*3.2. Non-Colluding Eavesdroppers*

In a non-colluding eavesdroppers scenario, we investigated the STP for traditional beamforming (*η* = 1) and AN beamforming (*η* = 1). When traditional beamforming is adopted, we derived the exact value of STP, whereas AN is introduced, and we calculated the lower bound of STP which is a rigorous assumption and common practice [16,37]. We denote the STP for traditional beamforming as *P*<sup>1</sup> and for AN beamforming as *P*2, respectively. Based on Equation (12), *P*<sup>1</sup> is given by:

$$P\_1 = \prod\_{E\_i \in \Phi} P(\frac{1 + \text{SNR}\_B}{1 + \text{SNR}\_{E\_i}} > 2^R) \stackrel{(b)}{=} E\_{\mathbb{S}\_B} \{ \exp(-2\pi\lambda\_p \int\_0^{R\_S} P(S\_E > \frac{S\_B L\_B}{2^R L\_E}) \rho d\rho) \},\tag{15}$$

where (*b*) holds for SNRB 1, SNRE 1 and the *probability generating functional lemma* (PGFL, ref. [43]) over PPP.

By denoting *u* = *kGtGr*(*η*0*Kλ*)2/128*π*3, we have *LE* = *u* <sup>1</sup> *d*2 . As *SE* ∼ *E*(1), the Equation (15) can finally be derived as:

$$P\_1 = \mathbb{E}\_{S\_B} \{ \exp(2\pi\lambda\_p((vR\_S + v^2)e^{-\frac{R\_S}{v}} - v^2)) \},\tag{16}$$

where *v* = *u*2*R*/*SBLB*.

Similarly to the calculation of *<sup>P</sup>*<sup>1</sup> and by denoting *<sup>β</sup>* <sup>=</sup> <sup>2</sup>*Rσ*<sup>2</sup> *n PLB* , *<sup>P</sup>*2(*CS* <sup>&</sup>gt; *<sup>R</sup>*) is given by:

$$P\_2 = \prod\_{E\_i \in \Phi} P(\frac{1 + \text{SNR}\_B}{1 + \text{SNR}\_{E\_i}} > 2^R) \stackrel{(c)}{=} E\_{S\_B} \{ \exp(\frac{-\pi R\_S^2 \lambda\_p}{(1 + \frac{S\_B \eta - \beta}{\beta \Phi})^{N-1}}) \}. \tag{17}$$

where (*c*) holds for SNRB 1 and the PGFL over PPP.

### *3.3. Colluding Eavesdroppers*

In colluding case, we denote the STP without AN as *P*<sup>3</sup> and with AN as *P*4, respectively. The STP *P*3(*CS* > *R*) is given by:

$$P\_3 = P(\frac{1 + \text{SNR}\_B}{1 + \sum \text{SNR}\_{E\_i}} > 2^R) = P(S\_B > \frac{2^R \sum\_{E\_i \in \Phi} S\_{E\_i} L\_{E\_i}}{L\_B}).\tag{18}$$

We let *<sup>I</sup>*<sup>1</sup> <sup>=</sup> <sup>∑</sup>*Ei*∈<sup>Φ</sup> *SEi LEi* and thus *P*<sup>3</sup> can be modified as:

$$P\_3 = \int\_0^\infty P(S\_B > p\_1 i) f\_{I\_1}(i) di \stackrel{(d)}{=} \sum\_{b=0}^{mN-1} m\_b p\_1^b (-1)^b \mathcal{L}^{(b)}\{f\_{I\_1}(i)\}(m p\_1),\tag{19}$$

where *fI*<sup>1</sup> (*i*) is the probability density function (PDF) of *I*<sup>1</sup> and *p*<sup>1</sup> = 2*R*/*LB*, (*d*) holds for *SB* ∼ *Gamma*(*mN*, *m*) and the complementary cumulative distribution function (CCDF) of *SB* is given by *F<sup>c</sup> SB* <sup>=</sup> *<sup>e</sup>*−*mx* <sup>∑</sup>*b*∈*<sup>B</sup> mbx<sup>b</sup>*, where *mb* <sup>=</sup> *<sup>m</sup><sup>b</sup> <sup>b</sup>*! and *b* ∼ (0, *mN* − 1). The Laplace transformation L{ *fI*<sup>1</sup> (*i*)}(*mp*1) of function *fI*<sup>1</sup> (*i*) is given by:

$$\mathcal{L}\{f\_{\rm I}\}(p\_{\rm I}) = \exp\{-2\pi\lambda\_{\rm p}p\_{\rm I}u\mathcal{R}\_{\rm S}\}(1+\frac{\mathcal{R}\_{\rm S}}{p\_{\rm I}u})^{2\pi\lambda p\_{\rm I}^2u^2}.\tag{20}$$

By adopting *Bruno's formula* [44], we can obtain the *n*-degree derivation of L{ *fI*<sup>1</sup> (*i*)}(*p*1) as:

$$\mathcal{L}^{(n)}\{f\_{l\_1}\}(p\_1) = \sum \frac{n!}{b\_1! \cdot b\_n!} e^{f(p\_1)} \prod\_{j=1}^n (\frac{f^{(j)}(p\_1)}{j!})^{b\_j} \,. \tag{21}$$

where the sum is over all the solutions *b*1, ···, *bn* ≥ 0 to *b*<sup>1</sup> + 2*b*<sup>2</sup> + ··· + *nbn* = *n*. By denoting *w* = *RS*/*u*, *c*<sup>1</sup> = 2*πλp*, *c*<sup>2</sup> = 1 + *<sup>w</sup> p*1 , *c*<sup>3</sup> = <sup>1</sup> *<sup>p</sup>*1+*<sup>w</sup>* <sup>−</sup> <sup>1</sup> *p*1 , *c*<sup>4</sup> = <sup>1</sup> *p*2 1 <sup>−</sup> <sup>1</sup> (*p*1+*w*)<sup>2</sup> , *<sup>f</sup>*(*p*1) and *f* (*j*)(*p*1) are given by:

$$f(p\_1) = c\_1(p\_1^2 \mu^2 \ln c\_2 - p\_1 \mu \mathcal{R}\_S),\tag{22a}$$

$$f^{(1)}(p\_1) = c\_1(2p\_1u^2 \ln c\_2 + p\_1^2u^2 c\_3 - uR\_S)\_\prime \tag{22b}$$

$$f^{(2)}(p\_1) = c\_1(2u^2 \text{lnc}\_2 + 4p\_1u^2c\_3 + p\_1^2u^2c\_4),\tag{22c}$$

$$f^{(j>2)}(p\_1) = c\_1 u^2 \sum\_{k k=0}^2 \mathbb{C}\_2^{k k} (-1)^{j-k k} \frac{j!}{(j-k k)} p\_1^{2-k k} (\frac{1}{p\_1^{j-k k}} - \frac{1}{(p\_1+w)^{j-k k}}).\tag{22d}$$

When AN beamforming is introduced, *P*4(*CS* > *R*) is given by:

$$P\_4 = \int\_0^\infty P(S\_B > p\_2 i) f\_{l\_2}(i) di = P(S\_B > \frac{\beta(1 + \sum\_{\tilde{c} \in \Phi} \phi \mathbb{S}\_{equal\_i})}{\eta}).\tag{23}$$

We let *<sup>I</sup>*<sup>2</sup> <sup>=</sup> <sup>1</sup> <sup>+</sup> <sup>∑</sup>*Ei*∈<sup>Φ</sup> *<sup>φ</sup>Sequali* and thus *<sup>P</sup>*<sup>4</sup> can be rewritten as:

$$P\_4 = \int\_0^\infty P(S\_B > p\_2i) f\_{l\_2}(i) di = \sum\_{b=0}^{mN-1} m\_b p\_2^b (-1)^b \mathcal{L}^{(b)}\{f\_{l\_2}(i)\}(mp\_2),\tag{24}$$

where *fI*<sup>2</sup> (*i*) is the PDF of *<sup>I</sup>*<sup>2</sup> and *<sup>p</sup>*<sup>2</sup> <sup>=</sup> *<sup>β</sup> <sup>η</sup>* . As long as we obtain L{ *fI*<sup>2</sup> (*i*)}(*mp*2), *P*<sup>4</sup> can be calculated. The Laplace transformation L{ *fI*<sup>2</sup> (*i*)}(*p*2) is given by:

$$\mathcal{L}\{f\_{l\_2}\}(p\_2) = \exp\{-p\_2 - N\_E + q\_1 q\_2\},\tag{25}$$

where *<sup>q</sup>*<sup>1</sup> <sup>=</sup> *exp*(*p*2*φ*)*EN*(*p*2*φ*), *EN*(*x*) = <sup>∞</sup> 1 *e*−*xt <sup>t</sup><sup>N</sup> dt* is the *N*-degree exponential integral and *q*<sup>2</sup> = *NE*(*N* − 1). As such, the *n*-degree of L{ *fI*<sup>2</sup> (*i*)}(*p*2) is given by:

$$\mathcal{L}^{(n)}\{f\_{l\_2}\}(p\_2) = \sum \frac{n!}{b\_1! \cdot b\_n!} e^{\mathcal{S}(p\_2)} \prod\_{j=1}^n (\frac{\mathcal{S}^{(j)}(p\_2)}{j!})^{b\_j} \,. \tag{26}$$

where *g*(*p*2) and *g*(*j*)(*p*2) are given by:

$$g(p\_2) = -p\_2 - N\_E + q\_1 q\_{2\prime} \tag{27a}$$

$$g^{(1)}(p\_2) = -1 + q\_2 \phi e^{k\phi} (E\_N - E\_{N-1})\_\prime \tag{27b}$$

$$\mathcal{S}^{(j\geq 2)}(p\_2) = q\_2 \phi^j e^{k\phi} \sum\_{j\bar{j}=0}^j \mathcal{C}\_j^{j\bar{j}} (-1)^{j\bar{j}} E\_{N-j\bar{j}}.\tag{27c}$$
