Multicast throughput for large scale cognitive networks

Abstract

In this paper, we focus on the achievable throughput of cognitive networks consisting of the primary ad hoc network (PaN) and the secondary ad hoc network (SaN). We construct PaN and SaN by placing nodes according to Poisson point processes of density n and m respectively over a unit square region. We directly study the multicast throughput of cognitive network to unify that of unicast and broadcast sessions. In order to ensure the priority of primary users in meanings of throughput, we design a metric called throughput decrement ratio (TDR) to measure the ratio of the throughput of PaN in presence of SaN to that of PaN in absence of SaN. Endowing PaN with the right to determine the threshold of the TDR, we propose multicast schemes based on TDMA and multihop routing for the two networks respectively and derive their achievable multicast throughput depending on the given threshold. Specially, we show when PaN has sparser density than SaN, to be specific, $n=o\left({\frac{m} {(\log m)^2}}\right),$ and if PaN only cares about the order of its throughput, SaN can simultaneously achieve the same order of the aggregated multicast throughput as it were a stand-alone network in absence of PaN.

Appendix A

1.1 Proofs for lemmas

Proof

Lemma 8: Consider any link $\langle v^p_i,v^p_j\rangle\in \Upxi_p(\digamma,\tau)$ for any $\tau \in {\mathcal{T}},$ where v^p _i and v^p _j locate on two adjacent cells respectively.

First, we bound I _pp (v ^p _i , v ^p _j ; τ). We notice that the transmitters in the eight closest cells are located at distance at least $(K-2)\sqrt{a_p}$ (i.e., $\sqrt{a_p}$ in Fig. 1) from the receiver. The 16 next closest cells are of distance at least $(2K-2)\sqrt{a_p}$ (i.e., $4\sqrt{a_p}$ in Fig. 1). By extending the sum of the interferences to the whole region, it can be upper bounded as follows:

$$ I_{pp}(v^p_i,v^p_j;\tau) \le \sum\limits_{i = 1}^{\lceil1/{a_p}\rceil} {8iP({a_p})^{\alpha/2} \ell((Ki - 2)\sqrt {a_p} )} \leq P\sum\limits_{i = 1}^n {8i(Ki - 2)^{ - \alpha }}. $$

For K ≥ 3 and α > 2, and by the Cauchy Test, we have

$$ \lim_{n\to \infty}I_{pp}(v^p_i,v^p_j;\tau )\leq 8P \int_{1}^{+\infty}{{\frac{x} {(Kx-2)^{\alpha}}}}dx = 8P\cdot C_1(K,\alpha), $$

where the constant C ₁(K, α) (depending on K and α) that is defined in Eq. (16). It is clear that $\lim\limits_{n\to \infty}{I_{pp}(v^p_i,v^p_j;\tau)}$ gets the minimum value when v ^p _j locates in the cell on the corner. Thus, we obtain, for any $\langle v^p_i,v^p_j\rangle{\in}\Upxi_p(\digamma,\tau),$ it holds that

$$ \lim_{n\to \infty}{I_{pp}(v^p_i,v^p_j;\tau)}\geq {\frac{1} {4}}\cdot8P\cdot C_1(K,\alpha)=2P\cdot C_1(K,\alpha). $$

(18)

Second, we upper bound I _sp (v ^p _i , v ^p _j ; τ). The preservation region centered on v ^p _j consists of K ² secondary cells, hence, for any slot τ, there must exist one cell out of the K ² cells that were to be scheduled in τ except it locates in the preservation region, we denote that cell as c _τ. We consider those cells containing the nodes in ${\mathcal{V}}_s(v^p_i,\tau)$. It can be seen that the secondary users in the eight closest cells to c _τ are far away from v ^p _j with distance at least of $\sqrt{a_s}(K-1)/2$. The secondary users in the 16 next closest cells to c _τ are at Euclidean distance at least $\sqrt{a_s}(K+(K-1)/2)$ between v ^p _j . Thus, the sum of the interferences can be upper-bounded as follows:

$$ \begin{aligned} I_{sp}(v^p_i,v^p_j;\tau) \le & \sum\limits_{i = 1}^{\lceil1/{a_s}\rceil} {8i\cdot P\cdot\theta(n)\cdot {a_s}^{{\frac{\alpha} {2}}} \cdot \ell\left(\left(K(i-1)+{\frac{K-1} {2}}\right)\sqrt {a_s} \right)} \\ \leq\;2P\cdot\theta(n)\cdot\sum\limits_{i = 1}^m {4i\left(Ki - {\frac{K+1} {2}}\right)^{ - \alpha }}. \\ \end{aligned} $$

Recall that $n=o\left({\frac{m} {(\log m)^2}}\right),$ we have

$$ \lim\limits_{n\to \infty} \sum\limits_{i = 1}^m {4i\left(Ki - {\frac{K+1} {2}}\right)^{ - \alpha}} \leq \int_{1}^{+\infty}{{\frac{4x} {(Kx-{\frac{k+1} {2}})^{\alpha}}}}d x. $$

The right integral can be calculated to be a constant C ₂(K, α) depending on K and α that is described by Eq. (16). Hence, we have

$$ \lim_{n\to \infty} I_{sp}(v^p_i,v^p_j;\tau)\le 2P \cdot\theta\cdot C_2(K, \alpha). $$

(19)

Third, we lower bound the signal received from the transmitter S(v ^p _i , v ^p _j ;τ). Since the node only transmits to a destination located in the adjacent cell, i.e., $|v^p_i-v^p_j|\leq\sqrt {5a_p}.$ Thus, the strength of the signal is bounded as:

$$ S(v^p_i,v^p_j;\tau) \ge P \cdot ({a_p})^{\alpha/2} \cdot (\sqrt{5a_p})^{-\alpha }=5^{ -{\frac{\alpha} {2}}}\cdot P. $$

(20)

Finally, taking the limit of the SINR, we can obtain

$$ R_p(v^p_i,v^p_j;\tau) \geq \log \left(1 + {\frac{5^{ -{\frac{\alpha} { 2}}}} {N_0/P + 8C_1(K,\alpha)+8\theta(n)C_2(K,\alpha)}}\right). $$

By Eqs. (18) and (19), $\Updelta(\digamma, K, P)\leq{\theta}\cdot{\frac{ C_2(K,\alpha)} {C_1(K,\alpha)}}$. Let $\lim\limits_{n\to \infty}\theta(n)=\theta \leq \Updelta \cdot {\frac{C_1(K,\alpha)} { C_2(K,\alpha)}},$we can get $\Updelta(\digamma, K, P){\leq}\triangle$. Hence,

$$ R_p(v^p_i,v^p_j;\tau) \geq \log \left(1 + {\frac{5^-{\frac{\alpha}{2}}} {N_0/P + 8(1+\Updelta)C_1(K,\alpha)}} \right). $$

Proof

Lemma 9 Given a primary cell c _t *, we define the number of multicast sessions (flows) that are routed through the nodes inside c _t * as a random variable X _t , and we finally consider the uniform upper bound of X _t , denoted by X, for every primary cell.

Define event A(k, t): Multicast session ${\mathcal{M}}_{k}$ passes through the cell c _t *. For any link $v_iv_j \in \hbox{EST}({\mathcal{U}}_k)$, Define event A ^h _ij (k, t): ${\mathcal{HP}}_{ij}$ passes through c _t *; and define event A ^v _ij (k, t): ${\mathcal{VP}}_{ij}$ passes through c _t *. Obviously, we have

$$ {\Pr}(A(k,t) )=\Pr\left(\bigcup\limits_{e_{ij} \in \Pi_k} (A^h_{ij}(k,t)\cup A^v_{ij}(k,t) )\right). $$

By union bounds, it holds that

$$ {\Pr}(A(k,t) )\le \sum\limits_{e_{ij} \in \Pi_k} (\Pr(A_{ij}^h (k,t)) + \Pr(A_{ij}^v (k,t)) ). $$

First, we give the bound of $\Pr(A(k,t))$. Based on the cell c _t *, we construct the region ${\mathcal{S}}^h_{ij}(k,t)$ of area $S^h_{ij}(k,t)=\sqrt{a_p}(2 | {{\mathcal{HP}}_{ij} } | + \sqrt {a_p}),$ as in Fig. 3. Similarly, we construct the region ${\mathcal{S}}^v_{ij}(k,t)$ area of $S^v_{ij}(k,t)=\sqrt{a_p}(2 | {{\mathcal{VP}}_{ij} } | + \sqrt {a_p})$. The following two propositions obviously hold.

Fig. 3

The construction of the region ${\mathcal{S}}^h_{ij}(k,t)$. Here |·| represents the Euclid length of a line segment or the Euclid distance between two nodes

Proposition 1

The Poisson pointv _i locates in the region${\mathcal{S}}^h_{ij}(k,t)$if the eventA ^h _ij (k, t) happens.

Proposition 2

The Poisson pointv _j locates in the region${\mathcal{S}}^v_{ij}(k,t)$if the eventA ^v _ij (k, t) happens.

Define event B ^h _ij (k, t) (or B ^v _ij (k, t)): A Poisson node locates in a region of area S ^h _ij (k, t) (or S ^v _ij (k, t)). Then, by Propositions 1 and 2, we have

$$ \Pr(A_{ij}^h (k,t))\leq \Pr(B_{ij}^h (k,t)), \Pr(A_{ij}^v (k,t))\leq \Pr(B_{ij}^v (k,t)). $$

Moreover, define event B(k, t): A Poisson node locates in a region of area $S(k,t)=\min\{1, \tilde{S}(k,t) \},$where

$$ \tilde{S}(k,t)=\sum \limits_{e_{ij} \in \Pi_k}(S^h_{ij}(k,t)+ S^v_{ij}(k,t)). $$

Hence, ${\Pr}(A(k,t) ){\leq}{\Pr}(B(k,t) )$.

For A(k, t) and B(k, t), define their indicator variables as ${\mathcal{I}}(A(K,t))$ and ${\mathcal{I}}(B(K,t)),$ where ${\mathcal{I}}(A)$ takes value 1 if event A happens, otherwise 0. Hence, $X_t=\sum\nolimits_{k=1}^{n_s}{\mathcal{I}}(A(K,t))$. Define $Y_t=\sum\nolimits_{k=1}^{n_s}{\mathcal{I}}(B(K,t)),$ then it represents the number of nodes in the region of area S(k, t) according to a p.p.p of density n _s . So it follows a Poisson distribution of mean λ = n _s S(k, t).

Next, we consider the upper bound of S(k, t). Recall that

$$ \tilde{S}(k,t) =\sum\limits_{e_{ij} \in \Pi_k}(2| {{\mathcal{HP}}_{ij} } | + \sqrt {a_p}+ 2| {{\mathcal{VP}}_{ij} } | + \sqrt {a_p})\sqrt {a_p}. $$

Since $| {{\mathcal{HP}}_{ij} } | + | {{\mathcal{VP}}_{ij} } | \le \sqrt 2 | {v_i-v_j } |$ and by Lemma 5, we have

$$\tilde{S}(k,t) \le 2\sqrt {a_{p} } \sum\limits_{{e_{{ij}} \in \Pi _{k} }} ( \sqrt 2 |v_{i} - v_{j} | + \sqrt {a_{p} } ) \le 2(n_{d} a_{p} + 4\sqrt {n_{d} a_{p} } ).$$

From the latest inequality, we obtain the uniform upper bound of S(k, t), independent of k, denoted as S. That is,

$$ S(k,t)\leq S= \min \{2(n_d {a_p}+ 4\sqrt {n_d a_p}), 1\}. $$

Hence, the upper bound of Y _t follows Poisson with λ = n _s S.

When n _s S = ω(log n), by union bounds and Lemma 4, we get

$$ \begin{aligned} \Pr(X \ge 2n_s S ) \le& \left({\frac{n} {\log n}}\right)\cdot\Pr(X_t\ge 2 n_s S ) \le \left({\frac{n} {\log n}}\right)\cdot\Pr(Y_t \ge 2n_s S )\\ \leq& \left({\frac{n} {\log n}}\right)\cdot{\frac{{ e^ {- n_s S}(e n_s S)^{2 n_s S}}} {(2 n_s S)^{(2 n_s S)}}} = \left({\frac{n} {\log n}}\right)\cdot({\frac{4} {e}})^{ - n_s S } = o({\frac{1} {\log n}})\to 0, (n\to\infty). \end{aligned} $$

When n _s S = O(logn), we can assume that $\lim\nolimits_{n\to \infty }{\frac{n_s S} {\log n}}=f,$ where f ≥ 0 is a constant. Let z = e ²max{f, 1}, by union bounds and Lemma 1, we have

$$ \begin{aligned} \Pr(X \ge z\log n ) \le& \left({\frac{n} {z\log n}}\right)\cdot\Pr(X_t\ge z\log n ) \le \left({\frac{n} {\log n}}\right)\cdot {\Pr}(Y_t\geq z\log n ) \\ \leq& \left({\frac{1} {\log n}}\right)\cdot\left({\frac{e^{{\frac{(1-f)} {z}}+1}f} {z}}\right)^{z\log n} \leq (e^{{\frac{1} {e^2}}-1})^{z\log n} \to 0, \hbox{as} n\to \infty. \\ \end{aligned} $$

To sum up the two cases, we can prove the lemma.

Proof

Lemma 10 For any link $\langle v_i^s, v_j^s\rangle \in \Upxi_s(\bar{\digamma}), v_j^s$ must locate in a primary cell, and it may be served if it is out of the preservation regions. However, there is possibly a time slot τ₀ in which the distance from a primary node $v_0^p \in {\mathcal{V}}_p(v_i^s, \tau_0)$ to v ^s _j is so close that a fatal interference is imposed on v ^s _j . The secondary transmission scheduling scheme can prevent this scenario from happening. Since for ${\mathcal{N}}_s(m)$ the same packet is transmitted along M time slots, we can guarantee that there exists a time slot τ out of M time slots in which the minimum distance to v ^s _j from all $v^p \in {\mathcal{V}}_p(v_i^s, \tau)$ is at least of $(M-{\frac{5} {2}})\sqrt{a_p}$. Hence,

$$\begin{aligned} I_{{ps}} (v_{i}^{s} ,v_{j}^{s} ;\tau ) \le & \sum\limits_{{i = 1}}^{\infty } {8iP(a_{p} )^{{{\frac{\alpha }{2}}}} \ell ((Ki - 2)\sqrt {a_{p} } )} \\ & + P(a_{p} )^{{{\frac{\alpha }{2}}}} \ell (\left( {M - {\frac{5}{2}}} \right)\sqrt {a_{p} } ). \\ \end{aligned}$$

So, we have

$$ I_{ps}(v^p_i,v^p_j;\tau )< 8P\cdot C_1+ (M-{\frac{5} {2}})^{-\alpha}\cdot P, $$

where C₁ = C₁(K, α) is the constant defined in Eq. (16). Similar to Lemma 8, we can obtain that I _ps (v^s _i , v^s _j ) < 8P · θ(n) · C₁. Since

$$ S(v^s_i,v^s_j;\tau )\geq P\cdot \theta(n) \cdot {a_s}^{{\frac{\alpha} {2}}}(5 a_s)^{{\frac{\alpha} {2}}})=P \cdot \theta(n) \cdot 5^{-{\frac{\alpha} {2}}}, $$

when $0<\lim\nolimits_{n {\to}\infty}\theta(n)<\infty,$ we have

$$ R_s(v^s_i,v^s_j;\tau)\geq {\frac{1} {MK^2}}\cdot\log \left( 1+{\frac{P\cdot \theta \cdot (\sqrt{5})^{-\alpha}}{N_0+P\cdot (6C_1(1+ \theta)+({\frac{2} {2M-5}})^{\alpha})}}\right ). $$

If θ(n) = o(1), we have

$$ R_s(v^s_i,v^s_j;\tau)\geq {\frac{P\cdot \theta(n) \cdot (\sqrt{5})^{-\alpha}} {MK^2N_0+P\cdot (6C_1(1+ \theta)+({\frac{2} {2M-5}})^{\alpha})}}. $$

Combining two cases in terms of θ(n), we complete the proof.

Proof

Lemma 12 Define the number of non-served sources as a random variable ξ _s ′. According to Lemma 11, it follows a Poisson distribution of mean

$$ \lambda^{\prime}_s \leq m_s\cdot S^{\prime}_{max}=K^2\mu \cdot m_s \cdot n\cdot {\frac{\log m} {m}}. $$

When $m_s \cdot n\cdot {\frac{\log m} {m}}=\omega(1),$ let ${\rho}^{\prime}_s(m)=2K^2 \cdot \mu \cdot n\cdot {\frac{\log m} {m}}$. By Lemma 4, we get

$$ \Pr(\xi^{\prime}_s \geq {\rho}^{\prime}_{s}(m)\cdot m_s) \leq ({e}/{4})^{ K^2\mu \cdot m_sn\cdot {\frac{\log m} {m}} } \to 0. $$

When $m_s \cdot n\cdot {\frac{\log m} {m}}=O(1),$ let ${\rho}^{\prime}_s(m)=(m_s)^{-\kappa_1},$ where 0 < κ₁ < 1 is a constant. By Lemma 4, there exists a constant κ₂ such that

$$ \Pr(\xi^{\prime}_s \geq (m_s)^{1-\kappa_1}) \leq \kappa_2 \cdot (e/(m_s)^{1-\kappa_1})^{(m_s)^{1-\kappa_1}} \to 0. $$

Combining the two cases in terms of the relations among m _s , m and n, we complete the proof.

Proof

Lemma 13 According to Lemma 11, the random variable ξ^d _k follows a Poisson distribution of mean $K^2\mu \cdot m_d\cdot n\cdot {\frac{\log m} {m}}$. By using Lemma 4 and union bounds, we separately commence our analysis in two cases according to the relations among m _d , m and n. When $m_d \cdot n \cdot {\frac{\log m} {m}}=\omega(\log m_s),$ it holds that,

$$ \Pr(\xi^d \geq 2K^2\mu m_d\cdot n\cdot {\frac{\log m} {m}}) \leq m_s\cdot \left({\frac{e} {4}}\right)^{ K^2\mu m_d\cdot n\cdot {\frac{m} {\log m}} } \to 0. $$

When $m_d \cdot n\cdot {\frac{\log m} {m}}=O(\log m_s),$ and

$$ \lim\limits_{n\to \infty }{\frac{m_d \cdot n\cdot \log m} {m\cdot\log m_s}}=f_1, f_1\geq 0 $$

Let z₁ = 3 + log(1 + K²μf₁), we have

$$ \Pr(\xi^d \geq e^{z_1}\log m_s) \le m_s\Pr(\Upxi^d_k \geq z_1\log m_s) \le m_s^{1+e^{z_1}-K^2\mu f_1-z_1e^{z_1}+e^{z_1}\log (1+K^2\mu f_1)} \to 0. $$

Combining the two cases, we obtain that

$$ \rho^{\prime}_d(m)= \left\{ \begin{array}{lll} O\left(n \cdot {\frac{\log m} {m}}\right) & \hbox{when} & m_d \cdot n \cdot {\frac{\log m} {m}}=\Upomega(\log m_s) \\ O\left({\frac{\log m_s} {m_d}}\right) & \hbox{when} & m_d \cdot n\cdot {\frac{\log m} {m}}=O(\log m_s)\\ \end{array} \right. $$

Thus, we have that ρ _d ′(m) = o(1) when m _d  = ω(log m _s ).

Proof

Lemma 14 Define the number of the failed multicast sessions as a random variable γ. Note that γ follows a Poisson distribution of mean

$$ \lambda^{\prime\prime}_s\leq m_d\cdot m_s \cdot K^2\cdot\mu \cdot n\cdot \log m/m. $$

When $m_d\cdot m_s \cdot n\cdot {\frac{\log m} {m}}=\omega(1),$ let ${\rho}^{\prime\prime}_s(m)=2m_d \cdot K^2\cdot\mu \cdot n\cdot {\frac{\log m} {m}}$. By Lemma 4, we have

$$ \Pr(\gamma \geq {\rho}^{\prime\prime}_s(m) \cdot m_s )\leq (e/4)^{m_dm_s \cdot K^2\cdot\mu \cdot n\cdot {\frac{\log m} {m}})}\to 0. $$

When $m_d\cdot m_s \cdot n\cdot {\frac{\log m} {m}}=O(1),$ let ${\rho}^{\prime\prime}_s(m)=(m_s)^{-\kappa_3},$ where 0 < κ₃ < 1 is a constant. By Lemma 4, there exists a constant κ₄ such that

$$ \Pr(\gamma \geq {\rho}^{\prime\prime}_s(m) \cdot m_s) \leq \kappa_4 \cdot (e/(m_s)^{1-\kappa_3})^{(m_s)^{1-\kappa_3}} \to 0, {\rm as\ } m, n \to \infty. $$

Combining the two cases, the lemma can be proved.

Proof

Lemma 16 As in N _p (n), to simplify the description, we define a sequence of sets of directed edges:

$$ \Uppi^{\prime}_k=\{e_{ij} | \langle v_i, v_j\rangle \in \hbox{EST}({\mathcal{U}}^{\prime}_k )\}, {\ for\ } k=1, 2, \ldots, m_s. $$

Given a secondary cell $\bar{c}^{*}_{t},$ we define the number of multicast flows that will be routed through the nodes inside this cell as a random variable Z _t , and we finally consider the uniform upper bound Z of Z _t for every secondary cell.

Define event D(k, t): Multicast session ${\mathcal{M}}_{k}$ passes through the cell $\bar{c}^*_{t}$. For any link $v_iv_j \in \hbox{EST}({\mathcal{U}}^{\prime}_k),$ Define event D ^h _ij (k, t): $\overline{{\mathcal{HP}}}_{ij}$ passes through $\bar{c}^*_{t} ;$ and event D ^v _ij (k, t): $\overline{{\mathcal{VP}}}_{ij}$ passes through $\bar{c}^*_{t}$. Obviously, we have

$$ {\Pr}(D(k,t) ) \leq\sum\limits_{e_{ij} \in \Pi^{\prime}_k} (\Pr(D_{ij}^h (k,t)) + \Pr(D_{ij}^v (k,t)). $$

We firstly give the bound of $\Pr(D(k,t) )$. Based on the cell $\bar{c}^{*}_{t},$ we construct the region ${\bar{\mathcal{S}}}^{h}_{ij}(k,t)$ of area

$$ \bar{S}^h_{ij}(k,t)=(2\mu K+1)\cdot \sqrt{a_s}\cdot(2 | {{\mathcal{HP}}_{ij} } | + (K+2)\sqrt {a_s}), $$

as in Fig. 4. Similarly, we construct the region ${\bar{\mathcal{S}}}^{v}_{ij}(k,t)$ of area

$$ \bar{S}^v_{ij}(k,t)=(2\mu K+1)\cdot \sqrt{a_s}\cdot(2 | {{\mathcal{VP}}_{ij} } | + (K+2)\sqrt {a_s}). $$

The following two propositions are obviously true.

Fig. 4

The construction of the region $\bar{{\mathcal{S}}}^h_{ij}(k,t)$. Here, $l_0=\mu K\sqrt{a_s}$ because the maximum size of any clusters is μ. Obviously, $l=(2\mu K+1)\cdot \sqrt{a_s}$

Proposition 3

The Poisson pointv _i locates in the region${\bar{\mathcal{S}}}^{h}_{ij}(k,t)$if the eventD ^h _ij (k, t) happens.

Proposition 4

The Poisson pointv _j locates in the region${\bar{\mathcal{S}}}^{v}_{ij}(k,t)$if the eventD ^v _ij (k, t) happens.

Then, we get that $Z_t=\sum\nolimits_{k=1}^{m_s}{\mathcal{I}}(D(K,t))$ follows a Poisson distribution of mean $\bar{\lambda}\leq m_s\cdot \bar{S},$ where

$$ \bar{S}=\min\left\{1, \sum\limits_{e_{ij} \in \Pi^{\prime}_k}{(\bar{S}^h_{ij}(k,t)+\bar{S}^v_{ij}(k,t))} \right\}. $$

By Lemma 5, we have

$$\begin{aligned} \bar{S} \le & 2(2\mu K + 1)\cdot\sqrt {a_{s} } \cdot\sum\limits_{{e_{{ij}} \in \Pi _{k}^{\prime } }} {\left( {\sqrt 2 |v_{i} v_{j} | + (K + 2)\sqrt {a_{s} } } \right)} \\ \le & 2(2\mu K + 1)\cdot(4\sqrt {m_{d} a_{s} } + (K + 2)m_{d} a_{s} ) = \hat{S}. \\ \end{aligned}$$

When $m_s \hat{S}=\omega(\log n),$ by union bounds and Lemma 4, we can obtain

$$ \Pr(Z \ge 2m_s \hat{S} ) \leq (m / \log m)(4/e)^{ - m_s \hat{S} } = o(1/\log m) \to 0. $$

When $m_s \hat{S}=O(\log m)$ and $\lim\nolimits_{n\to \infty }{\frac{m_s \hat{S}} {\log m}}=\bar{f}\geq 0$. Let $\bar{z}=e^2\max\{\bar{f},1\},$we obtain that

$$ \Pr(Z \ge \bar{z}\log n ) \leq (e^{{\frac{1} {e^2}}-1})^{\bar{z}\log n} \to 0. $$

Summing up these two cases, we complete the proof.