Robust Time-Inconsistent Stochastic Linear-Quadratic Control with Drift Disturbance

Abstract

This paper studies stochastic linear-quadratic control with a time-inconsistent objective and worst-case drift disturbance. We allow the agent to introduce disturbances to reflect her uncertainty about the drift coefficient of the controlled state process. We adopt a two-step equilibrium control approach to characterize the robust time-consistent controls, which can preserve the order of preference. Under a general framework allowing random parameters, we derive a sufficient condition for equilibrium controls using the forward-backward stochastic differential equation approach. We also provide analytical solutions to mean-variance portfolio problems for various settings. Our empirical studies confirm the improvement in portfolio’s performance in terms of out-of-sample Sharpe ratio by incorporating with robustness.

Appendices

Proofs of Results in Section 3

1.1 Proof of Lemma 3.5

Proof

Let k be a positive constant. For $j=1,...,d$,

$$\begin{aligned}&\Big \Vert \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon }{\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \left\{ \langle \varLambda ^\varepsilon (s;t),\eta \rangle \right\} ds - \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \left\{ \langle \varLambda (s;t),\eta \rangle \right\} ds \Big \Vert _2 \nonumber \\&\quad \le \Big \Vert \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon }{\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \Big \{ (C^j_s X^*_s + D^j_s u^*_s + \sigma ^j_s)' p(s;t) - \frac{h^*_j(X^*_s , u^*_s, s)}{\varPhi _j (X^*_s , u^*_s, s)} \Big \} ds \nonumber \\&\qquad - \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \Big \{ (C^j_s {\tilde{X}}^*_s + D^j_s u^{t,\varepsilon ,v}_s + \sigma ^j_s)' p^\varepsilon (s;t) - \frac{h^*_j({\tilde{X}}^*_s , u^{t,\varepsilon ,v}_s, s)}{\varPhi _j ({\tilde{X}}^*_s , u^{t,\varepsilon ,v}_s, s)} \Big \} ds\Big \Vert _2 \nonumber \\&\quad \le \lim _{\varepsilon \downarrow 0} \frac{k}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \Big \{ \Big \Vert (C^j_s {\tilde{X}}^*_s)' p^\varepsilon (s;t)- (C^j_s X^*_s)' p(s;t) \Big \Vert _2 + \Big \Vert (\sigma ^j_s)'(p^\varepsilon (s;t) - p(s;t)) \Big \Vert _2 \nonumber \\&\qquad + \Big \Vert (D^j_s u_s^{t,\varepsilon ,v})' p^\varepsilon (s;t) - \frac{h^*_j ({\tilde{X}}^*_s , u^{t,\varepsilon ,v}_s, s)}{\varPhi _j ({\tilde{X}}^*_s , u^{t,\varepsilon ,v}_s, s)} - (D^j_s u^*_s)' p(s;t) + \frac{h^*_j (X^*_s , u^*_s, s)}{\varPhi _j (X^*_s , u^*_s, s)} \Big \Vert _2 \Big \} ds. \end{aligned}$$

(A.1)

Note that $C^j,~j=1,\ldots ,d$ are essentially bounded. Let k be a positive constant, we have

$$\begin{aligned}&\lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \Big \Vert (C^j_s {\tilde{X}}^*_s)' p^\varepsilon (s;t)- (C^j_s X^*_s)' p(s;t) \Big \Vert _2 ds \\&\quad \le \lim _{\varepsilon \downarrow 0} \frac{k}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \Vert {\tilde{X}}^*_s\Vert _2 \Vert p^\varepsilon (s;t)-p(s;t)\Vert _2 + \Vert p(s;t)\Vert _2 \Vert {\tilde{X}}^*_s - X^*_s\Vert _2 ds \\&\quad \le \lim _{\varepsilon \downarrow 0} \Big ( \frac{k}{\varepsilon } \sqrt{\int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert {\tilde{X}}^*_s\Vert ^2_2ds} \sqrt{\int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert p^\varepsilon (s;t)-p(s;t)\Vert ^2_2ds} \\&\qquad + \frac{k}{\varepsilon } \sqrt{\int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert p(s;t)\Vert ^2_2 ds} \sqrt{\int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert {\tilde{X}}^*_s - X^*_s\Vert ^2_2 ds} \Big ). \end{aligned}$$

We have

$$\begin{aligned}&\frac{k}{\varepsilon } \sqrt{\int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert {\tilde{X}}^*_s\Vert ^2_2 ds} \sqrt{\int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert p^\varepsilon (s;t)-p(s;t)\Vert ^2_2 ds}\\&\quad \le k \sqrt{{\mathbb {E}}^{\mathbb {P}}_t\big [\sup _{s\in [t,T]} \Vert p^\varepsilon (s;t)-p(s;t)\Vert ^2_2\big ]}\sqrt{{\mathbb {E}}^{\mathbb {P}}_t\big [ \sup _{s\in [t,T]} \Vert {\tilde{X}}^*_s\Vert ^2_2 \big ]}. \end{aligned}$$

By the stability results of BSDEs, (see, e.g., Theorem 3.3, Chapter 7 in [38]), we have

$$\begin{aligned}&{\mathbb {E}}^{\mathbb {P}}_t\left[ \sup _{s\in [t,T]} \Vert p^\varepsilon (s;t)-p(s;t)\Vert ^2_2\right] \rightarrow 0,\quad {\mathbb {E}}^{\mathbb {P}}_t\big [\sup _{s\in [t,T]}\Vert {\tilde{X}}^*_s - X^*_s\Vert ^2_2\big ]=O(\varepsilon ), \quad \\&\quad {\mathbb {E}}^{\mathbb {P}}_t\big [ \sup _{s\in [t,T]} \Vert X^*_s\Vert ^2_2 \big ] < \infty . \end{aligned}$$

Hence, we have

$$\begin{aligned} k \sqrt{{\mathbb {E}}^{\mathbb {P}}_t\big [\sup _{s\in [t,T]} \Vert p^\varepsilon (s;t)-p(s;t) \Vert ^2_2\big ]}\sqrt{{\mathbb {E}}^{\mathbb {P}}_t\big [ \sup _{s\in [t,T]} \Vert {\tilde{X}}^*_s\Vert ^2_2 \big ]} \rightarrow 0. \end{aligned}$$

Other two terms in (A.1) can be proven in similar manner. Then the result follows. $\square$

1.2 Proof of Theorem 3.7

Proof

Let $X^{t,\varepsilon ,v}$ be the state process corresponding to $u^{t,\varepsilon ,v}, h^*$. Define $Y\equiv Y^{t,\varepsilon ,v}$ and $Z\equiv Z^{t,\varepsilon ,v}$ satisfying

$$\begin{aligned}&\left\{ \begin{array}{ll} dY_s=&{} \mu ^*_x (X^*_s, u^*_s, s)Y_sds + \sum _{j=1}^d[C_s^jY_s+D_s^j v{{\mathbf {1}}}_{s\in [t, t+\varepsilon )}] dW^{\mathbb {P}}_{js},\;\;s\in [t,T],\\ Y_t =&{} 0; \end{array}\right. \\&\left\{ \begin{array}{ll} dZ_s=&{} [\mu ^*_x (X^*_s, u^*_s, s) Z_s+ \delta \mu ^*(X^*_s, u^*_s, s) + \delta \mu ^*_x(X^*_s, u^*_s, s) Y_s]ds\\ &{}+ \frac{1}{2} \mu ^*_{xx} (X^*_s, u^*_s, s) Y_s Y_s ds \\ &{} + \sum _{j=1}^dC_s^jZ_sdW^{\mathbb {P}}_{js},\;\; s\in [t,T],\\ Z_t =&{} 0. \end{array}\right. \end{aligned}$$

Since Assumption 3.6 holds, by Lemma 1 in [23], we have the following moment estimates,

$$\begin{aligned} {\mathbb {E}}^{\mathbb {P}}_t\big [ \sup _{s\in [t,T]} \Vert X^{t,\varepsilon ,v}_s - X^*_s - Y^{t,\varepsilon ,v}_s - Z^{t,\varepsilon ,v}_s\Vert ^2_2 \big ]= & {} o(\varepsilon ^2), \quad \\ {\mathbb {E}}^{\mathbb {P}}_t\big [\sup _{s\in [t,T]} \Vert Y_s\Vert ^2_2\big ]= & {} O(\varepsilon ), \quad \\ {\mathbb {E}}^{\mathbb {P}}_t\big [\sup _{s\in [t,T]}\Vert Z_s\Vert ^2_2\big ]= & {} O(\varepsilon ^2). \end{aligned}$$

Furthermore, since $\mu ^*_x (X^*_s, u^*_s, s)$ is deterministic, we take conditional expectation on both sides of the SDE for Y, then ${\mathbb {E}}^{\mathbb {P}}_t[Y_s]$ satisfies an ODE with 0 as its unique solution. Therefore, ${\mathbb {E}}^{\mathbb {P}}_t[Y_s]=0, \; s\in [t,T]$.

Then, we obtain the following

$$\begin{aligned}&2[{\mathcal {L}}(t,X^*_t; u^{t,\varepsilon ,v},h^*({\tilde{X}}^*_s,u^{t,\varepsilon ,v}_s,s))-{\mathcal {L}}(t,X^*_t;u^*,h^*(X^*_s,u^*_s,s))]\\&\quad ={\mathbb {E}}^{\mathbb {P}}_t\int _t^T\Big [\langle 2 f^*_x, Y_s +Z_s \rangle + \langle f^*_{xx} (Y_s + Z_s), Y_s + Z_s \rangle \\&\qquad + \langle 2f^*_u, v \rangle {{\mathbf {1}}}_{s\in [t, t+\varepsilon )} + \langle f^*_{uu} v, v \rangle {{\mathbf {1}}}_{s\in [t, t+\varepsilon )} \Big ]ds\\&\qquad +{\mathbb {E}}^{\mathbb {P}}_t[ 2\langle GX^*_T-\nu {\mathbb {E}}^{\mathbb {P}}_t[X^*_T]-\mu _1 X_t^*-\mu _2, Y_T+Z_T \rangle ]\\&\qquad +{\mathbb {E}}^{\mathbb {P}}_t[\langle G(Y_T+Z_T),Y_T+Z_T \rangle ] + o(\varepsilon ). \end{aligned}$$

Using the definition of $({\tilde{p}}(\cdot ; t), {\tilde{k}}(\cdot ; t))$ and $({\tilde{P}}(\cdot ; t), {\tilde{K}}(\cdot ; t))$ in (3.7) and (3.8) and applying Itô’s lemma to the last two terms, we have

$$\begin{aligned}&{\mathbb {E}}^{\mathbb {P}}_t[\langle GX^*_T-\nu {\mathbb {E}}^{\mathbb {P}}_t[X^*_T]-\mu _1 X_t^*-\mu _2, Y_T+Z_T \rangle ]\\&\quad ={\mathbb {E}}^{\mathbb {P}}_t\int _t^T \Big [\langle - f^*_x, Y_s+Z_s \rangle + \langle (\mu ^*_u)' {\tilde{p}}(s;t)+\sum _{j=1}^d(D_s^j)'{\tilde{k}}^j(s;t), v {{\mathbf {1}}}_{s\in [t, t+\varepsilon )} \rangle \\&\qquad +\frac{1}{2} \langle \mu ^*_{xx} (X^*_s, u^*_s, s) Y_s {\tilde{p}}(s;t) , Y_s \rangle + \frac{1}{2} \langle \mu ^*_{uu}vv, {\tilde{p}}(s;t) \rangle {{\mathbf {1}}}_{s\in [t, t+\varepsilon )}\Big ]ds + o(\varepsilon ), \end{aligned}$$

and

$$\begin{aligned} {\mathbb {E}}^{\mathbb {P}}_t[\langle G(Y_T+Z_T), Y_T+Z_T \rangle ]&={\mathbb {E}}^{\mathbb {P}}_t\int _t^T\Big [ -\langle f^*_{xx}(Y_s+Z_s), Y_s+Z_s \rangle \\&\quad +\sum _{j=1}^d \langle (D_s^j)' {\tilde{P}}(s;t)D^j_s v, v \rangle {{\mathbf {1}}}_{s\in [t, t+\varepsilon )} \\&\quad - \langle \mu ^*_{xx}(Y_s+Z_s){\tilde{p}}(s;t), Y_s+Z_s \rangle \Big ]ds + o(\varepsilon ). \end{aligned}$$

After simplifications, we prove (3.11). $\square$

1.3 Proof of Theorem 3.9

Proof

Since $H^\varepsilon (s;t)\preceq 0$ and ${\tilde{H}}(s;t) \succeq 0$, from Theorems 3.2 and 3.7, $(h^*, u^*)$ is an equilibrium control pair if and only if

$$\begin{aligned} \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \varLambda ^\varepsilon (s;t)ds = 0, \quad \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } {\tilde{\varLambda }}(s;t)ds = 0. \end{aligned}$$

Using the notations in (3.13) and (3.14), we only need to show

$$\begin{aligned} \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \lambda (s;t)ds = 0. \end{aligned}$$

(A.2)

The first step is to prove

$$\begin{aligned} \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \lambda (s;t)ds = \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\int _t^{t+\varepsilon } \lambda (s;s)ds. \end{aligned}$$

(A.3)

The proof is analogous to Proposition 3.3 in [11]. Define $\varPsi (\cdot )$ as the solution of

$$\begin{aligned} d\varPsi (s) = \varPsi (s) \alpha '_sds, \; \varPsi (T) = I_n. \end{aligned}$$

(A.4)

$I_n$ is the $n\times n$ identity matrix. Since $\alpha _s$ is bounded and deterministic, $\varPsi (\cdot )$ is invertible, and $\varPsi (\cdot ), \varPsi ^{-1}(\cdot )$ are bounded.

Denote ${\hat{p}}(s;t) = \varPsi (s) p(s;t) + \nu {\mathbb {E}}^{\mathbb {P}}_t[X^*_T] + \mu _1 X_t^* + \mu _2$ and ${\hat{k}}^j(s;t) = \varPsi (s) k^j(s;t)$. Then

$$\begin{aligned} \left\{ \begin{array}{ll} d{\hat{p}} (s;t) &{}= -\Big [ \sum _{j=1}^d \varPsi (s) (\beta _s^j)' \varPsi ^{-1}(s) \hat{k}^j(s;t) + \varPsi (s)\gamma _s \Big ]ds \\ &{}\quad + \sum _{j=1}^d {\hat{k}}^j(s;t)dW^{\mathbb {P}}_{js},\;\;s\in [t,T],\\ {\hat{p}}(T;t) &{}= G X^*_T \end{array}\right. \end{aligned}$$

has a unique solution which does not depend on t, so we can denote ${\hat{p}}(s;t) = {\hat{p}}(s)$, ${\hat{k}}^j(s;t) = {\hat{k}}^j(s)$, therefore, $p(s;t) = \varPsi ^{-1}(s){\hat{p}}(s) + \varPsi ^{-1}(s) w_t$, where $w_t = - \nu {\mathbb {E}}^{\mathbb {P}}_t[X^*_T] - \mu _1 X_t^* - \mu _2$. Then $\lambda (s;t) = f_1(s) + f_2(s) w_t$, where

$$\begin{aligned}&f_1(s) \!=\! \lambda _0(s)\varPsi ^{-1}(s){\hat{p}}(s) \!+\! \sum ^d_{j=1} (\lambda ^j_s)' \varPsi ^{-1}(s)\hat{k}^j(s) \!+\! \lambda _{d+1}(s),\quad \!\! f_2(s) \!=\! \lambda _0(s) \varPsi ^{-1}(s). \end{aligned}$$

Since ${\mathbb {E}}^{\mathbb {P}}_t\left[ \sup _{s\in [t,T]} \Vert f_2(s)\Vert ^2_2 \right] < \infty$, we have

$$\begin{aligned}&\lim _{\varepsilon \downarrow 0} {\mathbb {E}}^{\mathbb {P}}_t\left[ \frac{1}{\varepsilon } \int _t^{t+\varepsilon } \Vert f_2(s) (w_s - w_t) \Vert _2 ds \right] \\&\quad \le \lim _{\varepsilon \downarrow 0} \sqrt{ \frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t[\Vert f_2(s)\Vert ^2_2] ds} \sqrt{\frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert w_s - w_t\Vert ^2_2 ds} \\&\quad \le \lim _{\varepsilon \downarrow 0} \sqrt{{\mathbb {E}}^{\mathbb {P}}_t\left[ \sup _{s\in [t,T]} \Vert f_2(s)\Vert ^2_2 \right] } \sqrt{\frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t\Vert w_s - w_t\Vert ^2_2 ds} = 0. \end{aligned}$$

Then (A.3) is proved.

For the equivalence between (3.12) and (A.2), we note that

• If (3.12) is true, then ${\mathbb {E}}^{\mathbb {P}}_t[\lambda (s;s)] = 0$. Therefore,

  $$\begin{aligned} \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t[\lambda (s;t)]ds = \lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t[\lambda (s;s)]ds = 0. \end{aligned}$$

• If $\lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t[\lambda (s;t)]ds = 0$, then $\lim _{\varepsilon \downarrow 0} \frac{1}{\varepsilon } \int _t^{t+\varepsilon } {\mathbb {E}}^{\mathbb {P}}_t[\lambda (s;s)]ds = 0$. By Lemma 3.5 in [9], the stochastic Lebesgue differentiation theorem, we have (3.12).

$\square$

Proofs of Results in Section 4

1.1 Proof of Lemma 4.3

Proof

With slightly abuse of notations, suppose there is another solution pair (X, h) and the corresponding adjoint process is p(s; t). Note $X_T - {\mathbb {E}}^{\mathbb {P}}_t[ X_T] - \mu _2 \in L^2_{{{\mathcal {F}}}_T}(\varOmega ; {\mathbb {R}}, {\mathbb {P}})$, by [38, Chapter 7, Theorem 2.2], we still have a unique adapted solution

$$\begin{aligned} p(s;t) = {\mathbb {E}}^{\mathbb {P}}[e^{\int ^T_s r_vdv} (X_T - {\mathbb {E}}^{\mathbb {P}}_t[X_T] - \mu _2) | {{\mathcal {F}}}_s]. \end{aligned}$$

(B.1)

Therefore, $p(t;t)= -\mu _2 e^{\int ^T_t r_s ds}$ and we have the same solution for $h_t = - \xi e^{\int ^T_t r_s ds} u^*_t = h^*_t$. $\square$

1.2 Proof of Lemma 4.5

Proof

Suppose there is another solution pair $(X, {\bar{u}})$, and ${\bar{u}}$ is admissible. Let

$$\begin{aligned} {\bar{p}}(s;t)&= {\tilde{p}}(s;t) - \Big [ {\tilde{M}}_s X_s + {\tilde{\varGamma }}_s^{(2)} - {\mathbb {E}}^{\mathbb {P}}_t[{\tilde{M}}_s X_s +{\tilde{\varGamma }}_s^{(3)}] \Big ],\quad \\ {\bar{k}}(s;t)&={\tilde{k}}(s;t) - \Big [ X_s {\tilde{U}}_s + {\tilde{M}}_s {\bar{u}}_s + {\tilde{\gamma }}^{(2)}_s \Big ]. \end{aligned}$$

By ${\tilde{\varLambda }}(t;t)=0$, we derive

$$\begin{aligned} {\bar{u}}_t&= - (2\alpha _t {\bar{p}}(t;t) + \alpha _t ({\tilde{\varGamma }}^{(2)}_t - {\tilde{\varGamma }}^{(3)}_t) + {\tilde{M}}_t )^{-1} \\&\quad \Big [ \theta _t {\bar{p}}(t;t) + \theta _t ({\tilde{\varGamma }}^{(2)}_t - {\tilde{\varGamma }}^{(3)}_t) + {\bar{k}}(t;t)+{\tilde{\gamma }}_t^{(2)} \Big ]. \end{aligned}$$

Then

$$\begin{aligned} d[{\tilde{M}}_sX_s] = [ -r{\tilde{M}}X + {\tilde{M}}(\theta +\alpha {\bar{u}})' {\bar{u}}]ds + {\tilde{M}} {\bar{u}}'dW^{\mathbb {P}}_s. \end{aligned}$$

Finally,

$$\begin{aligned} d {\bar{p}}(s;t)&= d {\tilde{p}}(s;t) - d \Big [ {\tilde{M}}_s X_s+{\tilde{\varGamma }}_s^{(2)} - {\mathbb {E}}^{\mathbb {P}}_t[{\tilde{M}}_s X_s + {\tilde{\varGamma }}_s^{(3)}] \Big ] \\&= -r {\bar{p}}(s;t) ds + {\bar{k}}(s;t)'dW^{\mathbb {P}}_s \\&\quad - [ {\tilde{M}}(\theta +\alpha {\bar{u}})' {\bar{u}} - {\tilde{M}}(\theta +\alpha u^*)' u^*]ds \\&\quad + {\mathbb {E}}^{\mathbb {P}}_t[ {\tilde{M}}(\theta +\alpha {\bar{u}})' {\bar{u}} - {\tilde{M}}(\theta +\alpha u^*)' u^*] ds. \end{aligned}$$

We take conditional expectation on both sides and notice that $r_s$ is deterministic, therefore ${\bar{p}}(t;t)=0$. So,

$$\begin{aligned} {\bar{u}}_t= & {} - [\alpha _t ({\tilde{\varGamma }}^{(2)}_t - {\tilde{\varGamma }}^{(3)}_t) + {\tilde{M}}_t ]^{-1} \Big [ \theta _t ({\tilde{\varGamma }}^{(2)}_t - {\tilde{\varGamma }}^{(3)}_t) + {\bar{k}}(t;t)+{\tilde{\gamma }}_t^{(2)} \Big ] \\= & {} u^*_t - [\alpha _t ({\tilde{\varGamma }}^{(2)}_t - {\tilde{\varGamma }}^{(3)}_t) + {\tilde{M}}_t ]^{-1} {\bar{k}}(t;t). \end{aligned}$$

Then ${\bar{k}}(t;t)$ should be essentially bounded, and

$$\begin{aligned} d {\bar{p}}(s;t)&= -r {\bar{p}}(s;t) ds + {\bar{k}}(s;t)'dW^{\mathbb {P}}_s \\&\quad + (\alpha ({\tilde{\varGamma }}^{(2)}_s - {\tilde{\varGamma }}^{(3)}_s) + {\tilde{M}})^{-1} {\tilde{M}} {\bar{k}}(s;s)'( \theta + 2 \alpha u^*) ds\\&\quad -\alpha {\tilde{M}} (\alpha ({\tilde{\varGamma }}^{(2)}_s - {\tilde{\varGamma }}^{(3)}_s) + {\tilde{M}})^{-2} {\bar{k}}(s;s)'{\bar{k}}(s;s) ds \\&\quad - {\mathbb {E}}^{\mathbb {P}}_t\Big [ (\alpha ({\tilde{\varGamma }}^{(2)}_s - {\tilde{\varGamma }}^{(3)}_s) + {\tilde{M}})^{-1} {\tilde{M}} {\bar{k}}(s;s)'(\theta + 2 \alpha u^*) \Big ] ds \\&\quad + {\mathbb {E}}^{\mathbb {P}}_t\Big [ \alpha {\tilde{M}} (\alpha ({\tilde{\varGamma }}^{(2)}_s - {\tilde{\varGamma }}^{(3)}_s) + {\tilde{M}})^{-2} {\bar{k}}(s;s)'{\bar{k}}(s;s) \Big ] ds. \end{aligned}$$

The existence of a solution is obvious. For uniqueness, consider $({\bar{p}}(\cdot ;t), {\bar{k}}(\cdot ; t) )$ in the space $L^2_{{\mathcal {F}}}(\varOmega ; \,C(t,T;{\mathbb {R}}), {\mathbb {P}})$ $\times$ $L^\infty _{{\mathcal {F}}}(t, T;\, {\mathbb {R}}^d)$.

Without loss of generality, let $r=0$. As ${\bar{k}}(s;t)$ does not depend on t, we denote ${\bar{k}}(s) = {\bar{k}}(s;s) = {\bar{k}}(s;t)$. Moreover, we introduce $a_1(s), a_2(s)$ to rewrite $d{\bar{p}}(s;t)$ as follows, noting that $a_1(s), a_2(s)$ are essentially bounded.

$$\begin{aligned} d {\bar{p}}(s;t)&= {\bar{k}}(s)'dW^{\mathbb {P}}_s + a'_1(s) {\bar{k}}(s) ds - a_2(s) {\bar{k}}(s)'{\bar{k}}(s) ds \\&\quad - {\mathbb {E}}^{\mathbb {P}}_t\big [ a'_1(s) {\bar{k}}(s) \big ] ds + {\mathbb {E}}^{\mathbb {P}}_t\big [ a_2(s) {\bar{k}}(s)'{\bar{k}}(s) \big ] ds. \end{aligned}$$

Define

$$\begin{aligned} p_0(s;t) = {\bar{p}}(s;t) - \int ^T_s {\mathbb {E}}^{\mathbb {P}}_t\big [ a'_1(v) {\bar{k}}(v) \big ] dv + \int ^T_s {\mathbb {E}}^{\mathbb {P}}_t\big [ a_2(v) {\bar{k}}(v)'{\bar{k}}(v) \big ] dv. \end{aligned}$$

Then

$$\begin{aligned}&d p_0 (s;t) = a'_1(s) {\bar{k}}(s) ds - a_2(s) {\bar{k}}(s)'{\bar{k}}(s) ds + {\bar{k}}(s)'dW^{\mathbb {P}}_s, \quad p_0(T;t) = 0. \end{aligned}$$

As ${\bar{p}}(\cdot ;t) \in L^2_{{\mathcal {F}}}(\varOmega ; \,C(t,T;{\mathbb {R}}), {\mathbb {P}})$, $p_0(\cdot ;t) \in L^2_{{\mathcal {F}}}(\varOmega ; \,C(t,T;{\mathbb {R}}), {\mathbb {P}})$.

Suppose there are two solutions $(p^{(1)}_0 , {\bar{k}}^{(1)})$ and $(p^{(2)}_0 , {\bar{k}}^{(2)})$. Denote $p_\varDelta (s;t) = p^{(1)}_0(s;t) - p^{(2)}_0(s;t)$, $k_\varDelta (s) = {\bar{k}}^{(1)}(s) - {\bar{k}}^{(2)}(s)$, then

$$\begin{aligned} dp_\varDelta (s;t) = a'_1(s)k_\varDelta (s) - a_2(s) k'_\varDelta (s) \big ({\bar{k}}^{(1)}(s) + {\bar{k}}^{(2)}(s) \big ) ds + k'_\varDelta (s) dW^{\mathbb {P}}_s, \quad p_\varDelta (T; t)=0. \end{aligned}$$

Applying Itô’s lemma to $\Vert p_\varDelta (s;t)\Vert ^2_2$ on s, taking expectation, and noting that ${\bar{k}}^{(1)}(s)$ and ${\bar{k}}^{(2)}(s)$ are essentially bounded, we have

$$\begin{aligned} {\mathbb {E}}^{\mathbb {P}}\Big [ \Vert p_\varDelta (s;t)\Vert ^2_2 \Big ] + {\mathbb {E}}^{\mathbb {P}}\Big [ \int ^T_s \Vert k_\varDelta (v) \Vert ^2_2 dv \Big ]&\le C {\mathbb {E}}^{\mathbb {P}}\Big [ \int ^T_s \Vert p_\varDelta (s; t)\Vert _2 \Vert k_\varDelta (v) \Vert _2 dv \Big ]\\&\le C{\mathbb {E}}^{\mathbb {P}}\Big [ \int ^T_s \Vert p_\varDelta (v; t) \Vert ^2_2 dv \Big ] \\&\quad + \frac{1}{2} {\mathbb {E}}^{\mathbb {P}}\Big [ \int ^T_s \Vert k_\varDelta (v)\Vert ^2_2 dv \Big ]. \end{aligned}$$

By Gronwall’s inequality, ${\mathbb {E}}^{\mathbb {P}}\big [ \Vert p_\varDelta (s;t)\Vert ^2_2 \big ] = 0$. Therefore, ${\bar{p}}(s;t) = 0,~{\bar{k}}(s;t) = 0$. $\square$

1.3 Proof of Proposition 4.7

Proof

A direct calculation shows

$$\begin{aligned} \tilde{F}^{(2)}_t= & {} r_t {\tilde{\varGamma }}^{(2)}_t + \frac{\xi \mu _2 - 1}{(\xi \mu _2 + 1)^2} \theta '_t {\tilde{\gamma }}^{(2)}_t - \frac{\xi }{(\xi \mu _2 + 1)^2} e^{-\int ^T_t r_s ds} \Vert {\tilde{\gamma }}^{(2)}_t \Vert ^2_2 \nonumber \\&+ \frac{\mu _2}{(\xi \mu _2 + 1)^2} e^{\int ^T_t r_s ds} \Vert \theta _t\Vert ^2_2. \end{aligned}$$

(B.2)

Since $\theta$ is essentially bounded and we can introduce a new probability measure ${\tilde{{\mathbb {Q}}}}$ under which ${\tilde{W}}$ is a standard Brownian motion,

$$\begin{aligned} {\tilde{W}}_t \triangleq - \int ^t_0 \frac{\xi \mu _2 - 1}{(\xi \mu _2 + 1)^2} \theta _s ds + W^{\mathbb {P}}_t. \end{aligned}$$

(B.3)

Under ${\tilde{{\mathbb {Q}}}}$, the driver of ${\tilde{\varGamma }}^{(2)}_t$ has no cross term $\theta '_t {\tilde{\gamma }}^{(2)}_t$ and

$$\begin{aligned} d\vartheta _s = \Big [b_{\vartheta }(s, \vartheta _s) + \frac{\xi \mu _2 - 1}{(\xi \mu _2 + 1)^2} \sigma _{\vartheta }(s) \theta _s \Big ]ds + \sigma _{\vartheta }(s)d{\tilde{W}}_s. \end{aligned}$$

(B.4)

Then it is straightforward to verify the conditions in [27, Theorem 2.5]. The Lipschitz constant in [27, Assumption (F.1)] is $K_b + \frac{|\xi \mu _2 - 1|\Vert \sigma _\vartheta \Vert _\infty }{(\xi \mu _2 + 1)^2}$. We can take constants in [27, Assumption (B.1)] as follows.

$$\begin{aligned} K_{f,y} = \Vert r\Vert _\infty , \quad l = 1, \quad \alpha = 0, \quad \beta = \frac{2\mu _2}{(\xi \mu _2 +1)^2} e^{\int ^T_0 r_s ds}, \quad \gamma = \frac{2\xi }{(\xi \mu _2 +1)^2}.\nonumber \\ \end{aligned}$$

(B.5)

In particular, [27, Assumption (B.1)(4)] becomes Condition (2) above. Therefore, we can apply [27, Theorem 2.5] under measure ${\tilde{{\mathbb {Q}}}}$. ${\tilde{\gamma }}^{(2)}_t$ is essentially bounded since $\theta _t$ is still essentially bounded under ${\tilde{{\mathbb {Q}}}}$. $\square$

1.4 Proof of Proposition 4.8

Proof

The proof is a direct application of [27, Proposition 3.1]. With the truncation function $\rho _M(\cdot )$ in [27, Proposition 3.1], our truncated driver (corresponding to $f_M$ in [27, Proposition 3.1]) is Lipschitz in $\vartheta$ with constant $\frac{|\xi \mu _2 - 1|}{(\xi \mu _2 + 1)^2} M + \frac{2\mu _2}{(\xi \mu _2 + 1)^2} e^{\int ^T_0 r_s ds} ( C_\theta + 1)$, Lipschitz in ${\tilde{\gamma }}^{(2)}$ with constant $\frac{|\xi \mu _2 - 1|}{(\xi \mu _2 + 1)^2} ( C_\theta + 1) + \frac{2\xi }{(\xi \mu _2 + 1)^2} M$, and Lipschitz in ${\tilde{\varGamma }}^{(2)}$ with constant $\Vert r\Vert _\infty$. The remaining proof follows exactly from [27, Proposition 3.1]. $\square$

1.5 Proof of Lemma 4.13

Proof

Suppose there is another solution pair (X, h) and the corresponding adjoint process is p(s; t). With the same idea as in the proof of Lemma 4.3, we have a unique adapted solution

$$\begin{aligned} p(s;t) = {\mathbb {E}}^{\mathbb {P}}[e^{\int ^T_s r_vdv} (X_T - {\mathbb {E}}^{\mathbb {P}}_t[X_T] - \mu _1 X^*_t) | {{\mathcal {F}}}_s]. \end{aligned}$$

(B.6)

Therefore, we have $h_t = h^*_t$.

$\square$

1.6 Proof of Lemma 4.14

Proof

Let $J=\frac{{\tilde{M}}}{{\tilde{N}}}$, $K=\frac{J}{{\tilde{M}}} {\tilde{U}}- \frac{J^2}{{\tilde{M}}} {\tilde{V}}$, then to prove the existence and uniqueness of $({\tilde{M}}, {\tilde{U}})$, $({\tilde{N}}, {\tilde{V}})$, we only need to show the existence and uniqueness of $({\tilde{M}}, {\tilde{U}}), (J, K)$. It is easy to show

$$\begin{aligned} \left\{ \begin{array}{rcl} d{\tilde{M}}_s&{}=&{} \big [-2(r+\alpha ){\tilde{M}} + \theta '\theta (1-\frac{1}{J}) {\tilde{M}} + \frac{\alpha ^2}{\xi } - \theta '\theta {\tilde{\varGamma }} \big ] ds \\ &{}&{}+ \big [ (2-\frac{1}{J}- \frac{\tilde{\varGamma }}{\tilde{M}} )\theta '{\tilde{U}} + \frac{{\tilde{U}}' {\tilde{U}}}{{\tilde{M}}} \big ]ds \\ &{} &{} + {\tilde{U}}_s 'dW_s^{\mathbb {P}},\quad {\tilde{M}}_T=1,\\ dJ_s&{}=&{}\big [\frac{\alpha ^2}{\xi } \frac{J}{{\tilde{M}}} + (1-\frac{1}{J}-\frac{{\tilde{\varGamma }}}{{\tilde{M}}}) \theta 'K + \frac{K'K}{J} \big ]ds + K_s' dW_s^{\mathbb {P}},\quad J_T=1. \end{array}\right. \end{aligned}$$

(B.7)

We consider ${\tilde{M}}^c = {\tilde{M}}\vee c$, $J^c = J \vee c$, where $c\le 1$ is a constant. The corresponding diffusion terms are denoted by ${\tilde{K}}^c$, ${\tilde{U}}^c$. Then,

$$\begin{aligned} \left\{ \begin{array}{rcl} d{\tilde{M}}^c_s&{}=&{} \big [-2(r+\alpha ){\tilde{M}}^c + \theta '\theta (1 -\frac{1}{J^c}) {\tilde{M}}^c + \frac{\alpha ^2}{\xi } - \theta '\theta {\tilde{\varGamma }} \big ]ds \\ &{}&{}+ \big [ (2-\frac{1}{J^c}- \frac{\tilde{\varGamma }}{\tilde{M}^c} )\theta '{\tilde{U}}^c + \frac{({\tilde{U}}^c)' {\tilde{U}}^c}{{\tilde{M}}^c} \big ]ds \\ &{} &{} + ({\tilde{U}}^c_s) 'dW_s^{\mathbb {P}},\quad {\tilde{M}}^c_T=1,\\ dJ^c_s&{}=&{}\big [ \frac{\alpha ^2 }{\xi } \frac{J^c}{{\tilde{M}}^c} + (1-\frac{1}{J^c}-\frac{{\tilde{\varGamma }}}{{\tilde{M}}^c}) \theta 'K^c + \frac{(K^c)'K^c}{J^c} \big ]ds + (K^c_s)' dW_s^{\mathbb {P}},\quad J^c_T=1. \end{array}\right. \end{aligned}$$

(B.8)

(B.8) is a standard quadratic BSDE system, there exists a solution pair $({\tilde{M}}^c, {\tilde{U}}^c) \in L^\infty _{{\mathcal {F}}}(0,T;{\mathbb {R}}) \times L^2_{{\mathcal {F}}}(0,T;{\mathbb {R}}^d, {\mathbb {P}})$ and $(J^c, K^c) \in L^\infty _{{\mathcal {F}}}(0,T;{\mathbb {R}}) \times L^2_{{\mathcal {F}}}(0,T;{\mathbb {R}}^d, {\mathbb {P}})$. ${\tilde{U}}^c\cdot W^{\mathbb {P}}, K^c\cdot W^{\mathbb {P}}$ are BMO martingales, see [16, 22]. Since $1-\frac{1}{J^c}\le 0$, by comparison principle for quadratic BSDE in [16], ${\tilde{M}}^c \ge {\hat{M}}^c$, where ${\hat{M}}^c$ is the solution to following BSDE,

$$\begin{aligned} d{\hat{M}}^c_s= & {} \bigg [ -2(r+\alpha ){\hat{M}}^c+ \frac{\alpha ^2}{\xi } +\left( 2-\frac{1}{J^c}- \frac{\tilde{\varGamma }}{\hat{M}^c} \right) \theta ' {\hat{U}}^c + \frac{({\hat{U}}^c)' {\hat{U}}^c}{\hat{M}^c} \bigg ]ds\nonumber \\&+ ({\hat{U}}^c_s) 'dW_s^{\mathbb {P}},\quad {\hat{M}}^c_T=1. \end{aligned}$$

(B.9)

Since $\big [(2-\frac{1}{J^c}- \frac{\tilde{\varGamma }}{\hat{M}^c})\theta + \frac{{\hat{U}}^c}{\hat{M}^c}\big ]\cdot W^{\mathbb {P}}$ is a BMO martingale, we can introduce a new probability measure ${\mathbb {Q}}$ and under ${\mathbb {Q}}$ define a new Brownian motion $W^{\mathbb {Q}}_s$ by $W^{\mathbb {Q}}_s = W^{\mathbb {P}}_s + \int ^s_0 (2-\frac{1}{J^c}- \frac{\tilde{\varGamma }}{\hat{M}^c} )\theta + \frac{{\hat{U}}^c}{\hat{M}^c} dt$. Then

$$\begin{aligned} \hat{M}^c_s= & {} {\mathbb {E}}^{\mathbb {Q}}_t \big [ e^{2\int ^T_s r_t+\alpha _t dt} - \int ^T_s \frac{\alpha ^2_v}{\xi } e^{2\int ^v_s r_t+\alpha _t dt} dv \big ] \\= & {} e^{2\int ^T_s r_t dt}\big [ e^{2\int ^T_s \alpha _t dt} - \mu ^2_1 \xi \int ^T_s e^{2\int ^v_s \alpha _t dt} dv \big ] \ge {\underline{l}} > 0. \end{aligned}$$

Therefore, ${\tilde{M}}^c \ge {\underline{l}}$, ${\underline{l}}$ does not depend on c. By comparison principle for quadratic BSDE in [16], since $\frac{\alpha ^2}{\xi {\tilde{M}}^c} \le \frac{\alpha ^2}{\xi {\underline{l}}}$ then $J^c_s \ge \exp (-\int ^T_s\frac{\alpha ^2_t}{\xi {\underline{l}}} dt)$. This lower bound also does not depend on c. Finally, let constant $0<c< \min \big \{\exp (-\int ^T_0 \frac{\alpha ^2_t}{\xi {\underline{l}}}dt), \;\; {\underline{l}}\big \}$, we have ${\tilde{M}} = {\tilde{M}}^c$, $J =J^c$, ${\tilde{U}} = {\tilde{U}}^c$, ${\tilde{K}} = {\tilde{K}}^c$. The existence of solutions is guaranteed.

Next, we prove the uniqueness. Since ${\tilde{M}}, J \ge c >0$, define $Y=\frac{1}{{\tilde{M}}}, Z=-\frac{{\tilde{U}}}{{\tilde{M}}^2}, G = \frac{1}{J}$, $E=-\frac{K}{J^2}$. Y, G are essentially bounded. Then

$$\begin{aligned} \left\{ \begin{array}{rcl} dY_s &{}=&{}\big [ [2(r+\alpha )-\theta '\theta ] Y + \theta '\theta YG +(\theta '\theta {\tilde{\varGamma }} - \frac{\alpha ^2}{\xi })Y^2 \big ]ds+\big [ 2\theta 'Z-\theta 'ZG-\theta 'ZY{\tilde{\varGamma }} \big ]ds \\ &{} &{} + Z'_s dW^{\mathbb {P}}_s,\quad Y_T=1,\\ dG_s&{}=&{}\big [ -\frac{\alpha ^2}{\xi } YG + \theta 'E-\theta 'EG- \theta 'E Y \tilde{\varGamma } \big ]ds + E_s' dW^{\mathbb {P}}_s,\quad G_T=1. \end{array}\right. \nonumber \\ \end{aligned}$$

(B.10)

Suppose there are two solutions $(Y^{(1)},Z^{(1)}),(G^{(1)},E^{(1)})$ and $(Y^{(2)},Z^{(2)}),(G^{(2)},E^{(2)})$. Denote ${\bar{Y}} = Y^{(1)} - Y^{(2)}, {\bar{Z}} = Z^{(1)} - Z^{(2)}, {\bar{G}} = G^{(1)} - G^{(2)}, {\bar{E}} = E^{(1)} - E^{(2)}$. Then

$$\begin{aligned} \left\{ \begin{array}{rcl} d{\bar{Y}}_s &{}=&{} \big [[2(r+\alpha )-\theta '\theta ] {\bar{Y}} + \theta '\theta [Y^{(1)}{\bar{G}}+G^{(2)}{\bar{Y}}] \big ]ds + (\theta '\theta {\tilde{\varGamma }} - \frac{\alpha ^2}{\xi }) {\bar{Y}}[Y^{(1)}+Y^{(2)}] ds\\ &{}&{}+\big [2\theta ' {\bar{Z}} - \theta '[Z^{(1)}{\bar{G}}+G^{(2)}{\bar{Z}}] \big ]ds - \theta '{\tilde{\varGamma }} [Z^{(1)}{\bar{Y}} + Y^{(2)}{\bar{Z}}]ds + {\bar{Z}}'_s dW^{\mathbb {P}}_s,\qquad {\bar{Y}}_T=0,\\ d{\bar{G}}_s&{}=&{} \big [ -\frac{\alpha ^2}{\xi } [Y^{(1)}{\bar{G}}+G^{(2)}{\bar{Y}}] + \theta '{\bar{E}}-\theta '[E^{(1)}{\bar{G}}+G^{(2)}{\bar{E}}] \big ]ds - \tilde{\varGamma } \theta ' [Y^{(1)}{\bar{E}} + E^{(2)}{\bar{Y}}]ds \\ &{} &{} + {\bar{E}}_s' dW^{\mathbb {P}}_s,\quad {\bar{G}}_T=0. \end{array}\right. \nonumber \\ \end{aligned}$$

(B.11)

Applying Itô’s lemma to $\Vert {\bar{G}}_s\Vert ^2_2+\Vert {\bar{Y}}_s\Vert ^2_2$ and taking conditional expectation, we have

$$\begin{aligned}&\Vert {\bar{G}}_s\Vert ^2_2 + {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{E}}_v \Vert ^2_2 dv \Big ] + \Vert {\bar{Y}}_s\Vert ^2_2 + {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{Z}}_v \Vert ^2_2 dv \Big ] \\&\quad \le C {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{G}}_v \Vert ^2_2 + \Vert {\bar{Y}}_v \Vert ^2_2 + \Vert {\bar{G}}_v \Vert _2 \Vert {\bar{Y}}_v\Vert _2 + \Vert {\bar{G}}_v \Vert _2 \Vert {\bar{E}}_v\Vert _2 \\&\qquad + \Vert E^{(1)}_v \Vert _2 \Vert {\bar{G}}_v\Vert ^2_2 + \Vert E^{(2)}_v \Vert _2 \Vert {\bar{G}}_v \Vert _2 \Vert {\bar{Y}}_v\Vert _2 \\&\qquad + \Vert {\bar{Y}}_v \Vert _2 \Vert {\bar{Z}}_v\Vert _2 + \Vert Z^{(1)}_v \Vert _2 \Vert {\bar{Y}}_v\Vert ^2_2 + \Vert Z^{(1)}_v \Vert _2 \Vert {\bar{G}}_v \Vert _2 \Vert {\bar{Y}}_v\Vert _2 dv \Big ]. \end{aligned}$$

By Hölder inequality, we have

$$\begin{aligned} {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert E^{(2)}_v \Vert _2 \Vert {\bar{G}}_v \Vert _2 \Vert {\bar{Y}}_v\Vert _2 dv \Big ]&\le C \sqrt{{\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert E^{(2)}_v \Vert ^2_2 dv \Big ] {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{G}}_v \Vert ^2_2 \Vert {\bar{Y}}_v\Vert ^2_2 dv \Big ]} \\&\le C \sqrt{ {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{G}}_v \Vert ^2_2 \Vert {\bar{Y}}_v\Vert ^2_2 dv \Big ]} \\&\le C \sqrt{ {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{G}}_v \Vert ^4_2 + \Vert {\bar{Y}}_v\Vert ^4_2 dv \Big ]}. \end{aligned}$$

Other terms can be treated in a similar way. Finally,

$$\begin{aligned}&\Vert {\bar{G}}_s\Vert ^2_2 + {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{E}}_v \Vert ^2_2 dv \Big ] + \Vert {\bar{Y}}_s\Vert ^2_2 + {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{Z}}_v \Vert ^2_2 dv \Big ] \\&\quad \le C{\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{G}}_v \Vert ^2_2 + \Vert {\bar{Y}}_v \Vert ^2_2 dv \Big ] + \frac{1}{2} {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{E}}_v \Vert ^2_2 dv \Big ] \\&\qquad + \frac{1}{2} {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{Z}}_v \Vert ^2_2 dv \Big ]\\&\qquad + C \sqrt{ {\mathbb {E}}^{\mathbb {P}}_s \Big [ \int ^T_s \Vert {\bar{G}}_v \Vert ^4_2 + \Vert {\bar{Y}}_v\Vert ^4_2 dv \Big ]}. \end{aligned}$$

Consider $s\in [T-\delta ,T]$ and denote $G_\delta = \Vert {\bar{G}}_. \Vert _{L^\infty _{{\mathcal {F}}}(T -\delta , T; {\mathbb {R}})}$, $Y_\delta = \Vert {\bar{Y}}_. \Vert _{L^\infty _{{\mathcal {F}}}(T -\delta , T; {\mathbb {R}})}$, we obtain

$$\begin{aligned} \Vert {\bar{G}}_s\Vert ^2_2 + \Vert {\bar{Y}}_s\Vert ^2_2\le & {} C \delta [G^2_\delta + Y^2_\delta ] + C\sqrt{\delta } \sqrt{G^4_\delta + Y^4_\delta } \\\le & {} C \delta [G^2_\delta + Y^2_\delta ] + C\sqrt{\delta } [G^2_\delta + Y^2_\delta ] \le C\sqrt{\delta } [G^2_\delta + Y^2_\delta ]. \end{aligned}$$

Let $I_\delta = G_\delta \vee Y_\delta$. Taking supremum on the left-hand side over $s\in [T-\delta ,T]$ yields

$$\begin{aligned} I^2_\delta \le C\sqrt{\delta } [G^2_\delta + Y^2_\delta ] \le C\sqrt{\delta } I^2_\delta . \end{aligned}$$

By choosing sufficiently small $\delta$ such that $C\sqrt{\delta } < 1$, we have $G^2_\delta = Y^2_\delta = 0$. The same steps are repeated on $[T-2\delta ,T-\delta ], [T-3\delta ,T-2\delta ],...$, until time 0 is reached. The uniqueness follows. $\square$

1.7 Proof of Lemma 4.16

Proof

Suppose there is another solution pair $(X, {\bar{u}})$, let

$$\begin{aligned} {\bar{p}}(s;t)&= {\tilde{p}}(s;t) - \big [ {\tilde{M}}_s X_s - {\tilde{\varGamma }}_s X_t - {\mathbb {E}}^{\mathbb {P}}_t[{\tilde{N}}_s X_s] \big ], \\ {\bar{k}}(s;t)&={\tilde{k}}(s;t) - \big [ X_s{\tilde{U}}_s + {\tilde{M}}_s {\bar{u}}_s \big ]. \end{aligned}$$

By ${\tilde{\varLambda }}(t;t)=0$, we derive

$$\begin{aligned} {\bar{u}}_t = - {\tilde{M}}_t^{-1} [ \theta _t {\bar{p}}(t;t) + {\bar{k}}(t;t)] + L_t X_t. \end{aligned}$$

(B.12)

Then

$$\begin{aligned} d[{\tilde{M}}_sX_s]&= \big [ -(r+\alpha ){\tilde{M}}X + \frac{\alpha ^2}{\xi } X - \big (\theta + \frac{{\tilde{U}}}{{\tilde{M}}}\big )'[ \theta _s {\bar{p}}(s;s) + {\bar{k}}(s;s)] \big ]ds \\&\quad + ({\tilde{U}} X + {\tilde{M}} {\bar{u}}) 'dW^{\mathbb {P}}_s. \end{aligned}$$

Finally, we can show

$$\begin{aligned} d {\bar{p}}(s;t)= & {} -(r+\alpha ) {\bar{p}}(s;t) ds + {\bar{k}}(s;t)'dW^{\mathbb {P}}_s+\big \{ \theta ' {\bar{k}}(s;s) \\&+ \frac{{\tilde{U}}'}{{\tilde{M}}} {\bar{k}}(s;s)\big \}ds - {\mathbb {E}}^{\mathbb {P}}_t\big \{ \theta ' {\bar{k}}(s;s) + \frac{{\tilde{V}}'}{{\tilde{M}}} {\bar{k}}(s;s)\big \}ds. \end{aligned}$$

This BSDE has the exact same form as in [11], then it admits a unique solution $({\bar{p}}(\cdot ;t), {\bar{k}}(\cdot ; t) )$ in the space $L^q_{{\mathcal {F}}}(\varOmega ; \,C(t,T;{\mathbb {R}}), {\mathbb {P}})$ $\times$ $L^q_{{\mathcal {F}}}(t, T;\, {\mathbb {R}}^d, {\mathbb {P}})$ for any $q \in (1,2)$. Therefore, ${\bar{p}}(s;t) = 0, {\bar{k}}(s;t) = 0$. $\square$

1.8 Proof of Proposition 4.18

Proof

Consider the BSDE (B.10) for (Y, G, Z, E). For notation simplicity, introduce functions $f^y$ and $f^g$ for the drivers to rewrite (B.10) as

$$\begin{aligned} \left\{ \begin{array}{rcl} dY_s &{}=&{}- f^y(s, \theta _s, Y_s, Z_s, G_s) ds + Z'_s dW^{\mathbb {P}}_s,\quad Y_T=1,\\ dG_s&{}=&{}- f^g(s, \theta _s, G_s, E_s, Y_s) ds + E_s' dW^{\mathbb {P}}_s,\quad G_T=1. \end{array}\right. \end{aligned}$$

(B.13)

Lemma 4.14 shows $0 < Y, G \le \frac{1}{c}$ for a constant $0< c < 1$. Indeed, c depends on the value of T. Therefore, we use notation c(T) to highlight this dependence. As in [27, Proposition 3.1], consider a truncation function $\rho _\kappa$ which is a smooth modification of the centered Euclidean ball of radius $\kappa$. Denote $(Y^\kappa , G^\kappa , Z^\kappa , E^\kappa )$ as the solution to the truncated BSDE

$$\begin{aligned} \left\{ \begin{array}{rcl} dY^\kappa _s &{}=&{}- f^y_\kappa (s, \theta _s, Y^\kappa _s, Z^\kappa _s, G^\kappa _s) ds + (Z^\kappa _s)' dW^{\mathbb {P}}_s,\quad Y^\kappa _T=1,\\ dG^\kappa _s&{}=&{} - f^g_\kappa (s, \theta _s, G^\kappa _s, E^\kappa _s, Y^\kappa _s) ds + (E^\kappa _s)' dW^{\mathbb {P}}_s,\quad G^\kappa _T=1, \end{array}\right. \end{aligned}$$

(B.14)

where $f^y_\kappa \triangleq f^y(\cdot , \varphi (\cdot ), \cdot , \rho _\kappa (\cdot ), \cdot )$ and $f^g_\kappa \triangleq f^g(\cdot , \varphi (\cdot ), \cdot , \rho _\kappa (\cdot ), \cdot )$ truncate on (Z, E). First consider that $b_\vartheta$ is differentiable. Then $(\vartheta , Y, G, Z, E)$ is differentiable with respect to $\zeta$ in (4.12) and

$$\begin{aligned} \nabla \vartheta _t&= I_d + \int ^t_0 \nabla b_\vartheta (s, \vartheta _s) \nabla \vartheta _s ds, \\ \nabla Y^\kappa _t&= \int ^T_t \big (\nabla _\vartheta f^y_\kappa \nabla \vartheta _s + \nabla _y f^y_\kappa \nabla Y^\kappa _s + \nabla _z f^y_\kappa \nabla Z^\kappa _s - \theta '_s \theta _s Y^\kappa _s \nabla G^\kappa _s + \theta '_s \rho _\kappa (Z^\kappa _s) \nabla G^\kappa _s \big )ds \\&\quad - \int ^T_t \nabla Z^\kappa _s dW^{\mathbb {P}}_s, \\ \nabla G^\kappa _t&= \int ^T_t \big (\nabla _\vartheta f^g_\kappa \nabla \vartheta _s + \nabla _g f^g_\kappa \nabla G^\kappa _s + \nabla _e f^g_\kappa \nabla E^\kappa _s + \frac{\alpha ^2_s}{\xi } G^\kappa _s \nabla Y^\kappa _s + \theta '_s \rho _\kappa (E^\kappa _s) {\tilde{\varGamma }}_s \nabla Y^\kappa _s \big )ds \\&\quad - \int ^T_t \nabla E^\kappa _s dW^{\mathbb {P}}_s, \end{aligned}$$

where $\nabla \vartheta _t = (\partial \vartheta _{it}/ \partial \zeta _{jt} )_{1 \le i, j \le d}$ etc are defined like the counterparts in [26, Theorem 3.1] and we omit arguments in $f^y_\kappa$ and $f^g_\kappa$ for simplicity. One difficulty is $\nabla Y^\kappa _t$ and $\nabla G^\kappa _t$ are coupled. The idea is to truncate $\nabla Y^\kappa _t$ in $\nabla G^\kappa _t$ and $\nabla G^\kappa _t$ in $\nabla Y^\kappa _t$. Without loss of generality, we use the same constant $\kappa$ and the truncation function is denoted as ${\bar{\rho }}_\kappa$. Denote $(\nabla {\bar{Y}}^\kappa , \nabla {\bar{G}}^\kappa , \nabla {\bar{Z}}^\kappa , \nabla {\bar{E}}^\kappa )$ as the solution to

$$\begin{aligned} \nabla {\bar{Y}}^\kappa _t&= \int ^T_t \big (\nabla _\vartheta f^y_\kappa \nabla \vartheta _s + \nabla _y f^y_\kappa \nabla {\bar{Y}}^\kappa _s + \nabla _z f^y_\kappa \nabla {\bar{Z}}^\kappa _s - \theta '_s \theta _s Y^\kappa _s {\bar{\rho }}_\kappa ( \nabla {\bar{G}}^\kappa _s) \\&\quad + \theta '_s \rho _\kappa (Z^\kappa _s) {\bar{\rho }}_\kappa (\nabla {\bar{G}}^\kappa _s) \big )ds \\&\quad - \int ^T_t \nabla {\bar{Z}}^\kappa _s dW^{\mathbb {P}}_s, \\ \nabla {\bar{G}}^\kappa _t&= \int ^T_t \big (\nabla _\vartheta f^g_\kappa \nabla \vartheta _s + \nabla _g f^g_\kappa \nabla {\bar{G}}^\kappa _s + \nabla _e f^g_\kappa \nabla {\bar{E}}^\kappa _s + \frac{\alpha ^2_s}{\xi } G^\kappa _s {\bar{\rho }}_\kappa (\nabla {\bar{Y}}^\kappa _s) \\&\quad + \theta '_s \rho _\kappa (E^\kappa _s) {\tilde{\varGamma }}_s {\bar{\rho }}_\kappa (\nabla {\bar{Y}}^\kappa _s) \big )ds \\&\quad - \int ^T_t \nabla {\bar{E}}^\kappa _s dW^{\mathbb {P}}_s. \end{aligned}$$

Note the BSDE above uses $(Y^\kappa , G^\kappa , Z^\kappa , E^\kappa )$. Since $\nabla _z f^y_\kappa$ and $\nabla _e f^g_\kappa$ are essentially bounded, we can apply Girsanov’s theorem to introduce a Brownian motion $W^y_t \triangleq W^{\mathbb {P}}_t - \int ^t_0 \nabla _z f^y_\kappa ds$ under measure ${\mathbb {Q}}^y$ and $W^g_t \triangleq W^{\mathbb {P}}_t - \int ^t_0 \nabla _e f^g_\kappa ds$ under ${\mathbb {Q}}^g$. Then

$$\begin{aligned} \nabla {\bar{Y}}^\kappa _t&= {\mathbb {E}}^{{\mathbb {Q}}^y}_t \Big [\int ^T_t e^{\int ^s_t \nabla _y f^y_\kappa du} \big (\nabla _\vartheta f^y_\kappa \nabla \vartheta _s - \theta '_s \theta _s Y^\kappa _s {\bar{\rho }}_\kappa ( \nabla {\bar{G}}^\kappa _s) + \theta '_s \rho _\kappa (Z^\kappa _s) {\bar{\rho }}_\kappa (\nabla {\bar{G}}^\kappa _s) \big )ds \Big ], \\ \nabla {\bar{G}}^\kappa _t&= {\mathbb {E}}^{{\mathbb {Q}}^g}_t \Big [ \int ^T_t e^{\int ^s_t \nabla _g f^g_\kappa du} \big (\nabla _\vartheta f^g_\kappa \nabla \vartheta _s + \frac{\alpha ^2_s}{\xi } G^\kappa _s {\bar{\rho }}_\kappa (\nabla {\bar{Y}}^\kappa _s) + \theta '_s \rho _\kappa (E^\kappa _s) {\tilde{\varGamma }}_s {\bar{\rho }}_\kappa (\nabla {\bar{Y}}^\kappa _s) \big )ds \Big ]. \end{aligned}$$

Since $\Vert \nabla \vartheta _t\Vert _2 \le e^{K_b T}$ and

$$\begin{aligned} \Vert \nabla _\vartheta f^y_\kappa \Vert _2&\le 2 (C_\theta + 1) \Big [ \frac{1}{c(T)} + \frac{1 + \Vert \tilde{\varGamma } \Vert _\infty }{c^2(T)} \Big ] + 2 \kappa + \frac{\kappa }{c(T)} + \frac{\kappa \Vert \tilde{\varGamma } \Vert _\infty }{c(T)} \triangleq C^y_\vartheta , \\ \Vert \nabla _y f^y_\kappa \Vert _2&\le 2(\Vert r\Vert _\infty + \Vert \alpha \Vert _\infty )+ (C_\theta + 1)^2 \Big [1 + \frac{1}{c(T)} + 2 \frac{\Vert \tilde{\varGamma } \Vert _\infty }{c(T)} \Big ] + 2\frac{\Vert \alpha \Vert ^2_\infty }{\xi c(T)} \\&\quad + \kappa \Vert \tilde{\varGamma } \Vert _\infty (C_\theta + 1) \triangleq C^y_y, \\ \Vert \nabla _\vartheta f^g_\kappa \Vert _2&\le \kappa + \frac{\kappa }{c(T)} + \frac{\kappa \Vert \tilde{\varGamma } \Vert _\infty }{c(T)} \triangleq C^g_\vartheta , \quad \Vert \nabla _g f^g_\kappa \Vert _2 \le \frac{\Vert \alpha \Vert ^2_\infty }{\xi c(T)} + \kappa (C_\theta + 1) \triangleq C^g_g. \end{aligned}$$

It is direct to show

$$\begin{aligned} \Vert \nabla {\bar{Y}}^\kappa _t \Vert _2&\le Te^{ C^y_y T} \left( C^y_\vartheta e^{K_b T} + \frac{\kappa (C_\theta + 1)^2}{c(T)} + (C_\theta + 1) \kappa ^2 \right) , \\ \Vert \nabla {\bar{G}}^\kappa _t \Vert _2&\le T e^{C^g_g T} \left( C^g_\vartheta e^{K_b T} + \frac{ \kappa \Vert \alpha \Vert ^2_\infty }{\xi c(T)} + \kappa ^2 (C_\theta + 1) \Vert {\tilde{\varGamma }} \Vert _\infty \right) . \end{aligned}$$

Then by setting T sufficiently small, the right hand side of the two inequalities can be smaller than $\kappa$. Therefore, the truncation ${\bar{\rho }}_\kappa$ is not binding and $(\nabla {\bar{Y}}^\kappa , \nabla {\bar{G}}^\kappa , \nabla {\bar{Z}}^\kappa , \nabla {\bar{E}}^\kappa ) =(\nabla Y^\kappa , \nabla G^\kappa , \nabla Z^\kappa , \nabla E^\kappa )$. With Malliavin calculus as in [26, Theorem 3.1], a version of $(Z^\kappa _t, E^\kappa _t)$ is given by $(\nabla Y^\kappa _t (\nabla \vartheta _t)^{-1}\sigma _\vartheta (t)$, $\nabla G^\kappa _t (\nabla \vartheta _t)^{-1}\sigma _\vartheta (t))$. Since $\Vert (\nabla \vartheta _t)^{-1}\sigma _\vartheta (t)\Vert _2 \le C_\sigma e^{K_b T}$, we can further select T small enough such that truncation $\rho _\kappa$ is also not binding. Then we derive that (Z, E) is essentially bounded and therefore ${\tilde{U}}$ is essentially bounded.

When $b_\vartheta$ is not differentiable, the result can be proved by a standard approximation and stability results for Lipschitz BSDEs as noted in [26, 27]. $\square$