Analytical pricing formulae for vulnerable vanilla and barrier options

Abstract

This paper proposes analytically vulnerable vanilla option pricing formulae that simultaneously consider the premature default, the correlation between the underlying asset and the issuer’s asset, and other outstanding debts of the issuer. Our pricing formulae can be easily extended to solve the problem of pricing vulnerable barrier options, which has been rarely studied before. We show that previous studies on pricing (non)-vulnerable vanilla options and barrier options are degenerate cases of our formulae. We conduct numerical experiments to analyze the relations among the financial/contract parameters and counterparty risk, and also empirically evaluate vulnerable vanilla warrants on the TAIEX issued by Capital Securities with detailed studies of parameter calibrations to examine the robustness of our approach.

Appendices

Appendix 1: Proof of Lemma 3.1

Note that ${\hat{B}}_1(t)$ and ${\hat{B}}_2(t)$ can be expressed as

$$\begin{aligned} {\hat{B}}_1(t)\equiv & {} \alpha t + W_1(t), \\ {\hat{B}}_2(t)\equiv & {} \beta t + \rho W_1(t)+\sqrt{1-\rho ^2}W_2(t), \end{aligned}$$

(33)

where $W_1(t)$ and $W_2(t)$ are independent standard Brownian motions. The joint probability density function of ${\hat{M}}_1(T)$ and ${\hat{B}}_1(T)$ is known in closed form as

$$\begin{aligned} f_{{\hat{M}}_1(T),{\hat{B}}_1(T)}(m, b) = {\left\{ \begin{array}{ll} \frac{2(b-2m)}{T\sqrt{2\pi T}} e^{\alpha b-\frac{1}{2}\alpha ^2 T-\frac{1}{2T}(2m-b)^2} &{} \text{ if } m\le b^- \\ 0 &{} \text{ otherwise } \end{array}\right. }, \end{aligned}$$

(34)

where $b^-$ stands for $\min (b, 0)$. For the derivation see for example the proof of Theorem 7.2.1 in Shreve (2004).

The joint cumulative distribution function of ${\hat{M}}_1(T)$, ${\hat{B}}_1(T)$, and ${\hat{B}}_2(T)$ can be derived as

$$\begin{aligned}&P\left\{ {\hat{M}}_1(T)<m_1,{\hat{B}}_1(T)<b_1,{\hat{B}}_2(T)<b_2\right\} \end{aligned}$$

(35)

$$\begin{aligned}& = {} P\left\{ {\hat{M}}_1(T)<m_1,{\hat{B}}_1(T)<b_1, W_2(T)<\frac{b_2 - \beta T + \rho \alpha T -\left( \rho \alpha T + \rho W_1(T)\right) }{\sqrt{1-\rho ^2}}\right\} \end{aligned}$$

(36)

$$\begin{aligned}& = {} P\left\{ {\hat{M}}_1(T)<m_1,{\hat{B}}_1(T)<b_1, W_2(T)<\frac{b_2 + (\rho \alpha - \beta ) T - \rho {\hat{B}}_1(T)}{\sqrt{1-\rho ^2}}\right\} \\& = {} \int _{-\infty }^{m_1}\int _{-\infty }^{b_1}\int _{-\infty }^{\frac{b_2 + (\rho \alpha - \beta ) T - \rho b}{\sqrt{1-\rho ^2}}}f_{{\hat{M}}_1(T),{\hat{B}}_1(T)}(m, b)f_{W_2(T)}(w)dw db dm, \end{aligned}$$

(37)

where the last inequality in Eq. (36) is obtained by substituting Eq. (33) into the last inequality in Eq. (35), and $f_{{\hat{M}}_1(T),{\hat{B}}_1(T)}$ and $f_{W_2(T)}$ in Eq. (37) denote the density function of $({\hat{M}}_1(T),{\hat{B}}_1(T)),$ and $W_2(T)$, respectively. By differentiating Eq. (37) with respect to $m_1$, $b_1$, and $b_2$, we obtain the joint density function of ${\hat{M}}_1(T)$, ${\hat{B}}_1(T)$, and ${\hat{B}}_2(T)$:

$$\begin{aligned} f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_2(T)}(m_1, b_1, b_2)& = {} \frac{\partial }{\partial m_1 \partial b_1 \partial b_2}\int _{-\infty }^{m_1}\int _{-\infty }^{b_1}\int _{-\infty }^{\frac{b_2 + (\rho \alpha - \beta ) T - \rho b}{\sqrt{1-\rho ^2}}}f_{{\hat{M}}_1(T),{\hat{B}}_1(T)}(m, b)f_{W_2(T)}(w)dw db dm\\& = {} \frac{1}{\sqrt{1-\rho ^2}}f_{{\hat{M}}_1(T),{\hat{B}}_1(T)}(m_1, b_1) f_{W_2(T)}\left( \frac{b_2 + (\rho \alpha - \beta )T - \rho b_1 }{\sqrt{1-\rho ^2}}\right) . \end{aligned}$$

By substituting Eq. (34) into the above formula, we obtain the joint density function illustrated in Eq. (14) in Lemma 3.1.

Appendix 2: Proof of Theorem 3.2

By substituting the joint probability density function Eq. (14) derived in Lemma 3.1 into the triple integral illustrated in Theorem 3.2, we have

$$\begin{aligned}&\int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty \int _{{\bar{m}}_1}^\infty e^{p b_1 + q b_2} f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_2(T)}(m_1, b_1, b_2) dm_1 db_1 db_2 \\& = {} \int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty \int _{{\bar{m}}_1}^{b_1^-} e^{p b_1 + q b_2} \frac{1}{\sqrt{1-\rho ^2}}\frac{2(b_1-2m_1)}{T\sqrt{2\pi T}}e^{\alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{1}{2T}(b_1-2m_1)^2}\frac{1}{\sqrt{2\pi T}}e^{-\frac{1}{2T(1-\rho ^2)}(b_2+(\rho \alpha -\beta )T-\rho b_1)^2} dm_1 db_1 db_2 \\& = {} \int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty e^{p b_1 + q b_2} \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{\sqrt{2\pi T}}e^{-\frac{1}{2T(1-\rho ^2)}(b_2+(\rho \alpha -\beta )T-\rho b_1)^2}\left( \int _{{\bar{m}}_1}^{b_1^-}\frac{2(b_1-2m_1)}{T\sqrt{2\pi T}}e^{\alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{1}{2T}(b_1-2m_1)^2}dm_1 \right) db_1 db_2. \end{aligned}$$

(38)

Note that the last integral in the above equation can be simplified as

$$\begin{aligned} \int _{{\bar{m}}_1}^{b_1^-}\frac{2(b_1-2m_1)}{T\sqrt{2\pi T}}e^{\alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{1}{2T}(b_1-2m_1)^2}dm_1& = {} \left[ \frac{1}{\sqrt{2\pi T}}e^{\alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{1}{2T}(b_1-2m_1)^2} \right] _{m_1 = {\bar{m}}_1}^{m_1 = b_1^-}\\& = {} \frac{1}{\sqrt{2\pi T}}e^{\alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{b_1^2}{2T}}\left( 1 - e^\frac{2{\bar{m}}_1(b_1-{\bar{m}}_1)}{T}\right) . \end{aligned}$$

By substituting the above simplification into Eq. (38), we have

$$\begin{aligned}&\int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty e^{p b_1 + q b_2} \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{2\pi T}e^{-\frac{1}{2T(1-\rho ^2)}(b_2+(\rho \alpha -\beta )T-\rho b_1)^2 + \alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{b_1^2}{2T}}\left( 1 - e^\frac{2{\bar{m}}_1(b_1-{\bar{m}}_1)}{T}\right) db_1 db_2 \\& = {} \int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty e^{p b_1 + q b_2} \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{2\pi T}e^{-\frac{1}{2T(1-\rho ^2)}(b_2+(\rho \alpha -\beta )T-\rho b_1)^2 + \alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{b_1^2}{2T}} db_1 db_2 \end{aligned}$$

(39)

$$\begin{aligned}&-e^{-\frac{2{\bar{m}}_1^2}{T}} \int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty e^{\left( p+\frac{2{\bar{m}}_1}{T}\right) b_1 + q b_2} \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{2\pi T}e^{-\frac{1}{2T(1-\rho ^2)}(b_2+(\rho \alpha -\beta )T-\rho b_1)^2 + \alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{b_1^2}{2T}} db_1 db_2. \end{aligned}$$

(40)

The aforementioned formulae can be expressed in terms of a linear combination of the tail probabilities of the standard bivariate normal distributions by a changing of variables as illustrated in the following lemma:

Lemma 7.1

Let $C_1$ and $C_2$ be two arbitrary constants. Then

$$\begin{aligned}&\int _{{\bar{b}}_2}^\infty \int _{{\bar{b}}_1}^\infty e^{C_1 b_1 + C_2 b_2} \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{2\pi T}e^{-\frac{1}{2T(1-\rho ^2)}(b_2+(\rho \alpha -\beta )T-\rho b_1)^2 + \alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{b_1^2}{2T}} db_1 db_2 \end{aligned}$$

(41)

$$\begin{aligned}& = {} e^{T(\frac{C_1^2}{2} + \rho C_1C_2 + \frac{C_2^2}{2} + \alpha C_1 + \beta C_2)} N_2\left( -\frac{ {\bar{b}}_1}{\sqrt{T}} + \sqrt{T}(\alpha + C_1 + \rho C_2) , -\frac{ {\bar{b}}_2}{\sqrt{T}}+\sqrt{T}(\beta + \rho C_1 + C_2), \rho \right) . \end{aligned}$$

(42)

Proof

By substituting $b_1 = -\sqrt{T} x + T(\alpha + C_1 + \rho C_2)$ and $b_2 = -\sqrt{T} y + T(\beta + \rho C_1 + C_2)$ into Eq. (41), we have

$$\begin{aligned}&e^{T(\frac{C_1^2}{2} + \rho C_1C_2 + \frac{C_2^2}{2} + \alpha C_1 + \beta C_2)}\int _{-\frac{ {\bar{b}}_2}{\sqrt{T}}+\sqrt{T}(\beta + \rho C_1 + C_2)}^{-\infty }\int _{ -\frac{ {\bar{b}}_1}{\sqrt{T}} + \sqrt{T}(\alpha + C_1 + \rho C_2) }^{-\infty } \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{2\pi }e^{-\frac{1}{2(1-\rho ^2)} \left( x^2 - 2\rho x y + y^2\right) } dx dy\\& = {} e^{T(\frac{C_1^2}{2} + \rho C_1C_2 + \frac{C_2^2}{2} + \alpha C_1 + \beta C_2)} \int ^{-\frac{ {\bar{b}}_2}{\sqrt{T}}+\sqrt{T}(\beta + \rho C_1 + C_2)}_{-\infty }\int ^{ -\frac{ {\bar{b}}_1}{\sqrt{T}} + \sqrt{T}(\alpha + C_1 + \rho C_2) }_{-\infty } \frac{1}{\sqrt{1-\rho ^2}}\frac{1}{2\pi }e^{-\frac{1}{2(1-\rho ^2)} \left( x^2 - 2\rho x y + y^2\right) } dx dy, \end{aligned}$$

which can be interpreted as a double integral over a bivariate normal density function. The result is equal to Eq. (42). $\square$

By substituting $C_1=p$ and $C_2=q$ into Eq.  (41), Eq. (39) can be simplified as

$$\begin{aligned} e^{T\left( \frac{p^2}{2} + \rho p q + \frac{q^2}{2} + \alpha p + \beta q\right) } N_2\left( -\frac{ {\bar{b}}_1}{\sqrt{T}} + \sqrt{T}(\alpha + p + \rho q) , -\frac{ {\bar{b}}_2}{\sqrt{T}}+\sqrt{T}(\beta + \rho p + q), \rho \right) . \end{aligned}$$

Again, by substituting $C_1 = p + \frac{2{\bar{m}}_1}{T}$, $C_2 = q$ into Eq. (41), Eq. (40) can be simplified as

$$\begin{aligned} - e^{T \left( \frac{p^2}{2}+ \rho p q+\frac{q^2}{2}+\alpha p+ \beta q\right) +2 {\bar{m}}_1(p +\rho q) + 2 \alpha {\bar{m}}_1}N_2\left( -\frac{ {\bar{b}}_1-2{\bar{m}}_1}{\sqrt{T}} + \sqrt{T}\left( \alpha + p + \rho q\right) , -\frac{ {\bar{b}}_2-2\rho {\bar{m}}_1}{\sqrt{T}}+\sqrt{T}\left( \beta + \rho p + q\right) , \rho \right) . \end{aligned}$$

By summing the above two terms, we obtain Theorem 3.2.

Appendix 3: Proof of Convergence of Blocks A and B in Eq. (16)

To show that Eq. (16) degenerates to the Black–Scholes call option pricing formula as default boundary $D \rightarrow 0$, it suffices to show that blocks A and B both converge to zero by applying the squeeze theorem and L’Hôpital’s rule described as follows. Recall that

$$\begin{aligned} l \equiv \frac{\log (\frac{D}{V_{0}})}{\sigma _{V}}, \quad d_{4} \equiv \frac{\log (\frac{V_{0}}{D}) + (r+\sigma _{V}^{2}/2)T }{\sigma _{V}\sqrt{T}} = \frac{- \frac{\log (\frac{D}{V_{0}})}{\sigma _{V}}}{\sqrt{T}} + \frac{(r+\sigma _{V}^{2}/2)T }{\sigma _{V}\sqrt{T}} \equiv -\frac{l}{\sqrt{T}} + C_{1}, \end{aligned}$$

where $C_{1}$ denotes the finite constant $\frac{(r+\sigma _{V}^{2}/2)T }{\sigma _{V}\sqrt{T}}$. Let us focus on the analysis of block B first. Note that

$$\begin{aligned}&0 \le e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_2\left( d_4 + \frac{2l}{\sqrt{T}} , d_2 + \frac{2\rho l}{\sqrt{T}} , \rho \right) < e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_2\left( d_4 + \frac{2l}{\sqrt{T}} , \infty , \rho \right) \\&\quad \le e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_1\left( d_4 + \frac{2l}{\sqrt{T}} \right) . \end{aligned}$$

To show that block B converges to zero as $D \rightarrow 0$ (i.e., $l \rightarrow -\infty$) by the squeeze theorem, it suffices to show that

$$\begin{aligned} \lim _{l\rightarrow -\infty } e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_1\left( d_4 + \frac{2l}{\sqrt{T}}\right) = 0 . \end{aligned}$$

(43)

The term in the normal cumulative distribution function $N_{1}(.)$ can be rewritten as

$$\begin{aligned} N_1\left( d_4 + \frac{2l}{\sqrt{T}} \right) = N_1\left( -\frac{l}{\sqrt{T}} + C_{1} + \frac{2l}{\sqrt{T}} \right)& = {} N_1\left( \frac{l}{\sqrt{T}} + C_{1} \right) \\& = {} \int _{-\infty }^{\frac{l}{\sqrt{T}} + C_{1}} \frac{1}{\sqrt{2\pi }}e^{-\frac{t^{2}}{2}}dt. \end{aligned}$$

Thus we have

$$\begin{aligned} \lim _{l\rightarrow -\infty } e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_1\left( d_4 + \frac{2l}{\sqrt{T}}\right)& = {} \lim _{l\rightarrow -\infty } e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_1\left( \frac{l}{\sqrt{T}} + C_{1}\right) \\& = {} \lim _{l\rightarrow -\infty } \frac{ \int _{-\infty }^{\frac{l}{\sqrt{T}} + C_{1}} \frac{1}{\sqrt{2\pi }}e^{-\frac{t^{2}}{2}}dt }{ e^{\frac{2l}{\sigma _V}(\sigma _V^2/2 - r)} } \\& = {} \lim _{l\rightarrow -\infty } \frac{ \int _{-\infty }^{\frac{l}{\sqrt{T}} + C_{1}} \frac{1}{\sqrt{2\pi }}e^{-\frac{t^{2}}{2}}dt }{ e^{C_{2}l} }, \end{aligned}$$

(44)

where $C_{2} \equiv \frac{2}{\sigma _V}(\sigma _V^2/2 - r)$. If $r-\sigma _V^2/2 > 0$, then $\lim _{l\rightarrow -\infty } e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}= 0$ and Eq. (43) is ensured. If $r-\sigma _V^2/2 =0$, then $\lim _{l\rightarrow -\infty } e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}= 1$ and Eq. (43) is ensured since $\lim _{l\rightarrow -\infty } N_1\left( d_4 + \frac{2l}{\sqrt{T}}\right) =0$. If $r-\sigma _V^2/2 < 0$, the LHS of Eq. (43) is an indeterminate form and L’Hôpital’s rule is applied in the first equality of the following derivations to solve the limit:

$$\begin{aligned} \lim _{l\rightarrow -\infty } \frac{ \int _{-\infty }^{\frac{l}{\sqrt{T}} + C_{1}} \frac{1}{\sqrt{2\pi }}e^{-\frac{t^{2}}{2}}dt }{ e^{C_{2}l} }& = {} \lim _{l\rightarrow -\infty } \frac{ \frac{1}{\sqrt{2\pi T}} e^{-\frac{ (\frac{l}{\sqrt{T}} + C_{1})^{2} }{2}} }{ C_{2}e^{C_{2}l} } \\& = {} \lim _{l\rightarrow -\infty } \frac{1}{C_{2}\sqrt{2\pi T}} e^{ \frac{ - \frac{l^2}{T}-2C_{1}\frac{l}{ \sqrt{T} } - C_{1}^{2} - 2C_{2}l}{2} } \\& = {} \lim _{l\rightarrow -\infty } \frac{1}{C_{2}\sqrt{2\pi T}} e^{ \frac{ - \frac{l^2}{T}- 2l(\frac{C_{1}}{ \sqrt{T} } + C_{2}) - C_{1}^{2}}{2} }. \end{aligned}$$

Recall that $C_{1}$ and $C_{2}$ are finite constants. Thus we have

$$\begin{aligned} \lim _{l\rightarrow -\infty } \frac{ - \frac{l^2}{T}- 2l(\frac{C_{1}}{ \sqrt{T} } + C_{2}) - C_{1}^{2}}{2} = -\infty . \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{l\rightarrow -\infty } \frac{1}{C_{2}\sqrt{2\pi T}} e^{ \frac{ - \frac{l^2}{T}- 2l(\frac{C_{1}}{ \sqrt{T} } + C_{2}) - C_{1}^{2}}{2} } = 0, \end{aligned}$$

ensuring

$$\begin{aligned} \lim _{l\rightarrow -\infty }e^{\frac{2l}{\sigma _V}(r-\sigma _V^2/2)}N_2\left( d_4 + \frac{2l}{\sqrt{T}} , d_2 + \frac{2\rho l}{\sqrt{T}} , \rho \right) = 0. \end{aligned}$$

A similar analysis can be applied to show that block A converges to zero by showing that

$$\begin{aligned}&0 \le \lim _{l\rightarrow -\infty }e^{ 2\rho \sigma _{S}l + \frac{2l}{\sigma _V}(r-\sigma _V^2/2) }N_2\left( d_4 + \frac{2l}{\sqrt{T}}+\sqrt{T}\rho \sigma _{S} , d_1 + \frac{2\rho l}{\sqrt{T}} , \rho \right) \\&\quad \le \lim _{l\rightarrow -\infty }e^{ 2\rho \sigma _{S}l + \frac{2l}{\sigma _V}(r-\sigma _V^2/2) }N_1\left( d_4 + \frac{2l}{\sqrt{T}}+\sqrt{T}\rho \sigma _{S} \right) =0. \end{aligned}$$

Appendix 4: Proof of Theorem 4.1

The triple integral on the left-hand side of Eq. (27) can be solved by first exchanging the upper and the lower limit of the outermost integral so that the resulting formula can be simplified by applying Theorem 3.2. We first substitute $-b_3$ for $b_2$ in the left-hand side of Eq. (27) to obtain

$$\begin{aligned}&\int ^{{\bar{b}}_2}_{-\infty }\int _{{\bar{b}}_1}^\infty \int _{{\bar{m}}_1}^\infty e^{p b_1 + q b_2} f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_2(T)}(m_1, b_1, b_2) dm_1 db_1 db_2 \\& = {} \int _{-{\bar{b}}_2}^{\infty }\int _{{\bar{b}}_1}^\infty \int _{{\bar{m}}_1}^\infty e^{p b_1 - q b_3} f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_2(T)}(m_1, b_1, -b_3) dm_1 db_1 db_3. \end{aligned}$$

(45)

According to Lemma 3.1, the density function $f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_2(T)}(m_1, b_1, -b_3)$ is obtained by substituting $-b_3$ for $b_2$ in Eq. (14) as

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{\sqrt{1-\rho ^2}}\frac{2(b_1-2m_1)}{T\sqrt{2\pi T}}e^{\alpha b_1 - \frac{1}{2}\alpha ^2 T-\frac{1}{2T}(b_1-2m_1)^2}\frac{1}{\sqrt{2\pi T}}e^{-\frac{1}{2T(1-\rho ^2)}(-b_3+(\rho \alpha -\beta )T-\rho b_1)^2}&{} \text{ if } m_1\le b_1^- \\ 0 &{} \text{ otherwise } \end{array}\right. }. \end{aligned}$$

(46)

This density function is exactly the same as the density function of $f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_3(T)}(m_1, b_1, b_3)$, where ${\hat{B}}_3(t)$ is defined as $-\beta t - \rho W_1(t) + \sqrt{1-\rho ^2}W_2(t)$. Thus, Eq. (45) can be rewritten as

$$\begin{aligned}&\int _{-{\bar{b}}_2}^{\infty }\int _{{\bar{b}}_1}^\infty \int _{{\bar{m}}_1}^\infty e^{p b_1 - q b_3} f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_2(T)}(m_1, b_1, -b_3) dm_1 db_1 db_3,\\& = {} \int _{-{\bar{b}}_2}^{\infty }\int _{{\bar{b}}_1}^\infty \int _{{\bar{m}}_1}^\infty e^{p b_1 - q b_3} f_{{\hat{M}}_1(T),{\hat{B}}_1(T),{\hat{B}}_3(T)}(m_1, b_1, b_3) dm_1 db_1 db_3. \end{aligned}$$

This triple integral can be further simplified by Theorem 3.2 to obtain the right-hand side of Eq. (27).

Appendix 5: Barrier option pricing formulae

The pricing formulae for vulnerable up-and-out calls $C_{UO}$, up-and-in puts $P_{UI}$, and down-and-in puts $P_{DI}$ under Merton’s structural credit risk model are given as follows:

$$\begin{aligned} C_{UO} =&S_0\left[ N_2\left( f_1+\frac{b}{\sqrt{T}} , -e_1, -\rho \right) - N_2\left( f_1+\frac{k^\prime }{\sqrt{T}}, -e_1, -\rho \right) \right] \\- & {} S_0 e^{2b\sigma _S - 2b\alpha _S}\left[ N_2\left( f_2+\frac{b}{\sqrt{T}} , -e_2, -\rho \right) - N_2\left( f_2+\frac{k^\prime }{\sqrt{T}}, -e_2, -\rho \right) \right] \\- & {} Ke^{-rT}\left[ N_2\left( f_3+\frac{b}{\sqrt{T}} , -e_3, -\rho \right) - N_2\left( f_3+\frac{k^\prime }{\sqrt{T}}, -e_3, -\rho \right) \right] \\+ & {} Ke^{-rT-2b\alpha _S}\left[ N_2\left( f_4+\frac{b}{\sqrt{T}} , -e_4, -\rho \right) - N_2\left( f_4+\frac{k^\prime }{\sqrt{T}}, -e_4, -\rho \right) \right] \\+ & {} \left( \frac{1-\alpha }{D}\right) V_0 S_0 e^{{\bar{\gamma }} T - 2\sigma _v\rho \alpha _S T + \frac{1}{2}\sigma _S^2 T + \rho \sigma _S\sigma _V T - \sigma _S\alpha _S T}\left[ N_2\left( f_5+\frac{b}{\sqrt{T}} , e_5, \rho \right) - N_2\left( f_5+\frac{k^\prime }{\sqrt{T}}, e_5, \rho \right) \right] \\- & {} \left( \frac{1-\alpha }{D}\right) V_0 S_0 e^{{\bar{\gamma }} T - 2\sigma _v\rho \alpha _S T + \frac{1}{2}\sigma _S^2 T + \rho \sigma _S\sigma _V T - \sigma _S\alpha _S T - 2\alpha _S b +2b\rho \sigma _V+2b\sigma _S} \\&\cdot \left[ N_2\left( f_6+\frac{b}{\sqrt{T}} , e_6, \rho \right) - N_2\left( f_6 + \frac{k^\prime }{\sqrt{T}}, e_6, \rho \right) \right] \\- & {} \left( \frac{1-\alpha }{D}\right) K V_0 e^{-2\sigma _V\rho \alpha _S T}\left[ N_2\left( f_7+\frac{b}{\sqrt{T}} , e_7, \rho \right) - N_2\left( f_7+\frac{k^\prime }{\sqrt{T}}, e_7, \rho \right) \right] \\+ & {} \left( \frac{1-\alpha }{D}\right) K V_0 e^{-2\sigma _V\rho \alpha _S T - 2\alpha _S b + 2b\rho \sigma _V}\left[ N_2\left( f_8+\frac{b}{\sqrt{T}} , e_8, \rho \right) - N_2\left( f_8+\frac{k^\prime }{\sqrt{T}}, e_8, \rho \right) \right] , \\ P_{UI} =&Ke^{-rT-2b\alpha _S}N_2\left( f_4 + \frac{k^{\prime }}{\sqrt{T}}, -e_4, -\rho \right) \\- & {} S_0 e^{{\bar{\gamma }} T - r T + \frac{1}{2}\sigma _S^2 T - \alpha _S\sigma _S T + 2b\sigma _S - 2\alpha _S b}N_2\left( f_2 + \frac{k^{\prime }}{\sqrt{T}}, -e_2, -\rho \right) \\+ & {} \left( \frac{1-\alpha }{D}\right) K V_0 e^{-2\sigma _V\rho \alpha _S T - 2\alpha _S b + 2b\rho \sigma _V}N_2\left( f_8 + \frac{k^{\prime }}{\sqrt{T}}, e_8, \rho \right) \\- & {} \left( \frac{1-\alpha }{D}\right) V_0 S_0 e^{{\bar{\gamma }} T - 2\sigma _V\rho \alpha _S T + \frac{1}{2}\sigma _S^2 T + \rho \sigma _S\sigma _V T - \sigma _S\alpha _S T-2\alpha _S b+2b\rho \sigma _V+2b\sigma _S}N_2\left( f_6 + \frac{k^{\prime }}{\sqrt{T}}, e_6, \rho \right) , \\ P_{DI} =&K e^{-rT}N_2\left( -h_3+\frac{b}{\sqrt{T}}, -g_3, -\rho \right) \\+ & {} K e^{-rT + 2b\alpha _S}\left[ N_2\left( h_4-\frac{b}{\sqrt{T}}, -g_4, \rho \right) - N_2\left( h_4-\frac{k^{\prime }}{\sqrt{T}}, -g_4, \rho \right) \right] \\- & {} S_0 N_2\left( -h_1+\frac{b}{\sqrt{T}}, -g_1, -\rho \right) \\- & {} S_0 e^{2b\sigma _S+2b\alpha _S b}\left[ N_2\left( h_2-\frac{b}{\sqrt{T}}, -g_2, \rho \right) - N_2\left( h_2-\frac{k^{\prime }}{\sqrt{T}}, -g_2, \rho \right) \right] \\+ & {} \left( \frac{1-\alpha }{D}\right) K V_0 N_2\left( -h_7+\frac{b}{\sqrt{T}}, g_7, \rho \right) \\+ & {} \left( \frac{1-\alpha }{D}\right) K V_0 e^{2\alpha _S b+2b\rho \sigma _V} \left[ N_2\left( h_8-\frac{b}{\sqrt{T}}, g_8, -\rho \right) - N_2\left( h_8-\frac{k^{\prime }}{\sqrt{T}}, g_8, -\rho \right) \right] \\- & {} \left( \frac{1-\alpha }{D}\right) V_0 S_0 e^{rT + \rho \sigma _S\sigma _V T}N_2\left( -h_5+\frac{b}{\sqrt{T}}, g_5, \rho \right) \\- & {} \left( \frac{1-\alpha }{D}\right) V_0 S_0 e^{rT + \rho \sigma _S\sigma _V T + 2\alpha _S b+2b\rho \sigma _V+2b\sigma _S}\left[ N_2\left( h_6-\frac{b}{\sqrt{T}}, g_6, -\rho \right) - N_2\left( h_6-\frac{k^{\prime }}{\sqrt{T}}, g_6, -\rho \right) \right] , \end{aligned}$$

(47)

where $\alpha _S \equiv \frac{1}{\sigma _S}(r - {\bar{\gamma }} -\sigma _S^2/2)$. Other variables are defined as follows:

$$\begin{aligned} \begin{array}{lllllll} e_1 &{}\equiv &{} e_3 -\sigma _S\rho \sqrt{T} &{}\qquad f_1 &{}\equiv &{} -\sigma _S\sqrt{T} + \alpha _S\sqrt{T}\\ \\ e_3 &{}\equiv &{} \frac{\ln \frac{D}{V_0}-\left( r-\frac{1}{2}\sigma _V^2 \right) }{\sigma _V\sqrt{T}} + 2\rho \alpha _S\sqrt{T} &{}\qquad f_3 &{}\equiv &{} \alpha _S\sqrt{T}\\ \\ e_5 &{}\equiv &{} e_3 -\sigma _S\rho \sqrt{T} - \sigma _V\sqrt{T} &{}\qquad f_5 &{}\equiv &{} -\sigma _S\sqrt{T} -\sigma _V\rho \sqrt{T} + \alpha _S\sqrt{T}\\ \\ e_7 &{}\equiv &{} e_3 - \sigma _V\sqrt{T} &{}\qquad f_7 &{}\equiv &{} -\sigma _V\rho \sqrt{T} + \alpha _S\sqrt{T}\\ \\ \\ \\ g_1 &{}\equiv &{} g_3 - \sigma _S\rho \sqrt{T} &{}\qquad h_1 &{}\equiv &{} \sigma _S\sqrt{T} + \alpha _S\sqrt{T} \\ \\ g_3 &{}\equiv &{} \frac{\ln \frac{D}{V_0}-\left( r-\frac{1}{2}\sigma _V^2\right) }{\sigma _V\sqrt{T}} &{}\qquad h_3 &{}\equiv &{} \alpha _S \sqrt{T}\\ \\ g_5 &{}\equiv &{} g_3 - \sigma _V\sqrt{T} - \sigma _S\rho \sqrt{T} &{}\qquad h_5 &{}\equiv &{} \sigma _S\sqrt{T} + \sigma _V\rho \sqrt{T} + \alpha _S\sqrt{T} \\ \\ g_7 &{}\equiv &{} g_3 - \sigma _V\sqrt{T} &{}\qquad h_7 &{}\equiv &{} \sigma _V\rho \sqrt{T} + \alpha _S\sqrt{T}\\ \\ \\ \\ e_i &{}\equiv &{} e_{i-1} -\frac{2b\rho }{\sqrt{T}},&{}\qquad f_i &{}\equiv &{} f_{i-1} -\frac{2b}{\sqrt{T}}, \\ \\ g_i &{}\equiv &{} g_{i-1} - \frac{2b\rho }{\sqrt{T}}, &{}\qquad h_i &{}\equiv &{} h_{i-1} + \frac{2b}{\sqrt{T}}\quad \forall i=2, 4, 6, 8. \end{array} \end{aligned}$$

Appendix 6: Proof of Eqs. (31) and (32)

To model the last 30-min average settlement price of TAIEX options, a “partial” Asian option pricing formula is derived to mimic the Black–Scholes formula. Specifically, the average index price over the 30-min interval (denoted by $\tau$) immediately prior to option maturity T is approximated by the geometric mean price $S_G$. Recall that the underlying TAIEX follows the lognormal diffusion process as defined in Eq. (2). $S_{G}$ can be expressed as the limit of $n+1$ equal-space discrete geometric average price over the time interval $[T-\tau ,T]$ as n approaches infinity as follows:

$$\begin{aligned} S_{G}& = {} \lim _{n \rightarrow \infty }\root n+1 \of {S_{T-\tau } \cdot S_{T-\frac{(n-1)\tau }{n}} \cdot \ldots \cdot S_{T} }\\& = {} S_{0} e^{ \frac{ \int _{T-\tau }^{T}(r-\frac{\sigma _{S}^{2}}{2})tdt +\sigma _{S}\int _{T-\tau }^{T}Z_S(t)dt}{\tau } } \\& = {} S_{0} e^{ (r-\frac{\sigma _{S}^{2}}{2})( T-\frac{\tau }{2} ) +\sigma _{S} \frac{\int _{T-\tau }^{T}Z_S(t)dt}{\tau } }. \end{aligned}$$

Note that term $\frac{\int _{T-\tau }^{T}Z_S(t)dt}{\tau }$ in the above equation can be derived as

$$\begin{aligned} \frac{\int _{T-\tau }^{T}Z_S(t)dt}{\tau }& = {} \lim _{n \rightarrow \infty }\frac{ Z_S(T-\tau ) + Z_S(T-\frac{(n-1)\tau }{n} ) + \ldots + Z_S(T) }{n+1} \\& = {} \lim _{n \rightarrow \infty }\frac{ Z_S(T-\tau ) + \left( Z_S(T-\tau )+ I_1 \right) + \ldots + \left( Z_S(T-\tau )+ I_1+ \ldots + I_n \right) }{n+1}\\& = {} \lim _{n \rightarrow \infty }\frac{(n+1)Z_S(T-\tau ) + nI_1 +(n-1)I_2 \ldots + I_n }{n+1}\\& = {} \lim _{n \rightarrow \infty } \left\{ Z_S(T-\tau ) + \frac{n}{n+1}I_1 + \ldots + \frac{1}{n+1}I_n\right\} , \end{aligned}$$

where $I_i$ is the value increment for $Z_S$: $Z_S(T-\frac{(n-i)\tau }{n}) - Z_S(T-\frac{(n-(i-1))\tau }{n})$, and $I_i {\mathop {\sim }\limits ^{iid}} N(0,\frac{\tau }{n})$. Clearly, $\frac{\int _{T-\tau }^{T}Z_S(t)dt}{\tau }$ follows a normal distribution, and the mean can be derived as

$$\begin{aligned} E\left[ \frac{\int _{T-\tau }^{T}Z_S(t)dt}{\tau }\right] = \lim _{n \rightarrow \infty } E\left[ Z_S(T-\tau ) + \frac{n}{n+1}I_1 + \ldots + \frac{1}{n+1}I_n \right] = 0. \end{aligned}$$

The variance is

$$\begin{aligned} Var\left[ \frac{\int _{T-\tau }^{T}Z_S(t)dt}{\tau }\right]& = {} \lim _{n \rightarrow \infty } Var\left[ Z_S(T-\tau ) + \frac{n}{n+1}I_1 + \ldots + \frac{1}{n+1}I_n \right] \\& = {} \lim _{n \rightarrow \infty } \left[ Var\left[ Z_S(T-\tau ) \right] + \frac{n^2}{(n+1)^2}Var\left[ I_1 \right] + \ldots + \frac{1}{(n+1)^2}Var\left[ I_n\right] \right] \\& = {} (T-\tau ) + \lim _{n \rightarrow \infty } \frac{(n^2+ \ldots + 1)}{(n+1)^2} \times \frac{\tau }{n}\\& = {} (T-\tau ) + \lim _{n \rightarrow \infty } \frac{n(n+1)(2n+1)\tau }{ 6n(n+1)^2 } \\& = {} (T-\tau ) + \frac{\tau }{3} \\& = {} T-\frac{2\tau }{3}. \end{aligned}$$

Consequently, the distribution of $S_{G}$ is

$$\begin{aligned} \ln S_{G} \sim N\left( \ln S_0 + (r-\frac{\sigma _{S}^{2}}{2})( T-\frac{\tau }{2} ), \sigma ^{2}_{S} (T-\frac{2\tau }{3}) \right) . \end{aligned}$$

Thus $S_{G}$ can be expressed as $e^{m+wQ}$, where

$$\begin{aligned} m& = {} \ln S_0 + (r-\frac{\sigma _{S}^{2}}{2})( T-\frac{\tau }{2} ), \\ w^2& = {} \sigma ^{2}_{S} (T-\frac{2\tau }{3}), \end{aligned}$$

and Q denotes a standard normal random variable. The pricing formula for the partial Asian option can be expressed as the vanilla option pricing formulae with the underlying asset price $S_G$ and the strike price K. Note that the probability density function for the standard normal random variable Q is

$$\begin{aligned} h(Q)& = {} \frac{1}{\sqrt{2\pi }}e^{-\frac{ Q^2 }{2} }. \end{aligned}$$

The option is in the money if

$$\begin{aligned} S_G=e^{m+wQ}\ge K \rightarrow Q\ge \frac{\ln K -m}{w}. \end{aligned}$$

The call option value $C_G$ can be expressed as

$$\begin{aligned} C_G& = {} e^{-rT} E\left[ (S_G-K)^+ \right] \\& = {} e^{-rT} \left( \int _{\frac{\ln K -m}{w}}^{\infty } (e^{m+wQ}-K) \frac{1}{\sqrt{2\pi }}e^{-\frac{Q^2}{2}}dQ \right) \\& = {} e^{-rT} \left( \int _{\frac{\ln K -m}{w}}^{\infty } e^{m+wQ} \frac{1}{\sqrt{2\pi }}e^{-\frac{Q^2}{2}}dQ\right) - Ke^{-rT} \int _{\frac{\ln K -m}{w}}^{\infty } \frac{1}{\sqrt{2\pi }}e^{-\frac{Q^2}{2}}dQ \\& = {} e^{-rT} \cdot e^{m+\frac{w^2}{2}} \int _{\frac{\ln K -m}{w}}^{\infty } \frac{1}{\sqrt{2\pi }}e^{-\frac{(Q-w)^2}{2}}dQ - Ke^{-rT} \int _{\frac{\ln K -m}{w}}^{\infty } \frac{1}{\sqrt{2\pi }}e^{-\frac{Q^2}{2}}dQ \\& = {} e^{-rT} \cdot e^{m+\frac{w^2}{2}} N\left( w - (\frac{\ln K -m}{w}) \right) - Ke^{-rT} N\left( \frac{m - \ln K }{w} \right) \\& = {} S_0 e^{-\frac{r}{2}\tau - \frac{\sigma _{S}^{2}}{12}\tau } N\left( d_5 \right) - Ke^{-rT} N(d_6), \end{aligned}$$

where

$$\begin{aligned} d_5& = {} \frac{ \ln \frac{S_0}{K} + (r- \frac{\sigma _{S}^{2}}{2} )(T-\frac{\tau }{2}) }{ \sigma _{S} \sqrt{ T-\frac{2}{3}\tau } } + \sigma _{S} \sqrt{ T-\frac{2}{3}\tau }, \\ d_6& = {} d_5 - \sigma _{S} \sqrt{ T-\frac{2}{3}\tau }. \end{aligned}$$

Similarly, mimicking the above derivations yields the put option value $P_G$

$$\begin{aligned} P_G& = {} e^{-rT} E\left[ (K-S_G)^+ \right] \\& = {} Ke^{-rT} N(-d_6) - S_0 e^{-\frac{r}{2}\tau - \frac{\sigma _{S}^{2}}{12}\tau } N\left( -d_5 \right) . \end{aligned}$$