voter model by $\xi ^n_t$. Let $m_n = n/\log n$ in $d=2$ and n in $d \ge 3$, and define a measure valued process by

$$\begin{aligned} X^n_t = \frac{1}{m_n} \sum _{y \in \xi ^n_t} \delta _y. \end{aligned}$$

We write $X^{n,0}_t$ when the initial state is $\xi ^n_0 = \{0\}$. Let $\mathbf{D}$ be the space of functions from $[0,\infty )$ into the space of finite measures on ${\mathbb {R}}^d$ that are right-continuous and have left limits in the weak topology. Of particular interest is Theorem 4 from Bramson et al. (2001):

Theorem 5

Assume $d \ge 2$ and let $\mathbf{N}_0$ be the excursion measure of super-Brownian motion on ${\mathbb {R}}^d$ with branching rate $2\beta _d$ and diffusion coefficient $\sigma ^2$. Let $\alpha > 0$ and let F be a bounded continuous function on $\mathbf{D}$ with $F(\omega )=0$ if $\omega _t=0$ for $t \ge \alpha$. Then

$$\begin{aligned} \lim _{n\rightarrow \infty } m_n EF(X^{n,0}_\cdot ) = \mathbf{N}_0(F). \end{aligned}$$

The excursion measure is defined by starting the super process from $\epsilon \delta _0$, multiplying the probability measure by $1/\epsilon$ and letting $\epsilon \rightarrow 0$. See Section 3 of Bramson et al. (2001) and references in that paper for more details. This is the super-process analogue of starting Brownian motion at $\epsilon$, killing it when it hits 0, considering the limit of $(1/\epsilon )$ times the probability measure, which defines Ito’s excursion measure. See Chapter XII of Revuz and Yor (1991) for a thorough treatment. In most cases the killed Brownian motion $\bar{B}_t$ dies out quickly but when $\epsilon <1$

$$\begin{aligned} (1/\epsilon ) P_\epsilon (\max _t \bar{B}_t > 1 ) = 1. \end{aligned}$$

Theorem 5 tells us that a rescaled version of the voter model behaves like the excursion measure of a super Brownian motion. Since the super Brownian motion has compact support it makes sense that the voter model should not be able to run much farther than the spatial scaling used in Theorem 5. This result is made precise in Merle (2008) which we then extend to establish Lemma 1.

Proof

We will first establish the result for the voter model. In particular, define $T_0$ to be the extinction time of the voter model started from a single site on $\mathbb {Z}^d$, and $p(t)=P(T_0>t)$. Recall from Bramson and Griffeath (1980a) the large t asymptotics

$$\begin{aligned} p(t)\sim {\left\{ \begin{array}{ll} \frac{1}{\beta _dt},&{}\quad d\ge 3\\ \frac{1}{\pi }\frac{\log t}{t},&{}\quad d=2. \end{array}\right. } \end{aligned}$$

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Based on these asymptotics it is easy to verify that

$$\begin{aligned} p\left( M\ell (s)\right) \sim s/M, \end{aligned}$$

which establishes the claim with regards to survival time. In order to establish the claim with regard to escape from the large box we use Claim 1 of Merle 2008 (p828):

Claim

There exists a positive $K_0$ and $K_2$ such that for any $\alpha >1$ and any $A\ge 1$

$$\begin{aligned} P\left( \sup _{t\le 2\alpha }\sup _{x\in \xi _t^0}|x|>A\sqrt{\alpha }\right) \le K_0p(\alpha )\exp (-K_2A) \end{aligned}$$

where $\xi _t^0$ is a voter model started with a single seed at the origin.

Now consider

$$\begin{aligned}&P\biggl (\sup _{t>0} \sup _{x\in \xi _t^0}|x|>M\sqrt{\ell (s)}\biggr ) \\&\quad \le P\left( \sup _{t\le 2\sqrt{M}\ell (s)}\sup _{x\in \xi _t^0}|x|>M\sqrt{\ell (s)}\right) +p\left( 2\sqrt{M}\ell (s)\right) \\&\quad \le p\left( 2\sqrt{M}\ell (s)\right) \left( K_0e^{-K_2\sqrt{M}}+1\right) . \end{aligned}$$

The result then follows for the subcritical biased voter model by comparison.

Appendix C: Proof of Theorem 2

Proof

If each new mutant was placed on its own graphical representation then the number of successful type 1 mutations by time $t/Nu_1s$ would converge to a Poisson with mean t. Recalling the definition of successful mutation given in (4), we divide space-time

$$\begin{aligned} \mathcal {B}_{L,t} = (\mathbb {Z}\,\hbox {mod}\,L)^d\times [0,t/(Nu_1s)] \end{aligned}$$

into boxes with time length $T= M_s\ell (s)$ and spatial side length $R = M_s\ell (s)^{1/2}$ where $M_s\rightarrow \infty$ sufficiently slowly such that, see (A0) and (A0’),

$$\begin{aligned}&u_1 R^d T = u_1\ell (s)^{(d+2)/2}M_s^{d+1}\rightarrow 0\\&K_s =L/R = L/(\ell (s)^{1/2}M_s)\hbox { is an integer and }K_s\rightarrow \infty . \end{aligned}$$

For $m \in {\mathbb {Z}}^d$ with $0\le m_i < K_s$ for $1\le i \le d$, and integers $n\ge 0$ let $B_{m,n}$ be the box with spatial side length R, time length T and lower left corner at Rm, Tn. Let $\bar{B}_{m,n}$ be the space time cube consisting of the $5^{d+1}$ boxes closest to $B_{m,n}$. If we have mutation in $B_{m,n}$ and another not in $\bar{B}_{m,n}$ then the success of these two mutations are independent events.

The expected number of successful mutations that occur in a box $B_{m,n}$ in $\mathcal {B}_{L,t}$, along with another mutation (successful or not) in $\bar{B}_{m,n}$ is

$$\begin{aligned}&\left( \frac{L}{R}\right) ^d \cdot \frac{t/Nu_1s}{T} \cdot (R^d T u_1 s) (5^d R^d Tu_1)\\&\quad = t 5^d R^d Tu_1 = t 5^d M_s^{d+1} \ell (s)^{d+2/2} \rightarrow 0 \end{aligned}$$

This shows that with high probability all the successful type 1 mutants act as if they are independent. Form this we see that the probability of no successful mutants in $\mathcal {B}_{L,t} \rightarrow e^{-t}$, which gives the desired result.

To prepare for later developments note that if we know that the fraction of the boxes occupied by successful type 1 mutations is small, the successful type 1 mutations will be a Poisson process.

Appendix D: Upper bounds on the subcritical BVM

In this lengthy section we bound the size of type 1 families that later die out. These results are needed for the proofs of Theorems 3 and 4 but the reader can skip them and still understand those proofs. The first step is to determine the effect on the process of conditioning it to die out.

Our first result relates unsuccessful type 1 clones to a subcritical biased voter process.

Lemma 2

Let $\xi ^0_t$ be the set of 1’s in a supercritical biased voter model with $\lambda =1+s$ on ${\mathbb {Z}}^d$ starting from $\xi ^0_0 = \{0\}$. Let $T_0$ be the time at which the process dies out. Let $\bar{\xi }^0_t$ be the biased voter model with $\bar{\xi }^0_0 = \{0\}$ and the roles of 1 and 0 interchanged, i.e., 1’s give birth at rate 1, and 0’s give birth at rate $\lambda$. Then

$$\begin{aligned} (\{ |\xi ^0_t| , t \le T_0\} \, \vert \, T_0 < \infty ) =_d \{|\bar{\xi }^0_t|, t \le T_0 \} \end{aligned}$$

Proof

If $\xi ^0_t = A$ with $|A|=k$ and $|\partial A| = \ell$ then $|\xi ^0_t|$ grows to size $k+1$ at rate $\lambda \ell$, and shrinks to size $k-1$ at rate $\ell$, so the transition probability of the embedded discrete time chain is

$$\begin{aligned} p(k,k+1) = \frac{\lambda }{1+\lambda } \qquad p(k,k-1) = \frac{1}{1+\lambda } \end{aligned}$$

(26)

If we let $\varphi (x)=\lambda ^{-x}$ then it is easy to check that

$$\begin{aligned} \varphi (k) = p(k,k+1)\varphi (k+1) + p(k,k-1) \varphi (k-1) \end{aligned}$$

hence if $a<x<b$ then

$$\begin{aligned} P_x(T_a<T_b) = \frac{\varphi (b)-\varphi (x)}{\varphi (b)-\varphi (a)} \qquad P_x(T_b<T_a) = \frac{\varphi (x)-\varphi (a)}{\varphi (b)-\varphi (a)} \end{aligned}$$

(27)

Let $a=0$, $x=1$, and $b\rightarrow \infty$ in the first formula

$$\begin{aligned} P_k(T_0<\infty ) = \lambda ^{-k}. \end{aligned}$$

(28)

If we condition a random walk with positive drift to hit 0 then the conditioned process has transition probability

$$\begin{aligned} {\bar{p}}(k,k+1) = \frac{p(k,k+1)\varphi (k+1)}{\varphi (k)} = \frac{1}{1+\lambda } \qquad {\bar{p}}(k,k-1) = \frac{\lambda }{1+\lambda } \end{aligned}$$

(29)

In words, the result is a random walk with the probabilities of up and down interchanged. Conditioning $\xi _t$ to hit 0 does not change the exponential holding times, and desired result follows.

Next we need to relate the behavior of the subcritical biased voter process, $\bar{\xi }^0_t$ to the behavior of the unbiased voter process (i.e., $s=0$ ) denoted by $\hat{\xi }^0_t$.

Lemma 3

Let $P_s$ be the law of the subcritical biased voter process $\bar{\xi }^0_t$ and let $P_0$ be the law of an unbiased voter process run at a sped up rate, so that the rate of events is $(2+s)/2d$ on each edge between a type 0 and type 1. The Radon-Nikodym derivative between these two measures until n steps is given by:

$$\begin{aligned} \frac{dP_s}{dP_0} = \left( \frac{2}{2+s} \right) ^{n_+-n_-} \left( \frac{4+4s}{4+4s+s^2} \right) ^{n_-} , \end{aligned}$$

(30)

where $n_+$ and $n_-$ are the number of jumps up and down respectively, and $n_-+n_+=n$.

Proof

Under $P_0$, the law of the voter model dictates that a type 1 next to a type 0 forces it to flip at rate $(1+s/2)/2d$ and a type 0 next to a type 1 also forces it to flip at rate $(1+s/2)/2d$. Under $P_s$, the law of the subcritical voter model dictates that a type 1 next to a type 0 forces it to flip at rate 1 / 2d and a type 0 next to a type 1 forces it to flip at rate $(1+s)/2d$. Thus when we look at the Radon-Nikodym derivative it will only depend on the embedded chain.

$$\begin{aligned} \frac{dP_s}{dP_0} = \left( \frac{2}{2+s} \right) ^{n_+} \left( \frac{2+2s}{2+s} \right) ^{n_-} \end{aligned}$$

(31)

where $n_+$ is the number of up jumps and $n_-$ is the number of down jumps. Rewriting this gives the statement of the Lemma.

For what follows it will be useful to establish a bound on the Radon-Nikodym derivative in the previous lemma.

Lemma 4

If $|\bar{\xi }_0|=O(1/s)$ as $s\rightarrow 0$ then $dP_s/dP_0=O(1)$ as $s\rightarrow 0$.

Proof

For ease of notation define $n_0=|\bar{\xi }_0|$. Since 0 is an absorbing state for the stochastic process $|\bar{\xi }_t|$ we know that $n_+-n_-\ge -(n_0+1)$. Rearranging the result in Lemma 3 and then applying the inequality $n_-\le n_++(n_0+1)$ we get that

$$\begin{aligned} \frac{dP_s}{dP_0}&=\left( 1+\frac{s}{2+s}\right) ^{n_-}\left( 1-\frac{s}{2+s}\right) ^{n_+}\\&\le \left[ \left( 1-\frac{s}{2+s}\right) \left( 1+\frac{s}{2+s}\right) \right] ^{n_+}\left( 1+\frac{s}{2+s}\right) ^{n_0+1}\\&\le \left( 1+\frac{s}{2+s}\right) ^{n_0+1}\\&= \exp \left[ (n_0+1)\log (1+s/(2+s))\right] \le \exp \left[ s(n_0+1)/(2+s)\right] =O(1). \end{aligned}$$

and we have proved the desired result. $\square$

Lemma 4 will allow us to derive results for the subcritical biased voter model from those for the ordinary voter model. The size of the voter model, when $|\xi ^0_t|>0$, is a time change of symmetric simple random walk, with jumps happening at two times the size of the boundary $|\partial \xi ^0_t|$, which is the number of nearest neighbor pairs with $x \in \xi ^0_t$ and $y\not \in \xi ^0_t$. The one dimensional case is easy because when $\xi ^0_t \ne \emptyset$ the boundary $|\partial \xi ^0_t|=2$. The key to the study of the process in $d \ge 2$ is the observation that there are constants $\beta _d$ so that

$$\begin{aligned} |\partial \xi ^0_t| \sim _p {\left\{ \begin{array}{ll} 2d\beta _d |\xi ^0_t| &{}\quad d\ge 3, \\ 4\beta _2 |\xi ^0_t|/\log (|\xi ^0_t|) &{}\quad d=2. \end{array}\right. } \end{aligned}$$

(32)

where $|\partial \xi ^0_t| \sim _p f(|\xi ^0_t|)$ means that when $|\xi ^0_t|$ is large, $|\partial \xi ^0_t|/f(|\xi ^0_t|)$ is close to 1 with high probability.

The intuition behind this result is that the voter model is dual to a collection of coalescing random walks, so in $d \ge 3$ neighbors of points in $\xi ^0_t$ will be unoccupied with probability $\approx \beta _d$, the probability two simple random walks started at 0 and $e_1=(1,0,\ldots 0)$ never hit. In dimension $d=2$, the recurrence of random walks implies that when $|\xi ^0_t|=k$ is large, most neighbors of points in $\xi ^0_t$ will be occupied, but due to the fat tail of the recurrence time sites will be vacant with probability $\sim \beta _2/\log k$, where $\beta _2=\pi$.

Let $V'_{n,s}(x)$ be the fraction of sites adjacent to x in state 0 at time s (with the prime indicating that we multiply this by $\log n$ in $d=2$, see page 196). A key step in the proof in Cox et al. (2000) is to show, see (I1) on page 202, that for nice test functions $\phi$

$$\begin{aligned} E \left[ \left( \int _0^T X^n_s( \{ V'_{n,s} - \beta _d \} \phi ^2) \, ds \right) ^{2} \right] \rightarrow 0 \end{aligned}$$

(33)

where $X^n_s(f)$ denote the integral of the function f against the measure $X^n_s$. The result in (33) shows that when we integrate in time and average in space (multiplying by a test function to localize the average) then (32) is true.

If one wants to give a rigorous proof of the estimates in this section, then small values of k, can be treated with the inequalities

$$\begin{aligned} C k^{(d-1)/d} \le |\partial \xi ^0_t| \le 2dk, \end{aligned}$$

and one can control large values of k using (33) and estimates such as (J1) and (J2) on page 208 of Cox et al. (2000). In order to avoid getting bogged down in technicalities we will suppose that

Assumption 1

Let k be a non-negative integer. If $|\bar{\xi }_t|=k$ then $|\partial \bar{\xi }_t|=\partial (k)$ where

$$\begin{aligned} \partial (k) = {\left\{ \begin{array}{ll} 2d\beta _d k &{}\quad d\ge 3 \\ 4\beta _2 k/\log k &{}\quad d=2. \end{array}\right. } \end{aligned}$$

(34)

Using Lemma 2 and Assumption 1, we can approximate the space-time volume of unsuccessful type 1 mutants. To this end we generalize the subcritical biased voter model defined in Lemma 2 to have arbitrary initial starting state and denote this process by $\bar{\xi }_t$. Furthermore, we denote the expectation operator conditioned on $|\bar{\xi }_0|=k$ by $E_k$, i.e. $E_k[\cdot ]\doteq E[\cdot \big | |\bar{\xi }_0|=k]$. Then we can prove the following upper bound on the space-time volume of unsuccessful type 1 mutants. Here and in other proofs in this section we will sometimes suppose in addition to $\lambda > 1$ that $\lambda \le 2$ in order to get rid of $\lambda$’s from the formulas.

Lemma 5

There is a positive constant C that depends only on d such that

$$\begin{aligned} E_1\left( \int _0^{T_0} |\bar{\xi }_t| \,dt \right) \le C\ell (s) \end{aligned}$$

where $\ell (s)$ was defined in (3)

Proof

Let $\bar{S}_n$ be the simple random walk that jumps according to ${\bar{p}}$ defined in (29) and note that $|\bar{\xi }_t|$ is a continuous time version of $\bar{S}_n$ that jumps approximately at rate $q(k)= (1+\lambda )\partial (k)$ when in state k and makes jumps according to ${\bar{p}}$. Let $T_k^+ = \min \{ n \ge 1 : S_n = k \}$. By considering the expected number of visits to k and the amount of time spent there on each one we have

$$\begin{aligned} E_1\left( \int _0^{T_0} |\bar{\xi }_t |\,dt \, \right) = \sum _{k=1}^\infty \frac{\bar{P}_1(T_k<T_0)}{\bar{P}_k(T_k^+ = \infty )} \cdot \frac{k}{q(k)} \end{aligned}$$

(35)

If $\bar{P}$ is the law of the conditioned chain with transition probability (29), and P is the law of the process with transition probability chain following (26) then by symmetry and (27),

$$\begin{aligned} \bar{P}_1(T_k<T_0) = P_{k-1}(T_0<T_k) = \frac{\lambda ^{-k} - \lambda ^{-(k-1)} }{ \lambda ^{-k} - 1 } = \frac{\lambda -1}{\lambda ^k-1}. \end{aligned}$$

(36)

Using symmetry and (27) again,

$$\begin{aligned} \bar{P}_k(T_k^+ = \infty )&= \frac{\lambda }{1+\lambda } \bar{P}_{k-1}(T_0<T_k) = \frac{\lambda }{1+\lambda } P_{1}(T_k<T_0)\nonumber \\&= \frac{\lambda }{1+\lambda } \cdot \frac{\lambda ^{-1}-1}{ \lambda ^{-k}-1 }= \frac{\lambda -1}{(1+\lambda )(1-\lambda ^{-k})}, \end{aligned}$$

(37)

so we have

$$\begin{aligned} \frac{\bar{P}_1(T_k<T_0)}{\bar{P}_k(T_k^+=\infty )} =\left( \frac{\lambda -1}{\lambda ^k-1}\right) \left( \frac{1+\lambda }{\lambda -1}\right) \lambda ^{-k}(\lambda ^{k}-1) = \lambda ^{-k}(1+\lambda ) \end{aligned}$$

and (35) becomes

$$\begin{aligned} E\left( \int _0^{T_0} |\bar{\xi }_t| \,dt \right) = (1+\lambda ) \sum _{k=1}^\infty \lambda ^{-k} \cdot \frac{k}{q(k)}. \end{aligned}$$

(38)

In one dimension $q(k)=2(1+\lambda )$, so doing some algebra and using the formula for the mean of the geometric distribution, two times the quantity in (38) is

$$\begin{aligned} \sum _{k=1}^\infty k \lambda ^{-k}&= \frac{1/\lambda }{(1-1/\lambda )} \sum _{k=1}^\infty k \lambda ^{-(k-1)}(1 - 1/\lambda )\nonumber \\&= \frac{1/\lambda }{(1-1/\lambda )^2} \le C s^{-2}. \end{aligned}$$

(39)

Since $q(k) \sim 2d (1+\lambda )\beta _d k$, in $d\ge 3$, so the quantity in (38) is

$$\begin{aligned} \frac{1}{2d\beta _d} \sum _{k=1}^\infty \lambda ^{-k} = \frac{1}{2d\beta _d} \times \frac{1/\lambda }{1-1/\lambda } \le Cs^{-1}. \end{aligned}$$

(40)

In $d=2$ we have $q(k) \sim 4(1+\lambda )\beta _2 k/\log k$ so the quantity in (38) is

$$\begin{aligned} \frac{1}{4\beta _2} \sum _{k=1}^\infty \lambda ^{-k} \log k \le Cs^{-1} \log (1/s). \end{aligned}$$

(41)

To see the last inequality note that for $k \le 1/s^2$, $\log k \le 2\log (1/s)$ and as $s\rightarrow 0$ we can ignore the contribution from $k > 1/s^2$.

Appendix E: Proof of Theorem 3

Proof

Let $A_t$ be the event that the first successful type 2 mutant comes from an unsuccessful type 1 family that arises before time $t/Nu_1s$. The expected number of such families is t / s. Using Lemma 5 to bound the space time volume covered by an unsuccessful mutation and ignoring the possibility that they overlap, we have

$$\begin{aligned} P(A_t) \le u_2s \cdot \frac{t}{s} \cdot C \ell (s) \rightarrow 0 \end{aligned}$$

(42)

by assumption (A3).

In view of Theorem 2, it suffices to show that $P(\sigma _2-\sigma _1 > \epsilon /Nu_1s) \rightarrow 0$ for any $\epsilon >0$. Since the expected number of successful type 1 mutations in a time interval of length $\epsilon /Nu_1s$ is $\le \epsilon$, this tells us that the possibility of a second successful type 1 mutation can also be ignored.

Under the simplified model, the total space-time volume occupied by descendants of the successful type 1 mutation up to time $\sigma _1+t$ is

$$\begin{aligned} g(t) = \frac{\gamma _d c_d^d t^{d+1}}{d+1}, \end{aligned}$$

assuming that at this time the diameter is $< L$. Let B(t) be the event that there is no successful type 2 mutation by time $\sigma _1+t$, and let

$$\begin{aligned} t_2 = \left( u_2 s \gamma _d c_d^d / (d+1)\right) ^{-1/(d+1)}. \end{aligned}$$

be chosen so that $g(t_2) = 1/u_2s$. Using the Poisson approximation

$$\begin{aligned} P( B(Mt_2) ) \sim \exp \left( - u_2s g(Mt_2) \right) \rightarrow \exp \left( - M^{d+1} \right) . \end{aligned}$$

This shows that if M is large, $\sigma _2 - \sigma _1 \le Mt_2$ with high probability. Now the diameter of the ball covered by the descendants of the successful type 1 at time $\sigma _1 + Mt_2$ is

$$\begin{aligned} O(c_d t_2) = O\left( (c_d / u_2 s)^{1/(d+1)}\right) \ll L \end{aligned}$$

by assumption (A1), so our computation of the volume is legitimate. Finally, the time

$$\begin{aligned} t_2 = O\left( (c_d^d u_2s)^{-1/(d+1)} \right) \ll 1/Nu_1s \end{aligned}$$

by assumption (A2). This shows that the time difference $\sigma _2-\sigma _1 = o(1/Nu_1s)$ and completes the proof.

Appendix F: Proof of Theorem 4

Proof

The number of unsuccessful mutations by time $t/Nu_1s$ is of order 1 / s, so as in the discussion of the previous theorem, (A3) implies that we can ignore the possibility that the successful type 2 comes from a type 1 family that dies out.

We next show that the probability of overlap between successful type 1 clones by time $t/(Nu_1s)$ is negligible. By time $t/(Nu_1s)$ the maximum possible radius of a successful type 1 clone is $r=(c_dt)/(Nu_1s)$. Let X and Y be two independent random variables distributed uniformly over the torus $({\mathbb {Z}}\,\hbox {mod}\,L)^d$, and note that it suffices to establish that

$$\begin{aligned} P(|X-Y|\le 2r)\ll 1 \end{aligned}$$

where $|\cdot |$ is the Euclidean norm. By rotating the torus we can suppose that $X=0$.

$$\begin{aligned} P(|Y|\le 2r)\le P(\Vert Y_1\Vert _\infty \le 2r)= (2r/L)^d. \end{aligned}$$

Then we use assumption (A1) to establish the desired result

$$\begin{aligned} r/L=\left( \frac{1}{N}\right) ^{1/d}\left( \frac{c_d}{Nu_1s}\right) =\frac{c_d}{N^{(d+1)/d}u_1s}\ll 1. \end{aligned}$$

Since we have shown that the fraction of the torus occupied by successful type 1 families is small, it follows from the proof of Theorem 2 that if we run time at rate $1/Nu_1s$ then type 1 mutations are a rate 1 Poisson process.

Denote the number of successful type 1 mutations to arrive by time t by M(t), and their arrival times by $T_1,\ldots ,$. Since we can know that overlaps occur with negligible probability we compute that

$$\begin{aligned} P(\sigma _2>t)=E\exp \left( -u_2s\sum _{j=1}^{M(t)}V_j(t)\right) , \end{aligned}$$

where

$$\begin{aligned} V_j(t)=\frac{\gamma _dc_d^d(s)(t-T_j)^{d+1}}{d+1}. \end{aligned}$$

We define

$$\begin{aligned} \phi (t)=\frac{1}{t}\int _0^{t}\exp \left( -\frac{u_2s\gamma _dc_d^d(s)}{d+1}\left( t-x\right) ^{d+1}\right) dx, \end{aligned}$$

so that $P(\sigma _2>t|M(t))=\left( \phi (t)\right) ^{M(t)}$. Next consider $t'=t/Nu_1s$, then

$$\begin{aligned} \phi (t')&=\frac{Nu_1s}{t}\int _0^{t/Nu_1s}\exp \left( -\frac{u_2s\gamma _dc_d^d(s)}{d+1}\left( t/Nu_1s-x\right) ^{d+1}\right) dx\\&=\frac{1}{t}\int _0^{t}\exp \left( -\frac{u_2s\gamma _dc_d^d(s)}{(d+1)(Nu_1s)^{d+1}}(t-x)^{d+1}\right) dx\\&\rightarrow \frac{1}{t}\int _0^t\exp \left[ -\frac{\gamma _d}{(d+1)I}y^{d+1}\right] dy\doteq \hat{\phi }(t). \end{aligned}$$

The limit in the previous display comes from the fact that $1/\varGamma \rightarrow 1/I$ as $s\rightarrow 0$.

Note that $M(t')$ is a Poisson random variable with mean t, so

$$\begin{aligned} P(\sigma _2>t')=\exp \left[ t(\phi (t')-1)\right] \rightarrow \exp \left[ t\left( \hat{\phi }(t)-1\right) \right] \end{aligned}$$

as $s\rightarrow 0$. $\square$

Appendix G: Proof of Theorem 1

Convergence to Branching Brownian Motion. This part of the proof is from Durrett and Zähle (1997). To make it easier to compare with their computation, we use their time scale which is 2d times as fast as the one used in this paper. When we are done we have to divide the speeds computed here by 2d. Let $\zeta _t$ be the coalescing random walk in which:

1. Particles jump at rate 2d to a randomly chosen neighboring site.

2. Particles give birth at rate s to a particle sent to a randomly chosen neighboring site.

3. If a particle lands on an occupied site (due to jump or a birth) then the two coalesce to 1.

If we let $\zeta ^B_t$ be the system starting with $\zeta ^B_0=B$ and let $\xi ^A_t$ be the biased voter model starting from $\xi ^A_0=A$ then the two systems satisfy the duality equation (24):

$$\begin{aligned} P( \xi ^A_t \cap B \ne \emptyset ) = P( \zeta ^B_t \cap A \ne \emptyset ). \end{aligned}$$

In $d \ge 3$ random walks are transient, so there is positive probability $\beta _d$ that an offspring will never coalesce with its parent. Durrett and Zähle (1997) show that in this case if time is run at rate 1 / s, and space is scaled by $1/\sqrt{s}$, the coalescing random walk converges to a branching Brownian motion $\chi _t$ in which

• 1. Particles perform independent Brownian motions run at rate 2

• 2. and give birth to new particles at rate $\beta _d$.

In order to achieve weak convergence they have to remove the particles that coalesce with their parents, because these result in temporary increases of the population that last (on the sped up time scale) for times of order s. To do this, they ignore the new born particles for time $\tau (s) = \sqrt{s}$ before we assign them mass 1.

In $d=2$ random walks are recurrent but the probability an offspring does not coalesce with its parent for time $>$ t is

$$\begin{aligned} \sim \pi /(\log t) \end{aligned}$$

(43)

see e.g., Lemma 3.1 of Zähle, Cox and Durrett (2005). To compensate for the fact that most particles coalesce with their parents, they run time at rate $h(s) = (1/s) \log (1/s)$ and scale space by $\sqrt{h(s)}$. Furthermore we ignore the new born particles for time $\tau (s) = 1/\sqrt{\log (1/s)}$ (on the sped up time scale) before we assign them mass 1. Note that on the sped-up time scale we create new particles at rate $sh(s)=\log (1/s)$, and we assign mass to only the fraction of those that survive for $\tau (s)$ units of time, which from (43) is $\sim \pi /\log (1/s)$. Therefore we assign mass to new particles at O(1) rate. Based on this the rescaled coalescing random walk converges to a branching Brownian motion $\chi _t$ in which

• 1. Particles perform independent Brownian motions run at rate 2

• 2. Give birth to new particles at rate $\pi$.

At this point if one ignores the detail of interchanging two limits Theorem 2 is obvious. If particles are born at rate

$$\begin{aligned} \rho = {\left\{ \begin{array}{ll} \beta _d &{}\quad d \ge 3 \\ \pi &{}\quad d = 2 \end{array}\right. } \end{aligned}$$

and perform Brownian motions with covariance matrix $\sigma ^2I$ then the mean measure at time t for the process started with a single particle at 0 at time 0 is

$$\begin{aligned} e^{wt} \frac{1}{(2\pi t)^{d/2}} e^{-|x|^2/2\sigma ^2 t} \end{aligned}$$

(44)

Ignoring the polynomial this is 1 when

$$\begin{aligned} |x| = t\sqrt{2\sigma ^2 w} \end{aligned}$$

Taking into account that time is run at rate 1 / s in $d\ge 3$ and at rate $(1/s)\log (1/s)$ in $d=2$, the desired result follows.

The lower bound on the speed follows from a block construction. The argument is almost the same as in Durrett and Zähle (2007), but we have to change some details to get a sharp result. Let $I=[-1,1]^d$, let $e_1$ be the first unit vector, let $v< \sqrt{4\rho }$, and for each m let $I_m = 2m(Lv)e_1 + I$. Let $\hat{\chi }_t$ be a modification of the branching Brownian motion in which particles are killed when they land outside $[-4Lv,4Lv]^d$. Calculations on page 1760 of Durrett and Zähle (2007) show that for any $\epsilon >0$, we can pick L large and then K large enough so that if there are at least K particles in $I_0$ in $\hat{\zeta }_0$ then with probability $\ge 1 - \epsilon$ we have $|\hat{\zeta }(L^2) \cap I_{1}| \ge K$ and $|\hat{\zeta }(L^2) \cap I_{-1}| \ge K$. Picking L large makes the mean of $|\hat{\zeta }(L^2) \cap I_{1}|$ large because of (44). Then picking K large controls the probability of deviations from the mean.

For integers $m \ge 0$ and n with $m+n$ even let $\theta (m,n)=1$ if $|\hat{\chi }(nL^2) \cap I_{m}| \ge K$. The result in the previous paragraph implies that $\chi$ dominates 1-dependent oriented percolation with density $1-\epsilon$. Let $r_n = \sup \{ m : \theta (m,n) = 1 \}$. A result in Durrett (1984), see (2) on page 1030, implies that if $\delta >0$ and $\epsilon < \epsilon (\delta )$ then on the set where the oriented percolation does not die out, $\liminf _{n\rightarrow \infty } r_n/n \ge 1-\delta$. This implies that for the rescaled process the edge speed is $\ge (1-\delta ) v$, which gives the lower bound.

The upper bound is proved by comparing the dual process on its original time scale with the branching process. Suppose first that $d\ge 3$. If we ignore newborn particles that will coalesce with their parents then we have a branching process in which particles are born at rate $w=\beta _ds$. We ignore coalescence events other than mother-daughter so we can project onto the x-axis to get a one-dimensional branching process $Z_t$ at time t. The number of particles to the right of ct at time t is

$$\begin{aligned} EZ_t(ct,\infty ) = e^{w t} P(S_t \ge ct) \end{aligned}$$

(45)

where $S_t$ is a random walks that takes steps that are $\pm 1$ with equal probability at rate 2. The steps have moment generating function

$$\begin{aligned} \phi (\theta ) = \frac{e^\theta + e^{-\theta }}{2} \end{aligned}$$

so the continuous time walk has

$$\begin{aligned} \psi _t(\theta ) = E\exp (\theta S_t) = \sum _{n=0}^\infty e^{-2t} \frac{[2t]^n}{n!} \phi (\theta )^n = \exp \left( 2 t(\phi (\theta )-1) \right) . \end{aligned}$$

As $\theta \rightarrow 0$, $\phi (\theta ) - 1 \sim \theta ^2/2$, so if $\delta >0$ and $0< \theta < \theta _0(\delta )$

$$\begin{aligned} \phi (\theta ) - 1 \le \frac{\theta ^2}{2-\delta }. \end{aligned}$$

Using Markov’s inequality,

$$\begin{aligned} P( S_t \ge (a+b)t ) \le e^{-\theta _a (a+b)t} \psi _t(\theta _a) \le \exp \left( - t \left[ \theta _a (a+b) - \frac{2\theta _a^2}{2-\delta } \right] \right) \end{aligned}$$

Changing notation $\theta = a/2$, the right-hand side is

$$\begin{aligned} = \exp \left( - t \left[ \frac{a(a+b)}{2} - \frac{a^2}{2(2-\delta )} \right] \right) = \exp \left( - t \left[ \frac{ab}{2} + \frac{a^2(1-\delta )}{2(2-\delta )} \right] \right) \end{aligned}$$

If we let $a = \sqrt{ 2w(2-\delta )/(1-2\delta )}$, which is just a little larger than $\sqrt{4w}$ then the above is

$$\begin{aligned} \le \exp \left( - t \left[ \frac{1-\delta }{1-2\delta } w + \frac{ab}{2} \right] \right) \end{aligned}$$

Using (45)

$$\begin{aligned} P( Z_t((a+b)t,\infty ))>0 ) \le \exp \left( - t \left[ \frac{\delta w}{1-2\delta } + \frac{ab}{2} \right] \right) . \end{aligned}$$

To bound the spread of the biased voter model, let $\Lambda = {\mathbb {Z}}^d - [-at,at]^d$, By duality

$$\begin{aligned} P( \xi _t^0 \cap \, [-at,at]^d \ne \emptyset )&= P( 0 \in \zeta ^\Lambda _t ) \\&\le \sum _{x\in \Lambda } P( 0 \in \zeta ^x_t) \le \sum _{k=at}^\infty c_d k^{d-1} P( Z_t(k,\infty ) >0 ) \le e^{-ct}. \end{aligned}$$

To prove the result in $d=2$ we will compare the dual process with a branching random walk, where if multiple offspring land on one site they are all retained. Modifying the construction of Durrett and Zähle (2007) we ignore new born particles for time 1 / s, and add them to the dual only if they have not collided with their parents. Let $Z_t$ denote the number of particles in the modified branching random walk by time t and let $m(t)=E[Z_t]$. From the result (43) we know that the fraction of newly created particles that are eventually added to the process is $\sim \pi /\log (1/s)$. Thus for $t>1/s$ we have

$$\begin{aligned} m(t)\le \exp \left( \frac{\pi s}{\log (1/s)}t\right) . \end{aligned}$$

If we set $w = \pi s/\log (1/s)$ and repeat the calculation for the case $d\ge 3$, we obtain the desired result in $d=2$. $\square$