# Systemic robustness: a mean-field particle system approach

Erhan Bayraktar111Department of Mathematics, University of Michigan. Email:
<EMAIL_ADDRESS>E. Bayraktar is partially supported by the National Science
Foundation under grant DMS-2106556 and by the Susan M. Smith chair. Gaoyue
Guo222Université Paris-Saclay, CentraleSupélec, MICS and CNRS FR-3487. Email:
<EMAIL_ADDRESS>Wenpin Tang333Department of Industrial
Engineering and Operations Research, Columbia University. Email:
<EMAIL_ADDRESS>W. Tang gratefully acknowledges financial support through
NSF grants DMS-2113779 and DMS-2206038, and through a start-up grant at
Columbia University. Yuming Paul Zhang444Department of Mathematics, Auburn
University. Email<EMAIL_ADDRESS>
###### Abstract

This paper is concerned with the problem of budget control in a large particle
system modeled by stochastic differential equations involving hitting times,
which arises from considerations of systemic risk in a regional financial
network. Motivated by Tang and Tsai (Ann. Probab., 46 (2018), pp. 1597–1650),
we focus on the number or proportion of surviving entities that never default
to measure the systemic robustness. First we show that both the mean-field
particle system and its limiting McKean-Vlasov equation are well-posed by
virtue of the notion of minimal solutions. We then establish a connection
between the proportion of surviving entities in the large particle system and
the probability of default in the limiting McKean-Vlasov equation as the size
of the interacting particle system $N$ tends to infinity. Finally, we study
the asymptotic efficiency of budget control in different economy regimes: the
expected number of surviving entities is of constant order in a negative
economy; it is of order $\sqrt{N}$ in a neutral economy; and it is of order
$N$ in a positive economy where the budget’s effect is negligible.

Key words: Budget control, drifted Brownian motion, hitting times, interacting
particle systems, large system limits, McKean-Vlasov equation, mean-field
interactions, systemic risk.

AMS 2010 Mathematics Subject Classification: 35K10, 60H30, 60K35, 91G80.

##  1 Introduction

Systemic risk in financial markets has emerged as a major research topic since
the 2008 financial crisis, see e.g. [14]. It refers to the risk that small
losses and defaults can escalate to cause an event affecting large parts of
the financial industry. The existing literature on modeling systemic risk
relies on network models [17, 22, 24], quantitative risk measures [1, 6], and
particle systems [5, 16, 25]. This paper is concerned with a particle system
approach with mean-field interactions to analyze systemic risk in a large
financial network.

Consider an interconnected financial system of $N$ homogeneous banks, and let
$X^{N,1}_{t},\ldots,X^{N,N}_{t}$ be their respective capital levels at time
$t\geq 0$. A bank becomes insolvent if its capital drops below some threshold,
which is set to be zero without loss of generality. In the absence of
defaults, the capital processes $(X_{t}^{N,i},\,t\geq 0)$ are supposed to
follow the stochastic differential equations (SDEs): for $i=1,\ldots,N$,

$\displaystyle dX^{N,i}_{t}=\beta\,dt+dB^{i}_{t},\quad t\geq 0,$ (1)

where $\beta\in\mathbb{R}$ stands for the regional economy (see e.g. [19]),
and $B^{1},\ldots,B^{N}$ are independent Brownian motions.

If $k\geq 1$ banks default simultaneously at some time, we assume that the
capital of each remaining bank suffers an immediate loss equal to $G(k/N)$,
which may cause further defaults. Here $G:[-1,1]\to\mathbb{R}$555The existing
research focuses on the case $\beta=0$ and $G(x)=\alpha x$ for $\alpha>0$,
while [25] considers the alternative case $G(x)=-\alpha\log(1-x)$ for
$\alpha>0$. is a continuous and non-decreasing function such that $G(0)=0$.
After the default event, the remaining capital processes continue to follow
the SDEs (1) until one of them hits zero, and so on. Rephrasing in
mathematical language, the capitals evolve as follows: for $i=1,\ldots,N$,

$\displaystyle X^{N,i}_{t}=Z^{N,i}+\beta t+B^{i}_{t}-G(L^{N}_{t})$ and
$\displaystyle
L^{N}_{t}:=\frac{1}{N}\sum_{i=1}^{N}\mathds{1}_{\\{\tau^{N}_{i}\leq
t\\}},\quad t\geq 0,$ (2)

where $Z^{N,1},\ldots,Z^{N,N}$ are the initial capitals as i.i.d. random
variables that are independent of Brownian motions $B^{1},\ldots,B^{N}$, and
$\tau^{N}_{i}:=\inf\\{t\geq 0:X^{N,i}_{t}\leq 0\\}$ is the default time of
bank $i$. Here $L_{t}^{N}$ denotes the fraction of insolvent banks up to and
including time $t$666From the practical viewpoint, we are interested in the
evolution of $X^{N,i}$ only until time $\tau^{N}_{i}$, i.e.
$(X^{N,i}_{t},\,0\leq t\leq\tau^{N}_{i})$, while we let the process $X^{N,i}$
diffuse freely after its default time for technical purposes.. Letting
$N\to\infty$, we get formally the McKean-Vlasov limit as follows:

$\displaystyle X_{t}=Z+\beta t+B_{t}-G(\Lambda_{t})$ and
$\displaystyle\Lambda_{t}:=\mathbb{P}(\tau\leq t),\quad t\geq 0,$ (3)

where $Z$ is a random variable (distributed as $Z^{N,1}$), $B$ is an
independent Brownian motion, and $\tau:=\inf\\{t\geq 0:X_{t}\leq 0\\}$. It is
known that the case $G(x)=\alpha x$ for $\alpha>0$ provides a probabilistic
interpretation of the supercooled Stefan problem and has given rise to a rich
body of literature, see Delarue et al. [10, 11, 12], Hambly et al. [18],
Cuchiero et al. [8], and Bayraktar et al. [3].

The system transitions between two regimes:

  * •

the well-behaved regime, in which the system spends most of its time, and
during which the default cascades are very small or do not appear at all;

  * •

the systemic crisis regime, which occurs rarely, and which is characterized by
a large group of banks defaulting in a short period of time.

In our setting, the _systemic failure_ occurs at $t_{sys}:=\\{t\geq
0:\Delta\Lambda_{t}:=\Lambda_{t}-\Lambda_{-t}>0\\}$ when a non-negligible
fraction of defaults occur in a short period and the system passes abruptly
from the well-behaved regime to the systemic crisis regime. The systemic event
has gained significant attention in the wake of financial crisis (see e.g.
[10, 25]), while we focus on the _systemic robustness_ , i.e. the proportion
of banks surviving forever. For $t\geq 0$, we denote by

$\displaystyle S^{N}_{t}:=\sum_{i=1}^{N}\mathds{1}_{\\{\tau^{N}_{i}>t\\}},$

the number of banks that survive until time $t$. Thus, $S^{N}_{\infty}$
(number of banks surviving forever) or $1-L^{N}_{\infty}$ (proportion of banks
surviving forever) encodes the system’s robustness.

Our first goal is to study the expected surviving proportion
$\mathbb{E}[S^{N}_{\infty}]/N$, which boils down to showing under suitable
conditions,

$\displaystyle\lim_{N\to\infty}\frac{\mathbb{E}[S^{N}_{\infty}]}{N}=\mathbb{P}(\tau=\infty).$
(4)

For $\beta\leq 0$, the identity (4) follows from a simple observation. As
Brownian motion hits each level in finite time with probability one, we have
for all $t\geq 0$,

$\displaystyle X^{N,i}_{t}\leq Z^{N,i}+B^{i}_{t}$ and $\displaystyle X_{t}\leq
Z+B_{t},$

which implies that
$\mathds{1}_{\\{\tau^{N}_{i}=\infty\\}}\leq\mathds{1}_{\\{Z^{N,i}+B^{i}_{t}>0,\,\forall
t\geq 0\\}}=0$ and
$\mathds{1}_{\\{\tau=\infty\\}}\leq\mathds{1}_{\\{Z+B_{t}>0,\,\forall t\geq
0\\}}=0$ with probability one. Hence,
$\mathbb{E}[S^{N}_{\infty}]=0=\mathbb{P}(\tau=\infty)$, and (4) holds. The
case $\beta>0$ is less obvious, and will be treated in Section 3. Note that
both the particle system (2) and the McKean-Vlasov equation (3) may have more
than one solution (see e.g. [11, Section 3.1]), which renders the statement of
our desired result (4) more subtle. Among the solutions to (2) and (3), we
need to choose suitable ones. To this end, we adopt the _minimal solution_ as
in [8], and more details will be provided in Section 2.2.

The next problem is to find a feasible way to compute the minimal solution to
the McKean-Vlasov equation (3). The idea is to approximate it by a sequence of
regularized problems. Note that

$\displaystyle\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\stackrel{{\scriptstyle\varepsilon\to
0}}{{\longrightarrow}}\mathds{1}_{\\{\tau>t\\}}$
$\displaystyle\Longrightarrow$
$\displaystyle\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\right]\stackrel{{\scriptstyle\varepsilon\to
0}}{{\longrightarrow}}\Lambda_{t}.$

For $\varepsilon>0$, we consider the $\varepsilon$-regularized equation:

$\displaystyle X^{\varepsilon}_{t}=Z+\beta
t+B_{t}-G\left(1-\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\right]\right),\quad
t\geq 0.$ (5)

We show in Theorem 5 that the equation (5) has a unique solution that
converges to the minimal solution to the McKean-Vlasov equation (3).

Finally, we consider the problem of controlling the systemic risk by _equity
injections_ from financial regulators’ perspective. Namely, granted a budget
that is normalized to one unit without loss of generality, the regulators aim
to divide and allocate it among all surviving banks in order to maximize the
expected number/proportion of banks that survive forever. An
$\mathbb{F}-$progressively measurable process
$\phi\equiv\big{(}(\phi^{1}_{t},\ldots,\phi^{N}_{t}),\,t\geq 0\big{)}$ taking
values in $[0,1]^{N}$ is called a strategy if

$\displaystyle\sum_{i=1}^{N}\phi^{i}_{t}\leq 1,\quad t\geq 0,$

where $\mathbb{F}\equiv\big{(}{\cal
F}_{t}:=\sigma(B^{1}_{s},\ldots,B^{N}_{s},\,s\in[0,t]),\,t\geq 0\big{)}$. Here
$\phi^{i}_{t}$ stands for the equity injected to bank $i$ at time $t$, and let
$\Phi_{N}$ be the collection of all strategies. So the capital processes
subject to the strategy $\phi\in\Phi_{N}$ evolve as follows: for
$i=1,\ldots,N$

$\displaystyle
X^{\phi,N,i}_{t}=Z^{N,i}+\int_{0}^{t}(\beta+\phi^{i}_{s})\mathrm{d}s+B^{i}_{t}-G(L^{\phi,N}_{t})$
and $\displaystyle
L^{\phi,N}_{t}:=\frac{1}{N}\sum_{i=1}^{N}\mathds{1}_{\\{\tau^{\phi,N}_{i}\leq
t\\}},\quad t\geq 0,$ (6)

with $\tau^{\phi,N}_{i}:=\inf\\{t\geq 0:X^{\phi,N,i}_{t}\leq 0\\}$.
Accordingly, the number of banks surviving up to time $t$ is

$\displaystyle
S^{\phi,N}_{t}:=\sum_{i=1}^{N}\mathds{1}_{\\{\tau^{\phi,N}_{i}>t\\}}.$

A similar budget control problem and its numerical solutions were studied in
[8]. Our goal, however, is to investigate the budget’s significance in the
large $N$ limit. To be more precise, we show in Theorem 6 that
$\sup_{\phi\in\Phi_{N}}\mathbb{E}[S_{\infty}^{\phi,N}]$ has different scaling
in $N$ depending on whether the regional economy $\beta<0$, $\beta=0$ or
$\beta>0$.

#### A word on notations.

For ease of presentation, we drop the superscript $N$ when the context is
clear, i.e.

$\displaystyle Z^{i}\equiv Z^{N,i},~{}X^{i}\equiv
X^{N,i},~{}\tau_{i}\equiv\tau^{N}_{i},~{}\mbox{etc.}$

Furthermore, we write when there is no risk of confusion

$\displaystyle X^{i}\equiv X^{\phi,i}\equiv
X^{\phi,N,i},~{}\tau_{i}\equiv\tau^{\phi}_{i}\equiv\tau^{\phi,N}_{i},~{}L^{\phi}\equiv
L^{\phi,N},~{}\mbox{etc.}$

The rest of the paper is organized as follows. We state the main results in
Section 2, and the proofs are given in Sections 3–5.

##  2 Preliminaries and main results

We provide measure theoretical background in Section 2.1, and then present the
main results in Section 2.2.

###  2.1 Preliminaries

Given a generic Polish space $E$, let ${\cal P}(E)$ (resp. ${\cal
P}_{\leq}(E)$) be the set of probability (sub-probability) measures on $E$. We
set ${\cal P}:={\cal P}(\mathbb{R}_{+})$ and ${\cal P}_{\leq}:={\cal
P}_{\leq}(\mathbb{R}_{+})$. ${\cal C}$ (resp. ${\cal D}$) stands for the space
of continuous (resp. càdlàg) functions on $[-1,\infty)$. Let ${\cal C}$ be
endowed with the compact convergence induced by the metric $d_{u}$, i.e.

$\displaystyle d_{u}(f,f^{\prime}):=\sum_{n\geq
1}\frac{1}{2^{n}}\big{(}\|f-f^{\prime}\|_{n}\wedge 1\big{)}\quad\mbox{with
}\|f-f^{\prime}\|_{n}:=\sup_{t\in[0,n]}|f_{t}-f^{\prime}_{t}|.$

As for ${\cal D}$, we use the Skorokhod $M_{1}$ topology which turns out to be
more convenient for our purpose. Denote by $d_{m}$ the corresponding metric
(see e.g. Whitt [28, Chapter 12] for a complete overview). Moreover, we
introduce the set

$\displaystyle\mathbb{M}:=\big{\\{}\ell\in{\cal D}\mbox{ is non-
decreasing:}\quad\lim_{t\to\infty}\ell_{t}=:\ell_{\infty}\leq 1\mbox{ and
}\ell_{t}=0,~{}\forall t\in[-1,0)\big{\\}}$

and the Lévy distance $d$ defined by

$\displaystyle
d(\ell,\ell^{\prime}):=\inf\big{\\{}\varepsilon>0:~{}\ell_{t+\varepsilon}+\varepsilon\geq\ell^{\prime}_{t}\geq\ell_{t-\varepsilon}-\varepsilon,~{}\forall
t\geq 0\big{\\}},\quad\ell,\ell^{\prime}\in\mathbb{M}.$

For each $t\geq 0$, define $d_{t}:\mathbb{M}\times\mathbb{M}\to\mathbb{R}_{+}$
by $d_{t}(\ell,\ell^{\prime}):=d(\ell_{t\wedge},\ell^{\prime}_{t\wedge})$, and
further $\hat{d}$ by

$\displaystyle\hat{d}(\ell,\ell^{\prime}):=\int_{0}^{\infty}e^{-t}\big{(}d_{t}(\ell,\ell^{\prime})\wedge
1\big{)}\mathrm{d}t,$

where $\ell_{t\wedge}\in\mathbb{M}$ is the function stopped at $t$, i.e.
$\ell_{t\wedge s}:=\ell_{\min(t,s)}$ for all $s\geq-1$. Lemma 1 summarizes the
properties of $d$ and $\hat{d}$, and its proof can be read from [28].

######  Lemma 1.

For every $\ell\in\mathbb{M}$, there exists a unique $\theta_{\ell}\in{\cal
P}_{\leq}$ such that $\ell_{t}=\theta_{\ell}((-\infty,t])$ for all $t\geq 0$.
For any $(\ell^{n})_{n\geq 1}\subset\mathbb{M}$ and $\ell\in\mathbb{M}$, we
have $(1)\Longleftrightarrow(2)\Longrightarrow(3)\Longleftrightarrow(4)$,
where

  * (1)

$\lim_{n\to\infty}d(\ell^{n},\ell)=0$;

  * (2)

$\theta_{\ell^{n}}$ converges weakly to $\theta_{\ell}$ in ${\cal P}_{\leq}$;

  * (3)

$\lim_{n\to\infty}\hat{d}(\ell^{n},\ell)=0$;

  * (4)

$\lim_{n\to\infty}\ell^{n}_{t}=\ell_{t}$ for all points of continuity $t\geq
0$ of $\ell$.

We have $(\mathbb{M},d)$ and $(\mathbb{M},\hat{d})$ are Polish spaces.
Moreover, $(\mathbb{M},\hat{d})$ is compact.

###  2.2 Main results

We start with a few vocabularies. Given independent random variables
$Z^{N,1},\ldots,Z^{N,N}$ and independent Brownian motions
$B^{1},\ldots,B^{N}$, $(X^{N,1},\ldots,X^{N,N},L^{N})$ is said to be a
solution to (2) if $L^{N}$ is $\mathbb{F}-$adapted and takes values in
$\mathbb{M}$, and the equations of (2) are satisfied with probability one.
Similarly, for a random variable $Z$ and an independent Brownian motion $B$,
the pair $(X,\Lambda)$ is said to be a solution to (3) if
$\Lambda\in\mathbb{M}$ and the equations of (3) hold. Note that
$(X^{N,1},\ldots,X^{N,N})$ (resp. $X$) is fully determined if $L^{N}$ (resp.
$\Lambda$) is fixed. Hence, we also say that $L^{N}$ (resp. $\Lambda$) is a
solution to (2) (resp. (3)) for the sake of simplicity.

As previously mentioned, neither (2) nor (3) can guarantee the uniqueness of
their solution. Thus, we introduce the minimal solution as follows. A solution
$(\underline{X}^{N,1},\ldots,\underline{X}^{N,N},\underline{L}^{N})$ (resp.
$(\underline{X},\underline{\Lambda})$) to (2) (resp. (3)) is called minimal if

$\displaystyle\underline{L}^{N}_{t}\leq L^{N}_{t}\quad\left(\mbox{resp.
}\underline{\Lambda}_{t}\leq\Lambda_{t}\right),\quad t\geq 0,$

holds for any solution $(X^{N,1},\ldots,X^{N,N},L^{N})$ (resp. $(X,\Lambda)$)
to (2) (resp. (3)). The minimal solutions, if exist, must be unique. Theorem 2
ensures the existence of minimal solutions, which yields the wellposedness of
(2) and (3). Its proof is given in Section 3.

######  Theorem 2.

Both (2) and (3) have a unique minimal solution. If $\beta>0$, we have
$\underline{\Lambda}_{\infty}<\Lambda_{\infty}$ for any solution $(X,\Lambda)$
to (3) that is different from the minimal solution.

From the financial perspective, all banks will default in finite time when
$\beta\leq 0$. If $\beta>0$, then $1-\underline{\Lambda}_{\infty}$ is the
largest proportion of banks surviving forever in our setting (3).

Next we turn to the asymptotic proportion of banks that survive forever. As
$G$ is continuous, it is uniformly continuous restricted on any compact
subset. Let $\alpha:[0,1]\to\mathbb{R}_{+}$ be the modulus of continuity of
$G$, i.e. $|G(z)-G(z^{\prime})|\leq\alpha(|z-z^{\prime}|)$ for all
$z,z^{\prime}\in[0,1]$. We also need the following technical assumption on the
initial distributions.

######  Assumption 3.

$Z^{N,1},\ldots,Z^{N,N}$ are i.i.d. random variables whose probability
distribution $\theta^{N}\in{\cal P}$ is such that $(\theta^{N},\,N\geq 1)$
converges weakly to $\theta$.

For any $b\in\mathbb{R}$, denote by $Q_{b}:\mathbb{R}\to\mathbb{R}$ the
translation, i.e. $Q_{b}(z):=z+b$. Theorem 4 relates the expected surviving
proportion $\mathbb{E}[S^{N}_{\infty}]/N$ to the minimal solution
$\underline{\Lambda}$ to the McKean-Vlasov equation (3). Its proof is given in
Section 4.1.

######  Theorem 4.

Assume that there exists $(\gamma_{N},\,N\geq 1)\subset\mathbb{R}_{+}$ such
that

$\displaystyle\lim_{N\to\infty}\gamma_{N}=\lim_{N\to\infty}\frac{1}{N\gamma_{N}^{2}}=0$
and $\displaystyle\theta^{N}=Q_{\alpha(\gamma_{N})}(\theta),\quad N\geq 1,$

where $Q_{b}(\theta)$ is the image measure of $\theta$ by $Q_{b}$. Taking the
minimal solution
$(X^{N,1},\ldots,X^{N,N},L^{N})=(\underline{X}^{N,1},\ldots,\underline{X}^{N,N},\underline{L}^{N})$,
it holds that

$\displaystyle\lim_{N\to\infty}\mathbb{E}\left[\left|\frac{\underline{S}^{N}_{\infty}}{N}-(1-\underline{\Lambda}_{\infty})\right|\right]=0.$

Theorem 5 shows that the unique solution to the $\varepsilon$-regularized
problem (5) converges to the minimal solution $\underline{\Lambda}$ to (3).
Its proof is given in Section 4.2.

######  Theorem 5.

Let the assumptions in Theorem 4 hold, and assume further that $G$ is
Lipschitz. For every $\varepsilon>0$, (5) has a unique solution
$X^{\varepsilon}$ such that $X^{\varepsilon}$ has continuous paths and
$\lim_{\varepsilon\to 0}X^{\varepsilon}_{t}=\underline{X}_{t}$ for all $t\geq
0$.

Finally, we consider the efficiency of equity injections. Similar to (2), the
equations (6) do not ensure the uniqueness of their solution in general.
Hence, we adopt the minimal solution as follows.
$(X^{\phi,N,1},\ldots,X^{\phi,N,N},L^{\phi,N})$ is said to be a solution to
(6) if $L^{\phi,N}$ is $\mathbb{F}-$adapted and takes values in $\mathbb{M}$,
and the equations of (6) are satisfied with probability one. We say that
$(\underline{X}^{\phi,N,1},\ldots,\underline{X}^{\phi,N,N},\underline{L}^{\phi,N})$
is a minimal solution to (6) if

$\displaystyle\underline{L}^{\phi,N}_{t}\leq L^{\phi,N}_{t},\quad t\geq 0,$

holds for any solution $(X^{\phi,N,1},\ldots,X^{\phi,N,N},L^{\phi,N})$ to (6).

######  Theorem 6.

For all $N\geq 1$ and $\phi\in\Phi_{N}$, (6) have a unique minimal solution.
Assume further that $G(x)=\alpha x$ for some $\alpha\in\mathbb{R}_{+}$.

  1. (i)

If $\beta<0$, then the budget is useless and the number of banks surviving
forever is finite, i.e.

$\displaystyle\sup_{\phi\in\Phi_{N}}\mathbb{E}[S^{\phi,N}_{\infty}]<-2/\beta,\quad\mbox{for
all }N\geq 1.$

Here we emphasize that $S^{\phi,N}_{\infty}$ corresponds to any solution to
(6).

  2. (ii)

If $\beta=0$ and $\theta$ is compactly supported, then the budget allows to
maintain the number of banks surviving forever of order $\sqrt{N}$, i.e.

$\displaystyle
0<\liminf_{N\to\infty}\left(\sup_{\phi\in\Phi_{N}}\mathbb{E}\left[\frac{\underline{S}^{\phi,N}_{\infty}}{\sqrt{N}}\right]\right)\leq\limsup_{N\to\infty}\left(\sup_{\phi\in\Phi_{N}}\mathbb{E}\left[\frac{\underline{S}^{\phi,N}_{\infty}}{\sqrt{N}}\right]\right)<\infty.$

  3. (iii)

If $\beta>0$ and $\alpha=0$, then the budget’s effect is negligible and the
proportion of banks surviving forever remains unchanged, i.e.

$\displaystyle\lim_{N\to\infty}\left(\sup_{\phi\in\Phi_{N}}\mathbb{E}\left[\frac{S^{\phi,N}_{\infty}}{N}\right]\right)=\lim_{N\to\infty}\mathbb{E}\left[\frac{S^{{\bf
0},N}_{\infty}}{N}\right]>0,$

where ${\bf 0}\in\Phi_{N}$ stands for the strategy $\big{(}(\phi^{1}_{t}\equiv
0,\ldots,\phi^{N}_{t}\equiv 0),\,t\geq 0\big{)}$.

The proof of Theorem 6 is given in Section 5. The theorem indicates that the
number of surviving banks scales differently with $\beta<0$ (negative
economy), $\beta=0$ (neutral economy), and $\beta>0$ (positive economy). Part
(i) implies that the number of surviving banks is of constant order (bounded
by $-2/\beta$) in a negative economy, so equity injections as economic
intervention are far from efficient. Part (ii) shows that in a neutral
economy, the number of surviving banks scales as $\sqrt{N}$, which is of the
same order as the “Up the River” model [2, 27]. Lower and upper bounds will
also be derived in Section 5.2. Part (iii) indicates that in a positive
economy with no systemic interactions, the budget control is not efficient in
the sense that the proportion of surviving banks is unchanged asymptotically
with or without equity injections. The general case where $\alpha\neq 0$ seems
to be challenging, and we conjecture that the budget’s effect is also
negligible.

##  3 Proof of Theorem 2

The idea of the proof is similar to [8], which relies on a fixed point
argument. Here we provide more details specific to our problem.

Let us introduce the operator $\Gamma:{\cal D}\to\mathbb{M}$ as follows. For
$\ell\in{\cal D}$, let $\Gamma[\ell]\in\mathbb{M}$ be defined by
$\Gamma[\ell]_{t}:=\mathbb{P}(\tau^{\ell}\leq t)$ for all $t\geq 0$, where

$\displaystyle X^{\ell}_{t}:=Z+\beta t+B_{t}-G(\ell_{t})$ and
$\displaystyle\tau^{\ell}:=\inf\big{\\{}t\geq 0:~{}X^{\ell}_{t}\leq
0\big{\\}}.$

We define similarly $\Gamma_{N}:{\cal S}({\cal D})\to{\cal S}(\mathbb{M})$,
where ${\cal S}({\cal D})$ (resp. ${\cal S}(\mathbb{M})$) denotes the set of
$\mathbb{F}-$adapted processes taking values in ${\cal D}$ (resp.
$\mathbb{M}$). For $L\in{\cal S}({\cal D})$, define

$\displaystyle\Gamma_{N}[L]_{t}:=\frac{1}{N}\sum_{i=1}^{N}\mathds{1}_{\\{\tau^{L}_{i}\leq
t\\}},$

where for $1\leq i\leq N$,

$\displaystyle X^{L,i}_{t}:=Z^{i}+\beta t+B^{i}_{t}-G(L_{t})$ and
$\displaystyle\tau^{L}_{i}:=\inf\big{\\{}t\geq 0:~{}X^{L,i}_{t}\leq
0\big{\\}}.$

For any $\ell,\ell^{\prime}\in{\cal D}$ (resp. $L,L^{\prime}\in{\cal S}({\cal
D})$), we write $\ell\preceq\ell^{\prime}$ (resp. $L\preceq L^{\prime}$) if
$\ell_{t}\leq\ell^{\prime}_{t}$ for all $t\geq-1$ (resp. $L_{t}\leq
L^{\prime}_{t}$ for all $t\geq-1$ almost surely). It is easy to verify that
$\Gamma,\Gamma_{N}$ are monotone with respect to $\preceq$, i.e.
$\ell\preceq\ell^{\prime}\Longrightarrow\Gamma[\ell]\preceq\Gamma[\ell^{\prime}]$
(resp. $L\preceq
L^{\prime}\Longrightarrow\Gamma_{N}[L]\preceq\Gamma_{N}[L^{\prime}])$. In
particular, for any $\ell\in\mathbb{M}$ (resp. $L\in{\cal S}(\mathbb{M})$),
$(X^{\ell},\ell)$ (resp. $(X^{L,1},\ldots,X^{L,N},L)$) is a solution to (3)
(resp. (2)) if and only if $\Gamma[\ell]=\ell$ (resp. $\Gamma_{N}[L]=L$).

Our goal is to prove that $\Gamma$ (resp. $\Gamma_{N}$) has a fixed point
corresponding to the minimal solution. To this end, we need the following
lemma.

######  Lemma 7.

For any $(\ell^{n},\,n\geq 1)\subset\mathbb{M}$ and $\ell\in\mathbb{M}$
satisfying $\lim_{n\to\infty}\hat{d}(\ell^{n},\ell)=0$, we have

$\displaystyle\limsup_{n\to\infty}\Gamma[\ell^{n}]_{t}\leq\Gamma[\ell]_{t},\quad\mbox{for
all }t\geq 0.$

###### Proof.

Denote $g^{n}:=G(\ell^{n})$ and $g:=G(\ell)$ for all $n\geq 1$. By Lemma 1,
there exist $\theta^{n},\theta\in{\cal P}_{\leq}$ such that
$\ell^{n}_{t}=\theta^{n}([0,t])$ and $\ell_{t}=\theta([0,t])$ for all $t\geq
0$. By the Portmanteau theorem, we get

$\displaystyle\limsup_{n\to\infty}\ell^{n}_{t}=\limsup_{n\to\infty}\theta^{n}([0,t])\leq\theta([0,t])=\ell_{t},$

which yields $\Gamma[\ell]_{t}=\mathbb{P}(\tau^{\ell}\leq
t)\geq\mathbb{P}(C)$, where $C:=\\{\exists s\in[0,t]:Z+\beta s+B_{s}\leq
h_{s}\\}$ and
$h_{t}:=G(\limsup_{n\to\infty}\ell^{n}_{t})=\limsup_{n\to\infty}g^{n}_{t}$.
Then it suffices to prove

$\displaystyle\limsup_{n\to\infty}\Gamma[\ell^{n}]_{t}=\limsup_{n\to\infty}\mathbb{P}(\tau^{\ell^{n}}\leq
t)\leq\mathbb{P}(C).$

For each $t\geq 0$, set $h^{n}_{t}:=\sup_{k\geq n}g^{k}_{t}$ so that
$g^{n}_{t}\leq h^{n}_{t}\downarrow h_{t}$. Hence,

$\displaystyle\limsup_{n\to\infty}\mathbb{P}(\tau^{\ell^{n}}\leq
t)\leq\lim_{n\to\infty}\mathbb{P}(\exists s\in[0,t]:Z+\beta s+B_{s}\leq
h^{n}_{s})=\mathbb{P}(A),$

where

$\displaystyle A:=\\{\forall n,~{}\exists s\in[0,t]:Z+\beta s+B_{s}\leq
h^{n}_{s}\\}=\\{\forall n\geq m,~{}\exists s\in[0,t]:Z+\beta s+B_{s}\leq
h^{n}_{s}\\}.$

Here $m$ can be any positive integer. Note that $h^{n}$ and $h$ are non-
decreasing by definition, but may not be right-continuous. It is enough to
show $\mathbb{P}(A\setminus C)=0$. Introducing an equivalent probability
measure $\mathbb{Q}$ defined by

$\displaystyle\left(\frac{d\mathbb{Q}}{d\mathbb{P}}\right)_{t}:=\exp(-\beta
B_{t}-\beta^{2}t/2).$

Then $\mathbb{P}(A\setminus C)=0\Longleftrightarrow\mathbb{Q}(A\setminus
C)=0$, and $(\beta s+B_{s})_{0\leq s\leq t}$ is a Brownian motion under
$\mathbb{Q}$. Thus, it suffices to prove for the case $\beta=0$. Define

$\displaystyle\tau^{n}:=\inf\\{s\in[0,t]:~{}Z+B_{s}\leq
h^{n}_{s}\\},\quad\forall n\geq 1.$

Clearly, $\tau^{n}\in[0,t]$ on $A$. By definition, $n\mapsto\tau^{n}$ is non-
decreasing as $h^{n}\downarrow h$, which implies the existence of
$\tau:=\lim_{n\to\infty}\tau^{n}$. In particular, $\tau\in[0,t]$ also holds on
$A$. In the following, we let the event $A$ occur and distinguish four cases.

  1. (i)

Suppose $\tau^{n}<\tau$ for all $n\geq 1$. By definition, there is some
$t_{n}\in[\tau^{n},\tau)$ such that $x+B_{t_{n}}\leq h^{n}_{t_{n}}\leq
h^{n}_{\tau}$. Hence,
$x+B_{\tau}=\lim_{n\to\infty}(x+B_{t_{n}})\leq\lim_{n\to\infty}h^{n}_{\tau}=h_{\tau}$.
Thus, we get

$\displaystyle A\cap\\{\tau^{n}<\tau,~{}\forall n\geq 1\\}\subseteq C.$ (7)

  2. (ii)

If $\tau^{n}=\tau=t$ for some $n$, then we have $x+B_{t}\leq h^{n}_{t}$, and
further $x+B_{t}\leq h_{t}$. Therefore,

$\displaystyle A\cap\\{\exists n:\tau^{n}=\tau=t\\}\subseteq C.$ (8)

  3. (iii)

If $\tau^{n}=\tau<t$ for some $n$, and $\tau$ is a point of discontinuity of
$h$, then we have $x+B_{\tau}=x+B_{\tau^{n}}\leq
h^{n}_{\tau^{n}+}=h^{n}_{\tau+}$. Define
$\hat{h}_{s}:=\lim_{n\to\infty}h^{n}_{s+}$, then
$x+B_{\tau}\leq\hat{h}_{\tau}$. Namely,

$\displaystyle x+B_{d}\leq\hat{h}_{d},$ (9)

at some point $d\in D$, where $D$ is the countable subset of points of
discontinuity of $h$. For all $0\leq u<s\leq t$, we have
$h_{s}=\lim_{n\to\infty}h^{n}_{s}\geq\lim_{n\to\infty}h^{n}_{u+}=\hat{h}_{u}$.
Now suppose that $C$ does not occur, which implies that $x+B_{s}>h_{s}$ for
all $s\in[0,t]$ and hence $x+B_{s}>\hat{h}_{d}$ for all $d\in D$ and
$s\in(d,t]$. By the local behavior of Brownian motion, for every $d\in[0,t)$
the set $\\{x+B_{d}\leq\hat{h}_{d},~{}x+B_{s}>\hat{h}_{d},\ \forall
s\in(d,t]\\}$ is negligible. We conclude by (9) that

$\displaystyle\mathbb{P}(A\cap\\{\exists n:\tau^{n}=\tau<t,~{}\tau\in
D\\}\setminus C)\leq\sum_{d\in D}\mathbb{P}(A\cap\\{\exists
n:\tau^{n}=\tau=d<t\\}\setminus C)=0.$ (10)

  4. (iv)

Finally, suppose $\tau^{n}=\tau<t$ for some $n$, and $\tau$ is a point of
continuity of $h$. For any $\varepsilon>0$, there exists $\delta>0$ such that
$h_{\tau+\delta}\leq h_{\tau}+\varepsilon$. Hence, for all $n$ large enough,

$\displaystyle x+B_{\tau}=x+B_{\tau^{n}}\leq
h^{n}_{\tau^{n}+\delta}=h^{n}_{\tau+\delta}\to h_{\tau+\delta}\leq
h_{\tau}+\varepsilon.$

As $\varepsilon$ is arbitrary, we get $x+B_{\tau}\leq h_{\tau}$. Therefore,

$\displaystyle A\cap\\{\exists n:\tau^{n}=\tau<t,~{}\tau\notin D\\}\subseteq
C.$ (11)

Combining (7), (8), (10) and (11), we conclude that $\mathbb{P}(A\setminus
C)=0$ and $\limsup_{n\to\infty}\mathbb{P}(\tau^{\ell^{n}}\leq
t)\leq\mathbb{P}(\tau^{\ell}\leq t)$. ∎

Next we show that $\Gamma:(\mathbb{M},\hat{d})\to(\mathbb{M},d)$ is
continuous.

######  Proposition 8.

For any $(\ell^{n},\,n\geq 1)\subset\mathbb{M}$ and $\ell\in\mathbb{M}$,
$\lim_{n\to\infty}d(\Gamma[\ell^{n}],\Gamma[\ell])=0$ holds when
$\lim_{n\to\infty}\hat{d}(\ell^{n},\ell)=0$.

###### Proof.

For simplicity, we write $X^{\ell^{n}}\equiv X^{n}$, $X^{\ell}\equiv X$,
$\tau^{\ell^{n}}\equiv\tau^{n}$ and $\tau^{\ell}\equiv\tau$. We first show
that $\lim_{n\to\infty}\Gamma[\ell^{n}]_{t}=\Gamma[\ell]_{t}$ for all $t\in
J$, where $J\subset\mathbb{R}_{+}$ denotes the collection of the points of
continuity of $\Gamma[\ell]$. By Lemma 7, we have
$\limsup_{n\to\infty}\Gamma[\ell^{n}]_{t}\leq\Gamma[\ell]_{t}$ for all $t\geq
0$, and it remains to prove
$\lim_{n\to\infty}\big{(}\Gamma[\ell]_{t}-\Gamma[\ell^{n}]_{t}\big{)}^{+}=0$
for $t\in J$. Write

$\displaystyle\big{(}\Gamma[\ell]_{t}-\Gamma[\ell^{n}]_{t}\big{)}^{+}\leq\mathbb{P}(\tau^{n}>t,\tau\leq
t)=\int_{[0,t]}\mathbb{P}(\tau^{n}>t|\tau=s)\mathrm{d}\Gamma[\ell]_{s}.$

As in the proof of Lemma 7, we denote $g^{n}:=G(\ell^{n})$ and $g:=G(\ell)$.
We split the integrand in its continuous and jump part, writing
$\Gamma[\ell]^{c}_{s}$ for the continuous part, which we estimate in the
following.

$\displaystyle\int_{[0,t]}\mathbb{P}(\tau^{n}>t|\tau=s)\mathrm{d}\Gamma[\ell]^{c}_{s}$
$\displaystyle=$ $\displaystyle\int_{[0,t]}\mathbb{P}(X^{n}_{u}>0,\forall
u\leq t|\tau=s)\mathrm{d}\Gamma[\ell]^{c}_{s}$ $\displaystyle\leq$
$\displaystyle\int_{[0,t]}\mathbb{P}(X^{n}_{s}+\beta(u-s)+B_{u}-B_{s}-(g^{n}_{u}-g^{n}_{s})>0,\forall
s\leq u\leq t|\tau=s)\mathrm{d}\Gamma[\ell]^{c}_{s}$ $\displaystyle\leq$
$\displaystyle\int_{[0,t]}\mathbb{P}(X^{n}_{s}+\beta(u-s)+B_{u}-B_{s}>0,\forall
s\leq u\leq t|\tau=s)\mathrm{d}\Gamma[\ell]^{c}_{s}.$

On the event $\\{\tau=s\\}$, we have $X_{s}\leq 0$, so $X^{n}_{s}\leq
X^{n}_{s}-X_{s}=g_{s}-g^{n}_{s}$. We conclude, using the reflection principle
(see e.g. [21, Section 2.8A]),

$\displaystyle\int_{[0,t]}\mathbb{P}(\tau^{n}>t|\tau=s)\mathrm{d}\Gamma[\ell]^{c}_{s}$
$\displaystyle\leq$
$\displaystyle\int_{[0,t]}\mathbb{P}(\beta(u-s)+B_{u}-B_{s}>g^{n}_{s}-g_{s},\forall
s\leq u\leq t|\tau=s)\mathrm{d}\Gamma[\ell]^{c}_{s}$ $\displaystyle=$
$\displaystyle\int_{[0,t]}\mathbb{P}\left(\inf_{s\leq u\leq
t}\big{(}\beta(u-s)+B_{u}-B_{s}\big{)}>g^{n}_{s}-g_{s}\right)\mathrm{d}\Gamma[\ell]^{c}_{s}$
$\displaystyle=$ $\displaystyle\int_{[0,t]}\mathbb{P}\left(\inf_{0\leq u\leq
t-s}\big{(}\beta
u+B_{u}\big{)}>g^{n}_{s}-g_{s}\right)\mathrm{d}\Gamma[\ell]^{c}_{s}$
$\displaystyle=$ $\displaystyle\int_{[0,t]}\left({\cal
N}\left(\frac{g_{s}-g^{n}_{s}+\beta(t-s)}{\sqrt{t-s}}\right)-e^{2\beta(g^{n}_{s}-g_{s})}{\cal
N}\left(\frac{g^{n}_{s}-g_{s}+\beta(t-s)}{\sqrt{t-s}}\right)\right)\mathrm{d}\Gamma[\ell]^{c}_{s},$

where ${\cal N}$ denotes the cumulative distribution function of standard
normal. Since $\lim_{n\to\infty}g^{n}_{s}=g_{s}$ holds in a co-countable set,
the integrand above converges $\Gamma[\ell]^{c}-$almost everywhere to $0$.
Consequently, the integral above vanishes as $n\to\infty$ by the dominated
convergence theorem.

For the integral with respect to the jump part, we get

$\displaystyle\int_{[0,t]}\mathbb{P}(\tau^{n}>t|\tau=s)\mathrm{d}\big{(}\Gamma[\ell]_{s}-\Gamma[\ell]^{c}_{s}\big{)}$
$\displaystyle=$ $\displaystyle\sum_{s\leq
t}\mathbb{P}(\tau^{n}>t|\tau=s)\Delta\Gamma[\ell]_{s}$ $\displaystyle=$
$\displaystyle\sum_{s<t}\mathbb{P}(\tau^{n}>t|\tau=s)\mathbb{P}(\tau=s)$
$\displaystyle=$ $\displaystyle\sum_{s<t}\mathbb{P}(\tau^{n}>t,\tau=s),$

where we may take the sum over $s<t$ because $t\in J$. For any $s>0$, we have
by the Portmanteau theorem

$\displaystyle\liminf_{n\to\infty}\ell^{n}_{s-}\geq\ell_{s-}$
$\displaystyle\Longrightarrow$
$\displaystyle\liminf_{n\to\infty}g^{n}_{s-}\geq g_{s-}.$

For $s<t$, we have
$\mathbb{P}(\tau^{n}>t,\tau=s)\leq\mathbb{P}(\forall\varepsilon\in(0,t-s):Z+\beta(s+\varepsilon)+B_{s+\varepsilon}\geq
g^{n}_{(s+\varepsilon)-},\tau=s)$. Taking the $\limsup$ yields, by Fatou’s
lemma,

$\displaystyle\limsup_{n\to\infty}\mathbb{P}(\tau^{n}>t,\tau=s)$
$\displaystyle\leq$
$\displaystyle\mathbb{P}(\forall\varepsilon\in(0,t-s):Z+\beta(s+\varepsilon)+B_{s+\varepsilon}\geq
g_{(s+\varepsilon)-},\tau=s)$ $\displaystyle\leq$
$\displaystyle\mathbb{P}(Z+\beta s+B_{s}\geq g_{s},\tau=s)$
$\displaystyle\leq$ $\displaystyle\mathbb{P}(Z+\beta s+B_{s}=g_{s})=0.$

Again by the dominated convergence theorem, we get that the sum converges to
zero as $n\to\infty$, which proves
$\lim_{n\to\infty}\Gamma[\ell^{n}]_{t}=\Gamma[\ell]_{t}$. Hence,
$\lim_{n\to\infty}\hat{d}(\Gamma[\ell^{n}],\Gamma[\ell])=0$.

To show $\lim_{n\to\infty}d(\Gamma[\ell^{n}],\Gamma[\ell])=0$, it remains to
show $\lim_{n\to\infty}\Gamma[\ell^{n}]_{\infty}=\Gamma[\ell]_{\infty}$. This
is straightforward when $\beta\leq 0$, and we assume $\beta>0$ in the
remaining of proof. For each $z\in\mathbb{R}$, denote by $p(z)$ the
probability that $z+\beta t+B_{t}$ never hits $(-\infty,0]$. We have
$p(z)=1-e^{-2\beta z^{+}}$, and further $\lim_{z\to\infty}p(z)=1$. For each
$t\geq 0$, we get

$\displaystyle\big{|}\Gamma[\ell^{n}]_{\infty}-\Gamma[\ell]_{\infty}\big{|}$
$\displaystyle=$
$\displaystyle\big{|}\mathbb{P}(\tau^{n}=\infty)-\mathbb{P}(\tau=\infty)\big{|}$
$\displaystyle\leq$
$\displaystyle\big{|}\mathbb{P}(\tau^{n}>t)-\mathbb{P}(\tau^{n}=\infty)\big{|}+\big{|}\mathbb{P}(\tau^{n}>t)-\mathbb{P}(\tau>t)\big{|}+\big{|}\mathbb{P}(\tau>t)-\mathbb{P}(\tau=\infty)\big{|}$
$\displaystyle=$
$\displaystyle\big{|}\mathbb{P}(\tau^{n}>t)-\mathbb{P}(\tau^{n}=\infty)\big{|}+\big{|}\Gamma[\ell^{n}]_{t}-\Gamma[\ell]_{t}\big{|}+\big{|}\mathbb{P}(\tau>t)-\mathbb{P}(\tau=\infty)\big{|}.$

Let us estimate the first and third terms.

$\displaystyle\big{|}\mathbb{P}(\tau^{n}>t)-\mathbb{P}(\tau^{n}=\infty)\big{|}$
$\displaystyle=$
$\displaystyle\mathbb{E}[\mathds{1}_{\\{\tau^{n}>t\\}}]-\mathbb{E}[\mathds{1}_{\\{\tau^{n}>t\\}}\mathds{1}_{\\{\tau^{n}=\infty\\}}]$
$\displaystyle=$
$\displaystyle\mathbb{E}[\mathds{1}_{\\{\tau^{n}>t\\}}]-\mathbb{E}\big{[}\mathbb{P}(\tau^{n}=\infty|\tau^{n}>t)\mathds{1}_{\\{\tau^{n}>t\\}}\big{]}.$

On the event $\\{\tau^{n}>t\\}$, we have

$\displaystyle\\{\tau^{n}=\infty\\}$ $\displaystyle=$
$\displaystyle\\{X^{n}_{t}+\beta(s-t)+(B_{s}-B_{t})-(g^{n}_{s}-g^{n}_{t})>0,\forall
s\geq t\\}$ $\displaystyle\supset$
$\displaystyle\\{X^{n}_{t}-c+\beta(s-t)+(B_{s}-B_{t})>0,\forall s\geq
t\\},\quad\mbox{with }c:=G(1),$

which yields by the Markov property
$\mathbb{P}(\tau^{n}=\infty|\tau^{n}>t)\geq p(X^{n}_{t}-c)$ and thus

$\displaystyle\big{|}\mathbb{P}(\tau^{n}>t)-\mathbb{P}(\tau^{n}=\infty)\big{|}=\mathbb{E}\big{[}\mathds{1}_{\\{\tau^{n}>t\\}}\big{(}1-\mathbb{P}(\tau^{n}=\infty|\tau^{n}>t)\big{)}\big{]}\leq\mathbb{E}\big{[}\mathds{1}_{\\{\tau^{n}>t\\}}\big{(}1-p(X^{n}_{t}-c)\big{)}\big{]}.$

Now we couple the process $X_{t}$ by $Y_{t}:=Z+\beta t+B_{t}$ for all $t\geq
0$, so $X^{n}_{t}\geq Y_{t}-c$. Let $\sigma:=\inf\\{t\geq 0:Y_{t}\leq 0\\}$,
and we have $\tau^{n}\leq\sigma$. Hence,

$\displaystyle\big{|}\mathbb{P}(\tau^{n}>t)-\mathbb{P}(\tau^{n}=\infty)\big{|}\leq\mathbb{E}\big{[}\mathds{1}_{\\{\tau^{n}>t\\}}\big{(}1-p(X^{n}_{t}-c)\big{)}\big{]}\leq\big{|}\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-2c)\big{)}\big{]}.$

Similarly, we can establish
$\big{|}\mathbb{P}(\tau>t)-\mathbb{P}(\tau=\infty)\big{|}\leq\big{|}\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-2c)\big{)}\big{]}$,
and thus

$\displaystyle\big{|}\mathbb{P}(\tau^{n}=\infty)-\mathbb{P}(\tau=\infty)\big{|}\leq
2\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-2c)\big{)}\big{]}+\big{|}\Gamma[\ell^{n}]_{t}-\Gamma[\ell]_{t}\big{|}.$

As $\lim_{t\to\infty}Y_{t}=\infty$ holds almost surely, we get by the
dominated convergence theorem,

$\displaystyle\lim_{t\to\infty}\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-2c)\big{)}\big{]}=0.$

For any $\varepsilon>0$, there exists $t_{\varepsilon}$ such that
$\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-2c)\big{)}\big{]}\leq\varepsilon$
for all $t\geq t_{\varepsilon}$. Fix an arbitrary $t\in J$ with
$t>t_{\varepsilon}$. Then we have

$\displaystyle\lim_{n\to\infty}\big{|}\Gamma[\ell^{n}]_{\infty}-\Gamma[\ell]_{\infty}\big{|}\leq
2\varepsilon+\lim_{n\to\infty}\big{|}\Gamma[\ell^{n}]_{t}-\Gamma[\ell]_{t}\big{|}=2\varepsilon,$

which yields the desired result. ∎

Now we prove Theorem 2.

###### Proof of Theorem 2.

(i) We start with the particle system (2). For each $n\geq 0$, denote by
$\Gamma_{N}^{(n)}$ the $n$-th iterate of the operator $\Gamma_{N}$. First we
show $\underline{L}^{N}:=\Gamma_{N}^{(N)}[{0}]$ is the minimal solution to
(2), where $0\in\mathbb{M}$ is the function that is identically equal to zero.
The monotonicity of $\Gamma_{N}$, combined with the fact
$0\preceq\Gamma_{N}[0]=\Gamma_{N}^{(1)}[0]$, yields
$\Gamma_{N}^{(1)}[0]\preceq\Gamma_{N}\big{[}\Gamma_{N}^{(1)}[0]\big{]}=\Gamma_{N}^{(2)}[0]$,
and thus by induction $\Gamma_{N}^{(n-1)}[0]\preceq\Gamma_{N}^{(n)}[0]$ for
all $n\geq 1$. Note that $\underline{\Lambda}^{N}\in\mathbb{M}$ only takes
values in $\\{k/N:k=0,\ldots,N\\}$. For $0\leq n\leq N$, define the stopping
times

$\displaystyle\sigma_{n}:=\inf\left\\{t\geq
0:~{}\Gamma_{N}^{(n)}[{0}]_{t}\geq\frac{n}{N}\right\\}\in[0,\infty].$

We claim that (12) holds for all $1\leq n\leq N$:

$\displaystyle\Gamma_{N}^{(n-1)}[{0}]_{t}=\Gamma_{N}^{(n)}[{0}]_{t},\quad\mbox{for
all }t<\sigma_{n}.$ (12)

Clearly, $\Gamma_{N}^{(0)}$ denotes the identity operator. For $n=1$, observe
that the first jump time $\sigma_{1}$ of $\Gamma_{N}^{(0)}[0]$ coincides with
the first time at which any of the drifted Brownian motions $Z^{i}+\beta
t+B^{i}_{t}$ hits zero, and $\Gamma_{N}^{(1)}[0]$ is equal to zero prior to
the time $\sigma_{1}$, regardless of $\sigma_{1}<\infty$ and
$\sigma_{1}=\infty$. This means that
$\Gamma_{N}^{(0)}[0]=0=\Gamma_{N}^{(1)}[0]$ holds for all $t<\sigma_{1}$. For
the inductive step, assume that (12) is true for all positive integers up to
$n$. Applying $\Gamma_{N}$ to both sides, we obtain

$\displaystyle\Gamma_{N}^{(n)}[{0}]_{t}=\Gamma_{N}^{(n+1)}[{0}]_{t},\quad\mbox{for
all }t<\sigma_{n}.$

We are done if $\sigma_{n}=\infty$, and it suffices to deal with the case
$\sigma_{n}<\infty$. We distinguish two cases. In the first case, we suppose
$\Gamma_{N}^{(n)}[{0}]_{\sigma_{n}}>n/N$. Therefore, we have
$\Gamma_{N}^{(n)}[{0}]_{\sigma_{n}}\geq(n+1)/N$, and hence
$\sigma_{n}=\sigma_{n+1}$, completing the inductive step. For the second case,
we have $\Gamma_{N}^{(n)}[{0}]_{\sigma_{n}}=n/N$, and for
$t\in(\sigma_{n},\sigma_{n+1})$,

$\displaystyle\frac{n}{N}=\Gamma_{N}^{(n)}[{0}]_{\sigma_{n}}\leq\Gamma_{N}^{(n)}[{0}]_{t}\leq\Gamma_{N}^{(n+1)}[{0}]_{t}<\frac{n+1}{N},$

which implies that $\Gamma_{N}^{(n)}[{0}]$ and $\Gamma_{N}^{(n+1)}[{0}]$ agree
on $(0,\sigma_{n+1})$, completing the inductive step. Next we prove that
$\underline{L}^{N}=\Gamma_{N}^{(N)}[{0}]$ solves the particle system. For
$n\geq 0$, repeatedly applying $\Gamma_{N}$, we get

$\displaystyle\Gamma_{N}^{(n)}[{0}]_{t}=\Gamma_{N}^{(n+1)}[{0}]_{t}=\cdots=\underline{L}^{N}_{t}=\Gamma_{N}[\underline{L}_{N}]_{t},\quad\mbox{for
all }t<\sigma_{n}.$

Choosing $n=N$, we obtain
$\underline{L}^{N}_{t}=\Gamma_{N}[\underline{L}_{N}]_{t}$ on $[0,\sigma_{N})$.
If $\sigma_{N}=\infty$, then
$\underline{L}^{N}=\Gamma_{N}[\underline{L}_{N}]$. Otherwise, recall that
$\underline{L}^{N}\preceq\Gamma_{N}[\underline{L}^{N}]$. So we see that
$1\leq\underline{L}^{N}_{\sigma_{N}}=\Gamma_{N}[\underline{L}^{N}]_{\sigma_{N}}\leq
1$, and thus $\underline{L}^{N}=\Gamma_{N}[\underline{L}^{N}]$. To prove the
minimality of $\underline{L}^{N}$, we pick an arbitrary solution $L$. By
definition, we have $0\preceq L$, and thus by iteration,

$\displaystyle\underline{L}^{N}=\Gamma_{N}^{(N)}[0]\preceq\Gamma_{N}^{(N)}[L]=L,$

which implies that $\underline{L}^{N}$ is indeed the minimal solution.

(ii) Next we turn to the McKean-Vlasov equation (3). Define the sequence
$(\Gamma^{(n)}[{0}],\,n\geq 1)\subset\mathbb{M}$, where $\Gamma^{(n)}$ denotes
the $n$-th iterate of $\Gamma$. By construction, we see that the sequence is
non-decreasing with respect to $\preceq$. We show in the following that
$\Gamma^{(n)}[{0}]\stackrel{{\scriptstyle\hat{d}}}{{\to}}\underline{\Lambda}\in\mathbb{M}$
exists, and $\underline{\Lambda}$ is the minimal solution to (3). By
definition, we have ${0}\preceq\Gamma[{0}]$. For each $t\geq 0$, the sequence
$\big{(}\Gamma^{(n)}[{0}]_{t},\,n\geq 1\big{)}$ is non-decreasing and lying in
$[0,1]$, which implies that we can define $\tilde{\Lambda}$ to be its
pointwise limit

$\displaystyle\tilde{\Lambda}_{t}:=\lim_{n\to\infty}\Gamma^{(n)}[{0}]_{t},\quad\mbox{for
all }t\geq 0.$

Clearly, $\tilde{\Lambda}$ is non-decreasing with
$0\leq\tilde{\Lambda}_{t}\leq 1$, so its modification
$\underline{\Lambda}_{t}:=\tilde{\Lambda}_{t+}$ lies in $\mathbb{M}$. This
implies that
$\Gamma^{(n)}[{0}]\stackrel{{\scriptstyle\hat{d}}}{{\to}}\underline{\Lambda}$,
and further by Proposition 8,

$\displaystyle\Gamma[\underline{\Lambda}]=\Gamma\left[\lim_{n\to\infty}\Gamma^{(n)}[{0}]\right]=\lim_{n\to\infty}\Gamma^{(n+1)}[{0}]=\underline{\Lambda}.$

So $\underline{\Lambda}$ solves (3). Suppose $\Lambda$ is another solution. By
definition, it holds that ${0}\preceq\Lambda$, and using the monotonicity of
$\Gamma$ this leads to $\Gamma[{0}]\preceq\Gamma[\Lambda]=\Lambda$. A
straightforward induction shows that $\Gamma^{(n)}[{0}]\preceq\Lambda$ for all
$n\geq 1$. If $t$ is a point of continuity of $\underline{\Lambda}$, then
$\underline{\Lambda}_{t}=\lim_{n\to\infty}\Gamma_{t}^{(n)}[{0}]\leq\Lambda_{t}$,
and further $\underline{\Lambda}\preceq\Lambda$ by the right continuity of
$\underline{\Lambda}$. This proves that $\underline{\Lambda}$ is the minimal
solution.

(iii) For any solution $(X,\Lambda)$ of (3) that is different to
$(\underline{X},\underline{\Lambda})$, there exist $v>u>0$ and $b>a>0$ such
that $\Lambda_{t}>b>a>\underline{\Lambda}_{t}$ for all $t\in[u,v]$. Recall
that $\underline{\tau}:=\inf\\{t\geq 0:\underline{X}_{t}\leq 0\\}$ and
$\tau:=\inf\\{t\geq 0:X_{t}\leq 0\\}$, and we want to show

$\displaystyle\underline{\Lambda}_{\infty}<\Lambda_{\infty}$
$\displaystyle\Longleftrightarrow$
$\displaystyle\mathbb{P}(\tau=\infty)<\mathbb{P}(\underline{\tau}=\infty),$

which is implied by the fact that
$\mathbb{P}(\tau<\infty=\underline{\tau})>0$. Denote by $Y_{t}:=Z+\beta
t+B_{t}$ for $t\geq 0$, then

$\displaystyle\\{\tau<\infty=\underline{\tau}\\}\supseteq\left\\{Y_{t}>\underline{\Lambda}_{t}:\forall
t\in[0,u]\right\\}\cap\left\\{Y_{t}\in[a,b]:\forall
t\in[0,u]\right\\}\cap\left\\{Y_{t}>\underline{\Lambda}_{t}:\forall
t\in[0,u]\right\\},$

which yields the desired result by the Markov property. ∎

##  4 Proof of Theorems 4 and 5

###  4.1 Proof of Theorem 4

Recall that ${\cal C}$ (resp. $\mathbb{M}$) is endowed with the metric $d_{u}$
(resp. $\hat{d}$). Let $\Xi:={\cal C}\times\mathbb{M}$ be endowed with the
product topology, and we denote the corresponding metric by
$d_{u}\otimes\hat{d}$. Then $\Xi$ is Polish. Define accordingly the map
$\iota:\Xi\to{\cal D}$ by $\iota(f,\ell):=f-G(\ell)$. For every $t\geq-1$,
denote by $\lambda_{t}$ the path functionals on ${\cal D}$:

$\displaystyle\lambda_{t}(x):=\mathds{1}_{\\{\tau(x)\leq t\\}}$ with
$\displaystyle\tau(x):=\inf\\{s\geq 0:x_{s}\leq 0\\}.$

Set $\lambda(x):=(\lambda_{t}(x),\,t\geq-1)\in\mathbb{M}$. We say that $x$
satisfies the crossing property if

$\displaystyle\inf_{0\leq s\leq h}x_{\tau+s}-x_{\tau}<0,\quad\mbox{for all
}h>0,$

where we abbreviate $\tau(x)$ by $\tau$. The lemma below summarizes the
properties of $\iota$ and $\lambda$, which is an adaptation from [25]. We
provide its proof for completeness.

######  Lemma 9.

  1. (i)

The embedding $\iota:(\Xi,d_{u}\otimes\hat{d})\to({\cal D},d_{m})$ is
continuous.

  2. (ii)

Let $x\in\iota(\Xi)$ satisfy the crossing property. Then for any sequence
$(x^{n},\,n\geq 1)\subset\iota(\Xi)$ converging to $x$ under $d_{m}$, we have
$\lim_{n\to\infty}\lambda_{t}(x^{n})=\lambda_{t}(x)$ for all $t\geq 0$ in a
co-countable set.

  3. (iii)

Assume that $(\eta^{n})_{n\geq 1}\subset{\cal P}(\Xi)$ is a convergent
sequence with limit $\eta$. Define $\nu^{n}:=\iota(\eta^{n})$ and
$\nu:=\iota(\eta)$. If $\nu-$almost every path satisfies the crossing
property, i.e.

$\displaystyle\nu\left(\left\\{x\in{\cal D}:~{}\inf_{0\leq s\leq
h}x_{\tau+s}-x_{\tau}<0,~{}\forall h>0\right\\}\right)=1,$

then

$\displaystyle\left(\int_{{\cal
D}}\lambda_{t}(x)\mathrm{d}\nu^{n}(x),\,t\geq-1\right)=:\langle\nu^{n},\lambda\rangle\stackrel{{\scriptstyle\hat{d}}}{{\to}}\langle\nu,\lambda\rangle:=\left(\int_{{\cal
D}}\lambda_{t}(x)\mathrm{d}\nu(x),\,t\geq-1\right).$

###### Proof.

(i) Let $(f^{n},\ell^{n})\to(f,\ell)$ in $\Xi$ and $T$ be any point of
continuity of $\ell$. Then we have $f^{n}\to f$ in ${\cal C}([-1,T])$, and
thus in ${\cal D}([-1,T])$. Moreover, as $G(\ell^{n})$ and $G(\ell)$ are non-
decreasing and $\lim_{n\to\infty}G(\ell^{n}_{t})=G(\ell_{t})$ for each point
of continuity $t$ of $G(\ell)$, [28, Corollary 12.5.1] shows that
$G(\ell^{n})\to G(\ell)$ in ${\cal D}([-1,T])$. The conclusion follows from
[28, Theorem 12.7.3].

(ii) Let $J$ be a co-countable set consisting of all points of continuity of
$x$. Recall that [28, Theorem 13.4.1] ensures $\lim_{n\to\infty}\inf_{-1\leq
s\leq t}x^{n}_{s}=\inf_{-1\leq s\leq t}x_{s}$ for all $t\in J$. Hence, for any
$t\in J$ such that $\inf_{-1\leq s\leq t}x_{s}\neq 0$ we have

$\displaystyle\lim_{n\to\infty}\lambda_{t}(x^{n})=\lim_{n\to\infty}{\mathds{1}}_{\\{\inf_{-1\leq
s\leq t}x^{n}_{s}\leq 0\\}}={\mathds{1}}_{\\{\inf_{-1\leq s\leq t}x_{s}\leq
0\\}}=\lambda_{t}(x).$

Now let $t\in J$ with $\inf_{-1\leq s\leq t}x_{s}=0$. Then $t\geq\tau(x)$ by
definition. If $t>\tau(x)$, then the crossing property implies that
$\inf_{-1\leq s\leq t}x_{s}<0$, which leads to a contraction. It follows that
the only point in $J$ where the convergence may fail is $\tau(x)$, so
$J\setminus\\{\tau(x)\\}$ is still a co-countable set on which we have the
desired convergence.

(iii) For ease of presentation, denote
$\langle\nu^{n},\lambda_{t}\rangle=:\ell^{n}_{t}$ and
$\ell^{n}:=(\ell^{n}_{t})_{t\geq 0}\in\mathbb{M}$. By the compactness of
$\mathbb{M}$, we may assume that $\ell^{n}_{t}\to\ell_{t}$ for some
$\ell\in\mathbb{M}$ at every point of continuity $t$ of $\ell$. Then it
suffices to prove $\ell=\langle\nu,\lambda\rangle$. Let $T>0$ and
$g:[0,T]\to\mathbb{R}$ be bounded and measurable. The dominated convergence
theorem implies

$\displaystyle\lim_{n\to\infty}\int_{0}^{T}\ell^{n}_{t}g_{t}\mathrm{d}t=\int_{0}^{T}\ell_{t}g_{t}\mathrm{d}t.$

By the Skorokhod representation theorem (see e.g. [20, Theorem 4.30]), we can
write

$\displaystyle\int_{0}^{T}\ell^{n}_{t}g_{t}\mathrm{d}t=\mathbb{E}\left[\int_{0}^{T}\lambda_{t}(Y^{n})g_{t}\mathrm{d}t\right],$

where $Y^{n}$ converges almost surely in ${\cal D}$ to $Y$ with ${\cal
L}(Y^{n})=\nu^{n}$ and ${\cal L}(Y)=\nu$. By (ii), it holds with probability
one that

$\displaystyle\lim_{n\to\infty}\int_{0}^{T}\lambda_{t}(Y^{n})g_{t}\mathrm{d}t=\int_{0}^{T}\lambda_{t}(Y)g_{t}\mathrm{d}t.$

Using the dominated convergence theorem again, we get

$\displaystyle\int_{0}^{T}\ell_{t}g_{t}\mathrm{d}t=\lim_{n\to\infty}\mathbb{E}\left[\int_{0}^{T}\lambda_{t}(Y^{n})g_{t}\mathrm{d}t\right]=\mathbb{E}\left[\int_{0}^{T}\lambda_{t}(Y)g_{t}\mathrm{d}t\right]=\int_{0}^{T}\langle\nu,\lambda_{t}\rangle
g_{t}\mathrm{d}t.$

It follows from the right-continuity of $\ell$ that
$\ell=\langle\nu,\lambda\rangle$ for all $t\leq T$. Since $T>0$ is arbitrary,
the claim follows. ∎

Let $(X^{N,1},\ldots,X^{N,N},L^{N})$ be a solution to (2). We define the
empirical measures $(\xi^{N},\,N\geq 1)$ and $(\mu^{N},\,N\geq 1)$ taking
values in ${\cal P}(\Xi)$ and ${\cal P}({\cal D})$ by

$\displaystyle\xi^{N}:=\frac{1}{N}\sum_{i=1}^{N}\delta_{(F^{N,i},L^{N})}$ and
$\displaystyle\mu^{N}:=\frac{1}{N}\sum_{i=1}^{N}\delta_{X^{N,i}},$ (13)

where $F^{N,i}:=(Z^{N,i}+\beta t+B^{i}_{t},\,t\geq 0)$. By definition, we have
$\mu^{N}=\iota(\xi^{N})$ for all $N\geq 1$. We need a few preliminary results
on the sequence of random measures $(\xi^{N},\,N\geq 1)$.

######  Lemma 10.

The sequence $(\xi^{N})_{N\geq 1}$ is tight under Assumption 3.

###### Proof.

The assumption yields the tightness of $(Z^{N,1}+\beta t+B^{1}_{t},\,t\geq
0)$, $N\geq 1$, and thus that of $((Z^{N,1}+\beta t+B^{1}_{t},\,t\geq
0),\,L^{N})$, $N\geq 1$ as $(\mathbb{M},\hat{d})$ is compact. By Sznitman’s
theorem (see e.g. [26, Proposition 2.2]), $(\xi^{N},\,N\geq 1)$ is tight. ∎

######  Proposition 11.

Let $\xi$ be any limit point of $(\xi^{N})_{N\geq 1}$ and $\mu:=\iota(\xi)$.

  1. (i)

For almost every realization $\omega$, if ${\cal L}((F,L))=\xi(\omega)$, then
$(F_{t}-F_{0}-\beta t,\,t\geq 0)$ is a Brownian motion with respect to the
filtration generated by $(F,L)$ In particular, $(F_{t}-F_{0}-\beta t,\,t\geq
0)$ is independent of $F_{0}$.

  2. (ii)

$\mu$ satisfies the crossing property with probability one.

###### Proof.

(i) For any $t>s\geq 0$, $n\in\mathbb{N}$, $0\leq s_{1}\leq\cdots\leq
s_{n}\leq s$ and continuous and bounded functions
$\phi_{k}:\mathbb{R}^{2}\to\mathbb{R}$, denote
$\Phi(f,\ell):=\Pi_{k=1}^{n}\phi_{k}(f_{s_{k}},\ell_{s_{k}})$ for all
$(f,\ell)\in\Xi$. We have

$\displaystyle\mathbb{E}\left[\left(\int_{\Xi}(f_{t}-f_{s}-\beta(t-s))G(f,\ell)\mathrm{d}\xi^{N}(f,\ell)\right)^{2}\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[\left(\frac{1}{N}\sum_{i=1}^{N}(B^{i}_{t}-B^{i}_{s})\Phi(W^{N,i},L^{N})\right)^{2}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{N}\mathbb{E}\left[\left((B^{1}_{t}-B^{1}_{s})\Phi(W^{N,1},L^{N})\right)^{2}\right]+\frac{N^{2}-N}{N^{2}}\mathbb{E}\left[(B^{1}_{t}-B^{1}_{s})(B^{2}_{t}-B^{2}_{s})\Phi(W^{N,1},L^{N})\Phi(W^{N,2},L^{N})\right]$
$\displaystyle=$
$\displaystyle\frac{1}{N}\mathbb{E}\left[\left((B^{1}_{t}-B^{1}_{s})\Phi(W^{N,1},L^{N})\right)^{2}\right]\leq\frac{\|\Phi\|_{\infty}(t-s)}{N}\longrightarrow
0,$

where $W^{N,i}=(W^{N,i}_{t},\,t\geq 0)$ with $W^{N,i}_{t}:=Z^{N,i}+\beta
t+B^{i}_{t}$. Thus, for almost every realization $\omega$, if
$(F,L)\sim\xi(\omega)$, then $(F_{t}-F_{0}-\beta t,\,t\geq 0)$ is a martingale
with respect to the filtration generated by $(F,L)$. Furthermore, repeating
the same arguments to the process $((F_{t}-F_{0}-\beta t)^{2}-t,\,t\geq 0)$,
we obtain that

$\displaystyle\mathbb{E}\left[\left(\int_{\Xi}[(f_{t}-f_{0}-\beta
t)-(f_{s}-f_{0}-\beta
s)-(t-s)]\Phi(f,\ell)\mathrm{d}\xi(f,\ell)\right)^{2}\right]=0,$

which implies that $((W_{t}-W_{0}-\beta t)^{2}-t,\,t\geq 0)$ is also a
martingale with respect to the filtration generated by $(W,L)$. We conclude by
Lévy’s characterization (see e.g. [20, Theorem 18.3]).

(ii) For simplicity, we assume that ${\cal L}(\mu^{N})$ converges to ${\cal
L}(\mu)$. Note that ${\cal L}(\mu)={\cal L}(\iota(\xi))$. For any fixed $h>0$,
we have

$\displaystyle\mathbb{E}\left[\mu\left(\left\\{x\in{\cal D}:\inf_{0\leq s\leq
h}x_{\tau+s}-x_{\tau}=0\right\\}\right)\right]$ $\displaystyle=$
$\displaystyle\mathbb{E}\left[\xi\left(\left\\{(f,\ell)\in\Xi:\inf_{0\leq
s\leq
h}f_{\tau+s}-f_{\tau}-(G(\ell_{\tau+s})-G(\ell_{\tau}))=0\right\\}\right)\right]$
$\displaystyle\leq$
$\displaystyle\mathbb{E}\left[\xi\left(\left\\{(f,\ell)\in\Xi:\inf_{0\leq
s\leq h}f_{\tau+s}-f_{\tau}=0\right\\}\right)\right].$

As for almost every realization $\omega$, if ${\cal L}((F,L))=\xi(\omega)$,
then $\tau=\tau(F-G(L))$ is a stopping time with respect to the filtration
generated by $(F,L)$. Since $F-F_{0}$ is a Brownian motion with respect to the
same filtration, the strong Markov property yields

$\displaystyle\mathbb{E}\left[\xi\left(\left\\{(f,\ell)\in\Xi:\inf_{0\leq
s\leq
h}f_{\tau_{0}+s}-f_{\tau_{0}}=0\right\\}\right)\right]=\mathbb{P}\left(\inf_{0\leq
s\leq h}F_{s}=0\right)=0.$

Hence,

$\displaystyle\mu\left(\left\\{x\in{\cal D}:\inf_{0\leq s\leq
h}x_{\tau_{0}+s}-x_{\tau_{0}}=0\right\\}\right)=0,$

holds almost surely. Repeating this reasoning for $h=h_{n}$ for a sequence
$(h_{n},\,n\geq 1)\subset\mathbb{R}_{+}$ such that $\lim_{n\to\infty}h_{n}=0$
yields the result. ∎

Following the arguments in [11], we have the following proposition which
establishes the convergence of random measures. (see also [8, 25]).

######  Proposition 12.

The sequence $(\xi^{N},\,N\geq 1)$ is tight under Assumption 3. Moreover, let
$\xi$ be any limit point. Then for almost every realization $\omega$,
$\xi(\omega)$ coincides with the law of $(F,\Lambda)$, where
$(F-G(\Lambda),\Lambda)$ is a solution to (3).

###### Proof.

The tightness of $(\xi^{N},\,N\geq 1)$ follows from Lemma 10. Without loss of
generality, we may assume the convergence of $(\xi^{N},\,N\geq 1)$ for ease of
presentation. Then we get $\mu^{N}\to\mu:=\iota(\xi)$ by the continuity of
$\iota$. Consider the maps

$\displaystyle t\mapsto\mathbb{E}[\langle\mu,\lambda_{t}\rangle]$ and
$\displaystyle
t\mapsto\mathbb{E}\left[\int_{\Xi}\ell_{t}\mathrm{d}\xi(f,\ell)\right],$

that are non-decreasing and thus have at most countably discontinuities.
Denote by $J$ the co-countable set of their points of continuity. Combining
Lemma 9 (iii) and Proposition 11 (iii), we deduce that
$\langle\mu^{N},\lambda\rangle\to\langle\mu,\lambda\rangle$ in $\mathbb{M}$
almost surely. For any $t\in J$, we have
$\mathbb{E}[\langle\mu,\lambda_{t}\rangle]=\mathbb{E}[\langle\mu,\lambda_{t-}\rangle]$,
and thus $\langle\mu,\lambda_{t}\rangle=\langle\mu,\lambda_{t-}\rangle$ almost
surely. Therefore, $\langle\mu^{N},\lambda\rangle\to\langle\mu,\lambda\rangle$
in $\mathbb{M}$ which yields

$\displaystyle\lim_{N\to\infty}\langle\mu^{N},\lambda_{t}\rangle=\langle\mu,\lambda_{t}\rangle.$
(14)

Let $\pi_{0}:\Xi\to\mathbb{R}$ be the projection defined by
$\pi(f,\ell)=f_{0}$. By definition, $\pi_{0}$ is continuous, and hence

$\displaystyle\lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^{N}\delta_{Z^{N,i}}=\lim_{N\to\infty}\pi_{0}(\xi^{N})=\pi_{0}(\xi),$

which implies that for almost every realization $\omega$, ${\cal
L}(\pi_{0}(\xi(\omega)))=\lim_{N\to\infty}\theta^{N}=\theta$. By Lemma 11 (i),
for almost every realization $\omega$, if ${\cal L}(F,L)=\xi(\omega)$, then
$(F_{t}-\beta t,\,t\geq 0)$ is Brownian motion with ${\cal L}(F_{0})=\theta$.
Hence, it remains to prove

$\displaystyle\Lambda_{t}=\mathbb{P}\big{(}\tau(F-G(L))\leq
t\big{)}=\langle\mu(\omega),\lambda_{t}\rangle,\quad\mbox{for all }t\geq 0.$

To this end, note for every $t\in J$,

$\displaystyle\mathbb{E}\left[\int_{\Xi}\big{|}|\ell_{t}-\langle\mu,\lambda_{t}\rangle|-|\ell_{t}-\langle\mu^{N},\lambda_{t}\rangle|\big{|}\mathrm{d}\xi^{N}(f,\ell)\right]\leq\mathbb{E}\left[|\langle\mu,\lambda_{t}\rangle-\langle\mu^{N},\lambda_{t}\rangle|\right]$

which vanishes as $N\to\infty$ by (14) and the dominated convergence theorem.
As $t\in J$, we have $\ell_{t-}=\ell_{t}$ for $\xi-$almost every $\ell$, and
thus the map $\mathbb{M}\ni\ell\to\ell_{t}\in\mathbb{R}_{+}$ is continuous
$\xi-$almost surely. It follows that

$\displaystyle\mathbb{E}\left[\int_{\Xi}|\ell_{t}-\langle\mu,\lambda_{t}\rangle|\mathrm{d}\xi(f,\ell)\right]$
$\displaystyle=$
$\displaystyle\lim_{N\to\infty}\mathbb{E}\left[\int_{\Xi}|\ell_{t}-\langle\mu,\lambda_{t}\rangle|\mathrm{d}\xi^{N}(f,\ell)\right]$
$\displaystyle\leq$
$\displaystyle\lim_{N\to\infty}\mathbb{E}\left[\int_{\Xi}|\ell_{t}-\langle\mu^{N},\lambda_{t}\rangle|\mathrm{d}\xi^{N}(f,\ell)\right]~{}~{}=~{}~{}0,$

where the last equality holds since $\langle\mu^{N},\lambda\rangle=\ell$ holds
for $\xi^{N}-$almost every $\ell\in\mathbb{M}$, almost surely. The conclusion
follows by letting $t$ range over a countable dense subset of $J$ and using
the right-continuity of $\ell$. ∎

Now we prove Theorem 4.

###### Proof of Theorem 4.

First, we show that any limit point $\xi$ is equal to ${\cal
L}(\underline{X},\underline{\Lambda})$ almost surely. Consider the perturbed
system: for $i=1,\ldots,N$,

$\displaystyle\tilde{X}_{t}^{L,i}=\tilde{Z}^{N,i}+\beta
t+B^{i}_{t}-G(L_{t}),\quad\mbox{for all }t\geq 0$

and

$\displaystyle\tilde{\Gamma}_{N}[L]_{t}:=\frac{1}{N}\sum_{i=1}^{N}{\mathds{1}}_{\\{\tilde{\tau}^{L}_{i}\leq
t\\}}$ and $\displaystyle\tilde{\tau}^{L}_{i}:=\inf\\{t\geq
0:\tilde{X}^{L,i}_{t}\leq 0\\},$

where ${\cal L}(\tilde{Z}^{N,i})=\theta$ for $1\leq i\leq N$. Taking
$L\equiv\underline{\Lambda}$, we have

$\displaystyle\tilde{X}^{\underline{\Lambda},i}_{t}=\tilde{Z}^{N,i}+\beta
t+B^{i}_{t}-G(\underline{\Lambda}_{t}),\quad\mbox{for all }t\geq 0,$

and $\tilde{\tau}^{\underline{\Lambda}}_{i}$ are i.i.d. random variables that
are identical in law to $\underline{\tau}:=\inf\\{t\geq
0:\underline{X}_{t}\leq 0\\}$. It follows from the inequality of Dvoretzky-
Kiefer-Wolfowitz [23] that,

$\displaystyle\mathbb{P}\left(\|\tilde{\Gamma}_{N}[\underline{\Lambda}]-\underline{\Lambda}\|_{\infty}>\gamma_{N}\right)\leq
2e^{-2N\gamma_{N}^{2}},\quad\mbox{for all }N\geq 1.$

For each $N\geq 1$, define
$A_{N}:=\\{\|\tilde{\Gamma}_{N}[\underline{\Lambda}]-\underline{\Lambda}\|_{\infty}\leq\gamma_{N}\\}$
and the operator $\hat{\Gamma}_{N}$ by

$\displaystyle\hat{\Gamma}_{N}[L]_{t}:=\frac{1}{N}\sum_{i=1}^{N}\mathds{1}_{\\{\hat{\tau}^{L}_{i}\leq
t\\}},$

where for $i=1,\ldots,N$,

$\displaystyle\hat{X}^{L,i}_{t}=\tilde{Z}^{N,i}+\alpha(\gamma_{N})+\beta
t+B^{i}_{t}-G(L_{t})$ and $\displaystyle\hat{\tau}^{L}_{i}=\inf\big{\\{}t\geq
0:~{}\hat{X}^{L,i}_{t}\leq 0\big{\\}}.$

In particular, ${\cal L}(\tilde{Z}^{N,i}+\alpha(\gamma_{N}))=\theta^{N}={\cal
L}(Z^{N,i})$ for $1\leq i\leq N$. On the set $A_{N}$, we have

$\displaystyle\underline{\Lambda}_{t}\geq\tilde{\Gamma}_{N}[\underline{\Lambda}]_{t}-\gamma_{N},\quad\mbox{for
all }t\geq 0.$

Hence,
$\hat{\Gamma}_{N}[0]-\gamma_{N}\preceq\tilde{\Gamma}_{N}[-\gamma_{N}]-\gamma_{N}\preceq\underline{\Lambda}$
holds as $-G(-\gamma_{N})-\alpha(\gamma_{N})\leq-G(0)=0$. Using the
monotonicity again, we get

$\displaystyle\hat{\Gamma}_{N}^{(2)}[0]-\gamma_{N}=\hat{\Gamma}_{N}[\hat{\Gamma}_{N}[0]]-\gamma_{N}\preceq\tilde{\Gamma}_{N}[\hat{\Gamma}_{N}[0]-\gamma_{N}]-\gamma_{N}\preceq\tilde{\Gamma}_{N}[\underline{\Lambda}]-\gamma_{N}\preceq\underline{\Lambda},$

where the second inequality still follows from the uniform continuity of $G$,
i.e.

$\displaystyle-G\big{(}\hat{\Gamma}_{N}[0]_{t}-\gamma_{N}\big{)}\leq\alpha(\gamma_{N})-G\big{(}\hat{\Gamma}_{N}[0]_{t}\big{)},\quad\mbox{for
all }t\geq 0.$

Repeating this arguments, we obtain
$\hat{\Gamma}_{N}^{(n)}[0]-\gamma_{N}\preceq\underline{\Lambda}$ for all
$n\geq 1$, and finally
$\underline{L}^{N}-\gamma_{N}\preceq\underline{\Lambda}$. Hence, it holds with
probability one that

$\displaystyle\limsup_{N\to\infty}\underline{L}^{N}\preceq\underline{\Lambda}.$
(15)

Take an arbitrary limit point $\xi$ of a convergent subsequence, still denoted
by $(\xi^{N},\,N\geq 1)_{N\geq 1}$ for simplicity. It follows that for almost
every realization $\omega$,

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}\delta_{\big{(}\underline{X}^{N,i}(\omega),\underline{L}^{N}(\omega)\big{)}}\to\xi(\omega)$
in $\displaystyle{\cal P}(\Xi).$

On the other hand, there exists a solution $(F,\Lambda)$ to (3) such that
$\xi(\omega)={\cal L}(F,\Lambda)$. Therefore,
$\underline{L}^{N}(\omega)\stackrel{{\scriptstyle\hat{d}}}{{\to}}\Lambda$ and

$\displaystyle\underline{\Lambda}_{t}\geq\lim_{N\to\infty}\underline{L}^{N}_{t}(\omega)=\Lambda_{t}\geq\underline{\Lambda}_{t},\quad\mbox{for
all }t\in J,$

where $J\subset\mathbb{R}_{+}$ is the set of all points of continuity of $L$.
Consequently, $\Lambda=\underline{\Lambda}$ and $\xi(\omega)={\cal
L}(\underline{X},\underline{\Lambda})$. In particular, the dominated
convergence theorem yields

$\displaystyle\lim_{N\to\infty}\mathbb{E}[\underline{L}^{N}_{t}]=\underline{\Lambda}_{t},\quad\mbox{for
all }t\in J.$

Set $c:=G(1)$. By symmetry, we get
$\mathbb{P}(\tau^{N}_{i}=\infty)=\mathbb{P}(\tau^{N}_{1}=\infty)$ for all
$1\leq i\leq N$. Thus,

$\displaystyle\mathbb{E}\left[S_{\infty}^{N}\right]=\frac{1}{N}\sum_{i=1}^{N}\mathbb{P}(\tau^{N}_{i}=\infty)=\mathbb{P}(\tau^{N}_{1}=\infty).$

Consider the two coupled processes below: for all $t\geq 0$,

$\displaystyle X^{N,1}_{t}=Z^{N,1}+\beta t+B^{1}_{t}-G(L^{N}_{t})$ and
$\displaystyle\hat{X}_{t}=(Z^{N,1}-\alpha(\gamma_{N}))+\beta
t+B^{1}_{t}-G(\Lambda_{t}).$

Set $\hat{\tau}:=\inf\\{t\geq 0:\hat{X}_{t}\leq 0\\}$. It suffices to show
$\lim_{N\to\infty}\mathbb{P}(\tau^{N}_{1}=\infty)=\mathbb{P}(\hat{\tau}=\infty)=\mathbb{P}(\tau=\infty)$.
For $z\in\mathbb{R}$, let $p(z):=\mathbb{P}(z+\beta t+B^{1}_{t}>0,\,\forall
t\geq 0)$. Then we have $\lim_{z\to\infty}p(z)=1$. For each $t\geq 0$, we get

$\displaystyle\big{|}\mathbb{P}(\tau^{N}_{1}=\infty)-\mathbb{P}(\hat{\tau}=\infty)\big{|}$
$\displaystyle\qquad\leq\big{|}\mathbb{P}(\tau^{N}_{1}>t)-\mathbb{P}(\tau^{N}_{1}=\infty)\big{|}+\big{|}\mathbb{P}(\tau^{N}_{1}>t)-\mathbb{P}(\hat{\tau}>t)\big{|}+\big{|}\mathbb{P}(\hat{\tau}>t)-\mathbb{P}(\hat{\tau}=\infty)\big{|}$
$\displaystyle\qquad=\big{|}\mathbb{P}(\tau^{N}_{1}>t)-\mathbb{P}(\tau^{N}_{1}=\infty)\big{|}+\big{|}\mathbb{E}[\underline{L}_{t}^{N}]-\underline{\Lambda}_{t}\big{|}+\big{|}\mathbb{P}(\hat{\tau}>t)-\mathbb{P}(\hat{\tau}=\infty)\big{|},$

where $\mathbb{P}(\tau^{N}_{1}>t)=1-\mathbb{E}[\underline{L}_{t}^{N}]$ is by
symmetry. Furthermore,

$\displaystyle\big{|}\mathbb{P}(\tau^{N}_{1}>t)-\mathbb{P}(\tau^{N}_{1}=\infty)\big{|}$
$\displaystyle=$
$\displaystyle\mathbb{E}[\mathds{1}_{\\{\tau^{N}_{1}>t\\}}]-\mathbb{E}[\mathds{1}_{\\{\tau^{N}_{1}>t\\}}\mathds{1}_{\\{\tau^{N}_{1}=\infty\\}}]$
$\displaystyle=$
$\displaystyle\mathbb{E}[\mathds{1}_{\\{\tau^{N}_{1}>t\\}}]-\mathbb{E}\big{[}\mathbb{P}(\tau^{N}_{1}=\infty|\tau^{N}_{1}>t)\mathds{1}_{\\{\tau^{N}_{1}>t\\}}\big{]}.$

On the event $\\{\tau^{N}_{1}>t\\}$, we have

$\displaystyle\\{\tau^{N}_{1}=\infty\\}$ $\displaystyle=$
$\displaystyle\\{X^{N,1}_{t}+\beta(s-t)+(B^{1}_{s}-B^{1}_{t})-(G(L^{N}_{s})-G(L^{N}_{t}))>0,\forall
s\geq t\\}$ $\displaystyle\supset$
$\displaystyle\\{X_{t}^{N,1}-c+\beta(s-t)+(B^{1}_{s}-B^{1}_{t})>0,\forall
s\geq t\\},$

which yields by the Markov property
$\mathbb{P}(\tau^{N}_{1}=\infty|\tau^{N}_{1}>t)\geq p(X^{N,1}_{t}-c)$, and
thus

$\displaystyle\big{|}\mathbb{P}(\tau^{N}_{1}>t)-\mathbb{P}(\tau^{N}_{1}=\infty)\big{|}$
$\displaystyle=$
$\displaystyle\mathbb{E}\big{[}\mathds{1}_{\\{\tau^{N}_{1}>t\\}}\big{(}1-\mathbb{P}(\tau^{N}_{1}=\infty|\tau^{N}_{1}>t)\big{)}\big{]}$
$\displaystyle\leq$
$\displaystyle\mathbb{E}\big{[}\mathds{1}_{\\{\tau^{N}_{1}>t\\}}\big{(}1-p(X^{N,1}_{t}-c)\big{)}\big{]}.$

Set $Y_{t}:=Z^{N,1}+\beta t+B^{1}_{t}$ for all $t\geq 0$ and
$\sigma:=\inf\\{t\geq 0:Y_{t}\leq 0\\}$. Then $X^{N,1}_{t}\geq Y_{t}-c$ and
$\tau^{N}_{1}\leq\sigma$. Hence,

$\displaystyle\big{|}\mathbb{P}(\tau^{N}_{1}>t)-\mathbb{P}(\tau^{N}_{1}=\infty)\big{|}\leq\mathbb{E}\big{[}\mathds{1}_{\\{\tau^{N}_{1}>t\\}}\big{(}1-p(X^{N,1}_{t}-c)\big{)}\big{]}\leq\big{|}\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-2c)\big{)}\big{]}.$

Similarly, we can establish
$\big{|}\mathbb{P}(\hat{\tau}>t)-\mathbb{P}(\hat{\tau}=\infty)\big{|}\leq\big{|}\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-\alpha(\gamma_{N})-2c)\big{)}\big{]}$,
and thus

$\displaystyle\big{|}\mathbb{P}(\tau_{1}=\infty)-\mathbb{P}(\hat{\tau}=\infty)\big{|}\leq
2\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-\alpha(1)-2c)\big{)}\big{]}+\big{|}\mathbb{E}[\underline{L}^{N}_{t}]-\underline{\Lambda}_{t}\big{|}.$

As $\lim_{t\to\infty}Y_{t}=\infty$ holds almost surely, we have by the
dominated convergence theorem,

$\displaystyle\lim_{t\to\infty}\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y_{t}-\alpha(1)-2c)\big{)}\big{]}=0.$

For any $\varepsilon>0$, there exists $t_{\varepsilon}$ such that
$\mathbb{E}\big{[}\mathds{1}_{\\{\sigma>t\\}}\big{(}1-p(Y(t)-\alpha(1)-2c)\big{)}\big{]}\leq\varepsilon$
for all $t\geq t_{\varepsilon}$. Fix an arbitrary $t\in J$ with
$t>t_{\varepsilon}$. Then we get

$\displaystyle\lim_{N\to\infty}\big{|}\mathbb{P}(\tau^{N}_{1}=\infty)-\mathbb{P}(\hat{\tau}=\infty)\big{|}\leq
2\varepsilon+\lim_{N\to\infty}\big{|}\mathbb{E}[\underline{L}_{t}^{N}]-\underline{\Lambda}_{t}\big{|}=2\varepsilon,$

which yields
$\lim_{N\to\infty}\mathbb{E}[\underline{S}_{\infty}^{N}]=1-\underline{\Lambda}_{\infty}$.
Take a sequence $(t_{m})_{m\geq 1}\subset J$ satisfying
$\lim_{m\to\infty}t_{m}=\infty$. Then

$\displaystyle\underline{S}_{\infty}^{N}=1-\underline{L}^{N}_{\infty}\leq
1-\underline{L}^{N}_{t_{m}}$ $\displaystyle\Longrightarrow$
$\displaystyle\limsup_{N\to\infty}\underline{S}_{\infty}^{N}\leq
1-\underline{\Lambda}_{t_{m}}.$

Letting $m\to\infty$, we deduce that
$\limsup_{N\to\infty}\underline{S}_{\infty}^{N}\leq
1-\underline{\Lambda}_{\infty}$. On the other hand, we get by Fatou’s lemma
that
$1-\underline{\Lambda}_{\infty}=\lim_{N\to\infty}\mathbb{E}[\underline{S}_{\infty}^{N}]\leq\mathbb{E}[\limsup_{N\to\infty}\underline{S}_{\infty}^{N}]$,
which implies
$\limsup_{N\to\infty}\underline{S}_{\infty}^{N}=1-\underline{\Lambda}_{\infty}$.
We conclude by applying Fatou’s lemma,

$\displaystyle\limsup_{N\to\infty}\mathbb{E}[|\underline{S}_{\infty}^{N}-1+\underline{\Lambda}_{\infty}|]\leq\mathbb{E}\left[\limsup_{N\to\infty}|\underline{S}_{\infty}^{N}-1+\underline{\Lambda}_{\infty}|\right]=0,$

which yields the desired result. ∎

###  4.2 Proof of Theorem 5

We aim to show that the regularized problem (5) is well-posed. Define the
operators $\Gamma^{\varepsilon}:{\cal D}\to{\cal D}$ as follows: for $f\in D$,
let $\Gamma^{\varepsilon}[f]\in{\cal D}$ be defined by

$\displaystyle\Gamma^{\varepsilon}[f]_{t}:=1-\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{f}_{s})^{-}\mathrm{d}s\right)\right],$

where

$\displaystyle X^{f}_{t}:=Z+\beta t+W_{t}-G(f_{t})$ and
$\displaystyle\tau^{f}:=\inf\\{t\geq 0:~{}X^{f}_{t}\leq 0\\}.$

It is easy to see that $\Gamma^{\varepsilon}$ takes values in ${\cal
C}\cap\mathbb{M}$. By definition, $\ell^{\varepsilon}\in{\cal
C}\cap\mathbb{M}$ is a solution to (5) if and only if $\ell^{\varepsilon}$ is
a fixed point of $\Gamma^{\varepsilon}$. We start with a few technical
results.

######  Lemma 13.

Let $f\in{\cal D}$ be non-decreasing and set

$A_{t}:=\left\\{\inf_{0\leq s\leq t}\big{(}W_{s}-f_{s}\big{)}=0\right\\}.$

Then $A_{t}$ is negligible for all $t>0$.

###### Proof.

Denote by $\tau:=\inf\\{s\geq 0:W_{s}\leq f_{s}\\}$ the hitting time. Then it
holds that

$\displaystyle
A_{t}=\Big{(}A_{t}\cap\\{\tau<t\\}\Big{)}\bigcup\Big{(}A_{t}\cap\\{\tau=t\\}\Big{)}=:B_{t}\cup
C_{t}.$

Clearly $C_{t}\subset\\{W_{t}=f_{t}\\}$. As for $B_{t}$, we have

$\displaystyle B_{t}$ $\displaystyle\subset$ $\displaystyle\\{\tau<t\mbox{ and
}W_{\tau}=f_{\tau}\mbox{ and }W_{s}\geq f_{s}\mbox{ for }s\in[\tau,t]\\}$
$\displaystyle\subset$ $\displaystyle\\{\tau<t\mbox{ and
}W_{\tau}=f_{\tau}\mbox{ and }W_{s}\geq f_{\tau}\mbox{ for
}s\in[\tau,t]\\}:=B_{t}^{\prime},$

where the second inclusion holds as $f$ is non-decreasing. We conclude by the
fact that there are infinite times $u\in[\tau,t]$ so that
$W_{u}<W_{\tau}=f_{\tau}$. ∎

######  Lemma 14.

Let $(f^{n})_{n\geq 1}\subset{\cal D}\cap\mathbb{M}$ be a sequence such that
$n\mapsto f^{n}_{t}$ is non-decreasing for all $t\geq 0$. Denote by $f$ its
pointwise limit. For every $x>0$,

$\displaystyle\lim_{n\to\infty}\mathbb{P}\left(\inf_{0\leq s\leq
t}(x+W_{s}-f^{n}_{s})\leq 0\right)=\mathbb{P}\left(\inf_{0\leq s\leq
t}(x+W_{s}-f_{s})\leq 0\right),$

holds for all $t\geq 0$.

###### Proof.

Denote by $\tau_{n}$ (resp. $\tau$) the first hitting time of $W-f^{n}$ (resp.
$W-f$) at $-x$. Then

$\displaystyle\mathbb{P}\left(\inf_{0\leq s\leq
t}(W_{s}-f^{n}_{s})\leq-x\right)=\mathbb{P}(\tau_{n}\leq t)$ and
$\displaystyle\mathbb{P}\left(\inf_{0\leq s\leq
t}(W_{s}-f_{s})\leq-x\right)=\mathbb{P}(\tau\leq t).$

By assumption, we have $\tau\leq\tau_{n}$, and thus

$\displaystyle 0\leq\mathbb{P}(\tau\leq t)-\mathbb{P}(\tau_{n}\leq
t)=\mathbb{P}(\tau\leq
t,\tau_{n}>t)=\int_{(0,t]}\mathbb{P}(\tau_{n}>t|\tau=s)\mathbb{P}(\tau\in\mathrm{d}s),$

where the last equality holds as $\tau>0$. Restricted on the set
$\\{\tau=s\\}$ for $s\in(0,t]$, we get

$\displaystyle\\{\tau_{n}>t\\}\subset\\{W_{s}-f_{s}^{n}>-x\\}\cap\\{W_{s}-f_{s}\leq-x\\}.$

This implies that

$\displaystyle\mathbb{P}(\tau_{n}>t|\tau=s)\leq\mathbb{P}(f_{s}-x<W_{s}\leq
f_{s}^{n}-x)\to 0\quad\mbox{as }n\to\infty,$

and we conclude by the dominated convergence theorem. ∎

Now we prove Theorem 5.

###### Proof of Theorem 5.

(i) If the existence holds, then any solution is a strong solution. Let
$X^{\varepsilon},Y^{\varepsilon}$ be two arbitrary solutions of (5). Observe
that $|X^{\varepsilon}_{t}-Y^{\varepsilon}_{t}|$ is deterministic and
satisfies

$\displaystyle|X^{\varepsilon}_{t}-Y^{\varepsilon}_{t}|\leq
C\mathbb{E}\left[\frac{1}{\varepsilon}\int_{0}^{t}|(X^{\varepsilon}_{s})^{-}-(Y^{\varepsilon}_{s})^{-}|\mathrm{d}s\right]\leq\frac{C}{\varepsilon}\int_{0}^{t}|X^{\varepsilon}_{s}-Y^{\varepsilon}_{s}|\mathrm{d}s,\quad\mbox{for
all }t\geq 0,$

where $C>0$ denotes the Lipschitz constant of $G$. This yields the uniqueness
by Gronwall’s inequality. To show the existence, we adopt the proof of the
fixed-point theorem. Note that the operator $\Gamma^{\varepsilon}$ preserves
the monotonicity, i.e. $f\preceq
g\Longrightarrow\Gamma^{\varepsilon}[f]\preceq\Gamma^{\varepsilon}[g]$. Denote
$\ell^{0}\equiv 0$ and set for $n\geq 1$,
$\ell^{n}:=\Gamma^{\varepsilon}[\ell^{n-1}]\in{\cal C}\cap\mathbb{M}$. Then we
get by induction $\ell^{n-1}\preceq\ell^{n}$ for all $n\geq 1$. Hence,
$n\mapsto\ell^{n}_{t}$ is non-increasing for all $t\geq 0$, and the pointwise
$\ell_{t}:=\lim_{n\to\infty}\ell^{n}_{t}\in[0,1]$ exists. Finally, rewriting
the equality

$\displaystyle\ell^{n+1}_{t}=G\left(1-\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}\big{(}Z+\beta
s+W_{s}-\alpha\ell^{n}_{s}\big{)}^{-}\mathrm{d}s\right)\right]\right),\quad\mbox{for
all }t\geq 0,$

we conclude the desired existence by the dominated convergence theorem.

(ii) Fix arbitrary $\delta<\varepsilon$. We claim that
$\ell^{\delta}\preceq\ell^{\varepsilon}$. Indeed, it is known from the proof
above that

$\displaystyle\ell^{\delta}=\lim_{n\to\infty}\Gamma^{\delta,n}[\ell^{0}]$ and
$\displaystyle\ell^{\varepsilon}=\lim_{n\to\infty}\Gamma^{\varepsilon,n}n[\ell^{0}],$

where $\Gamma^{\delta,n}$ (resp. $\Gamma^{\varepsilon,n}$) is the
$n-$composition of $\Gamma^{\delta}$ (resp. $\Gamma^{\varepsilon}$). By
definition, we have $\Gamma^{\varepsilon}(f)\preceq\Gamma^{\delta}(f)$ for all
$f\in{\cal D}$. In particular,
$\Gamma^{\varepsilon}(\ell^{0})\preceq\Gamma^{\delta}(\ell^{0})$. Furthermore,
we have

$\displaystyle\Gamma^{\varepsilon,n+1}[\ell^{0}]=\Gamma^{\varepsilon}\big{[}\Gamma^{\varepsilon,n}[\ell^{0}]\big{]}\preceq\Gamma^{\varepsilon}\big{[}\Gamma^{\delta,n}[\ell^{0}]\big{]}\preceq\Gamma^{\delta}\big{[}\Gamma^{\delta,n}[\ell^{0}]\big{]}=\Gamma^{\delta,n+1}[\ell^{0}],$

where the second inequality follows from the induction and the monotonicity of
$\Gamma^{\varepsilon}$. Therefore, we get
$\Gamma^{\varepsilon,n}[\ell^{0}]\preceq\Gamma^{\delta,n}[\ell^{0}]$ for every
$n\geq 1$, and $\ell^{\varepsilon}\preceq\ell^{\delta}$. Namely, for each
$t\geq 0$, $[0,1]\ni\varepsilon\mapsto\ell^{\varepsilon}_{t}\in[0,1]$ is non-
increasing, and we can define the pointwise limit
$\hat{\ell}_{t}:=\lim_{\varepsilon\to 0+}\ell^{\varepsilon}_{t}$. Thanks to
Proposition 15, $\hat{\ell}$ is a solution to (3) .

######  Proposition 15.

$\hat{\ell}$ is a solution to the McKean-Vlasov equation (3).

###### Proof.

Set $\hat{X}_{t}:=Z+\beta t+W_{t}-\alpha\hat{\ell}_{t}$ for $t\geq 0$. By
definition, we have

$\displaystyle X^{\varepsilon}_{t}=Z+\beta
t+W_{t}-G\left(1-\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\right]\right)$

and

$\displaystyle\left|X^{\varepsilon}_{t}-\left(Z+\beta
t+W_{t}-G\left(\mathbb{P}(\inf_{0\leq s\leq t}X^{\varepsilon}_{s}\leq
0)\right)\right)\right|$ $\displaystyle=$
$\displaystyle\left|G\left(1-\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\right]\right)-G\left(\mathbb{E}[\mathds{1}_{\\{\inf_{0\leq
s\leq t}X^{\varepsilon}_{s}\leq 0\\}}]\right)\right|$ $\displaystyle\leq$
$\displaystyle
C\left|1-\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\right]-\mathbb{E}[\mathds{1}_{\\{\inf_{0\leq
s\leq t}X^{\varepsilon}_{s}\leq 0\\}}]\right|$ $\displaystyle=$ $\displaystyle
C\left|\mathbb{E}\left[\mathds{1}_{\\{\inf_{0\leq s\leq
t}X^{\varepsilon}_{s}>0\\}}-\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\right]\right|$
$\displaystyle=$ $\displaystyle
C\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\mathds{1}_{\\{\inf_{0\leq
s\leq t}X^{\varepsilon}_{s}<0\\}}\right],$

where the last equality follows from Lemma 13. Applying the dominated
convergence theorem, we get

$\displaystyle\lim_{\varepsilon\to
0+}\mathbb{E}\left[\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{\varepsilon}_{s})^{-}\mathrm{d}s\right)\mathds{1}_{\\{\inf_{0\leq
s\leq t}X^{\varepsilon}_{s}<0\\}}\right]=0$

and thus $\hat{X}_{t}=Z+\beta t+W_{t}-\alpha\lim_{\varepsilon\to
0+}\mathbb{P}[\inf_{0\leq s\leq t}X^{\varepsilon}_{s}\leq 0]$, which yields
the desired result by Lemma 14. ∎

Recall that $\underline{\Lambda}$ is the minimal solution to (3). Then it
suffices to show $\ell^{\varepsilon}\preceq\underline{\Lambda}$ for every
$\varepsilon>0$. Recall that
$\underline{\Lambda}=\lim_{n\to\infty}\Gamma_{n}[\ell^{0}]$. Again, we prove
$\Gamma^{\varepsilon}[f]\preceq\Gamma[f]$ for all $f\in{\cal D}$. Indeed, we
rewrite

$\displaystyle\Gamma^{\varepsilon}[f]_{t}=G\left(\mathbb{E}\left[1-\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{f}_{s})^{-}\mathrm{d}s\right)\right]\right)$
and $\displaystyle\Gamma[f]_{t}=G\left(\mathbb{E}[\mathds{1}_{\\{\inf_{0\leq
s\leq t}X^{f}_{s}\leq 0\\}}]\right)$

and then compare the two random variables by distinguishing the following
cases:

  * •

On $\\{\inf_{0\leq s\leq t}X^{f}_{s}\leq 0\\}$, it holds

$\displaystyle
1-\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{f}_{s})^{-}\mathrm{d}s\right)\leq
1=\mathds{1}_{\\{\inf_{0\leq s\leq t}X^{f}_{s}\leq 0\\}};$

  * •

On $\\{\inf_{0\leq s\leq t}X^{f}_{s}>0\\}$, it holds

$\displaystyle
1-\exp\left(-\frac{1}{\varepsilon}\int_{0}^{t}(X^{f}_{s})^{-}\mathrm{d}s\right)=0=\mathds{1}_{\\{\inf_{0\leq
s\leq t}X^{f}_{s}\leq 0\\}}.$

This yields in particular
$\Gamma^{\varepsilon}[\ell^{0}]\preceq\Gamma[\ell^{0}]$. Then we have

$\displaystyle\Gamma^{\varepsilon,n+1}[\ell^{0}]=\Gamma^{\varepsilon}\big{[}\Gamma^{\varepsilon,n}[\ell^{0}]\big{]}\preceq\Gamma^{\varepsilon}\big{[}\Gamma^{n}[\ell^{0}]\big{]}\preceq\Gamma\big{[}\Gamma^{n}[\ell^{0}]\big{]}=\Gamma^{n+1}[\ell^{0}],$

where the second inequality follows from the induction and the monotonicity of
$\Gamma^{\varepsilon}$. Letting $n\to\infty$ and then $\delta\to 0$, we obtain
$\hat{\ell}\preceq\underline{\Lambda}$, which implies that
$\hat{\ell}=\underline{\Lambda}$ is the minimal solution of (3). ∎

##  5 Proof of Theorem 6

In this section, we analyze the effect of budget control. Following the same
lines as in the proof of Theorem 2, we can show that (6) has a unique minimal
solution, which yields its wellposedness. In what follows, we distinguish the
cases $\beta<0$, $\beta=0$ and $\beta>0$.

###  5.1 Case of $\beta<0$

###### Proof of Theorem 6 (i).

For each $\phi\in\Phi_{N}$, let
$(X^{\phi,N,1},\ldots,X^{\phi,N,N},L^{\phi,N})$ be an arbitrary solution to
(6). For each $t>0$, we have for $i=1,\ldots,N$,

$\displaystyle\\{\tau^{\phi}_{i}>t\\}=\left\\{\tau^{\phi}_{i}>t,\,\int_{0}^{t}\phi^{i}_{s}ds>-\frac{\beta
t}{2}\right\\}\bigcup\left\\{\tau^{\phi}_{i}>t,\,\int_{0}^{t}\phi^{i}_{s}ds\leq-\frac{\beta
t}{2}\right\\}.$

Hence,

$\displaystyle\sum_{i=1}^{N}{\mathds{1}}_{\\{\tau^{\phi}_{i}>t\\}}$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{N}{\mathds{1}}_{\\{\tau^{\phi}_{i}>t,\,\int_{0}^{t}\phi^{i}_{s}ds>-\frac{\beta
t}{2}\\}}+\sum_{i=1}^{N}{\mathds{1}}_{\\{\tau^{\phi}_{i}>t,\,\int_{0}^{t}\phi^{i}_{s}ds\leq-\frac{\beta
t}{2}\\}}$ $\displaystyle\leq$
$\displaystyle-\frac{2}{\beta}+\sum_{i=1}^{N}{\mathds{1}}_{\\{\tau^{\phi}_{i}>t,\,\int_{0}^{t}\phi^{i}_{s}ds\leq-\frac{\beta
t}{2}\\}}.$

On the event $\\{\tau^{\phi}_{i}>t,\,\int_{0}^{t}\phi^{i}_{s}ds\leq-\beta
t/2\\}$, we get

$\displaystyle X^{\phi,i}_{t}\leq Z^{i}+\frac{\beta
t}{2}+B^{i}_{t},\quad\mbox{for all }t\geq 0,$

and thus

$\displaystyle\left\\{\tau^{\phi}_{i}>t,\int_{0}^{t}\phi^{i}_{s}ds\leq-\frac{\beta
t}{2}\right\\}\subset\left\\{Z^{i}+B^{i}_{t}>-\frac{\beta t}{2}\right\\}.$

Consequently,

$\displaystyle\mathbb{E}[S^{\phi,N}_{\infty}]=\lim_{t\to\infty}\mathbb{E}\left[\sum_{i=1}^{N}{\mathds{1}}_{\\{\tau^{\phi}_{i}>t\\}}\right]\leq-\frac{2}{\beta}+\lim_{t\to\infty}N\mathbb{P}(Z+B_{t}>-\beta
t/2)=-\frac{2}{\beta}.$

∎

###  5.2 Case of $\beta=0$

Theorem 6 (ii) asserts that at criticality where $\beta=0$, the number of
surviving banks scales as $\sqrt{N}$ if the initial capital levels
$X^{N,i}_{0}$ are i.i.d. according to $\theta^{N}$, provided that $\theta$ is
compactly supported. For simplicity, we assume that $\theta^{N}=\delta_{1}$
without loss of generality. Since the model (6) is stochastically bounded from
above by the Up-the-River model via a natural coupling, the upper bound in
(30) follows easily from [27]. The main task is to establish the lower bound,
which boils down to a few lemmas.

Let $\tau^{N}_{(k)}$ be the first time that $k$ entities go bankrupt; that is,
$k$ members of $X^{N,1},\ldots X^{N,N}$ enter $(-\infty,0]$. The following
lemma implies that a significant fraction of banks being ruined cannot occur
in too short time.

######  Lemma 16.

Let $\varepsilon>0$ such that $\alpha\varepsilon<1$, and

$T_{\alpha,\varepsilon}:=\left(\frac{1-\alpha\varepsilon}{\widetilde{\mathcal{N}}^{-1}\left(\frac{1}{2}\varepsilon(1-\varepsilon)^{\frac{1-\varepsilon}{\varepsilon}}\right)}\right)^{2},$
(16)

where $\widetilde{\mathcal{N}}(x):=\mathbb{P}(B_{1}>x)$ is the tail
distribution of standard normal. Then for $T<T_{\alpha,\varepsilon}$, there is
$C>0$ such that

$\mathbb{P}\left(\tau^{N}_{(\varepsilon N)}<T\right)\leq CN^{-k},$ (17)

for any $k>0$, as $N\to\infty$.

###### Proof.

Let $\widetilde{X}^{N,i}_{t}:=(1-\alpha\varepsilon)+B^{i}_{t}$ be driven by
the same Brownian motion as $X^{N,i}$, and $\widetilde{\tau}^{N}_{(k)}$ be the
$k^{th}$ hitting time of
$\\{\widetilde{X}^{N,1},\ldots,\widetilde{X}^{N,N}\\}$ to $0$. By the coupling
of $\widetilde{X}^{N}$ and $X^{N}$, we get for $0\leq
t\leq\tau^{N}_{(\varepsilon N)}$,

$\frac{1}{N}\sum_{k=1}^{N}1_{\\{\tau^{N}_{k}\leq
t\\}}\leq\varepsilon\quad\mbox{and}\quad
X^{N,i}_{t}=1+B^{i}_{t}-\frac{\alpha}{N}\sum_{k=1}^{N}1_{\\{\tau^{N}_{k}\leq
t\\}}\geq\widetilde{X}^{N,i}_{t}.$

Consequently,

$\mathbb{P}\left(\tau^{N}_{(\varepsilon
N)}<T\right)\leq\mathbb{P}\left(\widetilde{\tau}^{N}_{(\varepsilon
N)}<T\right).$ (18)

We proceed to bounding $\mathbb{P}\left(\widetilde{\tau}^{N}_{(\varepsilon
N)}<T\right)$. Let $\mathcal{S}_{\varepsilon}$ be the set of $(\varepsilon
N)$-tuples of $\\{1,\ldots,N\\}$, so
$\\#\mathcal{S}_{\varepsilon}=\binom{N}{\varepsilon N}$ We have

$\displaystyle\mathbb{P}\left(\widetilde{\tau}^{N}_{(\varepsilon N)}<T\right)$
$\displaystyle=\mathbb{P}\left(\bigcup_{S\in\mathcal{S}_{\varepsilon}}\bigg{\\{}\max_{k\in
S}\widetilde{\tau}^{N}_{k}<T\bigg{\\}}\right)$
$\displaystyle\leq\binom{N}{\varepsilon
N}\mathbb{P}\left(\max_{k\leq\varepsilon
N}\widetilde{\tau}^{N}_{k}<T\right)=\binom{N}{\varepsilon
N}\mathbb{P}\left(\widetilde{\tau}^{N}_{1}<T\right)^{\varepsilon N},$ (19)

where the second inequality is by the union bound, and the third equality
follows from the fact that $\widetilde{X}^{N,i}$ are i.i.d. Note that

$\displaystyle\mathbb{P}\left(\widetilde{\tau}^{N}_{1}<T\right)$
$\displaystyle=\mathbb{P}\left(\inf_{0\leq t\leq
T}B^{i}_{t}\leq\alpha\varepsilon-1\right)=\mathbb{P}(|B_{T}|>1-\alpha\varepsilon)=2\widetilde{\mathcal{N}}\left(\frac{1-\alpha\varepsilon}{\sqrt{T}}\right),$

where the second equality is by the reflection principle $\inf_{0\leq t\leq
T}B^{i}_{t}\stackrel{{\scriptstyle d}}{{=}}-|B^{i}_{T}|$ (see [21, Section
2.8A]). As a result,

$\mathbb{P}\left(\widetilde{\tau}^{N}_{1}<T\right)^{\varepsilon
N}=\exp\left(N\varepsilon\log\bigg{(}2\widetilde{\mathcal{N}}\left(\frac{1-\alpha\varepsilon}{\sqrt{T}}\right)\bigg{)}\right).$
(20)

Moreover,

$\binom{N}{\varepsilon
N}\sim\exp\left(N(-\varepsilon\log\varepsilon-(1-\varepsilon)\log(1-\varepsilon))\right),\quad\mbox{as
}N\to\infty,$ (21)

where we only focus on the exponential, and neglect the polynomial factor of
$N$ in (21). Combining (5.2), (20) and (21) yields

$\mathbb{P}\left(\widetilde{\tau}^{N}_{(\varepsilon
N)}<T\right)\lesssim\exp\left(Nh_{\alpha,\varepsilon}(T)\right),$ (22)

where

$h_{\alpha,\varepsilon}(T):=\varepsilon\log\bigg{(}2\widetilde{\mathcal{N}}\left(\frac{1-\alpha\varepsilon}{\sqrt{T}}\right)\bigg{)}-\varepsilon\log\varepsilon-(1-\varepsilon)\log(1-\varepsilon).$
(23)

Observe that $T\to h_{\alpha,\varepsilon}(T)$ is increasing on $(0,\infty)$,
and

$\lim_{T\to
0}h_{\alpha,\varepsilon}(T)=-\infty,\quad\lim_{T\to\infty}h_{\alpha,\varepsilon}(T)=-\varepsilon\log\varepsilon-(1-\varepsilon)\log(1-\varepsilon)>0,$

so the unique root of $h_{\alpha,\varepsilon}(T)=0$ on $(0,\infty)$ is
$T=T_{\alpha,\varepsilon}$ defined by (16). Therefore,
$h_{\alpha,\varepsilon}(T)<0$ for $T<T_{\alpha,\varepsilon}$, and the bound
(17) follows from (18) and (22). ∎

The next lemma shows that many (most) of the banks surviving up to time
$T_{\alpha,\varepsilon}$ have suitably large capital levels.

######  Lemma 17.

Let $\delta>0$, and let

$Z_{\alpha,\varepsilon,\delta}:=\\#\left\\{k:X^{N,k}_{T_{\alpha,\varepsilon}}\geq\alpha+\delta\mbox{
and }X^{N,k}_{t}>0\mbox{ for }t\leq T_{\alpha,\varepsilon}\right\\},$ (24)

be the number of surviving banks whose final capital levels are above
$\alpha+\delta$ at time $T_{\alpha,\varepsilon}$. Let

$p_{\alpha,\,\varepsilon,\delta}:=\widetilde{\mathcal{N}}\left(\left(\frac{\alpha+\delta}{1-\alpha\varepsilon}-1\right)\widetilde{\mathcal{N}}^{-1}\left(\frac{1}{2}\varepsilon(1-\varepsilon)^{\frac{1-\varepsilon}{\varepsilon}}\right)\right).$
(25)

Then we have

$\mathbb{E}Z_{\alpha,\varepsilon,\delta}\geq(1-\varepsilon)p_{\alpha,\varepsilon.\delta}N,$
(26)

and for $\lambda<(1-\varepsilon)p_{\alpha,\varepsilon,\delta}$,

$\mathbb{P}(Z_{\alpha,\varepsilon,\delta}>\lambda N)\geq 1-CN^{-k},$ (27)

for any $k>0$, as $N\to\infty$.

###### Proof.

Let $\Omega_{\alpha,\varepsilon}:=\left\\{\tau^{N}_{(\varepsilon
n)}>T_{\alpha,\varepsilon}\right\\}$, so
$\mathbb{P}(\Omega^{c}_{\alpha,\varepsilon})\lesssim N^{-2}$ by Lemma 16. We
have

$\displaystyle\mathbb{E}Z_{\alpha,\varepsilon,\delta}$
$\displaystyle=\mathbb{E}\left(Z_{\alpha,\varepsilon,\delta}1_{\Omega_{\alpha,\varepsilon}}\right)+\mathbb{E}\left(Z_{\alpha,\varepsilon,\delta}1_{\Omega^{c}_{\alpha,\varepsilon}}\right)$
$\displaystyle=\mathbb{E}\left(Z_{\alpha,\varepsilon,\delta}\,|\,\Omega_{\alpha,\varepsilon}\right)+\mathcal{O}(N^{-1}).$
(28)

On the event $\Omega_{\alpha,\varepsilon}$, there are at least
$(1-\varepsilon)N$ surviving banks, the capital levels of which are
stochastically larger than Brownian motion starting at $1-\alpha\varepsilon$
(see [9]). Thus,
$(Z_{\alpha,\varepsilon,\delta}\,|\,\Omega_{\alpha,\varepsilon})$ is
stochastically larger than ${\rm
Bin}((1-\varepsilon)N,p^{\prime}_{\alpha,\,\varepsilon,\delta})$, where

$p^{\prime}_{\alpha,\,\varepsilon,\delta}:=\mathbb{P}\left(B_{T_{\alpha,\varepsilon}}\geq\alpha(1+\varepsilon)+\delta-1\right)=\widetilde{\mathcal{N}}\left(\frac{\alpha(1+\varepsilon)+\delta-1}{\sqrt{T_{\alpha,\varepsilon}}}\right).$

After simplification, we get
$p^{\prime}_{\alpha,\varepsilon,\delta}=p_{\alpha,\varepsilon,\delta}$.
Combining with (5.2) yields the lower bound (26). The bound (27) follows from
the stochastic comparison and the Chernorff bound. ∎

Now we prove Theorem 6 (ii) .

###### Proof of Theorem 6 (ii).

The difference between (6) and the Up-the-River model is that there is an
additional term $-\frac{\alpha}{N}\sum_{k=1}^{N}1_{\\{\tau^{N}_{k}\leq t\\}}$,
which drags the capital level down. Under the natural coupling of the model
(6) and the Up-the-River model, if a bank survives using a control in the
model (6), it will also survive using the same control in the Up-the-River
model. This implies

$\limsup_{N\to\infty}\frac{\underline{S}^{\phi,N}_{\infty}}{\sqrt{N}}\leq\limsup_{N\to\infty}\frac{S^{\tiny\mbox{UR}}_{N}}{\sqrt{N}},$

where $S^{\tiny\mbox{UR}}_{N}$ denotes the maximum number of surviving banks
in the Up-the-River model. By [27, Theorem 1.2], we have
$\lim_{N\to\infty}\frac{S^{\tiny\mbox{UR}}_{N}}{\sqrt{N}}=4/\sqrt{\pi}$, and
hence the upper bound in (30). Now we prove the lower bound. Fix $m>0$. We use
the following control strategy:

  * •

For each bank, exercise no control until it reaches the capital level $N/m$.

  * •

If the bank does reach the capital level $N/m$, and is the one of the first
$m$ banks achieving so, then exercise the control $1/m$ thereafter.

We write $\underline{S}^{\phi,N}\equiv S_{N}$ without any danger of confusion.
For $\lambda<(1-\varepsilon)p_{\alpha,\varepsilon,\delta}$, let
$\Omega_{\alpha,\varepsilon,\delta,\lambda}:=\\{Z_{\alpha,\varepsilon,\delta}>\lambda
N\\}$. By Lemma 17, we have
$\mathbb{P}(\Omega_{\alpha,\varepsilon,\delta,\lambda})\geq 1-N^{-2}$. On the
event $\Omega_{\alpha,\varepsilon,\delta,\lambda}$, there are (at least)
$\lambda N$ banks whose capital levels are above $\alpha+\delta$ at time
$T_{\alpha,\varepsilon}$. The capital levels of these banks after
$T_{\alpha,\varepsilon}$ are stochastically larger than Brownian motion
starting at $\delta$. Recall from [13, Theorem 7.5.3],

$\mathbb{P}\left(\delta+B_{t}\mbox{ hits }\frac{N}{m}\mbox{ before
}0\right)=\frac{\delta m}{N}.$

As a result, the number of banks whose capital levels reach $K/m$ is bounded
from below by

$\min\left(m,{\rm Bin}(\lambda N,\,\frac{\delta m}{N})\right).$

Also note that ([13, Exercise 7.5.2])

$\mathbb{P}\left(\frac{N}{m}-\alpha+B_{t}+\frac{t}{m}\mbox{ does not hit
}0\right)=1-\exp\left(-\frac{2}{m}\left(\frac{N}{m}-\alpha\right)\right).$

Therefore,

$\displaystyle\mathbb{E}S_{N}$
$\displaystyle\geq\left[1-\exp\left(-\frac{2}{m}\left(\frac{N}{m}-\alpha\right)\right)\right]\mathbb{E}\left[\min\left(m,{\rm
Bin}\bigg{(}\lambda N,\,\frac{\delta
m}{N}\bigg{)}\right)\right]+\mathcal{O}(N^{-1})$
$\displaystyle\stackrel{{\scriptstyle
m=\theta\sqrt{N}}}{{=}}\min(1,\lambda\delta)\,\theta\left(1-\exp\left(-\frac{2}{\theta^{2}}\right)\right)\sqrt{N}+o(1).$

By taking $\lambda\uparrow(1-\varepsilon)p_{\alpha,\varepsilon,\delta}$, we
get

$\liminf_{N\to\infty}\frac{\mathbb{E}S_{N}}{\sqrt{N}}\geq\min\left(1,(1-\varepsilon)\delta
p_{\alpha,\varepsilon,\delta}\right)\sup_{\theta}\left\\{\theta\left(1-\exp\left(-\frac{2}{\theta^{2}}\right)\right)\right\\}.$
(29)

Note that
$\sup_{\theta}\left\\{\theta\left(1-\exp\left(-\frac{2}{\theta^{2}}\right)\right)\right\\}\approx
0.9$, which is attained at $\theta\approx 1.26$. Further by taking
$\varepsilon=\min\left(\frac{1}{2},\frac{1}{2\alpha}\right)$ and
$\delta=\alpha$ yields

$c_{\alpha}\leq\liminf_{N\to\infty}\frac{\mathbb{E}S_{N}}{\sqrt{N}}\leq\limsup_{N\to\infty}\frac{\mathbb{E}S_{N}}{\sqrt{N}}\leq\frac{4}{\sqrt{\pi}},$
(30)

where $c_{\alpha}\approx 0.9\min(1,\rho_{\alpha})$ with

$\rho_{\alpha}:=\left\\{\begin{array}[]{ccl}\left(\alpha-\frac{1}{2}\right)\widetilde{\mathcal{N}}\left((4\alpha-1)\widetilde{\mathcal{N}}^{-1}\left(\frac{1}{4\alpha}\left(1-\frac{1}{2\alpha}\right)^{2\alpha-1}\right)\right)&\mbox{for}&\alpha>1\\\
\frac{1}{2}\alpha\widetilde{\mathcal{N}}\left(\frac{5\alpha-2}{2-\alpha}\,\widetilde{\mathcal{N}}^{-1}(1/8)\right)&\mbox{for}&\alpha\leq
1.\end{array}\right.$ (31)

∎

Note that the lower and upper bounds appearing in the proof above may not be
optimal. It is easy to see from the proof that the lower bound can be
numerically improved to $c^{\prime}_{\alpha}\approx
0.9\min(1,\rho^{\prime}_{\alpha})$ with

$\rho^{\prime}_{\alpha}:=\sup_{\varepsilon,\delta}\left\\{(1-\varepsilon)\delta
p_{\alpha,\varepsilon,\delta}\right\\}.$

But it does not seem that $\rho^{\prime}_{\alpha}$ is easily simplified, and
has a closed-form expression. It is also possible to refine the stochastic
comparisons in Lemmas 16 and 17 to further improve the bounds in (30).

###  5.3 Case of $\beta>0$

Finally, we consider the positive economy where the capital level of each bank
has a positive drift $\beta>0$:

$X^{N,i}_{t}=X^{N,i}_{0}+\beta
t+B_{t}^{i}-\frac{\alpha}{N}\sum_{k=1}^{N}1_{\\{\tau^{N}_{k}\leq t\\}}.$ (32)

The question is whether the budget control is helpful, and what is the optimal
control to maximize the number of surviving banks. As previously mentioned,
this problem is challenging.

To provide insights, we focus on the setting without interaction, i.e.
$\alpha=0$. In this case, the equations

$X_{t}^{\phi,N,i}=X^{N,i}_{0}+\beta t+B_{t}^{i}+\int_{0}^{t}\phi_{s}^{N,i}ds,$
(33)

have a unique solution, and we look for

$S_{N}=\max_{\phi}S^{\phi,N}:=\max_{\phi}\sum_{k=1}^{N}1_{\\{\tau^{\phi,N}_{k}=\infty\\}}.$

where $\tau^{\phi,N}_{k}:=\inf\\{t>0:X^{\phi,N,i}_{t}\leq 0\\}$. For $\phi=0$,
we have $X^{0,N,i}_{t}=X_{0}^{N,i}+\beta t+B^{i}_{t}$, which are i.i.d.
Brownian motion with drift $\beta>0$. By [13, Exercise 7.5.2],

$\mathbb{P}(X^{N,i}_{0}+\beta t+B^{i}_{t}\mbox{ does not hit
}0)=1-\mathbb{E}\left[e^{-2\beta X^{N,1}_{0}}\right].$ (34)

Thus, $\mathbb{E}S^{0,N}=\left(1-\mathbb{E}\left[e^{-2\beta
X^{N,i}_{0}}\right]\right)N$. We aim to prove that in the supercritical case
where $\beta>0$, the number of surviving banks scales as $N$, and moreover,
the unit control does not help. That is, for any $\phi\in\Phi_{N}$,

$\lim_{N\to\infty}\frac{\mathbb{E}S^{\phi,N}}{N}=\lim_{N\to\infty}\frac{\mathbb{E}S^{0,N}}{N}=1-\mathbb{E}\left[e^{-2\beta
X^{N,1}_{0}}\right].$

To this end, we need the following lemma.

######  Lemma 18.

Let $Y^{x}_{t}=x+\beta t+B_{t}$, and $\Psi(t,x):=\mathbb{P}\left(\inf_{s\leq
t}Y^{x}_{s}>0\right)$. Then

  1. 1.

$\Psi(t,x)$ solves the PDE

$\frac{\partial\Psi}{\partial t}=\beta\frac{\partial\Psi}{\partial
x}+\frac{1}{2}\frac{\partial^{2}\Psi}{\partial
t^{2}},\quad\Psi(t,0)=0,\quad\Psi(0,x)=1_{\\{x>0\\}}.$ (35)

  2. 2.

We have

$\Psi(t,x)=\Phi\left(\frac{x+\beta t}{\sqrt{t}}\right)-e^{-2\beta
x}\Phi\left(\frac{-x+\beta t}{\sqrt{t}}\right),$ (36)

where $\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{z^{2}}{2}}dz$
is the cumulative distribution function of standard normal.

###### Proof.

(1) Let $T_{x}:=\inf\\{t>0:Y^{x}_{t}=0\\}$ be the first time at which $Y^{x}$
hits $0$. By duality, we have

$\Psi(t,x):=\mathbb{P}\left(\inf_{s\leq
t}Y^{x}_{s}>0\right)=\mathbb{P}(T_{x}>t),$

which is the Green function of $Y^{x}$ absorbed at $0$. It follows from the
classical Feynman–Kac formula (see [15, Section 5.4]) that $\Psi$ solves the
initial-boundary problem (35).

(2) It is easy to check that the formula (36) satisfies the PDE (35). It can
also be read from [4], [7, Example 5.5]. ∎

###### Proof of Theorem 6 (iii).

First by the natural coupling, we have for any control $\phi\in\Phi_{N}$,

$\mathbb{E}S^{\phi,N}\geq\mathbb{E}S^{0,N}=\left(1-\mathbb{E}\left[e^{-2\beta
X^{N,1}_{0}}\right]\right)N.$

This implies that $\liminf_{N\to\infty}\mathbb{E}S_{N}/N\geq
1-\mathbb{E}\left[e^{-2\beta X^{N,1}_{0}}\right]$. Recall the definition of
$\Psi(t,x)$ in Lemma 18. Let $Z_{s}=\sum_{i=1}^{N}\Psi(t-s,X_{s}^{\phi,N,i})$,
$s\leq t$. Applying Itô’s formula, we get

$\displaystyle Z_{t}$
$\displaystyle=N\Psi(t,X^{N,i}_{0})+\sum_{i=1}^{N}\int_{0}^{t}\left(-\frac{\partial}{\partial
t}+\beta\frac{\partial}{\partial x}+\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\right)\Psi(t-s,X_{s}^{\phi,N,i})ds$ (37)
$\displaystyle\qquad\qquad\qquad+\int_{0}^{t}\sum_{i=1}^{N}\phi^{N,i}(s)\frac{\partial}{\partial
x}\Psi(t-s,X_{s}^{\phi,N,i})ds+\mbox{martingale}$
$\displaystyle=N\Psi(t,X^{N,i}_{0})+\int_{0}^{t}\sum_{i=1}^{N}\phi^{N,i}(s)\frac{\partial}{\partial
x}\Psi(t-s,X_{s}^{\phi,N,i})ds+\mbox{martingale},$

where the second equality follows from (35). Next by (36), we get

$\frac{\partial\Psi}{\partial x}(t,x)=p(t,x+\beta t)+e^{-2\beta x}p(t,x-\beta
t)+2\beta e^{-2\beta x}\Phi\left(\frac{-x+\beta t}{\sqrt{t}}\right),$ (38)

where $p(t,x):=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{x^{2}}{2t}\right)$ is
the heat kernel. Observe that for $x\geq 0$,

$p(t,x+\beta t)\lesssim t^{-\frac{1}{2}}e^{-\beta^{2}t},\quad e^{-2\beta
x}p(t,x-\beta t)\lesssim t^{-\frac{1}{2}}e^{-\beta^{2}t},\quad e^{-2\beta
x}\Phi\left(\frac{-x+\beta t}{\sqrt{t}}\right)\leq 1.$ (39)

By (38) and (39), we get

$\int_{0}^{t}\sum_{i=1}^{N}\phi^{N,i}(s)\frac{\partial}{\partial
x}\Psi(t-s,X_{s}^{\phi,N,i})ds\leq
C\int_{0}^{t}\frac{e^{-\beta^{2}(t-s)}}{\sqrt{t-s}}ds+2\beta t\leq
C^{\prime}+2\beta t.$ (40)

Combining (37) and (40) yields $\mathbb{E}S_{N}\leq\mathbb{E}Z_{t}\leq
N\mathbb{E}\Psi(t,X^{N,1}_{0})+2\beta t+C^{\prime}$ for any $t>0$. Fixing
$\varepsilon>0$ and taking $t=\frac{\varepsilon}{2\beta}N$, we get

$\limsup_{N\to\infty}\frac{\mathbb{E}S_{N}}{N}\leq\lim_{N\to\infty}\mathbb{E}\Psi\left(\frac{\varepsilon
N}{2\beta},X^{N,1}_{0}\right)+\varepsilon=1-\mathbb{E}\left[e^{-2\beta
X^{N,1}_{0}}\right]+\varepsilon.$ (41)

Since (41) holds for any $\varepsilon>0$, we conclude that
$\limsup_{N\to\infty}\mathbb{E}S_{N}/N\leq 1-\mathbb{E}\left[e^{-2\beta
X^{N,1}_{0}}\right]$. ∎

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