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 said to be 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. I am interested in the asymptotic order of


$$\sup\_{\phi} S^{\phi}\_n,$$


where the the supremum is taken over all allocation strategies. My question is whether one has $0<\alpha<1$ and $C>0$ s.t.


$$0<\liminf\_{n\to\infty}\frac{\sup\_{\phi} S^{\phi}\_n}{n^{\alpha}} \le \limsup\_{n\to\infty}\frac{\sup\_{\phi} S^{\phi}\_n}{n^{\alpha}}\le C\quad \left( \mbox{or}\quad 0<\liminf\_{n\to\infty}\frac{\mathbb E[\sup\_{\phi} S^{\phi}\_n]}{n^{\alpha}} \le \limsup\_{n\to\infty}\frac{\mathbb E[\sup\_{\phi} S^{\phi}\_n]}{n^{\alpha}}\le C\right)?$$


Any answers, comments or references are highly appreciated!



Number of drifted Brownian motions that never hit zero under allocation
I think its O(1).
$$\lbrace \tau^{\phi}\_i > t \rbrace = \lbrace \tau^{\phi}\_i > t , \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace \cup \lbrace \tau^{\phi}\_i > t,\int\_0^t \phi^i\_sds < \frac {\beta t} 2 \rbrace = $$,
so $$1\_{\lbrace \tau^{\phi}\_i > t \rbrace} \le 1\_{ \lbrace \tau^{\phi}\_i > t , \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace } + 1\_ {\lbrace \tau^{\phi}\_i > t,\int\_0^t \phi^i\_sds < \frac {\beta t} 2 \rbrace} $$ $$$$
$$\Sigma 1\_{ \lbrace \tau^{\phi}\_i > t , \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace } \le \Sigma 1\_{ \lbrace \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace } \le \frac 2 { \beta }$$ by $\Sigma \phi\_i < 1$ while $$\lbrace \tau^{\phi}\_i > t,\int\_0^t \phi^i\_sds < \frac {\beta t} 2 \rbrace \subset \lbrace W\_t > \frac {\beta t } 2 \rbrace$$ which has probability, say $e^{\frac {-\beta t } 2}$, maybe a little different but not materially so. Therefore the expected number of paths for which $\tau^{\phi}\_i > t $ is $\le \frac 2 { \beta } + n e^{\frac {-\beta t } 2}$ and t is at your disposal.


