Extending the Merton model with applications to credit value adjustment

Abstract

Following the global financial crisis, the measurement of counterparty credit risk has become an essential part of the Basel III accord with credit value adjustment being one of the most prominent components of this concept. In this study, we extend the Merton structural credit risk model for counterparty credit risk calculation in the context of calculating the credit value adjustment mainly by estimating the probability of default. We improve the Merton model in a variance-convoluted-gamma environment to include default dependence between counterparties through a linear factor decomposition framework. This allows one to tackle dependence through a systematic common component. Our set-up allows for easier, faster and more accurate fitting for the credit spread. Results confirm that use of the variance-gamma-convolution clearly solves the vanishing credit spread problem for short time-to-maturity or low leverage cases compared to a Brownian motion environment and its modifications.

Appendix

1.1 A.0: The Merton model

Merton (1974) assesses a company’s ability to meet its obligations by evaluating the credit risk of that company’s debt. The model assumes that the total value of the assets, A(t), follows a geometric Brownian motion:

$$\begin{aligned} dA(t)=r A(t)dt+\sigma _{A} A(t) dW(t), \end{aligned}$$

(A.1)

where r is the expected rate of return, $\sigma$ is the volatility of the assets, and W(t) is the Brownian motion at time t. The model further assumes that the financial market is frictionless so that the liquidation value is equal to firm value.¹⁵ Denoting the company’s total value of debt with maturity T by D and the total value of equity by E(t) where $t\le T$, from the fundamental accounting identity we have

$$\begin{aligned} E(T)=\text {max}(A(T)-D,0), \end{aligned}$$

(A.2)

which shows that the equity is an implicit call option on the company’s total value of assets with strike price D and maturity T (Gray & Malone, 2008). Therefore, one can use the B-S call option formula to calculate the market value of equity,

$$\begin{aligned} E(t)=A(t){\mathcal {N}}(d_1)-De^{-r(T-t)}{\mathcal {N}}(d_2), \end{aligned}$$

(A.3)

where

$$\begin{aligned} d_1= & {} \frac{\ln (A(t)/D)+(r+\frac{\sigma _{A}^2}{2})(T-t)}{\sigma \sqrt{T-t}}, \end{aligned}$$

(A.4)

$$\begin{aligned} d_2= & {} d_1-\sigma _{A} \sqrt{T-t}, \end{aligned}$$

(A.5)

and ${\mathcal {N}}(x)$ is the cumulative standard normal distribution function. Under this setup, the default probability at maturity T under the risk-neutral probability measure is

$$\begin{aligned} P(A(T)<D | A(t))={\mathcal {N}}(-d_2). \end{aligned}$$

(A.6)

1.2 A.1: Derivation of the VG model delta

Proof

First, the option value is written as usual:

$$\begin{aligned} {\mathbb {E}}^{}(e^{-r\tau }(S(T)-K)^{+})={\mathbb {E}}^{}(e^{-r\tau }(S(T)-K){\mathbb {I}}_{\left\{ S(T)>K\right\} }). \end{aligned}$$

(A.7)

Then the derivative of this expectation with respect to the stock price is:

$$\begin{aligned} \frac{\partial {\mathbb {E}}^{}((S(T)-K){\mathbb {I}}_{\left\{ S(T)>K\right\} })}{\partial S(t)}&={\mathbb {E}}^{S}({\mathbb {I}}_{\left\{ S(T)>K\right\} }|{\mathcal {F}}(t))+{\mathbb {E}}^{}(\delta ((S(T)-K)e^{-r\tau })S_{T}e^{-r\tau }|{\mathcal {F}}(t)) \nonumber \\&\quad -{\mathbb {E}}^{}(Ke^{-r\tau }\delta ((S(T)-K)e^{-r\tau })| {\mathcal {F}}(t)), \end{aligned}$$

(A.8)

where $\delta$ is the Dirac function which is the derivative of Heaviside function ${\mathbb {I}}_{\left\{ S_{T}>K\right\} }$. Since Dirac terms are equal to 1 when $S(T)=K$, this leads to the following result:

$$\begin{aligned} \frac{\partial {\mathbb {E}}^{}((S(T)-K){\mathbb {I}}_{\left\{ S(T)>K\right\} })}{\partial S}=\frac{\partial C}{\partial S}={\mathbb {E}}^{S}({\mathbb {I}}_{\left\{ S(T)>K\right\} }|{\mathcal {F}}(t))={\mathbb {Q}}^{S}(S(T)>K)=F^{S}(x) \end{aligned}$$

(A.9)

where $F^S(x)$ is the cumulative distribution function under ${\mathbb {Q}}^{S}$. $\square$

1.3 A.2: Sum of weighted independent gamma random variables with different scale and shape parameters

Proof

We start by writing,

$$\begin{aligned} \zeta _{1}= & {} c_{1}\gamma _{1}=\Lambda _{1},\\ \zeta _{2}= & {} c_{1}\gamma _{1}+c_{2}\gamma _{1}=\Lambda _{1}+\Lambda _{2}, \end{aligned}$$

where $\Lambda _{1}\sim Ga(\alpha _{1},c_{1}\sigma _{1})$ and $\Lambda _{2}\sim Ga(\alpha _{2},c_{2}\sigma _{2})$. Then using $\Lambda _{2}=\zeta _{2}-\Lambda _{1} =\zeta _{2}-\zeta _{1}$, we calculate the Jacobian matrix and the determinant,

$$\begin{aligned} \mathbf {\zeta }_{i,j}= & {} \begin{bmatrix} \frac{\partial \Lambda _1}{\partial \zeta _1} &{} \frac{\partial \Lambda _1}{\partial \zeta _2} \\[1ex] \frac{\partial \Lambda _2}{\partial \zeta _1} &{} \frac{\partial \Lambda _2}{\partial \zeta _2} \\[1ex] \end{bmatrix}=\begin{bmatrix} 1&{}0\\[1ex] -1&{}1 \end{bmatrix}, \\ |J|= & {} \begin{vmatrix} 1&0\\ -1&1 \end{vmatrix}=1. \end{aligned}$$

Then we can start by writing the density of $\zeta _{2}$ in line with the convolution of two random variables,

$$\begin{aligned} f(\zeta _{2},\alpha _{1},\alpha _{2},\sigma _{1},\sigma _{2},c_{1},c_{2})=\int _{0}^{\zeta _{2}}\frac{\Lambda _{1}^{\alpha _{1}-1}e^{\frac{-\Lambda _{1}}{c_{1}\sigma _{1}}+\frac{\Lambda _{1}}{c_{2}\sigma _{2 }}}\left( \zeta _{2}-\Lambda _{1}\right) ^{\alpha _{2}-1}e^{\frac{-\zeta _{2}}{c_{2}\sigma _{2}}}}{\Gamma (\alpha _{1})\Gamma (\alpha _{2})}d\Lambda _{1}, \end{aligned}$$

after some tedious algebra and modifying the boundary of the integral we obtain,

$$\begin{aligned}{} & {} f(\zeta _{2},\alpha _{1},\alpha _{2},\sigma _{1},\sigma _{2},c_{1},c_{2})\nonumber \\{} & {} \quad =\zeta _{2}^{\alpha _{2}+\alpha _{1}-1}e^{\frac{-\zeta _{2}}{c_{2}\sigma _{2}}}\underbrace{\int _{0}^{1}\frac{\Lambda _{1}^{\alpha _{1}-1}\left( 1-\Lambda _{1}\right) ^{\alpha _{2}-1}e^{-\zeta _{2}\Lambda _{1}\left( \frac{1}{c_{1}\sigma _{1}}-\frac{1}{c_{2}\sigma _{2 }}\right) }}{\Gamma (\alpha _{1})\Gamma (\alpha _{2})}d\Lambda _{1}}_{\frac{_{1}F\left( \alpha _{1},\alpha _{2}+\alpha _{1},\zeta _{2}(\frac{1}{c_{1}\sigma _{1}}-\frac{1}{c_{2}\sigma _{2}})\right) }{\Gamma \left( \alpha _{1}+\alpha _{2}\right) }}. \end{aligned}$$

(A.10)

We see that the integral in () can be written in terms of the confluent hypergeometric function of the second kind using Gradshteyn and Ryzhik (2007) (page 870, Equation$-$7.621-5).

Therefore final representation will be,

$$\begin{aligned}{} & {} f(\zeta _{2},\nu _{1},\nu _{2},\sigma _{1},\sigma _{2},c_{1},c_{2})=\zeta _{2}^{\alpha _{2} +\alpha _{1}-1}e^{\frac{-\zeta _{2}}{c_{2}\sigma _{2}}}(c_{1}\sigma _{1})^{-\alpha _{1}} (c_{2}\sigma _{2})^{-\alpha _{2}}\nonumber \\{} & {} \frac{_{1}F_{1}\left( \alpha _{1},\alpha _{2}+\alpha _{1}, \zeta _{2}(\frac{1}{c_{1}\sigma _{1}}-\frac{1}{c_{2}\sigma _{2}})\right) }{\Gamma \left( \alpha _{1}+\alpha _{2}\right) } \end{aligned}$$

(A.11)

The characteristic function will be straightforward to derive. First we write characteristic function,

$$\begin{aligned} \Phi _{\zeta _{2}}(u,\alpha _{1},\alpha _{2},\sigma _{1},\sigma _{2},c_{1},c_{2})= & {} \int _{0}^{\infty }e^{iu\zeta _{2}}\zeta _{2}^{\alpha _{2}+\alpha _{1}-1}e^{\frac{-\zeta _{2}}{c_{2}\sigma _{2}}}(c_{1}\sigma _{1})^{-\alpha _{1}}(c_{2}\sigma _{2})^{-\alpha _{2}}\nonumber \\{} & {} \times \frac{_{1}F_{1}\left( \alpha _{1},\alpha _{2}+\alpha _{1},\zeta _{2}(\frac{1}{c_{1}\sigma _{1}}-\frac{1}{c_{2}\sigma _{2}})\right) }{\Gamma \left( \alpha _{1}+\alpha _{2}\right) }d\zeta _{2}\nonumber \\= & {} \int _{0}^{\infty }e^{-\left( \frac{1}{c_{2}\sigma _{2}}-iu\right) \zeta _{2}}\zeta _{2}^{\alpha _{2}+\alpha _{1}-1}(c_{1}\sigma _{1})^{-\alpha _{1}}(c_{2}\sigma _{2})^{-\alpha _{2}} \nonumber \\{} & {} \times \frac{_{1}F_{1}\left( \alpha _{1},\alpha _{2}+\alpha _{1},\zeta _{2}(\frac{1}{c_{1}\sigma _{1}}-\frac{1}{c_{2}\sigma _{2}})\right) }{\Gamma \left( \alpha _{1}+\alpha _{2}\right) }d\zeta _{2} \end{aligned}$$

(A.12)

Then using Gradshteyn and Ryzhik (2007) (page 822, Equation 4) we can write,

$$\begin{aligned}{} & {} \Phi _{\zeta _{2}}(u,\alpha _{1},\alpha _{2},\sigma _{1},\sigma _{2},c_{1},c_{2})=D\left( 2c_{1}\sigma _{1}-c_{2}\sigma _{2} -iuC\right) ^{-\alpha _{1}-\alpha _{2}}\nonumber \\{} & {} \times F\left( \alpha _{2}, \alpha _{1}+\alpha _{2}, \alpha _{1}+\alpha _{2},\frac{k}{k-\left( \frac{1}{c_{2}\sigma _{2}}-iu \right) }\right) \end{aligned}$$

(A.13)

where $D=(c_{1}\sigma _{1})^{\alpha _{2}}(c_{2}\sigma _{2})^{\alpha _{1}}, k=\frac{1}{c_{1}\sigma _{1}}-\frac{1}{c_{2}\sigma _{2}}$ and $C=c_{1}\sigma _{1}c_{2}\sigma _{2}$. $\square$

1.4 A.3: VGC characteristic function and martingale correction factor

Proof

The characteristic function of the VGC¹⁶ random variable can be written as,

$$\begin{aligned} {\mathbb {E}}\left( e^{iuX(t)}\right)= & {} {\mathbb {E}}\left( {\mathbb {E}}\left( e^{iu\theta g+iu \sigma W(g)}|\gamma =g \right) \right) \\= & {} {\mathbb {E}}\left( e^{i\left( u\theta -\frac{u^{2}\sigma ^{2}}{2}\right) g}\right) \\= & {} D\left( 2c_{1}\sigma _{1}-c_{2}\sigma _{2} -iuC\theta +C\frac{u^{2}\sigma ^{2}}{2}\right) ^{-\alpha _{1}-\alpha _{2}}\\\times & {} F\left( \alpha _{2}, \alpha _{1}+\alpha _{2}, \alpha _{1}+\alpha _{2},\frac{k}{k-\left( \frac{1}{c_{2}\sigma _{2}}-iu\theta +\frac{u^{2}\sigma ^{2}}{2} \right) }\right) \end{aligned}$$

Then $\phi _{X}(-i)$, the natural logarithm of VGC characteristic function evaluated at $-i$, and also the martingale correction factor could be written as follows,

$$\begin{aligned} \phi _{X}(-i)= & {} \log (D)+\left( -\alpha _{1}-\alpha _{2}\right) \log \left( 2c_{1}\sigma _{1}-c_{2}\sigma _{2}+A\right) \nonumber \\{} & {} +\log \left( F\left( \alpha _{2}, \alpha _{1}+\alpha _{2}, \alpha _{1}+\alpha _{2},\frac{k}{k-\left( \frac{1}{c_{2}\sigma _{2}}+A \right) }\right) \right) \end{aligned}$$

(A.14)

where $A=C\theta -\frac{C\sigma ^{2}}{2}$ and $\kappa =\left( 2c_{1}\sigma _{1}-c_{2}\sigma _{2}-C\theta -\frac{C\sigma ^{2}}{2}\right)$ which is used in Eq. (39). $\square$

1.5 A.4: Multivariate gamma density

Proof

Let $Z_{1}\left( \nu _{1},\beta _{1}\right) =X_{1}+c_{1}Y$ and $Z_{2}\left( \nu _{2},\beta _{2}\right) =X_{2}+c_{2}Y$ be correlated random variable. Then their joint density can be shown first by re-defining the linear relationships,

$$\begin{aligned} Y= & {} Z_{0}, \\ X_{1}= & {} Z_{1}-c_{1}Z_{0}, \\ X_{2}= & {} Z_{2}-c_{2}Z_{0}. \end{aligned}$$

Then we define the Jacobian

$$\begin{aligned} \mathbf {\zeta }_{i,j}= & {} \begin{bmatrix} \frac{\partial X_1}{\partial Z_0} &{} \frac{\partial X_1}{\partial Z_1} &{} \frac{\partial X_1}{\partial Z_2} \\[1ex] \frac{\partial X_2}{\partial Z_0}&{} \frac{\partial X_2}{\partial Z_1} &{} \frac{\partial Y}{\partial Z_2} \\[1ex] \frac{\partial Y}{\partial Z_0}&{} \frac{\partial Y}{\partial Z_1} &{} \frac{\partial Y}{\partial Z_2} \end{bmatrix}=\begin{bmatrix} -c_{1}&{}1&{}0\\[1ex] -c_{2}&{}0&{}1\\[1ex] 1&{}0&{}0 \end{bmatrix}, \\ |J|= & {} \begin{vmatrix} -c_{1}&1&0\\ -c_{2}&0&1 \\ 1&0&0 \nonumber \end{vmatrix}=1. \end{aligned}$$

We write the density of $Z_{1},Z_{2}$ given that $z_{0},z_{1},z_{2}$ are all gamma distributed random variables with $Ga(\alpha _{0},\beta _{0})$ and $Ga(\alpha _{1},\beta _{1},\alpha _{1},\beta _{2})$ and then we integrate out $z_{0}$ to obtain $f(z_{1},z_{2})$. We begin by writing the component of the density (25),

$$\begin{aligned} f(z_{1},z_{2})= & {} \frac{z_{1}^{\alpha {1}-1}z_{2}^{\alpha {2}-1}e^{-\frac{z_{1}}{\beta _{1}}} e^{-\frac{z_{2}}{\beta _{2}}}\beta _{1}^{-\alpha _{1}}\beta _{2}^{-\alpha _{2}}}{\Gamma (\alpha _{0})\Gamma (\alpha _{1})\Gamma (\alpha _{2})}\int _0^{\min (z_{1},z_{2})}z_{0} ^{\alpha _{0}-1}e^{z_{0}\left( \frac{c_{1}}{\beta _{1}}+\frac{c_{2}}{\beta _{2}}-\frac{1}{\beta _{0}}\right) }\\{} & {} \left( 1-c_{1}\frac{z_{0}}{z_{1}}\right) ^{\alpha _{1}-1}\left( 1-c_{2}\frac{z_{0}}{z_{2}}\right) ^{\alpha _{2}-1}dz_{0}. \end{aligned}$$

Then we write the second component (26),

$$\begin{aligned} f(z_{1},z_{2})= & {} \frac{z_{1}^{\alpha {1}-1}z_{2}^{\alpha {2}-1}e^{-\frac{z_{1}}{\beta _{1}}}e^{-\frac{z_{2}}{\beta _{2}}}\beta _{1}^{-\alpha _{1}}\beta _{2}^{-\alpha _{2}}}{\Gamma (\alpha _{0})\Gamma (\alpha _{1})\Gamma (\alpha _{2})}\int _0^{\min (z_{1},z_{2})}z_{0}^{\alpha _{0}-1} e^{z_{0}\left( \frac{c_{1}}{\beta _{1}}+\frac{c_{2}}{\beta _{2}}-\frac{1}{\beta _{0}}\right) }\\{} & {} \left( 1-c_{1}\frac{z_{0}}{z_{2}}\right) ^{\alpha _{1}-1}\left( 1-c_{2}\frac{z_{0}}{z_{1}}\right) ^{\alpha _{2}-1}dz_{0}. \end{aligned}$$

Without loss of generality, assume that $z_{1}>z_{2}$, and if we set $u=c_{1}z_{0}$ together with the function defined in Humbert (1922) (page 79) yields

$$\begin{aligned} f(z_{1},z_{2})= & {} \frac{z_{1}^{\alpha _{1}-1}z_{2}^{\alpha _{2}-1}e^{-\frac{z_{1}}{\beta _{1}}}e^{-\frac{z_{2}}{\beta _{2}}}\beta _{1}^{-\alpha _{1}}\beta _{2}^{-\alpha _{2}}}{\Gamma (\alpha _{0})\Gamma (\alpha _{1})\Gamma (\alpha _{2})}\\{} & {} \times \underbrace{\int _0^{1}u^{\alpha _{0}-1}\left( 1-u\right) ^{\alpha _{1}-1} \left( 1-u\frac{c_{2}z_{1}}{z_{2}c_{1}}\right) ^{\alpha _{2}-1}e^{u\left( z_{1} \left( \frac{1}{\beta _{1}}+\frac{c_{2}}{\beta _{2}{c_{1}}}-\frac{1}{\beta _{0}c_{1}}\right) \right) }du} _{\frac{\Gamma (\alpha _{0})\Gamma (\alpha _{1})\Phi _{1}\left( \alpha _{0},1-\alpha _{2},\alpha _{0} +\alpha _{1};\frac{c_{2}z_{2}}{z_{1}c_{1}},z_{1}\left( \frac{1}{\beta _{1}}+\frac{c_{2}}{c_{1}\beta _{2}} -\frac{1}{\beta _{0}c_{1}} \right) \right) }{\Gamma (\alpha _{0}+\alpha _{2})}}. \end{aligned}$$

Finally, regarding two cases $z_{1}>z_{2}$ and $z_{1}<z_{2}$, we obtain respectively the densities,

$$\begin{aligned}{} & {} f(z_{1},z_{2},\alpha _{0},\alpha _{1},\alpha _{2},\beta _{0},\beta _{1},\beta _{2})=\frac{z_{1}^{\alpha _{1}-1}z_{2}^{\alpha _{2}-1}e^{\frac{-z{1}}{\beta _{1}}}e^{\frac{-z{2}}{\beta _{2}}}}{\Gamma (\alpha _{0})\Gamma (\alpha _{1})\Gamma (\alpha _{2})}B(\alpha _{0},\alpha _{1})\nonumber \\{} & {} \times \Phi _{1}\left( \alpha _{0},1-\alpha _{2},\alpha _{0}+\alpha _{1};\frac{c_{2}z_{2}}{c_{1}z_{1}},z_{1}\left( \frac{1}{\beta _{1}}+\frac{c_{2}}{c_{1}\beta _{2}}-\frac{1}{c_{1}\beta _{0}}\right) \right) , \end{aligned}$$

(A.15)

$$\begin{aligned}{} & {} f(z_{1},z_{2},\alpha _{0},\alpha _{1},\alpha _{2},\beta _{0},\beta _{1},\beta _{2})=\frac{z_{1}^{\alpha _{1}-1}z_{2}^{\alpha _{2}-1}e^{\frac{-z{1}}{\beta _{1}}}e^{\frac{-z{2}}{\beta _{2}}}}{\Gamma (\alpha _{0})\Gamma (\alpha _{1})\Gamma (\alpha _{2})}B(\alpha _{0},\alpha _{1})\nonumber \\{} & {} \times \Phi _{1}\left( \alpha _{0},1-\alpha _{1},\alpha _{0}+\alpha _{2};\frac{c_{1}z_{1}}{c_{2}z_{2}},z_{2}\left( \frac{c_{1}}{c_{2}\beta _{1}}+\frac{1}{\beta _{2}}-\frac{1}{c_{2}\beta _{0}}\right) \right) , \end{aligned}$$

(A.16)

where $\Phi _{1}$ is the confluent hypergeometric function of two variables in Humbert (1922).

Using the derivations above and applying them to the case of three correlated random variables $Z_{i},Z_{j},Z_{k}$ of the form $Z_{j}=X_{j}+c_{j}Y$, we obtain the following density where we have the condition (without loss of generality) $Z_{i}=\min (Z_{i},Z_{j},Z_{k})$,

$$\begin{aligned}{} & {} f(z_{i},z_{j},z_{k},\alpha _{0},\alpha _{i},\alpha _{j},\alpha _{k},\beta _{0},\beta _{i},\beta _{j},\beta _{k})\nonumber \\{} & {} \quad =\frac{z_{i}^{\alpha _{i}-1}z_{j}^{\alpha _{j}-1}z_{k}^{\alpha _{k}-1}e^{-\frac{z_{i}}{\beta _{i}}}e^{-\frac{z_{j}}{\beta _{j}}}e^{-\frac{z_{k}}{\beta _{k}}}\beta _{i}^{-\alpha _{i}}\beta _{j}^{-\alpha _{j}}\beta _{k}^{-\alpha _{k}}}{\Gamma (\alpha _{0})\Gamma (\alpha _{i})\Gamma (\alpha _{j})\Gamma (\alpha _{k})}\nonumber \\{} & {} \times \underbrace{\int _0^{1}u^{\alpha _{0}-1}\left( 1-u\right) ^{\alpha _{i}-1}\left( 1-u\frac{c_{i}z_{i}}{c_{j}z_{j}}\right) ^{\alpha _{j}-1}\left( 1-u\frac{z_{j}}{z_{k}}\right) ^{\alpha _{k}-1}e^{u\left( z_{i}\left( \frac{c_{i}}{c_{j}\beta _{i}}+\frac{1}{\beta _{j}}+\frac{1}{\beta _{k}}-\frac{1}{\beta _{0}c_{i}}\right) \right) }du}_{ \Phi _{2}\left( \alpha _{0},\alpha _{1},\alpha _{2},\alpha _{3},\beta _{0},\beta _{1},\beta _{2},\beta _{3},z_{1},z_{2},z_{3}\right) }.\nonumber \\ \end{aligned}$$

(A.17)

Furthermore, we can write $\Phi _{2}$ in terms of special functions. Starting with a Taylor expansion of $e^{x}$ and rewriting (A.17), we obtain

$$\begin{aligned}{} & {} \Phi _{2}\left( \alpha _{0},\alpha _{i},\alpha _{j},\alpha _{k},\beta _{0},\beta _{i},\beta _{j}, \beta _{k},z_{i},z_{j},z_{k}\right) =\sum _{m,n,l=0}^{\infty }F_{D}^{(2)}\\{} & {} \left( \alpha _{0}+m,\alpha _{j}-1,\alpha _{k}-1,\alpha _{i}+\alpha _{0}+m\right) \frac{\Gamma (\alpha _{0}+m)\Gamma (\alpha _{i})}{\Gamma (\alpha _{0}+\alpha _{i}+m)}\times \\{} & {} \frac{\left( z_{i}\left( \frac{c_{i}}{c_{j}\beta _{i}}+\frac{1}{\beta _{j}}+\frac{1}{\beta _{k}}-\frac{1}{\beta _{0}c_{i}}\right) \right) ^{m} (\alpha _{j}-1)^{n}(\alpha _{k}-1)^{l}}{m!n!l!}. \end{aligned}$$

Multiplying Eq. () by $\Gamma (\alpha _{0}+\alpha _{i})$ and dividing by $\Gamma (\alpha _{0})$, we obtain $\Phi _{2}$ in terms of a hypergeometric series,

$$\begin{aligned} \Phi _{2}\left( \alpha _{0},\alpha _{i},\alpha _{j},\alpha _{k},\beta _{0},\beta _{i},\beta _{j},\beta _{k},z_{i},z_{j},z_{k}\right) =\sum _{m=0}^{\infty }\sum _{n=0,l=0}^{\infty }\frac{(\alpha _{0})_{m+n+l}(\alpha _{j}-1)_{n}(\alpha _{k}-1)_{l}}{(\alpha _{0}+\alpha _{i})_{m+n+l}}\times \\ \frac{\left( z_{i}\left( \frac{c_{i}}{c_{j}\beta _{i}}+\frac{1}{\beta _{j}}+\frac{1}{\beta _{k}}-\frac{1}{\beta _{0}c_{i}}\right) \right) ^{m}\left( \frac{c_{i}z_{i}}{c_{j}z_{j}}\right) ^{n}\left( \frac{z_{j}}{z_{k}}\right) ^{l}}{m!n!l!}, \end{aligned}$$

where $(q)_{n}$ is the Pochhammer symbol and $F_{D}^{2}$ is the Lauricella function of $n=2$ and type D. $\square$

1.6 A.5: Approximate VGC model CDF

Proof

The approximate VGC model CDF can be obtained by using the fact that,

$$\begin{aligned} F(x)= & {} \int _{0}^{\infty }{\mathcal {N}}\left( \frac{(x-\theta g)}{\sqrt{g}}\right) \frac{g^{\alpha -1}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg,\nonumber \\= & {} I(m,n) \end{aligned}$$

(A.18)

$$\begin{aligned}= & {} \int _{0}^{\infty }\int _{-\infty }^{V}\left( {\mathcal {N}}_{n}\left( \frac{n(v)}{\sqrt{g}}-m(v) \sqrt{g}\right) n_{v}g^{-\frac{1}{2}} +{\mathcal {N}}_{m}\left( \frac{n(v)}{\sqrt{g}}-m(v) \sqrt{g}\right) m_{v}g^{\frac{1}{2}}\right) dv\nonumber \\{} & {} \frac{g^{\alpha -1}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg. \end{aligned}$$

(A.19)

Here we set $m=m(v)$, $n=n(v)$ and $m(V)=m$, $n(V)=n$. We will define the following integrals and parameterizations,

$$\begin{aligned} I^{1}(v)= & {} \int _{0}^{\infty }{\mathcal {N}}_{n}\left( \frac{n(v)}{\sqrt{g}}-m(v) \sqrt{g}\right) \frac{g^{\alpha -\frac{3}{2}}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg\\= & {} \int _{0}^{\infty }\frac{1}{\sqrt{2\pi }}\exp {\left( \frac{-\left( n-mg \right) ^{2}}{2g}\right) }\frac{g^{\alpha -\frac{3}{2}}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg\\ I^{2}(v)= & {} \int _{0}^{\infty }{\mathcal {N}}_{n}\left( \frac{n(v)}{\sqrt{g}}-m(v) \sqrt{g}\right) \frac{g^{\alpha -\frac{1}{2}}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg\\= & {} \int _{0}^{\infty }\frac{1}{\sqrt{2\pi }}\exp {\left( \frac{-\left( n-mg \right) ^{2}}{2g}\right) }\frac{g^{\alpha -\frac{1}{2}}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg \end{aligned}$$

Using Eq. (10) in Chaudhry and Syed (2002), we have

$$\begin{aligned} I^{1}(v)= & {} \int _{0}^{\infty }\frac{1}{\sqrt{2\pi }}\exp {\left( \frac{-\left( n-mg \right) ^{2}}{2g}\right) }\frac{g^{\alpha -\frac{3}{2}}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg\\= & {} \sqrt{\frac{2}{\pi }}e^{mn}\left( \frac{n}{\sqrt{\frac{2}{\beta }+m^{2}}} \right) ^{\alpha -\frac{1}{2}}K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \\ I^{2}(v)= & {} \int _{0}^{\infty }\frac{1}{\sqrt{2\pi }}\exp {\left( \frac{-\left( n-mg \right) ^{2}}{2g}\right) }\frac{g^{\alpha -\frac{1}{2}}e^{-\frac{g}{\beta }}\beta ^{-\alpha }}{\Gamma (\alpha )}dg\\= & {} \sqrt{\frac{2}{\pi }}e^{mn}\left( \frac{n}{\sqrt{\frac{2}{\beta }+m^{2}}} \right) ^{\alpha +\frac{1}{2}}K_{\alpha +\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \end{aligned}$$

In order to obtain F(x) defined in (A.18), we need to compute the following

$$\begin{aligned} I^{1}=\int _{-\infty }^{\infty }I^{1}(v)dv,\nonumber \\ I^{2}=\int _{-\infty }^{\infty }I^{2}(v)dv, \end{aligned}$$

(A.20)

since $F(x)=I^{1}+I^{2}$. Here, we need a variable transformation and a domain which keeps the Bessel function K(.) inside the integrals in Eq. (A.20) unchanged. We choose the domain $[-\infty ,n]$ and apply the following transformations:

$$\begin{aligned} m(v)=\frac{n\sqrt{m^{2}+\frac{2}{\beta }}}{v},\nonumber \\ n(v)=\sqrt{v^{2}-\frac{2}{\beta }}. \end{aligned}$$

(A.21)

After these transformations, we first obtain

$$\begin{aligned} \int _{-\infty }^{n}I^{1}(v)n_{v}dv= & {} K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \int _{-\infty }^{n}\nonumber \\{} & {} \exp {\left( \frac{m\sqrt{n^{2}+\frac{2}{\beta }}}{v}\sqrt{v^{2}-\frac{2}{\beta }}\right) }\left( \frac{m\sqrt{n^{2}+\frac{2}{\beta }}}{v^{2}}\right) ^{\alpha -1/2}\nonumber \\{} & {} \times v\left( v^{2}-2/\beta \right) ^{\frac{-1}{2}}n_{v}dv \end{aligned}$$

(A.22)

$$\begin{aligned} \int _{-\infty }^{n}I^{2}(v)m_{v}dv= & {} K_{\alpha +\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \int _{-\infty }^{m}\nonumber \\{} & {} \exp {\left( \frac{m\sqrt{n^{2}+\frac{2}{\beta }}}{v}\sqrt{v^{2}-\frac{2}{\beta }}\right) }\left( \frac{m\sqrt{n^{2}+\frac{2}{\beta }}}{v^{2}}\right) ^{\alpha +1/2}\nonumber \\{} & {} \times \frac{m\sqrt{n^{2}+\frac{2}{\beta }}}{v^2}m_{v}dv \end{aligned}$$

(A.23)

Then using the following,

$$\begin{aligned} u= & {} \sqrt{\frac{v^{2}-\frac{2}{\beta }}{v}},\\ v= & {} \sqrt{\frac{2}{\beta (1-u^{2})}},\\ dv= & {} \sqrt{2}u(1-u^{2})^{\frac{-3}{2}}, \end{aligned}$$

we obtain,

$$\begin{aligned} \int _{-\infty }^{n}I^{1}(v)n_{v}dv= & {} K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \int _{-1}^{\sqrt{\frac{n^{2}-\frac{2}{\beta }}{n}}}\\{} & {} \sqrt{2}e^{cu}\left( \frac{(1-u^{2})\beta }{2} \right) ^{\alpha -1/2}u\left( 1-u^{2} \right) ^{-1/2}du,\\ \int _{-\infty }^{n}I^{2}(v)m_{v}dv= & {} K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \int _{-1}^{\sqrt{\frac{n^{2}-\frac{2}{\beta }}{n}}}\\{} & {} \sqrt{2}e^{cu}\left( \frac{(1-u^{2})\beta }{2} \right) ^{\alpha -1/2}\left( 1-u^{2} \right) ^{-1/2}du. \end{aligned}$$

Then setting $q={\sqrt{\frac{n^{2}-\frac{2}{\beta }}{n}}}$, we obtain

$$\begin{aligned} I^{1}= & {} \int _{-\infty }^{n}I^{1}(v)n_{v}dv= K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \\ {}\times & {} \int _{0}^{1} \sqrt{2}e^{c\left( u(1+q)-1\right) }\left( \frac{1-\left( u(1+q)-1\right) ^{2})\beta }{2} \right) ^{\alpha -1/2}\\{} & {} \left( u(1+q)-1\right) \left( 1-\left( u(1+q)-1\right) ^{2} \right) ^{-1/2}du,\\= & {} K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \times \int _{0}^{1} u(1+q)^{\alpha }\left( 1-\frac{u(1+q)}{2} \right) ^{\alpha -1}e^{c\left( u(1+q)-1\right) }\frac{C^{\alpha +\frac{1}{2}}}{\sqrt{2\pi }}du\\ I^{2}= & {} \int _{-\infty }^{n}I^{2}(v)m_{v}dv=K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) \\{} & {} \times \int _{0}^{1} u(1+q)^{\alpha -1}\left( 1-\frac{u(1+q)}{2} \right) ^{\alpha -1}e^{c\left( u(1+q)-1\right) }\frac{C^{\alpha +\frac{1}{2}}}{\sqrt{2\pi }}du\\ \end{aligned}$$

Then in the context of confluent hypergeometric function of second kind (Humbert 1922, page 79) or Gradshteyn and Ryzhik (2007), formula 3.385), for $I^{1}$ we have

$$\begin{aligned} a=\alpha +1,\quad \beta =1-\alpha ,\quad \gamma =\alpha +2 \end{aligned}$$

for $I^{2}$, we have

$$\begin{aligned} a=\alpha ,\quad \beta =1-\alpha ,\quad \gamma =\alpha +1 \end{aligned}$$

After collecting all these and using the definition of the confluent hypergeometric function of the second kind we obtain,

$$\begin{aligned} F(x)= & {} I^{1}+I^{2}=e^{-C}\beta ^{\alpha }\frac{C^{\alpha +1/2}}{\sqrt{2\pi }}\Phi _{1}\left( \alpha ,1-\alpha ,\alpha +1,\frac{1+q}{2},c(1+q) \right) \\{} & {} \times \frac{K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) }{\alpha \Gamma (\alpha )}\\{} & {} -e^{-C}\beta ^{\alpha }\frac{C^{\alpha +1/2}}{\sqrt{2\pi }}\Phi _{1}\left( \alpha +1,1-\alpha ,\alpha +2,\frac{1+q}{2},c(1+q) \right) \\{} & {} \times \frac{K_{\alpha -\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) }{(1+\alpha )\Gamma (\alpha )}\\{} & {} +e^{-C}\beta ^{\alpha }\frac{C^{\alpha +1/2}}{\sqrt{2\pi }}\Phi _{1}\left( \alpha ,1-\alpha ,\alpha +1,\frac{1+q}{2},c(1+q) \right) \\{} & {} \times \frac{K_{\alpha +\frac{1}{2}}\left( |n|\sqrt{m^{2}+\frac{2}{\beta }} \right) }{\alpha \Gamma (\alpha )}. \end{aligned}$$

where $C=n\sqrt{m^{2}+\frac{2}{\beta }}$. $\square$