$\displaystyle=b(t_{0},m_{0})+\int^{s}_{0}\partial_{t}b(t_{0}+u,M^{m_{0}}_{s})du+\int^{m_{0}e^{as}}_{m_{0}}\partial_{m}b(t_{0},u)du.$

However, from (A.15) and (A.5) we have

$\displaystyle\partial_{m}b(t_{0},m)$
$\displaystyle\leq|\rho+1|(1+b_{\infty}(m))e^{aT},$
$\displaystyle\partial_{t}b(t_{0}+u,M^{m_{0}}_{s})$
$\displaystyle\leq\frac{Cb_{\infty}(M^{m_{0}}_{s})+(b_{\infty}(M^{m_{0}}_{s}))^{\frac{1}{\gamma}}}{(T-t_{0}-u)C_{1}(1-K^{\frac{1-\gamma}{\gamma}})},$

which used in (A.6) give

$\displaystyle X_{s}^{x_{0}}(\omega)\leq
b(t_{0},m_{0})+\int^{s}_{0}\frac{Cb_{\infty}(M^{m_{0}}_{s})+(b_{\infty}(M^{m_{0}}_{s}))^{\frac{1}{\gamma}}}{(T-t_{0}-u)C_{1}(1-K^{\frac{1-\gamma}{\gamma}})}du+\int^{m_{0}e^{as}}_{m_{0}}|\rho+1|(1+b_{\infty}(u))e^{aT}du$
$\displaystyle=b(t_{0},m_{0})+\frac{Cb_{\infty}(M^{m_{0}}_{s})+(b_{\infty}(M^{m_{0}}_{s}))^{\frac{1}{\gamma}}}{C_{1}(1-K^{\frac{1-\gamma}{\gamma}})}\int^{s}_{0}\frac{1}{T-t_{0}-u}du+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT}$
$\displaystyle\leq
b(t_{0},m_{0})+\frac{Cb_{\infty}(M^{m_{0}}_{s})+(b_{\infty}(M^{m_{0}}_{s}))^{\frac{1}{\gamma}}}{C_{1}(1-K^{\frac{1-\gamma}{\gamma}})}\frac{s}{T-t_{0}-s}+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT}$
$\displaystyle=b(t_{0},m_{0})+F(s,m_{0})\frac{s}{T-t_{0}-s}+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT},$

where we have used the monotonicity of $b_{\infty}$ in Lemma 4.3, and
$F(s,m_{0}):=\frac{Cb_{\infty}(m_{0}e^{as})+(b_{\infty}(m_{0}e^{as}))^{\frac{1}{\gamma}}}{C_{1}(1-K^{\frac{1-\gamma}{\gamma}})}$.
Since $b(t_{0},m_{0})=x_{0}$, then we have

$\displaystyle
x_{0}\exp\bigg{(}\int^{s}_{0}\Big{[}(\rho+1)(\beta-r+M_{u})+\mu_{1}-\frac{\sigma_{1}^{2}}{2}\Big{]}du+\int^{s}_{0}\sigma_{1}d\widehat{B}_{u}(\omega)\bigg{)}$
(A.30) $\displaystyle\leq
x_{0}+F(s,m_{0})\frac{s}{T-t_{0}-s}+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT}.$

By the law of the iterated logarithm (cf. Theorem 9.23 in Karatzas and Shreve
[27]), for all $\epsilon>0$ we have (along a sequence of times converging to
zero)

$\displaystyle\widehat{B}_{s}(\omega)\geq(1-\epsilon)\sqrt{2s\log(\log(\frac{1}{s}))},$

which combined with (A.6) yields (with
$\sigma_{1}=\rho\sigma_{y}-(\rho+1)\theta=\frac{\gamma\sigma_{y}-\theta}{1-\gamma}>0$
due to $\gamma>1$ and Assumption 4.1)

$\displaystyle
x_{0}e^{((\rho+1)(\beta-r)+\mu_{1}-\frac{\sigma_{1}^{2}}{2})s+(\rho+1)\int^{s}_{0}M_{u}du}e^{\sigma_{1}(1-\epsilon)\sqrt{2s\log(\log(\frac{1}{s}))}}$
$\displaystyle\leq
x_{0}+F(s,m_{0})\frac{s}{T-t_{0}-s}+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT}.$

On the other hand, since $e^{x}=1+x+O(x^{2})$ when $x\approx 0$, the last
display equation implies (for $s$ small enough) that

$\displaystyle
x_{0}\bigg{[}1+\sigma_{1}(1-\epsilon)\sqrt{2s\log(\log(\frac{1}{s}))}+((\rho+1)(\beta-r)+\mu_{1}-\frac{\sigma_{1}^{2}}{2})s+(\rho+1)\int^{s}_{0}M_{u}du\bigg{]}$
$\displaystyle\leq
x_{0}+F(s,m_{0})\frac{s}{T-t_{0}-s}+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT},$

which simplified gives

$\displaystyle
x_{0}\sigma_{1}(1-\epsilon)\sqrt{2s\log(\log(\frac{1}{s}))}+x_{0}((\rho+1)(\beta-r)+\mu_{1}-\frac{\sigma_{1}^{2}}{2})s+x_{0}(\rho+1)\int^{s}_{0}M_{u}du$
(A.31) $\displaystyle\leq
F(s,m_{0})\frac{s}{T-t_{0}-s}+(m_{0}e^{as}-m_{0})|\rho+1|(1+b_{\infty}(m_{0}e^{as}))e^{aT}.$

Then dividing by $s$ and letting $s\downarrow 0$, we obtain that the left
hand-side of the inequality in (A.6) is $\infty$ (since
$\sqrt{2s\log(\log(\frac{1}{s}))}/s\to\infty$ for $s\downarrow 0$), but the
right hand-side of the inequality in (A.6) is the constant
$\frac{F(0,m_{0})}{T-t_{0}}+m_{0}a|\rho+1|(1+b_{\infty}(m_{0}))e^{aT}$. Thus,
we reach a contradiction and
$\hat{\tau}(t,b(t,m),m)=0,\widehat{\mathbb{P}}$-a.s.

Step 2. In order to prove that $\hat{\tau}(t,x,m)\leq\tau^{*}(t,x,m),$ one can
finally use arguments as in the proof of Lemma 5.1 in De Angelis and Ekström
[11].

∎

###  A.7. Proof of Theorem 5.1

###### Proof.

The proof is organized in two steps.

Step 1: We show that $V\in C^{1,1,1,1}(\mathcal{O}^{\prime})$ and
$V_{ww},V_{yy},V_{wy}\in L^{\infty}_{loc}(\mathcal{O}^{\prime})$ and $V$ is a
solution in the a.e. sense to the HJB equation

$\displaystyle
0=\max\Big{\\{}\widehat{V}-V,\sup_{c,\pi}\Big{[}V_{t}+\frac{1}{2}\sigma^{2}\pi^{2}V_{ww}+(\pi(\mu-r)+rw-c+y)V_{w}+\sigma\pi\sigma_{y}yV_{wy}$
(A.32)
$\displaystyle+\frac{1}{2}\sigma_{y}^{2}y^{2}V_{yy}+\mu_{y}yV_{y}+amV_{m}+U_{1}(c)-(\beta+m)V\Big{]}\Big{\\}}.$

Step 1-(a): Firstly we show the regularity of $J$. From (4.1) we know that
$J(t,z,m,y)=zy\widetilde{J}(t,z^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}},m)$.
Therefore,

$\displaystyle J_{t}=zy\widetilde{J}_{t},\quad J_{m}=zy\widetilde{J}_{m},\quad
J_{z}=y\widetilde{J}+z^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}}[\frac{y}{1-\gamma}\widetilde{J}_{x}],\quad
J_{y}=z\widetilde{J}+z^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}}[\frac{z\gamma}{1-\gamma}\widetilde{J}_{x}],$
$\displaystyle
J_{z}=y\widetilde{J}+z^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}}[\frac{y}{1-\gamma}\widetilde{J}_{x}],\quad
J_{y}=z\widetilde{J}+z^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}}[\frac{z\gamma}{1-\gamma}\widetilde{J}_{x}],$
$\displaystyle
J_{yy}=z^{\frac{1}{1-\gamma}+1}y^{\frac{\gamma}{1-\gamma}-1}\frac{\gamma}{(1-\gamma)^{2}}\widetilde{J}_{x}+z^{\frac{2}{1-\gamma}+1}y^{\frac{2\gamma}{1-\gamma}-1}\frac{\gamma^{2}}{(1-\gamma)^{2}}\widetilde{J}_{xx},\
\text{in the a.e. sense},$ $\displaystyle
J_{zz}=z^{\frac{1}{1-\gamma}-1}y^{\frac{\gamma}{1-\gamma}+1}\frac{2-\gamma}{(1-\gamma)^{2}}\widetilde{J}_{x}+z^{\frac{2}{1-\gamma}-1}y^{\frac{2\gamma}{1-\gamma}+1}\frac{1}{(1-\gamma)^{2}}\widetilde{J}_{xx},\
\text{in the a.e. sense},$ (A.33) $\displaystyle
J_{zy}=\widetilde{J}+z^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}}\frac{\gamma-\gamma^{2}+1}{(1-\gamma)^{2}}\widetilde{J}_{x}+z^{\frac{2}{1-\gamma}}y^{\frac{2\gamma}{1-\gamma}}\frac{\gamma}{(1-\gamma)^{2}}\widetilde{J}_{xx},\
\text{in the a.e. sense}.$

Then we conclude that $J\in C^{1,1,1,1}(\mathcal{O})\cap
C^{1,2,1,2}(\overline{\mathcal{W}})$ with $J_{yy},J_{zz},J_{zy}\in
L^{\infty}_{loc}(\mathcal{O})$ due to Corollary 4.1.

Step 1-(b): Now we show the regularity of $V$. From (5.2), using that
$J_{z}(t,z^{*}(t,w,m,y),m,y)=-w-g(t)$ and
$J_{zy}(t,z^{*}(t,w,m,y),m,y)+J_{zz}z^{*}_{y}=-q(t)$, one has

(A.34) $\displaystyle V_{t}$
$\displaystyle=J_{t}(t,z^{*}(t,w,m,y),m,y)+J_{z}(t,z^{*}(t,w,m,y),m,y)z^{*}_{t}(t,w,m,y)$
$\displaystyle+(w+g(t))z^{*}_{t}(t,w,m,y)+g^{\prime}(t)z^{*}_{t}(t,w,m,y)=J_{t}(t,z^{*}(t,w,m,y),m,y)+g^{\prime}(t)z^{*}_{t}(t,w,m,y),$
$\displaystyle V_{m}$
$\displaystyle=J_{m}(t,z^{*}(t,w,m,y),m,y)+J_{z}(t,z^{*}(t,w,m,y),m,y)z^{*}_{m}(t,w,m,y)$
$\displaystyle+(w+g(t))z^{*}_{m}(t,w,m,y)=J_{m}(t,z^{*}(t,w,m,y),m,y),$
$\displaystyle V_{y}$
$\displaystyle=J_{y}(t,z^{*}(t,w,m,y),m,y)+J_{z}(t,z^{*}(t,w,m,y),m,y)z^{*}_{y}(t,w,m,y)$
$\displaystyle+(w+g(t))z^{*}_{y}(t,w,m,y)+z^{*}(t,w,m,y)q(t)=J_{y}(t,z^{*}(t,w,m,y),m,y)+z^{*}(t,w,m,y)q(t),$
$\displaystyle V_{yy}$
$\displaystyle=J_{yy}(t,z^{*}(t,w,m,y),m,y)+J_{yz}(t,z^{*}(t,w,m,y),m,y)z^{*}_{y}(t,w,m,y)+z_{y}^{*}(t,w,m,y)q(t),$
$\displaystyle V_{w}$
$\displaystyle=J_{z}(t,z^{*}(t,w,m,y),m,y)z^{*}_{w}(t,w,m,y)+z^{*}(t,w,m,y)+(w+g(t))z^{*}_{w}(t,w,m,y)=z^{*}(t,w,m,y),$
$\displaystyle V_{wy}$
$\displaystyle=z^{*}_{y}(t,w,m,y)=\frac{-q(t)-J_{zy}(t,z^{*}(t,w,m,y),m,y)}{J_{zz}(t,z^{*}(t,w,m,y),m,y)},\
\text{in the a.e. sense},$ $\displaystyle V_{ww}$
$\displaystyle=z_{w}^{*}(t,w,m,y)=-\frac{1}{J_{zz}(t,z^{*}(t,w,m,y),m,y)},\
\text{in the a.e. sense}.$

The proof is then completed due to Step 1-(a).

Step 1-(c): Now we characterize the optimal retirement time in the primal
variables and show that $V$ is a solution in the a.e. sense to the HJB
equation. Define $\widetilde{W}(t,z,m,y):=Q(z,m)-zg(t)$. Recalling that
$J\geq\widetilde{W}$ on $\mathcal{O}^{\prime}$ by (3.18), we notice that if
$J(t,z^{*}(t,w,m,y),m,y)=\widetilde{W}(t,z^{*}(t,w,m,y),m,y)$, then the
function $z\mapsto(J-\widetilde{W})(t,z,m,y)$ attains its minimum value $0$ at
$(t,z^{*}(t,w,m,y),m,y).$ Hence,

$\displaystyle
J_{z}(t,z^{*}(t,w,m,y),m,y)=\widetilde{W}_{z}(t,z^{*}(t,w,m,y),m,y)=-w-g(t).$

This means that $z^{*}(t,w,m,y)$ is a stationary point of the convex function
$z\mapsto\widetilde{W}(t,z,m,y)+z(w+g(t))$, so that

$\displaystyle\widetilde{W}(t,z^{*}(t,w,m,y),m,y)+(w+g(t))z^{*}(t,w,m,y)$
$\displaystyle=\min_{z}(\widetilde{W}(t,z,m,y)+z(w+g(t)))$
$\displaystyle=\min_{z}[{Q}(z,m)+zw]=\widehat{V}(w,m),$

by Theorem 3.1. Together with (5.2), we obtain $V(t,w,m,y)=\widehat{V}(w,m)$.

On the other hand, if $V(t,w,m,y)=\widehat{V}(w,m)$, then by (5.2) and Theorem
3.1

$\displaystyle J(t,z^{*}(t,w,m,y),m,y)+(w+g(t))z^{*}(t,w,m,y)$
$\displaystyle=\inf_{z}(\widetilde{W}(t,z,m,y)+z(w+g(t)))\leq\widetilde{W}(t,z^{*}(t,w,m,y),m,y)+(w+g(t))z^{*}(t,w,m,y).$

Hence, since $J\geq\widetilde{W}$ on $\mathcal{O}^{\prime}$,
$J(t,z^{*}(t,w,m,y),m,y)=\widetilde{W}(t,z^{*}(t,w,m,y),m,y)$.

Combining these two arguments we have that

$\displaystyle\\{(t,w,m,y)\in\mathcal{O}^{\prime}:V(t,w,m,y)=\widehat{V}(w,m)\\}$
$\displaystyle=\\{(t,w,m,y)\in\mathcal{O}^{\prime}:J(t,z^{*}(t,w,m,y),m,y)=\widetilde{W}(t,z^{*}(t,w,m,y),m,y)\\}$
$\displaystyle=\\{(t,x,m)\in\mathcal{U}:z^{*}y\widetilde{J}(t,x^{*}(t,w,m,y),m)=z^{*}y{Q}(t,x^{*}(t,w,m,y),m)-z^{*}yq(t)\\}$
$\displaystyle=\\{(t,x,m)\in\mathcal{U}:\widetilde{J}(t,x^{*}(t,w,m,y),m)=\widehat{W}(x^{*}(t,w,m,y),m)\\}$
$\displaystyle=\\{(t,x,m)\in\mathcal{U}:\widehat{J}(t,x^{*}(t,w,m,y),m)=0\\},$

where
$x^{*}(t,w,m,y):=(z^{*}(t,w,m,y))^{\frac{1}{1-\gamma}}y^{\frac{\gamma}{1-\gamma}}$.
This, together with (5.4), leads to express the optimal retirement time in the
original coordinates as

$\left\\{\begin{aligned} \tau^{*}(t,w,m,y)&=\inf\\{s\geq
0:W^{w}_{s}\geq\widehat{b}(t+s,M^{m}_{s},Y^{y}_{s})\\}\wedge(T-t)\\\
&=\inf\\{s\geq
0:V(t+s,W^{w}_{s},M^{m}_{s},Y^{y}_{s})=\widehat{V}(W^{w}_{s},M^{m}_{s})\\}\wedge(T-t).\end{aligned}\right.$

Due to the regularity of $V$ and the dual relations between $V$ and $J$ (cf.
Step 1-(b)), from Corollary 4.1 we then find that $V$ is a solution in the
a.e. sense to the HJB equation.

Step 2: Assuming that there exists a unique strong solution $W^{*}$ to the SDE
(2.4), when $\pi,c$ and $\tau$ are replaced by $\pi^{*},c^{*}$ and $\tau^{*}$,
respectively. From (5.2) we have $W^{*}_{s}=-J_{z}(s,Z_{s},M_{s},Y_{s})-g(s)$,
where $Z_{s}$ is the solution to Equation (3.8) with the initial condition
$Z_{t}=z^{*}$. Since $J_{z}<0$ by Proposition 5.1, it is easy to verify that
$(c^{*},\pi^{*},\tau^{*})\in\mathcal{A}(t,w,m,y)$, in particular,
$W^{*}_{s}>-g(s)$. Finally, a standard verification argument leads to the
result. ∎

##  Appendix B Two auxiliary results

######  Lemma B.1.

Let $w>0$ be given, let $c\geq 0$ be a consumption process satisfying

$\displaystyle\mathbb{E}_{w,m}\bigg{[}\int_{t}^{\infty}\xi_{s,t}c_{s}ds\bigg{]}=w.$

Then, there exists a portfolio process $\pi$ such that the pair $(c,\pi)$ is
admissible and

$\displaystyle W^{c,\pi,\tau}_{s}>0,\ \text{for}\ s\geq\tau.$

###### Proof.

The proof is similar to Theorem 3.3.5 in Karatzas and Shreve [28], and we thus
omit details.

∎

######  Lemma B.2.

For any $\tau\in\mathcal{S}$, let $w>-g(t)$ be given, let $c\geq 0$ be a
consumption process. For any $\mathcal{F}_{\tau}$-measurable random variable
$\phi$ with $\mathbb{P}[\phi>-g(\tau)]=1$ such that

$\displaystyle\mathbb{E}_{t,w,m,y}\bigg{[}\xi_{\tau,t}(\phi+g(\tau))+\int_{t}^{\tau}\xi_{s,t}c_{s}ds\bigg{]}=w+g(t),$

there exists a portfolio process $\pi$ such that the pair $(c,\pi)$ is
admissible and

$\displaystyle W^{c,\pi,\tau}_{s}>-g(s),\ \text{for}\ s\leq\tau,\
\phi=W^{c,\pi,\tau}_{\tau}.$

###### Proof.

The proof is similar to Lemma 6.3 in Karatzas and Wang [29], and we thus omit
details.

∎

##  Appendix C Recursive integration method

The numerical method is inspired by Huang et al. [22] and Jeon et al. [24].
Here we only consider the case $\gamma>1$. We rewrite (4.27) as

(C.1) $\displaystyle 0=\int^{T-t}_{0}e^{-\kappa
s}\widehat{\mathbb{E}}_{\bar{b}(0),m}\Big{[}\Big{(}(1-K^{\frac{1-\gamma}{\gamma}}){U}^{*}_{1}(X_{s})+1\Big{)}\mathds{1}_{\\{X_{s}\leq\bar{b}(s)\\}}\Big{]}ds,$

where $\bar{b}(s):=b(t+s,M_{s})$. Now we compute the expectation inside the
integral in the right-hand side of (C.1),

$\displaystyle\widehat{\mathbb{E}}_{\bar{b}(0),m}\Big{[}\Big{(}(1-K^{\frac{1-\gamma}{\gamma}}){U}^{*}_{1}(X_{s})+1\Big{)}\mathds{1}_{\\{X_{s}\leq\bar{b}(s)\\}}\Big{]}$
$\displaystyle=(1-K^{\frac{1-\gamma}{\gamma}})\frac{\gamma}{1-\gamma}\widehat{\mathbb{E}}_{\bar{b}(0),m}\Big{[}X_{s}^{\frac{\gamma-1}{\gamma}}\mathds{1}_{\\{X_{s}\leq\bar{b}(s)\\}}\Big{]}+\widehat{\mathbb{P}}_{\bar{b}(0),m}[X_{s}\leq\bar{b}(s)]$
$\displaystyle=(1-K^{\frac{1-\gamma}{\gamma}})\frac{\gamma}{1-\gamma}\widehat{\mathbb{E}}_{\bar{b}(0),m}\Big{[}{\bar{b}(0)}^{\frac{\gamma-1}{\gamma}}H_{s}N(s,m)\mathds{1}_{\\{X_{s}\leq\bar{b}(s)\\}}\Big{]}+\widehat{\mathbb{P}}_{\bar{b}(0),m}[X_{s}\leq\bar{b}(s)]$
$\displaystyle=(1-K^{\frac{1-\gamma}{\gamma}})\frac{\gamma}{1-\gamma}{\bar{b}(0)}^{\frac{\gamma-1}{\gamma}}N(s,m)\widetilde{\mathbb{P}}_{\bar{b}(0),m}[X_{s}\leq\bar{b}(s)]+\widehat{\mathbb{P}}_{\bar{b}(0),m}[X_{s}\leq\bar{b}(s)].$

By direct computations we have

$\displaystyle\widehat{\mathbb{P}}_{\bar{b}(0),m}[X_{s}\leq\bar{b}(s)]$
$\displaystyle=\Phi\bigg{(}\frac{\log\frac{\bar{b}(s)}{\bar{b}(0)}-\int^{s}_{0}[(\rho+1)(\beta-r+me^{au})+\mu_{1}-\frac{\sigma_{1}^{2}}{2}]du}{\sigma_{1}\sqrt{s}}\bigg{)}=\Phi(d^{1}(s,\frac{\bar{b}(s)}{\bar{b}(0)})),$
$\displaystyle\widetilde{\mathbb{P}}_{\bar{b}(0),m}[X_{s}\leq\bar{b}(s)]$
$\displaystyle=\Phi\bigg{(}\frac{\log\frac{\bar{b}(s)}{\bar{b}(0)}-\int^{s}_{0}[(\rho+1)(\beta-r+me^{au})+\mu_{1}+\frac{(\gamma-2)\sigma_{1}^{2}}{2\gamma}]du}{\sigma_{1}\sqrt{s}}\bigg{)}=\Phi(d^{2}(s,\frac{\bar{b}(s)}{\bar{b}(0)})),$

where

$\displaystyle d^{1}(s,y):$ $\displaystyle=\frac{\log
y-\int^{s}_{0}[(\rho+1)(\beta-r+me^{au})+\mu_{1}-\frac{\sigma_{1}^{2}}{2}]du}{\sigma_{1}\sqrt{s}},$
$\displaystyle d^{2}(s,y):$ $\displaystyle=\frac{\log
y-\int^{s}_{0}[(\rho+1)(\beta-r+me^{au})+\mu_{1}+\frac{(\gamma-2)\sigma_{1}^{2}}{2\gamma}]du}{\sigma_{1}\sqrt{s}},$

with $\Phi(\cdot)$ being the cumulative distribution function of a standard
normal random variable. Then the integral equation (C.1) can be converted to

$\displaystyle 0=\int^{T-t}_{0}e^{-\kappa
s}\bigg{(}(1-K^{\frac{1-\gamma}{\gamma}})\frac{\gamma}{1-\gamma}{\bar{b}(0)}^{\frac{\gamma-1}{\gamma}}N(s,m)\Phi\Big{(}d^{2}(s,\frac{\bar{b}(s)}{\bar{b}(0)})\Big{)}+\Phi\Big{(}d^{1}(s,\frac{\bar{b}(s)}{\bar{b}(0)})\Big{)}\bigg{)}ds.$

Letting $\xi:=T-t$ and $\bar{b}^{*}(\xi)=\bar{b}(T-t-\xi)$, we can show that
$\bar{b}^{*}(\xi)$ satisfies the following integral equation:

(C.2) $\displaystyle
0=\int^{\xi}_{0}G(\xi,s,\bar{b}^{*}(\xi),\bar{b}^{*}(\xi-s))ds,$

where

$\displaystyle G(\xi,s,\bar{b}^{*}(\xi),\bar{b}^{*}(\xi-s)):$
$\displaystyle=e^{-\kappa
s}\bigg{(}(1-K^{\frac{1-\gamma}{\gamma}})\frac{\gamma}{1-\gamma}{\bar{b}^{*}(\xi)}^{\frac{\gamma-1}{\gamma}}N(s,m)\Phi\Big{(}d^{2}(s,\frac{\bar{b}^{*}(\xi-s)}{\bar{b}^{*}(\xi)})\Big{)}$
$\displaystyle+\Phi\Big{(}d^{1}(s,\frac{\bar{b}^{*}(\xi-s)}{\bar{b}^{*}(\xi)})\Big{)}\bigg{)}.$

In order to solve the above integral equation (C.2), the recursive iteration
method proceeds as follows.

We divide the interval $[0,\xi]$ into $n$ subintervals with end points
$\xi_{j},j=0,1,2,...,n$ where $\xi_{0}=0,\xi_{n}=\xi$ and
$\Delta\xi=\frac{\xi}{n}$. Let $\bar{b}^{*}_{j}$ denote the numerical
approximation to $\bar{b}^{*}(\xi_{j}),j=0,1,...,n$. For $\xi=\xi_{1}$, by the
trapezoidal rule, integral equation (C.2) is approximated by

(C.3) $\displaystyle
0=\frac{\Delta\xi}{2}[G(\xi_{1},\xi_{0},\bar{b}^{*}_{1},\bar{b}^{*}_{1})+G(\xi_{1},\xi_{1},\bar{b}^{*}_{1},\bar{b}^{*}_{0})].$

Since $\bar{b}^{*}_{0}=b(T,M^{m}_{T-t})=L$, the only unknown in (C.3) is
$\bar{b}^{*}_{1}$. We can solve the algebraic equation (C.3) by applying the
bisection method. Similarly, for $\xi=\xi_{2}$, we have

(C.4) $\displaystyle
0=\frac{\Delta\xi}{2}[G(\xi_{2},\xi_{0},\bar{b}^{*}_{2},\bar{b}^{*}_{2})+2G(\xi_{2},\xi_{1},\bar{b}^{*}_{2},\bar{b}^{*}_{1})+G(\xi_{2},\xi_{2},\bar{b}^{*}_{2},\bar{b}^{*}_{0})].$

Since $\bar{b}_{1}^{*}$ is known from previous step, equation (C.4) can be
solved for $\bar{b}_{2}^{*}$ by the same procedure. Hence, for
$\bar{b}_{k}^{*},k=2,3,...,n$, we can obtain $\bar{b}_{k}^{*}$ recursively as
the solution of the following algebraic equation,

$\displaystyle
0=\frac{\Delta\xi}{2}[G(\xi_{k},\xi_{0},\bar{b}^{*}_{k},\bar{b}^{*}_{k})+2\sum^{k-1}_{j=1}G(\xi_{k},\xi_{k-j},\bar{b}^{*}_{k},\bar{b}^{*}_{j})+G(\xi_{k},\xi_{k},\bar{b}^{*}_{k},\bar{b}^{*}_{0})].$

Now from the values of $\\{\bar{b}^{*}_{i}\\}_{i=1}^{n}$, $\widehat{J}$ in
(4.26) can be approximated by

$\displaystyle\widehat{J}(\xi,x,m)\approx\widehat{J}_{n}(\xi,x,m):=\frac{\Delta\xi}{2}[G(\xi_{n},\xi_{0},x,\bar{b}^{*}_{n})+2\sum^{n-1}_{j=1}G(\xi_{n},\xi_{n-j},x,\bar{b}^{*}_{j})+G(\xi_{n},\xi_{n},x,\bar{b}^{*}_{0})].$

As shown by Huang et al. [22], for sufficiently large number of subintervals
$n$, the approximated free boundary $\bar{b}^{*}_{n}$ converges to
$\bar{b}^{*}(\xi)$, and therefore, $\widehat{J}_{n}$ converges to
$\widehat{J}$ as well.

## Acknowledgments

Funded by the Deutsche Forschungsgemeinschaft (DFG, German Research
Foundation) – Project-ID 317210226 – SFB 1283. The work of Shihao Zhu was also
supported by the program of China Scholarships Council.

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