# 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 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.

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

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.
• Thanks for your answer. The observation $\{\tau^{\phi}_i>t\}\subset \{X^{\phi,i}_t> 0\}$ is amazing Oct 20 '21 at 7:57
I claim this is far from being an answer. For any $$\alpha>0$$, consider $$X_t=1-\alpha t+ W_t$$ for $$t\ge 0$$ and denote $$\tau:=\inf\{t\ge 0: X_t\le 0\}$$. It is known that $${\bf 1}_{\{\tau=\infty\}}=0$$ almost surely as $$\mathbb E[{\bf 1}_{\{\tau=\infty\}}]=\mathbb P[\tau=\infty]=0$$.
Therefore, $$S^{\phi}_n=0$$ almost surely if $$\beta\ge 1$$. For general $$\beta$$, it is known that for each $$t\ge 0$$, the number of processes with allocation greater than $$\beta /2$$ is at mostly equal to $$2/\beta$$. I believe that the number of processes that never hit zero is thus less than $$C/\beta$$ for some constant $$C$$, but I don't know whether this intuition is correct or not