**5. Asmussen's Embedding Approach for Solving Kolmogorov's Integro-Differential Equation with Phase-Type Jumps**

One of the most convenient approaches to ge<sup>t</sup> rid of the integral term in (29) is a probabilistic transformation which gets rid of the jumps as in Asmussen (1995), when the downward phase-type jumps have a survival function

$$\bar{F}\_{\mathbb{C}}(\mathbf{x}) = \int\_{\mathbf{x}}^{\infty} f\_{\mathbb{C}}(\boldsymbol{\mu}) d\boldsymbol{u} = \vec{\beta}e^{\mathbf{R}\cdot\mathbf{x}}\mathbf{1},$$

where *B* is a *n* × *n* stochastic generating matrix (nonnegative off-diagonal elements and nonpositive row sums), *β* = (*β*1, ... , *βn*) is a row probability vector (with nonnegative elements and ∑*nj*=<sup>1</sup> *βj* = 1), and **1** = (1, 1, ..., 1) is a column probability vector.

The density is *fC*(*x*) = *βe*<sup>−</sup>*Bxb*, where *b* = (−*<sup>B</sup>*)**<sup>1</sup>** is a column vector, and the Laplace transform is

$$
\hat{b}(s) = \vec{\beta}(sI - \mathbf{B})^{-1}\mathbf{b}.
$$

Asmussen's approach Asmussen (1995); Asmussen et al. (2002) replaces the negative jumps by segments of slope −1, embedding the original spectrally negative Lévy process into a continuous Markov modulated Lévy process. For the new process we have auxiliary unknowns *Ai*(*x*) representing ruin or survival probabilities (or, more generally, Gerber-Shiu functions) when starting at *x* conditioned on a phase *i* with drift downwards (i.e., in one of the "auxiliary stages of artificial time" introduced by changing the jumps to segments of slope −1). Let **A** denote the column vector with components *A*1, ... , *An*. The Kolmogorov integro-differential equation turns then into a system of ODE's, due to the continuity of the embedding process.

$$
\begin{pmatrix}
\Psi'\_q(\mathbf{x}) \\
\mathbf{A}'(\mathbf{x})
\end{pmatrix} = \begin{pmatrix}
\frac{\lambda+q}{c(\mathbf{x})} & -\frac{\lambda}{c(\mathbf{x})} \vec{\beta} \\
\mathbf{b} & \mathbf{B}
\end{pmatrix} \begin{pmatrix}
\Psi\_q(\mathbf{x}) \\
\mathbf{A}(\mathbf{x})
\end{pmatrix}, \ \mathbf{x} \ge 0. \tag{41}
$$

For the ruin probability with exponential jumps of rate *μ* for example, there is only one downward phase, and the system is:

$$
\begin{pmatrix}
\Psi'\_q(\mathbf{x}) \\
 A'(\mathbf{x})
\end{pmatrix} = \begin{pmatrix}
\frac{\lambda+q}{\mathfrak{c}(\mathbf{x})} & -\frac{\lambda}{\mathfrak{c}(\mathbf{x})} \\
\mu & -\mu
\end{pmatrix} \begin{pmatrix}
\Psi\_q(\mathbf{x}) \\
 A(\mathbf{x})
\end{pmatrix} \ge 0. \tag{42}
$$

For survival probabilities, one only needs to modify the boundary conditions—see the following section.

*5.1. Exit Problems for the Segerdahl-Tichy process, with q* = 0

Asmussen's approach is particular convenient for solving exit problems for the Segerdahl-Tichy process.

**Example 1. The eventual ruin probability***.* **When** *q* = 0, *the system for the ruin probabilities with x* ≥ 0 *is:*

$$\begin{cases} \Psi'(\mathbf{x}) = \frac{\lambda}{c(\mathbf{x})} \left( \Psi(\mathbf{x}) - A(\mathbf{x}) \right), & \Psi(\infty) \quad = A(\infty) = 0 \\ A'(\mathbf{x}) = \mu \left( \Psi(\mathbf{x}) - A(\mathbf{x}) \right), & A(0) = 1 \end{cases} \tag{43}$$

*This may be solved by subtracting the equations. Putting*

$$K(x) = e^{-\mu x + \int\_0^x \frac{\lambda}{c(v)} dv},$$

*we find:*

$$\begin{cases} \Psi(\mathbf{x}) - A(\mathbf{x}) &= (\Psi(0) - A(0))K(\mathbf{x}), \\ A(\mathbf{x}) &= \mu(A(0) - \Psi(0)) \int\_{\mathcal{X}}^{\infty} K(v) dv, \end{cases} \tag{44}$$

*whenever <sup>K</sup>*(*v*) *is integrable at* ∞*.*

> *The boundary condition A*(0) = 1 *implies that* 1 − Ψ(0) = 1 *μ* ∞0 *<sup>K</sup>*(*v*)*dv and*

$$\begin{aligned} A(\mathbf{x}) &= \mu (1 - \Psi(0)) \int\_{x}^{\infty} K(v) dv = \frac{\int\_{x}^{\infty} K(v) dv}{\int\_{0}^{\infty} K(v) dv}, \\ \Psi(\mathbf{x}) - A(\mathbf{x}) &= -\frac{K(\mathbf{x})}{\mu \int\_{0}^{\infty} K(v) dv}. \end{aligned}$$

*Finally,*

$$\Psi(\mathbf{x}) = A(\mathbf{x}) + (\Psi(\mathbf{x}) - A(\mathbf{x})) = \frac{\mu \int\_{\mathcal{X}}^{\infty} K(v) dv - K(\mathbf{x})}{\mu \int\_{0}^{\infty} K(v) dv} \rho$$

*and for the survival probability* Ψ*,*

$$\begin{aligned} \mathbf{F}(\mathbf{x}) &= \frac{\mu \int\_0^\mathbf{x} K(v) dv + K(\mathbf{x})}{\mu \int\_0^\infty K(v) dv} := \mathbf{F}(0) \mathbf{W}(\mathbf{x}) = \frac{\mathbf{W}(\mathbf{x})}{\mathbf{W}(\infty)}, \\ \mathbf{W}(\mathbf{x}) &= \mu \int\_0^\mathbf{x} K(v) dv + K(\mathbf{x}), \end{aligned} \tag{45}$$

*where* Ψ(0) = 1 **<sup>W</sup>**(∞) *by plugging* **W**(0) = 1 *in the first and last terms in* (45)*. We may also rewrite* (45) *as:*

$$\overline{\mathbf{V}}(\mathbf{x}) = \frac{1 + \int\_0^\mathbf{x} \mathbf{w}(v) dv}{1 + \int\_0^\infty \mathbf{w}(v) dv} \Leftrightarrow \mathbf{V}(\mathbf{x}) = \frac{\int\_x^\infty \mathbf{w}(v) dv}{1 + \int\_0^\infty \mathbf{w}(v) dv}, \mathbf{w}(\mathbf{x}) := \mathbf{W}'(\mathbf{x}) = \frac{\lambda \mathbf{K}(\mathbf{x})}{\mathbf{c}(\mathbf{x})} \tag{46}$$

*Note that* **w**(*x*) > 0 *implies that the scale function* **<sup>W</sup>**(*x*) *is nondecreasing.*

**Example 2.** *For the two sided exit problem on* [*a*, *b*]*, a similar derivation yields the scale function*

$$\mathbf{W}(\mathbf{x}, a) = \mu \int\_{a}^{\mathbf{x}} \frac{K(v)}{K(a)} dv + \frac{K(\mathbf{x})}{K(a)} = 1 + \frac{1}{K(a)} \int\_{a}^{\mathbf{x}} \mathbf{w}(y) dy.$$

*with scale derivative derivative* **<sup>w</sup>**(*<sup>x</sup>*, *a*) = 1 *<sup>K</sup>*(*a*) **<sup>w</sup>**(*x*)*, where* **w**(*x*) *given by* (46) *does not depend on a. Indeed, the analog of* (44) *is:*

$$\begin{cases} \overline{\mathbf{F}}^b(\mathbf{x}, a) - A^b(\mathbf{x}) = \overline{\mathbf{F}}^b(a, a) \frac{K(\mathbf{x})}{K(a)}, \\\\ A^b(\mathbf{x}) = \mu \overline{\mathbf{F}}^b(a, a) \int\_a^\mathbf{x} \frac{K(\upsilon)}{K(a)} d\upsilon, \end{cases}$$

*implying by the fact that* <sup>Ψ</sup>*<sup>b</sup>*(*b*, *a*) = 1 *that*

$$\begin{split} \mathbf{V}^{b}(\mathbf{x},a) &= \mathbf{V}^{b}(a,a) \left( \frac{\mathbf{K}(\mathbf{x})}{\mathbf{K}(a)} + \mu \int\_{a}^{x} \frac{\mathbf{K}(v)}{\mathbf{K}(a)} d\upsilon \right) = \frac{\mathbf{W}(\mathbf{x},a)}{\mathbf{W}(b,a)} = \frac{1 + \frac{1}{\mathbf{K}(a)} \int\_{a}^{x} \mathbf{w}(u) du}{1 + \frac{1}{\mathbf{K}(a)} \int\_{a}^{b} \mathbf{w}(u) du} \Leftrightarrow\\ \mathbf{V}^{b}(\mathbf{x},a) &= \frac{\int\_{x}^{b} \mathbf{w}(u) du}{\mathbf{K}(a) + \int\_{a}^{b} \mathbf{w}(u) du} \Leftrightarrow\\ \mathbf{V}^{b}(\mathbf{x},a) &:= - (\mathbf{V}^{b})'(\mathbf{x},a) = \frac{\mathbf{w}(x)}{\mathbf{K}(a) + \int\_{a}^{b} \mathbf{w}(u) du} = \mathbf{w}(x,a) \frac{\overline{\mathbf{V}}(a,a)}{\overline{\mathbf{V}}(b,a)}. \end{split} \tag{47}$$

**Remark 7.** *The definition adopted in this section for the scale function* **<sup>W</sup>**(*<sup>x</sup>*, *a*) *uses the normalization* **<sup>W</sup>**(*<sup>a</sup>*, *a*) = 1*, which is only appropriate in the absence of Brownian motion.*

**Problem 4.** *Extend the equations for the survival and ruin probability of the Segerdahl-Tichy process in terms of the scale derivative* **<sup>w</sup>***q, when q* > 0*. Essentially, this requires obtaining*

$$T\_{\emptyset}(\mathbf{x}) = E\_{\mathbf{x}}\left[\mathbf{e}^{-\mathbf{q}\left[T\_{a,-}\min T\_{b,+}\right]}\right]$$

### **6. Revisiting Segerdahl's Process via the Scale Derivative/Integrating Factor Approach, When** *q* **= 0**

Despite the new scale derivative/integrating factor approach, we were not able to produce further explicit results beyond (33), due to the fact that neither the scale derivative, nor the integral of the integrating factor are explicit when *q* > 0 (this is in line with Avram et al. (2010)). (33) remains thus for now an outstanding, not well-understood exception.

### **Problem 5.** *Are there other explicit first passage results for Segerdahl's process when q* > 0*?*

In the next subsections, we show that via the scale derivative/integrating factor approach, we may rederive well-known results for *q* = 0.

### *6.1. Laplace Transforms of the Eventual Ruin and Survival Probabilities*

For *q* = 0, both Laplace transforms and their inverses are explicit, and several classic results may be easily checked. The scale derivative may be obtained using Proposition 1 and Γ(*λ*˜ + 1, *v*) = *e* −*vv<sup>λ</sup>* ˜ + *λ*Γ(*λ*˜ , *v*) with *v* = *c*˜(*s* + *μ*). We find

$$
\hat{\mathbf{w}}(s,a) = \frac{\varepsilon^{\sharp \mu} (\bar{\varepsilon}\mu)^{-\lambda} \Gamma(\lambda + 1, \bar{\varepsilon}(s + \mu))}{\varepsilon^{-\sharp s} (1 + s/\mu)^{\bar{\lambda}}} - 1 = 1 + \lambda \varepsilon^{\nu} (\nu)^{-\bar{\lambda}} \Gamma(\bar{\lambda}, \nu) - 1 = \bar{\lambda} \mathcal{U}(1, 1 + \bar{\lambda}, \bar{\varepsilon}(s + \mu))
$$

$$
\implies \mathbf{w}(\mathbf{x}, a) = \frac{\bar{\lambda}}{\bar{\varepsilon}} \left(1 + \frac{\mathbf{x}}{\bar{\varepsilon}}\right)^{\bar{\lambda} - 1} e^{-\mu \mathbf{x}},
\tag{48}
$$

which checks (46). Using again **w** (*s*) = *c*˜*<sup>I</sup>*(*y*) *<sup>I</sup>*(*y*) − 1 yields the ruin and survival probabilities:

$$\begin{aligned} s\widehat{\overline{\Psi}}(s) &= \frac{\int\_s^\infty \widetilde{\varepsilon}\overline{\Psi}(0)I(\boldsymbol{y})d\boldsymbol{y}}{I(\boldsymbol{s})} = \overline{\Psi}(0)(\widehat{\mathbf{w}}(\boldsymbol{s}) + 1) \\\ s\widehat{\Psi}(s) &= \frac{\int\_s^\infty (\widetilde{\varepsilon}\Psi(0) - \frac{\overline{\lambda}}{\overline{y} + \overline{\mu}})I(\boldsymbol{y})d\boldsymbol{y}}{I(\boldsymbol{s})} = \overline{\Psi}(0)(\widehat{\mathbf{w}}(\boldsymbol{s}) + 1) - \widehat{\mathbf{w}}(\boldsymbol{s}). \end{aligned}$$

Letting *s* → 0 yields

$$\Psi(0) = \frac{\hat{\mathbf{w}}(0)}{\hat{\mathbf{w}}(0) + 1} = \frac{\bar{\lambda}\mathcal{U}(1, 1 + \bar{\lambda}, \mu\bar{\varepsilon})}{\mu\bar{\varepsilon}\,\mathcal{U}(1, 2 + \bar{\lambda}, \mu\bar{\varepsilon})} = \frac{\bar{\lambda}\Gamma(\bar{\lambda}, \varepsilon\mu)}{\Gamma(\bar{\lambda} + 1, \bar{\varepsilon}\mu)} \Leftrightarrow$$

$$\Psi(0) = \frac{\lim\_{s \to 0} s\hat{\overline{\Psi}}(s)}{\hat{\mathbf{w}}(0) + 1} = \frac{\overline{\Psi}(\infty)}{1 + \bar{\lambda}\,\mathcal{U}(1, 1 + \bar{\lambda}, \mu\bar{\varepsilon})} = \frac{1}{\mu\bar{\varepsilon}\,\mathcal{U}(1, 2 + \bar{\lambda}, \mu\bar{\varepsilon})}\tag{49}$$

For the survival probability, we finally find

$$s\hat{\Psi}(s) = \Psi(0)(1 + \hat{\mathbf{w}}(s)) = \frac{1 + \bar{\lambda}\mathcal{U}(1, 1 + \bar{\lambda}, \mu(\vec{\varepsilon} + s))}{1 + \bar{\lambda}\mathcal{U}(1, 1 + \bar{\lambda}, \mu\vec{\varepsilon})} = \frac{\vec{\varepsilon}(\mu + s)\mathcal{U}(1, 2 + \bar{\lambda}, \mu(\vec{\varepsilon} + s))}{\vec{\varepsilon}\mu\mathcal{U}(1, 2 + \bar{\lambda}, \mu\vec{\varepsilon})} \ge \frac{\vec{\varepsilon}(\mu + s)\mathcal{U}(1, 2 + \bar{\lambda}, \mu\vec{\varepsilon})}{\vec{\varepsilon}\mu\mathcal{U}(1, 2 + \bar{\lambda}, \mu\vec{\varepsilon})} \ge \frac{\vec{\varepsilon}(\mu + s)\mathcal{U}(1, 2 + \bar{\lambda}, \mu\vec{\varepsilon})}{\vec{\varepsilon}\mu\mathcal{U}(1, 2 + \bar{\lambda}, \mu\vec{\varepsilon})}$$

which checks with the Laplace transform of the Segerdahl result (53).

### *6.2. The Eventual Ruin and survival probabilities*

These may also be obtained directly by integrating the explicit scale derivative **<sup>w</sup>**(*<sup>x</sup>*, *a*) = *λ*˜ *c*˜ 1 + *x c*˜*λ*˜ −1 *e*<sup>−</sup>*μ<sup>x</sup>* (48) Indeed,

$$\int\_{\mu}^{\infty} \mathbf{w}(\mathbf{x})d\mathbf{x} = \int\_{\mu}^{\infty} \frac{\tilde{\lambda}}{\tilde{c}} \left(1 + \frac{\chi}{\tilde{c}}\right)^{\tilde{\lambda} - 1} e^{-\mu x} d\mathbf{x} = \tilde{\lambda} e^{\mu \tilde{c}} \int\_{1 + \frac{\mu}{\tilde{c}}}^{\infty} y^{\tilde{\lambda} - 1} e^{\mu \tilde{c}y} dy$$

$$\tilde{\lambda} = \tilde{\lambda} e^{\mu \tilde{c}} \frac{1}{(\mu \tilde{c})^{\tilde{\lambda}}} \int\_{\mu(\tilde{c} + \mu)}^{\infty} t^{\tilde{\lambda} - 1} e^{-t} dt = \tilde{\lambda} e^{\mu \tilde{c}} (\mu \tilde{c})^{-\tilde{\lambda}} \Gamma(\tilde{\lambda}, \mu(\tilde{c} + \mu)),$$

where <sup>Γ</sup>(*η*, *x*) = ∞ *x t <sup>η</sup>*−1*e*<sup>−</sup>*tdt* is the incomplete gamma function. The ruin probability is Segerdahl (1955), (Paulsen and Gjessing 1997, ex. 2.1):

$$\Psi(\mathbf{x}) = \lambda \frac{\exp(\mu \vec{\varepsilon}) (\mu \vec{\varepsilon})^{-\bar{\lambda}} \Gamma(\bar{\lambda}, \mu(\vec{\varepsilon} + \mathbf{x}))}{1 + \bar{\lambda} \exp(\mu \vec{\varepsilon}) (\mu \vec{\varepsilon})^{-\bar{\lambda}} \Gamma(\bar{\lambda}, \mu \vec{\varepsilon})} = \lambda \frac{e^{-\mu x} (1 + \mathbf{x}/\varepsilon)^{\bar{\lambda}} \, \mathrm{U} (1, 1 + \bar{\lambda}, \mu(\vec{\varepsilon} + \mathbf{x}))}{1 + \bar{\lambda} \mathrm{U} (1, 1 + \bar{\lambda}, \mu \vec{\varepsilon})}$$

$$= \frac{\lambda}{\mu \vec{\varepsilon}} \frac{e^{-\mu x} (1 + \mathbf{x}/\varepsilon)^{\bar{\lambda}} \, \mathrm{U} (1, 1 + \bar{\lambda}, \mu(\vec{\varepsilon} + \mathbf{x}))}{\mathrm{U} (1, 2 + \bar{\lambda}, \mu \vec{\varepsilon})} = \frac{\lambda \Gamma(\bar{\lambda}, \mu(\vec{\varepsilon} + \mathbf{x}))}{\Gamma(\bar{\lambda} + 1, \bar{\varepsilon} \mu)},\tag{50}$$

where we used

$$\mathcal{U}(1,1+\tilde{\lambda},v) = e^v v^{-\tilde{\lambda}} \Gamma(\tilde{\lambda},v) \tag{51}$$

and

$$\mathbf{1} + \mathbf{\tilde{\lambda}} \mathbf{U} (\mathbf{1}, \mathbf{1} + \mathbf{\tilde{\lambda}}, \mathbf{v}) = \mathbf{v} \mathbf{U} (\mathbf{1}, \mathbf{2} + \mathbf{\tilde{\lambda}}, \mathbf{v}), \tag{52}$$

which holds by integration by parts.

> A simpler formula holds for the rate of ruin *ψ*(*x*) and its Laplace transform

$$\begin{split} \boldsymbol{\Psi}(\mathbf{x}) &= -\mathbf{P}'(\mathbf{x}) = \frac{\mathbf{w}(\mathbf{x})}{1 + \int\_0^\infty \mathbf{w}(\mathbf{x})d\mathbf{x}} = \frac{\tilde{\boldsymbol{\lambda}}}{\Gamma(\tilde{\boldsymbol{\lambda}} + 1, \tilde{\boldsymbol{\varepsilon}}\boldsymbol{\mu})} \boldsymbol{\mu}(\boldsymbol{\mu}(\boldsymbol{\varepsilon} + \mathbf{x}))^{\tilde{\boldsymbol{\lambda}} - 1} \boldsymbol{\varepsilon}^{-\boldsymbol{\mu}(\boldsymbol{\varepsilon} + \mathbf{x})} = \boldsymbol{\varepsilon}^{-\boldsymbol{\mu}\boldsymbol{\varepsilon}} \boldsymbol{\gamma}\_{\boldsymbol{\lambda}, \boldsymbol{\mu}}(\mathbf{x} + \boldsymbol{\varepsilon}) \Leftrightarrow \\ \boldsymbol{\hat{\Psi}}(\boldsymbol{s}) &= \overline{\mathbf{F}}(0) \boldsymbol{\hat{\mathbf{w}}}(\mathbf{s}) = \begin{cases} \frac{\tilde{\boldsymbol{\lambda}} \boldsymbol{\varepsilon} \boldsymbol{\mu}(1, \mathbf{1} + \tilde{\boldsymbol{\lambda}}\boldsymbol{\varepsilon}(\mathbf{s} + \boldsymbol{\mu}))}{\tilde{\boldsymbol{\varepsilon}} \boldsymbol{\mu} \boldsymbol{\varepsilon} \boldsymbol{\mu}(1, 2 + \tilde{\boldsymbol{\lambda}}\boldsymbol{\varepsilon})}, & \boldsymbol{c} > 0 \\ (1 + \boldsymbol{s}/\mu)^{-\tilde{\boldsymbol{\lambda}}}, & \boldsymbol{c} = 0 \end{cases} \end{split} \tag{53}$$

where *γ* denotes a (shifted) Gamma density. Of course, the case *c* > 0 simplifies to a Gamma density when moving the origin to the "absolute ruin" point −*c*˜ = −*c r*,, i.e., by putting *y* = *x* + *c*˜,*Yt* = *Xt* + *c*˜, where the process *Yt* has drift rate *rYt*.

**Problem 6.** *Find a relation between the ruin derivative ψq*(*x*) = <sup>−</sup>Ψ*q*(*x*) *and the scale derivative* **<sup>w</sup>***q*(*x*) *when q* > 0*.*

### **7. Further Details on the Identities Used in the Proof of Theorem 2**

We recall first some continuity and differentiation relations needed here Abramowitz and Stegun (1965)

**Proposition 2.** *Using the notation M* = *<sup>M</sup>*(*<sup>a</sup>*, *b*, *<sup>z</sup>*), *<sup>M</sup>*(*a*+) = *<sup>M</sup>*(*a* + 1, *b*, *<sup>z</sup>*), *<sup>M</sup>*(+, +) = *<sup>M</sup>*(*a* + 1, *b* + 1, *<sup>z</sup>*)*, and so on, the Kummer and Tricomi functions satisfy the following identities:*

$$bM + (a - b)M(b +) = aM(a +) \tag{13.4.3}$$

$$b\left(M(a+)-M\right) = zM\left(+,+\right) \tag{13.4.4}$$

$$lL(b-a)lI + zlI(b+2) = (z+b)lI(b+1)\tag{13.4.16}$$

$$\mathcal{U} + a \, \mathcal{U} \, (+, +) = \mathcal{U} \, (b+) \tag{13.4.17}$$

$$
\mathcal{U} + (b - a - 1)\mathcal{U}(a+1) = z\mathcal{U}(+,+)\tag{13.4.18}
$$

*(see corresponding equations in Abramowitz and Stegun (1965)).*

$$
\mathcal{U}\mathcal{U}' = -a\mathcal{U}\left(+, +\right), \quad \mathcal{M}' = \frac{a}{b}\mathcal{M}\left(+, +\right). \tag{54}
$$

**Proposition 3.** *The functions Ki*(*q*˜, *λ*˜ , *z*) *defined by* (32) *satisfy the identities*

$$K\_1'(\vec{q}, n, z) \quad = \ (\vec{q} + \vec{\lambda})e^{-z}z^{\mathfrak{q} + \tilde{\lambda} - 1} \ M\left(\vec{q}, \vec{q} + \vec{\lambda}, z\right) = (\vec{q} + \vec{\lambda})K\_1(\vec{q} - 1, \vec{\lambda}, z) \tag{55}$$

$$K\_2'(\vec{q}, n, z) \quad = \ -e^{-z}z^{\mathfrak{q} + \bar{\lambda} - 1} \llcorner (\vec{q}, \vec{q} + \vec{\lambda}, z) = -K\_2(\vec{q} - 1, n, z) \tag{56}$$

$$K\_2(\bar{q}, n, z) = \int\_z^\infty (y - z)^{\bar{q}} (y)^{n - \eta - 1} e^{-y} dy \tag{57}$$

**Proof:** For the first identity, note, using (Abramowitz and Stegun 1965, 13.4.3, 13.4.4), that

$$\begin{split} \frac{e^{z}}{z\bar{q}+\bar{\lambda}-1}K\_{1}'(z) &= \begin{pmatrix} \bar{q}+\bar{\lambda}-z)M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right)+z\frac{\bar{q}+1}{\bar{q}+\bar{\lambda}+1}M\left(\bar{q}+2,\bar{q}+2+\bar{\lambda},z\right) \\ &+ \end{pmatrix} \\ &= \begin{pmatrix} (\bar{q}+\bar{\lambda})M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right) \\ &+ \end{pmatrix} \\ &+ \begin{array}{c} \frac{z}{\bar{q}+\bar{\lambda}+1}\left((\bar{q}+1)M\left(\bar{q}+2,\bar{q}+2+\bar{\lambda},z\right)-(\bar{q}+\bar{\lambda}+1)M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right)\right) \\ &= \begin{pmatrix} (\bar{q}+\bar{\lambda})M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right) - \frac{z}{\bar{q}+\bar{\lambda}+1}\bar{\lambda}M\left(\bar{q}+1,\bar{q}+2+\bar{\lambda},z\right) \\ &+ \end{pmatrix} \\ &= \begin{pmatrix} (\bar{q}+\bar{\lambda})M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right) - \bar{\lambda}\left(M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right) - M\left(\bar{q},\bar{q}+1+\bar{\lambda},z\right)\right) \\ &+ \end{pmatrix} \\ &= \begin{pmatrix} \bar{q}M\left(\bar{q}+1,\bar{q}+1+\bar{\lambda},z\right) + \bar{\lambda}M\left(\bar{q},\bar{q}+1+\bar{\lambda},z\right) \\ \end{pmatrix} \end{split}$$

The second formula may be derived similarly using 13.4.17, or by considering the function

$$\,\_z\acute{U}(\vec{q}+1,\vec{q}+1+\vec{\lambda},\mu) := \Gamma(q+1)K\_2(z) = \int\_z^{\infty} (s-z)^{\frac{q}{q}} (s)^{\tilde{\lambda}-1} e^{-\mu s} ds$$

appearing in the numerator of the last form of (57). An integration by parts yields

$$\begin{aligned} &\,\_z\hat{\mathcal{U}}'(\vec{q}+1,\vec{q}+1+\vec{\lambda},1) = \int\_z^\infty (s-z)^{\vec{q}} \frac{d}{dz} [(s)^{\vec{\lambda}-1} e^{-s}] ds \\ &= (\vec{\lambda}-1)\_z \hat{\mathcal{U}}(\vec{q}+1,\vec{q}+\vec{\lambda},1) - \_z \hat{\mathcal{U}}(\vec{q}+1,\vec{q}+\vec{\lambda}+1,1), \implies \\ &\,\_z\hat{\mathcal{U}}'\_2(\vec{q}+1,\vec{\lambda},z) = e^{-z} z^{\vec{q}+\vec{\lambda}-1} \left( (\vec{\lambda}-1) \mathcal{U}(\vec{q}+1,\vec{q}+\vec{\lambda},z) - \mathcal{U}(\vec{q}+1,\vec{q}+\vec{\lambda}+1,z) \right) \end{aligned}$$

and the result follows by (Abramowitz and Stegun 1965, 13.4.18.)<sup>11</sup>

The third formula is obtained by the substitution *y* = *z*(*<sup>t</sup>* + <sup>1</sup>).
