Appendix A: Proof of Lemma 2.1
Define
$$\begin{aligned} X(t):=\int ^{(T-t)^\theta }_0 g_2(u)dW'(u). \end{aligned}$$
Then, using (2.2) and (2.3), we obtain
$$\begin{aligned} X(t)&=X(0)+\int ^t_0 G(u)d\alpha (u)+\int ^t_0 g_1(u)d\hat{W}(u)\nonumber \\ {}&\quad +\int ^t_0 \left\{ \dot{G}(u)+g_1^2(u) \right\} \alpha (u)du. \end{aligned}$$
(A.1)
In particular, X is a semimartingale in the filtration \({\mathcal {G}}\). From (2.2) recall that W is also a semimartingale in the filtration \({\mathcal {G}}\). Now, for each filtration \(\mathcal {K}\) we denote by \(\langle X_1, X_2 \rangle ^{\mathcal {K}}\) the quadratic covariation process of \(X_1, X_2\) with respect to the filtration \(\mathcal {K}\). Using Theorem 23 (ii) of Chapter II in [17] and adopting a partition \(0=t_0<t_1< \cdots <t_n=t\) we have
$$\begin{aligned} \langle W(\cdot ), X(\cdot ) \rangle ^{\mathcal {G}}_t&=\left\langle W(\cdot ), \int ^{(T-(\cdot ))^\theta }_0 g_2(u)dW'(u)\right\rangle ^{ \mathcal {F}}_t\nonumber \\&=-\lim _{n\rightarrow \infty }\sum _{k=1}^n (W(t_k)-W(t_{k-1}))\left( \int ^{(T-t_{k-1})^\theta }_{(T-t_{k})^\theta } g_2(s)dW'(s) \right) \\&\qquad in \ prob. \nonumber \end{aligned}$$
(A.2)
Now, we compute, for \(\eta \in \mathbb {R}\)
$$\begin{aligned}&E\left[ {\mathrm {e}}^{-i \eta \sum _{k=1}^n (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))}\right] \\&\quad =\prod _{k=1}^n E\left[ {\mathrm {e}}^{-i \eta (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))}\right] \\&\quad = \displaystyle \prod _{k=1}^n E\left[ E\left[ {\mathrm {e}}^{-i \eta (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))} \Bigr |X(t_{k-1}), X(t_k) \right] \right] \\&\quad =\prod _{k=1}^nE \left[ \frac{1}{\sqrt{2\pi (t_k-t_{k-1})}} \int ^{\infty }_{-\infty } {\mathrm {e}}^{-i \eta (X(t_{k-1})-X(t_k)) x-\frac{x^2}{2(t_k-t_{k-1})}}dx \right] \\&\quad =\prod _{k=1}^nE \left[ {\mathrm {e}}^{-\frac{1}{2}(t_k-t_{k-1}) \eta ^2(X(t_{k-1})-X(t_k))^2} \right] \\&\quad =\prod _{k=1}^n\left\{ 1+\eta ^2 (t_k-t_{k-1}) \int ^{(T-t_{k-1})^\theta }_{(T-t_{k})^\theta } g^2_2(s)ds \right\} ^{-1/2}. \end{aligned}$$
Hence, we see that
$$\begin{aligned} E\left[ {\mathrm {e}}^{-i \eta \sum _{k=1}^n (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))}\right] \rightarrow 1 \ \mathrm{as} \ n\rightarrow \infty . \end{aligned}$$
Then, \(\sum _{k=1}^n (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))\) converges in distribution to 0 (constant). And, \(\sum _{k=1}^n (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))\) converges to 0 in probability. We see that there exists a subsequence \(\{n_\ell \}\) such that \(\sum _{k=1}^{n_\ell } (W(t_k)-W(t_{k-1}))(X(t_{k-1})-X(t_k))\) converges to 0 almost surely. Using (A.2), we see that
$$\begin{aligned} \langle W(\cdot ), X(\cdot ) \rangle ^{\mathcal {G}}_t=0, \end{aligned}$$
and
$$\begin{aligned} \langle \hat{W}(\cdot ), X(\cdot ) \rangle ^{\mathcal {G}}_t=0. \end{aligned}$$
(A.3)
For \(s<t\) we have
$$\begin{aligned} E\left[ X(t)-X(s) \Bigr | \mathcal {G}_s \right]&=-E\left[ \int ^{(T-s)^\theta }_{(T-t)^\theta }g_2(u)dW'(s)\biggr | \mathcal {G}_s \right] \\&=-E\left[ \int ^{(T-s)^\theta }_{(T-t)^\theta }g_2(u)dW'(u)\biggr | \int ^T_s g_1(u)dW(u)\right. \\ {}&\quad +\left. \int ^{(T-s)^\theta }_0 g_2(u)dW'(u) \right] \\&=-\int ^{(T-s)^\theta }_{(T-t)^\theta }g^2_2(u)du \cdot \frac{1}{G(s)} \left( \int ^T_s g_1(u)dW(u)\right. \\ {}&\quad +\left. \int ^{(T-s)^\theta }_0 g_2(u)dW'(u) \right) \\&=-\int ^{(T-s)^\theta }_{(T-t)^\theta }g^2_2(u)du \cdot \alpha (s) \\&=-\int ^{t}_{s}f^2(u)du \cdot \alpha (s). \end{aligned}$$
Moreover, we have
$$\begin{aligned}&E\left[ X(t)-X(s)+\int ^t_s \frac{f^2(u)}{G(u)} \left( \int ^T_u g_1(v)dW(v)\right. \right. \nonumber \\ {}&\quad +\left. \left. \int ^{(T-u)^\theta }_0 g_2(v)dW'(v) \right) du \biggr |\mathcal {G}_s \right] \\&= -\int ^{t}_{s}f^2(u)du \cdot \alpha (s)+ \int ^t_s \frac{f^2(u)}{G(u)}E\left[ \int ^T_u g_1(v)dW(v)\right. \nonumber \\ {}&\quad +\left. \int ^{(T-u)^\theta }_0 g_2(v)dW'(v) \biggr |\mathcal {G}_s \right] du \nonumber \\&=\quad -\int ^{t}_{s}f^2(u)du \cdot \alpha (s)+ \int ^t_s \frac{f^2(u)}{G(u)} \nonumber \\&\cdot E\left[ \int ^T_u g_1(v)dW(v)+\int ^{(T-u)^\theta }_0 g_2(v)dW'(v) \biggr |\int ^T_s g_1(v)dW(v)\right. \nonumber \\ {}&\quad +\left. \int ^{(T-s)^\theta }_0 g_2(v)dW'(v) \right] du \nonumber \\&= -\int ^{t}_{s}f^2(u)du \cdot \alpha (s)+ \int ^t_s \frac{f^2(u)}{G(u)}\cdot \frac{G(u)}{G(s)}\left( \int ^T_s g_1(v)dW(v)\right. \nonumber \ \\&\quad + \left. \int ^{(T-s)^\theta }_0 g_2(v)dW'(v) \right) du \nonumber \\&\quad -\int ^{t}_{s}f^2(u)du \cdot \alpha (s)+\int ^{t}_{s}f^2(u)du \cdot \alpha (s) \nonumber \\&=0. \nonumber \end{aligned}$$
(A.4)
Setting
$$\begin{aligned} M^X(t):=X(t)+\int ^t_0 f(s)^2 \alpha (s) ds, \end{aligned}$$
and using (A.4), we see that \(\left\{ M^X(t); t\in [0, T) \right\} \) is a \(\mathcal {G}_t\)-martingale. Then, we have
$$\begin{aligned} \langle M^X(\cdot )\rangle ^{\mathcal {G}}_t=\langle X(\cdot ) \rangle ^{\mathcal {G}}_t = \langle X(\cdot ) \rangle ^{\mathcal {F}}_t=\int ^t_0f^2(s)ds. \end{aligned}$$
Then, \(\left\{ \widehat{B}(t); t\in [0, T) \right\} \) defined by
$$\begin{aligned} \widehat{B}(t):=\int ^t_0 \frac{1}{f(s)}dM^X(s) \end{aligned}$$
is a \(\mathcal {G}_t\)-Wiener process. And, we have
$$\begin{aligned} \langle \widehat{W}(\cdot ), \widehat{B}(\cdot ) \rangle ^{\mathcal {G}}_t=\int ^t_0 \frac{1}{f(s)}d\langle \widehat{W}(\cdot ), M^X(\cdot ) \rangle ^{\mathcal {G}}_s=\int ^t_0 \frac{1}{f(s)}d \langle \widehat{W}(\cdot ), X(\cdot ) \rangle ^{\mathcal {G}}_s=0, \end{aligned}$$
where the last equality comes from (A.3). Hence, \(\widehat{W}\) and \(\widehat{B}(\cdot )\) are mutually independent. On the other hand, we recall
$$\begin{aligned} X(t)=X(0)-\int ^t_0 f^2(s) \alpha (s)ds +\int ^t_0 f(s) d\widehat{B}(s). \end{aligned}$$
and
$$\begin{aligned} G(t)\alpha (t)=\int ^T_t g_1(s)d\widehat{W}(s)+\int ^T_t g^2_1(s)\alpha (s)ds+X(t). \end{aligned}$$
Then, we have
$$\begin{aligned} d\alpha (t)=\frac{1}{G(t)}\left\{ -g_1(t)d\widehat{W}(t)+f(t)\widehat{B}(t) \right\} . \end{aligned}$$
(A.5)
Appendix B: Proof of Lemma 3.1
(1) This is obtained immediately.
(2) Now, we perform the change of variables \(G_{1}(s)=g_{1}(s)(\dot{g}_{1}(s))^{-1}\). Then, the equation (3.11) becomes
$$\begin{aligned} 1-\dot{G}_{1}(s)=\frac{a+\sigma ^{2}+\{(1-\gamma )\sigma ^{2}-2a\} \theta (T-s)^{\theta -1}}{a}. \end{aligned}$$
The general solution of the above equation is (for a suitable constant \(C_1\))
$$\begin{aligned} G_{1}(s)=C_{1}-\frac{\sigma ^{2}}{a}s+\frac{(1-\gamma ) \sigma ^{2}-2a}{a}\left( (T-s)^{\theta }-T^{\theta }\right) . \end{aligned}$$
From here, one obtains that for a fixed \(t_{0}\in [0,T]\) and a constant \(C_{2}\), the solution is
$$\begin{aligned} g_{1}(t)=C_{2}\exp \left( \int _{t_{0}}^{t}\left( C_{1}-\frac{\sigma ^{2}s}{a}+\frac{(1-\gamma )\sigma ^{2}-2a}{a}\left( (T-s)^{\theta }-T^{\theta }\right) \right) ^{-1}ds\right) . \end{aligned}$$
In order to determine the constants we take \(t_{0}=0\), then one obtains that \(C_{2}=\sigma \) due to the condition \(g_{1}(0)=\sigma \). Moreover, if we take \(C_{1}\) as
$$\begin{aligned} C_{1}:=\frac{\sigma ^{2}T}{a}+\frac{(1-\gamma )\sigma ^{2}-2a}{a}T^{\theta }, \end{aligned}$$
then we obtain (3.13).
Further, under (H2), we observe \(\dot{g}_1(t)>0\) and that
$$\begin{aligned} \sigma \le g_{1}(t) \le \sigma \exp \left\{ \frac{a}{(1-\gamma )\sigma ^{2}-2a}\int ^{t}_{0}\frac{du}{(T-u)^{\theta }}\right\} . \end{aligned}$$
Hence, we have (3.13). And we obtain \(\int _{0}^{T}\frac{\left| g_{1}(s)\right| ^{2}}{G(s)}ds\) immediately.
3. Note that \(G(t)=\int ^{T}_{t}\left\{ 1+\theta (T-u)^{\theta -1}\right\} g^{2}_{1}(u)du\). From (3.15) we have
$$\begin{aligned} \dot{P}(t)&=\frac{1}{\gamma }\left[ \dot{G}(t)-\frac{(1-\gamma )\sigma ^{2}-2a}{a}\frac{G(t)}{g^{4}_{1}(t)} \left[ \left\{ \ddot{g}_{1}(t)G(t)+2\dot{g}_{1} (t)\dot{G}(t)\right\} g_{1}(t)\right. \right. \\ {}&\quad \left. \left. -3\dot{g}^{2}_{1}(t)G(t)\right] \right] \end{aligned}$$
Therefore, we observe that
$$\begin{aligned}&\dot{P}(t)+\frac{\gamma }{G^{2}(t)}\left\{ -\dot{G}(t)+ \frac{\gamma \sigma ^{2}g^{2}_{1}(t)}{(1-\gamma )\sigma ^{2}-2a}\right\} P^{2}(t) \\&\qquad -\frac{2\gamma }{(1-\gamma ) \sigma ^{2}-2a}\frac{\sigma ^{2}g^{2}_{1}(t)}{G(t)}P(t)+ \frac{\sigma ^{2} g^{2}_{1}(t)}{(1-\gamma )\sigma ^{2}-2a} \\&\quad = \dot{P}(t)+\frac{\sigma ^{2}g^{2}_{1}(t)}{(1-\gamma ) \sigma ^{2}-2a} \left\{ -\frac{\gamma P(t)}{G(t)}+1\right\} ^{2}-\frac{\gamma \dot{G}(t)}{G^{2}(t)}P^{2}(t)\\&\quad =\frac{1}{\gamma }\left[ \dot{G}(t)- \frac{(1-\gamma )\sigma ^{2}-2a}{a} \frac{G(t)}{g^{4}_{1}(t)} \left\{ \left( \ddot{g}_{1}(t)G(t)+2\dot{g}_{1} (t)\dot{G}(t)\right) g_{1}(t)- 3\dot{g}^{2}_{1}(t)G(t)\right\} \right] \\&\qquad +\frac{\sigma ^{2}\{(1-\gamma ) \sigma ^{2}-2a\}}{a^{2}}\frac{\dot{g}^{2}_{1} (t)G^{2}(t)}{g^{4}_{1}(t)}- \frac{\dot{G}(t)}{\gamma }\left\{ 1- \frac{(1-\gamma )\sigma ^{2}-2a}{a} \frac{\dot{g}_{1}(t)G(t)}{g^{3}_{1}(t)}\right\} ^{2}\\&\quad =-\frac{(1-\gamma ) \sigma ^{2}-2a}{\gamma a^{2}}\frac{G^{2}(t)}{g^{6}_{1}(t)}\left[ a \ddot{g}_{1}(t)g^{3}_{1}(t)-(3a+\gamma \sigma ^{2})\dot{g}^{2}_{1}(t)g^{2}_ {1}(t)+\{(1-\gamma )\sigma ^{2}\right. \\&\qquad -\left. 2a\}\dot{g}^{2}_{1}(t)\dot{G}(t)\right] \\&\quad =\frac{(1-\gamma )\sigma ^{2}-2a}{\gamma a^{2}}\frac{G^{2}(t)}{g^{4}_{1} (t)}\left\{ -a\ddot{g}_{1}(t)g_{1}(t)+\left[ (a+\sigma ^{2})+\{ (1-\gamma )\sigma ^{2}\right. \right. \\&\qquad -\left. \left. 2a\}\theta (T-t)^{\theta -1}\right] \dot{g}^{2}_{1}(t)\right\} \\&\quad =0. \end{aligned}$$
Next, from the proof of (2), recall that
$$\begin{aligned} G_{1}(t)=\frac{g_{1}(t)}{\dot{g}_{1}(t)}=\frac{1}{a}\left[ \sigma ^{2}(T-t)+\{(1-\gamma )\sigma ^{2}-2a\}(T-t)^{\theta }\right] . \end{aligned}$$
(B.1)
Then, we have
$$\begin{aligned} P(t)&=\frac{G(t)}{\gamma }\left[ 1-\frac{(1-\gamma )\sigma ^{2}-2a}{\sigma ^{2}(T-t)+\{(1-\gamma )\sigma ^{2}-2a\} (T-t)^{\theta }}\cdot \frac{G(t)}{g^{2}_{1}(t)} \right] \\&=\frac{G(t)}{[\sigma ^{2}(T-t)+\{ (1-\gamma )\sigma ^{2}-2a\}(T-t)^{\theta }]g^{2}_{1}(t) } \kappa (t), \nonumber \end{aligned}$$
(B.2)
where \(\kappa (t)\) is defined by
$$\begin{aligned} \kappa (t):=\frac{1}{\gamma } \left( [\sigma ^{2}(T-t)+\{(1-\gamma )\sigma ^{2}-2a\}(T-t)^{\theta }]g^{2}_{1}(t) -\{(1-\gamma )\sigma ^{2}-2a\}G(t) \right) . \end{aligned}$$
Then, since \(\dot{\kappa }(t)=-\sigma ^2 g^2_1(t)<0\) for \(t\in [0,T]\) and \(\kappa (T)=0\), we see that \(\kappa (t)\ge 0\) for \(t\in [0,T]\). Hence, \(P(t)\ge 0\) for \(t\in [0,T]\). In particular, \(P(t)> 0\) for \(t\in [0,T)\).
Note that under (H2), \(\dot{g}_1(t)>0\) holds. Therefore, we observe
$$\begin{aligned} \frac{G(t)}{g^{2}_{1}(t)} \ge (T-t)+(T-t)^{\theta }, \end{aligned}$$
and from (3.14) and (B.2) we see that there is \(K_T>0\) such that
$$\begin{aligned} P(t)&\le \frac{G(t)}{\gamma } \frac{(\gamma \sigma ^2+2a)(T-t)}{ \sigma ^{2}(T-t)+\{(1-\gamma )\sigma ^{2}-2a\}(T-t)^{\theta }}\\&\le K_T(T-t). \end{aligned}$$
Appendix C: Proof of Lemma 3.2
Set
$$\begin{aligned} \xi _{t}:=-\frac{\gamma }{(1-\gamma )\sigma ^{2}-2a}\frac{\sigma ^{2}g^{2}_{1}(t)}{G(t)} +\frac{\gamma }{G^{2}(t)}\left\{ -\dot{G}(t)+\frac{\gamma \sigma ^{2}g^{2}_{1}(t)}{(1-\gamma )\sigma ^{2}-2a}\right\} P(t). \end{aligned}$$
From (3.15) we have
$$\begin{aligned} \xi _{t}&\le K_T \frac{\{1+(T-t)^{\theta -1}\}(T-t)}{\{T-t+(T-t)^{1-\theta }\}^2}\le \frac{K_{T}}{(T-t)^{\theta }}, \end{aligned}$$
and
$$\begin{aligned} \int ^{s}_{t} \xi _{u}du \le K_{T}T^{1-\theta } \quad \text {for} \ s\ge t. \end{aligned}$$
Moreover, from (3.17) we observe that
$$\begin{aligned} \left| -\frac{\gamma P(t)}{G(t)}+1 \right|&= K_T \frac{T-t}{T-t +(T-t)^{1-\theta }}+1 \le K_{T}. \end{aligned}$$
Noting that by using the variations of constants method we have
$$\begin{aligned} Q(t)=\int _{t}^{T}{\mathrm {e}}^{\int ^{s}_{t}\xi _{u}du} \frac{(\mu -r)\sigma g_{1}(s)}{(1-\gamma )\sigma ^{2}-2a} \left\{ -\frac{\gamma P(s)}{G(s)}+1\right\} ds, \end{aligned}$$
we obtain \(|Q(t)|\le K_{T}(T-t)\).
As for the estimation of R(t), we have
$$\begin{aligned} |R(t)|&\le K_{T}\int ^{T}_{t}\left\{ \left| \frac{\dot{G}(u)}{G^{2}(u)} \right| |P(t)|+\left| \frac{\dot{G}(u)}{G^{2}(u)}\right| |Q(u)|^{2}+ \left| \frac{Q(u)}{G(u)}\right| ^{2}+1 \right\} du\\&\le K_{T} (T-t)^{1-\theta }. \end{aligned}$$
Moreover, from (B.2) we have, for \(t\in [0, T)\)
$$\begin{aligned} \frac{Q^{2}(t)}{P(t)}&=\frac{\gamma [\sigma ^{2}(T-t)+\{(1-\gamma )\ \sigma ^{2}-2a\}(T-t)^{\theta }]g^{2}_{1}(t) Q^{2}(t)}{G(t)[(\sigma ^{2}(T-t)+ \{(1-\gamma )\sigma ^{2}-2a\}(T-t)^{\theta })g^{2}_{1}(t)-\{(1-\gamma ) \sigma ^{2}-2a\}G(t)]}\\&\le K_{T}J(t), \end{aligned}$$
where
$$\begin{aligned} J(t):=\frac{\gamma (T-t)^{2}}{[\sigma ^{2}(T-t)+\{(1-\gamma )\sigma ^{2}-2a\} (T-t)^{\theta }]g^{2}_{1}(t)-\{(1-\gamma )\sigma ^{2}-2a\}G(t)}. \end{aligned}$$
Here, we observe
$$\begin{aligned} \lim _{t\rightarrow T}J(t)&=\lim _{t\uparrow T}\frac{-2\gamma (T-t)}{(-\gamma \sigma ^{2}+2a)g^{2}_{1}(t)+2[\sigma ^{2}(T-t)+\{(1-\gamma ) \sigma ^{2}-2a\}(T-t)^{\theta }]\dot{g}_{1}(t)g_{1}(t)}\\&=\lim _{t\uparrow T}\frac{-2\gamma (T-t)}{(-\gamma \sigma ^{2}-2a)g^{2}_{1}(t)+2a g^{2}_{1}(t)}\\&=\frac{2}{\sigma ^{2}}\lim _{t \uparrow T}\frac{T-t}{g^{2}_{1}(t)}, \end{aligned}$$
where the second inequality follows from (B.1). Hence, we see that J(t)\(=\mathcal {O}\left( (T-t)\right) \) as \(t\uparrow T\).
Appendix D: The martingale property of \(\{ Z_{t}(\widehat{\pi })\}_{t\in [0,T)}\)
Lemma D.1
Use (3.13). Let \(h_i(t, \alpha ) (i=1,2)\) be continuous functions satisfying \(h_i(t, \alpha )\le C(1+|\alpha |)\). Define \(\rho _t\) by
$$\begin{aligned} \rho _t:=\mathcal {E}\left( \int ^{(\cdot )}_{0}h_1(s,\alpha (s) ) d\widehat{W}(s)+\int ^{(\cdot )}_{0}h_2(s,\alpha (s)) d\widehat{B}(s)\right) _{t}. \end{aligned}$$
Then, \(\{ \rho _{t}\}_{t\in [0,T)}\) is a \(\mathcal {G}_t\)-martingale.
Proof
The proof follows similar arguments as for Lemma 4.1.1 in [2]. Recall that
$$\begin{aligned} d&\rho _t=\rho _t \left\{ h_1(t,\alpha (t) ) d\widehat{W}(t)+h_2(t,\alpha (t)) d\widehat{B}(t)\right\} \end{aligned}$$
(D.1)
For \(s\le t\) we set \(\rho _{t,s}:=\rho _t \rho _s^{-1}\). Let \(\epsilon >0\) be arbitrary. Then, we have
$$\begin{aligned}&d\left( \frac{\rho _{t,s}}{1+\epsilon \rho _{t,s}} \right) = d\overline{N}_{t,s}-\overline{A}_{t,s} dt, \end{aligned}$$
(D.2)
where \(\overline{N}_{t,s}\) and \(\overline{A}_{t,s}\) are defined by
$$\begin{aligned} \overline{N}_{t,s}&:=\int ^{t}_{s} \frac{\rho _{u,s}}{\left( 1+\epsilon \rho _{u,s}\right) ^2} \left\{ h_1(u,\alpha (u)) d\widehat{W}(u)+h_2(u,\alpha (u)) d\widehat{B}(u)\right\} ,\\ \overline{A}_{t,s}&:=\frac{\epsilon \rho _{t,s}^2}{(1+\epsilon \rho _{t,s})^3}\left\{ h_1^2(t,\alpha (t))+h_2^2(t,\alpha (t)) \right\} . \end{aligned}$$
If we can check
$$\begin{aligned} E\left[ \rho _{t,s} \alpha ^2(t)\right] \le K_{t,T}, \end{aligned}$$
(D.3)
we see that
$$\begin{aligned} E\left[ |\overline{N}_{t,s}|^2 \right]&\le \overline{C}_{\epsilon } E\left[ \int ^{t}_{s} \left\{ \rho _{u,s}+ \rho _{u,s} \alpha ^2(u)\right\} du \right] \\&\le \overline{C}_{\epsilon }(1+K_{t,T})T, \end{aligned}$$
and that \(\overline{N}_{t,s}\) is a square-integrate martingale. Here, we use \(E\left[ \rho _{t,s} \right] \le 1\) Therefore, integrating (D.2) on [s, t] and taking expectation for both sides, we have
$$\begin{aligned} E\left[ \frac{\rho _{t,s}}{1+\epsilon \rho _{t,s}} \Bigr | \mathcal {G}_s \right]&=\frac{1}{1+\epsilon }-E\left[ \int ^t_s \overline{A}_{u,s} du \Bigr | \mathcal {G}_s \right] \\&=\frac{1}{1+\epsilon }-E\left[ \int ^t_s \overline{A}_{u,s} du \right] . \nonumber \end{aligned}$$
(D.4)
Here, we observe the following :
-
\(\displaystyle \overline{A}_{u,s} \rightarrow 0 \ a.e.\ (u, \omega ) \in [s, T] \times \Omega .\) as \(\epsilon \rightarrow 0\),
-
\(\displaystyle \overline{A}_{u,s} \le K \left\{ \rho _{u,s}+ \rho _{u,s} \alpha ^2(u)\right\} \).
Hence, from (D.3) and the dominated convergence theorem, we have
$$\begin{aligned} E\left[ \int ^t_s \overline{A}_{u,s} du \right] \rightarrow 0 \ \text {as} \ \epsilon \rightarrow 0. \end{aligned}$$
Meanwhile, since \(E\left[ \rho _{t,s} \right] \le 1\),
$$\begin{aligned} E\left[ \frac{\rho _{t,s}}{1+\epsilon \rho _{t,s}} \Bigr | \mathcal {G}_s \right] \rightarrow E \left[ \rho _{t,s} | \mathcal {G}_s \right] \ \text {as} \ \epsilon \rightarrow 0. \end{aligned}$$
Letting \(\epsilon \rightarrow 0\) in (D.4), we have \(E \left[ \rho _{t,s} | \mathcal {G}_s \right] =1\), or equivalently
$$\begin{aligned} E \left[ \rho _{t} | \mathcal {G}_s \right] =\rho _{s}. \end{aligned}$$
Finally, we shall prove (D.3). Recalling that
$$\begin{aligned} d\langle \alpha (\cdot )\rangle _t=\frac{1}{G(t)}\left\{ g_1^2(t)d \widehat{W}(t)+f^2(t)\right\} =-\frac{\dot{G}(t)}{G^2(t)}dt, \end{aligned}$$
we have
$$\begin{aligned} d\alpha ^2(t)=\frac{2\alpha (t)}{G(t)}\left\{ -g_1(t)d\widehat{W}(t)+ f(t)d\widehat{B}(t) \right\} -\frac{\dot{G}(t)}{G^2(t)}dt. \end{aligned}$$
(D.5)
From (D.2) and (D.5) we have
$$\begin{aligned} d\{\rho _{t,s} \alpha ^2(t)\}&= \frac{\rho _{t,s}}{G(t)} \left\{ -\frac{\dot{G} (t)}{G(t)}+2\alpha (t)\{-g_1(t)h_1(t, \alpha (t))+f(t)h_2(t, \alpha (t)) \} \right\} dt\\&\quad +\rho _{t,s} \alpha (t) \left[ \left( -\frac{2g_1(t)}{G(t)}+ \alpha (t)h_1(t,\alpha (t))\right) d\widehat{W}(t)+\left( \frac{2f(t)}{G(t)}+ \alpha (t)h_2(t,\alpha (t))\right) d\widehat{B}(t) \right] . \end{aligned}$$
Hence, for \(\epsilon >0\) we have
$$\begin{aligned}&d\left( \frac{\rho _{t,s} \alpha ^2(t)}{1+\epsilon \rho _{t,s} \alpha ^2(t)} \right) = d\check{N}_{t,s}+\check{A}_{t,s} dt, \end{aligned}$$
(D.6)
where
$$\begin{aligned} \check{N}_{t,s}&:=\int ^t_s \frac{\rho _{u,s} \alpha (u)}{\{1+\epsilon \rho _{u,s} \alpha ^2(u)\}^2} \left[ \left( -\frac{2g_1(u)}{G(u)}+ \alpha (u)h_1(u,\alpha (u))\right) d\widehat{W}(u) \right. \\&\quad \left. +\left( \frac{2f(u)}{G(u)}+\alpha (u)h_2(u,\alpha (u))\right) d\widehat{B}(u) \right] \end{aligned}$$
is the local-martingale part and \((\check{A}_t)_{t\ge 0}\) is the bounded-variation part, which satisfies
$$\begin{aligned} \check{A}_{t,s} \le \frac{\rho _{t,s} }{\{1+\epsilon \rho _{t,s} \alpha ^2(t)\}^2}\left[ -\frac{\dot{G}(t)}{G^2(t)}+\frac{2\alpha (t)}{G(t)}\{-g_1(t)h_1(t, \alpha (t))+f(t)h_2(t, \alpha (t)) \} \right] . \end{aligned}$$
Here, observing that \(\dot{g}_1(t)>0\), and that
$$\begin{aligned} G(t)=\int ^T_t \{1+\theta (T-u)^{\theta -1}\}g_1^2(u)du \ge \{T-t+(T-t)^{\theta }\} g_1^2(t), \end{aligned}$$
we have
$$\begin{aligned} E[\check{N}_{t,s}^2]&\le E\left[ \int ^t_s \frac{\rho _{u,s}^2 \alpha ^2(u)}{\{1+\epsilon \rho _{u,s} \alpha ^2(u)\}^4} \left[ \left( -\frac{2g_1(u)}{G(u)}+\alpha (u)h_1(u,\alpha (u))\right) ^2 \right. \right. \\&\quad \left. \left. +\left( \frac{2f(u)}{G(u)}+\alpha (u)h_2(u,\alpha (u))\right) ^2 \right] du\right] \\&\le C_\epsilon \left( 1+\int ^t_s\left\{ -\frac{\dot{G}(u)}{G^2(u)}+ 1 \right\} E[\rho _{u,s}]du + \int ^t_s E[\alpha ^2(u)]du \right) \\&\le C_\epsilon \left( 1+\frac{1}{G(t)} \right) . \end{aligned}$$
Hence, \(\check{N}\) is a square-integrable martingale. Moreover, we have
$$\begin{aligned} \check{A}_{t,s} \le C\left\{ 1+\frac{1}{(T-t)^{1+\theta }}\rho _{t,s}+ \frac{1}{(T-t)^\theta }+\frac{1}{(T-t)^\theta }\frac{\rho _{t,s} \alpha ^2(t)}{1+\epsilon \rho _{t,s} \alpha ^2(t)} \right\} . \end{aligned}$$
Then, we have
$$\begin{aligned} E\left[ \frac{\rho _{t,s} \alpha ^2(t)}{1+\epsilon \rho _{t,s} \alpha ^2(t)} \right] \le C_T \left\{ 1+ \frac{1}{(T-t)^{\theta }}+\frac{1}{(T-t)^\theta }\int ^t_s E\left[ \frac{\rho _{u,s} \alpha ^2(s)}{1+\epsilon \rho _{u,s} \alpha ^2(s)} \right] ds \right\} . \end{aligned}$$
From Gronwall’s inequality we have
$$\begin{aligned} E\left[ \frac{\rho _{t,s} \alpha ^2(t)}{1+\epsilon \rho _{t,s} \alpha ^2(t)} \right] \le C_T \left\{ 1+\frac{1}{(T-t)^{\theta }} \right\} , \end{aligned}$$
and, by Fatou’s lemma, we obtain (D.3).
Appendix E: Proof of (3.28)
Now we shall compute
$$\begin{aligned} E^{(\widehat{\pi })}\left[ {\mathrm {e}}^{-\gamma \overline{u}(T_{n_k}, \alpha (T_{n_k}))}\biggl | \mathcal {G}_0\right] . \end{aligned}$$
Under \(P^{(\widehat{\pi })}\), \(\widehat{W}^{(\widehat{\pi })}(t)\) and \(\widehat{B}^{(\widehat{\pi })}(t)\) defined by
$$\begin{aligned} d\widehat{W}^{(\widehat{\pi })}(t)&=d\widehat{W}(t)-\gamma \left[ \left( -\left\{ 1+\frac{\gamma \sigma ^2}{(1-\gamma )\sigma ^2-2a}\right\} \frac{g_1(t)}{G(t)} P(t) \right. \right. \\ {}&\quad +\left. \left. \frac{\sigma ^2 g_1(t)}{(1-\gamma )\sigma ^2-2a}\right) \alpha (t) \right. \\&\quad \left. -\left\{ 1+\frac{\gamma \sigma ^2}{(1-\gamma )\sigma ^2-2a}\right\} \frac{g_1(t)}{G(t)} Q(t) +\frac{\sigma (\mu -r)}{(1-\gamma )\sigma ^2-2a} \right] dt \end{aligned}$$
and
$$\begin{aligned} d\widehat{B}^{(\widehat{\pi })}(t)=d\widehat{B}(t)-\gamma \frac{f(t)}{G(t)}\{P(t)\alpha (t)+Q(t) \}dt \end{aligned}$$
are \({\mathcal {G}}_t\)-Wiener processes and \(\alpha (t)\) satisfies
$$\begin{aligned} d\alpha (t)=\frac{1}{G(t)}\left\{ -g_1(t)d\widehat{W}^{(\widehat{\pi })}(t)+ f(t)d\widehat{B}^{(\widehat{\pi })}(t)\right\} + \left\{ b_1(t)\alpha (t)+b_0(t)\right\} dt, \end{aligned}$$
where \(b_0(t)\) and \(b_1(t)\) are defined by
$$\begin{aligned} b_0(t):=-\gamma \left\{ \frac{\dot{G}(t)}{G^2(t)}-\frac{\gamma \sigma ^2}{(1-\gamma )\sigma ^2-2a}\frac{g^2_1(t)}{G(t)} \right\} Q(t)- \frac{\gamma \sigma (\mu -r)}{(1-\gamma )\sigma ^2-2a}\frac{g_1(t)}{G(t)},\\ b_1(t):=-\gamma \left\{ \frac{\dot{G}(t)}{G^2(t)}-\frac{\gamma \sigma ^2}{(1-\gamma )\sigma ^2-2a}\frac{g^2_1(t)}{G(t)} \right\} P(t)-\frac{\gamma \sigma ^2}{(1-\gamma )\sigma ^2-2a} \frac{g^2_1(t)}{G(t)}. \end{aligned}$$
Define
$$\begin{aligned} m(t)&:=E[\alpha (t)|{\mathcal {G}}_0],\\ V(t)&:=E[\{\alpha (t)-m(t)\}^2 |{\mathcal {G}}_0]. \end{aligned}$$
Then, we have
$$\begin{aligned} m(t)&={\mathrm {e}}^{\int ^t_0 b_1(s)ds} \alpha (0)+\int ^t_0 {\mathrm {e}}^{\int ^t_s b_1(u)du}b_0(s)ds,\\ V(t)&=-\int ^t_0 {\mathrm {e}}^{2\int ^t_s b_1(u)du}\frac{\dot{G}(s)}{G^2(s)}ds. \end{aligned}$$
Here we observe the following :
-
\(\displaystyle |b_1(t)|\le K_T \left\{ \frac{1}{(T-t)^\theta }+1\right\} \)
-
\(\displaystyle |b_0(t)|\le K_T \left\{ \frac{1}{(T-t)^\theta }+1 \right\} \),
-
\(\displaystyle |m(t)| \le K_T \left\{ 1+|\alpha (0)| \right\} \),
-
\(\displaystyle |V(t)| \le \frac{K_T}{(T-t)^\theta }.\)
Hence, we have
$$\begin{aligned}&E^{(\widehat{\pi })}\left[ {\mathrm {e}}^{-\gamma \overline{u}(T_{n_k}, \alpha (T_{n_k}))}\biggl | \mathcal {G}_0\right] =\frac{1}{\sqrt{2\pi }}\int _{{\mathbb {R}}}{\mathrm {e}}^{-\gamma \bar{u}(T_{n_k}, m(T_{n_k})+\sqrt{V(T_{n_k})}x)}{\mathrm {e}}^{-\frac{x^2}{2}} dx \\&\quad =\frac{1}{\sqrt{2\pi }}\int _{{\mathbb {R}}} {\mathrm {e}}^{-\frac{1}{2}\{1+\gamma P(T_{n_k})V(T_{n_k}) \} \left\{ x+\gamma \frac{P(T_{n_k})m(T_{n_k})+Q(T_{n_k})}{1+\gamma P(T_{n_k})V(T_{n_k})} \sqrt{V(T_{n_k})} \right\} ^2 }dx\\&\qquad \cdot {\mathrm {e}}^{-\gamma \left\{ \frac{1}{2}P(T_{n_k})m^2(T_{n_k})+Q(T_{n_k}) m(T_{n_k})+R(T_{n_k})\right\} +\frac{\gamma ^2}{2}\frac{\{P(T_{n_k})m(T_{n_k})+Q(T_{n_k})\}^2}{1+\gamma P(T_{n_k})V(T_{n_k})} V(T_{n_k}) }\\&\quad =\frac{1}{\sqrt{1+\gamma P(T_{n_k})V(T_{n_k})}} \\&\qquad \cdot {\mathrm {e}}^{-\gamma \left\{ \frac{1}{2}P(T_{n_k})m^2(T_{n_k})+ Q(T_{n_k})m(T_{n_k})+R(T_{n_k})\right\} +\frac{\gamma ^2}{2}\frac{\{P(T_{n_k})m(T_{n_k})+Q(T_{n_k})\}^2}{1+\gamma P(T_{n_k})V(T_{n_k})} V(T_{n_k})}, \end{aligned}$$
which implies (3.28).
