Appendix
Proof of Proposation 1
1) First, notice that
$$\begin{aligned} X(T)&=X(t)+\mu (T-t)+\sigma B_1(T-t),\\ Y(T)&=e^{-\beta (T-t)}Y(t)+\int _t^T e^{-\beta (T-s)} dL_1(s),\\ U(T)&=U(t)+\alpha (T-t)+\xi B_2(T-t), \end{aligned}$$
and
$$\begin{aligned} V(T)=e^{-\kappa (T-t)}V(t)+\int _t^T e^{-\kappa (T-s)} dL_2(s). \end{aligned}$$
Due to the \({\mathcal {F}}(t)\)-adaptedness of X(t), Y(t), U(t), and V(t), the independent increment property of Lévy processes and the independence between \(B_i\) and \(L_i\), for \(i=1, 2\), we obtain
$$\begin{aligned} F(t,T)&=F^{EL}(t,T)-h_R F^{GA}(t,T)\\&={\mathbb {E}}^{{\mathbb {Q}}}[S^{EL}(T)|{\mathcal {F}}(t)]-h_R{\mathbb {E}}^{{\mathbb {Q}}}[S^{GA}(T)|{\mathcal {F}}(t)]\\&=\varLambda ^{EL}(T){\mathbb {E}}^{{\mathbb {Q}}}[\exp \{X(T)+Y(T)\}|{\mathcal {F}}(t)]\\&\quad -h_R\varLambda ^{GA}(T){\mathbb {E}} ^{{\mathbb {Q}}}[\exp \{U(T)+V(T)\}|{\mathcal {F}}(t)]\\&=\varLambda ^{EL}(T) \exp \{X(t)+\mu (T-t)\\&\quad +e^{-\beta (T-t)}Y(t)\}{\mathbb {E}}^{{\mathbb {Q}}}\big [\exp \{\sigma B_1(T-t)\}\big ] {\mathbb {E}}^{{\mathbb {Q}}} \big [\exp \{\int _t^T e^{-\beta (T-s)} dL_1(s)\}\big ] \\&\quad -h_R\varLambda ^{GA}(T)\exp \{U(t)+\alpha (T-t)\\&\quad +e^{-\kappa (T-t)}V(t)\}{\mathbb {E}}^{{\mathbb {Q}}}\big [\exp \{\xi B_2(T-t)\}\big ] {\mathbb {E}}^{{\mathbb {Q}}}\big [\exp \{\int _t^T e^{-\kappa (T-s)} dL_2(s)\}\big ]\\&=\varLambda ^{EL}(T)\exp \Big \{\mu (T-t)+\frac{1}{2}\sigma ^2(T-t)\\&\quad +\int _t^T\phi _1(e^{-\beta (T-s)})ds+X(t)+e^{-\beta (T-t)}Y(t)\Big \}\\&\quad -h_R\varLambda ^{GA}(T)\exp \Big \{\alpha (T-t)\\&\quad +\frac{1}{2}\xi ^2(T-t)+\int _t^T\phi _2(e^{-\kappa (T-s)})ds+U(t)+e^{-\kappa (T-t)}V(t)\Big \}. \end{aligned}$$
This proves the result of first part.
II) Applying Itô’s formula to\(F^{EL}(t,T)\) and \(F^{GA}(t,T)\), gives
$$\begin{aligned} \frac{dF^{EL}(t,T)}{F^{EL}(t^-,T)}&=-\phi _1(e^{-\beta (T-t)})dt+\sigma dB_1(t)+\int _{{\mathbb {R}}}\big (\exp \{e^{-\beta (T-t)} z_1\}-1\big ){\bar{N}}_1(dt,dz_1)\\&\qquad +\int _{|z_1|\le 1} \big (\exp \{e^{-\beta (T-t)} z_1\}-1-e^{-\beta (T-t)}z_1\big )\ell _1(dz_1) dt \Big ),\\ \frac{dF^{GA}(t,T)}{F^{GA}(t^-,T)}&=-\phi _2(e^{-\kappa (T-t)})dt+\xi dB_2(t)+\int _{{\mathbb {R}}}\big (\exp \{e^{-\kappa (T-t)} z_2\}-1\big ){\bar{N}}_2(dt,dz_2) \\&\qquad +\int _{|z_2|\le 1} \big (\exp \{e^{-\kappa (T-t)} z_2\}-1-e^{-\kappa (T-t)}z_2\big )\ell _2(dz_2) dt. \end{aligned}$$
According to \(F(t^-,T)=F^{EL}(t^-,T)-h_R F^{GA}(t^-,T)\), the second part of proposition is established. \(\square \)
Proof of Theorem 2
The uniqueness follows from the Local Lipschitz property (16) and Itô isometry. Let \(F_1(t,T)\) and \(F_2(t,T)\) be solutions with initial values \(Z, {\bar{Z}}\) respectively. For our purposes here we only need the case \(Z={\bar{Z}}\).
Put \({\mathscr {A}}_i(s,T)=A_i(s,F_1(s,T))-A_i(s,F_2(s,T)), {\mathscr {C}}_i(s,T)=C_i(s,F_1(s,T))-C_i(s,F_2(s,T))\) and \({\mathscr {D}}_i(s,z_i,T)=D_i(s,z_i,F_1(s,T))-D_i(s,z_i,F_2(s,T))\), for \(i=1,2\). Then
$$\begin{aligned}&{\mathbb {E}}^{{\mathbb {Q}}}[|F_1(t,T)-F_2(t,T)|^2]\\&\quad ={\mathbb {E}}^{{\mathbb {Q}}}\Big [\Big (Z-{\bar{Z}}+ \sum _{i=1}^{2}\int _0^t {\mathscr {A}}_i ds+\sum _{i=1}^{2}\int _0^t {\mathscr {C}}_i dB_i(s)+\sum _{i=1}^{2}\int _0^t {\mathscr {D}}_i {\bar{N}}(ds,dz_i)\Big )^2\Big ]\\&\quad \le 4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big |Z-{\bar{Z}}\big |^2\big ]+4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big (\sum _{i=1}^2 \int _0^t{\mathscr {A}}_ids\big )^2\big ]\\&\qquad +4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big (\sum _{i=1}^2 \int _0^t{\mathscr {C}}_idB_i(s)\big )^2\big ]+4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big (\sum _{i=1}^2 \int _0^t\int _{{\mathbb {R}}}{\mathscr {D}}_i{\bar{N}}(ds,dz_i)\big )^2\big ]\\&\quad \le 4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big |Z-{\bar{Z}}\big |^2\big ]+8\sum _{i=1}^2{\mathbb {E}}^{{\mathbb {Q}}}\big [\big ( \int _0^t{\mathscr {A}}_ids\big )^2\big ]\\&\qquad +8\sum _{i=1}^2{\mathbb {E}}^{{\mathbb {Q}}}\big [\big ( \int _0^t{\mathscr {C}}_idB_i(s)\big )^2\big ]+8\sum _{i=1}^2 {\mathbb {E}}^{{\mathbb {Q}}}\big [\big (\int _0^t\int _{{\mathbb {R}}}{\mathscr {D}}_i{\bar{N}}(ds,dz_i)\big )^2\big ]\\&\quad \le 4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big |Z-{\bar{Z}}\big |^2\big ]+8t\sum _{i=1}^2{\mathbb {E}}^{{\mathbb {Q}}}\big [ \int _0^t{\mathscr {A}}_i^2ds\big ]\\&\qquad +8\sum _{i=1}^2{\mathbb {E}}^{{\mathbb {Q}}}\big [ \int _0^t{\mathscr {C}}_i^2ds\big ]+8\sum _{i=1}^2 {\mathbb {E}}^{{\mathbb {Q}}}\big [\int _0^t\int _{{\mathbb {R}}}{\mathscr {D}}_i^2\ell _i(dz_i)ds\big ]\\&\quad \le 4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big |Z-{\bar{Z}}\big |^2\big ]+8(t+2)K^2\int _0^t {\mathbb {E}}^{{\mathbb {Q}}}\big [|F_1(s,T)-F_2(s,T)|^2\big ]ds. \end{aligned}$$
Then, the following function
$$\begin{aligned} \nu (t)={\mathbb {E}}^{{\mathbb {Q}}}\big [\big |F_1(t,T)-F_2(t,T)\big |^2\big ], \quad 0\le t \le T, \end{aligned}$$
satisfies
$$\begin{aligned} \nu (t)\le \gamma +\chi \int _0^t\nu (s)ds, \end{aligned}$$
where
$$\begin{aligned} \gamma =4{\mathbb {E}}^{{\mathbb {Q}}}\big [\big |Z-{\bar{Z}}\big |^2\big ] ~~~\textit{and}~~~ \chi =8(t+2)K^2. \end{aligned}$$
Applying Gronwall inequality, gives
$$\begin{aligned} \nu (t)\le \gamma e^{\chi t} \end{aligned}$$
By continuity of \(t\longrightarrow \big |F_1(t,T)-F_2(t,T)\big |\) it follows that
$$\begin{aligned} {\mathbb {P}}^{\mathbb {Q}}\big (\big |F_1(t,T)-F_2(t,T)\big |=0~~~~\textit{for all}~~ t\in [0,T]\big )=1, \end{aligned}$$
and the uniqueness of solution is proved. \(\square \)
Proof of Theorem 3
First, from Proposition 1, we have
$$\begin{aligned} F(\tau ,T)&= \frac{F^{EL}(0,T)}{F^{EL}(0,T)}\varLambda ^{EL}(T)\exp \Big \{\mu (T-\tau )+\frac{1}{2}\sigma ^2(T-\tau )\\&\quad +\int _\tau ^T\phi _1(e^{-\beta (T-s)})ds+X(\tau )+e^{-\beta (T-\tau )}Y(\tau )\Big \} \\&\quad -h_R\frac{F^{GA}(0,T)}{F^{GA}(0,T)}\varLambda ^{GA}(T)\exp \Big \{\alpha (T-\tau )+\frac{1}{2}\xi ^2(T-\tau )\\&\quad +\int _\tau ^T\phi _2(e^{-\kappa (T-s)})ds+U(\tau )+e^{-\kappa (T-\tau )}V(\tau )\Big \}. \end{aligned}$$
But,
$$\begin{aligned} e^{-\beta (T-\tau )}Y(\tau )&= e^{-\beta T} y+\int _0^\tau e^{-\beta (T-s)} dL_1(s),\\ e^{-\kappa (T-\tau )}V(\tau )&= e^{-\kappa T} v+\int _0^\tau e^{-\kappa (T-s)} dL_2(s),\\ X(\tau )&=x+\mu \tau +\sigma B_1(\tau ),\\ U(\tau )&=u+\alpha \tau +\xi B_2(\tau ). \end{aligned}$$
Therefore, we can write
$$\begin{aligned} F(\tau ,T)&= Z_1(f^{EL})e^{\sigma B_1(\tau )-\frac{1}{2}\sigma ^2 \tau }-Z_2(f^{GA})e^{\xi B_2(\tau )-\frac{1}{2}\xi ^2 \tau }. \end{aligned}$$
Then the spark-spread option price is expressed as follows
$$\begin{aligned}&C(0,\tau ,T,f^{EL},f^{GA}) =e^{-r T}{\mathbb {E}}^{\mathbb {Q}}\Big [\max (F(\tau ,T),0)\Big ]\\&\quad =e^{-r T}{\mathbb {E}}^{\mathbb {Q}}\Big [{\mathbb {E}}^{\mathbb {Q}}\big [\max \Big (Z_1(f^{EL})e^{\sigma B_1(\tau )-\frac{1}{2}\sigma ^2 \tau }-Z_2(f^{GA})e^{\xi B_2(\tau )-\frac{1}{2}\xi ^2 \tau }, 0\Big )\big |{\mathcal {F}}^Z(f^{EL},f^{GA})\big ]\Big ] . \end{aligned}$$
where \({\mathcal {F}}^Z:={\mathcal {F}}^Z(f^{EL},f^{GA})=\sigma \big ((Z_1, Z_2)_t,~ t\le \tau \big )\). Since \(L =(L_1, L_2)'\) is independent of \(B=(B_1, B_2)'\), \({\mathcal {F}}^Z\) is independent of B. Hence the inner expectation can be computed as follows
$$\begin{aligned}&\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{+\infty } \max \Big (Z_1(f^{EL})e^{\sigma \sqrt{\tau }g-\frac{1}{2}\sigma ^2\tau }-Z_2(f^{GA})e^{\xi \sqrt{\tau }g-\frac{1}{2}\xi ^2\tau }, 0\Big )e^{-\frac{1}{2}g^2}dg\\&=\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{d(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s))}Z_1(f^{EL})e^{-\frac{1}{2}(g+\sigma \sqrt{\tau })^2}dg\\&\quad -\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{d(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s))} Z_2(f^{GA})e^{-\frac{1}{2}(g+\xi \sqrt{\tau })^2}dg\\&=Z_1(f^{EL})\varPhi (d(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s))+\sigma \sqrt{\tau }) \\&\quad -Z_2(f^{GA})\varPhi (d(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s))+\xi \sqrt{\tau }), \end{aligned}$$
where
$$\begin{aligned}&d(f^{EL},f^{GA},\omega _1,\omega _2)=\Big (\ln \big (\frac{f^{EL}}{h_R f^{GA}}\big )-\frac{1}{2}(\sigma ^2-\xi ^2)\tau \\&\quad -\int _0^{\tau } \big (\phi _1(e^{-\beta (T-s)})-\phi _2(e^{-\kappa (T-s)})\big )ds+\omega _1-\omega _2\Big )\Big ((\xi -\sigma )\sqrt{\tau }\Big )^{-1}. \end{aligned}$$
\(\square \)
Proof of Theorem 4
We have
$$\begin{aligned}&\complement (0,\tau ,T,f^{EL},f^{GA})=e^{-rT}{\mathbb {E}}^{\mathbb {Q}}\big [\max (F(\tau ,T),0)\big ]\\&\quad =\frac{e^{-rT}}{\sqrt{2\pi }}\int _{-\infty }^\infty \big (\max (f^{EL} e^{\sigma \sqrt{\tau }g-\frac{1}{2}\sigma ^2\tau }-h_R f^{GA} e^{\xi \sqrt{\tau }g-\frac{1}{2}\xi ^2\tau },0)\big )e^{-\frac{1}{2}g^2}\\&\quad =\frac{e^{-rT}}{\sqrt{2\pi }}\int _{-\infty }^{d(f^{EL},f^{GA})}\big (f^{EL} e^{-\frac{1}{2}(g+\sigma \sqrt{\tau })^2}-h_R f^{GA} e^{-\frac{1}{2}(g+\xi \sqrt{\tau })^2}\big )dg\\&\quad =e^{-rT}\Big (f^{EL}\varPhi (d(f^{EL},f^{GA})+\sigma \sqrt{\tau }))-h_Rf^{GA}\varPhi (d(f^{EL},f^{GA})+\xi \sqrt{\tau }))\Big ), \end{aligned}$$
where
$$\begin{aligned} d(f^{EL},f^{GA})=\frac{\ln (\frac{f^{EL}}{h_R f^{GA}})-\frac{1}{2}\tau (\sigma ^2-\xi ^2)}{(\xi -\sigma )\sqrt{\tau }}. \end{aligned}$$
\(\square \)
Proof of Lemma 1
We have
$$\begin{aligned} d_1(f^{EL},f^{GA},0,0)&= d_1(f^{EL},f^{GA})+\frac{1}{(\sigma -\xi )\sqrt{\tau }}\int _0^\tau (\phi _1(e^{-\beta (T-s)}-\phi _2(e^{-\kappa (T-s)}))ds. \end{aligned}$$
Using the mean value theorem, there is a random variable g such that
$$\begin{aligned} \varPhi (d_1(f^{EL},f^{GA},0,0))-\varPhi (d_1(f^{EL},f^{GA}))&= \varPhi '(g)(d_1(f^{EL},f^{GA},0,0)- d_1(f^{EL},f^{GA})). \end{aligned}$$
But,
$$\begin{aligned} \varPhi '(g)&=\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}g^2}\le \frac{1}{\sqrt{2\pi }}<1. \end{aligned}$$
Then
$$\begin{aligned}&\varPhi (d_1(f^{EL},f^{GA},0,0))-\varPhi (d_1(f^{EL},f^{GA}))<d_1(f^{EL},f^{GA},0,0)-d_1(f^{EL},f^{GA})\nonumber \\&\quad =\frac{1}{(\sigma -\xi )\sqrt{\tau }}\int _0^\tau (\phi _1(e^{-\beta (T-s)}-\phi _2(e^{-\kappa (T-s)}))ds. \end{aligned}$$
(26)
Now, we evaluate the log-moment generating function \(\phi _1(e^{-\beta (T-s)}\). For notational simplicity, let \(\gamma (s) = e^{-\beta (T - s)}\). From Eq. (7), we have
$$\begin{aligned} \phi _1(e^{-\beta (T-s)})=\ln {\mathbb {E}}[e^{\gamma (s)L_1(1)}]. \end{aligned}$$
Applying Itô’s formula to \(e^{\gamma (s)L_1(1)}\), gives
$$\begin{aligned} e^{\gamma (s)L_1(1)}&=\int _{|z_1|<1}(e^{\gamma (s)z_1}-1-z_1\gamma (s))\ell _1(dz_1)+\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1){\bar{N}}_1(dz_1)\\&=\int _{|z_1|<1}(e^{\gamma (s)z_1}-1-z_1\gamma (s))\ell _1(dz_1)\\&\qquad +\int _{|z_1|<1}(e^{\gamma (s)z_1}-1)(N_1(dz_1)-\ell _1(dz_1))+\int _{|z_1|>1}(e^{\gamma (s)z_1}-1)N_1(dz_1)\\&=\int _{|z_1|<1}-z_1\gamma (s)\ell _1(dz_1)+\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1)N_1(dz_1)\\&=\int _{|z_1|<1}-z_1\gamma (s)\ell _1(dz_1)+\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1)(\overset{\sim }{N}_1(dz_1)+\ell _1(dz_1))\\&=\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1-z_1\gamma (s)1_{|z_1|<1})\ell _1(dz_1)+\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1)\overset{\sim }{N}_1(dz_1) \end{aligned}$$
Note that \(\overset{\sim }{N}_1\) is a martingale process. Thus we have
$$\begin{aligned} {\mathbb {E}}[e^{\gamma (s)L_1(1)}]=\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1-z_1\gamma (s)1_{|z_1|<1})\ell _1(dz_1). \end{aligned}$$
By definition of the log-moment generating function
$$\begin{aligned} \phi _1(e^{-\beta (T-s)})=\ln \big (\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1-z_1\gamma (s)1_{|z_1|<1})\ell _1(dz_1)\big ). \end{aligned}$$
(27)
For \(|z_1| \ge 1\), we obtain
$$\begin{aligned} |e^{\gamma (s)z_1}-1|&\le \sum _{n=1}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\\&=\gamma (s)|z_1| \sum _{n=2}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\\&\le \gamma (s)|z_1| \sum _{n=0}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\\&=\gamma (s)|z_1| e^{\gamma (s)|z_1|}. \end{aligned}$$
If \(|z_1| < 1\), the series representation of the exponential function gives
$$\begin{aligned} |e^{\gamma (s)z_1}-1-z_1\gamma (s)|&\le \sum _{n=2}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\\&\le \gamma ^2(s)|z_1|^2\sum _{n=3}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\\&\le \gamma ^2(s)|z_1|^2\sum _{n=0}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\\&\le \gamma ^2(s)|z_1|^2\sum _{n=0}^{\infty }\frac{(\gamma (s))^n}{n!}\\&= \gamma ^2(s)|z_1|^2e^{\gamma (s)}. \end{aligned}$$
Hence, using the definition of \(\gamma (s),\)
$$\begin{aligned}&\int _{{\mathbb {R}}}\Big |e^{\gamma (s)z_1}-1-z_1\gamma (s)1_{|z_1|<1}\Big |\ell _1(dz_1)\\&\quad =\int _{|z_1|\ge 1}\Big |e^{\gamma (s)z_1}-1\Big |\ell _1(dz_1)+\int _{|z_1|<1}\Big |e^{\gamma (s)z_1}-1-z_1\gamma (s)1_{|z_1|<1}\Big |\ell _1(dz_1)\\&\quad \le \gamma (s)\int _{|z_1|\ge 1}|z_1|e^{\gamma (s)|z_1|}\ell _1(dz_1)+\gamma ^2(s)e^{\gamma (s)}\int _{|z_1|<1}z_1^2\ell _1(dz_1)\\&\quad =e^{-\beta (T-s)}\int _{|z_1|\ge 1}|z_1|e^{e^{-\beta (T-s)}|z_1|}\ell _1(dz_1)+e^{-2\beta (T-s)}e^{e^{-\beta (T-s)}}\int _{|z_1|<1}z_1^2\ell _1(dz_1)\\&\quad \le e^{-\beta (T-s)}\int _{|z_1|\ge 1}e^{2|z_1|}\ell _1(dz_1)+e^{-2\beta (T-s)+1}\int _{|z_1|<1}z_1^2\ell _1(dz_1)\\&\quad \le e^{-\beta (T-s)}\Big (\int _{|z_1|\ge 1}e^{2|z_1|}\ell _1(dz_1)+e^{1}\int _{|z_1|<1}z_1^2\ell _1(dz_1)\Big ). \end{aligned}$$
From Eq. (27), we have
$$ \begin{gathered} \phi _{1} \left( {e^{{ - \beta (T - s)}} } \right) \le \ln \left( {\int_{\mathbb{R}} | e^{{\gamma (s)z_{1} }} - 1 - z_{1} \gamma (s)1_{{|z_{1} | < 1}} |\ell \left( {dz_{1} } \right)} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; \le \ln \left( {e^{{ - \beta (T - s)}} \left( {\int_{{|z_{1} | \ge 1}} {e^{{2|z_{1} |}} } \ell _{1} (dz_{1} ) + e^{1} \int_{{|z_{1} | < 1}} {_{1}^{2} \ell \left( {dz_{1} } \right)} } \right)} \right)\, \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; \le e^{{ - \beta (T - s)}} \left( {\int_{{|z_{1} | \ge 1}} {e^{{2|z_{1} |}} } \ell _{1} (dz_{1} ) + e^{1} \int_{{|z_{1} | < 1}} {z_{1}^{2} \ell (dz_{1} )} } \right) \hfill \\ \end{gathered} $$
Therefore, there exists a constant \(C > 0\), such that
$$\begin{aligned} \phi _1(e^{-\beta (T-s)})\le C e^{-\beta (T-s)}. \end{aligned}$$
Similarly, we get
$$\begin{aligned} \phi _2(e^{-\kappa (T-s)})\le C e^{-\kappa (T-s)}. \end{aligned}$$
By absolute value inequalities
$$\begin{aligned} \Big |\phi _1(e^{-\beta (T-s)})-\phi _2(e^{-\kappa (T-s)})\Big |&\le \Big | C e^{-\beta (T-s)}- C e^{-\kappa (T-s)}\Big |\\&\le C e^{-\beta (T-s)}+ C e^{-\kappa (T-s)}. \end{aligned}$$
Using the inequality (26), we obtain
$$\begin{aligned}&\Big |\varPhi (d_1(f^{EL},f^{GA},0,0))-\varPhi (d_1(f^{EL},f^{GA}))\Big |\\&\quad \le \frac{1}{|\sigma -\xi |\sqrt{\tau }}\int _0^\tau (C e^{-\beta (T-s)}+C e^{-\kappa (T-s)})ds\\&\quad =\frac{1}{|\sigma -\xi |\sqrt{\tau }}\Big (\frac{C e^{-\beta (T-\tau )}(1-e^{-\beta \tau })}{\beta }+\frac{C e^{-\kappa (T-\tau )}(1-e^{-\kappa \tau })}{\kappa }\Big ). \end{aligned}$$
\(\square \)
Proof of Lemma 2
Consider the new probability measure \(\overset{\sim }{{\mathbb {Q}}}\) with Radon-Nikodym derivative
$$\begin{aligned} \frac{d\overset{\sim }{{\mathbb {Q}}}}{d{\mathbb {Q}}}=\exp \Big \{\int _0^\tau e^{-\beta (T-s)}dL_1(s)-\int _0^\tau \phi _1(e^{-\beta (T-s)})ds\Big \}. \end{aligned}$$
This is an Esscher transform, turning the Lévy process \(L_1\) into a independent increment process. Then the logarithmic-moment generating function of \(\int _0^\tau e^{-\beta (T-s)}ds\) under \(\overset{\sim }{{\mathbb {Q}}}\) is as follows
$$\begin{aligned} \phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}}(\theta _1)&=\ln {\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big [\exp \Big \{\theta _1\int _0^\tau e^{-\beta (T-s)}dL_1(s)\Big \}\Big ]\\&=\ln \Big ({\mathbb {E}}^{{\mathbb {Q}}}\Big [\exp \big \{\theta _1\int _0^\tau e^{-\beta (T-s)}dL_1(s)\big \}\\&\quad \exp \big \{\int _0^\tau e^{-\beta (T-s)}dL_1(s)\big \}\Big ]\exp \big \{-\int _0^\tau \phi _1(e^{-\beta (T-s)})ds\big \}\Big )\\&=\ln \Big ({\mathbb {E}}^{{\mathbb {Q}}}\Big [\exp \big \{(\theta _1+1)\int _0^\tau e^{-\beta (T-s)}dL_1(s)\big \}\Big ]\Big )-\int _0^\tau \phi _1(e^{-\beta (T-s)})ds\\&=\int _0^\tau \big (\phi _1((\theta _1+1)e^{-\beta (T-s)})- \phi _1(e^{-\beta (T-s)})\big )ds. \end{aligned}$$
Applying Cauchy-Schwarz inequality, yields
$$\begin{aligned}&\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [\frac{Z_1(f^{EL})}{f^{EL}}\varPhi (d_1(f^{EL},f^{GA}, \int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\Big ]\\&\qquad -\varPhi (d_1(f^{EL},f^{GA},0,0))\Big |\\&\quad =\Big |{\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big [\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\Big ]\\&\qquad -\varPhi (d_1(f^{EL},f^{GA},0,0))\Big |\\&\quad \le {\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big [\Big |\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\\&\qquad -\varPhi (d_1(f^{EL},f^{GA},0,0))\Big |\Big ]. \end{aligned}$$
Using the mean value theorem, there is a random variable g such that
$$\begin{aligned}&\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\\&\quad -\varPhi (d_1(f^{EL},f^{GA},0,0))\\&\quad =\varPhi '(g)\big (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\\&\quad \int _0^\tau e^{-\kappa (T-s)}dL_2(s)))-d_1(f^{EL},f^{GA},0,0)\big ). \end{aligned}$$
Note that
$$\begin{aligned} \varPhi '(g)<1. \end{aligned}$$
Thus
$$\begin{aligned}&\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))-\varPhi (d_1(f^{EL},f^{GA},0,0))\\&\quad <d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s))-d_1(f^{EL},f^{GA},0,0). \end{aligned}$$
Since
$$\begin{aligned}&d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s))=d_1(f^{EL},f^{GA},0,0)\\&\quad +\frac{1}{(\xi -\sigma )\sqrt{\tau }}\Big (\int _0^\tau e^{-\beta (T-s)}dL_1(s)\\&\quad -\int _0^\tau e^{-\kappa (T-s)}dL_2(s)\Big ). \end{aligned}$$
From Cauchy-Schwarz inequality we have
$$\begin{aligned}&{\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big |\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))-\varPhi (d_1(f^{EL},f^{GA},0,0))\Big |\\&\quad \le \frac{1}{|\xi -\sigma |\sqrt{\tau }}\Big ({\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big |\int _0^\tau e^{-\beta (T-s)}dL_1(s)-\int _0^\tau e^{-\kappa (T-s)}dL_2(s)\Big |\Big )\\&\quad \le \frac{1}{|\xi -\sigma |\sqrt{\tau }}\Big ({\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big |\int _0^\tau e^{-\beta (T-s)}dL_1(s)-\int _0^\tau e^{-\kappa (T-s)}dL_2(s)\Big |^2\Big )^{\frac{1}{2}}. \end{aligned}$$
Jensen’s inequality implies
$$\begin{aligned}&{\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big |\int _0^\tau e^{-\beta (T-s)}dL_1(s)-\int _0^\tau e^{-\kappa (T-s)}dL_2(s)\Big |^2\nonumber \\&\quad \le 2\Big ({\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\big [(\int _0^\tau e^{-\beta (T-s)}dL_1(s))^2\big ]+{\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\big [(\int _0^\tau e^{-\kappa (T-s)}dL_2(s))^2\big ]\Big ). \end{aligned}$$
(28)
From basic probability theory
$$\begin{aligned}&{\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\big [(\int _0^\tau e^{-\beta (T-s)}dL_1(s))^2\big ]=\frac{\partial ^2}{\partial \theta _1^2}e^{\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}}(\theta _1)}\Big |_{\theta _1=0}\nonumber \\&\quad =(\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{''}(0)e^{\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}}(0)}+(\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{'}(0))^2e^{\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}}(0)}\nonumber \\&\quad =(\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{''}(0)+((\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{'}(0))^2. \end{aligned}$$
(29)
Denoting \(\gamma (s)=e^{-\beta (T-s)}\), it follows,
$$\begin{aligned} (\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{'}(0)=\int _0^\tau e^{-\beta (T-s)} \phi ^{'}_1(e^{-\beta (T-s)})ds. \end{aligned}$$
(30)
We have
$$\begin{aligned} \phi _1(\gamma (s))=\int _{{\mathbb {R}}}(e^{\gamma (s)z_1}-1-\gamma (s)z_1 1_{|z_1|<1})\ell _1(dz_1). \end{aligned}$$
Thus
$$\begin{aligned} \phi '_1(\gamma (s))&=\frac{\partial }{\partial h}\int _{{\mathbb {R}}}(e^{h z_1}-1-h z_1 1_{|z_1|<1})\ell _1(dz_1)\Big |_{h=\gamma (s)}\\&=\int _{{\mathbb {R}}}(z_1 e^{\gamma (s)z_1}-z_1 1_{|z_1|<1})\ell _1(dz_1)\\&=\int _{|z_1|<1}z_1 (e^{\gamma (s)z_1}-1)\ell _1(dz_1)+\int _{|z_1|\ge 1}z_1 e^{\gamma (s)z_1}\ell _1(dz_1). \end{aligned}$$
As
$$\begin{aligned} \big |e^{\gamma (s)z_1}-1\big |&\le \gamma (s)|z_1|e^{\gamma (s)|z_1|}\\&\le |z_1|e^1, \end{aligned}$$
for \(|z_1| < 1\), while for \(|z_1| \ge 1\)
$$\begin{aligned} |z_1|e^{\gamma (s)|z_1|}\le e^{2|z_1|}, \end{aligned}$$
it follows that
$$\begin{aligned} |\phi '_1(\gamma (s))|\le e^1\int _{|z_1|<1}z_1^2\ell (dz_1)+\int _{|z_1|\ge 1}e^{2|z_1|}\ell _1(dz_1)\le C, \end{aligned}$$
where C is a positive constant.
Therefore, from expression (30) we have
$$\begin{aligned} |(\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{'}(0)|\le C\int _0^\tau e^{-\beta (T-s)}ds=\frac{C}{\beta }(1-e^{-\beta (T-s)}) \end{aligned}$$
(31)
On the other hand
$$\begin{aligned} (\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{''}(0)=\int _0^\tau e^{-2\beta (T-s)}\phi _1^{''}(e^{-\beta (T-s)})ds. \end{aligned}$$
Then
$$\begin{aligned} \phi {''}_1(\gamma (s))&=\frac{\partial }{\partial h}\int _{{\mathbb {R}}}(z_1e^{h z_1}- z_1 1_{|z_1|<1})\ell _1(dz_1)\Big |_{h=\gamma (s)} \\&=\int _{{\mathbb {R}}} z_1^2 e^{\gamma (s)z_1}\ell _1(dz_1) \\&=\int _{{\mathbb {R}}}(z_1^2 e^{\gamma (s)z_1} 1_{|z_1|<1}+z_1^2 e^{\gamma (s)z_1} 1_{|z_1|\ge 1}) \ell _1(dz_1). \end{aligned}$$
For \(|z_1|<1\), the series representation of the exponential function gives
$$\begin{aligned} |e^{\gamma (s)z_1}| \le \sum _{n=0}^{\infty }\frac{(\gamma (s)|z_1|)^n}{n!}\le \sum _{n=0}^{\infty }\frac{(\gamma (s))^n}{n!}=e^{\gamma (s)}\le e^1. \end{aligned}$$
While if \(|z_1 | \ge 1 \), we have
$$\begin{aligned} z_1^2e^{\gamma (s)z_1}\le z_1^2 e^{z_1}\le e^{3z_1}. \end{aligned}$$
Therefore from the condition on the Lévy measure in inequality (6), we get
$$\begin{aligned} |\phi {''}_1(e^{-\beta (T-s)})|\le C, \end{aligned}$$
where C is a positive constant and obtain
$$\begin{aligned} |(\phi _1^{\beta ,\overset{\sim }{{\mathbb {Q}}}})^{''}(0)|\le C\int _0^\tau e^{-2\beta (T-s)}ds=\frac{C}{2\beta }(1-e^{-2\beta (T-\tau )}). \end{aligned}$$
(32)
According to expression (29) and inequalities (31), (32) we have
$$\begin{aligned} {\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big [\big (\int _0^\tau e^{-\beta (T-s)}dL_1(s)\big )^2\Big ]\le \frac{C^2}{\beta ^2}(1-e^{-\beta (T-\tau )})^2+\frac{C}{2\beta }(1-e^{-2\beta (T-\tau )}). \end{aligned}$$
(33)
Similarly, we get
$$\begin{aligned} {\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big [\big (\int _0^\tau e^{-\kappa (T-s)}dL_1(s)\big )^2\Big ]\le \frac{C^2}{\kappa ^2}(1-e^{-2\kappa (T-\tau )})^2+\frac{C}{2\kappa }(1-e^{-2\kappa (T-\tau )}). \end{aligned}$$
(34)
Combining inequalities (28), (33), and (34) get
$$\begin{aligned}&{\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big [\big (\int _0^\tau e^{-\beta (T-s)}dL_2(s)-\int _0^\tau e^{-\kappa (T-s)}dL_2(s)\big )^2\Big ]\\&\quad \le \frac{C^2}{\beta ^2}(1-e^{-\beta (T-\tau )})^2+\frac{C}{2\beta }(1-e^{-2\beta (T-\tau )})\\&\quad +\frac{C^2}{\kappa ^2}(1-e^{-\kappa (T-\tau )})^2+\frac{C}{2\kappa }(1-e^{-2\kappa (T-\tau )}). \end{aligned}$$
Thus
$$\begin{aligned}&\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [\frac{Z_1(f^{EL})}{f^{EL}}\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\\&\quad \int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\Big ]-\varPhi (d_1(f^{EL},f^{GA},0,0))\Big |\\&\quad \le {\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big |\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\\&\quad \int _0^\tau e^{-\kappa (T-s)}dL_2(s)))-\varPhi (d_1(f^{EL},f^{GA},0,0))\Big |\\&\quad \le \frac{1}{|\xi -\sigma |\sqrt{\tau }}\Big ({\mathbb {E}}^{\overset{\sim }{{\mathbb {Q}}}}\Big |\int _0^\tau e^{-\beta (T-s)}dL_1(s)\\&\quad -\int _0^\tau e^{-\kappa (T-s)}dL_2(s)\Big |^2\Big )^{\frac{1}{2}}\\&\quad \le \frac{1}{|\xi -\sigma |\sqrt{\tau }}\Big (\frac{C^2}{\beta ^2}(1-e^{-\beta (T-\tau )})^2+\frac{C}{2\beta }(1-e^{-2\beta (T-\tau )})\\&\quad +\frac{C^2}{\kappa ^2}(1-e^{-\kappa (T-\tau )})^2+\frac{C}{2\kappa }(1-e^{-2\kappa (T-\tau )})\Big )^{\frac{1}{2}}. \end{aligned}$$
\(\square \)
Proof of Theorem 5
Applying triangle inequality and Lemmas 1-3, yield
$$\begin{aligned}&|C(0,\tau ,T,f^{EL},f^{GA})-\complement (0,\tau ,T,f^{EL},f^{GA})|\\&\quad =e^{-rT}\Big |\Big ({\mathbb {E}}^{{\mathbb {Q}}}\Big [Z_1(f^{EL})\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\\&\quad \int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\Big ]-f^{El}\varPhi (d_1(f^{EL},f^{GA}))\Big )\\&\qquad -\Big ({\mathbb {E}}^{{\mathbb {Q}}}\Big [Z_2(f^{GA})\varPhi (d_2(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\Big ]\\&\qquad -h_Rf^{GA}\varPhi (d_2(f^{EL},f^{GA}))\Big )\Big |\\&\le e^{-rT}\Big (\Big |\Big ({\mathbb {E}}^{{\mathbb {Q}}}\Big [Z_1(f^{EL})\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\\&\qquad -f^{El}\varPhi (d_1(f^{EL},f^{GA}))\Big ]\Big |\\&\qquad +\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [Z_2(f^{GA})\varPhi (d_2(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\\&\qquad -h_Rf^{GA}\varPhi (d_2(f^{EL},f^{GA}))\Big ]\Big |\Big )\\&\quad \le e^{-rT}f^{EL}\Big (\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [\frac{Z_1(f^{EL})}{f^{EL}}\varPhi (d_1(f^{EL},f^{GA},\int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\\&\qquad -\varPhi (d_1(f^{EL},f^{GA},0,0))\Big ]\Big |\\&\qquad +\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [\varPhi (d_1(f^{EL},f^{GA},0,0))-\varPhi (d_1(f^{EL},f^{GA}))\Big ]\Big |\Big )\\&\qquad +e^{-rT}h_Rf^{GA}\Big (\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [\frac{Z_2(f^{GA})}{h_R f^{GA}}\varPhi (d_2(f^{EL},f^{GA},\\&\quad \int _0^\tau e^{-\beta (T-s)}dL_1(s),\int _0^\tau e^{-\kappa (T-s)}dL_2(s)))\\&\qquad -\varPhi (d_2(f^{EL},f^{GA},0,0))\Big ]\Big |\\&\qquad +\Big |{\mathbb {E}}^{{\mathbb {Q}}}\Big [\varPhi (d_2(f^{EL},f^{GA},0,0))-\varPhi (d_2(f^{EL},f^{GA}))\Big ]\Big |\Big ) \\&\quad \le \Big (\frac{e^{-rT}}{|\xi -\sigma |\sqrt{\tau }}\big (\frac{C^2}{\beta ^2}(1-e^{-\beta (T-\tau )})^2+\frac{C}{2\beta }(1-e^{-2\beta (T-\tau )})\\&\qquad +\frac{C^2}{\kappa ^2}(1-e^{-\kappa (T-\tau )})^2+\frac{C}{2\kappa }(1-e^{-2\kappa (T-\tau )})\big )^{\frac{1}{2}}\\&\qquad +\frac{1}{|\sigma -\xi |\sqrt{\tau }}\big (\frac{C e^{-\beta (T-\tau )}(1-e^{-\beta \tau })}{\beta }+\frac{C e^{-\kappa (T-\tau )}(1-e^{-\kappa \tau })}{\kappa }\big )\Big )\Big (f^{EL}+h_Rf^{GA}\Big ). \end{aligned}$$
The right hand side of last inequality tend to zero as \(T\rightarrow \infty \). \(\square \)
Proof of Theorem 6
Let us \(C(t, \tau , T, x, y)\) and V(t) are respectively the discounted spark-spread option and the discounted self-financing portfolio. such that the variation of self-financing portfolio is
$$\begin{aligned} dV (t) = \varPsi (t)e^{-rt} dF(t, T). \end{aligned}$$
From no-arbitrage pricing theory, \(dC(t, \tau , T, x, y)\) and dV(t) are martingale. Thus from Proposition 1, and applying Itô’s formula to \(C(t, \tau , T, x, y)\) and V(t), we have
$$\begin{aligned}&dC(t, \tau , T, x, y)=e^{-rt}\Big (\sigma F^{EL}(t,T) C_x(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))d\overset{\sim }{B}_1(t)\\&\quad -h_R \xi F^{GA}(t,T)C_y(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))d\overset{\sim }{B}_2(t)\\&\quad +\int _{\mathbb {R}}\big (C(t,\tau ,T,F^{EL}(t,T)\exp \{z_1e^{-\beta (T-t)}\},F^{GA}(t,T))\\&\quad -C(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))\big )\overset{\sim }{N}_1(dt,dz_1)\\&\quad -h_R \int _{\mathbb {R}}\big (C(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T)\exp \{z_2e^{-\kappa (T-t)}\})\\&\quad -C(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))\big )\overset{\sim }{N}_2(dt,dz_2)\Big ),\\&dV(t)=\varPsi (t) e^{-rt} \Big (F^{EL}(t,T)\big (\sigma d\overset{\sim }{B}_1(t)\\&\quad +\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-s)}\}-1)\overset{\sim }{N}_1(dt,dz_1)\big )\\&\quad -h_R F^{GA}(t,T)\big (\xi d\overset{\sim }{B}_2(t)+\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-s)}\}-1)\overset{\sim }{N}_2(dt,dz_2)\big )\Big ), \end{aligned}$$
where \(\overset{\sim }{B_1}\) and \(\overset{\sim }{B_2}\) are two independent Brownian motions under \({\mathbb {Q}}\).
Suppose that \(V (0)= C(0, \tau , T, F^{EL}(0, T),F^{GA}(0,T))\). The hedging error is
$$\begin{aligned} {\mathcal {E}}(\varPsi ):= V(t, \tau , T, F^{EL}(t, T),F^{GA}(t,T))- C(t, \tau , T, F^{EL}(t, T),F^{GA}(t,T)), \end{aligned}$$
for \(\tau \le T\). Thus, according to the Itô isometry property for stochastic integration obtain
$$\begin{aligned}&{\mathbb {E}}^{{\mathbb {Q}}}\big [{\mathcal {E}}^2(\varPsi )\big ]=\int _0^\tau {\mathbb {E}}^{{\mathbb {Q}}}\\&\quad \big [(\sigma e^{-rs}F^{EL}(s,T))^2 \big (\varPsi (s)\\&\quad -C_x(s,\tau ,T,F^{EL}(s,T),F^{GA}(s,T))\big )^2\big ]ds\\&\quad + \int _0^\tau {\mathbb {E}}^{{\mathbb {Q}}}\big [(h_R \xi e^{-rs}F^{GA}(s,T))^2 \\&\quad \big (C_y(s,\tau ,T,F^{EL}(s,T),F^{GA}(s,T))-\varPsi (s)\big )^2\big ]ds\\&\quad +\int _0^\tau \int _{\mathbb {R}}{\mathbb {E}}^{{\mathbb {Q}}}\Big [e^{-2rs}\Big (\varPsi (s)F^{EL}(s,T)\\&\quad \big (\exp \{z_1 e^{-\beta (T-s)}\} -1\big ) +C(s,\tau ,T,F^{EL}(s,T),F^{GA}(s,T))\\&\quad -C(s,\tau ,T,F^{EL}(s,T)\exp \{z_1e^{-\beta (T-s)}\},F^{GA}(s,T))\Big )^2\ell _1(dz_1)\Big ]ds\\&\quad +\int _0^\tau \int _{\mathbb {R}}{\mathbb {E}}^{{\mathbb {Q}}}\Big [h_R^2e^{-2rs}\Big (-\varPsi (s)F^{GA}(s,T)\\&\quad \big (\exp \{z_2 e^{-\kappa (T-s)}\} -1\big ) -C(s,\tau ,T,F^{EL}(s,T),F^{GA}(s,T))\\&\quad +C(s,\tau ,T,F^{EL}(s,T),F^{GA}(s,T)\exp \{z_2e^{-\kappa (T-s)}\})\Big )^2\ell _2(dz_2)\Big ]ds. \end{aligned}$$
The first order condition for the minimizer of this quadratic expression solves
$$\begin{aligned}&\varPsi (t)\Big ((\sigma F^{EL}(t,T))^2+(h_R\xi F^{GA}(t,T))^2+(F^{EL}(t,T))^2\\&\quad \int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1)\\&\quad +(h_R F^{GA}(t,T))^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )\\&\quad =(\sigma F^{EL}(t,T))^2C_x(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))\\&\quad +(h_R\xi (F^{GA}(t,T))^2C_y(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T)) \\&\quad +F^{EL}(t,T)\int _{\mathbb {R}}\big (C(t,\tau ,T,F^{EL}(t,T)\\&\quad \exp \{z_1e^{-\beta (T-t)}\},F^{GA}(t,T))-C(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))\big ) \\&\quad \big (\exp \{z_1e^{-\beta (T-t)}\}-1\big )\ell _1(dz_1)\\&\quad +h^2_R F^{GA}(t,T)\int _{\mathbb {R}}\big (C(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T)\exp \{z_2e^{-\kappa (T-t)}\})\\&\quad -C(t,\tau ,T,F^{EL}(t,T),F^{GA}(t,T))\big ) \\&\quad \big (\exp \{z_2e^{-\kappa (T-t)}\}-1\big )\ell _2(dz_2) \end{aligned}$$
and the proof is completed. \(\square \)
Proof of Theorem 7
Applying triangle inequality, gives
$$\begin{aligned}&|\varPsi (t)-\big ((\sigma x)^2\complement _x+(h_R\xi y)^2\complement _y\big )|\nonumber \\&\le \Big ((\sigma x)^2|C_x-\complement _x|+(h_R\xi y)^2 |C_y- \complement _y|\nonumber \\&\quad +x\int _{\mathbb {R}}\big |C^\beta -C\big |\big |\exp \{z_1e^{-\beta (T-t)}\}-1\big |\ell _1(dz_1)\nonumber \\&\quad +h^2_R y\int _{\mathbb {R}}\big |C^\kappa -C\big |\big |\exp \{z_2e^{-\kappa (T-t)}\}-1\big |\ell _2(dz_2)\Big )\times \nonumber \\&\quad \Big ((\sigma x)^2+(h_R\xi y)^2+x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1)\nonumber \\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}, \end{aligned}$$
(35)
where
$$\begin{aligned}&C:=C(t,\tau ,T,x,y),~ \complement :=\complement (t,\tau ,T,x,y),~ C^\beta :=C(t,\tau ,T,x\exp \{z_1e^{-\beta (T-t)}\},y),\\&\quad C^\kappa :=C(t,\tau ,T,x,y\exp \{z_2e^{-\kappa (T-t)}\}). \end{aligned}$$
We have
$$\begin{aligned}&(\sigma x)^2|C_x-\complement _x|\Big ((\sigma x)^2+(h_R\xi y)^2+x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1)\\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}\\&\le e^{-r(T-t)}\Big ( {\mathbb {E}}^{\mathbb {Q}}\Big |\frac{Z_1(x)}{x}\varPhi (d_1(x,y,\omega _1,\omega _2)) -\varPhi (d_1(x,y))\Big |\\&+\frac{1}{|\xi -\sigma |\sqrt{\tau -t}} {\mathbb {E}}^{\mathbb {Q}}\Big |\frac{Z_1(x)}{x}\varPhi '(d_1(x,y,\omega _1,\omega _2)) -\varPhi '(d_1(x,y))\Big |\\&+\frac{h_Ry}{x|\xi -\sigma |\sqrt{\tau -t}} {\mathbb {E}}^{\mathbb {Q}}\Big |\frac{Z_2(y)}{h_R y}\varPhi '(d_2(x,y,\omega _1,\omega _2)) -\varPhi '(d_2(x,y))\Big |\Big ). \end{aligned}$$
Similar to proof of Lemma 2 for each of the statements on right hand side, we obtain
$$\begin{aligned}&(\sigma x)^2|C_x-\complement _x|\Big ((\sigma x)^2+(h_R\xi y)^2+x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1) \nonumber \\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}=0\quad as \quad T\rightarrow 0. \end{aligned}$$
(36)
Similarly, the following expression holds
$$\begin{aligned}&(h_R \xi y)^2|C_y-\complement _y|\Big ((\sigma x)^2+(h_R\xi y)^2+x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1) \nonumber \\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}=0\quad as \quad T\rightarrow 0. \end{aligned}$$
(37)
By mean value theorem, we have
$$\begin{aligned}&x\int _{\mathbb {R}}\big |C^\beta -C\big |\big |\exp \{z_1e^{-\beta (T-t)}\}-1\big |\ell _1(dz_1)\Big ((\sigma x)^2+(h_R\xi y)^2\\&\quad +x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1) \\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}\\&=\frac{1}{\sigma ^2}\int _{\mathbb {R}}\big |C_x (t,\tau ,g,y)\big |\big |\exp \{z_1e^{-\beta (T-t)}\}-1\big |^2\ell _1(dz_1), \end{aligned}$$
for some \(g\ge 0\).
It can be easily shown that
$$\begin{aligned} |C_x(t,\tau ,T,g,y)|\le e^{-r(T-t)}\Big (1+\frac{g+h_R y}{g|\xi -\sigma |\sqrt{\tau -t}}\Big ). \end{aligned}$$
On the other hand, for \(K\le 1\),
$$\begin{aligned} |e^{Kz_1-1}|\le K|z_1|\sum _{n=1}^\infty \frac{K^{n-1}|z_1|^{n-1}}{n!}\le K|z_1|\sum _{n=0}^\infty \frac{|z_1|^n}{n!}=K|z_1|e^{|z_1|}. \end{aligned}$$
If K is replaced with \(e^{-\beta (T-t)}\), we have
$$\begin{aligned}&\int _{\mathbb {R}}\Big (e^{-\beta (T-t)}-1\Big )^2\ell _1(dz_1)\le e^{-\beta (T-t)}\int _{\mathbb {R}}|z_1|^2 e^{2|z_1|}\ell _1(dz_1)\\&\quad \le e^{-2\beta (T-t)}\Big (e^2\int _{|z_1|\le 1}|z_1|^2\ell _1(dz_1)+e^2\int _{|z_1|> 1}z_1^2e^{2z_1}\ell _1(dz_1)\Big ). \end{aligned}$$
From the exponential moment condition on \(\ell _1(dz_1)\) and the condition that \(\ell _1\) is a Lévy measure, following inequality is obtained
$$\begin{aligned} \int _{\mathbb {R}}\Big (e^{-\beta (T-t)}-1\Big )^2\ell _1(dz_1)\le C e^{-2\beta (T-t)}, \end{aligned}$$
for a constant \(C > 0\).
Thus
$$\begin{aligned}&x\int _{\mathbb {R}}\big |C^\beta -C\big |\big |\exp \{z_1e^{-\beta (T-t)}\}-1\big |\ell _1(dz_1) \nonumber \\&\quad \Big ((\sigma x)^2+(h_R\xi y)^2+x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1) \nonumber \\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}=0 \quad as \quad T\rightarrow \infty . \end{aligned}$$
(38)
Similarly
$$\begin{aligned}&h_r^2 y\int _{\mathbb {R}}\big |C^\kappa -C\big |\big |\exp \{z_2e^{-\kappa (T-t)}\}-1\big |\ell _2(dz_2)\Big ((\sigma x)^2+(h_R\xi y)^2 \nonumber \\&\quad +x^2\int _{\mathbb {R}}(\exp \{z_1e^{-\beta (T-t)}\}-1)^2\ell _1(dz_1) \nonumber \\&\quad +(h_R y)^2\int _{\mathbb {R}}(\exp \{z_2e^{-\kappa (T-t)}\}-1)^2\ell _2(dz_2)\Big )^{-1}=0 \quad as \quad T\rightarrow \infty . \end{aligned}$$
(39)
Combining Eqs. (35)–(39) gives the result. \(\square \)