Appendix
Proof of Proposition 2.1
We first focus on the case \(j\le t\le T\). Note that given the information at time t, V(j), \(\Lambda (j)\) and \(\sum _{k=1}^{j-1} \Lambda (k)\) are all known. Therefore, we obtain that
$$\begin{aligned} f(t)= & {} E_{t}\Big [\exp \Big \{\phi _1 \ln S(T) +\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) \Big \}\Big ]\nonumber \\= & {} e^{\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) }E_{t}\Big [\exp \Big \{\phi _1 \ln S(T) \Big \}\Big ]. \end{aligned}$$
In addition, at time T, S(T) is also known, it follows that
$$\begin{aligned} f(T)= & {} e^{\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) }E_{T}\Big [\exp \Big \{\phi _1 \ln S(T) \Big \}\Big ]\nonumber \\= & {} e^{\phi _1 \ln S(T) +\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) }, \end{aligned}$$
which in turn implies that
$$\begin{aligned} A_0(T)=A_1(T)=A_2(T)=0. \end{aligned}$$
According to the law of iterated expectations, we have that
$$\begin{aligned}&E_{t}\Big [\exp \Big \{\phi _1 \ln S(T) \Big \}\Big ]\\ \quad&= {} E_t\Big [E_{t+1}\Big [\exp \Big \{\phi _1 \ln S(T) \Big \}\Big ]\Big ]\\ \quad&= E_t\Big [\exp \Big \{ \phi _1\ln S(t+1)+A_0(t+1)+A_1(t+1)h_m(t+2)+A_2(t+1)h_s(t+2) \Big \}\Big ]. \end{aligned}$$
Substituting the dynamics of \(\ln S(t+1)\), \(h_m(t+2)\) and \(h_s(t+2)\) yields that
$$\begin{aligned}&E_{t}\Big [\exp \Big \{\phi _1 \ln S(T) \Big \}\Big ]\\ &\quad= E_t\Big [\exp \Big \{ \phi _1\ln S(t+1)+A_0(t+1)+A_1(t+1)h_m(t+2)+A_2(t+1)h_s(t+2) \Big \}\Big ] \\ \quad&= E_t\Big [\exp \Big \{ \phi _1\ln S(t)+\phi _1 r-\frac{1}{2}\phi _1 h_s(t+1)+\phi _1\sqrt{h_s(t+1)}Z_s(t+1)\\ &\qquad-\frac{1}{2}\phi _1\beta _s^2h_m(t+1)+\phi _1\beta _s\sqrt{h_m(t+1)}Z_m(t+1)+ A_0(t+1)\\ &\qquad+A_1(t+1)\Big (w_m+b_m h_m(t+1)+a_m(Z_m(t+1)-c_m\sqrt{h_m(t+1)})^2\Big )\\ &\qquad+ A_2(t+1)\Big (w_s+b_s h_s(t+1)+a_s(Z_s(t+1)-c_s\sqrt{h_s(t+1)})^2\Big )\Big \}\Big ]. \end{aligned}$$
Using the fact that \(Ee^{a(Z+b)^2}=e^{-\frac{1}{2}\ln (1-2a)+\frac{ab^2}{1-2a}}\) with Z being a standard normal variable and some algebra shows that
$$\begin{aligned} A_0(t)&= \phi _1 r+A_0(t+1)+ w_m A_1(t+1)+w_sA_2(t+1)-\frac{1}{2}\ln (1-2a_mA_1(t+1))\\ \quad&-\frac{1}{2}\ln (1-2a_sA_2(t+1)),\\ A_1(t)&= b_mA_1(t+1)-\frac{1}{2}\phi _1\beta _s^2 +\phi _1\beta _sc_m-\frac{1}{2}c_m^2+\frac{\frac{1}{2}(\phi _1\beta _s-c_m)^2}{1-2a_mA_1(t+1)},\\ A_2(t)&= b_sA_2(t+1)-\frac{1}{2}\phi _1+\phi _1c_s-\frac{1}{2}c_s^2+\frac{\frac{1}{2}(\phi _1-c_s)^2}{1-2a_sA_2(t+1)}. \end{aligned}$$
Hence, \(A_0(t)\), \(A_1(t)\) and \(A_2(t)\) (\(j\le t\le T\)) can be obtained recursively with terminal conditions \(A_0(T)=A_1(T)=A_2(T)=0\) and the above expressions.
In what follows, we turn to the case \(t< j\). Applying the law of iterated expectations to f(t) yields that
$$\begin{aligned} f(t)&= E_{t}\Big [\exp \Big \{\phi _1 \ln S(T) +\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) \Big \}\Big ] \\ &= E_{t}\Big [E_{t+1} \Big [ \exp \Big \{\phi _1 \ln S(T) +\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) \Big \}\Big ]\Big ] \\ &= E_{t}\Big [f(t+1)\Big ] \\ &= E_{t}\Big [\exp \Big \{\phi _2\ln V(t+1) +\phi _4\sum _{k=1}^{t+1} \Lambda (k) +\phi _1\ln S(t+1)+B_0(t+1) \\ &\quad+B_1(t+1)h_m(t+2)+B_2(t+1)h_s(t+2)+B_3(t+1)h_v(t+2) + B_4(t+1)\Lambda (t+2)\Big \}\Big ]. \end{aligned}$$
Substituting the dynamics of \(\ln V(t+1)\), \(\Lambda (t+2)\), \(\ln S(t+1)\), \(h_m(t+2)\), \(h_s(t+2)\) and \(h_v(t+2)\) yields that
$$\begin{aligned} f(t)&= E_{t}\Big [\exp \Big \{\phi _2\ln V(t+1) +\phi _4\sum _{k=1}^{t+1} \Lambda (k) +\phi _1\ln S(t+1)+B_0(t+1) \\ &\quad+B_1(t+1)h_m(t+2)+B_2(t+1)h_s(t+2)+B_3(t+1)h_v(t+2)+ B_4(t+1)\Lambda (t+2)\Big \}\Big ]\\ &= E_t\Big [\exp \Big \{ \phi _2\ln V(t)+\phi _2 r-\frac{1}{2}\phi _2 h_v(t+1)+\phi _2\sqrt{h_v(t+1)}Z_v(t+1)\\&-\frac{1}{2}\phi _2\beta _v^2h_m(t+1)+\phi _2\beta _v\sqrt{h_m(t+1)}Z_m(t+1)+\phi _4\sum _{k=1}^{t+1} \Lambda (k)\\ &\quad +\phi _1\ln S(t)+\phi _1 r-\frac{1}{2}\phi _1 h_s(t+1)+\phi _1\sqrt{h_s(t+1)}Z_s(t+1) \\ &\quad-\frac{1}{2}\phi _1\beta _s^2h_m(t+1)+\phi _1\beta _s\sqrt{h_m(t+1)}Z_m(t+1)+ B _0(t+1)\\ \quad&+B_1(t+1)\Big (w_m+b_m h_m(t+1)+a_m(Z_m(t+1)-c_m\sqrt{h_m(t+1)})^2\Big )\\ &\quad+ B_2(t+1)\Big (w_s+b_s h_s(t+1)+a_s(Z_s(t+1)-c_s\sqrt{h_s(t+1)})^2\Big )\\ &\quad+B_3(t+1)\Big (w_v+b_v h_v(t+1)+a_v(Z_v(t+1)-c_v\sqrt{h_v(t+1)})^2\Big )\\ &\quad+ B_4(t+1)\Big (w_{\lambda }+b_{\lambda } \Lambda (t+1)+a_\lambda (Z_{m}(t+1))^2+ c_\lambda (Z_{v}(t+1))^2\Big )\Big \}\Big ]. \end{aligned}$$
Rearranging terms implies that
$$\begin{aligned} f(t)&= E_t\Big [\exp \Big \{ \phi _2\ln V(t)+\phi _4\sum _{k=1}^{t} \Lambda (k) +\phi _1\ln S(t)+ B _0(t+1)+ (\phi _2+\phi _1) r\\ &\quad+w_m B_1(t+1)+ w_sB_2(t+1)+w_vB_3(t+1)+w_{\lambda }B_4(t+1)\\ &\quad+(b_mB_1(t+1)-\frac{1}{2}\phi _2\beta _v^2-\frac{1}{2}\phi _1\beta _s^2) h_m(t+1)\\ &\quad+(b_vB_3(t+1)-\frac{1}{2}\phi _2) h_v(t+1)+(b_sB_2(t+1)-\frac{1}{2}\phi _1) h_s(t+1)\\ &\quad+ (b_{\lambda } B_4(t+1)+\phi _4)\Lambda (t+1) +\Phi _m+\Phi _s+\Phi _v\Big \}\Big ], \end{aligned}$$
where
$$\begin{aligned} \Phi _s&= \phi _1\sqrt{h_s(t+1)}Z_s(t+1)+a_sB_2(t+1)(Z_s(t+1)-c_s\sqrt{h_s(t+1)})^2,\\ \Phi _v&= \phi _2\sqrt{h_v(t+1)}Z_v(t+1)+a_v B_3(t+1)(Z_v(t+1)-c_v\sqrt{h_v(t+1)})^2\\ \quad&+c_\lambda B_4(t+1) (Z_{v}(t+1))^2,\\ \Phi _m&= (\phi _2\beta _v+ \phi _1\beta _s )\sqrt{h_m(t+1)}Z_m(t+1)+ a_mB_1(t+1)(Z_m(t+1)-c_m\sqrt{h_m(t+1)})^2 \\ &\quad+ a_\lambda B_4(t+1)(Z_{m}(t+1))^2. \end{aligned}$$
In order to obtain the explicit expression of f(t), we only need to calculate \(E_t[e^{\Phi _m+\Phi _s+\Phi _v}]=E_t[e^{\Phi _m}]E_t[e^{\Phi _s}]E_t[e^{\Phi _v}] \). Note that \(\Phi _s\), \(\Phi _m\) and \(\Phi _v\) have similar forms and all can be obtained based on the following form,
$$\begin{aligned} E[\exp \{ \mu _1 \sqrt{h}Z +\mu _2(Z-\mu _3\sqrt{h})^2+\mu _4Z^2\}], \end{aligned}$$
where \(\mu _1\), \(\mu _2\), \(\mu _3\) and \(\mu _4\) are all constants and Z is a standard normal variable. Using the fact that \(Ee^{a(Z+b)^2}=e^{-\frac{1}{2}\ln (1-2a)+\frac{ab^2}{1-2a}}\), we have that
$$\begin{aligned}&E[\exp \{ \mu _1 \sqrt{h}Z +\mu _2(Z-\mu _3\sqrt{h})^2+\mu _4Z^2\}] \\ &\quad= E[\exp \{ ( \mu _2 +\mu _4)Z^2-2(\mu _2\mu _3-\mu _1/2 ) Z\sqrt{h}+\mu _2\mu _3^2 h \}] \\ &\quad= E[\exp \{ ( \mu _2 +\mu _4)\Big ( Z- \frac{\mu _2\mu _3-\mu _1/2 }{\mu _2 +\mu _4}\sqrt{h}\Big )^2-\frac{(\mu _2\mu _3-\mu _1/2 )^2}{\mu _2 +\mu _4}h+\mu _2\mu _3^2 h \}] \\ &\quad= e^{\mu _2\mu _3^2h-\frac{(\mu _2\mu _3-\mu _1/2 )^2}{\mu _2 +\mu _4}h} E[\exp \{ ( \mu _2 +\mu _4)\Big ( Z- \frac{\mu _2\mu _3-\mu _1/2 }{\mu _2 +\mu _4}\sqrt{h}\Big )^2 \}] \\ &\quad= \exp \{ \mu _2\mu _3^2h-\frac{(\mu _2\mu _3-\mu _1/2 )^2}{\mu _2 +\mu _4}h-\frac{1}{2}\ln (1-2( \mu _2 +\mu _4 )) +\frac{( \mu _2 +\mu _4 )(\frac{\mu _2\mu _3-\mu _1/2 }{\mu _2 +\mu _4})^2}{1-2( \mu _2 +\mu _4 )}h \} \\ &\quad= \exp \{ -\frac{1}{2}\ln (1-2( \mu _2 +\mu _4 )) +\Big (\mu _2\mu _3^2-\frac{(\mu _2\mu _3-\mu _1/2 )^2}{\mu _2 +\mu _4} +\frac{( \mu _2 +\mu _4 )(\frac{\mu _2\mu _3-\mu _1/2 }{\mu _2 +\mu _4})^2}{1-2( \mu _2 +\mu _4 )}\Big )h \} \\ &\quad= \exp \{ -\frac{1}{2}\ln (1-2( \mu _2 +\mu _4 )) +\Big (\mu _2\mu _3^2 +\frac{2(\mu _2\mu _3-\mu _1/2 )^2}{1-2( \mu _2 +\mu _4 )}\Big )h \}. \end{aligned}$$
(A.1)
Therefore, we can write f(t) in the following form
$$\begin{aligned} f(t)&= \exp \Big \{\phi _2\ln V(t)+\phi _4\sum _{k=1}^{t} \Lambda (k) +\phi _1\ln S(t)+B_0(t) \\ &\quad+B_1(t)h_m(t+1)+B_2(t)h_s(t+1)+B_3(t)h_v(t+1)+B_4(t+1) \Lambda (t+1) \Big \}, \end{aligned}$$
where
$$\begin{aligned} B_0(t)&= B_0(t+1)+ (\phi _2+\phi _1) r +w_m B_1(t+1)+w_sB_2(t+1)+w_vB_3(t+1)+w_{\lambda }B_4(t+1)\\ &\quad-\frac{1}{2}\ln (1-2 (a_mB_1(t+1)+a_\lambda B_4(t+1) ))-\frac{1}{2}\ln (1-2 a_sB_2(t+1)))\\ &\quad-\frac{1}{2}\ln (1-2 (a_vB_3(t+1)+c_\lambda B_4(t+1) )),\\ B_1(t)&= b_mB_1(t+1)-\frac{1}{2}\phi _2\beta _v^2-\frac{1}{2}\phi _1\beta _s^2+ a_mc_m^2B_1(t+1)\\ &\quad+\frac{2(a_mc_mB_1(t+1)-(\phi _2\beta _v+ \phi _1\beta _s )/2)^2 }{1-2(a_mB_1(t+1)+a_\lambda B_4(t+1) )},\\ B_2(t)&= b_sB_2(t+1)-\frac{1}{2}\phi _1 + a_sc_s^2B_2(t+1)+\frac{2(a_sc_sB_2(t+1)- \phi _1/2)^2 }{1-2a_sB_2(t+1)},\\ B_3(t)&= b_vB_3(t+1)-\frac{1}{2}\phi _2+ a_vc_v^2B_3(t+1)+\frac{2(a_vc_vB_3(t+1)- \phi _2/2)^2 }{1-2(a_vB_3(t+1)+c_\lambda B_4(t+1) )},\\ B_4(t)&= b_{\lambda }B_4(t+1)+\phi _4. \end{aligned}$$
Now we need the terminal conditions of \(B_{k}(t),\ k=0,1,2,3,4\) (\(t< j\)). In other words, we need to determine the values of \(B_{k}(j-1),\ k=0,1,2,3,4\). Actually, we already have the expression of f(j) from the case \(j\le t\le T\) we previously considered,
$$\begin{aligned} f(j)&= \exp \Big \{\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) +\phi _1\ln S(j)+A_0(j)\\ &\quad+A_1(j)h_m(j+1)+A_2(j)h_s(j+1)\Big \}. \end{aligned}$$
According to the law of iterated expectations, we have that
$$\begin{aligned} f(j-1)&= E_{j-1}\Big [f(j)\Big ]\\= & {} E_{j-1}\Big [\exp \Big \{\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) +\phi _1\ln S(j)+A_0(j) \\ &\quad+A_1(j)h_m(j+1)+A_2(j)h_s(j+1)\Big \}\Big ]. \end{aligned}$$
Substituting the dynamics of \(\ln V(j)\), \(\ln S(j)\), \(h_m(j+1)\), and \(h_s(j+1)\) and using (A.1) imply that
$$\begin{aligned} f(j-1)&= E_{j-1}\Big [\exp \Big \{\phi _2\ln V(j) +\phi _3 \Lambda (j) +\phi _4\sum _{k=1}^{j-1} \Lambda (k) +\phi _1\ln S(j)+A_0(j) \\ &\quad+A_1(j)h_m(j+1)+A_2(j)h_s(j+1)\Big \}\Big ] \\ &= E_t\Big [\exp \Big \{ \phi _2\ln V(j-1)+\phi _2 r-\frac{1}{2}\phi _2 h_v(j)+\phi _2\sqrt{h_v(j)}Z_v(j)\\ &\quad-\frac{1}{2}\phi _2\beta _v^2h_m(j)+\phi _2\beta _v\sqrt{h_m(j)}Z_m(j)+\phi _3 \Lambda (j)+\phi _4\sum _{k=1}^{j-1} \Lambda (k)\\ &\quad+\phi _1\ln S(j-1)+\phi _1 r-\frac{1}{2}\phi _1 h_s(j)+\phi _1\sqrt{h_s(j)}Z_s(j)\\ &\quad-\frac{1}{2}\phi _1\beta _s^2h_m(j)+\phi _1\beta _s\sqrt{h_m(j)}Z_m(j)+ A_0(j)\\ &\quad+A_1(j)\Big (w_m+b_m h_m(j)+a_m(Z_m(j)-c_m\sqrt{h_m(j)})^2\Big )\\ &\quad+ A_2(j)\Big (w_s+b_s h_s(j)+a_s(Z_s(j)-c_s\sqrt{h_s(j)})^2\Big )\Big \}\Big ]\\ &= \exp \Big \{\phi _2\ln V(j-1)+\phi _4\sum _{k=1}^{j-1} \Lambda (k) +\phi _1\ln S(j-1)+B_0(j-1) \\ &\quad+B_1(j-1)h_m(j)+B_2(j-1)h_s(j)+B_3(j-1)h_v(j)+B_4(j-1) \Lambda (j) \Big \}, \end{aligned}$$
where
$$\begin{aligned} B_0(j-1)&= A_0(j)+ (\phi _2+\phi _1) r +w_m A_1(j)+w_sA_2(j)\\ &\quad-\frac{1}{2}\ln (1-2 a_mA_1(j))-\frac{1}{2}\ln (1-2 a_sA_2(j)),\\ B_1(j-1)&= b_mA_1(j)-\frac{1}{2}\phi _2\beta _v^2-\frac{1}{2}\phi _1\beta _s^2+ a_mc_m^2A_1(j) +\frac{2(a_mc_mA_1(j)-(\phi _2\beta _v+ \phi _1\beta _s )/2)^2 }{1-2a_mA_1(j)},\\ B_2(j-1)&= b_sA_2(j)-\frac{1}{2}\phi _1 + a_sc_s^2A_2(j)+\frac{2(a_sc_sA_2(j)- \phi _1/2)^2 }{1-2a_sA_2(j)},\\ B_3(j-1)&= -\frac{1}{2}\phi _2+\frac{1}{2}\phi _2^2,\\ B_4(j-1)&= \phi _3. \end{aligned}$$
This completes the proof of the proposition. \(\square \)
Proof of Theorem 2.2
First, we deal with the term \(E\Big [(S(T)-K)^+\Big ]\). Recall the definition of \(f(t;\phi _1,\phi _2,\phi _3,\phi _4)\) and note that \(f(0;i\phi _1,0,0,0)\) is the characteristic function of \(\ln S(T)\) under Q. From standard probability theory (see, e.g., Kendall and Stuart 1977), we can obtain the distribution function of \(\ln S(T)\), that is,
$$\begin{aligned} Q(\ln S(T)\le x)=\frac{1}{2}-\frac{1}{\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 x}f(0;i\phi _1,0,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1, \end{aligned}$$
which in turn implies that
$$\begin{aligned} Q(\ln S(T)\ge x)&= 1- Q(\ln S(T)\le x) \\ &= \frac{1}{2}+\frac{1}{\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 x}f(0;i\phi _1,0,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1. \end{aligned}$$
(A.2)
The term \(E\Big [(S(T)-K)^+\Big ]\) can be derived after introducing a new probability measure \(Q_1\) defined by
$$\begin{aligned} Q_1(O)=\frac{E\Big [I(O)S(T)\Big ]}{E\Big [S(T)\Big ]}, \end{aligned}$$
for any event \(O\in \mathscr {F}_{T}\). Obviously, the characteristic function of \(\ln S(T)\) under \(Q_1\) is given by
$$\begin{aligned} E^{Q_1}\Big [e^{i\phi _1 \ln S(T)}\Big ]= \frac{f(0;1+i\phi _1,0,0,0)}{f(0;1,0,0,0)}. \end{aligned}$$
In addition, under \(Q_1\), it holds that
$$\begin{aligned} Q_1(\ln S(T)\ge x)= & {} \frac{1}{2}+\frac{1}{\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 x}f(0;1+i\phi _1,0,0,0)/f(0;1,0,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1. \end{aligned}$$
(A.3)
Hence, (A.2) and (A.3) imply that
$$\begin{aligned} E\Big [(S(T)-K)^+\Big ]&= E\Big [(S(T)-K)^+\Big ] \\ &= E\Big [(S(T)-K)I(\ln S(T)\ge \ln K)\Big ]\\ &= E\Big [S(T)I(\ln S(T)\ge \ln K)\Big ] -K E\Big [I(\ln S(T)\ge \ln K)\Big ] \\ &= E[S(T)] E^{Q_1}\Big [I(\ln S(T)\ge \ln K)\Big ] -K E\Big [I(\ln S(T)\ge \ln K)\Big ] \\ &= E[S(T)]Q_1(\ln S(T)\ge \ln K) -K Q(\ln S(T)\ge \ln K) \Big ) \\ &= \frac{1}{2}f(0;1,0,0,0)+\frac{1}{\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;1+i\phi _1,0,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1 \\ &\quad- \frac{K}{2}-\frac{K}{\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1,0,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1, \end{aligned}$$
(A.4)
where in the last equality we used (A.2) and (A.3).
Next, we focus on the term \(E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( V(j)<L) (S(T)-K)^+ \Big ]\). We rewrite it as follows:
$$\begin{aligned}&E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( V(j)<L) (S(T)-K)^+ \Big ]\\ &\quad= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}I( V(j)<L, \ln S(T)\ge \ln K)\Big ] \\ &\quad-K E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( V(j)<L, \ln S(T)\ge \ln K)\Big ] \\ &= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\quad-K E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( -\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ]. \end{aligned}$$
In the following, we deal with the two parts in the above equality separately. To this end, we define a new probability measure
$$\begin{aligned} Q_2(O)=\frac{E\Big [I(O)e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]}, \end{aligned}$$
for any event \(O\in \mathscr {F}_{T}\). Under \(Q_2\), we have the joint characteristic function of \(-\ln V(j)\) and \(\ln S(T)\) as follows:
$$\begin{aligned} E^{Q_2}\Big [e^{i\phi _2(-\ln V(j))+ i\phi _1\ln S(T)}\Big ]&= E\Big [ \frac{e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]} e^{i\phi _2(-\ln V(j))+ i\phi _1\ln S(T)}\Big ] \\ &= \frac{1}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]} E\Big [e^{(i\phi _1+1)\ln S(T) -i\phi _2\ln V(j)-\sum _{k=1}^{j} \Lambda (k) }\Big ] \\ &= \frac{f(0;i\phi _1+1,-i\phi _2,-1,-1)}{f(0;1,0,-1,-1)}. \end{aligned}$$
By inverting the characteristic function, we have that
$$\begin{aligned} \Pi _{j,1}:= & {} E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\= & {} E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]E^{Q_2}\Big [I(-\ln V(j)>-\ln L, \ln S(T)\ge \ln K)\Big ]\nonumber \\= & {} f(0;1,0,-1,-1)Q_2\Big (-\ln V(j)> -\ln L, \ln S(T)\ge \ln K\Big )\nonumber \\= & {} \frac{1}{4}f(0;1,0,-1,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1+1,0,-1,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&\ +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;1,-i\phi _2,-1,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&\ -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1+1,-i\phi _2,-1,-1)}{ \phi _1 \phi _2}\Big ]\nonumber \\&\ -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1+1,i\phi _2,-1,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$
(A.5)
Likewise, we work under \(\bar{Q}_2\) defined by
$$\begin{aligned} \bar{Q}_2(O)=\frac{E\Big [I(O)e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}\Big ]}, \end{aligned}$$
for any event \(O\in \mathscr {F}_{T}\), and obtain that
$$\begin{aligned} \Pi _{j,2}&:= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &= f(0;0,0,-1,-1)\bar{Q}_2\Big (-\ln V(j)> -\ln L, \ln S(T)\ge \ln K\Big ) \\ &= \frac{1}{4}f(0;0,0,-1,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1,0,-1,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1 \\ &\quad +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;0,-i\phi _2,-1,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2 \\ &\quad -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1,-i\phi _2,-1,-1)}{ \phi _1 \phi _2}\Big ] \\ &\quad -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1,i\phi _2,-1,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$
(A.6)
Hence, it holds that
$$\begin{aligned}&E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( V(j)<L) (S(T)-K)^+ \Big ]\\ \quad= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\qquad-K E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( -\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\quad= \Pi _{j,1} -K\Pi _{j,2}. \end{aligned}$$
Similarly,
$$\begin{aligned}&E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)}I( V(j)<L) (S(T)-K)^+ \Big ]\\ \quad&= E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)+\ln S(T)}I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\qquad-K E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)}I( -\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\quad= \Pi _{j,3} -K\Pi _{j,4}, \end{aligned}$$
where
$$\begin{aligned} \Pi _{j,3}&:= \frac{1}{4}f(0;1,0,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1+1,0,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1 \\ &\quad +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;1,-i\phi _2,0,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2 \\ &\quad -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1+1,-i\phi _2,0,-1)}{ \phi _1 \phi _2}\Big ] \\ &\quad -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1+1,i\phi _2,0,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2, \end{aligned}$$
(A.7)
and
$$\begin{aligned} \Pi _{j,4}&:= \frac{1}{4}f(0;0,0,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1,0,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1 \\ &\quad +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;0,-i\phi _2,0,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2 \\ &\quad -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1,-i\phi _2,0,-1)}{ \phi _1 \phi _2}\Big ] \\ &\quad -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1,i\phi _2,0,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$
(A.8)
Note that \(\Pi _{j,3}\) and \(\Pi _{j,4}\) have similar forms as \(\Pi _{j,1}\) and \(\Pi _{j,2}\), and they can be obtained by replacing \(f(0;\cdot ,\cdot ,-1,\cdot )\) in \(\Pi _{j,1}\) and \(\Pi _{j,2}\) with \(f(0;\cdot ,\cdot ,0,\cdot )\), respectively.
We next calculate \(E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( V(j)<L) V(j) (S(T)-K)^+ \Big ]\). Note that \(E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)}I( V(j)<L) V(j) (S(T)-K)^+ \Big ]\) can be calculated in a similar way. To this end, define another probability measure \(Q_3\) as follows:
$$\begin{aligned} Q_3(O)=\frac{E\Big [I(O)e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)+\ln V(j)}\Big ]}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)+\ln V(j)}\Big ]}, \end{aligned}$$
for any event \(O\in \mathscr {F}_{T}\). The joint characteristic function of \(-\ln V(j)\) and \(\ln S(T)\) under \(Q_3\) is
$$\begin{aligned} E^{Q_3}\Big [e^{i\phi _2(-\ln V(j))+ i\phi _1\ln S(T)}\Big ]&= E\Big [ \frac{e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)+\ln V(j)}}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)+\ln V(j)}\Big ]} e^{i\phi _2(-\ln V(j))+ i\phi _1\ln S(T)}\Big ] \\ &= \frac{f(0;i\phi _1+1,-i\phi _2+1,-1,-1)}{f(0;1,1,-1,-1)}. \end{aligned}$$
By inverting the characteristic function, we obtain that
$$\begin{aligned} \Pi _{j,5}&:= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)}V(j)I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)+\ln V(j)}\Big ]E^{Q_3}\Big [I(-\ln V(j)>-\ln L, \ln S(T)\ge \ln K)\Big ] \\ &= \frac{1}{4}f(0;1,1,-1,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1+1,1,-1,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1 \\ &\quad +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;1,-i\phi _2+1,-1,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2 \\ &\quad -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1+1,-i\phi _2+1,-1,-1)}{ \phi _1 \phi _2}\Big ] \\ &\quad -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1+1,i\phi _2+1,-1,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2, \end{aligned}$$
(A.9)
and
$$\begin{aligned} \Pi _{j,6}&:= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}V(j)I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln V(j)}\Big ]E^{\bar{Q}_3}\Big [I(-\ln V(j)>-\ln L, \ln S(T)\ge \ln K)\Big ] \\ &= \frac{1}{4}f(0;0,1,-1,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1,1,-1,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1 \\ &\quad +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;0,-i\phi _2+1,-1,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2 \\ &\quad -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1,-i\phi _2+1,-1,-1)}{ \phi _1 \phi _2}\Big ] \\ &\quad -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1,i\phi _2+1,-1,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2, \end{aligned}$$
(A.10)
where \(\bar{Q}_3(O)\) is defined by
$$\begin{aligned} \bar{Q}_3(O)=\frac{E\Big [I(O)e^{-\sum _{k=1}^{j} \Lambda (k)+\ln V(j)}\Big ]}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln V(j)}\Big ]}, \end{aligned}$$
for any event \(O\in \mathscr {F}_{T}\). Therefore, we have that
$$\begin{aligned}&E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)}I( V(j)<L) V(j) (S(T)-K)^+ \Big ] \\ &\quad= E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)} V(j)I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\quad -K E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)} V(j)I( -\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ]\\ &= \Pi _{j,5} -K\Pi _{j,6}. \end{aligned}$$
Finally, we calculate \(E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)}I( V(j)<L) V(j) (S(T)-K)^+ \Big ]\) under \(Q_4\) and \(\bar{Q}_4\) defined by
$$\begin{aligned} Q_4(O)&= \frac{E\Big [I(O)e^{-\sum _{k=1}^{j-1} \Lambda (k)+\ln S(T)+\ln V(j)}\Big ]}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln S(T)+\ln V(j)}\Big ]},\\ \bar{Q}_4(O)&= \frac{E\Big [I(O)e^{-\sum _{k=1}^{j-1} \Lambda (k)+\ln V(j)}\Big ]}{E\Big [e^{-\sum _{k=1}^{j} \Lambda (k)+\ln V(j)}\Big ]}, \end{aligned}$$
for any event \(O\in \mathscr {F}_{T}\). Following along similar arguments we obtain
$$\begin{aligned}&E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)}I( V(j)<L) V(j) (S(T)-K)^+ \Big ] \\ &= E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)+\ln S(T)} V(j)I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &\quad-K E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)} V(j)I( -\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ]\\ &= \Pi _{j,7} -K\Pi _{j,8}, \end{aligned}$$
where
$$\begin{aligned} \Pi _{j,7}&:= E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)+\ln S(T)}V(j)I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ] \\ &=\frac{1}{4}f(0;1,1,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1+1,1,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\\ &\quad +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;1,-i\phi _2+1,0,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2 \\ &\quad -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1+1,-i\phi _2+1,0,-1)}{ \phi _1 \phi _2}\Big ] \\ &\quad -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1+1,i\phi _2+1,0,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$
(A.11)
and
$$\begin{aligned} \Pi _{j,8}:= & {} E\Big [e^{-\sum _{k=1}^{j-1} \Lambda (k)}V(j)I(-\ln V(j)> -\ln L, \ln S(T)\ge \ln K)\Big ]\nonumber \\= & {} \frac{1}{4}f(0;0,1,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K}f(0;i\phi _1,1,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&\ +\frac{1}{2\pi }\int _{0}^\infty \mathrm{Re}\Big [\frac{e^{i \phi _2 \ln L}f(0;0,-i\phi _2+1,0,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&\ -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;i\phi _1,-i\phi _2+1,0,-1)}{ \phi _1 \phi _2}\Big ]\nonumber \\&\ -\mathrm{Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;i\phi _1,i\phi _2+1,0,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$
(A.12)
Note that \(\Pi _{j,7}\) and \(\Pi _{j,8}\) can be obtained by replacing \(f(0;\cdot ,\cdot ,-1,\cdot )\) in \(\Pi _{j,5}\) and \(\Pi _{j,6}\) with \(f(0;\cdot ,\cdot ,0,\cdot )\), respectively. This completes the proof of the theorem. \(\square \)
