Dynamic Investment and Counterparty Risk

Abstract

We introduce a dynamic optimization framework in which collateral is used to mitigate losses arising at counterparty’s default. The investor faces two sources of risk: the default risk of the entity referencing the traded credit swap security, and counterparty risk generated from the default event of the trading counterparty. We show that the value function of the control problem coincides with the classical solution of a nonlinear dynamic programming equation. We provide an explicit characterization of the optimal investment strategy, and show that the investor does not trade if counterparty risk is sufficiently high. These findings suggest that moving credit swap trades into well-designed clearinghouses may stimulate economic activities.

Appendices

Appendix 1: Price Function: Closed-Form Expressions and Comparisons

Lemma 7.1

The CDS price function $\Gamma (t,\mathbf{z})$, $(t,\mathbf{z})\in [0,T_1]\times \mathcal{S}$, admits the following closed-form:

• $\Gamma (t,(1,0))=\Gamma (t,(1,1))=L$, for all $t\in [0,T_1]$.

• The price function $\Gamma (t,(0,1))$ is given by

  $$\begin{aligned} \Gamma (t,(0,1)) = \int _t^{T_1} \left( Lh_{1}(s,(0,1))-\nu \right) e^{-\int _t^s(r+h_{1}(u,(0,1)))\mathrm {d}u}\mathrm {d}s,\ \ \ \ t\in [0,T_1].\nonumber \\ \end{aligned}$$

  (58)

• The price function $\Gamma (t,(0,0))$ is given by

  $$\begin{aligned} \Gamma (t,(0,0))&= \int _t^{T_1} \big (Lh_{1}(s,(0,0))-\nu + h_{C}(s,(0,0))\Gamma (s,(0,1))\big )\nonumber \\&\quad \times e^{-\int _t^s(r+h_{1}(u,(0,0))+h_{C}(u,(0,0)))\mathrm {d}u}\mathrm {d}s,\ \ \ t\in [0,T_1]. \end{aligned}$$

  (59)

  Here $\Gamma (t,(0,1))$, $t\in [0,T_1]$, is given by (58).

Proof

Let $(t,\mathbf{z})\in [0,T_1]\times \mathcal{S}$. We decompose $\Gamma (t,\mathbf{z})$ as $\Gamma (t,\mathbf{z}) = L\Gamma ^{(1)}(t,\mathbf{z}) - \nu \Gamma ^{(2)}(t,\mathbf{z})$, where

$$\begin{aligned} \Gamma ^{(1)}(t,\mathbf{z})&:= \mathbb {E} \left[ H_1(T_1)e^{-\int _t^{T_1} r(1-H_1(u))\mathrm {d}u}\Big | \mathbf{H}(t)=\mathbf{z}\right] ,\ \ \ \mathrm{and}\nonumber \\ \Gamma ^{(2)}(t,\mathbf{z})&:= \mathbb {E} \left[ \int _t^{T_1}e^{-\int _t^u r\mathrm {d}s}(1-H_1(u))\mathrm {d}u \Big | \mathbf{H}(t)=\mathbf{z}\right] . \end{aligned}$$

Using the Feymann-Kac’s formula, it follows that $\Gamma ^{(1)}$ and $\Gamma ^{(2)}$ satisfy the following equations. For $\mathbf{z}=(z_1,z_C)\in \mathcal{S}$,

$$\begin{aligned} \left( \frac{\partial }{\partial t} + \mathcal{A}\right) \Gamma ^{(1)}(t,\mathbf{z}) = r(1-z_1)\Gamma ^{(1)}(t,\mathbf{z}),\ \ \ \ \ \Gamma ^{(1)}(T_1,\mathbf{z})=z_1, \end{aligned}$$

(60)

and

$$\begin{aligned} \left( \frac{\partial }{\partial t} + \mathcal{A}\right) \Gamma ^{(2)}(t,\mathbf{z}) + (1-z_1) = r\Gamma ^{(2)}(t,\mathbf{z}),\ \ \ \ \ \Gamma ^{(2)}(T_1,\mathbf{z})=0, \end{aligned}$$

(61)

where the operator $\mathcal{A}$ is defined by (1). From the representation given in (5), we have $\Gamma (t,(1,0))=\Gamma (t,(1,1))=L$. Using this, along with Eqs. (60) and (61), we have that $\Gamma (t,(0,1))$ satisfies the equation:

$$\begin{aligned} \Gamma '(t,(0,1)) + h_{1}(t,(0,1))\left( L-\Gamma (t,(0,1))\right) - r \Gamma (t,(0,1)) -\nu =0 \end{aligned}$$

(62)

with terminal condition $\Gamma (T_1,(0,1))=0$. Solving (62), we obtain (58). Using such an expression, along with Eqs. (60) and (61), we deduce that $\Gamma (t,(0,0))$ satisfies

$$\begin{aligned}&\Gamma '(t,(0,0)) + h_{1}(t,(0,0))\left( L-\Gamma (t,(0,0))\right) \nonumber \\&\qquad + h_{C}(t,(0,0))\left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) - r \Gamma (t,(0,0)) -\nu =0 \end{aligned}$$

(63)

with terminal condition $\Gamma (T_1,(0,0))=0$. Solving it, we obtain (59). $\square$

Remark 7.2

Under assumptions (A1) and (A2), we have $Lh_{1}(t,(0,1))-\nu>Lh_{1}(t,(0,0))-\nu >0$ for all $t\in [0,T_1]$. Then, $\Gamma (t,(0,1))>0$ for all $t\in [0,T_1)$ and $\Gamma (T_1,(0,1))=0$ from (58). Moreover, it follows from (59) that $\Gamma (t,(0,0))>0$ for all $t\in [0,T_1)$ and $\Gamma (T_1,(0,0))=0$. Further, using (58), Under assumptions (A1) and (A2), for all $t\in [0,T_1)$,

$$\begin{aligned} \Gamma (t,(0,1))\ge & {} \frac{L\inf _{u\in [0,T_1]}h_{1}(u,(0,1))-\nu }{r+\sup _{u\in [0,T_1]}{h}_{1}(u,(0,1))}\left[ 1-e^{-(r+(T_1-t)\sup _{u\in [0,T_1]}{h}_{1}(u,(0,1)))}\right] \\> & {} 0 \nonumber . \end{aligned}$$

Then, it holds that, for all $t\in [0,T_1)$,

$$\begin{aligned} \Gamma (t,(0,0))&\ge \frac{L\inf _{u\in [0,T_1]}{h}_{1}(u,(0,0))-\nu }{r+\sup _{u\in [0,T_1]}{h}_{1}(u,(0,0))+\sup _{u\in [0,T_1]}{h}_{C}(u,(0,0))}\nonumber \\&\quad \times \left[ 1-e^{-(r+\sup _{u\in [0,T_1]}{h}_{1}(u,(0,0))+(T_1-t)\sup _{u\in [0,T_1]}{h}_{C}(u,(0,0)))}\right] >0 \nonumber . \end{aligned}$$

On the other hand, using (58), it follows that, for all $t\in [0,T_1)$,

$$\begin{aligned} \Gamma (t,(0,1))&\le L\int _t^{T_1} (r+h_{1}(s,(0,1)))e^{-\int _t^s(r+h_{1}(u,(0,1)))\mathrm {d}u}\mathrm {d}s'\nonumber \\&=\left[ 1-e^{-\int _t^{T_1}(r+h_{1}(s,(0,1)))\mathrm {d}s}\right] L\nonumber \\&\le \left[ 1-e^{-(r+T_1\sup _{u\in [0,T_1]}h_{1}(u,(0,1)))}\right] L=:\delta L, \end{aligned}$$

(64)

where $0<\delta < 1$.

Next, we compare price functions $\Gamma (t,(0,0))$ and $\Gamma (t,(0,1))$. Since the first is the price in the state where both CDS reference entity and counterparty are alive, while the second is the price in the state where the counterparty is defaulted, we expect that $\Gamma (t,(0,1)) > \Gamma (t,(0,0))$. This is because after the counterparty defaults, the default risk of the reference entity increases due to contagion effects. The next lemma shows this formally.

Lemma 7.3

Under the assumption (A1), for all $T\in (0,T_1)$, it holds that $\Gamma (t,(0,1))>\Gamma (t,(0,0))$ for $t\in [0,T]$.

Proof

Recall $\Gamma (t,(0,1))$ and $\Gamma (t,(0,0))$ satisfying Eqs. (62) and (63), respectively. Set $E(t):=\Gamma (t,(0,1))-\Gamma (t,(0,0))$. Then we obtain the following ODE:

$$\begin{aligned}&E'(t) - \big (r + h_{1}(t,(0,0)) + h_{C}(t,(0,0))\big ) E(t) + (h_{1}(t,(0,1))-h_{1}(t,(0,0)))\\&\quad \times \big (L-\Gamma (t,(0,1))\big ) = 0 \end{aligned}$$

with terminal condition $E(T_1)=\Gamma (T_1,(0,1))-\Gamma (T_1,(0,0))=0$. Solving it, we obtain the explicit solution:

$$\begin{aligned} E(t)= & {} \int _t^{T_1}(h_{1}(s,(0,1))-h_{1}(s,(0,0)))\\&\times \big (L-\Gamma (s,(0,1))\big )e^{-\int _t^{s}(r + h_{1}(u,(0,0)) + h_{C}(u,(0,0)))\mathrm {d}u}\mathrm {d}s. \end{aligned}$$

Since $\Gamma (t,(0,1))<L$ for all $t\in [0,T_1]$ and $h_{1}(t,(0,1))-h_{1}(t,(0,0))>0$ for all $t\in [0,T_1]$ by the assumption (A1), it implies that $E(t)>0$ for all $t\in [0,T]$. This completes the proof. $\square$

Appendix 2: Proofs Related to Section 3

Proof of Lemma 3.2

Notice that CDS reference entity and counterparty cannot default simultaneously in our model. By virtue of (4), we have

$$\begin{aligned} C_{\tau _C}&= (1-H_1(\tau _C))\Gamma \big (\tau _C,(H_1(\tau _C),H_C(\tau _C))\big )\\&= (1-H_1(\tau _C-))\Gamma \big (\tau _C,(H_1(\tau _C-),1)\big )\nonumber \\&= (1-H_1(\tau _C-))\Gamma \big (\tau _C-,(H_1(\tau _C-),1)\big )=\tilde{C}_{\tau _C-}, \end{aligned}$$

where the last step follows from the continuity of the price function $t-\!\!\!\rightarrow \Gamma (t,(z_1,1))$ for each fixed default state $z_1\in \{0,1\}$. This completes the proof of the lemma. $\square$

Proof of Lemma 3.3

Using the price dynamics of CDS given by (6), it follows from (17) that, for $\psi \in \mathcal{U}_0$,

$$\begin{aligned} \frac{\mathrm {d}V_t^{\psi }}{V_{t-}^{\psi }}&= \bigg \{\pi _{B}(t)r + \pi _M(t)r_M + \psi (t)\Big [r\Gamma (t,\mathbf{H}(t)) - h_{1}(t,\mathbf{H}(t))\big (L-\Gamma (t,\mathbf{H}(t))\big )\nonumber \\&\qquad -\big (\Gamma (t,\mathbf{H}^C(t))-\Gamma (t,\mathbf{H}(t))\big )h_{C}(t,\mathbf{H}(t))\Big ]\bigg \}\mathrm {d}t\nonumber \\&\qquad + \psi (t)\big (L-\Gamma (t,\mathbf{H}(t-))\big )\mathrm {d}H_1(t)\\&\quad \quad + \psi (t)\big [\Gamma (t,\mathbf{H}^C(t-))-\Gamma (t,\mathbf{H}(t-))\big ]\mathrm {d}H_C(t)\nonumber \\&\qquad -\big \{\psi (t) \tilde{C}_{t-} - \alpha \psi ^+(t)C_{t-}\big \}^+\mathrm {d}H_C(t). \end{aligned}$$

Since, for $\psi \in \mathcal{U}_0$, $r\left[ \psi (t)\Gamma _{}(t,\mathbf{H}(t-)) + \pi _B(t) + \pi _M(t)\right] = r$, in terms of CDS price (4), we may rewrite the admissible control $\pi _M(t)$ as $\pi _M(t) = -\alpha \psi ^+(t)\Gamma (t,\mathbf{H}(t-))$, and hence

$$\begin{aligned} \pi _B(t) = 1 - \psi (t)\Gamma (t,\mathbf{H}(t-)) + \alpha \psi ^+(t)\Gamma (t,\mathbf{H}(t-)). \end{aligned}$$

(65)

Thus we obtain

$$\begin{aligned}&r\left[ \psi (t)\Gamma (t,\mathbf{H}(t-))+\pi _{B}(t)\right] + \pi _M(t)r_M&\\&= r + \pi _M(t)(r_M-r)= r- \psi ^+(t)\alpha \Gamma (t,\mathbf{H}(t-))(r_M-r), \end{aligned}$$

and hence yielding that the dynamics of the wealth is given by (18) using Lemma 3.2. $\square$

Appendix 3: Proofs Related to Section 4

Proof of Lemma 4.1

Recall (22). Notice that $\Gamma (t,(0,0))>0$ by Remark 7.2, and Lemma 7.3 yields that

$$\begin{aligned}&{\Gamma }(t,(0,1))-\alpha \Gamma (t,(0,0))\\&\quad ={\Gamma }(t,(0,1))-\Gamma (t,(0,0)) + (1-\alpha )\Gamma (t,(0,0))>0,\ \ \ \forall \ t\in [0,T], \end{aligned}$$

since $\alpha \in [0,1]$. Then, for $(\psi ,t)\in \mathbb {R}\times [0,T]$, it holds that

$$\begin{aligned}&\big \{\psi {\Gamma }(t,(0,1)) - \psi ^+\alpha \Gamma (t,(0,0))\big \}^+\nonumber \\&\quad =\left\{ \begin{array}{l@{\quad }l} 0, &{} \mathrm{if}\ \psi \le 0,\\ \psi \big ({\Gamma }(t,(0,1))-\alpha \Gamma (t,(0,0))\big ), &{} \mathrm{if}\ \psi >0 \end{array}\right. \nonumber \\&\quad = \psi ^+\big ({\Gamma }(t,(0,1))-\alpha \Gamma (t,(0,0))\big ). \end{aligned}$$

(66)

Substitute (66) into the second operator in (22) to conclude that

$$\begin{aligned}&\mathcal{H}(\psi ;t,B):=\gamma \bigg \{r -\psi ^+\alpha \Gamma (t,(0,0))(r_M-r)- \psi \Big [(L-\Gamma (t,(0,0)))h_1(t,(0,0))\nonumber \\&\qquad +(\Gamma (t,(0,1))-\Gamma (t,(0,0)))h_{C}(t,(0,0))\Big ]\bigg \}B\nonumber \\&\qquad + \Big \{\big [1+\psi (L-\Gamma (t,(0,0)))\big ]^{\gamma }B(t,(1,0)) - B(t,(0,0))\Big \}h_{1}^{ \mathbb {P} }(t,(0,0))\nonumber \\&\qquad + \bigg \{\Big [1+\psi \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) -\psi ^+\big ({\Gamma }(t,(0,1)) - \alpha \Gamma (t,(0,0))\big )\Big ]^{\gamma }\nonumber \\&\quad \quad \times B(t,(0,1))-B\bigg \}h_{C}^{ \mathbb {P} }(t,(0,0)). \end{aligned}$$

(67)

Then the decomposition (25) follow from (67) by considering $\psi \le 0$ and $\psi >0$ respectively. $\square$

Proof of Lemma 4.2

From (26) and (27), we have

$$\begin{aligned} {g}_1(\psi ;t,B)&= - \left[ (L-\Gamma (t,(0,0))) h_{1}(t,(0,0))+(\Gamma (t,(0,1))\right. \nonumber \\&\quad \left. -\Gamma (t,(0,0)))h_{C}(t,(0,0))\right] B+ \left[ 1+\psi (L-\Gamma (t,(0,0)))\right] ^{\gamma -1}\nonumber \\&\quad \times (L-\Gamma (t,(0,0)))h_{1}^{ \mathbb {P} }(t,(0,0))B(t,(1,0))\nonumber \\&\quad + \left[ 1+\psi \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) \right] ^{\gamma -1}\nonumber \\&\quad \times \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) h_{C}^{ \mathbb {P} }(t,(0,0))B(t,(0,1)),\nonumber \\ {g}_2(\psi ;t,B)&= - \left[ \alpha \Gamma (t,(0,0))(r_M-r) + (L-\Gamma (t,(0,0)))h_{1}(t,(0,0))\right. \nonumber \\&\quad \left. +(\Gamma (t,(0,1))-\Gamma (t,(0,0)))h_{C}(t,(0,0))\right] B\nonumber \\&\quad + \left[ 1+\psi (L-\Gamma (t,(0,0)))\right] ^{\gamma -1}(L-\Gamma (t,(0,0)))h_{1}^{ \mathbb {P} }(t,(0,0))B(t,(1,0))\nonumber \\&\quad - \left[ 1-\psi (1-\alpha )\Gamma (t,(0,0))\right] ^{\gamma -1}(1-\alpha )\Gamma (t,(0,0))h_{C}^{ \mathbb {P} }(t,(0,0))B(t,(0,1)). \end{aligned}$$

(68)

For fixed $(t,B)\in [0,T]\times \mathbb {R}_+$, we deduce from (68) that for all $\psi >-\frac{1}{L-\Gamma (t,(0,0))}$, the function $\psi -\!\!\!\rightarrow {g}_{1}(\psi ;t,B)$ is continuous and decreasing. Notice that the following limits hold:

$$\begin{aligned}&\lim _{\psi \downarrow -\frac{1}{L-\Gamma (t,(0,0))}}{g}_{1}(\psi ;t,B) =+\infty ,\nonumber \\&\lim _{\psi \uparrow +\infty } {g}_{1}(\psi ;t,B) = - \Big [(L-\Gamma (t,(0,0)))h_{1}(t,(0,0))+(\Gamma (t,(0,1))\nonumber \\&-\Gamma (t,(0,0)))h_{C}(t,(0,0))\Big ]B<0,\nonumber \end{aligned}$$

with the above inequality following from Lemma 7.3. Applying Intermediate Value Theorem we then obtain a unique finite solution $\psi _{1}^{foc}>-\frac{1}{L-\Gamma (t,(0,0))}$ satisfying Eq. (28). Further, notice that the derivative $\frac{\partial {g}_{1}(\psi ;t,B)}{\partial \psi }<0$ in the desired domain of $\psi$. Then, in light of Kumagai (1980)’s implicit function theorem, we also have that $\psi _{1}^{foc}$, viewed as a function of (t, B) is $C^1$ in (t, B).

We next consider ${g}_{2}(\psi ;t,B)$. For all $\psi$ satisfying $-\frac{1}{L-\Gamma (t,(0,0))}<\psi <\frac{1}{(1-\alpha )\Gamma (t,(0,0))}$, we deduce from (68) that $\psi -\!\!\!\rightarrow {g}_{2}(\psi ;t,B)$ is continuous and decreasing. Moreover, the following limits hold:

$$\begin{aligned} \lim _{\psi \downarrow -\frac{1}{L-\Gamma (t,(0,0))}} {g}_{2}(\psi ;t,B) =+\infty ,\ \ \mathrm{and}\ \ \lim _{\psi \uparrow \frac{1}{(1-\alpha )\Gamma (t,(0,0))}}{g}_{2}(\psi ;t,B) = -\infty . \end{aligned}$$

Applying Intermediate Value Theorem, we deduce that there exists a unique finite solution

$$\begin{aligned} \psi _2^{foc}\in \left( -\frac{1}{L-\Gamma (t,(0,0))},\frac{1}{(1-\alpha )\Gamma (t,(0,0))}\right) \end{aligned}$$

satisfying Eq. (28). Further, in light of Kumagai (1980)’s implicit function theorem, we also have that $\psi _{2}^{foc}$ viewed as a function of (t, B) is $C^1$ in (t, B), given that the derivative $\frac{\partial {g}_{2}(\psi ;t,B)}{\partial \psi }<0$ in the desired domain of $\psi$.

Next, we prove that $\psi _{2}^{foc}(t,B)<\psi _{1}^{foc}(t,B)$. Notice that $-\alpha \Gamma (t,(0,0))(r_M-r)\le 0$ since $r_M \ge r$. Moreover, for $\psi \le \frac{1}{(1-\alpha )\Gamma (t,(0,0))}$, it holds that

$$\begin{aligned} - \Big [1-\psi (1-\alpha )\Gamma (t,(0,0))\Big ]^{\gamma -1}(1-\alpha )\Gamma (t,(0,0))h_{C}^{ \mathbb {P} }(t,(0,0))B(t,(0,1))\le 0, \end{aligned}$$

since $\alpha \in [0,1]$, where we set $\frac{1}{0}=\infty$. Additionally, for all $\psi >-\frac{1}{L-\Gamma (t,(0,0))}$ using Lemma 7.3 we obtain

$$\begin{aligned}&\Big [1+\psi \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) \Big ]^{\gamma -1}\\&\quad \times \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) h_{C}^{ \mathbb {P} }(t,(0,0))B(t,(0,1))>0. \end{aligned}$$

Comparing the expressions of ${g}_1(\psi ;t,B)$ and ${g}_2(\psi ;t,{B})$ defined in (68), and taking the above established inequalities into account, it follows that for all $-\frac{1}{L-\Gamma (t,(0,0))}<\psi \le \frac{1}{(1-\alpha )\Gamma (t,(0,0))}$ and $(t,{B})\in [0,T]\times \mathbb {R}_+$, ${g}_2(\psi ;t,{B})<{g}_1(\psi ;t,{B})$. This implies that $\psi _{2}^{foc}(t,{B})<\psi _{1}^{foc}(t,{B})$ for all $(t,{B})\in [0,T]\times \mathbb {R}_+$. If $\alpha =1$, then ${g}_2(\psi ;t,{B})$ is reduced to

$$\begin{aligned} {g}_2(\psi ;t,{B})&= - \Big [\Gamma (t,(0,0))(r_M-r) + (L-\Gamma (t,(0,0)))h_{1}(t,(0,0))\\&\quad +(\Gamma (t,(0,1))-\Gamma (t,(0,0)))h_{C}(t,(0,0))\Big ]B\\&\quad + \big [1+\psi (L-\Gamma (t,(0,0)))\big ]^{\gamma -1}\nonumber \\&\quad \times (L-\Gamma (t,(0,0)))h_{1}^{ \mathbb {P} }(t,(0,0))B(t,(1,0)). \end{aligned}$$

Solving for ${g}_2(\psi _2^{foc};t,{B})=0$ yields (29). $\square$

Proof of Lemma 4.3

From (68), we deduce that $\psi -\!\!\!\rightarrow {g}_1(\psi ;t,{B})$ is continuous and decreasing if $\psi >-\frac{1}{L-\Gamma (t,(0,0))}$. Further, we may decompose ${g}_1(\psi ;t,{B})$ as ${g}_1(\psi ;t,{B}) = \tilde{g}_1(\psi ;t,{B}) + \hat{g}_1(\psi ;t,{B})$, where the functions in $(\psi ,t,{B})\in \mathbb {R}\times [0,T]\times \mathbb {R}_+$ are defined by

$$\begin{aligned} \tilde{g}_1(\psi ;t,{B})&:= - \left( L-\Gamma (t,(0,0))\right) h_{1}(t,(0,0)){B}+\left[ 1+\psi \left( L-\Gamma (t,(0,0))\right) \right] ^{\gamma -1}\nonumber \\&\quad \quad \times \left( L-\Gamma (t,(0,0))\right) h_{1}^{ \mathbb {P} }(t,(0,0))B(t,(1,0)),\ \ \ \ \mathrm{and}\nonumber \\ \hat{g}_1(\psi ;t,{B})&:=- \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) h_{C}(t,(0,0))B\nonumber \\&\qquad +\left[ 1+\psi \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) \right] ^{\gamma -1}\nonumber \\&\quad \quad \times \left( \Gamma (t,(0,1))-\Gamma (t,(0,0))\right) h_{C}^{ \mathbb {P} }(t,(0,0))B(t,(0,1)). \end{aligned}$$

It is easy to check that $\tilde{g}_1(\tilde{\psi }_{1}(t,{B});t,{B})=\hat{g}_1(\hat{\psi }_{1}(t,{B});t,{B})=0$. Next, we prove that

$$\begin{aligned} {g}_{1}\left( \max \left\{ \tilde{\psi }_{1}(t,{B}),\hat{\psi }_{1}(t,{B})\right\} ;t,{B}\right) \le 0. \end{aligned}$$

(69)

First we consider the case where $\tilde{\psi }_{1}(t,{B})\le \hat{\psi }_{1}(t,{B})$. In this case, ${g}_{1}(\hat{\psi }_{1}(t,{B});t,{B})=\tilde{g}_{1}(\hat{\psi }_{1}(t,{B});t,{B})$, since $\hat{g}_{1}(\hat{\psi }_{1}(t,{B});t,{B})=0$. Notice that the function $\psi -\!\!\!\rightarrow \tilde{g}_{1}(\psi ;t,{B})$ is continuous and decreasing on $\psi >-\frac{1}{L-\Gamma (t,(0,0))}$. Then by the assumption that $\tilde{\psi }_{1}(t,{B})\le \hat{\psi }_{1}(t,{B})$, it follows that $\tilde{g}_{1}(\tilde{\psi }_{1}(t,{B});t,{B})\ge \tilde{g}_{1}(\hat{\psi }_{1}(t,{B});t,{B})$. Since $\tilde{g}_{1}(\tilde{\psi }_{1}(t,{B});t,{B})=0$, we obtain that

$$\begin{aligned} {g}_{1}\left( \max \left\{ \tilde{\psi }_{1}(t,{B}),\hat{\psi }_{1}(t,{B})\right\} ;t,{B}\right) =\tilde{g}_{1}(\hat{\psi }_{1}(t,{B});t,{B})\le 0, \end{aligned}$$

yielding the inequality (69). Using a symmetric argument, we can establish the validity of (69) when $\tilde{\psi }_{1}(t,{B})>\hat{\psi }_{1}(t,{B})$ and we omit the proof details here. Then the estimate (30) follows immediately from the inequality (69).

Proof of Lemma 4.7

A direct computation results in the limits (31) and (32). Using the implicit function theorem and Lemma 4.2, we have

since $0<\gamma <1$. Similarly, it also holds that $\frac{\partial \psi _{2}^{foc}(t,B)}{\partial B}<0$, where we used $B(t,(1,0))=B(t,(0,1))=\frac{1}{\gamma }e^{\gamma r(T-t)}$. As for $\psi _{2}^{foc}$, the expressions in (32) follow from the definition of ${g}_2(\psi ;t,{B})$ given in (68). This ends the proof of the lemma. $\square$

Appendix 3: Proofs Related to Section 5

Proof of Lemma 5.1

Using (64) and Lemma 7.3, we have $0<\Gamma (t,(0,0))<\Gamma (t,(0,1))<\delta L$ for all $t\in [0,T]$, where $\delta \in (0,1)$. This yields $0<\frac{1}{L-\Gamma (t,(0,0))}<\frac{1}{L-\Gamma (t,(0,1))}<\frac{1}{(1-\delta )L}$, for all $t\in [0,T]$. Further using Theorem 4.4, the inequality (30) proved in Lemma 4.3, the estimates of $\Gamma (t,(0,1))$ and $\Gamma (t,(0,0))$ given in Remark 7.2 and that $B(t,(1,1))=\frac{1}{\gamma }e^{\gamma r(T-t)}$, $t\in [0,T]$, it follows that there exist constants $K_1,\bar{K}_1>0$ so that

$$\begin{aligned} \left| \psi ^*(t,B)\right| \le K_1 + \bar{K}_1 B^{\frac{1}{\gamma -1}},\ \ \ \ \ \ (t,B)\in [0,T]\times \mathbb {R}_+. \end{aligned}$$

(70)

Using the inequality $0<\Gamma (t,(0,0))<\Gamma (t,(0,0))< L$ for all $t\in [0,T]$, we obtain that for all $(t,B)\in [0,T]\times \mathbb {R}_+$,

for some constants $K,\bar{K}>0$ following from the estimate (70). Similarly, using the estimate (70), we deduce the existence of constants $K_1,K,\bar{K}>0$ so that

$$\begin{aligned} {C}(t,\psi ^*(t,B))&\le \big [1+|\psi ^*(t,B)|(L-\Gamma (t,(0,0)))\big ]^{\gamma }h_{1}^{ \mathbb {P} }(t,(0,0))B(t,(1,1))\nonumber \\&\qquad +\big [1+|\psi ^*(t,B)|(\Gamma (t,(0,1))-\Gamma (t,(0,0)))\nonumber \\&\quad \quad +|\psi ^*(t,B)|(\Gamma (t,(0,1))-\alpha \Gamma (t,(0,0)))\big ]^{\gamma }\nonumber \\&\qquad \times h_{C}^{ \mathbb {P} }(t,(0,0))B(t,(1,1))\nonumber \\&\le \bigg \{\big [1+2L|\psi ^*(t,B)|\big ]^{\gamma }\sup _{t\in [0,T]}h_1^{ \mathbb {P} }(t,(0,0)) +\big [1+4L|\psi ^*(t,B)|\big ]^{\gamma }\nonumber \\&\quad \quad \times \sup _{t\in [0,T]}h_1^{ \mathbb {P} }(t,(0,0))\bigg \}B(0,(1,1))\nonumber \\&\le K_1 + K_1\left| \psi ^*(t,B)\right| ^{\gamma }\le K + \bar{K}B^{\frac{\gamma }{\gamma -1}}, \end{aligned}$$

where $B(0,(1,1))=\frac{1}{\gamma }e^{\gamma r T}$, and we also used the inequality $(x+y)^{\gamma }\le x^{\gamma }+y^{\gamma }$ for all $x,y\in \mathbb {R}_+$ and $\gamma \in (0,1)$. This concludes the proof of the lemma. $\square$

Proof of Lemma 5.2

First we notice that $\psi ^*(t,B)$ is the optimum. Then, we have, for all $(t,B)\in [0,T]\times \mathbb {R}_+$,

$$\begin{aligned} A(t,\psi ^*(t,B))B + C(t,\psi ^*(t,B))\ge A(t,0)B + C(t,0). \end{aligned}$$

(71)

Correspondingly, we introduce the following equation:

$$\begin{aligned} u'(t) + A(t,0)u(t) + C(t,0) =0,\ \ \ t\in [0,T) \end{aligned}$$

(72)

with terminal condition $u(T)=a_0\in (0,\frac{1}{\gamma }]$. Recall $A(t,\psi )$ and $C(t,\psi )$ defined in (41). Since $\psi -\!\!\!\rightarrow \psi ^+$ is Lipschitz-continuous, we have that $\psi -\!\!\!\rightarrow A(t,\psi )$ is Lipschitz-continuous and $\psi -\!\!\!\rightarrow C(t,\psi )$ is locally Lipschitz-continuous uniformly in $t\in [0,T]$. By Lemma 4.9, we have that the continuous coefficients $B-\!\!\!\rightarrow {A}(t,\psi ^*(t,B))$ and $B-\!\!\!\rightarrow {C}(t,\psi ^*(t,B))$ are locally Lipschitz-continuous uniformly in $t\in [0,T]$. Since B(t) is a solution to Eq. (40), then it must be $C^1$ on [0, T], and hence it is bounded on [0, T]. Using the comparison theorem for first-order ODEs along with the inequality (71) and observing that $u(T)=a_0\le \frac{1}{\gamma }=B(T)$, we have that $B(t)\ge u(t)$ for all $t\in [0,T]$.

We next prove the existence of a constant $\eta >0$ so that $u(t)\ge \eta$ for all $t\in [0,T]$. First, we have

$$\begin{aligned} A(t,0)= & {} \gamma r -h_{1}^{ \mathbb {P} }(t,(0,0)) - h_{C}^{ \mathbb {P} }(t,(0,0)),\ \ \ \mathrm{and}\\ C(t,0)= & {} \big (h_{1}^{ \mathbb {P} }(t,(0,0))+h_{C}^{ \mathbb {P} }(t,(0,0))\big )B(t,(1,1)). \end{aligned}$$

Then, for all $t\in [0,T]$, it holds that $C(t,0)>0$, and

$$\begin{aligned} A(t,0) \ge \gamma r-\sup _{t\in [0,T]}h_{1}^{ \mathbb {P} }(t,(0,0)) - \sup _{t\in [0,T]}h_{C}^{ \mathbb {P} }(t,(0,0))=:\eta _{0}. \end{aligned}$$

(73)

Recall Eq. (72). We have its solution $(u(t);\ t\in [0,T])$ admits the following form, for all $t\in [0,T]$,

$$\begin{aligned} u(t)&= a_0e^{\int _t^TA(s,0)\mathrm {d}s} + \int _t^T C(s,0)e^{\int _t^sA(\iota ,0)\mathrm {d}\iota }\mathrm {d}s \ge a_0e^{\int _t^TA(s,0)\mathrm {d}s}\ge a_0e^{\eta _0(T-t)}. \end{aligned}$$

In what follows, define the positive constant

$$\begin{aligned} \eta :=\left\{ \begin{array}{ll} a_0, &{}\quad \mathrm{if}\ \eta _{0}\ge 0,\\ a_0e^{\eta _{0}T}, &{}\quad \mathrm{if}\ \eta _{0}<0. \end{array}\right. \end{aligned}$$

Then, it holds that $u(t)\ge \eta$ for all $t\in [0,T]$. This implies that $B(t)\ge u(t)\ge \eta$, for all $t\in [0,T]$. $\square$