Pricing commodity-linked bonds with stochastic convenience yield, interest rate and counterparty credit risk: application of Mellin transform methods

Abstract

This paper investigates the effects of the spot underlying commodity price, stochastic convenience yield, interest rate and counterparty credit risk on the pricing of the commodity-linked bonds. The stochastic factors or state variables in the model are the spot price of the underlying commodity follows geometrical Brownian motion process with a stochastic drift, the net convenience yield and the short-term interest rate are formulated as a mean-reverting Ornstein–Uhlenbeck stochastic process and the value of the firm issuing the bonds follows a geometrical Brownian motion process. Furthermore, we develop the two- and three-factor(I, II) pricing models for valuing the commodity-linked bonds. Closed-form pricing formulas of the commodity-linked bonds are derived based on the Mellin transform techniques, which are simply provided with standard (bivariate) normal cumulative distribution function so that the pricing and hedging of the commodity-linked bonds can be computed very accurately and rapidly. At last, numerical analysis compares the results of this four pricing models with realistic parameter values and demonstrates how the spot underlying commodity price, convenience yield, interest rate and counterparty credit risk affect the values of the commodity-linked bonds.

Appendices

Review of the Mellin transform

To obtain help in solving the PDE (11), (28) and (46) with given terminal condition, we first summarize the definition and some basic properties of without proof for readers who are unfamiliar with the double Mellin transforms. The interested reader can refer to Erdelyi et al. (1954) and Sneddon (1972) as well.

Definition 1

(Definition of the Mellin transform and inverse transform) The Mellin transform $\hat{g}(\omega )$ of a complex-valued function g(x) defined over positive reals is

$$\begin{aligned} \mathcal {M}_{x}[g(x);\omega ]:=\hat{g}(\omega )=\int _{0}^{\infty }g(x)x^{\omega -1}dx, \end{aligned}$$

with $\omega$ is complex number. Then the function g(x) can be recovered from its Mellin transform by the inverse Mellin transformation formula

$$\begin{aligned} g(x):=\mathcal {M}_{x}^{-1}[\hat{g}(\omega )]=\frac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }\hat{g}(\omega )x^{-\omega }d\omega , \end{aligned}$$

with $a<Re(\omega )$ and $a<c_{1}<b$ exist.

Definition 2

(Definition of the double Mellin transform and inverse transform) The double Mellin transform $\hat{g}(\omega _{1},\omega _{2})$ of a complex-valued function g(x, y) defined over positive reals is

$$\begin{aligned} \mathcal {M}_{x,y}[g(x,y);\omega _{1},\omega _{2}]:=\hat{g}(\omega _{1},\omega _{2})=\int _{0}^{\infty }\int _{0}^{\infty }g(x,y)x^{\omega _{1}-1}y^{\omega _{2}-1}dxdy, \end{aligned}$$

with $\omega _{1}$ and $\omega _{2}$ are complex numbers. Then the function g(x, y) can be recovered from its Mellin transform by the inverse Mellin transformation formula

$$\begin{aligned} g(x,y):=\mathcal {M}_{x,y}^{-1}[\hat{g}(\omega _{1},\omega _{2})]=\frac{1}{(2\pi i)^{2}}\int _{c_{1}-i\infty }^{c_{1}+i\infty }\int _{c_{2}-i\infty }^{c_{2}+i\infty }\hat{g}(\omega _{1},\omega _{2})x^{-\omega _{1}}y^{-\omega _{2}}d\omega _{1}d\omega _{2}, \end{aligned}$$

with $a<Re(\omega _{1}), Re(\omega _{2})<b$ and $a<c_{1},c_{2}<b$ exist.

Proposition 1

(Basic properties of the Mellin transform) Suppose that there exists a double Mellin transform of f(x, y). Then the following relations hold:

$$\begin{aligned} \mathcal {M}_{x,y}\left( x\frac{\partial ^{2}}{\partial x }f(x,y);\omega _{1},\omega _{2}\right)&=-\omega _{1}\hat{f}(\omega _{1},\omega _{2}),\\ \mathcal {M}_{x,y}\left( y\frac{\partial ^{2}}{\partial y }f(x,y);\omega _{1},\omega _{2}\right)&=-\omega _{2}\hat{f}(\omega _{1},\omega _{2}),\\ \mathcal {M}_{x,y}\left( xy\frac{\partial ^{2}}{\partial x \partial y}f(x,y);\omega _{1},\omega _{2}\right)&=\omega _{1}\omega _{2}\hat{f}(\omega _{1},\omega _{2}),\\ \mathcal {M}_{x,y}\left( x^{2}\frac{\partial ^{2}}{\partial x^{2}}f(x,y);\omega _{1},\omega _{2}\right)&=\omega _{1}(\omega _{1}+1) \hat{f}(\omega _{1},\omega _{2}),\\ \mathcal {M}_{x,y}\left( y^{2}\frac{\partial ^{2}}{\partial y^{2}}f(x,y);\omega _{1},\omega _{2}\right)&=\omega _{2}(\omega _{2}+1) \hat{f}(\omega _{1},\omega _{2}). \end{aligned}$$

Proposition 2

(Convolution of the Mellin transform) Let f(x) and g(x) be locally integrable functions on positive reals. $\hat{f}(\omega )$ and $\hat{g}(\omega )$ are two Mellin transforms of the functions f(x) and g(x), respectively. Then, the Mellin convolution is given by the inverse Mellin transform of $\hat{f}(\omega _{1})\hat{g}(\omega _{1})$ as follows:

$$\begin{aligned} f(x)\vee g(x):&=\frac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }\hat{f}(\omega )\hat{g}(\omega )x^{-\omega }d\omega \\&=\int _{0}^{\infty }f\left( \frac{x}{\omega }\right) g(x)\frac{d\omega }{\omega }. \end{aligned}$$

Proposition 3

(Convolution of the double Mellin transform) Let f(x, y) and g(x, y) be locally integrable functions on positive reals. $\hat{f}(\omega _{1},\omega _{2})$ and $\hat{g}(\omega _{1},\omega _{2})$ are two double Mellin transforms of the functions f(x, y) and g(x, y), respectively. Then, the double Mellin convolution is given by the inverse Mellin transform of $\hat{f}(\omega _{1},\omega _{2})\hat{g}(\omega _{1},\omega _{2})$ as follows:

$$\begin{aligned} f(x,y)\vee g(x,y):&=\frac{1}{(2\pi i)^{2}}\int _{c_{1}-i\infty }^{c_{1}+i\infty }\int _{c_{2}-i\infty }^{c_{2}+i\infty }\hat{f}(\omega _{1},\omega _{2})\hat{g}(\omega _{1},\omega _{2})x^{-\omega _{1}}y^{-\omega _{2}}d\omega _{1}d\omega _{2}\\&=\int _{0}^{\infty }\int _{0}^{\infty }f(\frac{x}{\omega _{1}},\frac{y}{\omega _{2}})g(x,y)\frac{d\omega _{1}}{\omega _{1}}\frac{d\omega _{2}}{\omega _{2}}. \end{aligned}$$

Proposition 4

(Inverse Mellin transform of exponential function) Given complex numbers $\alpha$ and $\beta$ with $Re(\alpha )\ge 0$, let $f(x)=\frac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }\hat{f}(s)x^{-s}ds$, where $\hat{f}(s)=e^{\alpha (s+\beta )^{2}}$. Then

$$\begin{aligned} f(x)=\frac{1}{2\sqrt{\pi \alpha }}x^{\beta }e^{-\frac{(\ln x)^{2}}{4\alpha }} \end{aligned}$$

holds.

Proof of Theorem 3.1

Proof

First of all, let $x=\frac{\ln \left( \frac{S}{u}\right) }{\sqrt{2E(\tau )}}$. By applying the change of variable from u to x , Eq. 20 becomes

$$\begin{aligned} B(\tau ,S,\delta )&=-\frac{1}{\sqrt{2\pi }}\int _{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}^{-\infty }(Se^{-\sqrt{2E(\tau )}x}+F-K)e^{c^{*}-\frac{1}{2}x^{2}+\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}x}dx\nonumber \\&\quad -\frac{1}{\sqrt{2\pi }}\int _{\infty }^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}Fe^{c^{*}-\frac{1}{2}x^{2}+\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}x}dx\nonumber \\&=\frac{S}{\sqrt{2\pi }}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{-\infty }e^{-\frac{1}{2}\left( x+\frac{2E(\tau )-G_{2}(\tau )}{\sqrt{2E(\tau )}}\right) ^{2}+E(\tau )-G_{2}(\tau )-\tau r}dx+\frac{F-K}{\sqrt{2\pi }}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{-\infty }\nonumber \\&\qquad e^{-\frac{1}{2}\left( x-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right) ^{2}-\tau r}dx+\frac{F}{\sqrt{2\pi }}\int _{\frac{\ln (\frac{S}{K})}{\sqrt{2E(\tau )}}}^{\infty }e^{-\frac{1}{2}\left( x-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right) ^{2}-\tau r}dx\nonumber \\&=P(\tau ,\delta )S\frac{1}{\sqrt{2\pi }}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{-\infty }e^{-\frac{1}{2}(x+\frac{2E(\tau )-G_{2}(\tau )}{\sqrt{2E(\tau )}})^{2}}dx-e^{-\tau r}K\frac{1}{\sqrt{2\pi }}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{-\infty }\nonumber \\&\qquad e^{-\frac{1}{2}\left( x-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right) ^{2}}dx+e^{-\tau r}F. \end{aligned}$$

(48)

Here

$$\begin{aligned} P(\tau ,\delta )&=e^{E(\tau )-G_{2}(\tau )-\tau r}\\&=e^{\left( \theta ^{*}+\gamma _{1}-\frac{\sigma _{\delta }^{2}}{2\kappa ^{2}}\right) (H(\tau )-\tau )-\frac{\sigma _{\delta }^{2}}{4\kappa }H^{2}(\tau )-\delta H(\tau )}\\&=A(\tau )\exp (-\delta H(\tau )), \end{aligned}$$

where

$$\begin{aligned} A(\tau )&=\exp \left\{ \left( \theta ^{*}+\gamma _{1}-\frac{\sigma _{\delta }^{2}}{2\kappa ^{2}}\right) (H(\tau )-\tau )-\frac{\sigma _{\delta }^{2}}{4\kappa }H^{2}(\tau )\right\} \\ H(\tau )&=\frac{1-e^{-\kappa \tau }}{\kappa }. \end{aligned}$$

Then, $B(S,\tau )$ is given by the formula

$$\begin{aligned} B(S,\tau )&=P(\tau ,\delta )S\frac{1}{\sqrt{2\pi }}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{-\infty }e^{-\frac{1}{2}(x'_{1})^{2}}dx-e^{- r\tau }K\frac{1}{\sqrt{2\pi }}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{-\infty }\nonumber \\&\qquad e^{-\frac{1}{2}(x'_{2})^{2}}dx+e^{-r\tau }F\nonumber \\&=P(\tau ,\delta )S\mathcal {N}(d_{1})-e^{-r\tau }K\mathcal {N}(d_{2})+e^{-r\tau }F, \end{aligned}$$

(49)

where $d_{1}$ and $d_{2}$ are given by

$$\begin{aligned} d_{1}&=\frac{\ln \left( \frac{S}{K}\right) +r\tau +\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(t)dt+\ln (P(\delta ,\tau ))}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\\ d_{2}&=\frac{\ln \left( \frac{S}{K}\right) +r\tau -\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(t)dt+\ln (P(\delta ,\tau ))}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}}. \end{aligned}$$

$\square$

Proof of Lemma 4.1

Proof

For $\tau >0$, because $E(\tau )>0$, so the inequality $G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}>0\Leftrightarrow 4E(\tau )G_{1}(\tau )>G^{2}(\tau )$. And $E(\tau )=\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(t)dt$ and $G_{1}(\tau )=\frac{1}{2}\sigma _{v}^{2}\tau =\frac{1}{2}\int ^{\tau }_{0}\sigma _{v}^{2}dt$. From the Cauchy–Schwartz inequality, we have

$$\begin{aligned} 4E(\tau )G_{1}(\tau )&=\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(t)dt\frac{1}{2}\sigma _{v}^{2}\tau =\frac{1}{2}\int ^{\tau }_{0}\sigma _{v}^{2}dt\\&\ge \left( \int ^{\tau }_{0}\hat{\sigma }_{s}(t)\sigma _{v}dt\right) ^{2}\\&=\left( \int ^{\tau }_{0}\sqrt{\sigma _{v}^{2}(\sigma _{s}^{2}-2\rho _{s\delta }\sigma _{s}\sigma _{\delta }H(t)+\sigma ^{2}_{\delta }H^{2}(t))}dt\right) ^{2}, \end{aligned}$$

and we also know $G(\tau )=\rho _{sv}\sigma _{s}\sigma _{v}\tau -\gamma _{2}(\tau -H(\tau ))=\int ^{\tau }_{0}\rho _{sv}\sigma _{s}\sigma _{v}-\rho _{\delta v}\sigma _{\delta }\sigma _{v}H(t)dt$. Therefore $4E(\tau )G_{1}(\tau )>G^{2}(\tau )$ is satisfied if and only if

$$\begin{aligned} \nabla&=\sigma _{v}^{2}\left( \sigma _{s}^{2}-2\rho _{s\delta }\sigma _{s}\sigma _{\delta }H(t)+\sigma ^{2}_{\delta }H^{2}(t))-(\rho _{sv}\sigma _{s}\sigma _{v}-\rho _{\delta v}\sigma _{\delta }\sigma _{v}H(t)\right) ^{2}\\&=\sigma _{v}^{2}\left[ (1-\rho _{\delta v}^{2})\sigma _{\delta }^{2}H^{2}(t)+2(\rho _{sv}\rho _{\delta v}-\rho _{s\delta })\sigma _{s}\sigma _{\delta }H(t)+(1-\rho _{sv}^{2})\sigma _{s}^{2}\right] >0. \end{aligned}$$

If we consider $\nabla$ as a quadratic equation of $\sigma _{\delta }H(t)$, then $\nabla >0$ is satisfied if and only if $\sigma _{v}^{2}>0$, $1-\rho _{\delta v}^{2}>0$ and

$$\begin{aligned} \Delta&=(2(\rho _{sv}\rho _{\delta v}-\rho _{s\delta })\sigma _{s})^{2}-4(1-\rho _{\delta v}^{2})(1-\rho _{sv}^{2})\sigma _{s}^{2}\\&=4\sigma _{s}^{2}\left[ \rho _{sv}^{2}+\rho _{\delta v}^{2}+\rho _{s\delta }^{2}-2\rho _{sv}\rho _{\delta v}\rho _{s\delta }-1\right] <0 \end{aligned}$$

are satisfied. Because $\sigma _{v}^{2}>0$ for $\tau >0$ is satisfied, and according to conditions (25), the two later conditions are obviously true. Consequently, $G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}>0$ has been verified. $\square$

Proof of Theorem 4.1

Proof

First of all, let $y=\frac{\ln \left( \frac{V}{w}\right) }{\sqrt{2G_{1}(\tau )}}$ and $\rho =\frac{G(\tau )}{2\sqrt{E(\tau )G_{1}(\tau )}}$. By applying the change of variables from u and w to x and y, $I_{B}^{1}(t,S,V,\delta )$ of Eq. 41 becomes

$$\begin{aligned}&I_{B}^{1}(t,S,V,\delta )=\frac{1}{2\pi }\sqrt{\frac{G_{1}(\tau )}{{G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}}}}\int ^{-\infty }_{\frac{\ln (\frac{S}{K})}{\sqrt{2E(\tau )}}}\int ^{-\infty }_{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2G_{1}(\tau )}}} e^{\eta }\exp \left\{ -\frac{(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )(G_{2}(\tau )-\sqrt{2E(\tau )}x)}{2E(\tau )})^{2}}{4(G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )})}\right\} \nonumber \\&\qquad (Se^{-\sqrt{2E(\tau )}x}+F-K)\exp \Bigg \{\frac{y\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )(G_{2}(\tau )-\sqrt{2E(\tau )}x)}{2E(\tau )})}{2(G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )})}\nonumber \\&\quad +\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}x\Bigg \}e^{-\frac{1}{2}x^{2}}e^{-\frac{G_{1}(\tau )y^{2}}{2\left( G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}\right) }}dxdy\nonumber \\&=\frac{S}{2\pi \sqrt{1-\rho ^{2}}} \int ^{-\infty }_{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}\int ^{-\infty }_{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2G_{1}(\tau )}}}\exp \Bigg \{\eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) -\Bigg [\sqrt{2E(\tau )}-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\nonumber \\&\quad +\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) \frac{G(\tau )}{\sqrt{2E(\tau )}}}{2G_{1}(\tau )(1-\rho ^{2})} \Bigg ]x +\frac{y\sqrt{2G_{1}(\tau )}\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) }{2G_{1}(\tau )(1-\rho ^{2})}\nonumber \\&\quad +\frac{\rho xy}{(1-\rho ^{2})}-\frac{(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\Bigg \}dxdy\nonumber \\&\quad +\frac{F-K}{2\pi \sqrt{1-\rho ^{2}}} \int ^{-\infty }_{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}\int ^{-\infty }_{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2G_{1}(\tau )}}}\exp \Bigg \{\eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) -\Bigg [\frac{(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}\nonumber \\&\quad \cdot \frac{G(\tau )}{\sqrt{2E(\tau )}}-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\Bigg ]x+\frac{y\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}+\frac{\rho xy}{(1-\rho ^{2})}\nonumber \\&\quad -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\Bigg \}dxdy. \end{aligned}$$

(50)

To evaluate the first term in Eq. 50, we introduce an auxiliary function

$$\begin{aligned}&I_{B}^{10}(\tau ,S,V,\delta )=\frac{S}{2\pi \sqrt{1-\rho ^{2}}}\int ^{-\infty }_{\frac{\ln (\frac{S}{K})}{\sqrt{2E(\tau )}}}\int ^{-\infty }_{\frac{\ln (\frac{V}{D^{*}})}{\sqrt{2G_{1}(\tau )}}} \exp \left\{ \eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) -\left[ \sqrt{2E(\tau )}-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right. \right. \nonumber \\&\left. \quad +\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) \frac{G(\tau )}{\sqrt{2E(\tau )}}}{2G_{1}(\tau )(1-\rho ^{2})} \right] x+\frac{y\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}\nonumber \\&\quad \left. +\frac{\rho xy}{(1-\rho ^{2})}-\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\right\} dxdy. \end{aligned}$$

(51)

Then the exponent of the integrand of Eq. 51 can be expressed as

$$\begin{aligned} M_{1}-\frac{1}{2(1-\rho ^{2})}[(x+a_{1})^{2}+(y+b_{1})^{2}]+\frac{\rho }{1-\rho ^{2}}(x+a_{1})(y+b_{1}), \end{aligned}$$

where $a_{1}$, $b_{1}$ and $M_{1}$ are given by

$$\begin{aligned} a_{1}&=\frac{2E(\tau )-G_{2}(\tau )}{\sqrt{2E(\tau )}},\\ b_{1}&=\frac{G(\tau )-G_{1}(\tau )-G_{3}(\tau )}{\sqrt{2G_{1}(\tau )}},\\ M_{1}&=\eta -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}+\frac{a_{1}^{2}+b^{2}_{1}}{2(1-\rho ^{2})}-\frac{\rho a_{1}b_{1}}{1-\rho ^{2}}\\&=E(\tau )-G_{2}(\tau )-\tau r\\&=\ln (P(\delta , \tau )). \end{aligned}$$

Similarly, to evaluate the second term in Eq. 50, we introduce an auxiliary function

$$\begin{aligned} I_{B}^{11}(\tau ,S,V,\delta )&=\frac{F-K}{2\pi \sqrt{1-\rho ^{2}}}\int ^{-\infty }_{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}\int ^{-\infty }_{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2B_{1}(\tau )}}} \exp \left\{ \eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) -\left[ \frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) }{2G_{1}(\tau )(1-\rho ^{2})}\right. \right. \nonumber \\&\left. \quad \cdot \frac{G(\tau )}{\sqrt{2E(\tau )}}-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right] x+\frac{y\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}+\frac{\rho xy}{2E(\tau )G_{1}(\tau )(1-\rho ^{2})}\nonumber \\&\quad \left. -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\right\} dxdy. \end{aligned}$$

(52)

Then the exponent of the integrand of Eq. 52 can be expressed as

$$\begin{aligned} M_{2}-\frac{1}{2(1-\rho ^{2})}[(x+a_{2})^{2}+(y+b_{2})^{2}]+\frac{\rho }{1-\rho ^{2}}(x+a_{2})(y+b_{2}), \end{aligned}$$

where $a_{2}$, $b_{2}$ and $M_{2}$ are given by

$$\begin{aligned} a_{2}&=-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}},\\ b_{2}&=-\frac{G_{1}(\tau )+G_{3}(\tau )}{\sqrt{2G_{1}(\tau )}},\\ M_{2}&=\eta -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}+\frac{a_{2}^{2}+b^{2}_{2}}{2(1-\rho ^{2})}-\frac{\rho a_{2}b_{2}}{1-\rho ^{2}}\\&=-\tau r. \end{aligned}$$

Then, $I_{B}^{1}(\tau ,S,V,\delta )$ is given by the formula

$$\begin{aligned} \begin{array}{lll} I_{B}^{1}(\tau ,S,V,\delta )&{}= P(\delta ,\tau )S\frac{1}{2\pi \sqrt{1-\rho ^{2}}}\int _{-\infty }^{a'_{1}}\int _{-\infty }^{a'_{2}}e^{-\frac{1}{2(1-\rho ^{2})}(x_{1}^{'2}-2\rho x'_{1}y'_{1}+y_{1}^{'2})}dx'_{1}dy'_{1}\\ &{}\qquad +e^{-r\tau }(F-K)\frac{1}{2\pi \sqrt{1-\rho ^{2}}}\int _{-\infty }^{b'_{1}}\int _{-\infty }^{b'_{2}}e^{-\frac{1}{2(1-\rho ^{2})}(x_{2}^{'2}-2\rho x'_{2}y'_{2}+y_{2}^{'2})}dx'_{2}dy'_{2}\\ &{}=P(\delta ,\tau )S\mathcal {N}_{2}(a'_{1},b'_{1},\rho )+e^{-r\tau }(F-K)\mathcal {N}_{2}(a'_{2},b'_{2},\rho ), \end{array} \end{aligned}$$

(53)

where

$$\begin{aligned} a'_{1}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P(\delta ,\tau )+\tau r+\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\\ b'_{1}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}-\frac{1}{2}\sigma _{v}^{2}\right) \tau +\rho _{sv}\sigma _{s}\sigma _{v}-\gamma _{2}(\tau -H(\tau ))}{\sigma _{v}\sqrt{\tau }},\\ a'_{2}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P(\delta ,\tau )+\tau r-\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\\ b'_{2}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}-\frac{1}{2}\sigma _{v}^{2}\right) \tau }{\sigma _{v}\sqrt{\tau }}. \end{aligned}$$

As with the same procedure for $I_{B}^{1}(\tau ,S,V,\delta )$, then $I_{B}^{2}(\tau ,S,V,\delta )$, $I_{B}^{3}(\tau ,S,V,\delta )$ and $I_{B}^{4}(\tau ,S,V,\delta )$ of Eq. 41 are given respectively,

$$\begin{aligned} \begin{array}{lll} I_{B}^{2}(\tau ,S,V,\delta )&{}=\frac{F}{2\pi \sqrt{1-\rho ^{2}}}\int ^{\infty }_{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}\int ^{-\infty }_{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2B_{1}(\tau )}}} \exp \left\{ \eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) -\left[ \frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) }{2G_{1}(\tau )(1-\rho ^{2})}\right. \right. \\ &{}\left. \quad \cdot \frac{G(\tau )}{\sqrt{2E(\tau )}}-\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right] x+\frac{y\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}+\frac{\rho xy}{2E(\tau )G_{1}(\tau )(1-\rho ^{2})}\\ &{}\left. \quad -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\right\} dxdy\\ &{}=e^{-r\tau }F\mathcal {N}_{2}(-c'_{2},d'_{2},-\rho ), \end{array} \end{aligned}$$

(54)

where

$$\begin{aligned} c'_{2}&=a'_{2}, d'_{2}=b'_{2}. I_{B}^{3}(\tau ,S,V,\delta )\nonumber \\&\quad =\frac{(1-\alpha )V}{D}\frac{1}{2\pi }\sqrt{\frac{G_{1}(\tau )}{{G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}}}}\int ^{-\infty }_{\frac{\ln (\frac{S}{K})}{\sqrt{2E(\tau )}}}\int ^{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2G_{1}(\tau )}}}_{\infty } \exp \left\{ -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )(G_{2}(\tau )-\sqrt{2E(\tau )}x)}{2E(\tau )}\right) ^{2}}{4\left( G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}\right) }\right\} \nonumber \\&\qquad e^{\eta -\sqrt{2G_{1}(\tau )}y}\left( Se^{-\sqrt{2E(\tau )}x}+F-K\right) \exp \Bigg \{\frac{y\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )(G_{2}(\tau )-\sqrt{2E(\tau )}x)}{2E(\tau )})}{2\left( G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )}\right) }\nonumber \\&\qquad +\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}x\Bigg \}e^{-\frac{1}{2}x^{2}}e^{-\frac{G_{1}(\tau )y^{2}}{2(G_{1}(\tau )-\frac{G^{2}(\tau )}{4E(\tau )})}}dxdy\nonumber \\&\quad =\frac{(1-\alpha )V}{D}S\frac{1}{2\pi \sqrt{1-\rho ^{2}}}\int ^{-\infty }_{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}\int ^{\frac{\ln (\frac{V}{D^{*}})}{\sqrt{2G_{1}(\tau )}}}_{\infty } \exp \Bigg \{\eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) -\Bigg [\sqrt{2E(\tau )}\nonumber \\&\qquad -\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}+ \frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) \frac{G(\tau )}{\sqrt{2E(\tau )}}}{2G_{1}(\tau )(1-\rho ^{2})} \Bigg ]x+[\frac{\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}\nonumber \\&\qquad -\sqrt{2G_{1}(\tau )}]y+\frac{\rho xy}{2E(\tau )G_{1}(\tau )(1-\rho ^{2})}-\frac{(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\Bigg \}dxdy\nonumber \\&\qquad +\frac{(1-\alpha )V}{D}\frac{F-K}{2\pi \sqrt{1-\rho ^{2}}}\int ^{-\infty }_{\frac{\ln (\frac{S}{K})}{\sqrt{2E(\tau )}}}\int ^{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2G_{1}(\tau )}}}_{\infty } \exp \left\{ \eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) +\left[ \frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\right. \right. \nonumber \\&\qquad \left. -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) }{2G_{1}(\tau )(1-\rho ^{2})}\frac{G(\tau )}{\sqrt{2E(\tau )}}\right] x+\left[ \frac{\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}\right. \nonumber \\&\qquad \left. \left. -\sqrt{2G_{1}(\tau )}\right] y+\frac{\rho xy}{2E(\tau )G_{1}(\tau )(1-\rho ^{2})}-\frac{(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\right\} dxdy\nonumber \\&\quad =\frac{(1-\alpha )V}{D}(P'(\delta ,\tau )Se^{(r+\rho _{sv}\sigma _{s}\sigma _{v}-\lambda _{v}\sigma _{v})\tau }\mathcal {N}_{2}(e'_{1},-f'_{1},-\rho )\nonumber \\&\qquad +e^{-\lambda _{v}\sigma _{v}\tau }(F-K)\mathcal {N}_{2}(e'_{2},-f'_{2},-\rho )), \end{aligned}$$

(55)

where

$$\begin{aligned} P'(\delta ,\tau )&=A'(\tau )\exp (-\delta H(\tau )), \nonumber \\ A'(\tau )&=\exp \left\{ \left( \theta ^{*}+\gamma _{1}+\gamma _{2}-\frac{\sigma _{\delta }^{2}}{2\kappa ^{2}}\right) (H(\tau )-\tau )-\frac{\sigma _{\delta }^{2}}{4\kappa }H^{2}(\tau )\right\} ,\nonumber \\ e'_{1}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P'(\delta ,\tau )+\tau r+\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds+\rho _{sv}\sigma _{s}\sigma _{v}\tau }{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\nonumber \\ f'_{1}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}+\frac{1}{2}\sigma _{v}^{2}\right) \tau +\rho _{sv}\sigma _{s}\sigma _{v}-\gamma _{2}(\tau -H(\tau ))}{\sigma _{v}\sqrt{\tau }},\nonumber \\ e'_{2}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P'(\delta ,\tau )+\tau r-\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds+\rho _{sv}\sigma _{s}\sigma _{v}\tau }{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\nonumber \\ f'_{2}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}+\frac{1}{2}\sigma _{v}^{2}\right) \tau }{\sigma _{v}\sqrt{\tau }}. \nonumber \\ I_{B}^{4}(\tau ,S,V,\delta )&=\frac{(1-\alpha )V}{D}\frac{F}{2\pi \sqrt{1-\rho ^{2}}}\int ^{\frac{\ln \left( \frac{S}{K}\right) }{\sqrt{2E(\tau )}}}_{\infty }\int ^{\frac{\ln \left( \frac{V}{D^{*}}\right) }{\sqrt{2G_{1}(\tau )}}}_{\infty } \exp \Bigg \{\eta -\frac{1}{2(1-\rho ^{2})}(x^{2}+y^{2}) +\Bigg [\frac{G_{2}(\tau )}{\sqrt{2E(\tau )}}\nonumber \\&\quad -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) }{2G_{1}(\tau )(1-\rho ^{2})}\frac{G(\tau )}{\sqrt{2E(\tau )}}\Bigg ]x+[\frac{\sqrt{2G_{1}(\tau )}(G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )})}{2G_{1}(\tau )(1-\rho ^{2})}\nonumber \\&\quad -\sqrt{2G_{1}(\tau )}]y+\frac{\rho xy}{2E(\tau )G_{1}(\tau )(1-\rho ^{2})} -\frac{\left( G_{1}(\tau )+G_{3}(\tau )-\frac{G(\tau )G_{2}(\tau )}{2E(\tau )}\right) ^{2}}{4G_{1}(\tau )(1-\rho ^{2})}\Bigg \}dxdy\nonumber \\&=\frac{(1-\alpha )V}{D}e^{-\lambda _{v}\sigma _{v}\tau }F\mathcal {N}_{2}(-g'_{2},-h'_{2},\rho ), \end{aligned}$$

(56)

where

$$\begin{aligned} g'_{2}=e'_{2},h'_{2}=f'_{2}. \end{aligned}$$

Finally, we can recombine $I_{B}^{1}(\tau ,S,V,\delta )$, $I_{B}^{2}(\tau ,S,V,\delta )$, $I_{B}^{3}(\tau ,S,V,\delta )$ and $I_{B}^{4}(\tau ,S,V,\delta )$ in the following formula:

$$\begin{aligned} B(S,V,\delta ,\tau )&=P(\delta ,\tau )S\mathcal {N}_{2}(a'_{1},b'_{1},\rho )+e^{-r\tau }(F-K)\mathcal {N}_{2}(a'_{2},b'_{2},\rho )\nonumber \\&\quad +\,e^{-r\tau }F\mathcal {N}_{2}(-c'_{2},d'_{2},-\rho )\nonumber \\&\quad +\,\frac{(1-\alpha )V}{D}(P'(\delta ,\tau )Se^{(r-\lambda _{1}\sigma _{v}+\rho _{sv}\sigma _{s}\sigma _{v})\tau }\mathcal {N}_{2}(e'_{1},-f'_{1},-\rho )\nonumber \\&\quad +\,e^{-\lambda _{v}\sigma _{v}\tau }(F-K)\mathcal {N}_{2}(e'_{2},-f'_{2},-\rho ))\nonumber \\&\quad +\,\frac{(1-\alpha )V}{D}e^{-\lambda _{v}\sigma _{v}\tau }F\mathcal {N}_{2}(-g'_{2},-h'_{2},\rho ), \end{aligned}$$

(57)

where $\mathcal {N}_{2}$ is the standard bivariate normal cumulative distribution as

$$\begin{aligned} \mathcal {N}_{2}(a,b,\rho )=\frac{1}{2\pi \sqrt{1-\rho ^{2}}}\int _{-\infty }^{a}\int _{-\infty }^{b}e^{-\frac{1}{2(1-\rho ^{2})}(x^{2}-2\rho xy+y^{2})}dxdy, \end{aligned}$$

and

$$\begin{aligned} \rho&=\frac{(\rho _{sv}\sigma _{s}\sigma _{v}-\gamma _{2})\tau +\gamma _{2}H(\tau )}{\sigma _{v}\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds\tau }},\\ a'_{1}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P(\delta ,\tau )+r\tau +\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\\ b'_{1}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}+\rho _{sv}\sigma _{s}\sigma _{v}-\gamma _{2}-\frac{1}{2}\sigma _{v}^{2}\right) \tau +\gamma _{2}H(\tau )}{\sigma _{v}\sqrt{\tau }},\\ a'_{2}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P(\delta ,\tau )+r\tau -\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(S)dS}},\\ b'_{2}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}-\frac{1}{2}\sigma _{v}^{2}\right) \tau }{\sigma _{v}\sqrt{\tau }},\\ e'_{1}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P'(\delta ,\tau )+(r+\rho _{sv}\sigma _{s}\sigma _{v})\tau +\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\\ f'_{1}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}+\rho _{sv}\sigma _{s}\sigma _{v}-\gamma _{2}+\frac{1}{2}\sigma _{v}^{2}\right) \tau +\gamma _{2}H(\tau )}{\sigma _{v}\sqrt{\tau }},\\ e'_{2}&=\frac{\ln \left( \frac{S}{K}\right) +\ln P'(\delta ,\tau )+ (r+\rho _{sv}\sigma _{s}\sigma _{v})\tau -\frac{1}{2}\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}{\sqrt{\int ^{\tau }_{0}\hat{\sigma }^{2}_{s}(s)ds}},\\ f'_{2}&=\frac{\ln \left( \frac{V}{D^{*}}\right) +\left( r-\lambda _{v}\sigma _{v}+\frac{1}{2}\sigma _{v}^{2}\right) \tau }{\sigma _{v}\sqrt{\tau }},\\ c'_{2}&=a'_{2}, d'_{2}=b'_{2},g'_{2}=e'_{2},h'_{2}=f'_{2}. \end{aligned}$$

$\square$