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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!

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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.

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  • $\begingroup$ Thanks for your answer. The observation $\{\tau^{\phi}_i>t\}\subset \{X^{\phi,i}_t> 0\}$ is amazing $\endgroup$
    – GJC20
    Oct 20 '21 at 7:57
  • $\begingroup$ By the way, I have a similar question on whether the unit allocation is significant for positively drifted Brownian motions. Could you please take a look? The question is posted at mathoverflow.net/questions/406609/… $\endgroup$
    – GJC20
    Oct 20 '21 at 8:13
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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

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