Branching Brownian motion with selection. Two models, the same rightmost particle velocity?

Question

A free branching brownian motion is define as follows: At time $t=0$ a single particle at the origin start evolving like a brownian motion and after an exponential time of mean $1$ the particle splits in two and each of this particles evolve independently as their father.

Now we introduce two models for the branching brownian motion with selection. Let $N \in \mathbb{N}$:

1) At time $t=0$, $N$ particles at the origin start moving as a brownian motion and after an exponential time of mean $N$ we choose uniformly a particle and split it in two an then erase the leftmost particle. The remaining particles evolve the same way.

2) Fix $\delta > 0$ and run $N$ independent free branching brownian motions for $\delta$ units of time and erase the leftmost particles in order to keep $N$ rightmost particles. The remaining particles evolves as their fathers.

For both cases one can show (using kingman subadditive ergodic theorem) that there exist $v_N$ and $v_{N,\delta}$ such that right most particles $x^N(t)$ and $x_\delta^N(n)$ respectively satisfy $$\frac{x^N(t)}{t} \stackrel{t\to\infty}{\longrightarrow} v_N \quad \text{and}\quad \frac{x^N_\delta(n)}{n\delta} \stackrel{n\to\infty}{\longrightarrow} v_{N,\delta}$$ with a simple coupling one can show that $v_{N,\delta} \geq v_N$ and $v_{N,\delta}$ decreases as $\delta$ does. I'm wondering if $v_{N,\delta} = v_N$, $v_{N,\delta} > v_N$ or $v_{N,\delta} \stackrel{\delta \to 0}{\longrightarrow} v_N$. Any idea will be appreciated.

Answer 1

Claim 1: $\nu_{N, \delta} \ne \nu_N$ as $\delta \to 0$.

Claim 2: If the mean waiting time for each free branching Brownian motion in the second branching process with selection is changed to $N^2$ (in place of 1), then we have that $\nu_{N, \delta} \to \nu_N$ as $\delta \to 0$.

Why?

I'll focus on the second claim, since the first claim obviously follows from the second. With the amendment given in the second claim, the transition rules of the two branching processes with selection are indistinguishable when $\delta$ is sufficiently small. Indeed, note that if $\delta$ is sufficiently small, then deletions of extraneous particles happen instantaneously. Also, recall that the minimum of $N$ independent exponentials each with mean $N^2$ is also exponential, but with mean $N$. Moreover, the probability that a given index achieves this minimum is $1/N$. Thus, referring to the second branching process: (i) the random time it takes before any particle in this process splits is exponentially distributed with mean $N$; and (ii) the probability that a given particle in this process splits is $1/N$. As we said, as long as $\delta$ is sufficiently small, one can ensure that the leftmost particle is deleted before the next splitting occurs with arbitrarily high probability. Since exponential distributions are memoryless, this rule iterates. In particular, the waiting time until the next splitting occurs is once again exponential with mean $N$ and the probability that a given particle splits is once again uniform over the number of particles. To finish, just note that the transition rule we just described is no different from the transition rule of the first branching process with selection.

ADD

To be sure, the embedded chain and random splitting times in the first branching process with selection are obtained by iterating the following procedure.

Given the current time $t_0$ and and current position of the $N$ particles $(x_0, \cdots, x_N)$, determine the next time $t_1$ and position as follows:

1. Generate an exponentially distributed random variable $\delta t$ with mean $N$, and update time via $t_1=t_0+\delta t$.
2. Evolve $N$ iid Brownian motions $(B_1, \cdots, B_N)$ over $[t_0, t_1)$ with position at $t_0$ given by $$ (B_1(t_0), \cdots , B_N(t_0))=(x_0, \cdots, x_N) $$
3. Generate a random number $k$ uniformly in the range $\{1, \cdots, N\}$.
4. Order $(B_1(t_1), \cdots, B_{k-1}(t_1), B_k(t_1), B_k(t_1), B_{k+1}(t_1), \cdots , B_N(t_1))$ in ascending order (from smallest to largest): $(B_{(1)}(t_1), \cdots, B_{(N+1)}(t_1))$ and output $(B_{(2)}(t_1), \cdots, B_{(N+1)}(t_1))$.

Note that the amended second branching process with selection follows exactly this update rule, but with the interpretation that (Step 1) corresponds to the minimum of $N$ iid exponentials each with mean $N^2$, that (Step 3) corresponds to the selection of which free branching process splits, and that (Step 4) is a consequence of $\delta$ (the deterministic time between successive deletions of extraneous particles) being extremely small.