Joint Influences of Erroneous Channel Information, Fading Severity, Jamming Suppression Error, Energy Harvesting Non-Linearity on Underlay Relaying Networks

Abstract

This paper considers an underlay relaying network where a cognitive relay performs decode-and-forward operation to aid communication between a cognitive source-destination pair. Transmit powers of cognitive users are subject to both interference and peak transmit power constraints. To stimulate support, the relay merely spends power scavenged from source signals with realistic non-linear energy harvester for its operation. Additionally, to secure secret transmission of both the source and the relay, this paper proposes to merge jamming signals with secret ones before their transmission. Relied on practical observations that jamming signals and channel state information at legitimate receivers may not be suppressed and estimated perfectly, respectively, this paper analyzes system performance under inaccuracies of jamming suppression and channel estimation. Purposively, this paper analyzes security-reliability trade-off of underlay relaying networks under joint practical influences of erroneous channel information, Nakagami-m fading severity, jamming suppression error, energy harvesting non-linearity, and power constraints of cognitive users. Towards this end, explicit expressions of intercept and outage probabilities are first derived. Then, they are used to quickly generate numerous results so as to withdraw interesting findings. More specifically, the proposed jamming strategy experienced the trade-off between security and reliability as well as reliability/security saturation at high peak transmit power. Additionally, channel estimation error, fading severity, jamming suppression error, and energy harvesting non-linearity significantly impact system performance. Nonetheless, system parameters can be set properly to achieve different security levels and prevent absolute outage.

Appendices

Appendix A: Formulas for ${{{\hat{\Lambda }}}_1}$, ${{{\hat{\Lambda }}}_2}$, ${{{\hat{\Lambda }}}_3}$

This appendix derives formulas for ${{{\hat{\Lambda }}}_1}$, ${{{\hat{\Lambda }}}_2}$, and ${{{\hat{\Lambda }}}_3}$ in (22), (23), and (24), respectively, to complete numerical evaluation for the intercept probability.

Invoking $P_r$ in (9) to further simplify (20) as

(36)

where and

(37)

Applying [69, eq. (8.352.1)] to expand $\gamma (\cdot ,\cdot )$ into series yields

(38)

where ${{\bar{H}}} = \frac{{\beta _{rp}^{{m_{rp}}}{e^{ - {{\bar{U}}}}}}}{{\Gamma \left( {{m_{rp}}} \right) }}$, , and . It is reminded that ${\left( \star \right) }$ is obtained by using binomial expansion while ${\left( \circ \right) }$ is achieved by using the pdf of ${{{{\tilde{g}}}_{rp}}}$ and the definition of the upper incomplete Gamma function $\Gamma (\cdot ,\cdot )$ in [69].

Since in (2) takes two distinct values corresponding to two different conditions, (36) is solved by firstly considering the following two scenarios and then applying total probability law to obtain its explicit form.

Scenario 1:

This scenario happens as ${g_{sr}} \le \frac{{{\phi }}}{{{P_s}}}$. By taking the expectation over ${g_{sr}} \le \frac{{{\phi }}}{{{P_s}}}$, one deduces and denotes in (36) for this scenario as

(39)

where

(40)

(41)

(42)

Now, , , and are derived to solve . First of all, expanding $\gamma (\cdot ,\cdot )$ into series yields

(43)

Then, using binomial expansion and performing some manipulations yield

(44)

where

$$\begin{aligned} {{\mathcal {A}}}\left( {a,b,k} \right) = \int \limits _0^a {\frac{{{e^{ - b/x}}}}{{{x^k}}}{f_{{g_{sr}}}}\left( x \right) dx}. \end{aligned}$$

(45)

Using the pdf of $g_{sr}$ and applying series expansion for ${{e^{ - {\nu _{sr}}x}}}$, one obtains an explicit form of ${{\mathcal {A}}}\left( {a,b,k} \right)$ as

$$\begin{aligned} \begin{aligned} {{\mathcal {A}}}\left( {a,b,k} \right)&= \int \limits _0^a {\frac{{{e^{ - b/x}}}}{{{x^k}}}\frac{{\nu _{sr}^{{m_{sr}}}}}{{\Gamma \left( {{m_{sr}}} \right) }}{x^{{m_{sr}} - 1}}{e^{ - {\nu _{sr}}x}}dx} \\&= \frac{{\nu _{sr}^{{m_{sr}}}}}{{\Gamma \left( {{m_{sr}}} \right) }}\int \limits _0^a {\frac{{{e^{ - b/x}}}}{{{x^k}}}{x^{{m_{sr}} - 1}}\left( {\sum \limits _{j = 0}^\infty {\frac{{{{\left( { - {\nu _{sr}}x} \right) }^j}}}{{j!}}} } \right) dx} \\&\mathop = \limits ^{\left( \bullet \right) } \frac{{\nu _{sr}^{{m_{sr}}}}}{{\Gamma \left( {{m_{sr}}} \right) }}\sum \limits _{j = 0}^\infty {\frac{{{{\left( { - {\nu _{sr}}} \right) }^j}}}{{j!}}} \Theta \left( {\frac{1}{a},b,{m_{sr}} + j - k} \right) . \end{aligned} \end{aligned}$$

(46)

In ${\left( \bullet \right) }$, variable change $y=1/x$ is applied to reduce the last integral to the form of the function $\Theta (\cdot ,\cdot ,\cdot )$ as

$$\begin{aligned} \begin{aligned} \Theta \left( {c,v,b} \right)&= \int \limits _c^\infty {\frac{{{e^{ - vy}}}}{{{y^{b + 1}}}}dy} \\&= {\left( { - 1} \right) ^{b + 1}}\frac{{{{\mathrm{Ei}}}\left( { - cv} \right) }}{{b!{v^{ - b}}}} + \frac{{{e^{ - cv}}}}{{{c^b}}}\sum \limits _{k = 0}^{b - 1} {\frac{{{{\left( { - cv} \right) }^k}}}{{\prod \limits _{t = 0}^k {\left( {b - t} \right) } }}}, \end{aligned}\nonumber \\ \end{aligned}$$

(47)

where ${\mathrm{Ei}}(\cdot )$ is the exponential integral [69].

Secondly, using the pdf of $g_{sr}$ and applying [69, eq. (8.352.2)] to expand $\Gamma (\cdot ,\cdot )$ into series yield

(48)

Finally, is solved by recalling that and are in a similar form:

(49)

Scenario 2:

This scenario occurs as ${g_{sr}} > \frac{{{\phi }}}{{{P_s}}}$. By taking the expectation over this region, one deduces and denotes in (36) for this scenario as

(50)

where

(51)

Conditioned on these two scenarios, total probability law reduces (36) to

(52)

To continue the proof, plugging (18) into (17) and (52) into (19), one deduces $\Lambda _1$ and $\Lambda _2$ as

$$\begin{aligned} {\Lambda _1} = \left\{ {\begin{array}{ll} {\frac{{\gamma \left( {{m_{se}},{\tilde{H}} + {{\bar{J}}}/{P_s}} \right) }}{{\Gamma \left( {{m_{se}}} \right) }}},&{}{\quad {{{{\bar{\Delta }}} }_e}G < 1}\\ 1,&{}{\quad {{{{\bar{\Delta }}} }_e}G \ge 1} \end{array}} \right. \end{aligned}$$

(53)

and

(54)

Now, inserting (53) and (54) into (16), one obtains (21) where

$$\begin{aligned} {{\hat{\Lambda }}_1} = {{\tilde{\Lambda }}_{21}} + {{\tilde{\Lambda }}_{22}} + {\bar{Q}}{{\hat{\Lambda }}_{1b}} - {{\bar{H}}}\sum \limits _{u = 0}^{{m_{re}} - 1} {\sum \limits _{v = 0}^u {{{\bar{K}}}} } {{\tilde{\Lambda }}_{23}}, \end{aligned}$$

(55)

$$\begin{aligned} {{{\hat{\Lambda }}}_2} = {{\mathbb {E}}_{{P_s}}}\left\{ {\frac{{\gamma \left( {{m_{se}},{\tilde{H}} + {{\bar{J}}}/{P_s}} \right) }}{{\Gamma \left( {{m_{se}}} \right) }}} \right\} , \end{aligned}$$

(56)

$$\begin{aligned} {{{\hat{\Lambda }}}_3} = {{{{\bar{\Lambda }}}}_{21}} + {{{{\bar{\Lambda }}}}_{22}} + {\bar{Q}}{{{\tilde{\Lambda }}}_{2b}} - {{\bar{H}}}\sum \limits _{u = 0}^{{m_{re}} - 1} {\sum \limits _{v = 0}^u {{{\bar{K}}}} } {{{{\bar{\Lambda }}}}_{23}}, \end{aligned}$$

(57)

with

(58)

(59)

(60)

(61)

(62)

(63)

(64)

(65)

Before deriving (56) and (58)-(65) to obtain explicit forms of ${{{\hat{\Lambda }}}_1}$, ${{{\hat{\Lambda }}}_2}$, and ${{{\hat{\Lambda }}}_3}$ in (22), (23), and (24), respectively, one needs to find the pdf of $P_s$. Toward this end, one derives firstly the cdf of $P_s$ as

(66)

The first derivative of ${F_{{P_s}}}\left( x \right)$ yields the pdf of $P_s$ as

(67)

Now, we are back to deriving (56) and (58)-(65). Firstly, using in (44) yields the explicit form of ${{{\tilde{\Lambda }}}_{21}}$ as

(68)

where

$$\begin{aligned} {{{\tilde{\Lambda }}}_{21a}} = {{\mathbb {E}}_{{P_s}}}\left\{ {\frac{{\gamma \left( {{m_{se}},{\tilde{H}} + {{\bar{J}}}/{P_s}} \right) }}{{\Gamma \left( {{m_{se}}} \right) }}\frac{{\gamma \left( {{m_{sr}},{\nu _{sr}}{\phi }/{P_s}} \right) }}{{\Gamma \left( {{m_{sr}}} \right) }}} \right\} ,\nonumber \\ \end{aligned}$$

(69)

and

(70)

Before deriving ${{{\tilde{\Lambda }}}_{21a}}$ and ${\mathcal {B}}(\cdot ,\cdot )$ in an explicit form, the following integral must be solved first:

$$\begin{aligned} {{\mathcal {D}}}\left( {{{\bar{J}}},q} \right) = {{\mathbb {E}}_{{P_s}}}\left\{ {\frac{{{e^{ - {{\bar{J}}}/{P_s}}}}}{{P_s^q}}} \right\} . \end{aligned}$$

(71)

Using (67) and appropriate variable change, ${{\mathcal {D}}}\left( {{{\bar{J}}},q} \right)$ is represented in an explicit form as

(72)

Therefore, using explicit form of ${\mathcal {A}}(\cdot ,\cdot ,\cdot )$ in (46) and performing some simplifications, one expresses ${\mathcal {B}}(\cdot ,\cdot )$ as

(73)

where .

Then, using series expansion for $\gamma (\cdot ,\cdot )$ and binomial expansion to solve ${{{\tilde{\Lambda }}}_{21a}}$ as

$$\begin{aligned} {{{\tilde{\Lambda }}}_{21a}}= & {} {{\mathbb {E}}_{{P_s}}}\left\{ \left[ {1 - {e^{ - \left( {{\tilde{H}} + \frac{{{{\bar{J}}}}}{{{P_s}}}} \right) }}\sum \limits _{l = 0}^{{m_{se}} - 1} {\frac{1}{{l!}}{{\left( {{\tilde{H}} + \frac{{{{\bar{J}}}}}{{{P_s}}}} \right) }^l}} } \right] \right. \nonumber \\&\times \left. \left[ {1 - {e^{ - \frac{{{\nu _{sr}}{\phi }}}{{{P_s}}}}}\sum \limits _{k = 0}^{{m_{sr}} - 1} {\frac{1}{{k!}}{{\left( {\frac{{{\nu _{sr}}{\phi }}}{{{P_s}}}} \right) }^k}} } \right] \right\} \nonumber \\= & {} 1 - \sum \limits _{k = 0}^{{m_{sr}} - 1} {\frac{{{{\mathcal {D}}}\left( {{\nu _{sr}}{\phi },k} \right) }}{{k!{{\left( {{\nu _{sr}}{\phi }} \right) }^{ - k}}}}} - {e^{ - {\tilde{H}}}}\sum \limits _{l = 0}^{{m_{se}} - 1} \sum \limits _{q = 0}^l \frac{{{\mathbb {C}}_l^q{{{{\bar{J}}}}^q}}}{{l!{{{\tilde{H}}}^{q - l}}}}\nonumber \\&\times \left[ {{{\mathcal {D}}}\left( {{{\bar{J}}},q} \right) - \sum \limits _{k = 0}^{{m_{sr}} - 1} {\frac{{{{\mathcal {D}}}\left( {{{\bar{J}}} + {\nu _{sr}}{\phi },k + q} \right) }}{{k!{{\left( {{\nu _{sr}}{\phi }} \right) }^{ - k}}}}} } \right] . \nonumber \\ \end{aligned}$$

(74)

Secondly, using in (48) and in (49), one deduces

(75)

and

$$\begin{aligned} {{\tilde{\Lambda }}_{23}} = \Gamma \left( {{m_{rp}} + v} \right) \sum \limits _{i = 0}^{{m_{rp}} + v - 1} {\frac{1}{{i!}}} {\left( {\frac{{{{\bar{L}}}}}{D}} \right) ^i}{{\mathcal {B}}}\left( {{{\bar{L}}},i} \right) . \end{aligned}$$

(76)

Thirdly, ${{{\hat{\Lambda }}}_{1b}}$ is solved as

$$\begin{aligned} \begin{aligned} {{{\hat{\Lambda }}}_{1b}}&= {{\mathbb {E}} _{{P_s}}}\left\{ {\frac{{\gamma \left( {{m_{se}},{\tilde{H}} + {{\bar{J}}}/{P_s}} \right) }}{{\Gamma \left( {{m_{se}}} \right) }}\left[ {1 - \frac{{\gamma \left( {{m_{sr}},{\nu _{sr}}{\phi }/{P_s}} \right) }}{{\Gamma \left( {{m_{sr}}} \right) }}} \right] } \right\} \\&= {{{{\hat{\Lambda }}}_2} - {{{\tilde{\Lambda }}}_{21a}}}, \end{aligned}\nonumber \\ \end{aligned}$$

(77)

where ${{{\hat{\Lambda }}}_2}$, which is defined in (56), is solved in an explicit form by using series expansion for $\gamma (\cdot ,\cdot )$ and binomial expansion as

$$\begin{aligned} \begin{aligned} {{{\hat{\Lambda }}}_2}&= {{\mathbb {E}}_{{P_s}}}\left\{ {1 - {e^{ - \left( {{\tilde{H}} + \frac{{{{\bar{J}}}}}{{{P_s}}}} \right) }}\sum \limits _{u = 0}^{{m_{se}} - 1} {\frac{1}{{u!}}{{\left( {{\tilde{H}} + \frac{{{{\bar{J}}}}}{{{P_s}}}} \right) }^u}} } \right\} \\&= {{\mathbb {E}}_{{P_s}}}\left\{ {1 - {e^{ - {\tilde{H}}}}\sum \limits _{u = 0}^{{m_{se}} - 1} {\sum \limits _{q = 0}^u {\frac{{{\mathbb {C}}_u^q{{{{\bar{J}}}}^q}}}{{u!{{{\tilde{H}}}^{q - u}}}}} \frac{{{e^{ - {{\bar{J}}}/{P_s}}}}}{{P_s^q}}} } \right\} \\&= 1 - {e^{ - {\tilde{H}}}}\sum \limits _{u = 0}^{{m_{se}} - 1} {\sum \limits _{q = 0}^u {\frac{{{\mathbb {C}}_u^q{{{{\bar{J}}}}^q}}}{{u!{{{\tilde{H}}}^{q - u}}}}} {{\mathcal {D}}}\left( {{{\bar{J}}},q} \right) }. \end{aligned} \end{aligned}$$

(78)

Fourthly, using in (44) yields

(79)

where

$$\begin{aligned} {{{{\bar{\Lambda }}}}_{21a}} = {{\mathbb {E}}_{{P_s}}}\left\{ {\frac{{\gamma \left( {{m_{sr}},{\nu _{sr}}{\phi }/{P_s}} \right) }}{{\Gamma \left( {{m_{sr}}} \right) }}} \right\} , \end{aligned}$$

(80)

and

(81)

By using series expansion for $\gamma (\cdot ,\cdot )$ and performing some simplifications, ${{{{\bar{\Lambda }}}}_{21a}}$ is computed as

$$\begin{aligned} {{{{\bar{\Lambda }}}}_{21a}} = 1 - \sum \limits _{u = 0}^{{m_{sr}} - 1} {\frac{{{{\left( {{\nu _{sr}}{\phi }} \right) }^u}}}{{u!}}{{\mathcal {D}}}\left( {{\nu _{sr}}{\phi },u} \right) }. \end{aligned}$$

(82)

Also, using the explicit form of $\mathcal {A}(\cdot ,\cdot ,\cdot )$ in (46) and executing basic manipulations yield

(83)

Fifthly, using in (48) and in (49), one obtains

(84)

and

(85)

Finally, ${{{\tilde{\Lambda }}}_{2b}}$ is solved as

$$\begin{aligned} \begin{aligned} {{{\tilde{\Lambda }}}_{2b}}&= {{\mathbb {E}}_{{P_s}}}\left\{ {1 - \frac{{\gamma \left( {{m_{sr}},{\nu _{sr}}{\phi }/{P_s}} \right) }}{{\Gamma \left( {{m_{sr}}} \right) }}} \right\} \\&= {1 - {{{{\bar{\Lambda }}}}_{21a}}}. \end{aligned} \end{aligned}$$

(86)

Appendix B: Explicit Expression for

This appendix derives the explicit expression for in (35) to complete numerical evaluation for the outage probability.

Since in (2) accepts two different values corresponding to two conditions, in (34) is derived by firstly considering the following two scenarios and then applying total probability law to reach its explicit form.

Scenario 1:

This scenario happens as ${g_{sr}} \le \frac{{{\phi }}}{{{P_s}}}$. Incorporating this condition with in (34) yields a feasible region of . By taking the expectation of in (30) over , one deduces and denotes for this scenario as

(87)

where

(88)

and

(89)

Using series expansions for $\Gamma (\cdot ,\cdot )$ and $\gamma (\cdot ,\cdot )$ as well as binomial expansion, one achieves

(90)

where

(91)

At first, we solve ${\bar{{\mathcal {K}}}}$. By variable changes $t = \sqrt{y}$ and then $y = t\sqrt{2{\beta _{sr}}/\left( {1 - \psi } \right) }$, one obtains

(92)

where ${{{\mathcal {Q}}}_{{m}}}\left( { \cdot , \cdot } \right)$ is the Marcum-Q function in [64, eq. (4.60)].

Inserting (92) into (91) and then applying a simple representation for $\mathcal {Q}_m(\cdot , \cdot )$ in [70, eq. (6)], it follows that ${{\mathcal {K}}}\left( {a,l} \right)$ is expressed in an explicit form after some basic simplifications:

$$\begin{aligned} {{\mathcal {K}}}\left( {a,l} \right)= & {} \sum \limits _{q = 0}^\omega \sum \limits _{j = 0}^\infty {\tilde{V}}\Gamma \left( {{m_{sr}} + q,{\hat{B}} + \frac{{{\hat{C}}}}{{{P_s}}}} \right) \Theta \nonumber \\&\quad \times \left( {\frac{{{P_s}}}{{{\phi }}},a,{m_{sr}} + q + j - l} \right) , \end{aligned}$$

(93)

where the parameter $\omega$ represents the accuracy-complexity trade-off when simplifying $\mathcal {Q}_m(\cdot , \cdot )$, and ${\tilde{V}} = \frac{{\nu _{sr}^{{m_{sr}} + q + j}{{\left( { - 1} \right) }^j}{\psi ^q}{{\left( {1 - \psi } \right) }^{ - q - j}}{\omega ^{1 - 2q}}\Gamma \left( {\omega + q} \right) }}{{j!\Gamma \left( {{m_{sr}}} \right) \Gamma \left( {q + 1} \right) \Gamma \left( {\omega - q + 1} \right) \Gamma \left( {{m_{sr}} + q} \right) }}$.

Again, using series expansion for $\Gamma (\cdot ,\cdot )$ yields

(94)

Scenario 2:

This scenario occurs as ${g_{sr}} > \frac{{{\phi }}}{{{P_s}}}$. Incorporating this condition with in (34) yields a feasible region of $\left( g_{sr}, \tilde{g}_{sr}\right)$ as . By taking the expectation of in (30) over , one deduces and denotes for this scenario as

(95)

where denotes computed at .

Following the same steps as deriving $\mathcal {K}(\cdot ,\cdot )$, one obtains

$$\begin{aligned} {{\mathcal {L}}} = \sum \limits _{q = 0}^\omega {{{\bar{V}}}\Gamma \left( {{m_{sr}} + q,{\hat{B}} + \frac{{{\hat{C}}}}{{{P_s}}}} \right) } \Gamma \left( {{m_{sr}} + q,\frac{{{{\bar{B}}}}}{{{P_s}}}} \right) , \end{aligned}$$

(96)

where ${{\bar{B}}} = \frac{{{\nu _{sr}}\phi }}{{1 - \psi }}$ and ${{\bar{V}}} = \frac{{\Gamma \left( {\omega + q} \right) {\omega ^{1 - 2q}}{{\left( {1 - \psi } \right) }^{{m_{sr}}}}{\psi ^q}}}{{\Gamma \left( {{m_{sr}}} \right) \Gamma \left( {q + 1} \right) \Gamma \left( {\omega - q + 1} \right) \Gamma \left( {{m_{sr}} + q} \right) }}$.

Conditioned on these two scenarios, total probability law yields . Therefore, in (35) reduces to

(97)

where

$$\begin{aligned} {{{\hat{\Upsilon }}}_1}= & {} {{\mathbb {E}} _{{P_s}}}\left\{ {{{{{\bar{\Upsilon }}}}_1}} \right\} , \end{aligned}$$

(98)

$$\begin{aligned} {{{\hat{\Upsilon }}}_2}= & {} {{\mathbb {E}}_{{P_s}}}\left\{ {{{{{\bar{\Upsilon }}}}_2}} \right\} , \end{aligned}$$

(99)

$$\begin{aligned} \hat{{\mathcal {L}}}= & {} {{\mathbb {E}}_{{P_s}}}\left\{ {\Gamma \left( {{m_{sr}} + q,{\hat{B}} + \frac{{{\hat{C}}}}{{{P_s}}}} \right) \Gamma \left( {{m_{sr}} + q,\frac{{{{\bar{B}}}}}{{{P_s}}}} \right) } \right\} .\nonumber \\ \end{aligned}$$

(100)

Now, we proceed to derive ${{{\hat{\Upsilon }}}_1}$, ${{{\hat{\Upsilon }}}_2}$, and $\hat{{\mathcal {L}}}$ to complete in (97), eventually completing $\Upsilon$ in (35). Firstly, inserting (90) into (98) yields

(101)

where

$$\begin{aligned} {{\mathcal {H}}}\left( {a,l} \right) = {{\mathbb {E}}_{{P_s}}}\left\{ {P_s^{ - l}{{\mathcal {K}}}\left( {\frac{a}{{{P_s}}},l} \right) } \right\} . \end{aligned}$$

(102)

Plugging (93) into (102), it follows that

$$\begin{aligned} {{\mathcal {H}}}\left( {a,l} \right) = \sum \limits _{q = 0}^\omega {\sum \limits _{j = 0}^\infty {{\tilde{V}}{{\mathcal {W}}}\left( {{m_{sr}} + q + j} \right) } }, \end{aligned}$$

(103)

where

$$\begin{aligned} {{\mathcal {W}}}\left( c \right) = {{\mathbb {E}}_{{P_s}}}\left\{ {P_s^{ - l}\Gamma \left( {{m_{sr}} + q,{\hat{B}} + \frac{{{\hat{C}}}}{{{P_s}}}} \right) \Theta \left( {\frac{{{P_s}}}{{{\phi }}},\frac{a}{{{P_s}}},c - l} \right) } \right\} .\nonumber \\ \end{aligned}$$

(104)

Using series expansion for $\Gamma (\cdot ,\cdot )$ and the explicit form of $\Theta (\cdot ,\cdot ,\cdot )$ in (47), one achieves an explicit form of ${{\mathcal {W}}}\left( c \right)$ as

(105)

where

(106)

Likewise, inserting (94) into (99) yields

$$\begin{aligned} {{{\hat{\Upsilon }}}_2} = \sum \limits _{k = 0}^{{m_{rp}} + v - 1} {\frac{1}{{k!}}{{\left( {\frac{{{\hat{L}}}}{D}} \right) }^k}} {{\mathcal {H}}}\left( {\frac{{{\hat{L}}}}{D},k} \right) . \end{aligned}$$

(107)

Finally, using series expansion for $\Gamma (\cdot ,\cdot )$ and binomial expansion, one obtains the following after some basic manipulations:

$$\begin{aligned} \hat{{\mathcal {L}}}= & {} {\left[ {\Gamma \left( {{m_{sr}} + q} \right) } \right] ^2}{e^{ - {\hat{B}}}}\sum \limits _{k = 0}^{{m_{sr}} + q - 1} \sum \limits _{a = 0}^k \sum \limits _{i = 0}^{{m_{sr}} + q - 1} \nonumber \\&\quad {\frac{{{\mathbb {C}}_k^a{{{\hat{C}}}^a}{{{{\bar{B}}}}^i}}}{{k!i!{{{\hat{B}}}^{a - k}}}}{{\mathcal {D}}}\left( {{{\bar{B}}} + {\hat{C}},a + i} \right) } . \end{aligned}$$

(108)