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Consider the usual branching Brownian motion with dyadic branching, where a particle starts from origin and travels according to the law of a standard $1$ dimensional Brownian motion. After an $exp(1)$ time, it dies and gives birth to two identical particles who evolve independent of each other and have the same law as their parent and this process goes on.

Let $\mathcal E_t[m, \infty)$ denote the number of particles that are inside the set $(m,\infty)$ at time $t$. I want to show that for fixed $n$ (could be large enough), $$\small \mathbb P(\mathcal E_t[m, \infty)=n) \ge \mathbb P(\mathcal E_t[m, \infty)=n+1)$$ as $m \to \infty$. This seems very obvious heuristically, but I am unable to write a formal proof. Any help, idea or suggestion in this direction will be extremely helpful.

Amir
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L--
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  • is the branching time deterministic fixed at $e$, or is it $e$ on average distributed as an exponential distribution? – LPZ Aug 09 '24 at 08:56
  • Branching time here is a random variable, exponentially distributed with mean $1$. – L-- Aug 09 '24 at 15:37
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    By $P(\cdots)\ge P(\cdots)$ as $m\to\infty$, are you saying you want to show that is true for sufficiently large $m$? – Varun Vejalla Aug 09 '24 at 19:14

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