这是用户在 2025-4-6 23:56 为 https://app.immersivetranslate.com/pdf-pro/949183d1-37d4-4fe9-bbbc-c5040780334c/?isTrial=true 保存的双语快照页面,由 沉浸式翻译 提供双语支持。了解如何保存?
We observe that Y H Y H Y^(H)Y^{H} and Y ~ K 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 T b T a 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} Y ~ K 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 T b T a 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 Y ~ K 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 K R H R 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 Y ~ K 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 K R H R 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
x R , # { j H R : Y R H ( j ) x } # { j K R : Y ~ R K ( j ) x } x R , # j H R : Y R H ( j ) x # j K R : Y ~ R K ( j ) x AA x inR,#{j <= H_(R):Y_(R)^(H)(j) >= x} <= #{j <= K_(R): widetilde(Y)_(R)^(K)(j) >= x}\forall x \in \mathbb{R}, \#\left\{j \leq H_{R}: Y_{R}^{H}(j) \geq x\right\} \leq \#\left\{j \leq K_{R}: \widetilde{Y}_{R}^{K}(j) \geq x\right\}
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 Y ~ K 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 Y ~ K 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
x R , # { j H R : Y R H ( j ) x } # { j K R : Y ~ R K ( j ) x } x R , # j H R : Y R H ( j ) x # j K R : Y ~ R K ( j ) x AA x inR,#{j <= H_(R):Y_(R)^(H)(j) >= x} <= #{j <= K_(R): widetilde(Y)_(R)^(K)(j) >= x}\forall x \in \mathbb{R}, \#\left\{j \leq H_{R}: Y_{R}^{H}(j) \geq x\right\} \leq \#\left\{j \leq K_{R}: \widetilde{Y}_{R}^{K}(j) \geq x\right\}
Finally, if R = S { T a , T b } R = S T a , T b 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 Y ~ K 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 { T a , T b } R = S T a , T b 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 Y ~ K 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 K s H s K s H_(s) <= K_(s)H_{s} \leq K_{s} for every 0 s t 0 s 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 Y ~ K Y ~ K widetilde(Y)^(K)\widetilde{Y}^{K}. Thus there is a.s. a finite sequence of times ( R k ) R k (R_(k))\left(R_{k}\right) smaller than t t tt such that Y H Y H Y^(H)Y^{H} or Y ~ K 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 ] R k , R k + 1 [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 K s H s K s H_(s) <= K_(s)H_{s} \leq K_{s} 对于每个 0 s t 0 s t 0 <= s <= t0 \leq s \leq t .由于 H H HH 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} Y ~ K 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 Y H Y H Y^(H)Y^{H} Y ~ K Y ~ K widetilde(Y)^(K)\widetilde{Y}^{K} 在每次 时被修改的有限时间 ( R k ) R k (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 ] R k , R k + 1 [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 N k N k inNk \in \mathbb{N}, there exists c k R c k R c_(k)inRc_{k} \in \mathbb{R} such that
引理 7.5。对于任何 k N k N k inNk \in \mathbb{N} ,存在 c k R c k R c_(k)inRc_{k} \in \mathbb{R} 这样的
lim t + Y t k ( 1 ) t = lim t + Y t k ( k ) t = c k a.s. lim t + Y t k ( 1 ) t = lim t + Y t k ( k ) t = c k  a.s.  lim_(t rarr+oo)(Y_(t)^(k)(1))/(t)=lim_(t rarr+oo)(Y_(t)^(k)(k))/(t)=c_(k)quad" a.s. "\lim _{t \rightarrow+\infty} \frac{Y_{t}^{k}(1)}{t}=\lim _{t \rightarrow+\infty} \frac{Y_{t}^{k}(k)}{t}=c_{k} \quad \text { a.s. }
Moreover, if Y 0 k ( 1 ) = Y 0 k ( 2 ) = = Y 0 k ( k ) = 0 Y 0 k ( 1 ) = Y 0 k ( 2 ) = = 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 ) = = Y 0 k ( k ) = 0 Y 0 k ( 1 ) = Y 0 k ( 2 ) = = 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
c k = inf t > 0 E [ Y t ( 1 ) ] t = sup t > 0 E [ Y t ( k ) ] t . c k = inf t > 0 E Y t ( 1 ) t = sup t > 0 E Y t ( k ) t . c_(k)=i n f_(t > 0)(E[Y_(t)(1)])/(t)=s u p_(t > 0)(E[Y_(t)(k)])/(t).c_{k}=\inf _{t>0} \frac{\mathbf{E}\left[Y_{t}(1)\right]}{t}=\sup _{t>0} \frac{\mathbf{E}\left[Y_{t}(k)\right]}{t} .
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 ) ) Y t k ( 1 ) (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 ) ) Y t k ( 1 ) (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 s t ) X s , t , 0 s t {: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 s t ) X s , t , 0 s t {:X_(s,t),0 <= s <= t)\left.X_{s, t}, 0 \leq s \leq t\right) 是满足
0 s t u , X s , u X s , t + X t , u a.s h 0 , ( X s + h , t + h , 0 s t ) = ( d ) ( X s , t , 0 s t ) h 0 , ( X s + h , t + h , 0 s t ) is independent of ( X s , t , 0 s t h ) A > 0 , t 0 , A t E ( X 0 , t ) < E ( | sup 0 s t 1 X s , t | ) < 0 s t u , X s , u X s , t + X t , u  a.s  h 0 , X s + h , t + h , 0 s t = ( d ) X s , t , 0 s t h 0 , X s + h , t + h , 0 s t  is independent of  X s , t , 0 s t h A > 0 , t 0 , A t E X 0 , t < E sup 0 s t 1 X s , t < {:[AA0 <= s <= t <= u","X_(s,u) <= X_(s,t)+X_(t,u)" a.s "],[AA h >= 0","(X_(s+h,t+h),0 <= s <= t)=^((d))(X_(s,t),0 <= s <= t)],[AA h >= 0","(X_(s+h,t+h),0 <= s <= t)" is independent of "(X_(s,t),0 <= s <= t <= h)],[EE A > 0","AA t >= 0","-At <= E(X_(0,t)) < oo],[E(|s u p_(0 <= s <= t <= 1)X_(s,t)|) < oo]:}\begin{aligned} & \forall 0 \leq s \leq t \leq u, X_{s, u} \leq X_{s, t}+X_{t, u} \text { a.s } \\ & \forall h \geq 0,\left(X_{s+h, t+h}, 0 \leq s \leq t\right) \stackrel{(d)}{=}\left(X_{s, t}, 0 \leq s \leq t\right) \\ & \forall h \geq 0,\left(X_{s+h, t+h}, 0 \leq s \leq t\right) \text { is independent of }\left(X_{s, t}, 0 \leq s \leq t \leq h\right) \\ & \exists A>0, \forall t \geq 0,-A t \leq \mathbf{E}\left(X_{0, t}\right)<\infty \\ & \mathbf{E}\left(\left|\sup _{0 \leq s \leq t \leq 1} X_{s, t}\right|\right)<\infty \end{aligned}
then γ := lim t 1 t E ( X 0 , t ) γ := lim t 1 t E X 0 , t 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 inf t 0 E ( X 0 , t ) t inf t 0 E X 0 , t 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 γ := lim t 1 t E ( X 0 , t ) γ := lim t 1 t E X 0 , t 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 inf t 0 E ( X 0 , t ) t inf t 0 E X 0 , t 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
lim t 1 t X 0 , t = γ a.s. and in L 1 lim t 1 t X 0 , t = γ  a.s. and in  L 1 lim_(t rarr oo)(1)/(t)X_(0,t)=gammaquad" a.s. and in "L^(1)\lim _{t \rightarrow \infty} \frac{1}{t} X_{0, t}=\gamma \quad \text { a.s. and in } L^{1}
We construct on the same probability space a family ( Y s , t k ( j ) , 0 s t , j k ) Y s , t k ( j ) , 0 s t , j k (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 0 , ( Y s , s + t k , t 0 ) s 0 , Y s , s + t k , t 0 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 s t ) Y s , t k ( 1 ) , 0 s t (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 s t , j k ) Y s , t k ( j ) , 0 s t , j k (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 0 , ( Y s , s + t k , t 0 ) s 0 , Y s , s + t k , t 0 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 s t ) Y s , t k ( 1 ) , 0 s t (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 , , N k N 1 , , 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 0 s 0 s >= 0s \geq 0, we set Y s , s k ( 1 ) = Y s , s k ( 2 ) = = Y s , s k ( k ) = 0 Y s , s k ( 1 ) = Y s , s k ( 2 ) = = 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 N t j N t j 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 , k Y s , k Y_(s,∙)^(k)Y_{s, \bullet}^{k} creates a new child, and the leftmost particle is erased.
N 1 , , N k N 1 , , 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 0 s 0 s >= 0s \geq 0 ,我们设置 Y s , s k ( 1 ) = Y s , s k ( 2 ) = = Y s , s k ( k ) = 0 Y s , s k ( 1 ) = Y s , s k ( 2 ) = = 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 N t j N t j N t j N_(t)^(j)!=N_(t-)^(j)N_{t}^{j} \neq N_{t-}^{j} t t t-t- 时间存活的 j j jj 第 th 个最大粒子 会 Y s , k Y s , 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 0 , ( Y s , s + t k , t 0 ) s 0 , Y s , s + t k , t 0 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 , k Y s , k Y_(s,∙)^(k)Y_{s, \bullet}^{k} is measurable with respect to the Poisson processes ( N t + s j N s j , t s , j k ) N t + s j N s j , t s , j k (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 u v s ) Y u , v k , 0 u v s (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 0 , ( Y s , s + t k , t 0 ) s 0 , Y s , s + t k , t 0 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 , k Y s , k Y_(s,∙)^(k)Y_{s, \bullet}^{k} 是 Poisson 过程 ( N t + s j N s j , t s , j k ) N t + s j N s j , t s , j k (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 u v s ) Y u , v k , 0 u v s (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 t s t s <= ts \leq t this process couples the k k kk-branching random walks ( Y s , t + h k , h 0 ) Y s , t + h k , h 0 (Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) and ( Y t , t + h k , h 0 ) Y t , t + h k , h 0 (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 k j k j <= kj \leq k and h 0 h 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 t s t s <= ts \leq t 过程,耦合 k k kk -分支随机游走 ( Y s , t + h k , h 0 ) Y s , t + h k , h 0 (Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) ( Y t , t + h k , h 0 ) Y t , t + h k , h 0 (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 k j k j <= kj \leq k h 0 h 0 h >= 0h \geq 0 ,都有
Y s , t + h k ( j ) Y t , t + h k ( j ) + Y s , t k ( 1 ) a.s. Y s , t + h k ( j ) Y t , t + h k ( j ) + Y s , t k ( 1 )  a.s.  Y_(s,t+h)^(k)(j) <= Y_(t,t+h)^(k)(j)+Y_(s,t)^(k)(1)quad" a.s. "Y_{s, t+h}^{k}(j) \leq Y_{t, t+h}^{k}(j)+Y_{s, t}^{k}(1) \quad \text { a.s. }
Indeed, the k k kk-branching random walk ( Y s , t + h k , h 0 ) Y s , t + h k , h 0 (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 0 ) Y t , t + h k + Y s , t ( 1 ) , h 0 (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 ) 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) <= 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 0 ) Y s , t + h k , h 0 (Y_(s,t+h)^(k),h >= 0)\left(Y_{s, t+h}^{k}, h \geq 0\right) k k kk 分支随机游走 ( Y t , t + h k + Y s , t ( 1 ) , h 0 ) Y t , t + h k + Y s , t ( 1 ) , h 0 (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 ) 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) <= 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)。
To prove the last two conditions, we observe that Y s , ( 1 ) Y s , ( 1 ) Y_(s,∙)(1)Y_{s, \bullet}(1) increases by at most 1 at each time one of the Poisson processes jumps. Moreover, t Y s , t t Y s , t t|->Y_(s,t)t \mapsto Y_{s, t} is non-decreasing, thus, for all t 0 t 0 t >= 0t \geq 0,
为了证明最后两个条件,我们观察到每次其中一个泊松进程跳跃时, Y s , ( 1 ) Y s , ( 1 ) Y_(s,∙)(1)Y_{s, \bullet}(1) 最多增加 1。此外, t Y s , t t Y s , t t|->Y_(s,t)t \mapsto Y_{s, t} 是非递减的,因此,对于所有 t 0 t 0 t >= 0t \geq 0
E ( | sup 0 s t 1 Y s , t ( 1 ) | ) k and 0 E ( Y 0 , t ( 1 ) ) < E sup 0 s t 1 Y s , t ( 1 ) k  and  0 E Y 0 , t ( 1 ) < E(|s u p_(0 <= s <= t <= 1)Y_(s,t)(1)|) <= k quad" and "quad0 <= E(Y_(0,t)(1)) < oo\mathbf{E}\left(\left|\sup _{0 \leq s \leq t \leq 1} Y_{s, t}(1)\right|\right) \leq k \quad \text { and } \quad 0 \leq \mathbf{E}\left(Y_{0, t}(1)\right)<\infty
proving both (7.8) and (7.9).
证明 (7.8) 和 (7.9)。

As a result, by Kingman’s sub-additive ergodic theorem, setting
因此,根据 Kingman 的子加性遍历定理,将
c k = lim t E [ Y t ( 1 ) ] t = inf t > 0 E [ Y t ( 1 ) ] t c k = lim t E Y t ( 1 ) t = inf t > 0 E Y t ( 1 ) t c_(k)=lim_(t rarr oo)(E[Y_(t)(1)])/(t)=i n f_(t > 0)(E[Y_(t)(1)])/(t)c_{k}=\lim _{t \rightarrow \infty} \frac{\mathbf{E}\left[Y_{t}(1)\right]}{t}=\inf _{t>0} \frac{\mathbf{E}\left[Y_{t}(1)\right]}{t}
we have lim t Y 0 , t ( 1 ) t = c k lim t Y 0 , t ( 1 ) t = c k lim_(t rarr oo)(Y_(0,t)(1))/(t)=c_(k)\lim _{t \rightarrow \infty} \frac{Y_{0, t}(1)}{t}=c_{k} a.s.
我们有 lim t Y 0 , t ( 1 ) t = c k lim t Y 0 , t ( 1 ) t = c k lim_(t rarr oo)(Y_(0,t)(1))/(t)=c_(k)\lim _{t \rightarrow \infty} \frac{Y_{0, t}(1)}{t}=c_{k} A.S.

With the same construction, one can observe that ( Y s , t k ( k ) , 0 s t ) Y s , t k ( k ) , 0 s t (Y_(s,t)^(k)(k),0 <= s <= t)\left(Y_{s, t}^{k}(k), 0 \leq s \leq t\right) is a super-additive sequence, satisfying similar integrability assumptions as Y s , t ( 1 ) Y s , t ( 1 ) Y_(s,t)(1)Y_{s, t}(1). Therefore, setting
使用相同的构造,可以观察到 是一个 ( Y s , t k ( k ) , 0 s t ) Y s , t k ( k ) , 0 s t (Y_(s,t)^(k)(k),0 <= s <= t)\left(Y_{s, t}^{k}(k), 0 \leq s \leq t\right) 超级加法序列,满足与 Y s , t ( 1 ) Y s , t ( 1 ) Y_(s,t)(1)Y_{s, t}(1) 类似的可积性假设。因此,设置
d k = lim t E [ Y 0 , t ( k ) ] t = sup t > 0 E [ Y 0 , t ( 1 ) ] t , d k = lim t E Y 0 , t ( k ) t = sup t > 0 E Y 0 , t ( 1 ) t , d_(k)=lim_(t rarr oo)(E[Y_(0,t)(k)])/(t)=s u p_(t > 0)(E[Y_(0,t)(1)])/(t),d_{k}=\lim _{t \rightarrow \infty} \frac{\mathbf{E}\left[Y_{0, t}(k)\right]}{t}=\sup _{t>0} \frac{\mathbf{E}\left[Y_{0, t}(1)\right]}{t},
we have lim t Y 0 , t ( k ) t = d k lim t Y 0 , t ( k ) t = d k lim_(t rarr oo)(Y_(0,t)(k))/(t)=d_(k)\lim _{t \rightarrow \infty} \frac{Y_{0, t}(k)}{t}=d_{k} a.s. As Y s , t ( k ) Y s , t ( 1 ) Y s , t ( k ) Y s , t ( 1 ) Y_(s,t)(k) <= Y_(s,t)(1)Y_{s, t}(k) \leq Y_{s, t}(1), we have d k c k d k c k d_(k) <= c_(k)d_{k} \leq c_{k}. We now prove these two quantities to be equal.
我们有 lim t Y 0 , t ( k ) t = d k lim t Y 0 , t ( k ) t = d k lim_(t rarr oo)(Y_(0,t)(k))/(t)=d_(k)\lim _{t \rightarrow \infty} \frac{Y_{0, t}(k)}{t}=d_{k} A.S.作为 Y s , t ( k ) Y s , t ( 1 ) Y s , t ( k ) Y s , t ( 1 ) Y_(s,t)(k) <= Y_(s,t)(1)Y_{s, t}(k) \leq Y_{s, t}(1) ,我们有 d k c k d k c k d_(k) <= c_(k)d_{k} \leq c_{k} 。我们现在证明这两个量是相等的。
We define a sequence of hitting times ( T n , n 0 ) T n , n 0 (T_(n),n >= 0)\left(T_{n}, n \geq 0\right) by setting T 0 = 0 T 0 = 0 T_(0)=0T_{0}=0, and T n + 1 T n + 1 T_(n+1)T_{n+1} is the first time after time T n T n T_(n)T_{n} where the last k k kk children are born from the same particle, and that this particle was the rightmost particle before the series of branching events. The probability that the next k k kk branching events are as such is 1 / k k > 0 1 / k k > 0 1//k^(k) > 01 / k^{k}>0, therefore T n < T n < T_(n) < ooT_{n}<\infty a.s. Moreover, by definition, we have Y 0 , T n ( 1 ) = Y 0 , T n ( k ) Y 0 , T n ( 1 ) = Y 0 , T n ( k ) Y_(0,T_(n))(1)=Y_(0,T_(n))(k)Y_{0, T_{n}}(1)=Y_{0, T_{n}}(k) a.s., all particles being at the same position at that time As a result, we have
我们通过设置 T 0 = 0 T 0 = 0 T_(0)=0T_{0}=0 来定义一系列撞击时间 ( T n , n 0 ) T n , n 0 (T_(n),n >= 0)\left(T_{n}, n \geq 0\right) ,并且 T n + 1 T n + 1 T_(n+1)T_{n+1} 是最后一次 k k kk 从同一粒子中诞生的最后一个 T n T n T_(n)T_{n} 子粒子,并且该粒子是一系列分支事件之前最右边的粒子。下一个 k k kk 分支事件的概率是 1 / k k > 0 1 / k k > 0 1//k^(k) > 01 / k^{k}>0 ,因此 T n < T n < T_(n) < ooT_{n}<\infty a.s.此外,根据定义,我们有 Y 0 , T n ( 1 ) = Y 0 , T n ( k ) Y 0 , T n ( 1 ) = Y 0 , T n ( k ) Y_(0,T_(n))(1)=Y_(0,T_(n))(k)Y_{0, T_{n}}(1)=Y_{0, T_{n}}(k) a.s.,所有粒子在那个时候都处于同一位置,因此,我们有
lim inf t Y 0 , t ( 1 ) Y 0 , t ( k ) t = 0 a.s. lim inf t Y 0 , t ( 1 ) Y 0 , t ( k ) t = 0  a.s.  l i m   i n f_(t rarr oo)(Y_(0,t)(1)-Y_(0,t)(k))/(t)=0quad" a.s. "\liminf _{t \rightarrow \infty} \frac{Y_{0, t}(1)-Y_{0, t}(k)}{t}=0 \quad \text { a.s. }
proving that c k = d k c k = d k c_(k)=d_(k)c_{k}=d_{k}, and that Y 0 , t ( 1 ) t Y 0 , t ( 1 ) t (Y_(0,t)(1))/(t)\frac{Y_{0, t}(1)}{t} and Y 0 , t ( k ) t Y 0 , t ( k ) t (Y_(0,t)(k))/(t)\frac{Y_{0, t}(k)}{t} have the same limit.
证明 c k = d k c k = d k c_(k)=d_(k)c_{k}=d_{k} , 和 that Y 0 , t ( 1 ) t Y 0 , t ( 1 ) t (Y_(0,t)(1))/(t)\frac{Y_{0, t}(1)}{t} 和 具有相同的 Y 0 , t ( k ) t Y 0 , t ( k ) t (Y_(0,t)(k))/(t)\frac{Y_{0, t}(k)}{t} 限制。

Finally, we consider a k k kk-branching random walk Y k Y k Y^(k)Y^{k} starting from an arbitrary initial configuration. After a finite amount of time t t tt, the process contains k k kk particles. From that point on, the process can be bounded from above and from below by k k kk-branching random walks starting with k k kk particles at position Y t k ( 1 ) Y t k ( 1 ) Y_(t)^(k)(1)Y_{t}^{k}(1) and Y t k ( k ) Y t k ( k ) Y_(t)^(k)(k)Y_{t}^{k}(k) respectively. Therefore, by the previous results, we also obtain
最后,我们考虑从任意初始配置开始的 k k kk -branching 随机游走 Y k Y k Y^(k)Y^{k} 。在有限的时间 t t tt 之后,该过程包含 k k kk 粒子。从那时起,该过程可以通过 k k kk -分支随机游走从上方和下方界定,分别从 k k kk 位置 Y t k ( 1 ) Y t k ( 1 ) Y_(t)^(k)(1)Y_{t}^{k}(1) 处的 Y t k ( k ) Y t k ( k ) Y_(t)^(k)(k)Y_{t}^{k}(k) 粒子开始。因此,通过前面的结果,我们还得到了
lim s Y s k ( 1 ) s = lim s Y s k ( k ) s = c k a.s. lim s Y s k ( 1 ) s = lim s Y s k ( k ) s = c k  a.s.  lim_(s rarr oo)(Y_(s)^(k)(1))/(s)=lim_(s rarr oo)(Y_(s)^(k)(k))/(s)=c_(k)quad" a.s. "\lim _{s \rightarrow \infty} \frac{Y_{s}^{k}(1)}{s}=\lim _{s \rightarrow \infty} \frac{Y_{s}^{k}(k)}{s}=c_{k} \quad \text { a.s. }
completing the proof.  完成校对。

7.4 End of the proof of Lemma 7.3
7.4 引理证明结束 7.3

In this section, we use Lemma 7.4 to compare the asymptotic behaviour of the continuous-time branching random walk with selection Y k Y k Y^(k)Y^{k} with a discrete-time branching random walk with selection. This latter model being well-studied, we are able to deduce Lemma 7.3 from it. The discrete-time branching random walk with selection of the rightmost k k kk individuals was introduced by Brunet and Derrida in [8] to study noisy FKPP equations. In that article, they conjecture that the cloud of particles drifts at speed v k v k v_(k)v_{k}, that satisfies
在本节中,我们使用 Lemma 7.4 来比较带选择的连续时间分支随机游走与带选择的 Y k Y k Y^(k)Y^{k} 离散时间分支随机游走的渐近行为。后一个模型经过充分研究,我们能够从中推断出引理 7.3。Brunet 和 Derrida 在 [8] 中引入了选择最 k k kk 右边个体的离散时间分支随机游走,用于研究嘈杂的 FKPP 方程。在那篇文章中,他们推测粒子云以 的速度 v k v k v_(k)v_{k} 漂移,这满足
v k v = χ ( log k + 3 log log k + o ( log log k ) ) 2 , as k , v k v = χ ( log k + 3 log log k + o ( log log k ) ) 2 ,  as  k , v_(k)-v=-(chi)/((log k+3log log k+o(log log k))^(2))," as "k rarr oo,v_{k}-v=-\frac{\chi}{(\log k+3 \log \log k+o(\log \log k))^{2}}, \text { as } k \rightarrow \infty,
for some explicit constants v R v R v inRv \in \mathbb{R} and χ > 0 χ > 0 chi > 0\chi>0.
对于某些显式常量 v R v R v inRv \in \mathbb{R} χ > 0 χ > 0 chi > 0\chi>0 .

We now describe more precisely the discrete-time k k kk-branching random walk. Let k N k N k inNk \in \mathbb{N} and M M M\mathcal{M} be the law of a point process on Z Z Z\mathbb{Z}. The system starts with k k kk particles on Z Z Z\mathbb{Z}. At each integer time n n nn, every particle dies while giving birth to offspring. The children of a given individual are positioned around their parent according to an i.i.d. point process with law M M M\mathcal{M}. Among all the children of
我们现在更精确地描述了离散时间 k k kk 分支随机游走。设 k N k N k inNk \in \mathbb{N} M M M\mathcal{M} 是 上的点过程的 Z Z Z\mathbb{Z} 定律。系统从 k k kk 上的 Z Z Z\mathbb{Z} 粒子开始。在每个整数时间 n n nn ,每个粒子在生下后代时都会死亡。给定个体的子代根据 i.i.d. 点过程与定律 M M M\mathcal{M} .在所有的子代中