We observe that $Y^H$$Y^H$Y^(H)Y^{H} and ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} are constant processes until time $R$$R$RR, that $R>0$$R>0$R > 0R>0 a.s. and that $T_a\ne T_b$$T_a\ne T_b$T_(a)!=T_(b)T_{a} \neq T_{b} a.s.
我们观察到 $Y^H$$Y^H$Y^(H)Y^{H} 和 ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} 是恒定的过程，直到时间 $R$$R$RR ，那个 $R>0$$R>0$R > 0R>0 a.s. 和那个 $T_a\ne T_b$$T_a\ne T_b$T_(a)!=T_(b)T_{a} \neq T_{b} a.s.

One of three things can happen at time $R$$R$RR. Firstly, if $R=T_a$$R=T_a$R=T_(a)R=T_{a}, there is a reproduction event in ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} but not in $Y^H$$Y^H$Y^(H)Y^{H}. If we rank in a non-increasing order these new particles, they again satisfy the partial ordering. Moreover, as $H_R\le K_R$$H_R\le K_R$H_(R) <= K_(R)H_{R} \leq K_{R}, applying the selection procedure to both models preserves this partial ordering, therefore
以下三种情况之一可能发生在 time $R$$R$RR .首先，如果 ，则 $R=T_a$$R=T_a$R=T_(a)R=T_{a} 中存在 ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} 繁殖事件，但在 中没有 $Y^H$$Y^H$Y^(H)Y^{H} 。如果我们以非递增顺序对这些新粒子进行排序，它们将再次满足部分排序。此外，由于 $H_R\le K_R$$H_R\le K_R$H_(R) <= K_(R)H_{R} \leq K_{R} ，将选择过程应用于两个模型会保留此部分排序，因此

If $R=T_b$$R=T_b$R=T_(b)R=T_{b}, then there is a reproduction event in $Y^H$$Y^H$Y^(H)Y^{H} and ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K}. We use the same point process to construct the child of the particle that reproduces in each process. Once again, ranking in a non-increasing order these new particles, then applying the selection, we have
如果 ，则 $R=T_b$$R=T_b$R=T_(b)R=T_{b} 中存在 和 ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} 中的 $Y^H$$Y^H$Y^(H)Y^{H} 重现事件。我们使用相同的 point 过程来构建在每个过程中再现的粒子的子对象。再一次，以非递增的顺序对这些新粒子进行排序，然后应用选择，我们有

Finally, if $R=S\notin \{T_a,T_b\}$$R=S\notin \left\{T_a,T_b\right\}$R=S!in{T_(a),T_(b)}R=S \notin\left\{T_{a}, T_{b}\right\}, the maximal size of at least one of the populations is modified. Even if this implies the death of some particles in $Y^H$$Y^H$Y^(H)Y^{H} and/or ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K}, the property (7.3) is preserved at time $R$$R$RR.
最后，如果 ，则 $R=S\notin \{T_a,T_b\}$$R=S\notin \left\{T_a,T_b\right\}$R=S!in{T_(a),T_(b)}R=S \notin\left\{T_{a}, T_{b}\right\} 至少修改一个总体的最大大小。即使这意味着 and/or ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} 中的 $Y^H$$Y^H$Y^(H)Y^{H} 某些粒子死亡，属性 （7.3） 也会在 time $R$$R$RR 保留。

Now fix $t>0$$t>0$t > 0t>0 and assume that $H_s\le K_s$$H_s\le K_s$H_(s) <= K_(s)H_{s} \leq K_{s} for every $0\le s\le t$$0\le s\le t$0 <= s <= t0 \leq s \leq t. As $H$$H$HH and $K$$K$KK are integer-valued càdlàg processes, they attain their maxima on compact sets. Therefore, they are both a.s. finite on the interval $[0,t]$$[0,t]$[0,t][0, t], so the number of particles is a.s. finite in both processes $Y^H$$Y^H$Y^(H)Y^{H} and ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K}. Thus there is a.s. a finite sequence of times $\left(R_k\right)$$\left(R_k\right)$(R_(k))\left(R_{k}\right) smaller than $t$$t$tt such that $Y^H$$Y^H$Y^(H)Y^{H} or ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} is modified at each time $R_k$$R_k$R_(k)R_{k}. Using this coupling on each time interval of the form $[R_k,R_{k+1}]$$\left[R_k,R_{k+1}\right]$[R_(k),R_(k+1)]\left[R_{k}, R_{k+1}\right] yields (7.3).
现在修复 $t>0$$t>0$t > 0t>0 并假设 $H_s\le K_s$$H_s\le K_s$H_(s) <= K_(s)H_{s} \leq K_{s} 对于每个 $0\le s\le t$$0\le s\le t$0 <= s <= t0 \leq s \leq t .由于 $H$$H$HH 和 $K$$K$KK 是整数值 càdlàg 进程，它们在紧集上达到最大值。因此，它们在区间 $[0,t]$$[0,t]$[0,t][0, t] 上都是 a.s. 有限的，因此粒子的数量在过程 $Y^H$$Y^H$Y^(H)Y^{H} 和 ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} 中都是 a.s. 有限的。因此，存在一个小于 $t$$t$tt 或 $Y^H$$Y^H$Y^(H)Y^{H} ${\tilde{Y}}^K$${\tilde{Y}}^K$widetilde(Y)^(K)\widetilde{Y}^{K} 在每次 时被修改的有限时间 $\left(R_k\right)$$\left(R_k\right)$(R_(k))\left(R_{k}\right) 序列 $R_k$$R_k$R_(k)R_{k} 。在形式的 $[R_k,R_{k+1}]$$\left[R_k,R_{k+1}\right]$[R_(k),R_(k+1)]\left[R_{k}, R_{k+1}\right] 每个时间间隔上使用此耦合可生成 （7.3）。

Using this lemma, we can prove that the cloud of particles in a $k$$k$kk-BRW drifts at linear speed $c_k$$c_k$c_(k)c_{k}. Note that by the coupling described in the proof of Lemma 7.1, this result can be obtained as a consequence of Theorem 1.1. However, we believe the following proof to be of independent interest, as it can be generalized to more diverse continuous-time branching random walks with selection.
使用这个引理，我们可以证明 $k$$k$kk -BRW 中的粒子云以线性速度 $c_k$$c_k$c_(k)c_{k} 漂移。请注意，通过引理 7.1 证明中描述的耦合，这个结果可以作为定理 1.1 的结果得到。然而，我们认为以下证明具有独立的兴趣，因为它可以推广到更多样化的连续时间分支随机游走。

Lemma 7.5. For any $k\in \mathbb{N}$$k\in \mathbb{N}$k inNk \in \mathbb{N}, there exists $c_k\in \mathbb{R}$$c_k\in \mathbb{R}$c_(k)inRc_{k} \in \mathbb{R} such that
引理 7.5。对于任何 $k\in \mathbb{N}$$k\in \mathbb{N}$k inNk \in \mathbb{N} ，存在 $c_k\in \mathbb{R}$$c_k\in \mathbb{R}$c_(k)inRc_{k} \in \mathbb{R} 这样的

Moreover, if $Y_0^k(1)=Y_0^k(2)=\dots =Y_0^k(k)=0$$Y_0^k(1)=Y_0^k(2)=\dots =Y_0^k(k)=0$Y_(0)^(k)(1)=Y_(0)^(k)(2)=dots=Y_(0)^(k)(k)=0Y_{0}^{k}(1)=Y_{0}^{k}(2)=\ldots=Y_{0}^{k}(k)=0, we have
此外，如果 $Y_0^k(1)=Y_0^k(2)=\dots =Y_0^k(k)=0$$Y_0^k(1)=Y_0^k(2)=\dots =Y_0^k(k)=0$Y_(0)^(k)(1)=Y_(0)^(k)(2)=dots=Y_(0)^(k)(k)=0Y_{0}^{k}(1)=Y_{0}^{k}(2)=\ldots=Y_{0}^{k}(k)=0 ，我们有

The proof of this lemma is adapted from [5, Proposition 2].
这个引理的证明改编自 [5， 命题 2]。
Proof. We prove that $(Y_t^k(1))$$\left(Y_t^k(1)\right)$(Y_(t)^(k)(1))\left(Y_{t}^{k}(1)\right) is a sub-additive process. We then use Kingman’s sub-additive ergodic theorem (see [19, Theorem 4] and [18, Theorem 9.14]),
证明。我们证明这是一个 $(Y_t^k(1))$$\left(Y_t^k(1)\right)$(Y_(t)^(k)(1))\left(Y_{t}^{k}(1)\right) 子加法过程。然后，我们使用 Kingman 的子加性遍历定理（参见 [19，定理 4] 和 [18，定理 9.14]），
stating that if ( $X_{s,t},0\le s\le t)$$\left.X_{s,t},0\le s\le t\right)${:X_(s,t),0 <= s <= t)\left.X_{s, t}, 0 \leq s \leq t\right) is a càdlàg family of random variables satisfying
表示 if （ $X_{s,t},0\le s\le t)$$\left.X_{s,t},0\le s\le t\right)${:X_(s,t),0 <= s <= t)\left.X_{s, t}, 0 \leq s \leq t\right) 是满足

then $\gamma :=\underset{t\to \mathrm{\infty }}{lim}\frac{1}{t}\mathbf{E}\left(X_{0,t}\right)$$\gamma :=\underset{t\to \mathrm{\infty }}{lim} \frac{1}{t}\mathbf{E}\left(X_{0,t}\right)$gamma:=lim_(t rarr oo)(1)/(t)E(X_(0,t))\gamma:=\lim _{t \rightarrow \infty} \frac{1}{t} \mathbf{E}\left(X_{0, t}\right) exists, is finite and is equal to $\underset{t\ge 0}{inf}\frac{\mathbf{E}\left(X_{0,t}\right)}{t}$$\underset{t\ge 0}{inf} \frac{\mathbf{E}\left(X_{0,t}\right)}{t}$i n f_(t >= 0)(E(X_(0,t)))/(t)\inf _{t \geq 0} \frac{\mathbf{E}\left(X_{0, t}\right)}{t} (by sub-additivity), and
然后 $\gamma :=\underset{t\to \mathrm{\infty }}{lim}\frac{1}{t}\mathbf{E}\left(X_{0,t}\right)$$\gamma :=\underset{t\to \mathrm{\infty }}{lim} \frac{1}{t}\mathbf{E}\left(X_{0,t}\right)$gamma:=lim_(t rarr oo)(1)/(t)E(X_(0,t))\gamma:=\lim _{t \rightarrow \infty} \frac{1}{t} \mathbf{E}\left(X_{0, t}\right) 存在，是有限的并且等于 $\underset{t\ge 0}{inf}\frac{\mathbf{E}\left(X_{0,t}\right)}{t}$$\underset{t\ge 0}{inf} \frac{\mathbf{E}\left(X_{0,t}\right)}{t}$i n f_(t >= 0)(E(X_(0,t)))/(t)\inf _{t \geq 0} \frac{\mathbf{E}\left(X_{0, t}\right)}{t} （通过子加法），并且

We construct on the same probability space a family $(Y_{s,t}^k(j),0\le s\le t,j\le k)$$\left(Y_{s,t}^k(j),0\le s\le t,j\le k\right)$(Y_(s,t)^(k)(j),0 <= s <= t,j <= k)\left(Y_{s, t}^{k}(j), 0 \leq s \leq t, j \leq k\right), such that for all $s\ge 0,(Y_{s,s+t}^k,t\ge 0)$$s\ge 0,\left(Y_{s,s+t}^k,t\ge 0\right)$s >= 0,(Y_(s,s+t)^(k),t >= 0)s \geq 0,\left(Y_{s, s+t}^{k}, t \geq 0\right) is a $k$$k$kk-branching random walk, and $(Y_{s,t}^k(1),0\le s\le t)$$\left(Y_{s,t}^k(1),0\le s\le t\right)$(Y_(s,t)^(k)(1),0 <= s <= t)\left(Y_{s, t}^{k}(1), 0 \leq s \leq t\right) is sub-additive.
我们在相同的概率空间上构造一个族 $(Y_{s,t}^k(j),0\le s\le t,j\le k)$$\left(Y_{s,t}^k(j),0\le s\le t,j\le k\right)$(Y_(s,t)^(k)(j),0 <= s <= t,j <= k)\left(Y_{s, t}^{k}(j), 0 \leq s \leq t, j \leq k\right) ，使得 for all $s\ge 0,(Y_{s,s+t}^k,t\ge 0)$$s\ge 0,\left(Y_{s,s+t}^k,t\ge 0\right)$s >= 0,(Y_(s,s+t)^(k),t >= 0)s \geq 0,\left(Y_{s, s+t}^{k}, t \geq 0\right) 是一个 $k$$k$kk 分支随机游走，并且 $(Y_{s,t}^k(1),0\le s\le t)$$\left(Y_{s,t}^k(1),0\le s\le t\right)$(Y_(s,t)^(k)(1),0 <= s <= t)\left(Y_{s, t}^{k}(1), 0 \leq s \leq t\right) 是 sub-addive 的。

Let $N^1,\dots ,N^k$$N^1,\dots ,N^k$N^(1),dots,N^(k)N^{1}, \ldots, N^{k} be $k$$k$kk i.i.d. Poisson processes with unit intensity. For all $s\ge 0$$s\ge 0$s >= 0s \geq 0, we set $Y_{s,s}^k(1)=Y_{s,s}^k(2)=\cdots =Y_{s,s}^k(k)=0$$Y_{s,s}^k(1)=Y_{s,s}^k(2)=\cdots =Y_{s,s}^k(k)=0$Y_(s,s)^(k)(1)=Y_(s,s)^(k)(2)=cdots=Y_(s,s)^(k)(k)=0Y_{s, s}^{k}(1)=Y_{s, s}^{k}(2)=\cdots=Y_{s, s}^{k}(k)=0, i.e. particles start at position 0 at time $s$$s$ss. Then the process evolves as follows: at each time $t$$t$tt such that $N_t^j\ne N_{t-}^j$$N_t^j\ne N_{t-}^j$N_(t)^(j)!=N_(t-)^(j)N_{t}^{j} \neq N_{t-}^{j}, the $j$$j$jj th largest particle alive at time $t-$$t-$t-t- in $Y_{s,\bullet }^k$$Y_{s,\bullet }^k$Y_(s,∙)^(k)Y_{s, \bullet}^{k} creates a new child, and the leftmost particle is erased.
设 $N^1,\dots ,N^k$$N^1,\dots ,N^k$N^(1),dots,N^(k)N^{1}, \ldots, N^{k} i.i.d $k$$k$kk .具有单位强度的 Poisson 过程。对于所有 $s\ge 0$$s\ge 0$s >= 0s \geq 0 ，我们设置 $Y_{s,s}^k(1)=Y_{s,s}^k(2)=\cdots =Y_{s,s}^k(k)=0$$Y_{s,s}^k(1)=Y_{s,s}^k(2)=\cdots =Y_{s,s}^k(k)=0$Y_(s,s)^(k)(1)=Y_(s,s)^(k)(2)=cdots=Y_(s,s)^(k)(k)=0Y_{s, s}^{k}(1)=Y_{s, s}^{k}(2)=\cdots=Y_{s, s}^{k}(k)=0 ，即粒子在时间 $s$$s$ss 上从位置 0 开始。然后，该过程将按如下方式发展：在每次 $t$$t$tt ， $N_t^j\ne N_{t-}^j$$N_t^j\ne N_{t-}^j$N_(t)^(j)!=N_(t-)^(j)N_{t}^{j} \neq N_{t-}^{j} 在 $t-$$t-$t-t- 时间存活的 $j$$j$jj 第 th 个最大粒子 会 $Y_{s,\bullet }^k$$Y_{s,\bullet }^k$Y_(s,∙)^(k)Y_{s, \bullet}^{k} 创建一个新的子粒子，而最左边的粒子将被擦除。

By definition, we observe that for all $s\ge 0,(Y_{s,s+t}^k,t\ge 0)$$s\ge 0,\left(Y_{s,s+t}^k,t\ge 0\right)$s >= 0,(Y_(s,s+t)^(k),t >= 0)s \geq 0,\left(Y_{s, s+t}^{k}, t \geq 0\right) is a $k$$k$kk-branching random walk starting with $k$$k$kk particles at position 0 at time 0 . In particular, (7.6) is satisfied. Moreover, $Y_{s,\bullet }^k$$Y_{s,\bullet }^k$Y_(s,∙)^(k)Y_{s, \bullet}^{k} is measurable with respect to the Poisson processes $(N_{t+s}^j-N_s^j,t\ge s,j\le k)$$\left(N_{t+s}^j-N_s^j,t\ge s,j\le k\right)$(N_(t+s)^(j)-N_(s)^(j),t >= s,j <= k)\left(N_{t+s}^{j}-N_{s}^{j}, t \geq s, j \leq k\right), therefore is independent of $(Y_{u,v}^k,0\le u\le v\le s)$$\left(Y_{u,v}^k,0\le u\le v\le s\right)$(Y_(u,v)^(k),0 <= u <= v <= s)\left(Y_{u, v}^{k}, 0 \leq u \leq v \leq s\right). This shows (7.7), i.e. that this process is ergodic.
根据定义，我们观察到 for all $s\ge 0,(Y_{s,s+t}^k,t\ge 0)$$s\ge 0,\left(Y_{s,s+t}^k,t\ge 0\right)$s >= 0,(Y_(s,s+t)^(k),t >= 0)s \geq 0,\left(Y_{s, s+t}^{k}, t \geq 0\right) 是一个 $k$$k$kk 分支随机游走，从 $k$$k$kk 时间 0 位置 0 的粒子开始。特别是 （7.6） 得到满足。此外， $Y_{s,\bullet }^k$$Y_{s,\bullet }^k$Y_(s,∙)^(k)Y_{s, \bullet}^{k} 是 Poisson 过程 $(N_{t+s}^j-N_s^j,t\ge s,j\le k)$$\left(N_{t+s}^j-N_s^j,t\ge s,j\le k\right)$(N_(t+s)^(j)-N_(s)^(j),t >= s,j <= k)\left(N_{t+s}^{j}-N_{s}^{j}, t \geq s, j \leq k\right) 可测量的，因此 独立于 $(Y_{u,v}^k,0\le u\le v\le s)$$\left(Y_{u,v}^k,0\le u\le v\le s\right)$(Y_(u,v)^(k),0 <= u <= v <= s)\left(Y_{u, v}^{k}, 0 \leq u \leq v \leq s\right) 。这表明 （7.7），即这个过程是遍历的。

Moreover, one can observe that the construction described here is the same as the one given in the proof of Lemma 7.4. Therefore, for all $s\le t$$s\le t$s <= ts \leq t this process couples the $k$$k$kk-branching random walks $(Y_{s,t+h}^k,h\ge 0)$$\left(Y_{s,t+h}^k,h\ge 0\right)$(Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) and $(Y_{t,t+h}^k,h\ge 0)$$\left(Y_{t,t+h}^k,h\ge 0\right)$(Y_(t,t+h)^(k),h >= 0)\left(Y_{t, t+h}^{k}, h \geq 0\right) in such a way that for all $j\le k$$j\le k$j <= kj \leq k and $h\ge 0$$h\ge 0$h >= 0h \geq 0, one has
此外，可以观察到这里描述的结构与引理 7.4 证明中给出的结构相同。因此，对于所有这些 $s\le t$$s\le t$s <= ts \leq t 过程，耦合 $k$$k$kk -分支随机游走 $(Y_{s,t+h}^k,h\ge 0)$$\left(Y_{s,t+h}^k,h\ge 0\right)$(Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) ， $(Y_{t,t+h}^k,h\ge 0)$$\left(Y_{t,t+h}^k,h\ge 0\right)$(Y_(t,t+h)^(k),h >= 0)\left(Y_{t, t+h}^{k}, h \geq 0\right) 并且以这种方式，对于所有 $j\le k$$j\le k$j <= kj \leq k 和 $h\ge 0$$h\ge 0$h >= 0h \geq 0 ，都有

Indeed, the $k$$k$kk-branching random walk $(Y_{s,t+h}^k,h\ge 0)$$\left(Y_{s,t+h}^k,h\ge 0\right)$(Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) is coupled with the $k$$k$kk branching random walk $(Y_{t,t+h}^k+Y_{s,t}(1),h\ge 0)$$\left(Y_{t,t+h}^k+Y_{s,t}(1),h\ge 0\right)$(Y_(t,t+h)^(k)+Y_(s,t)(1),h >= 0)\left(Y_{t, t+h}^{k}+Y_{s, t}(1), h \geq 0\right) which starts with $k$$k$kk particles at position $Y_{s,t}(1)$$Y_{s,t}(1)$Y_(s,t)(1)Y_{s, t}(1). In particular, we have $Y_{s,u}^k(1)\le Y_{s,t}^k(1)+Y_{t,u}^k(1)$$Y_{s,u}^k(1)\le Y_{s,t}^k(1)+Y_{t,u}^k(1)$Y_(s,u)^(k)(1) <= Y_(s,t)^(k)(1)+Y_(t,u)^(k)(1)Y_{s, u}^{k}(1) \leq Y_{s, t}^{k}(1)+Y_{t, u}^{k}(1) a.s., proving (7.5).
事实上， $k$$k$kk 分支随机游走 $(Y_{s,t+h}^k,h\ge 0)$$\left(Y_{s,t+h}^k,h\ge 0\right)$(Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) 与 $k$$k$kk 分支随机游走 $(Y_{t,t+h}^k+Y_{s,t}(1),h\ge 0)$$\left(Y_{t,t+h}^k+Y_{s,t}(1),h\ge 0\right)$(Y_(t,t+h)^(k)+Y_(s,t)(1),h >= 0)\left(Y_{t, t+h}^{k}+Y_{s, t}(1), h \geq 0\right) 相耦合，后者从 $k$$k$kk 位置 $Y_{s,t}(1)$$Y_{s,t}(1)$Y_(s,t)(1)Y_{s, t}(1) 的粒子开始。特别是，我们有 $Y_{s,u}^k(1)\le Y_{s,t}^k(1)+Y_{t,u}^k(1)$$Y_{s,u}^k(1)\le Y_{s,t}^k(1)+Y_{t,u}^k(1)$Y_(s,u)^(k)(1) <= Y_(s,t)^(k)(1)+Y_(t,u)^(k)(1)Y_{s, u}^{k}(1) \leq Y_{s, t}^{k}(1)+Y_{t, u}^{k}(1) a.s.，证明 （7.5）。