This is a continuation of [Number of drifted Brownian motions that never hit zero under allocation](https://mathoverflow.net/questions/406559/number-of-drifted-brownian-motions-that-never-hit-zero-under-allocation)


For each $n\ge 1$, consider $X^i\_t=1+\beta t + W^i\_t$ for $i=1,\ldots n$ and $t\ge 0$, where $\beta>0$ and $(W^i\_t)\_{t\ge 0}$ are independent Brownian motions. $\phi\equiv \big((\phi^1\_t)\_{t\ge 0},\ldots, (\phi^n\_t)\_{t\ge 0}\big)$ is called an allocation strategy if every $(\phi^i\_t)\_{t\ge 0}$ is progressively measurable w.r.t. the Brownian filtration $\big(\mathcal F\_t:=\sigma(W^1\_s,\ldots, W^n\_s, s\le t)\big)\_{t\ge 0}$,


$$\phi^i\_t\ge 0 \quad\mbox{ and }\quad \sum\_{i=1}^n\phi^i\_t\le 1,\quad \forall t\ge 0.$$


Denote


$$X^{\phi,i}\_t:=X^i\_t+\int\_0^t \phi^i\_sds \quad \mbox{and} \quad \tau^{\phi}\_i:=\inf\{t\ge 0: X^{\phi,i}\_t\le 0\}.$$


Let $S^{\phi}\_n:=\sum\_{1\le i\le n}{\bf 1}\_{\{\tau^{\phi}\_i=\infty\}}$ be the number of $X^{\phi,i}$ that never hits zero. Clearly,


$$\frac{\mathbb E[S^{\bf 0}]}{n}~=~\mathbb P[X^1\_t>0, \forall t\ge 0]~=~1-e^{-\beta},$$


where $\bf 0$ stands for the strategy with $\phi^i\equiv 0$ for $i=1,\ldots, n$. Can we can show


$$\lim\_{n\to\infty}\frac{\mathbb E[S^{\phi}]}{n}~~=~~1-e^{-\beta}$$


for all the strategies $\phi$? Any answers, comments or references are highly appreciated!



Does fixed allocation increase the proportion of positively drifted Brownian motions surviving forever?
This is not an answer to your question, but a similar result related to the case $\beta=0$ can be found in [Optimal surviving strategy for drifted Brownian motions with absorption](https://projecteuclid.org/journals/annals-of-probability/volume-46/issue-3/Optimal-surviving-strategy-for-drifted-Brownian-motions-with-absorption/10.1214/17-AOP1211.full) (with arxiv version <https://arxiv.org/pdf/1512.04493.pdf>).


More precisely, the (asymptotic) optimal strategy is given by


$$\phi^i\_t:=\begin{cases} 
 1 & \mbox{if } \tau^{\phi}\_i>t \mbox{ and } X^{\phi,i}\_{t}=Z\_t \\
 0 & \mbox{otherwise} 
 \end{cases}
,\quad \mbox{where } Z\_t:=\min\left\{X\_{t}^{\phi,i}:~ \tau^{\phi}\_i>t\right\}.$$


Adopting this strategy, $\mathbb E[S^{\phi}]=O(\sqrt{n})$. Namely, for the any strategy the number of processes never hitting zero is of order $\sqrt{n}$. Therefore, I believe your claim is true and it can be shown using the same arguments of Proposition 2.6 (see also Lemmas 2.1, 2.4 and 2.5).


