$L_{t}=\int_{0}^{t}\kappa^{\varepsilon}(t-s)L_{s}\mathop{}\\!\mathrm{d}{s}+\left[1-\int_{0}^{t}\kappa^{\varepsilon}(t-s)\mathop{}\\!\mathrm{d}{s}\right]L_{t}+\int_{0}^{t}\kappa^{\varepsilon}(t-s)(L_{t}-L_{s})\mathop{}\\!\mathrm{d}{s}.$

Observe

$\left[1-\int_{0}^{t}\kappa^{\varepsilon}(t-s)\mathop{}\\!\mathrm{d}{s}\right]L_{t}\leqslant\frac{2K\varepsilon^{1-\gamma}}{1-\gamma}\qquad\textnormal{and}\qquad\int_{0}^{t}\kappa^{\varepsilon}(t-s)(L_{t}-L_{s})\mathop{}\\!\mathrm{d}{s}\leqslant\frac{K\varepsilon^{1-\gamma}}{1-\gamma}.$

Therefore

$L_{t}=\int_{0}^{t}\kappa^{\varepsilon}(t-s)L_{s}\mathop{}\\!\mathrm{d}{s}+\Psi^{\varepsilon}(t)\qquad\text{
where
}|\Psi^{\varepsilon}(t)|\leqslant\frac{3K\varepsilon^{1-\gamma}}{1-\gamma}\,\forall\,t\in[0,\,t_{0}].$
(4.3)

Step 3: Comparison between the delayed loss and instantaneous loss. By Lemma
3.7, we have that $L\geqslant L^{\varepsilon}$, therefore by following in the
same spirit as [16, Proposition 3.1] we have

$\displaystyle 0\leqslant L_{t}-L_{t}^{\varepsilon}\leqslant
c_{1}\int_{0}^{t}\frac{L_{u}-\mathfrak{L}_{u}^{\varepsilon}}{\sqrt{t-u}}L^{\prime}_{u}\mathop{}\\!\mathrm{d}u\leqslant
c_{1}\int_{0}^{t}\int_{0}^{u}\kappa^{\varepsilon}(u-s)\frac{L_{s}-L_{s}^{\varepsilon}}{\sqrt{t-u}}L^{\prime}_{u}\mathop{}\\!\mathrm{d}s\mathop{}\\!\mathrm{d}u+c_{1}\int_{0}^{t}\frac{\Psi^{\varepsilon}(s)L^{\prime}_{s}}{\sqrt{t-s}}\mathop{}\\!\mathrm{d}s,$

where $c_{1}=\alpha\sqrt{2/\pi}$ and the second inequality follows by (4.3).
As $L\in\mathcal{S}(\gamma,\,K,\,t_{0})$, we have

$0\leqslant\left|L_{t}-L_{t}^{\varepsilon}\right|\leqslant
Kc_{1}\int_{0}^{t}\int_{0}^{u}\frac{\kappa^{\varepsilon}(u-s)\left|L_{s}-L_{s}^{\varepsilon}\right|}{u^{\gamma}\sqrt{t-u}}\mathop{}\\!\mathrm{d}s\mathop{}\\!\mathrm{d}u+Kc_{1}\int_{0}^{t}\frac{\left|\Psi^{\varepsilon}(s)\right|}{s^{\gamma}\sqrt{t-s}}\mathop{}\\!\mathrm{d}s.$

By (4.3), we may find a constant $C_{K,t_{0},\alpha}$ such that the second
term above is bounded by $C_{K,t_{0},\alpha}\varepsilon^{1-\gamma}$.
Therefore,

$\displaystyle 0\leqslant|L_{t}-{L}^{\varepsilon}_{t}|\leqslant
Kc_{1}\int_{0}^{t}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+C_{K,t_{0},\alpha}\varepsilon^{1-\gamma},$
(4.4)

where

$\rho^{\varepsilon}(t,s)=\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)}{u^{\gamma}\sqrt{t-u}}\mathop{}\\!\mathrm{d}u.$

Step 4: Bounds on $\rho^{\varepsilon}(t,s)$
As $\rho^{\varepsilon}$ depends on $t$ and $s$, we may not immediately apply
Gronwall’s lemma or any of its generalisations. Hence we construct upper
bounds to relax the dependence of $\rho^{\varepsilon}$ on $t$ and $s$ via the
function $\kappa$ and this allows us to apply a generalisation of Gronwall’s
lemma. Recall $t\geqslant s$, hence in the case when $t-s\leqslant\varepsilon$

$\displaystyle\rho^{\varepsilon}(t,s)$
$\displaystyle=\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)}{u^{\gamma}\sqrt{t-u}}\mathop{}\\!\mathrm{d}u=\int_{0}^{\frac{t-s}{\varepsilon}}\frac{\kappa(\tilde{u})}{(\varepsilon\tilde{u}+s)^{\gamma}\sqrt{t-s-\varepsilon\tilde{u}}}\mathop{}\\!\mathrm{d}\tilde{u}\leqslant\frac{\left\lVert\kappa\right\rVert_{L^{\infty}}}{s^{\gamma}\varepsilon^{1/2}}\int_{0}^{\frac{t-s}{\varepsilon}}\frac{\mathop{}\\!\mathrm{d}\tilde{u}}{\sqrt{\frac{t-s}{\varepsilon}-\tilde{u}}}$
$\displaystyle=\frac{2\left\lVert\kappa\right\rVert_{L^{\infty}}(t-s)^{1/2}}{s^{\gamma}\varepsilon}\leqslant\frac{2\left\lVert\kappa\right\rVert_{L^{\infty}}}{s^{\gamma}(t-s)^{1/2}},$

where we used the substitution $\tilde{u}=(u-s)\varepsilon^{-1}$. In the case
when ${t-s}>\varepsilon$, as the support of $\kappa^{\varepsilon}$ is in
$[0,\,\varepsilon]$

$\displaystyle\rho^{\varepsilon}(t,s)$
$\displaystyle=\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)}{u^{\gamma}\sqrt{t-u}}\mathop{}\\!\mathrm{d}u=\int^{s+\varepsilon}_{s}\frac{\kappa^{\varepsilon}(u-s)}{u^{\gamma}\sqrt{t-u}}\mathop{}\\!\mathrm{d}u$
$\displaystyle\leqslant\frac{\left\lVert\kappa\right\rVert_{L^{\infty}}}{s^{\gamma}\varepsilon}\int^{s+\varepsilon}_{s}\frac{\mathop{}\\!\mathrm{d}u}{\sqrt{t-u}}=\frac{2\left\lVert\kappa\right\rVert_{L^{\infty}}}{s^{\gamma}}\left[\frac{(t-s)^{1/2}-(t-s-\varepsilon)^{1/2}}{\varepsilon}\right].$

Step 5: Gronwall type argument
Now that we have sufficiently decoupled $\kappa$ from $\rho^{\varepsilon}$, we
may put (4.4) into a form where we may apply a generalised Gronwall Lemma. By
step 4 case 1 and (4.4), we have for $t\leqslant\varepsilon$

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|\leqslant
Kc_{1}\int_{0}^{t}2\left\lVert\kappa\right\rVert_{L^{\infty}}s^{-\gamma}(t-s)^{-1/2}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+C_{K,t_{0},\alpha}\varepsilon^{1-\gamma}.$

By the second case of step 4 and (4.4), we have for $t>\varepsilon$

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|$ $\displaystyle\leqslant
Kc_{1}\int_{0}^{t-\varepsilon}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+Kc_{1}\int_{t-\varepsilon}^{t}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+C_{K,t_{0},\alpha}\varepsilon^{1-\gamma}$
$\displaystyle\leqslant
2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t-\varepsilon}\left[\frac{(t-s)^{1/2}-(t-s-\varepsilon)^{1/2}}{\varepsilon}\right]s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s$
$\displaystyle\phantom{\leqslant}+2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{t-\varepsilon}^{t}(t-s)^{-1/2}s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+C_{K,t_{0},\alpha}\varepsilon^{1-\gamma}$
$\displaystyle\leqslant
2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\sum_{j\geqslant
2}\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s$
$\displaystyle\phantom{\leqslant}+2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t}(t-s)^{-1/2}s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+C_{K,t_{0},\alpha}\varepsilon^{1-\gamma},$

where the last line follows from applying Taylor’s Theorem and the Monotone
Convergence Theorem. We note
$\tilde{C}_{j}\vcentcolon=(2j-2)!/[j!(j-1)!2^{2j-1}]$ is summable. Now we turn
our attention onto the expression in the penultimate line. In the case when
$\varepsilon<t\leqslant 2\varepsilon$,

$\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}s^{-\gamma}\mathop{}\\!\mathrm{d}s\leqslant\tilde{C}_{j}\varepsilon^{j-1}\varepsilon^{\frac{-2j+1}{2}}\int_{0}^{t-\varepsilon}s^{-\gamma}\mathop{}\\!\mathrm{d}s\leqslant\tilde{C}_{j}\varepsilon^{\frac{-1}{2}}\int_{0}^{\varepsilon}s^{-\gamma}\mathop{}\\!\mathrm{d}s\leqslant\frac{\tilde{C}_{j}\varepsilon^{\frac{1}{2}-\gamma}}{1-\gamma}$

where the first inequality follows from the fact that $(-2j+1)/2<0$ as
$j\geqslant 2$ and $t-s\in[\varepsilon,\,t]$ for $s\in[0,\,t-\varepsilon]$. In
the case when $t>2\varepsilon$, we observe that

$\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{\varepsilon}(t-s)^{\frac{-2j+1}{2}}s^{-\gamma}\mathop{}\\!\mathrm{d}s\leqslant\tilde{C}_{j}\varepsilon^{j-1}\varepsilon^{\frac{-2j+1}{2}}\int_{0}^{\varepsilon}s^{-\gamma}\mathop{}\\!\mathrm{d}s\leqslant\frac{\tilde{C}_{j}\varepsilon^{\frac{1}{2}-\gamma}}{1-\gamma},$

and

$\displaystyle\tilde{C}_{j}\varepsilon^{j-1}\int_{\varepsilon}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}s^{-\gamma}\mathop{}\\!\mathrm{d}s$
$\displaystyle\leqslant\tilde{C}_{j}\varepsilon^{j-1}\varepsilon^{-\gamma}\int_{\varepsilon}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s$
$\displaystyle=\tilde{C}_{j}\varepsilon^{j-1-\gamma}\frac{2}{2j-3}\left.(t-s)^{\frac{-2j+3}{2}}\right|_{s=\varepsilon}^{t-\varepsilon}$
$\displaystyle\leqslant\frac{2\tilde{C}_{j}\varepsilon^{j-1-\gamma}\varepsilon^{\frac{-2j+3}{2}}}{2j-3}=\frac{2\tilde{C}_{j}\varepsilon^{1/2-\gamma}}{2j-3}\leqslant\frac{2\tilde{C}_{j}\varepsilon^{1/2-\gamma}}{1-\gamma}.$

Therefore, we have shown that

$\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}s^{-\gamma}\mathop{}\\!\mathrm{d}s\leqslant\frac{3\tilde{C}_{j}\varepsilon^{1/2-\gamma}}{1-\gamma}$

for all $t>\varepsilon$. As $L$ and ${L}^{\varepsilon}$ are bounded by $1$, we
have independent of $t$ being greater or less than $\varepsilon$,

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|$ $\displaystyle\leqslant
2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t}(t-s)^{-1/2}s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+\frac{12Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\varepsilon^{1/2-\gamma}\sum_{j\geqslant
2}\tilde{C}_{j}}{1-\gamma}+C_{K,t_{0},\alpha}\varepsilon^{1-\gamma}$
$\displaystyle=2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t}(t-s)^{-1/2}s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+C_{K,t_{0},\alpha,\gamma}\varepsilon^{1/2-\gamma},$

for any $t\in[0,\,t_{0}]$. Lastly, by Proposition A.10, using
$\tilde{\beta}=1/2$ and $\tilde{\alpha}=1-\gamma$, then
$\tilde{\alpha}+\tilde{\beta}-1>0$ as $\gamma<1/2$ and

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|$ $\displaystyle\leqslant
C_{K,t_{0},\alpha,\gamma}\varepsilon^{1/2-\gamma}\sum_{n\geqslant
0}(2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}})^{n}C_{n}t_{0}^{n(1/2-\gamma)}$
$\displaystyle=C_{K,t_{0},\alpha,\gamma}\varepsilon^{\beta/2}\sum_{n\geqslant
0}(2Kc_{1}t_{0}^{\beta/2}\left\lVert\kappa\right\rVert_{L^{\infty}})^{n}C_{n},$

where the last equality follows from the fact that $\gamma=(1-\beta)/2$. This
completes the proof. ∎

The works of Fasano et al. [14, 12], Di Benedetto et al. [11], and Chayes et
al. [5, 6] extensively investigate the supercooled Stefan cooling problem,
focusing on the existence of a unique solution without blow-ups for all time
or until the entire liquid freezes. Recently, Delarue et al. [10] established
global uniqueness for the system described in (4.1), under the condition that
the density of the initial condition $X_{0-}$ undergoes only a finite number
of changes in monotonicity. In fact, under this assumption, the loss is
continuously differentiable on $(0,\,\gamma)$ for some $\gamma>0$. Moreover,
if the initial density has sufficient regularity, the loss will be
continuously differentiable from the start. Motivated by these results, we
next investigate the rate of convergence when the loss function is
differentiable.

######  Proposition 4.2.

Suppose we have a unique physical solution $(X,L)$ to (4.1) such that
$L\in\mathcal{C}^{1}([0,\,t_{\mathrm{explode}}))$ for some
$t_{\mathrm{explode}}\in(0,\,\infty]$. Then for any
$t_{0}\in(0,\,t_{\mathrm{explode}})$, there exists a constant
$\tilde{K}=\tilde{K}(t_{0})$ such that

$\sup\limits_{s\in[0,\,t_{0}]}|L_{s}-{L}_{s}^{\varepsilon}|\leqslant\tilde{K}\varepsilon^{1/2}.$

###### Proof.

See appendix. ∎

###  4.2 Numerical simulations

Lastly, we investigate the convergence rate of the smoothed loss function
towards the singular loss function through numerical simulations. The
aforementioned estimates, for the case without common noise, provide insights
into the pace at which the smoothed system will approach the singular system,
prior to the decline in regularity of the singular loss function. The proofs
employed in the analysis utilised relatively crude upper bounds, prompting the
question of whether the obtained rates are optimal.

To the best of our knowledge, there is no existing literature on the
regularity of the loss process in the presence of common noise. Consequently,
the theoretical methods employed earlier may not be applicable in this
scenario. Nevertheless, we can still explore the convergence rate in this
context as well. We consider the simplest setting with common noise,

$\begin{cases}X_{t}^{\varepsilon}=X_{0-}+(1-\rho^{2})^{1/2}W_{t}+\rho
W_{t}^{0}-\alpha\mathfrak{L}_{t}^{\varepsilon},\\\
\tau^{\varepsilon}=\inf{\\{t\geqslant 0\,:\,X_{t}^{\varepsilon}\leqslant
0\\}},\\\ L_{t}^{\varepsilon}=\mathbb{P}\left[\tau^{\varepsilon}\leqslant
t\mid\mathcal{F}_{t}^{W^{0}}\right],\\\
\mathfrak{L}_{t}^{\varepsilon}=\int_{0}^{t}\kappa^{\varepsilon}(t-s)L_{s}^{\varepsilon}\mathop{}\\!\mathrm{d}s,\end{cases}\begin{cases}X_{t}=X_{0-}+(1-\rho^{2})^{1/2}W_{t}+\rho
W_{t}^{0}-\alpha{L}_{t},\\\ \tau=\inf{\\{t\geqslant 0\,:\,X_{t}\leqslant
0\\}},\\\ L_{t}=\mathbb{P}\left[\tau\leqslant
t\mid\mathcal{F}^{W^{0}}_{t}\right],\end{cases}$ (4.5)

where $\rho\in[0,\,1)$ is a fixed constant. We propose a numerical scheme that
employs a particle system approximation to compute both the limiting and
smoothed loss functions. Instead of employing numerical integration to compute
the mollified loss of $X^{\varepsilon}$, the system will feel the impulse from
a particle hitting the boundary at a random time in the future sampled from a
random variable whose probability density function is the mollification
kernel. The scheme is given in Algorithm 1.

Require : $N\,-\,$ number of interacting particles

Require : $n\,-\,$ number of time steps: $0<t_{1}<t_{2}<\ldots<t_{n}$

Require : $\varepsilon\,-\,$ the strength of the delay

1 Draw one sample of $W^{0}$

2 Draw $N$ samples of $X_{0-}$, $W$, and $\varsigma$ (r.v. with distribution
$\kappa^{\varepsilon}(t)\mathop{}\\!\mathrm{d}t$)

3 for  _$i=1:n$ _ do

4
$\hat{L}^{\varepsilon}_{t_{i}}=\frac{1}{N}\sum_{m=1}^{N}\mathbbm{1}_{(-\infty,0]}(\min_{t_{j}<t_{i}}\\{\hat{X}_{t_{j}}^{(m)}\\})$

5 for  _$k=1:N$ _ do

6 Update
$\hat{X}_{t_{i}}^{(k)}=X_{0-}^{(k)}+(1-\rho^{2})^{1/2}W_{t_{i}}^{(k)}+\rho
W^{0}_{t_{i}}-\frac{\alpha}{N}\sum_{m=1}^{N}\mathbbm{1}_{(-\infty,0]}(\min_{t_{j}<t_{i}-\varsigma^{(m)}}\\{\hat{X}_{t_{j}}^{(m)}\\})$

7

8 end for

9

10 end for

Algorithm 1 Discrete time Monte Carlo scheme for simulation of the smoothed
loss process with common noise

By setting $\rho$ to zero, the algorithm approximates the loss in the setting
without common noise. To compute the limiting loss function we set $\varsigma$
to zero. In the case when $\rho=0$ and $\varsigma=0$, we recover the numerical
scheme proposed in [20, 21]. In the numerical experiments below, we employed
$10^{6.5}$ particles and used a uniform time discretisation of size
$\Delta_{t}\vcentcolon=\min_{i}\\{\varepsilon_{i}\\}/10$, where
$\\{\varepsilon_{i}\\}_{i}$ is the set of delay values used for the rate of
convergence plots, so that $t_{i}=i\times\Delta_{t}$ in Algorithm 1.

Overall, given sufficient regularity of the loss function, a rate of
convergence close to $1$ is observed. In the other cases studied with Hölder
initial data, with the possibility of there being a jump after the test
interval, and with common noise, the rate of convergence appears to be between
$1/2$ and $1$. See Appendix B for further analysis regarding the rate of
convergence and further examples exploring how $\Delta_{t}$ affects the
estimated rate.

###  4.3 Initial density vanishing at zero and no discontinuity or common
noise

Two different initial conditions were examined in our experimental analysis,
and no discontinuity was observed in either case. In the first simulation, we
set $X_{0-}$ to follow a uniform distribution on $[0.25,0.35]$, with $\alpha$
assigned a value of $0.5$. In the second scenario, $X_{0-}$ was generated from
a gamma distribution with parameters $\left(2.1,\frac{1}{2}\right)$, with
$\alpha$ was set to $1.3$. Interestingly, the data from Fig. 1(b) indicate a
convergence rate of $1$ in both cases. This exceeds the predicted convergence
rate of $1/2$.

(a) $X_{0-}\sim^{d}\operatorname{Uniform}[0.25,\,0.35],\,\alpha=0.5$

(b) Rate of Convergence Figure 1: Initial density vanishing at zero with no
discontinuity and common noise

(c) $X_{0-}\sim^{d}\Gamma(2.1,\frac{1}{2}),\,\alpha=1.3$

(d) Rate of Convergence

###  4.4 Setting with discontinuity and without common noise

To simulate a setting where we would see a systemic event, we changed the
parameters of the Gamma distribution such that most of the mass will be near
the boundary and made $\alpha$ sufficiently large. In Fig. 2(c), we conducted
simulations using two different initial conditions. In the first case, we set
$X_{0-}$ to follow a Gamma distribution with parameters $(1.2,0.5)$ and
assigned $\alpha$ a value of 0.9. In the second case, $X_{0-}$ was generated
from a Gamma distribution with parameters $(1.4,0.5)$, and $\alpha$ was set to
2. Within this particular setup, we observe a convergence rate between $1/2$
and $1$ prior to the occurrence of the first jump. The rate appears to be
unaffected by the characteristics of the density of $X_{0-}$ near the
boundary, despite the theoretical estimates relying on such information.
Moreover, the theoretical estimates consistently predicted a convergence rate
strictly below $1/2$ in all scenarios involving an initial condition of this
form, in contrast to our empirical results, which indicate a convergence rate
greater than $1/2$.

(a) $X_{0-}\sim^{d}\Gamma(1.2,0.5),\,\alpha=0.9$

(b) $t_{max}=0.0001$

(c) Rate of Convergence Figure 2: Initial density vanishing at zero with
discontinuity and no common noise

(d) $X_{0-}\sim^{d}\Gamma(1.4,0.5),\,\alpha=2$

(e) $t_{max}=0.0001$

(f) Rate of Convergence

###  4.5 Simulations with common noise

Similar to the previous subsections, we conducted two experiments. In both
experiments, we assigned $X_{0-}$ a uniform distribution over the interval
$[0.25,0.35]$, set $\alpha$ to $0.5$ and $\rho$ to $0.5$. This initial
condition is the same as in Section 4.3. However, we used different common
noise paths for each experiment. In the first simulation, the common noise
path increases to $1$ over the time domain. This led to the loss process
becoming rougher than the loss in the previous setting. In the second
simulation, the common noise path decreases to $-1$. This induces a systemic
event due to the rapid loss of mass. Despite the differences between the
scenarios, we observed a similar rate of convergence between $1/2$ and $1$ as
illustrated in Fig. 3(c).

(a) Common Noise Path

(b) $X_{0-}\sim^{d}\operatorname{Uniform}[0.25,\,0.35],\,\alpha=0.5$

(c) Rate of Convergence Figure 3: Initial density vanishing at zero with
common noise

(d) Common Noise Path

(e) $X_{0-}\sim^{d}\operatorname{Uniform}[0.25,\,0.35],\,\alpha=0.5$

(f) Rate of Convergence

## Acknowledgement

This research has been supported by the EPSRC Centre for Doctoral Training in
Mathematics of Random Systems: Analysis, Modelling and Simulation
(EP/S023925/1). AP thanks Philipp Jettkant for discussions on this material.

##  Appendix A Technical lemmas

######  Lemma A.1.

Suppose that
$\tilde{P}^{\varepsilon_{n}}=\operatorname{Law}(\tilde{X}^{\varepsilon_{n}},\,\tilde{\mathcal{Y}}^{\varepsilon_{n}})$
converges weakly in
$\mathcal{P}(D_{\mathbb{R}}\times\mathcal{C}_{\mathbb{R}})$ where
$\tilde{X}^{\varepsilon_{n}},$ and $\tilde{\mathcal{Y}}^{\varepsilon_{n}}$ is
the extension of ${X}^{\varepsilon_{n}}$ and
${\mathcal{Y}}^{\varepsilon_{n}}$. Let $X^{*}$ and $Y^{*}$ be the canonical
processes on $D_{\mathbb{R}}\times\mathcal{C}_{\mathbb{R}}$ such that for
$(\eta,\,\omega)\in D_{\mathbb{R}}\times\mathcal{C}_{\mathbb{R}}$,
$X^{*}(\eta,\,\omega)=\eta$ and $Y^{*}(\eta,\,\omega)=\omega$. Then $Y^{*}$ is
a martingale with respect to the filtration generated by $(X^{*},\,Y^{*})$
with quadratic variation

$\left\langle Y^{*}\right\rangle_{t}=\begin{cases}\begin{aligned}
&0&\qquad&t\in[-1,\,0)\\\
&\int_{0}^{t}\sigma(s,\,X^{*}_{s})^{2}\mathop{}\\!\mathrm{d}s&\qquad&t\in[0,\,T]\\\
&\int_{0}^{t}\sigma(s,\,X_{s}^{*})^{2}\mathop{}\\!\mathrm{d}s+(t-T)&\qquad&t\in(T,\,\bar{T}]\end{aligned}\end{cases}$

###### Proof.

Set $\tilde{P}^{*}$ to be the limit point of
$(\tilde{P}^{\varepsilon_{n}})_{n\geqslant 0}$ and

$\mathbbm{T}^{\tilde{P}^{*}}\vcentcolon=\left\\{t\in[-1,\,\bar{T}]\,;\,\tilde{P}^{*}(\eta_{t}=\eta_{t-})=1\right\\}.$

Now for any $s_{0},\,t_{0}\in\mathbbm{T}^{\tilde{P}^{*}}$ with $s_{0}<t_{0}$
and $\\{s_{i}\\}_{i=1}^{k}\subset[-1,\,s_{0}]\cap\mathbbm{T}^{\tilde{P}^{*}}$,
we define the function

$F\;:\;D_{\mathbb{R}}\times\mathcal{C}_{\mathbb{R}}\to\mathbb{R},\quad\eta,\omega\mapsto(\omega_{t_{0}}-\omega_{s_{0}})\prod_{i=1}^{k}f_{i}(\eta_{s_{i}},\omega_{s_{i}}),$

for arbitrary
$f_{i}\in\mathcal{C}_{b}(D_{\mathbb{R}}\times\mathcal{C}_{\mathbb{R}})$. In
order to show that $Y^{*}$ is a martingale, it is sufficient to show that
$\mathbb{E}^{\tilde{P}^{*}}[F(X^{*},\,Y^{*})]=0.$

As $\tilde{P}^{\varepsilon_{n}}\implies\tilde{P}^{*}$, then by Skohorod’s
Representation Theorem, see [2, Theorem 7.6], there are
$((x^{n},\,y^{n}))_{n\geqslant 1}$ and $(x,\,y)$ defined on the same
background space such that $(x^{n},\,y^{n})$ converges to $(x,\,y)$ almost
surely in
$(D_{\mathbb{R}},\,\textnormal{M1})\times(\mathcal{C}_{\mathbb{R}},\,\left\lVert\cdot\right\rVert_{\infty})$
with $\operatorname{Law}(x^{n},\,y^{n})=\tilde{P}^{\varepsilon_{n}}$ and
$\operatorname{Law}(x,y)=\tilde{P}^{*}$. Now for any $p>1$,

$\mathbb{E}\left[\left|F(x^{n},\,y^{n})\right|\right]\leqslant
C\mathbb{E}\left[\underset{s\leqslant\bar{T}}{\sup}|\tilde{\mathcal{Y}}_{s}^{\varepsilon_{n}}|\right]\leqslant
C,$

where $C$ is a constant that changes from line to line and depends only on
$p,\,\sigma,\,T$ and the $f_{i}$’s but is independent of $\varepsilon$.
Therefore $F(x^{n},\,y^{n})$ is uniformly $L^{p}$ bounded. For
$t\in\\{t_{0},\,s_{0},\,s_{1},\,\ldots,\,s_{k}\\}$, it is an almost sure
continuity point of $x$. Therefore by the properties of M1-convergence,
$(x^{n}_{t},\,y^{n}_{t})$ converges to $(x_{t},\,y_{t})$ almost surely. Hence
we have almost sure convergence of $F(x^{n},\,y^{n})$ to $F(x,\,y)$. Vitali’s
Convergence Theorem states that almost sure convergence and uniform
integrability implies convergence of means, hence

$\mathbb{E}^{\tilde{P}^{*}}[F(X^{*},\,Y^{*})]=\mathbb{E}[F(x,\,y)]=\underset{n\xrightarrow[]{}\infty}{\lim}\mathbb{E}[F(x^{n},\,y^{n})]=\underset{n\xrightarrow[]{}\infty}{\lim}\mathbb{E}[F(\tilde{X}^{\varepsilon_{n}},\,\tilde{\mathcal{Y}}^{\varepsilon_{n}})]=0,$

where the last inequality follows from the fact
$\mathbb{E}[F(\tilde{X}^{\varepsilon_{n}},\,\tilde{\mathcal{Y}}^{\varepsilon_{n}})]=0$
for all $n$ as $\tilde{\mathcal{Y}}^{\varepsilon_{n}}$ is a martingale.
Therefore by a monotone class theorem argument, $Y^{*}$ is a continuous local
martingale.

Recall $x^{n}\xrightarrow[]{}x$ almost surely in
$(D_{\mathbb{R}},\,\textnormal{M1})$, hence we have pointwise convergence at
the continuity points of $x$, see [32, Theorem 12.5.1]. As $\sigma$ is in
$\mathcal{C}^{1,\,2}$ by 2.2, there is a set of full probability such that
$\sigma(s,x^{n}_{s})\xrightarrow[]{}\sigma(s,\,x_{s})$ for a set of $s$’s that
have full Lebesgue measure in $[0,\,T]$. Furthermore, as $\sigma$ is bounded,
by the Bounded Convergence Theorem

$\int_{0}^{t}\sigma(s,\,x^{n}_{s})\mathop{}\\!\mathrm{d}s\xrightarrow[]{}\int_{0}^{t}\sigma(s,\,x_{s})\mathop{}\\!\mathrm{d}s$
(A.1)

almost surely for any $t\in[0,\,T]$. Set

$\left\langle Y\right\rangle_{t}=\begin{cases}\begin{aligned}
&0&\qquad&t\in[-1,\,0)\\\
&\int_{0}^{t}\sigma(s,\,X_{s})^{2}\mathop{}\\!\mathrm{d}s&\qquad&t\in[0,\,T]\\\
&\int_{0}^{t}\sigma(s,\,X_{s})^{2}\mathop{}\\!\mathrm{d}s+(t-T)&\qquad&t\in(T,\,\bar{T}]\end{aligned}\end{cases}$

where $Y$ is anyone of $Y^{*},\,y^{n}$ or
$\tilde{\mathcal{Y}}^{\varepsilon_{n}}$ and $X$ is the respective
$X^{*},\,x^{n}$ or $\tilde{X}^{\varepsilon_{n}}$. Employing Eq. A.1,

$F(x^{n},\,(y^{n})^{2}-\langle
y^{n}\rangle)\xrightarrow[]{}F(x,\,y^{2}-\langle
y\rangle)\qquad\textnormal{almost surely.}$

Also by above and the boundness of $\sigma$ by 2.2,
$F(x^{n},\,(y^{n})^{2}-\langle y^{n}\rangle)$ is uniformly $L^{p}$ bounded in
$n$. Hence by Vitali’s Convergence Theorem,

$\displaystyle\mathbb{E}^{\tilde{P}^{*}}\left[F(X^{*},\,(Y^{*})^{2}-\langle
Y^{*}\rangle)\right]$ $\displaystyle=\mathbb{E}\left[F(x,\,y^{2}-\langle
y\rangle)\right]$
$\displaystyle=\underset{n\xrightarrow[]{}\infty}{\lim}\mathbb{E}\left[F(x^{n},\,(y^{n})^{2}-\langle
y^{n}\rangle)\right]$
$\displaystyle=\underset{n\xrightarrow[]{}\infty}{\lim}\mathbb{E}\left[F(\tilde{X}^{\varepsilon_{n}},\,(\tilde{\mathcal{Y}}^{\varepsilon_{n}})^{2}-\langle\tilde{\mathcal{Y}}^{\varepsilon_{n}}\rangle)\right]=0$

where the last inequality follows from the fact that
$(\tilde{\mathcal{Y}}^{\varepsilon_{n}})^{2}-\langle\tilde{\mathcal{Y}}^{\varepsilon_{n}}\rangle$
is true martingale from the boundness of $\sigma$. This completes the proof. ∎

######  Lemma A.2.

Consider the process $Z_{t}=M_{t}+tX$ for $t\in[-1,\,\bar{T}]$ where $M_{t}$
is a continuous local martingale with $c_{M}(t-s)\leqslant\langle
M\rangle_{t}-\langle M\rangle_{s}\leqslant C_{m}(t-s)$ for any $0\leqslant
s<t$ almost surely and $X$ is a non-negative random variable such that
$\mathbb{E}[X]<\infty$. Then for any stopping time $\tau$ where $\tau\geqslant
0$ almost surely, then

$\mathbb{P}\left[\underset{s\in(\tau,(\tau+h)\wedge\bar{T})}{\inf}\\{Z_{s}-Z_{\tau}\\}\geqslant
0,\,\tau<\bar{T}\right]=0,$

for and $h>0$.

###### Proof.

In the case when $M$ is simply a Brownian motion, the result readily follows
by the Stong Markov Property and standard properties of Brownian motion. As
$M$ is a continuous local martingale, we may view it as a (random) time-
changed Brownian motion. We exploit this fact to show the claim. To begin, fix
a $\Delta\in(0,h)$, $\lambda>0$ and set
$\bar{\tau}\vcentcolon=\tau\wedge(\bar{T}-\Delta)$. Then conditioning on the
event
$E\vcentcolon=\\{\tau\leqslant\bar{T}-\Delta\\}\cap\\{X\leqslant\lambda\\}$
and its complement

$\begin{split}\mathbb{P}\left[\underset{s\in(\tau,(\tau+h)\wedge\bar{T})}{\inf}\\{Z_{s}-Z_{\tau}\\}>-\Delta,\,\tau<\bar{T}\right]&\leqslant\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{M_{s}-M_{\bar{\tau}}+(s-\bar{\tau})\lambda\\}>-\Delta\right]\\\
&\qquad+\mathbb{P}\left[E^{\complement},\,\tau<\bar{T}\right]\end{split}$
(A.2)

Focussing on the first term, we observe

$\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{M_{s}-M_{\bar{\tau}}+(s-\bar{\tau})\lambda\\}>-\Delta\right]\leqslant\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{M_{s}-M_{\bar{\tau}}\\}>-\Delta(1+\lambda)\right]$

By the Dubins-Schwarz Theorem, $M$ is a time-changed Brownian motion.
Therefore there exists a Brownian Motion $B$ such that

$\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{M_{s}-M_{\bar{\tau}}\\}>-\Delta(1+\lambda)\right]=\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{B_{\langle
M\rangle_{s}-\langle M\rangle_{\bar{\tau}}}\\}>-\Delta(1+\lambda)\right]$

Now as $\bar{\tau}=\tau\wedge(\bar{T}-\Delta)>0$ almost surely, $\langle
M\rangle_{s}-\langle M\rangle_{\bar{\tau}}\geqslant c_{M}(s-\bar{\tau})$ for
any $s>\bar{\tau}$ almost surely. So

$\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{B_{\langle
M\rangle_{s}-\langle
M\rangle_{\bar{\tau}}}\\}>-\Delta(1+\lambda)\right]\leqslant\mathbb{P}\left[\underset{s\in(0,\Delta)}{\inf}\\{B_{c_{M}s}\\}>-\Delta(1+\lambda)\right]$

By the reflection principle of Brownian motion, we have

$\displaystyle\mathbb{P}\left[\inf_{s\leqslant\Delta}B_{c_{M}s}\leqslant-\Delta(1+\lambda)\right]$
$\displaystyle=2\mathbb{P}\left[B_{\gamma
c_{M}}\leqslant-\Delta(1+\lambda)\right]$
$\displaystyle=2(2\pi)^{-1/2}\int_{-\infty}^{-\Delta^{1/2}c_{M}^{-1/2}(1+\lambda)}e^{\frac{-y^{2}}{2}}\mathop{}\\!\mathrm{d}y\geqslant
1-2\Delta^{1/2}(1+\lambda)(2\pi c_{M})^{-1/2}.$

In conclusion, we have shown

$\mathbb{P}\left[\underset{s\in(\bar{\tau},\bar{\tau}+\delta)}{\inf}\\{M_{s}-M_{\bar{\tau}}+(s-\bar{\tau})\lambda\\}>-\Delta\right]\leqslant\mathbb{P}\left[\underset{s\in(0,\Delta)}{\inf}\\{B_{c_{M}s}\\}>-\Delta(1+\lambda)\right]\leqslant
2\Delta^{1/2}(1+\lambda)(2\pi c_{M})^{-1/2}$

Setting $\lambda=\Delta^{-1/4}$, then by continuity of measure and the above,
the expression in (A.2) converges to 0 as we send $\Delta$ to zero. This
completes the proof. ∎

######  Lemma A.3 (Convergence of Stopping Times).

Consider a sequence of functions $(z^{n})_{n\geqslant 1}$ in $D_{\mathbb{R}}$
converging towards some $z\in D_{\mathbb{R}}$ with respect to the
$M1$-topology. We assume that $z$ has the following crossing property:

$\forall\,h>0\;\tau_{0}(z)<\bar{T}\qquad\implies\qquad\inf_{s\in(\tau_{0}(z),(\tau_{0}(z)+h)\wedge\bar{T})}\left\\{z_{s}-z_{\tau_{0}(z)}\right\\}<0$
(A.3)

where $\tau_{0}$ is defined as in Eq. 2.13 and $\Delta z_{t}\leqslant 0$ for
all $t\in[-1,\bar{T}]$. Then we have

$\lim_{n\to\infty}\tau_{0}(z^{n})=\tau_{0}(z)$

###### Proof.

The proof is composed of two steps. We shall show that
$\limsup_{n\to\infty}\tau_{0}(z^{n})\leqslant\tau_{0}(z)\leqslant\liminf_{n\to\infty}\tau_{0}(z^{n})$.
Hence we will have equality and the claim follows.

Step 1: $\limsup_{n\to\infty}\tau_{0}(z^{n})\leqslant\tau_{0}(z)$
We define the set of continuity points of $z$ to be
$\mathbbm{T}^{z}\vcentcolon=\\{t\in[-1,\bar{T}];z_{t}=z_{t-}\\}$. We remark
that $\mathbbm{T}^{z}$ is co-countable by [2, Section 13]. As $\tau_{0}(z)<T$,
by (A.3) for any fixed $m\in\mathbb{N}$ there exists a
$t\in(\tau_{0}(z),(\tau_{0}(z)+1/m)\wedge\bar{T})\cap\mathbbm{T}^{z}$ such
that $z_{t}<0$. Now, as $t$ is a continuity point of $z$, by [32, Theorem
12.5.1], we have that $z^{n}_{t}\xrightarrow[]{}z_{t}$ in $\mathbb{R}$ as
$n\xrightarrow[]{}\infty$. Therefore for large $n$, $z_{t}^{n}<0$ hence

$\limsup_{n\xrightarrow[]{}\infty}z_{t}^{n}\leqslant
t\leqslant\tau_{0}(z)+\frac{1}{m}.$

As $m\in\mathbb{N}$ was arbitrary, the claim follows.

Step 2: $\liminf_{n\to\infty}\tau_{0}(z^{n})\geqslant\tau_{0}(z)$
As $z^{n}\xrightarrow[]{}z$ in the $M1$-topology, we may find a sequence of
parametric representations $((u^{n},r^{n}))_{n\geqslant 1}$ of
$(z^{n})_{n\geqslant 1}$ which converges uniformly to a parametric
representation $(u,r)$ of $z$, see [32, Theorem 12.5.1]. Therefore, we may
find a $s^{n}\in[0,1]$ such that
$(u^{n}_{s^{n}},r^{n}_{s^{n}})=(z^{n}_{\tau_{0}(z^{n})},\tau_{0}(z^{n}))$. By
step 1, since $\tau_{0}(z)<\bar{T}$, we have have

$\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})\leqslant\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})\leqslant\tau_{0}(z)<T.$

Therefore by the finiteness of
$\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})$ and compactness of $[0,1]$,
we may find a subsequence $n_{k}$ such that
$\tau_{0}(z^{n_{k}})\xrightarrow[]{}\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})$
and $s_{n_{k}}\xrightarrow[]{}s$ for some $s\in[0,1]$. By the uniform
convergence of the parametric representations

$\displaystyle\liminf_{k\xrightarrow[]{}\infty}z^{n_{k}}_{\tau_{0}(z^{n_{k}})}$
$\displaystyle=\liminf_{k\xrightarrow[]{}\infty}u^{n_{k}}_{s^{n_{k}}}=u_{s},$
$\displaystyle\liminf_{k\xrightarrow[]{}\infty}\tau_{0}(z^{n_{k}})$
$\displaystyle=\liminf_{k\xrightarrow[]{}\infty}r^{n_{k}}_{s^{n_{k}}}=r_{s}.$

As $r_{s}=\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})$, we may find
$\gamma\in[0,1]$ such that $u_{s}=\gamma
z_{(\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n}))-}+(1-\gamma)z_{\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})}$.
We also note $u_{s}\leqslant 0$ as
$\liminf_{k\xrightarrow[]{}\infty}z^{n_{k}}_{\tau_{0}(z^{n_{k}})}\leqslant 0$.
Lastly as $\Delta z_{t}\leqslant 0$ for all $t$, we have
$z_{\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})}\leqslant 0$. Therefore,
$\tau_{0}(z)\leqslant{\liminf_{n\xrightarrow[]{}\infty}\tau_{0}(z^{n})}$. This
completes the proof. ∎

######  Lemma A.4 (Functional Continuity II).

Let $\mu\in\mathcal{P}(D_{\mathbb{R}})$ be any measure such that

$\mu\left(\underset{s\in(\tau_{0}(\eta),\,(\tau_{0}(\eta)+h)\wedge\bar{T})}{\inf}\\{\eta_{s}-\eta_{\tau_{0}(\eta)}\\}\geqslant
0,\,\tau_{0}(\eta)<T\right)=0,$ (A.4)

for any $h>0$. Then for any sequence of measures $(\mu^{n})_{n\geqslant 1}$
such that $\mu^{n}\implies\mu$ in
$(\mathcal{P}(D_{\mathbb{R}}),\mathfrak{T}_{\text{M1}}^{\text{wk}})$, we have

$\nu^{\mu^{n}}_{t}\vcentcolon=\mu^{n}(\eta_{t}\in\cdot,\,\tau_{0}(\eta)>t)\implies\nu^{\mu}_{t}\vcentcolon=\mu(\eta_{t}\in\cdot,\,\tau_{0}(\eta)>t),$

in $\mathbf{M}_{\leqslant 1}(\mathbb{R})$, the space of sub-probability
measures on $\mathbb{R}$ endowed with the topology of weak convergence, for
and $t\geqslant 0$ such that $\mu(\eta_{t}=\eta_{t-})=1$ and
$\mu(\tau_{0}(\eta)=t)=0$.

###### Proof.

The proof is an application of the Continuous Mapping Theorem, [2, Theorem
2.7]. We only need to construct $\mu$-almost sure continuous maps.

Step 1: Projection of measures from $D_{\mathbb{R}}$ to
$\mathbb{R}\times\\{0,\,1\\}$
Consider the map

$(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})\,:\,D_{\mathbb{R}}\to\mathbb{R}\times\\{0,1\\},\qquad\eta\mapsto(\eta_{t},\,\mathbbm{1}_{\\{\tau_{0}(\eta)>t\\}})$
(A.5)

(A.5) is a $\mu$-almost sure continuous map. Choose a $\eta\in D_{\mathbb{R}}$
such that $\eta_{t}=\eta_{t-},\,\tau_{0}(\eta)\neq t$ and is in the complement
of the event in (A.4). Such $\eta$’shave full measure under $\mu$.
M1-convergence implies pointwise convergence at continuity points, [32,
Theorem 12.5.1], therefore $X_{t}$ is M1-continuous for every such $\eta$.
Also by Lemma 2.11, since (A.4) holds, $\tau_{0}$ is an M1-continuous map at
$\eta$. As $\tau_{0}(\eta)\neq t$, $\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}}$ is
M1-continuous at $\eta$. Hence,
$(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})$ is a $\mu$-almost sure
continuous map. By the Continuous Mapping Theorem,

$(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})^{\\#}\mu^{n}\implies(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})^{\\#}\mu\quad\text{in}\quad\mathcal{P}(D_{\mathbb{R}}\times\\{0,\,1\\}).$

Step 2: Weak convergence of sub-probability measures
For any $f\in\mathcal{C}_{b}(\mathbb{R})$, define the map

$\hat{f}\,:\,\mathbb{R}\times\\{0,\,1\\}\to\mathbb{R},\qquad(x,y)\mapsto
f(x)\mathbbm{1}_{\\{1\\}}(y).$

It is clear $\hat{f}\in\mathcal{C}_{b}(\mathbb{R}\times\\{0,\,1\\})$. By step
1,

$\langle(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})^{\\#}\mu^{n},\hat{f}\rangle\xrightarrow[]{}\langle(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})^{\\#}\mu,\hat{f}\rangle.$

But by definition,
$\langle(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})^{\\#}\mu^{n},\hat{f}\rangle=\nu^{\mu^{n}}_{t}(f)$
and
$\langle(X_{t},\,\mathbbm{1}_{\\{\tau_{0}(\cdot)\\}})^{\\#}\mu,\hat{f}\rangle=\nu^{\mu}_{t}(f)$.
So $\nu^{\mu^{n}}_{t}(f)\xrightarrow[]{}\nu^{\mu}_{t}(f)$. The conclusion now
follows by Portmanteau’s Theorem. ∎

######  Lemma A.5 (Weak convergence of sub-probability measures).

Suppose that
$\tilde{\mathbf{P}}^{\varepsilon_{n}}\implies\tilde{\mathbf{P}}^{*}$ on
$(\mathcal{P}(D_{\mathbb{R}}),\mathfrak{T}_{\text{M1}}^{\text{wk}})$ for a
positive sequence $(\varepsilon_{n})_{n\geqslant 1}$ which converges to zero.
Set

$\mathbbm{T}\vcentcolon=\left\\{t\in[-1,\bar{T}]\;:\;\mathbb{E}\left[\tilde{\mathbf{P}}^{*}(\eta_{t}=\eta_{t-})\right]=1,\,\mathbb{E}\left[\tilde{\mathbf{P}}^{*}(\tau_{0}(\eta)=t)\right]=0\right\\}.$

Then for any $t\in\mathbbm{T}$,

$\bm{\nu}_{t}^{\varepsilon_{n}}\implies\bm{\nu}_{t}^{*}\vcentcolon=\tilde{\mathbf{P}}^{*}(\eta_{t}\in\cdot,\,\tau_{0}(\eta)>t)\quad\text{in}\quad\mathbf{M}_{\leqslant
1}(\mathbb{R}).$

###### Proof.

By definition of $\mathbbm{T}$ and Lemma 2.9. for any $t\in\mathbbm{T}$ there
is a set of $\mu$’s of full $\operatorname{Law}(\mathbf{P}^{*})$-measure such
that

$\mu(\eta_{t}=\eta_{t-}),\,\mu(\tau_{0}(\eta)=t)=0,\,\mu\left(\underset{s\in(\tau_{0}(\eta),\,(\tau_{0}(\eta)+h)\wedge
T)}{\inf}\\{\eta_{s}-\eta_{\tau_{0}(\eta)}\\}\geqslant
0,\,\tau_{0}(\eta)<T\right)=0.$ (A.6)

As $\tilde{\mathbf{P}}^{\varepsilon_{n}}\implies\tilde{\mathbf{P}}^{*}$, by
Skohorod’s Representation Theorem there exists a $(\mathbf{Q}^{n})_{n\geqslant
1},\,\mathbf{Q}^{*}$ such that $\mathbf{Q}^{n}\xrightarrow[]{}\mathbf{Q}^{*}$
almost surely,
$\operatorname{Law}(\mathbf{Q}^{n})=\operatorname{Law}(\tilde{\mathbf{P}}^{\varepsilon_{n}})$,
$\operatorname{Law}(\mathbf{Q}^{*})=\operatorname{Law}(\tilde{\mathbf{P}}^{*})$
and $\mathbf{Q}^{*}$ satisfies (A.6) almost surely. Set

$\nu_{t}^{\mathbf{Q}^{n}}=\mathbf{Q}^{n}(\eta_{t}\in\cdot,\,\tau_{0}(\eta)>t),\quad\text{and}\quad\nu_{t}^{\mathbf{Q}^{*}}=\mathbf{Q}^{*}(\eta_{t}\in\cdot,\,\tau_{0}(\eta)>t)$

By Lemma A.4,
$\nu_{t}^{\mathbf{Q}^{n}}\xrightarrow[]{}\nu_{t}^{\mathbf{Q}^{*}}$ almost
surely in $\mathbf{M}_{\leqslant 1}(\mathbb{R})$. Now, for any
$F\in\mathcal{C}_{b}(\mathbf{M}_{\leqslant 1}(\mathbb{R}))$, by the Dominated
Convergence Theorem

$\underset{n\xrightarrow[]{}\infty}{\lim}\mathbb{E}\left[F(\bm{\nu}_{t}^{\varepsilon_{n}})\right]=\underset{n\xrightarrow[]{}\infty}{\lim}\mathbb{E}\left[F(\nu_{t}^{\mathbf{Q}^{n}})\right]=\mathbb{E}\left[F(\nu_{t}^{\mathbf{Q}^{*}})\right]=\mathbb{E}\left[F(\bm{\nu}_{t}^{*})\right].$

The result now follows by Portmanteau’s Theorem. ∎

######  Lemma A.6 (Functional Continuity III).

Let $\mu\in\mathcal{P}(D_{\mathbb{R}})$ be any measure such that

$\mu\left(\underset{s\in(\tau_{0}(\eta),\,(\tau_{0}(\eta)+h)\wedge
T)}{\inf}\\{\eta_{s}-\eta_{\tau_{0}(\eta)}\\}\geqslant
0,\,\tau_{0}(\eta)<\bar{T}\right)=0,$ (A.7)

for any $h>0$ and let $g(t,x,\nu)$ be any function satisfying 2.2 i. Then
$\int_{0}^{t}g(s,\eta_{s}^{n},\nu^{\mu^{n}}_{s})\mathop{}\\!\mathrm{d}s$
converges to $\int_{0}^{t}g(s,\eta_{s},\nu^{\mu}_{s})\mathop{}\\!\mathrm{d}s$
for any $t\geqslant 0$ whenever $(\eta^{n},\mu^{n})\xrightarrow[]{}(\eta,\mu)$
in
$(\mathcal{P}(D_{\mathbb{R}}),\mathfrak{T}_{\text{M1}}^{\text{wk}})\times(D_{\mathbb{R}},\textnormal{M1})$
along a sequence for which $\sup_{n\geqslant
1}\langle\mu^{n},\,\sup_{s\leqslant\bar{T}}\left|\tilde{\eta}_{s}\right|^{p}\rangle<\infty$
for some $p>1$ and any $t\geqslant 0$. For any measure
$m\in\mathcal{P}(D_{\mathbb{R}}),\,\nu_{s}^{m}\vcentcolon=m(\tilde{\eta}_{s}\in\cdot,\,\tau_{0}(\tilde{\eta})>s)$.

###### Proof.

By 2.2,

$\left|g(s,\eta_{s}^{n},\nu_{s}^{\mu^{n}})\right|\leqslant
C(1+\sup_{n\geqslant 1}\,\sup_{s\leqslant
T}\left|\eta_{s}^{n}\right|+\sup_{n\geqslant
1}\langle\mu^{n},\,\sup_{s\leqslant T}\left|\tilde{\eta}_{s}\right|\rangle)$
(A.8)

The R.H.S of (A.8) is finite because $\eta^{n}\xrightarrow[]{}\eta$ in
$(D_{\mathbb{R}},\textnormal{M1})$ and by assumption. So it is sufficient to
show $g(s,\eta_{s}^{n},\nu_{s}^{\mu^{n}})$ converges to
$g(s,\eta_{s},\nu_{s}^{\mu})$ on a set of full Lebesgue measure. The
conclusion then follows by the Dominated Convergence Theorem.

Choose an
$s\in\mathbbm{T}^{\mu}\vcentcolon=\left\\{t\in[-1,\bar{T}]\;:\;\mu(\eta_{t}=\eta_{t-})=1,\,\mu(\tau_{0}(\eta)=t)=0\right\\}$.
The by Lemma A.4, $\nu_{s}^{\mu^{n}}\implies\nu_{s}^{\mu}$. By Skohorod’s
Representation Theorem, there exists a $(X^{n})_{n\geqslant 1},\,X$ defined on
a common probability space such that
$\operatorname{Law}(X^{n})-\nu_{s}^{\mu^{n}},\,\operatorname{Law}(X)-\nu_{s}^{\mu}$
and $X^{n}\xrightarrow[]{}X$ almost surely in $\mathbb{R}$. For any
$\psi\in\mathcal{C}(\mathbb{R})$ with
$\left\lVert\psi\right\rVert_{\operatorname{Lip}}\leqslant 1$ and
$\left|\psi(0)\right|\leqslant 1$,

$\left|\langle\right|\nu_{s}^{\mu^{n}}-\nu^{\mu},\,\psi\rangle=\left|\mathbb{E}\left[\psi(X^{n})-\psi(X)\right]\right|\leqslant\mathbb{E}\left|X^{n}-X\right|.$

By assumption,

$\mathbb{E}\left[\left|X^{n}\right|^{p}\right]=\langle\nu_{s}^{\mu^{n}},\,\left|\cdot\right|^{p}\rangle\leqslant\langle\mu^{n},\,\sup_{s\leqslant
T}\left|\tilde{\eta}_{s}\right|^{p}\rangle<\infty,$

uniformly in $n$ for some $p>1$. So $\left|X^{n}-X\right|$ is uniformly
$L^{p}$-bounded and converges to zero almost surely. Therefore, by Vitali’s
Convergence Theorem,

$\underset{n\xrightarrow[]{}\infty}{\lim}d_{0}(\nu_{s}^{\mu^{n}},\,\nu_{s}^{\mu})=0$
(A.9)

Lastly by 2.2,

$\left|g(s,\eta_{s}^{n},\nu_{s}^{\mu^{n}})-g(s,\eta_{s},\nu_{s}^{\mu})\right|\leqslant
C\left|\eta_{s}^{n}-\eta_{s}\right|+C(1+\left|\eta_{s}\right|+\langle\nu_{s}^{\mu^{n}},\,\left|\cdot\right|\rangle)d_{0}(\nu_{s}^{\mu^{n}},\,\nu_{s}^{\mu}).$

As $s\in\mathbbm{T}^{\mu}$, the first term converges to zero as
$\eta^{n}\xrightarrow[]{}\eta$ in $(D_{\mathbb{R}},\textnormal{M1})$. By
assumption and (A.9), the second term converges to zero as
$n\xrightarrow[]{}\infty$. This completes the proof. ∎

######  Lemma A.7.

Fix any $t<T$. There is a constant $C>0$ independent of $\varepsilon$ and $t$
such that for any $\gamma<1\wedge(T-t)$ we have

$\mathbb{P}\left[\bm{\nu}_{t}^{\varepsilon}[0,\alpha_{t}z+C\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t})]\geqslant
z,\;\forall\;z\leqslant
L_{t+\gamma}^{\varepsilon}-L_{t}^{\varepsilon}-C\gamma^{1/3}\right]\geqslant
1-C\gamma^{1/3}$

###### Proof.

To begin, fix a $\gamma>0$ such that $\gamma<1\wedge(T-t)$ and fix a
$z\in\mathbb{R}$. Then we define the event

$E_{1}^{z}\vcentcolon=\Big{\\{}X_{t}^{\varepsilon}-\gamma
C(1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}])-\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|-\alpha_{t}z-\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})-(\alpha_{t+\gamma}-\alpha_{t})\leqslant
0,\,\tau^{\varepsilon}>t\Big{\\}}$

where $C$ is the constant from the linear growth condition on $b$. Now fix
$x\leqslant L_{t+\gamma}^{\varepsilon}-L_{t}^{\varepsilon}$. By the continuity
of the loss process, [17, Theorem 2.4], there exists a $s\leqslant\gamma$ such
that $x=L_{t+s}^{\varepsilon}-L_{t}^{\varepsilon}$. Employing the integration
by parts formula, we observe for any $u\in[t,t+s]$

$\displaystyle\int_{t}^{u}\alpha_{v}\mathop{}\\!\mathrm{d}\mathfrak{L}_{v}^{\varepsilon}$
$\displaystyle=\alpha_{u}\mathfrak{L}_{u}^{\varepsilon}-\alpha_{t}\mathfrak{L}_{t}^{\varepsilon}-\int_{t}^{u}\alpha^{\prime}(v)\mathfrak{L}_{v}^{\varepsilon}\mathop{}\\!\mathrm{d}v\leqslant\alpha_{u}\mathfrak{L}_{u}^{\varepsilon}-\alpha_{t}\mathfrak{L}_{t}^{\varepsilon}$
$\displaystyle\leqslant\alpha_{t+s}{L}_{t+s}^{\varepsilon}-\alpha_{t}\mathfrak{L}_{t}^{\varepsilon}\leqslant\alpha_{t}{L}_{t+s}^{\varepsilon}+(\alpha_{t+s}-\alpha_{t})-\alpha_{t}\mathfrak{L}_{t}^{\varepsilon},$

where to establish upper bounds we use the fact that $\alpha$ is non-negative
and non-decreasing, and $\mathfrak{L}_{v}^{\varepsilon}\leqslant
L_{v}^{\varepsilon}\leqslant 1$ for any $v\geqslant 0$. Therefore for any
$u\in[t,t+s]$

$\displaystyle X_{u}^{\varepsilon}$
$\displaystyle=X_{t}^{\varepsilon}+(X_{u}^{\varepsilon}-X_{t}^{\varepsilon})=X_{t}^{\varepsilon}+\int_{t}^{u}b(v,X_{v}^{\varepsilon},\bm{\nu}_{v}^{\varepsilon})\mathop{}\\!\mathrm{d}v+\mathcal{Y}_{u}-\mathcal{Y}_{t}-\int_{t}^{u}\alpha_{v}\mathop{}\\!\mathrm{d}\mathfrak{L}_{v}^{\varepsilon}$
$\displaystyle\leqslant X_{t}^{\varepsilon}-\gamma C(1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}])-\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|$
$\displaystyle\qquad-\\{\alpha_{t}{L}_{t+s}^{\varepsilon}+(\alpha_{t+s}-\alpha_{t})-\alpha_{t}\mathfrak{L}_{t}^{\varepsilon}\pm
L_{t}^{\varepsilon}\\}$ $\displaystyle\leqslant X_{t}^{\varepsilon}-\gamma
C(1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}])-\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|$
$\displaystyle\qquad-\alpha_{t}x-(\alpha_{t+s}-\alpha_{t})-\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})$

Therefore as $L^{\varepsilon}$ is $W^{0}$ measurable and conditioning on
$W^{0}$ fixes $L^{\varepsilon}$, we have

$\mathbb{P}\left[\left.E_{1}^{x}\right|W^{0}\right]\geqslant\mathbb{P}\left[\left.\underset{t\leqslant
u\leqslant t+s}{\inf}X_{u}^{\varepsilon}\leqslant
0,\tau^{\varepsilon}>t\right|W^{0}\right]=L_{t+s}^{\varepsilon}-L_{t}^{\varepsilon}=x.$

Now, for any fixed $z\leqslant
L_{t+\gamma}^{\varepsilon}-L_{t}^{\varepsilon}-2\gamma^{1/3}$, set
$z_{0}=z+2\gamma^{1/3}$. We define the event

$E_{2}\vcentcolon=\Big{\\{}\gamma C(1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}])+\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|\geqslant\gamma^{1/3}\Big{\\}}$

Then on the event $E_{1}^{z_{0}}\cap E_{2}^{\complement}$

$\displaystyle X_{t}^{\varepsilon}-\alpha_{t}z_{0}$
$\displaystyle\leqslant\gamma C(1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}])+\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t})$
$\displaystyle\leqslant\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t}).$

Therefore on the same event

$X_{t}^{\varepsilon}-\alpha_{t}z=X_{t}^{\varepsilon}-\alpha_{t}z_{0}+2\alpha_{t}\gamma^{1/3}\leqslant(1+2\alpha_{t})\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t})$

Consequently, we deduce

$\displaystyle\bm{\nu}_{t}^{\varepsilon}[0,\alpha_{t}z+(1+2\alpha_{t})\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t})]$
$\displaystyle\geqslant\mathbb{P}\left[\left.E_{1}^{z_{0}}\cap
E_{2}^{\complement}\right|W^{0}\right]$
$\displaystyle\geqslant\mathbb{P}\left[\left.E_{1}^{z_{0}}\right|W^{0}\right]-\mathbb{P}\left[\left.E_{2}\right|W^{0}\right]$
$\displaystyle\geqslant z_{0}-\mathbb{P}\left[\left.E_{2}\right|W^{0}\right].$

Therefore if we have control over the mass
$\mathbb{P}\left[\left.E_{2}\right|W^{0}\right]$, we may estimate the mass
with respect to $\bm{\nu}_{t}$ that is near the boundary. Therefore, defining
the event
$E_{3}\vcentcolon=\\{\mathbb{P}\left[\left.E_{2}\right|W^{0}\right]\leqslant\gamma^{1/3}\\}$
we deduce on $E_{3}$

$\bm{\nu}_{t}^{\varepsilon}[0,\alpha_{t}z+(1+2\alpha_{t})\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t})]\geqslant
z_{0}-\gamma^{1/3}\geqslant z.$

The last inequality follows from the fact that $z_{0}=z+2\gamma^{1/3}$. Now we
only need to find a $C$ independent of $\varepsilon,\,\gamma$ and $t$ such
that $\mathbb{P}[E_{3}^{\complement}]\leqslant C\gamma^{1/3}$. By application
of Markov’s inequality twice

$\displaystyle\mathbb{P}\left[E_{3}^{\complement}\right]$
$\displaystyle\leqslant\gamma^{-1/3}\mathbb{P}\left[E_{2}\right]$
$\displaystyle\leqslant\gamma^{-1/3}\mathbb{P}\left[\gamma
C(1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}])\geqslant
2^{-1}\gamma^{1/3}/2\right]$
$\displaystyle\qquad+\gamma^{-1/3}\mathbb{P}\left[\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|\geqslant
2^{-1}\gamma^{1/3}/2\right]$ $\displaystyle\leqslant
2C\gamma^{1/3}\mathbb{E}\left[1+\underset{u\leqslant
t+\gamma}{\sup}\left|X_{u}^{\varepsilon}\right|+\mathbb{E}[\underset{u\leqslant
t+\gamma}{\sup}\left.\left|X_{u}^{\varepsilon}\right|\right|W^{0}]\right]+2^{8}\gamma^{-3}\mathbb{E}\left[\underset{u\leqslant\gamma}{\sup}\left|\mathcal{Y}^{\varepsilon}_{t+u}-\mathcal{Y}^{\varepsilon}_{t}\right|^{8}\right]$
$\displaystyle\leqslant c_{1}(\gamma^{1/3}+\gamma)\leqslant
2c_{1}\gamma^{1/3},$

where $c_{1}$ depends on the constant from Proposition 2.6, the constant from
Burkholder-Davis-Gundy to bound the second term and the uniform bounds on
$\sigma$, but is notably independent of $\varepsilon$. Therefore, setting
$C=\max\\{1+2\alpha(T),c_{1}\\}$ completes the proof. ∎

######  Lemma A.8.

Suppose that
$\tilde{\mathbf{P}}^{\varepsilon_{n}}\implies\tilde{\mathbf{P}}^{*}$ on
$(\mathcal{P}(D_{\mathbb{R}}),\mathfrak{T}_{\text{M1}}^{\text{wk}})$ for a
positive sequence $(\varepsilon_{n})_{n\geqslant 1}$ which converges to zero.
Set

$\mathbbm{T}\vcentcolon=\left\\{t\in[-1,\bar{T}]\;:\;\mathbb{E}\left[\tilde{\mathbf{P}}^{*}(\eta_{t}=\eta_{t-})\right]=1,\,\mathbb{E}\left[\tilde{\mathbf{P}}^{*}(\tau_{0}(\eta)=t)\right]=0\right\\}.$

Then for any $t\in\mathbbm{T}\cap[0,T)$ and $\gamma>0$ such that
$t+\gamma\in\mathbbm{T}\cap[0,T)$ we have

$\mathbb{P}\left[\bm{\nu}_{t}[0,\alpha(t)z+C\gamma^{1/3}+\alpha(t+\gamma)-\alpha(t)]\geqslant
z,\;\forall\;z\leqslant L_{t+\gamma}-L_{t}-C\gamma^{1/3}\right]\geqslant
1-C\gamma^{1/3}$

###### Proof.

As $\tilde{\mathbf{P}}^{\varepsilon_{n}}\implies\tilde{\mathbf{P}}^{*}$, by
employing Skohorod’s Representation Theorem, there exists a
$(\bm{\mu}^{n})_{n\geqslant 1}$ and $\bm{\mu}$ such that
$\operatorname{Law}(\bm{\mu}^{n})=\operatorname{Law}(\tilde{\mathbf{P}}^{\varepsilon_{n}})$,
$\operatorname{Law}(\bm{\mu})=\operatorname{Law}(\tilde{\mathbf{P}}^{*})$ and
$\bm{\mu}^{n}\xrightarrow[]{}\bm{\mu}$ almost surely in
$(\mathcal{P}(D_{\mathbb{R}}),\mathfrak{T}_{\textnormal{M1}}^{\text{wk}})$. As
$\operatorname{Law}(\mathbf{P}^{*})$-almost every measure $\mu$ satisfy Eq.
A.4 by Lemma 2.9, then by Lemma A.4
$\nu_{t}^{\bm{\mu}^{n}}\xrightarrow[]{}\nu_{t}^{\bm{\mu}}$ almost surely for
any $t\in\mathbbm{T}$. Furthermore for any $t\in\mathbbm{T}\cap[0,T)$, by
Lemma 2.11 and Corollary 2.12, we have

$\bm{\mu}^{n}(\tau_{0}(\eta)\leqslant
t)\xrightarrow[]{}\bm{\mu}(\tau_{0}(\eta)\leqslant
t)\quad\textnormal{and}\quad\int_{0}^{t}\kappa^{\varepsilon_{n}}(t-s)\bm{\mu}^{n}(\tau_{0}(\eta)\leqslant
s)\mathop{}\\!\mathrm{d}s\xrightarrow[]{}\int_{0}^{t}\kappa^{\varepsilon_{n}}(t-s)\bm{\mu}(\tau_{0}(\eta)\leqslant
s)\mathop{}\\!\mathrm{d}s$

almost surely. Therefore for simplicity and notational convenience for the
remainder of this proof, we may suppose

$\displaystyle\bm{\nu}_{t}^{\varepsilon_{n}}\xrightarrow[]{}\bm{\nu}_{t}$
$\displaystyle\textnormal{a.s. in }\mathbf{M}_{\leqslant 1}(\mathbb{R}),$
$\displaystyle L_{t}^{\varepsilon_{n}}\xrightarrow[]{}L_{t}$
$\displaystyle\textnormal{a.s. in }\mathbb{R},$
$\displaystyle\mathfrak{L}_{t}^{\varepsilon_{n}}\xrightarrow[]{}{L}_{t}$
$\displaystyle\textnormal{a.s. in }\mathbb{R}.$

Recall by Lemma A.7,

$\mathbb{P}\left[\bm{\nu}_{t}^{\varepsilon}[0,\alpha_{t}z+C\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon}-\mathfrak{L}_{t}^{\varepsilon})+(\alpha_{t+\gamma}-\alpha_{t})]\geqslant
z,\;\forall\;z\leqslant
L_{t+\gamma}^{\varepsilon}-L_{t}^{\varepsilon}-C\gamma^{1/3}\right]\geqslant
1-C\gamma^{1/3},$

for any $\varepsilon>0$. It is well known that the Levy-Prokhorov metric,
$d_{L}$, metricizes weak convergence, [29, Theorem 1.11]. Fixing
$\delta_{1},\,\delta_{2},\,\delta_{3},\,\delta_{4}>0$, we define the event

$A_{n}\vcentcolon=\left\\{\left|L_{t}^{\varepsilon_{n}}-L_{t}\right|<\delta_{1},\left|L_{t+\gamma}^{\varepsilon_{n}}-L_{t+\gamma}\right|<\delta_{2},\left|\mathfrak{L}_{t}^{\varepsilon_{n}}-L_{t}^{\varepsilon_{n}}\right|<\delta_{3},d_{L}(\bm{\nu}_{t}^{\varepsilon_{n}},\bm{\nu}_{t})<\delta_{4}\right\\}$

Therefore

$\displaystyle 1-C\gamma^{1/3}$
$\displaystyle\leqslant\mathbb{P}\left[\bm{\nu}_{t}^{\varepsilon_{n}}[0,\alpha_{t}z+C\gamma^{1/3}+\alpha_{t}(L_{t}^{\varepsilon_{n}}-\mathfrak{L}_{t}^{\varepsilon_{n}})+(\alpha_{t+\gamma}-\alpha_{t})]\geqslant
z,\;\forall\;z\leqslant
L_{t+\gamma}^{\varepsilon_{n}}-L_{t}^{\varepsilon_{n}}-C\gamma^{1/3}\right]$
$\displaystyle\quad+\mathbb{P}[A_{n}^{\complement}]$
$\displaystyle\leqslant\mathbb{P}\left[\delta_{4}+\bm{\nu}_{t}(-\delta_{4},\alpha_{t}z+C\gamma^{1/3}+\alpha_{t}\gamma_{3}+(\alpha_{t+\gamma}-\alpha_{t})+\delta_{4})\geqslant
z,\;\forall\;z\leqslant
L_{t+\gamma}-L_{t}-C\gamma^{1/3}-\delta_{1}-\delta_{2}\right]$
$\displaystyle\quad+\mathbb{P}[A_{n}^{\complement}]$

Sending $\varepsilon_{n}\xrightarrow[]{}0$, then
$\mathbb{P}[A_{n}^{\complement}]\xrightarrow[]{}0$ by the Dominated
Convergence Theorem as we have almost sure convergence. Lasts by sending
$\delta_{1}\xrightarrow[]{}0,\,\delta_{2}\xrightarrow[]{}0,\,\delta_{3}\xrightarrow[]{}0,\,\delta_{4}\xrightarrow[]{}0$
one at a time and in order, then by employing continuity of measure we may
conclude

$\mathbb{P}\left[\bm{\nu}_{t}[0,\alpha_{t}z+C\gamma^{1/3}+(\alpha_{t+\gamma}-\alpha_{t})]\geqslant
z,\;\forall\;z\leqslant L_{t+\gamma}-L_{t}-C\gamma^{1/3}\right]\geqslant
1-C\gamma^{1/3}$

∎

######  Lemma A.9.

Suppose that
$\tilde{\mathbf{P}}^{\varepsilon_{n}}\implies\tilde{\mathbf{P}}^{*}$ on
$(\mathcal{P}(D_{\mathbb{R}}),\mathfrak{T}_{\text{M1}}^{\text{wk}})$ for a
positive sequence $(\varepsilon_{n})_{n\geqslant 1}$ which converges to zero.
Then we have

$L_{t}\leqslant\inf\left\\{x\geqslant
0\,;\,\bm{\nu}_{t-}[0,\,\alpha(t)x]<x\right\\}$ (A.10)

almost surely for any $t\in[0,\,T)$.

###### Proof.

It is clear that Lemma A.9 holds for any $t\in\mathbbm{T}$. Hence we must only
show the upper bound for $t\not\in\mathbbm{T}$. We first consider the case
when $t\in(0,T)\cap\mathbbm{T}^{\complement}$. The case when $t=0$ will be
treated separately. Now as $\mathbbm{T}$ is dense in $[0,T]$, we may find a
$(t_{n})_{n\geqslant 1},\,(t_{n}+\gamma_{n})_{n\geqslant 1}\subset\mathbbm{T}$
such that $t_{n}\uparrow t$, $t_{n}+\gamma_{n}\downarrow t$ and
$\gamma_{n}<2^{-3n}$. Now by the Borel-Cantelli Lemma, we have a set of full
measure such that

$\bm{\nu}_{t_{n}}[0,\alpha_{t_{n}}z+C\gamma_{n}^{1/3}+(\alpha_{{t_{n}}+\gamma_{n}}-\alpha_{t_{n}})]\geqslant
z,\;\forall\;z\leqslant L_{{t_{n}}+\gamma_{n}}-L_{t_{n}}-C\gamma_{n}^{1/3},$
(A.11)

for all (possibly stochastic) $n$ large. Furthermore, by the dominated
convergence theorem, we have

$\bm{\nu}_{t_{n}}\xrightarrow[]{}\bm{\nu}_{t-},$ (A.12)

$\operatorname{Law}(\tilde{\mathbf{P}}^{*})$-almost surely in
$\mathbf{M}_{\leqslant 1}(\mathbb{R})$ as for any
$\phi\in\mathcal{C}_{b}(\mathbb{R})$

$\lim_{n\xrightarrow[]{}\infty}\bm{\nu}_{t_{n}}(\phi)=\lim_{n\xrightarrow[]{}\infty}\int_{D_{\mathbb{R}}}\phi(\eta_{t_{n}})\mathbbm{1}_{\\{\tau_{0}(\eta)>t_{n}\\}}\mathop{}\\!\mathrm{d}\tilde{\mathbf{P}}^{*}(\eta)=\int_{D_{\mathbb{R}}}\phi(\eta_{t-})\mathbbm{1}_{\\{\tau_{0}(\eta)\geqslant
t\\}}\mathop{}\\!\mathrm{d}\tilde{\mathbf{P}}^{*}(\eta).$

So on an event of full $\operatorname{Law}(\mathbf{P}^{*})$ measure where Eq.
A.11 and Eq. A.12 holds, by Portmanteau Theorem for any $\gamma>0$, $z<\Delta
L_{t}$

$\bm{\nu}_{t-}[0,\,\alpha(t)z+\gamma]\geqslant\limsup_{n\xrightarrow[]{}\infty}\bm{\nu}_{t_{n}}[0,\,\alpha(t)z+\gamma]\geqslant\limsup_{n\xrightarrow[]{}\infty}\bm{\nu}_{t_{n}}[0,\alpha_{t_{n}}z+C\gamma_{n}^{1/3}+(\alpha_{{t_{n}}+\gamma_{n}}-\alpha_{t_{n}})]\geqslant
z.$

This holds as $z<\Delta L_{t}$ and
$L_{{t_{n}}+\gamma_{n}}-L_{t_{n}}-C\gamma_{n}^{1/3}\xrightarrow[]{}\Delta
L_{t}$. Sending $\gamma$ to zero shows the claim for every $t>0$.

In the case when $t=0$, we have by Lemma A.7

$\mathbb{P}\left[\bm{\nu}_{0-}^{\varepsilon}[0,\alpha_{0}z+C\gamma^{1/3}+(\alpha_{\gamma}-\alpha_{0})]\geqslant
z,\;\forall\;z\leqslant L_{\gamma}^{\varepsilon}-C\gamma^{1/3}\right]\geqslant
1-C\gamma^{1/3}.$

As $\bm{\nu}^{\varepsilon}_{0-}=\nu_{0-}$, where $\nu_{0-}$ is a deterministic
measure, and $\bm{nu}_{0-}$ is almost surely distributed as $\nu_{0-}$ almost
surely, then we have by Lemma A.8

$\mathbb{P}\left[\bm{\nu}_{0-}[0,\alpha(0)z+C\gamma^{1/3}+\alpha(\gamma)-\alpha(0)]\geqslant
z,\;\forall\;z\leqslant L_{\gamma}-C\gamma^{1/3}\right]\geqslant
1-C\gamma^{1/3}.$

for $\gamma\in\mathbbm{T}$. Now the rest of the proof follows similar
arguments as above by choosing $\gamma_{n}\in\mathbbm{T}$ such that
$\gamma_{n}\downarrow 0$. ∎

######  Proposition A.10 (Gronwall Type Inequality I).

Suppose $a,\,\tilde{\alpha},\,\tilde{\beta}\in\mathbb{R}_{+}$ such that
$a\geqslant 0,\,0<\tilde{\beta}<1,\,\tilde{\alpha}>0$. Suppose $g(t)$ is a
nonnegative, nondecreasing continuous function defined on $0\leqslant t<T$,
$g(t)\leqslant M$ (constant), and suppose $u(t)$ is nonnegative and bounded on
$0\leqslant t<T$ with

$u(t)\leqslant
a+g(t)\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}u(s)\mathop{}\\!\mathrm{d}{}s$

on this interval. Then

$u(t)\leqslant a\left[1+\sum_{n\geqslant
1}g_{t}^{n}C_{n}t^{n(\tilde{\alpha}+\tilde{\beta}-1)}\right],\quad 0\leqslant
t<T,$

where

$\displaystyle C_{0\phantom{+1}}$ $\displaystyle\vcentcolon=1,$ $\displaystyle
C_{1\phantom{+1}}$ $\displaystyle\vcentcolon=B(\tilde{\alpha},\tilde{\beta}),$
$\displaystyle C_{n+1}$
$\displaystyle\vcentcolon=B\left((n+1)\tilde{\alpha}+n\tilde{\beta}-n,\tilde{\beta}\right)C_{n},$
$\displaystyle B(\tilde{\alpha},\,\tilde{\beta})$
$\displaystyle\vcentcolon=\int_{0}^{1}(1-\tilde{s})^{\tilde{\beta}-1}\tilde{s}^{\tilde{\alpha}-1}\mathop{}\\!\mathrm{d}{}\tilde{s}.$

###### Proof.

Let
$B\phi_{t}=g_{t}\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}\phi_{s}\mathop{}\\!\mathrm{d}{}s$,
$t\geqslant 0$, for localling integrable functions $\phi$. Then
$u_{t}\leqslant a(t)+Bu_{t}$ implies

$u_{t}\leqslant\sum_{k=0}^{n-1}B^{k}a+B^{n}u_{t}.$

Let us prove that

$B^{n}(1)_{t}\leqslant_{n}t^{n(\tilde{\alpha}+\tilde{\beta}-1)}g_{t}^{n}$
(A.13)

and $B^{n}u_{t}\rightarrow 0$ as $n\rightarrow+\infty$ for each t in
$0\leqslant t<T$.
Step 1: $B^{n}(1)_{t}\leqslant
C_{n}t^{n(\tilde{\alpha}+\tilde{\beta}-1)}g_{t}^{n}$. For $n=1$,

$\displaystyle B(1)_{t}$
$\displaystyle=g_{t}\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}\mathop{}\\!\mathrm{d}{}s,\quad\text{set
}\tilde{s}=s/t$
$\displaystyle=g_{t}\int_{0}^{1}t^{\tilde{\beta}-1}(1-\tilde{s})^{\tilde{\beta}-1}t^{\tilde{\alpha}-1}\tilde{s}^{\tilde{\alpha}-1}t\mathop{}\\!\mathrm{d}{}\tilde{s}$
$\displaystyle=g_{t}t^{\tilde{\alpha}+\tilde{\beta}-1}B(\tilde{\alpha},\,\tilde{\beta}).$

Now, suppose the claim is true for $n=k$, then for $n=k+1$

$\displaystyle B^{k+1}(1)_{t}$
$\displaystyle=g_{t}\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}B^{k}(1)_{s}\mathop{}\\!\mathrm{d}{}s,$
$\displaystyle\leqslant
g_{t}\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}s^{k(\tilde{\alpha}+\tilde{\beta}-1)}g_{s}^{k}C_{k}\mathop{}\\!\mathrm{d}{}s,\quad\text{by
above}$ $\displaystyle\leqslant
C_{k}g_{t}^{k+1}\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}s^{k(\tilde{\alpha}+\tilde{\beta}-1)}\mathop{}\\!\mathrm{d}{}s,\quad\text{set
}\tilde{s}=s/t$
$\displaystyle=C_{k}g_{t}^{k+1}\int_{0}^{1}t^{\tilde{\beta}-1}(1-\tilde{s})^{\tilde{\beta}-1}t^{\tilde{\alpha}-1}\tilde{s}^{\tilde{\alpha}-1}t^{k(\tilde{\alpha}+\tilde{\beta}-1)}\tilde{s}^{k(\tilde{\alpha}+\tilde{\beta}-1)}t\mathop{}\\!\mathrm{d}{}\tilde{s},$
$\displaystyle=C_{k+1}g^{k+1}_{t}t^{(k+1)(\tilde{\alpha}+\tilde{\beta}-1)}.$

Hence the claim is true by the principle of induction.
Step 2: We observe that $B$ is monotone, that is if
$\phi_{1}\leqslant\phi_{2}\,\forall t\in[0,\,T]$, then by the nonnegativity of
$g$ we have $B(\phi_{1})\leqslant B(\phi_{2})$. Also by the linearity of
integration, we see also that $B$ is a linear operator. Therefore,

$B(u)_{t}=g(t)\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}u(s)\mathop{}\\!\mathrm{d}{}s\leqslant\left\lVert
u\right\rVert_{L^{\infty}}g(t)\int_{0}^{t}(t-s)^{\tilde{\beta}-1}s^{\tilde{\alpha}-1}\mathop{}\\!\mathrm{d}{}s=\left\lVert
u\right\rVert_{L^{\infty}}B(1)_{t}$

Therefore, by linearity, monotonicity and step 1

$B^{n}(u)_{t}\leqslant\left\lVert
u\right\rVert_{L^{\infty}}B^{n}(1)_{t}\leqslant\left\lVert
u\right\rVert_{L^{\infty}}C_{n}g_{t}^{n}t^{n(\tilde{\alpha}+\tilde{\beta}-1)}$

Step 3: Summability of $C_{n}$. By Gautschi’s inequality, [15], we have that
for all $x>0$ and $s\in(0,\,1)$

$x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s}$

Therefore,

$\displaystyle\frac{C_{n+1}}{C_{n}}$
$\displaystyle=B\left((n+1)\tilde{\alpha}+n\tilde{\beta}-n,\,\tilde{\beta}\right),$
$\displaystyle=\frac{\Gamma\left((n+1)\tilde{\alpha}+n\tilde{\beta}-n\right)\Gamma(\tilde{\beta})}{\Gamma\left((n+1)\tilde{\alpha}+(n+1)\tilde{\beta}-n\right)},$
$\displaystyle=\Gamma(\tilde{\beta})\left[\frac{\Gamma\left((n+1)\tilde{\alpha}+(n+1)\tilde{\beta}-(n+1)+1-\tilde{\beta}\right)}{\Gamma\left((n+1)\tilde{\alpha}+(n+1)\tilde{\beta}-(n+1)+1\right)}\right],$
$\displaystyle\leqslant\Gamma(\tilde{\beta})(n+1)^{-\tilde{\beta}}(\tilde{\alpha}+\tilde{\beta}-1)^{-\tilde{\beta}}\quad\text{
by Gautschi's Inequality}.$

Hence $C_{n+1}/C_{n}\rightarrow 0$ and $n\rightarrow+\infty$. Hence by the
ratio test, we have that $C_{n}$ is summable.
Step 4: Summability of $B^{n}(u)_{t}$. By step $3$, we have that

$\frac{C_{n+1}\left\lVert
u\right\rVert_{L^{\infty}}t^{(n+1)(\tilde{\alpha}+\tilde{\beta}-1)}g_{t}^{n+1}}{C_{n}\left\lVert
u\right\rVert_{L^{\infty}}t^{n(\tilde{\alpha}+\tilde{\beta}-1)}g_{t}^{n}}=\left\lVert
u\right\rVert_{L^{\infty}}t^{\tilde{\alpha}+\tilde{\beta}-1}g_{t}\frac{C_{n+1}}{C_{n}}\xrightarrow[]{n\rightarrow+\infty}0$

Therefore by the ratio test then the comparison test, we have that
$B^{n}(u)_{t}$ is summable. Hence $B^{n}(u)_{t}\rightarrow 0$ as
$n\rightarrow+\infty$.
Step 5: As $u_{t}\leqslant a+B(u)_{t}$, then it is clear by the Principle of
Induction, by using the monotonicity and linearity of $B$, we have
$u_{t}\leqslant\sum_{j=1}^{N-1}aB^{j}(1)_{t}+B^{N}(u)_{t}$. Hence taking
limiting as $N\xrightarrow[]{}\infty$, by step 1 and step 4, we conclude
$u_{t}\leqslant\sum_{j\geqslant
0}aC_{j}t^{j(\tilde{\alpha}+\tilde{\beta}-1)}g_{t}^{j}$. The proof is now
complete. ∎

###### Proof of Proposition 4.2.

This proof is analogous to that of Proposition 4.1. Most of the details have
been skipped for brevity. Choose $t_{0}\in(0,\,t_{\mathrm{explode}})$.

Step 1: Regularity of $L$ and decomposition into integral form. As
$L\in\mathcal{C}^{1}([0,\,t_{\mathrm{explode}}))$, by the Fundamental Theorem
of Calculus we have $L_{t}-L_{s}\leqslant K(t-s)$ for any
$t,s\in[0,\,t_{\mathrm{explode}})$ with $t>s$ and $K=\sup_{u\leqslant
t_{0}}|L^{\prime}_{u}|$. Now we may write $L$ as

$L_{t}=\int_{0}^{t}\kappa^{\varepsilon}(t-s)L_{s}\mathop{}\\!\mathrm{d}{s}+\left[1-\int_{0}^{t}\kappa^{\varepsilon}(t-s)\mathop{}\\!\mathrm{d}{s}\right]L_{t}+\int_{0}^{t}\kappa^{\varepsilon}(t-s)(L_{t}-L_{s})\mathop{}\\!\mathrm{d}{s}.$

Observe

$\left[1-\int_{0}^{t}\kappa^{\varepsilon}(t-s)\mathop{}\\!\mathrm{d}{s}\right]L_{t}\leqslant{2K\varepsilon}\qquad\textnormal{and}\qquad\int_{0}^{t}\kappa^{\varepsilon}(t-s)(L_{t}-L_{s})\mathop{}\\!\mathrm{d}{s}\leqslant{K\varepsilon}.$

Therefore

$L_{t}=\int_{0}^{t}\kappa^{\varepsilon}(t-s)L_{s}\mathop{}\\!\mathrm{d}{s}+\Psi^{\varepsilon}(t)\qquad\text{
where
}|\Psi^{\varepsilon}(t)|\leqslant{3K\varepsilon}\,\quad\forall\,t\in[0,\,t_{0}].$

Step 2: Comparison between the delayed loss and the instantaneous loss. As in
Proposition 4.1, we have

$0\leqslant\left|L_{t}-L_{t}^{\varepsilon}\right|\leqslant
Kc_{1}\int_{0}^{t}\int_{0}^{u}\frac{\kappa^{\varepsilon}(u-s)\left|L_{s}-L_{s}^{\varepsilon}\right|}{\sqrt{t-u}}\mathop{}\\!\mathrm{d}s\mathop{}\\!\mathrm{d}u-+Kc_{1}\int_{0}^{t}\frac{\left|\Psi^{\varepsilon}(s)\right|}{\sqrt{t-s}}.$

Note as $\left|\Psi^{\varepsilon}(t)\right|\leqslant 3K\varepsilon$ for all
$t\in[0,\,t_{0}]$, we see that the second term above is bounded above by
$C_{K,t_{0}}\varepsilon$. Therefore,

$\displaystyle 0\leqslant|L_{t}-{L}^{\varepsilon}_{t}|$
$\displaystyle\leqslant\;Kc_{1}\int_{0}^{t}\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)\left|L_{s}-L_{s}^{\varepsilon}\right|}{\sqrt{t-u}}\mathop{}\\!\mathrm{d}u\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon$
$\displaystyle=\;Kc_{1}\int_{0}^{t}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon,$
(A.14)

where

$\rho^{\varepsilon}(t,s)=\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)}{\sqrt{t-u}}\mathop{}\\!\mathrm{d}u.$

Step 3: Bounds on $\rho^{\varepsilon}(t,s)$. As in Proposition 4.1, the
presence of $\kappa$ in $\rho^{\varepsilon}$ makes the function to general to
do any analysis, hence we shall construct polynomial bounds on
$\rho^{\varepsilon}$. Then we may be able to apply generalised versions of
Gronwall’s Lemma. Recall $t\geqslant s$, therefore

Case 1: ${t-s}\leqslant\varepsilon$

$\displaystyle\rho^{\varepsilon}(t,s)$
$\displaystyle=\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)}{\sqrt{t-u}}\mathop{}\\!\mathrm{d}u\qquad\qquad\textrm{let
$\tilde{u}=\frac{u-s}{\varepsilon}$}$
$\displaystyle=\int_{0}^{\frac{t-s}{\varepsilon}}\frac{\kappa(\tilde{u})}{\sqrt{t-s-\varepsilon\tilde{u}}}\mathop{}\\!\mathrm{d}\tilde{u}\leqslant\frac{\left\lVert\kappa\right\rVert_{L^{\infty}}}{\varepsilon^{1/2}}\int_{0}^{\frac{t-s}{\varepsilon}}\frac{\mathop{}\\!\mathrm{d}\tilde{u}}{\sqrt{\frac{t-s}{\varepsilon}-\tilde{u}}}$
$\displaystyle=\frac{2\left\lVert\kappa\right\rVert_{L^{\infty}}(t-s)^{1/2}}{\varepsilon}\leqslant\frac{2\left\lVert\kappa\right\rVert_{L^{\infty}}}{(t-s)^{1/2}}$

Case 2: ${t-s}>\varepsilon$
As the support of $\kappa^{\varepsilon}$ is in $[0,\,\varepsilon]$

$\displaystyle\rho^{\varepsilon}(t,s)$
$\displaystyle=\int_{s}^{t}\frac{\kappa^{\varepsilon}(u-s)}{\sqrt{t-u}}\mathop{}\\!\mathrm{d}u=\int^{s+\varepsilon}_{s}\frac{\kappa^{\varepsilon}(u-s)}{\sqrt{t-u}}\mathop{}\\!\mathrm{d}u$
$\displaystyle\leqslant\frac{\left\lVert\kappa\right\rVert_{L^{\infty}}}{\varepsilon}\int^{s+\varepsilon}_{s}\frac{\mathop{}\\!\mathrm{d}u}{\sqrt{t-u}}={2\left\lVert\kappa\right\rVert_{L^{\infty}}}\left[\frac{(t-s)^{1/2}-(t-s-\varepsilon)^{1/2}}{\varepsilon}\right].$

Step 4: Gronwall type argument. Now that we have sufficiently simplified
$\rho^{\varepsilon}$, we may put Eq. A.14 into a form where we may apply
Gronwall’s inequality. By step 4 case 1 and (A.14), we have for
$t\leqslant\varepsilon$

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|\leqslant
Kc_{1}\int_{0}^{t}2\left\lVert\kappa\right\rVert_{L^{\infty}}(t-s)^{-1/2}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon.$

By step 4 case 2 and (A.14), we have for $t>\varepsilon$

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|$ $\displaystyle\leqslant
Kc_{1}\int_{0}^{t-\varepsilon}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+Kc_{1}\int_{t-\varepsilon}^{t}|L_{s}-{L}_{s}^{\varepsilon}|\rho^{\varepsilon}(t,s)\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon$
$\displaystyle\leqslant
2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t-\varepsilon}\left[\frac{(t-s)^{1/2}-(t-s-\varepsilon)^{1/2}}{\varepsilon}\right]|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s$
$\displaystyle\phantom{\leqslant}+2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{t-\varepsilon}^{t}(t-s)^{-1/2}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon$
$\displaystyle\leqslant
2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\sum_{j\geqslant
2}\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s$
$\displaystyle\phantom{\leqslant}+2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t}(t-s)^{-1/2}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon,$

where the second term in the last line is the higher order terms from
employing Taylor’s Theorem. By applying the Monotone Convergence Theorem, we
have swapped integrals and sums. We shall now turn our attention onto the
expression

$\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s.$

We shall proceed in two cases.
Case 1: $\varepsilon<t\leqslant 2\varepsilon$

$\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s\leqslant\tilde{C}_{j}\varepsilon^{j-1}\varepsilon^{\frac{-2j+1}{2}}\int_{0}^{t-\varepsilon}\mathop{}\\!\mathrm{d}s={\tilde{C}_{j}\varepsilon^{\frac{1}{2}}}$

where the first inequality follows from the fact that $(-2j+1)/2<0$ as
$j\geqslant 2$ and $t-s\in[\varepsilon,\,t]$ for $s\in[0,\,t-\varepsilon]$.
Case 2: $t>2\varepsilon$
We observe that

$\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s\leqslant\tilde{C}_{j}\varepsilon^{j-1}\varepsilon^{\frac{-2j+1}{2}}\int_{0}^{\varepsilon}\mathop{}\\!\mathrm{d}s={\tilde{C}_{j}\varepsilon^{\frac{1}{2}}}$

and

$\displaystyle\tilde{C}_{j}\varepsilon^{j-1}\int_{\varepsilon}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s$
$\displaystyle=\tilde{C}_{j}\varepsilon^{j-1}\frac{2}{2j-3}\left.(t-s)^{\frac{-2j+3}{2}}\right|_{s=\varepsilon}^{t-\varepsilon}$
$\displaystyle=\frac{2\tilde{C}_{j}\varepsilon^{j-1}}{2j-3}\left[\varepsilon^{\frac{-2j+3}{2}}-(t-\varepsilon)^{\frac{-2j+3}{2}}\right]$
$\displaystyle\leqslant\frac{2\tilde{C}_{j}\varepsilon^{j-1}\varepsilon^{\frac{-2j+3}{2}}}{2j-3}$
$\displaystyle=\frac{2\tilde{C}_{j}\varepsilon^{1/2}}{2j-3}\leqslant{2\tilde{C}_{j}\varepsilon^{1/2}},$

where the upper bound in the last inequality follows from the fact that
$j\geqslant 2$. Therefore we have shown Therefore, we have shown that

$\displaystyle\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s$
$\displaystyle=\tilde{C}_{j}\varepsilon^{j-1}\int_{0}^{\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s+\tilde{C}_{j}\varepsilon^{j-1}\int_{\varepsilon}^{t-\varepsilon}(t-s)^{\frac{-2j+1}{2}}\mathop{}\\!\mathrm{d}s$
$\displaystyle\leqslant{3\tilde{C}_{j}\varepsilon^{1/2}},$

for all $t>\varepsilon$. As $L$ and ${L}^{\varepsilon}$ are bounded by $1$, so
independent of $t$ being greater or less than $\varepsilon$ we have

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|$ $\displaystyle\leqslant
2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t}(t-s)^{-1/2}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+{12Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\varepsilon^{1/2}\sum_{j\geqslant
2}\tilde{C}_{j}}+C_{K,t_{0}}\varepsilon$
$\displaystyle=2Kc_{1}\left\lVert\kappa\right\rVert_{L^{\infty}}\int_{0}^{t}(t-s)^{-1/2}s^{-\gamma}|L_{s}-{L}_{s}^{\varepsilon}|\mathop{}\\!\mathrm{d}s+C_{K,t_{0}}\varepsilon^{1/2},$

for any $t\in[0,\,t_{0}]$. Lastly by Proposition A.10 using
$\tilde{\beta}=1/2$ and $\tilde{\alpha}=1$, then
$\tilde{\alpha}+\tilde{\beta}-1>0$ as $\gamma<1/2$ and

$\displaystyle|L_{t}-{L}^{\varepsilon}_{t}|$
$\displaystyle=C_{K,t_{0}}\varepsilon^{1/2}\sum_{n\geqslant
0}(2Kc_{1}t_{0}^{1/2}\left\lVert\kappa\right\rVert_{L^{\infty}})^{n}C_{n}.$

This completes the proof. ∎

##  Appendix B Further numerical analysis

In Section 4.2, we considered 6 examples to compare the theoretical rate of
convergence with that obtained in practice. The parameters used for each
simulation are given in Table 1.

Simulation |  CC1111Continuous case 1 |  CC2222Continuous case 2 |  DC1333Discontinuous case 1 |  DC2444Discontinuous case 2 |  CNC1555Common noise case 1: with increasing path |  CNC 2666Common noise case 2: with decreasing path
---|---|---|---|---|---|---
Initial Condition | $\operatorname{Unif}[0.25,\,0.35]$ | $\Gamma(2.1,0.5)$ | $\Gamma(1.2,\,0.5)$ | $\Gamma(1.4,0.5)$ | $\operatorname{Unif}[0.25,\,0.35]$ | $\operatorname{Unif}[0.25,\,0.35]$
$\alpha$ | $0.5$ | $1.3$ | $0.9$ | $2$ | $0.5$ | $0.5$
$\Delta_{t}$ | $10^{-6}$ | $10^{-6}$ | $10^{-9}$ | $10^{-9}$ | $10^{-6}$ | $10^{-6}$
$t_{\operatorname{max}}$ | $0.1$ | $0.1$ | $10^{-4}$ | $10^{-4}$ | $0.1$ | $2\times 10^{-2}$
Table 1: Parameters of numerical simulations in Section 4.2

With the chosen parameters, we generated the convergence graphs in Fig. 1(b),
Fig. 2(c) and Fig. 3(c) from $\operatorname{Error}(\varepsilon_{n})$, where
$\varepsilon_{n}\vcentcolon=\varepsilon\times\Delta^{n}$, with $\varepsilon$
and $\Delta$ as positive constants, and $\operatorname{Error}$ is the
corresponding difference between the smoothed and limiting loss functions.
Assuming a power law relationship between the error and the parameter
$\varepsilon$,

$\operatorname{Error}(\varepsilon)\approx A\varepsilon^{\beta},$

where $A$ and $\beta$ are constants, we performed a linear regression on
$\operatorname{Log}\operatorname{Error}(\varepsilon)$ versus
$\operatorname{Log}\varepsilon$, which determined the line of best fit shown
in the plots. The slope, shown in Table 2, represents our best estimate for
the rate of convergence for each specific setting.

Setting | CC1 | CC2 | DC1 | DC2 | CNC1 | CNC2
---|---|---|---|---|---|---
Rate | $1.0202$ | $0.9635$ | $0.9295$ | $0.8126$ | $0.7621$ | $0.8144$
Table 2: Gradient of the regression line

To assess if the estimated slope corresponds to an asymptotic value, we also
conducted an alternative analysis of the rate of convergence. By computing the
ratio between two consecutive errors, we observe

$\frac{\operatorname{Error}(\varepsilon_{n+1})}{\operatorname{Error}(\varepsilon_{n})}\approx\frac{A\Delta^{n\beta+\beta}\varepsilon^{\beta}}{A\Delta^{n\beta}\varepsilon^{\beta}}\approx\Delta^{\beta},$

and taking logarithms with base $\Delta$, we may deduce

$\operatorname{Log}_{\Delta}\left(\frac{\operatorname{Error}(\varepsilon_{n+1})}{\operatorname{Error}(\varepsilon_{n})}\right)\approx\beta.$

By using the relationship that
$\operatorname{Log}_{a}b=\operatorname{Log}_{c}b/\operatorname{Log}_{c}a$ for
any $a,\,b,\,c>0$, we obtain approximate expressions for $\beta$ as follows:

$\small\beta_{n}:=\operatorname{Log}_{\Delta}\left(\frac{\operatorname{Error}(\varepsilon_{n+1})}{\operatorname{Error}(\varepsilon_{n})}\right)=\frac{\operatorname{Log}({\operatorname{Error}(\varepsilon_{n+1})})-\operatorname{Log}({\operatorname{Error}(\varepsilon_{n})})}{\operatorname{Log}(\Delta)}=\frac{\operatorname{Log}({\operatorname{Error}(\varepsilon_{n+1})})-\operatorname{Log}({\operatorname{Error}(\varepsilon_{n})})}{\operatorname{Log}(\varepsilon_{n+1})-\operatorname{Log}(\varepsilon_{n})}.$
(B.1)

where $\operatorname{Log}$ represents the logarithm with respect to any base.

| $\varepsilon_{n}$
---|---
Simulation | $n=1$ | $n=2$ | $n=3$ | $n=4$ | $n=5$ | $n=6$ | $n=7$ | $n=8$ | $n=9$
CC1 | $0.9395$ | $0.9596$ | $1.0348$ | $0.9428$ | $0.9649$ | $1.0884$ | $1.0954$ | $1.1309$ | $0.9805$
CC2 | $0.9315$ | $1.0231$ | $0.9101$ | $0.8885$ | $0.9012$ | $0.5896$ | $1.3951$ | $1.1744$ | $1.0025$
DC1 | $0.9195$ | $1.0594$ | $0.8032$ | $1.0674$ | $1.2587$ | $0.9238$ | $0.3265$ | $1.5408$ | $0.0566$
DC2 | $0.5304$ | $0.7907$ | $0.5235$ | $1.1918$ | $0.7092$ | $0.6909$ | $0.8841$ | $1.2225$ | $0.5127$
CNC1 | $0.7646$ | $0.7054$ | $0.8223$ | $0.8060$ | $0.9489$ | $0.4754$ | $0.8792$ | $0.6219$ | $0.8243$
CNC2 | $0.7258$ | $0.7915$ | $0.8005$ | $0.8219$ | $0.8305$ | $0.7787$ | $0.7749$ | $0.8241$ | $1.0809$
Table 3: Gradient between adjacent points in the Log-Log plots in Fig. 1(b),
Fig. 2(c) and Fig. 3(c)

From Table 3, it is evident that in the cases of CC1 and CC2, the rate of
convergence approaches $1$ asymptotically. However, for all other scenarios,
there appears to be no distinct pattern or clear convergence of the gradients.
Nevertheless, the gradients generally lie between $1/2$ and $1$.

Furthermore, we investigated the sensitivity of the convergence rate analysis
to the choice of $\Delta_{t}$. It is clear that for meaningful approximations
to the smoothed system it is needed that $\Delta_{t}$ is sufficiently small
compared to $\epsilon$, which necessitates extremely small time steps and
makes the simulation of the particle systems computationally costly. To assess
whether $\Delta_{t}$ is sufficiently small, we generated in Fig. 4(c) and
Table 4 rate of convergence plots with different values of $\Delta_{t}$. For
each $\Delta_{t}$, we selected several values of $\varepsilon$ that were
uniformly spaced (after taking logarithms) within the interval
$[\Delta_{t}\times 10^{-2.5},\,\Delta_{t}\times 10^{-1}]$. The findings
indicate that the estimated rate of convergence remains consistent with
respect to variations in $\Delta_{t}$.

| $\Delta_{t}$
---|---
Simulation | $10^{-4.5}$ | $10^{-5}$ | $10^{-5.5}$ | $10^{-6}$ | $10^{-7.5}$ | $10^{-8}$ | $10^{-8.5}$ | $10^{-9}$
CC1 | $0.836$ | $0.944$ | $0.987$ | $0.967$ | $-$ | $-$ | $-$ | $-$
CC2 | $0.988$ | $1.014$ | $1.039$ | $0.970$ | $-$ | $-$ | $-$ | $-$
DC1 | $-$ | $-$ | $-$ | $-$ | $0.821$ | $0.940$ | $0.968$ | $0.909$
DC2 | $-$ | $-$ | $-$ | $-$ | $0.539$ | $0.829$ | $0.728$ | $0.850$
CNC1 | $0.775$ | $0.750$ | $0.875$ | $0.751$ | $-$ | $-$ | $-$ | $-$
CNC2 | $0.685$ | $0.793$ | $0.806$ | $0.830$ | $-$ | $-$ | $-$ | $-$
Table 4: Gradient of the line of best fit in Fig. 4(c) (if plotted)

(a) CC1

(b) DC1

(c) CNC1 Figure 4: Sensitivity of rate of convergence with respect to changes
in $\Delta_{t}$

(d) CC2

(e) DC2

(f) CNC2

Finally, we investigated a scenario where the initial condition is Hölder
continuous near the boundary without any observed jump discontinuity in the
simulations, specifically, $X_{0-}\sim^{d}\Gamma(1.5,2)$ with $\alpha=1.3$. By
[10, Theorem 1.1], the limiting loss function is $1/2$-Hölder continuous at
$0$. The rate of convergence appears to be between $1/2$ and $1$ in this
setting.

$\Delta_{t}$ | $10^{-4.5}$ | $10^{-5}$ | $10^{-5.5}$ | $10^{-6}$
---|---|---|---|---
Gradient | $0.913$ | $0.863$ | $0.798$ | $0.66$
Table 5: Gradient of the line of best fit in Fig. 5

(a) $X_{0-}\sim^{d}\Gamma(1.5,2),\,\alpha=1.3$

(b) Rate of Convergence

Figure 5: Sensitivity of rate of convergence with respect to changes in
$\Delta_{t}$

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