Valuing fade-in options with default risk in Heston–Nandi GARCH models

Abstract

In this paper, we present a pricing model to value fade-in options with default risk, where the underlying asset price is driven by the Heston–Nandi GARCH process and is correlated with the intensity process. The explicit pricing formulae are obtained, which contain pricing formulae of vanilla European options with/without default risk as special cases. Finally, a comparative analysis of the impacts of default risk is provided.

Appendix

Proof of Proposition 2.1

We first derive the closed form of $f(t;t_i,\phi _1,\phi _2,\phi _3)$ for $t_i\le t\le T$, that is,

$$\begin{aligned}&f(t;t_i,\phi _1,\phi _2,\phi _3) \\&\quad = \exp \Big \{\phi _1\ln S(t)+\phi _2\ln S(t_i)+\phi _3\sum _{u=1}^t\lambda (u)+A_0(t;t_i,\phi _1,\phi _2,\phi _3)\\&\qquad +A_1(t;t_i,\phi _1,\phi _2,\phi _3)h(t+1)+A_2(t;t_i,\phi _1,\phi _2,\phi _3)\lambda (t+1)\Big \}. \end{aligned}$$

In what follows, the more parsimonious notations f(t), $A_0(t)$, $A_1(t)$ and $A_2(t)$ are used. From the definition of $f(t;t_i,\phi _1,\phi _2,\phi _3)$, it is obvious that

$$\begin{aligned}&f(T;t_i,\phi _1,\phi _2,\phi _3)\\&\quad =e^{\phi _1\ln S(T)+\phi _2\ln S(t_i)+\phi _3\sum _{u=1}^T\lambda (u)}, \end{aligned}$$

implying that

$$\begin{aligned} A_0(T)=A_1(T)=A_2(T)=0. \end{aligned}$$

According to the law of iterated expectations and the dynamics of the underlying asset and the intensity process, we have that

$$\begin{aligned} f(t)= & {} E_t\Big [f(t+1)\Big ]\\= & {} E_t\Big [\exp \Big \{\phi _1\ln S(t+1)+\phi _2\ln S(t_i)+\phi _3\sum _{u=1}^{t+1}\lambda (u)+A_0(t+1)\\&\quad +\,A_1(t+1)h(t+2)+A_2(t+1)\lambda (t+2)\Big \}\Big ]\\= & {} \exp \Big \{\phi _1\ln S(t)+\phi _2\ln S(t_i)+\phi _3\sum _{s=1}^{t}\lambda (s)\\&\quad +\,\phi _1 r+A_0(t+1)+w_sA_1(t+1)+w_{\lambda }A_2(t+1)\\&\quad +\,(b_sA_1(t+1)+\beta _\lambda A_2(t+1)-\frac{1}{2}\phi _1) h(t+1)+(b_{\lambda } A_2(t+1)\\&\quad +\,\phi _3)\lambda (t+1)\Big \}\\&\times E_t\Big [\exp \Big \{ \phi _1\sqrt{h(t+1)}Z_s(t+1)\\&\quad +\,a_sA_1(t+1)\Big (Z_s(t+1)-c_s\sqrt{h(t+1)}\Big )^2 \\&\quad +\,a_{\lambda }A_2(t+1)\Big (Z_{\lambda }(t+1)-c_\lambda \sqrt{\lambda (t+1)}\Big )^2 \Big \}\Big ]. \end{aligned}$$

The expectations in the above expression can be derived using the following fact,

$$\begin{aligned} Ee^{a(Z+b)^2}=e^{-\frac{1}{2}\ln (1-2a)+\frac{ab^2}{1-2a}}, \end{aligned}$$

where Z is a standard normal variable. Indeed, we have that

$$\begin{aligned}&E_t\Big [\exp \Big \{ \phi _1\sqrt{h(t+1)}Z_s(t+1)+a_sA_1(t+1)\Big (Z_s(t+1)-c_s\sqrt{h(t+1)}\Big )^2 \\&\quad \quad \quad +\,a_{\lambda }A_2(t+1)\Big (Z_{\lambda }(t+1)-c_\lambda \sqrt{\lambda (t+1)}\Big )^2 \Big \}\Big ]\\&\quad = E_t\Big [\exp \Big \{ \phi _1\sqrt{h(t+1)}Z_s(t+1)+a_sA_1(t+1)\Big (Z_s(t+1)-c_s\sqrt{h(t+1)}\Big )^2\Big \}\Big ] \\&\qquad \times E_t\Big [\exp \Big \{a_{\lambda }A_2(t+1)\Big (Z_{\lambda }(t+1)-c_\lambda \sqrt{\lambda (t+1)}\Big )^2 \Big \}\Big ]\\&\quad = E_t\Big [e^{ a_sA_1(t+1)\Big (Z_s(t+1)-(c_s-\frac{\phi _1}{2a_sA_1(t+1)})\sqrt{h(t+1)}\Big )^2+ a_sA_1(t+1)\Big (c_s^2-(c_s-\frac{\phi _1}{2a_sA_1(t+1)})^2\Big )h(t+1)}\Big ] \\&\qquad \times E_t\Big [\exp \Big \{a_{\lambda }A_2(t+1)\Big (Z_{\lambda }(t+1)-c_\lambda \sqrt{\lambda (t+1)}\Big )^2 \Big \}\Big ]\\&\quad = e^{-\frac{1}{2}\ln (1-2a_sA_1(t+1))+\frac{a_sA_1(t+1)}{1-2a_sA_1(t+1)}(c_s-\frac{\phi _1}{2a_sA_1(t+1)})^2 h(t+1)+ a_sA_1(t+1)\Big (c_s^2-(c_s-\frac{\phi _1}{2a_sA_1(t+1)})^2\Big )h(t+1)}\Big ] \\&\qquad \times \exp \Big \{-\frac{1}{2}\ln (1-2a_{\lambda }A_2(t+1))+\frac{a_{\lambda }c^2_\lambda A_2(t+1)}{1-2a_{\lambda }A_2(t+1)} \lambda (t+1) \Big \}. \end{aligned}$$

By now, we have obtained the closed forms of $A_0(t)$, $A_1(t)$ and $A_2(t)$ for $t_i\le t\le T$ as follows,

$$\begin{aligned} A_0(t)= & {} \phi _1 r+A_0(t+1)+w_sA_1(t+1)+w_{\lambda }A_2(t+1)-\frac{1}{2}\ln (1-2a_sA_1(t+1))\\&-\,\frac{1}{2}\ln (1-2a_{\lambda }A_2(t+1)),\\ A_1(t)= & {} b_sA_1(t+1)+\beta _\lambda A_2(t+1) \\&\quad -\frac{1}{2}\phi _1+ a_sc_s^2A_1(t+1)+ \frac{1}{2}(\phi _1-a_sc_sA_1(t+1) )^2,\\ A_2(t)= & {} b_{\lambda } A_2(t+1)+\phi _3 \\&\quad +\frac{a_{\lambda }c^2_\lambda A_2(t+1)}{1-2a_{\lambda }A_2(t+1)}, \end{aligned}$$

Now we turn to derive the closed form of $f(t;t_i,\phi _1,\phi _2,\phi _3)$ for $0\le t \le t_i$, that is,

$$\begin{aligned} f(t;t_i,\phi _1,\phi _2,\phi _3)= & {} \exp \Big \{(\phi _1+\phi _2)\ln S(t)+\phi _3\sum _{u=1}^t\lambda (u)+B_0(t;t_i,\phi _1,\phi _2,\phi _3)\\&+B_1(t;t_i,\phi _1,\phi _2,\phi _3)h(t+1)+B_2(t;t_i,\phi _1,\phi _2,\phi _3)\lambda (t+1)\Big \}. \end{aligned}$$

Similarly, in the following the more parsimonious notations f(t), $B_0(t)$, $B_1(t)$ and $B_2(t)$ will be used. Actually, we have obtained the expression of $f(t_i;t_i,\phi _1,\phi _2,\phi _3)$, i.e.,

$$\begin{aligned}&f(t_i;t_i,\phi _1,\phi _2,\phi _3) \\&\quad =\exp \Big \{(\phi _1+\phi _2)\ln S(t_i)+\phi _3\sum _{u=1}^{t_i}\lambda (u)+A_0(t_i) +A_1(t_i)h(t_i+1)+A_2(t_i)\lambda (t_i+1)\Big \}, \end{aligned}$$

implying that

$$\begin{aligned} B_0(t_i)=A_0(t_i),\ \ B_1(t_i)=A_1(t_i),\ \ B_2(t_i)=A_2(t_i). \end{aligned}$$

Once again, based on the law of iterated expectations and the dynamics of the underlying asset and the intensity process, one gets that

$$\begin{aligned} f(t)= & {} E_t\Big [f(t+1)\Big ]\\= & {} E_t\Big [\exp \Big \{(\phi _1+\phi _2)\ln S(t+1)+\phi _3\sum _{u=1}^{t+1}\lambda (u)+B_0(t+1)\\&\quad +\,B_1(t+1)h(t+2)+B_2(t+1)\lambda (t+2)\Big \}\Big ]\\= & {} \exp \Big \{(\phi _1+\phi _2)\ln S(t)+\phi _3\sum _{s=1}^{t}\lambda (s)+(\phi _1+\phi _2) r+B_0(t+1)\\&\quad +\,w_sB_1(t+1)+w_{\lambda }B_2(t+1)\\&\quad +\,(b_sB_1(t+1)+\beta _\lambda B_2(t+1)\\&\quad -\,\frac{1}{2}(\phi _1+\phi _2)) h(t+1)+(b_{\lambda } B_2(t+1)+\phi _3)\lambda (t+1)\Big \}\\&\times E_t\Big [\exp \Big \{ (\phi _1+\phi _2)\sqrt{h(t+1)}Z_s(t+1)\\&\quad +\, a_sB_1(t+1)\Big (Z_s(t+1)-c_s\sqrt{h(t+1)}\Big )^2 \\&\quad +\, a_{\lambda }B_2(t+1)\Big (Z_{\lambda }(t+1)-c_\lambda \sqrt{\lambda (t+1)}\Big )^2 \Big \}\Big ]. \end{aligned}$$

Obviously, the above expectations have similar forms as we have derived before, and hence we have the following results for $0\le t \le t_i$,

$$\begin{aligned} B_0(t)= & {} (\phi _1+\phi _2)r+B_0(t+1)+w_sB_1(t+1)\\&\quad +w_{\lambda }B_2(t+1)-\frac{1}{2}\ln (1-2a_sB_1(t+1))\\&-\,\frac{1}{2}\ln (1-2a_{\lambda }B_2(t+1)),\\ B_1(t)= & {} b_sB_1(t+1)+\beta _\lambda B_2(t+1)\\&-\,\frac{1}{2}(\phi _1+\phi _2)++ a_sc_s^2B_1(t+1)+ \frac{1}{2}(\phi _1+\phi _2-a_sc_sB_1(t+1) )^2,\\ B_2(t)= & {} b_{\lambda } B_2(t+1)+\phi _3+\frac{a_{\lambda }c^2_\lambda B_2(t+1)}{1-2a_{\lambda }B_2(t+1)}. \end{aligned}$$

This completes the proof of Proposition 2.1. $\square$

Proof of Proposition 2.2

Recall that the prices of fade-in European options without default risk are given by

$$\begin{aligned} F_0= & {} e^{-r T}E\Big [\Big (S(T)-K\Big )^+\frac{1}{N}\sum _{i=1}^N I\Big (S(t_i)\in [L, H]\Big )\Big ]. \end{aligned}$$

(A.1)

Rewrite $F_0$ in the following form,

$$\begin{aligned} F_0= & {} e^{-r T}E\Big [\Big (S(T)-K\Big )^+\frac{1}{N}\sum _{i=1}^N I\Big (S(t_i)\in [L, H]\Big )\Big ]\\= & {} e^{-r T}\frac{1}{N}\sum _{i=1}^N E\Big [\Big (S(T)-K\Big )^+ I\Big (S(t_i)\in [L, H]\Big )\Big ]\\= & {} e^{-r T}\frac{1}{N}\sum _{i=1}^N E\Big [\Big (S(T)-K\Big )^+ \Big (I(S(t_i)\ge L)-I(S(t_i)\ge H)\Big ) \Big ]\\= & {} e^{-r T}\frac{1}{N}\sum _{i=1}^N E\Big [\Big (S(T)-K\Big )^+ I(S(t_i)\ge L)-\Big (S(T)-K\Big )^+I(S(t_i)\ge H) \Big ]\\= & {} e^{-r T}\frac{1}{N}\sum _{i=1}^N \Big (\pi _{i,1}-K*\pi _{i,2} -\pi _{i,3}+K*\pi _{i,4} \Big ), \end{aligned}$$

where

$$\begin{aligned} \pi _{i,1}= & {} E\Big [S(T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln L)\Big ],\\ \pi _{i,2}= & {} E\Big [I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln L)\Big ],\\ \pi _{i,3}= & {} E\Big [S(T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln H)\Big ],\\ \pi _{i,4}= & {} E\Big [I(\ln S(T)\ge K,\ \ln S(t_i)\ge \ln H)\Big ]. \end{aligned}$$

Note that $f(t;t_i,\phi _1,\phi _2,0)$ is the joint characteristic function of $\ln S(T)$ and $\ln S(t_i)$. By inverting the joint characteristic function [see, e.g., Kendall and Stuart (1977)], one gets that

$$\begin{aligned} \pi _{i,2}= & {} Q(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln L)\nonumber \\= & {} \frac{1}{4}+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln L}f(0;t_i,0,i\phi _2,0)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&-\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;t_i,i\phi _1,i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;t_i,i\phi _1,-i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2, \end{aligned}$$

(A.2)

and

$$\begin{aligned} \pi _{i,4}= & {} Q(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln H)\nonumber \\= & {} \frac{1}{4}+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&\ +\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln H}f(0;t_i,0,i\phi _2,0)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&\ -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln H}f(0;t_i,i\phi _1,i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln H}f(0;t_i,i\phi _1,-i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$

(A.3)

Using the change of measure technique [see, e.g., Heston and Nandi (2000), Wang (2018) and Liang and Wang (2021)], one obtains that

$$\begin{aligned} \pi _{i,1}= & {} E\Big [S(T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln L)\Big ]\nonumber \\= & {} \frac{1}{4} f(0;t_i,1,0,0)+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1+1,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&\ +\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln L}f(0;t_i,1,i\phi _2,0)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&\ -\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;t_i,i\phi _1+1,i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;t_i,i\phi _1+1,-i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2, \end{aligned}$$

(A.4)

and

$$\begin{aligned} \pi _{i,3}= & {} E\Big [S(T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln H)\Big ]\nonumber \\= & {} \frac{1}{4} f(0;t_i,1,0,0)+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1+1,0,0)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln H}f(0;t_i,1,i\phi _2,0)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&-\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln H}f(0;t_i,i\phi _1+1,i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln H}f(0;t_i,i\phi _1+1,-i\phi _2,0)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$

(A.5)

Therefore, we have obtained the closed form of $F_0$, i.e.,

$$\begin{aligned} F_0= & {} e^{-r T}\frac{1}{N}\sum _{i=1}^N \Big (\pi _{i,1}-K*\pi _{i,2} -\pi _{i,3}+K*\pi _{i,4} \Big ). \end{aligned}$$

In what follows, we turn to derive the closed form of $FD_0$. We first rewrite it in the following way,

$$\begin{aligned} FD_0= & {} (1-\alpha )e^{-r T}E\Big [I\Big (\tau >T\Big )\Big (S(T)-K\Big )^+\frac{1}{N}\sum _{i=1}^N I\Big (S(t_i)\in [L, H]\Big )\Big ]\nonumber \\&+\alpha e^{-r T}E\Big [\Big (S(T)-K\Big )^+\frac{1}{N}\sum _{i=1}^N I\Big (S(t_i)\in [L, H]\Big )\Big ]\nonumber \\= & {} (1-\alpha )e^{-r T}\frac{1}{N}\sum _{i=1}^N \Big (\pi _{i,5}-K*\pi _{i,6} -\pi _{i,7}+K*\pi _{i,8} \Big )\nonumber \\&+\,\alpha e^{-r T}\frac{1}{N}\sum _{i=1}^N \Big (\pi _{i,1}-K*\pi _{i,2} -\pi _{i,3}+K*\pi _{i,4} \Big ), \end{aligned}$$

(A.6)

where

$$\begin{aligned} \pi _{i,5}= & {} E\Big [I(\tau>T)S(T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln L)\Big ],\\ \pi _{i,6}= & {} E\Big [I(\tau>T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln L)\Big ],\\ \pi _{i,7}= & {} E\Big [I(\tau>T)S(T)I(\ln S(T)\ge \ln K,\ \ln S(t_i)\ge \ln H)\Big ],\\ \pi _{i,8}= & {} E\Big [I(\tau >T)I(\ln S(T)\ge K,\ \ln S(t_i)\ge \ln H)\Big ]. \end{aligned}$$

Note that $\pi _{i,5}$-$\pi _{i,8}$ have the same form as $\pi _{i,1}$-$\pi _{i,4}$ except for a term $I(\tau >T)$. As mentioned before, we can deal with this term using the change of measure technique. Hence, we have that

$$\begin{aligned} \pi _{i,5}= & {} \frac{1}{4} f(0;t_i,1,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1+1,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln L}f(0;t_i,1,i\phi _2,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&-\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;t_i,i\phi _1+1,i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;t_i,i\phi _1+1,-i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2,\nonumber \\ \end{aligned}$$

(A.7)

and

$$\begin{aligned} \pi _{i,7}= & {} \frac{1}{4} f(0;t_i,1,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1+1,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln H}f(0;t_i,1,i\phi _2,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&-\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln H}f(0;t_i,i\phi _1+1,i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln H}f(0;t_i,i\phi _1+1,-i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2.\nonumber \\ \end{aligned}$$

(A.8)

It is noted that by replacing $f(0;t_i,1,0,0)$ and $f(0;t_i,\cdot ,\cdot ,0)$ with $f(0;t_i,1,0,-1)$ and $f(0;t_i,\cdot ,\cdot ,-1)$ we obtain the closed form of $\pi _{i,5}$ and $\pi _{i,7}$.

Likewise, we can we can obtain the closed form of $\pi _{i,6}$ and $\pi _{i,8}$ given below,

$$\begin{aligned} \pi _{i,6}= & {} \frac{1}{4} f(0;t_i,0,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln L}f(0;t_i,0,i\phi _2,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&-\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln L}f(0;t_i,i\phi _1,i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln L}f(0;t_i,i\phi _1,-i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2, \end{aligned}$$

(A.9)

and

$$\begin{aligned} \pi _{i,8}= & {} \frac{1}{4}f(0;t_i,0,0,-1)+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _1 \ln K} f(0;t_i,i\phi _1,0,-1)}{ i\phi _1 }\Big ]{\mathrm {d}}\phi _1\nonumber \\&+\frac{1}{2\pi }\int _{0}^\infty \text {Re}\Big [\frac{e^{-i \phi _2 \ln H}f(0;t_i,0,i\phi _2,-1)}{ i\phi _2 }\Big ]{\mathrm {d}}\phi _2\nonumber \\&-\frac{1}{2\pi ^2}\int _{0}^\infty \int _0^\infty \Big (\text {Re}\Big [\frac{e^{-i \phi _1 \ln K-i \phi _2 \ln H}f(0;t_i,i\phi _1,i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\nonumber \\&-\text {Re}\Big [\frac{e^{-i \phi _1 \ln K+i \phi _2 \ln H}f(0;t_i,i\phi _1,-i\phi _2,-1)}{ \phi _1 \phi _2}\Big ]\Big ){\mathrm {d}}\phi _1{\mathrm {d}}\phi _2. \end{aligned}$$

(A.10)

This completes the proof of Proposition 2.2. $\square$