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This is a continuation of 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!

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2 Answers 2

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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 (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).

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  • $\begingroup$ Thanks for the reference. I will check the paper $\endgroup$
    – GJC20
    Oct 23, 2021 at 11:54
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This is a partial answer where I believe Girsanov's theorem may provides some insights but I can not prove one inequality.

Introduce $\mathbb Q$ by

$$\frac {d\mathbb Q}{d\mathbb P}\Big |_{\mathcal F_t} := \Pi_{i=i}^n \exp\left(-\int_0^t\phi^i_sdW^i_s-\frac{1}{2}\int_0^t(\phi^i_s)^2ds\right), \quad \forall t>0.$$

Then for each $1\le i\le n$, $(B^i_t:=W^i_t+\int_0^t\phi^i_sds)_{t\ge 0}$ turns to be a Brownian motion under $\mathbb Q$. Hence,

\begin{eqnarray*} \big|\mathbb P[\tau^{\phi}_i>t] - \mathbb Q[\tau^{\phi}_i>t]\big| &=& \left| \mathbb E^{\mathbb Q}\left[{\bf 1}_{\{\tau^{\phi}_i>t\}}\left(\frac {d\mathbb P}{d\mathbb Q}\Big |_{\mathcal F_t}-1\right)\right]\right|\\ &\le & \mathbb Q[\tau^{\phi}_i>t]^{1/2} \mathbb E^{\mathbb Q}\left[\left(\frac {d\mathbb P}{d\mathbb Q}\Big |_{\mathcal F_t}-1\right)^{2}\right]^{1/2} \\ &\le & \mathbb E^{\mathbb Q}\left[\left(\frac {d\mathbb P}{d\mathbb Q}\Big |_{\mathcal F_t}-1\right)^{2}\right]^{1/2}. \end{eqnarray*}

As $\mathbb Q[\tau^{\phi}_i>t]=\mathbb P[\tau^{\bf 0}_i>t]=\mathbb P[\tau^{\bf 0}_1>t]$ and

\begin{eqnarray*} \frac {d\mathbb P}{d\mathbb Q}\Big |_{\mathcal F_t}=\Pi_{i=i}^n \exp\left(\int_0^t\phi^i_sdW^i_s+\frac{1}{2}\int_0^t(\phi^i_s)^2ds\right)=\Pi_{i=i}^n \exp\left(\int_0^t\phi^i_sdB^i_s-\frac{1}{2}\int_0^t(\phi^i_s)^2ds\right), \end{eqnarray*}

one obtains

\begin{eqnarray*} \left|\frac{1}{n}\sum_{i=1}^n \mathbb P[\tau^{\phi}_i>t] - \frac{1}{n}\sum_{i=1}^n\mathbb Q[\tau^{\phi}_i>t]\right| &\le& \frac{1}{n}\sum_{i=1}^n\left| \mathbb P[\tau^{\phi}_i>t] -\mathbb Q[\tau^{\phi}_i>t]\right|\\ &\le& \mathbb E^{\mathbb Q}\left[\left(\Pi_{i=i}^n \exp\left(\int_0^t\phi^i_sdB^i_s-\frac{1}{2}\int_0^t(\phi^i_s)^2ds\right)-1\right)^2\right]^{1/2}, \end{eqnarray*}

which yields by letting $t\to\infty$

$$\left|\frac{\mathbb E\left[S^{\phi}\right]}{n}-(1-e^{-\beta})\right| \le \lim_{t\to\infty}\mathbb E^{\mathbb Q}\left[\left(\Pi_{i=i}^n \exp\left(\int_0^t\phi^i_sdB^i_s-\frac{1}{2}\int_0^t(\phi^i_s)^2ds\right)-1\right)^2\right]^{1/2}.$$

Then the claim follows if we can prove

$$\lim_{n\to\infty} \lim_{t\to\infty}\mathbb E^{\mathbb Q}\left[\left(\Pi_{i=i}^n \exp\left(\int_0^t\phi^i_sdB^i_s-\frac{1}{2}\int_0^t(\phi^i_s)^2ds\right)-1\right)^2\right]=0\quad\quad\quad\quad(\ast).$$

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  • $\begingroup$ Thanks for the idea. I will think about that $\endgroup$
    – GJC20
    Oct 22, 2021 at 6:18
  • $\begingroup$ Change of measure seems appealing to solve this problem, while I think the Holder inequality ($p=q=2$ is used here) is too rough for the estimation. Indeed, let us take the homogeneous strategy with $\phi^i\equiv 1/n$. Then $B^1,\ldots, B^n$ are independent and thus the expection appearing in $(\ast)$ can be explicitly computed and the value is $e^{t/n}-1$ which increases to $\infty$ as $t\to\infty$ $\endgroup$
    – GJC20
    Oct 22, 2021 at 18:09

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