Distributed Change Detection via Average Consensus over Networks

Abstract

Distributed change-point detection has been a fundamental problem when performing real-time monitoring using sensor networks.   We propose a distributed detection algorithm, where each sensor only exchanges CUSUM statistic with their neighbors based on the average consensus scheme, and an alarm is raised when local consensus statistic exceeds a prespecified global threshold. We provide theoretical performance bounds showing that the performance of the fully distributed scheme can match the centralized algorithms under some mild conditions. Numerical experiments demonstrate the good performance of the algorithm, especially, in detecting asynchronous changes.

Appendix

For simplicity, we first indeed the sensors from 1 to N. Use vector $\mathbf{L}^t$ to represent $\left( L(\mathbf{x}_1^{t}),\ldots ,L(\mathbf{x}_N^{t})\right) ^\mathrm{T}$, vector $\mathbf{y}^t$ to represent $\left( y_1^t,\ldots ,y_N^t\right) ^\mathrm{T}$ and vector $\mathbf{z}^t$ to represent $\left( z_1^t,\ldots ,z_N^t\right) ^\mathrm{T}$. Now, our algorithm can be rewritten as

$$\begin{aligned}&\mathbf{y}^{t+1} = ( \mathbf{y}^t + \mathbf{L}^{t+1})^+, \mathbf{z}^{t+1} = \mathbf{W}(\mathbf{z}^t + \mathbf{y}^{t+1}-\mathbf{y}^t),&T_s = \inf \big \{t>0: \Vert \mathbf{z}^t \Vert _{\infty } \ge b\big \}. \end{aligned}$$

(3)

First, we prove some useful lemmas before reaching the main results.

Remark 1

Since $\mathbf{W}{} \mathbf{1} = \mathbf{1}$, $\mathbf{W}^\mathrm{T}=\mathbf{W}$ and $z^0_v = y^0_v=0$, simple proof by m.i. can verify

$$\begin{aligned} \sum _{v\in \mathcal {V}} z_v^t = \sum _{v\in \mathcal {V}} y_v^t \text{ holds } \text{ for } \text{ all } t, \end{aligned}$$

(4)

Lemma 1

(Hoeffding Inequality) Let $X_i$ be independent, mean zero, $\sigma _i^2$-sub-Gaussian random variables. Then for $K>0$, $\mathbb {P}(\sum _{i=1}^{n} X_{n}\ge K)\le \exp \left( -\frac{K^2}{2\sum _{i=1}^{n}\sigma _i^2}\right) .$

Lemma 2

Consider a sequence of random variables $X_k {\mathop {\sim }\limits ^{i.i.d.}}\mathcal {P}$, for $k=1,2,\ldots ,t$. $\mathcal {P}$ is a sub-Gaussian distribution and its mean and variance are defined as $\mu _1<0$ and $\sigma _1$, respectively. Given $K>0$ large enough, we have

$$\begin{aligned} \sum _{k=1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right) < -\frac{2K}{\mu _1} \exp \left( \frac{2K\mu _1}{\sigma _1^{2}}\right) . \end{aligned}$$

Proof

Case 1. For $0 < t \le [-\frac{2K}{\mu _1}]$, by Hoeffding Inequality, we have

$$\begin{aligned} \sum _{k=1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right) < \sum _{k=1}^{t} \mathrm{exp}\left( -\frac{1}{2}(\frac{K-k\mu _1}{\sqrt{k}\sigma _1})^{2}\right) . \end{aligned}$$

(5)

Using $\frac{K-k\mu _1}{\sqrt{k}} \ge 2\sqrt{-K\mu _1}$ and $t \le -\frac{2K}{\mu _1}$, we obtain

$$\begin{aligned} \sum _{k=1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right) < -\frac{2K}{\mu _1} \mathrm{exp}\left( \frac{2K\mu _1}{\sigma _1^{2}}\right) . \end{aligned}$$

(6)

Case 2. For $[-\frac{2K}{\mu _1}]+1 \le t$, by (6), we have

$$\begin{aligned} \sum _{k=1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q}> K\right) < -\frac{2K}{\mu _1} \mathrm{exp}\left( \frac{2K\mu _1}{\sigma _1^{2}}\right) + \sum _{k=[-\frac{2K}{\mu _1}]+1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right) . \end{aligned}$$

(7)

Utilizing Hoeffding Inequality and $k\ge [-\frac{2K}{\mu _1}]+1$, we obtain

$$\begin{aligned} \mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right) <\mathrm{exp}\left( -\frac{1}{2}(\frac{K-k\mu _1}{\sqrt{k}\sigma _1})^{2}\right) \le \mathrm{exp}\left( \frac{9K\mu _1}{4\sigma _1^2}\right) . \end{aligned}$$

(8)

Besides, for $k \ge [-\frac{2K}{\mu _1}]+1$, we have

$$\begin{aligned} \frac{ \mathrm{exp}\left( -\frac{1}{2}(\frac{K-(k+1)\mu _1}{\sqrt{k+1}\sigma _1})^{2}\right) }{ \mathrm{exp}\left( -\frac{1}{2}(\frac{K-k\mu _1}{\sqrt{k}\sigma _1})^{2}\right) } = \mathrm{exp}\left( -\frac{\mu _1^2}{2\sigma _1^2}+\frac{K^2}{2k(k+1)\sigma _1^2}\right) < \mathrm{exp}\left( -\frac{3\mu _1^2}{8\sigma _1^2}\right) . \end{aligned}$$

(9)

Then, from Hoeffding Inequality, (8) and (9), we derive

$$\begin{aligned}&\sum _{k=[-\frac{2K}{\mu _1}]+1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right)< \sum _{k=[-\frac{2K}{\mu _1}]+1}^{t}\mathrm{exp}\left( -\frac{1}{2}(\frac{K-k\mu _1}{\sqrt{k}\sigma _1})^{2}\right) \\<&\sum _{k=[-\frac{2K}{\mu _1}]+1}^{t} \mathrm{exp}\left( \frac{9K\mu _1}{4\sigma _1^2}\right) \times \mathrm{exp}\left( -\frac{3\mu _1^2}{8\sigma _1^2} \left( k-[-\frac{2K}{\mu _1}]-1\right) \right) <\frac{ \mathrm{exp}\left( \frac{9K\mu _1}{4\sigma _1^2}\right) }{1-\mathrm{exp}\left( -\frac{3\mu _1^2}{8\sigma _1^2}\right) }.\nonumber \end{aligned}$$

(10)

From (10), we know that the second term on the RHS of (7) is a small quantity compared with the first term provided K large enough, so we can neglect it to obtain

$$\begin{aligned} \sum _{k=1}^{t}\mathbb {P}\left( \sum _{q=1}^{k} X_{q} > K\right) < -\frac{2K}{\mu _1} \mathrm{exp}\big (\frac{2K\mu _1}{\sigma _1^{2}}\big ). \end{aligned}$$

(11)

Note that $\mathbb E_{f_1} [L(x_j^t)] = \mu _1 < 0$, $\mathbb E_{f_2} [L(x_j^t)] = \mu _2 > 0$, $\text{ Var }_{f_1} [L(x_j^t)] = \sigma _1^2$, $\text{ Var }_{f_2} [L(x_j^t)] = \sigma _2^2$.

Given $\varepsilon >0$ and $p>0$, Define event

$$B(\varepsilon ,p)= \{ |L(\mathbf{x}_i^t)|< \varepsilon b, \text{ for } i=1,\ldots ,N \text{ and } t=1,\ldots ,p \},$$

where b is the prespecified threshold in detection. Besides, we use $\{T_s = p\}$ to represent the event that our algorithm detects the change at $t=p$. We have the following lemma

Lemma 3

For any $t \le p$, we have

$$\begin{aligned} \{T_s = t\}\wedge B(\varepsilon ,p) \subset \big \{ \frac{\sum _{j=1}^{N} y_{j}^{t}}{N}> (1- \frac{ \sqrt{N}\varepsilon \lambda _{2} }{1- \lambda _{2}})b \big \} \wedge B(\varepsilon ,p). \end{aligned}$$

Proof

Note that, by eigen-decomposition, $\mathbf{W} = \frac{1}{N} \mathbf{1} \mathbf{1}^\intercal + \sum _{j=2}^N \lambda _j w_j w_j^\intercal .$ Throughout the proof, we assume under the condition that $B(\varepsilon ,p)$ occurs. First, by the recursive form of our algorithm in (3), the result in (4) and the definition of $B(\varepsilon ,p)$, for any sensor j, we have

$$\begin{aligned}&|z_{j}^{t} - \frac{\sum _{i=1}^{N} y_{i}^{t}}{N} | = | z_{j}^{t} - \frac{\sum _{i=1}^{N} z_{i}^{t}}{N} | \le \Vert \mathbf{z}^t - \frac{\sum _{i=1}^{N} z_{i}^{t}}{N} \mathbf{1}\Vert _2 = \Vert \sum _{k=1}^{t} (\mathbf{W}^{t-k+1}-\frac{1}{N}{} \mathbf{1}{} \mathbf{1}^\mathrm{T})(\mathbf{y}^{k}-\mathbf{y}^{k-1}) \Vert _{2}\\ \le&\sum _{k=1}^{t} \lambda _{2}^{t-k+1} \Vert \mathbf{y}^{k}-\mathbf{y}^{k-1} \Vert _{2} \le \sum _{k=1}^{t} \lambda _{2}^{t-k+1} \Vert \mathbf{L}^{k} \Vert _{2} \le \sum _{k=1}^{t} \lambda _{2}^{t-k+1} \sqrt{N}\varepsilon b \le \frac{ \sqrt{N}\varepsilon \lambda _{2}b}{1- \lambda _{2}}, \end{aligned}$$

where $\lambda _2$ is the second largest eigenvalue modulus of $\mathbf{W}$. If $\{T_s = t\}$ happens, then $z_{j}^{t}>b$ holds for some j, which, together with the inequality above, leads to $\frac{\sum _{j=1}^{N} y_{j}^{t}}{N}> (1- \frac{ \sqrt{N}\varepsilon \lambda _{2}}{1- \lambda _{2}})b.$

Lemma 4

Assume a sequence of independent random variables $Y_1,\ldots ,Y_N$. Take any integer $M>N$ and let

$$\begin{aligned} C(M,N) {=}\big \{(i_1,\ldots ,i_N): i_j \in \mathbb {N} \ \text{ and }\ M-N \le \sum _{j=1}^{N}i_{j} \le M \big \}. \end{aligned}$$

Then we have

$$\begin{aligned} \mathbb {P}\left( \sum _{j=1}^{N} Y_j> K \right) \le \sum _{C(M,N)} \prod _{j=1}^{N}\mathbb {P}\left( Y_j > \frac{i_j K}{M} \right) . \end{aligned}$$

Proof

$\forall (y_1,\ldots ,y_N) \in \{(Y_1,\ldots ,Y_N): \sum _{j=1}^{N} Y_{j}> K\}$, take $i_j = \left[ \frac{y_j M}{ \sum _{j=1}^{N} y_{j}}\right]$, for $j=1,\ldots ,N$, where $\left[ x\right]$ denotes the largest integer smaller than x. We can verify that $M-N \le \sum _{j=1}^{N} i_j \le M$ and $y_{j} > i_j K/M$. To see $M-N \le \sum _{j=1}^{N} i_j$, notice that $i_j\ge \frac{y_jM}{\sum _{j=1}^N y_j}-1$.Therefore, we have

$$\begin{aligned} \big \{(Y_1,\ldots ,Y_N):\ \sum _{j=1}^{N} Y_{j}> K\big \} \subset \bigcup _{ C(M,N)} \left\{ (Y_1,\ldots ,Y_N): \ Y_{j} > \frac{i_j K}{M} \text{ for } j=1,\ldots ,N \right\} . \end{aligned}$$

Since $Y_{j}$’s are independent with each other, we obtain

$$\begin{aligned} \mathbb {P}\left( \sum _{j=1}^{N} Y_{j}> K \right) \le \sum _{C(M,N)} \prod _{j=1}^{N}\mathbb {P}\left( Y_{j} > \frac{i_j K}{M} \right) . \end{aligned}$$

1.1 5.1 Proof of Theorem 1

First, we calculate the probability that our algorithm stops within time p. The value of p is to be specified later.

$$\begin{aligned} \nonumber&\sum _{t=1}^{p} \mathbb {P}(T_s=t) = \sum _{t=1}^{p} \left( \mathbb {P}\left( \{T_s=t\} \wedge B(\varepsilon ,p)\right) + \mathbb {P}\left( \{T_s=t\} \wedge \bar{B}(\varepsilon ,p)\right) \right) \\ \nonumber \le&\sum _{t=1}^{p} \mathbb {P}\left( \{T_s=t\}\wedge B(\varepsilon ,p)\right) + \mathbb {P}\left( \bar{B}(\varepsilon ,p)\right) \\ \le&\sum _{t=1}^{p} \mathbb {P}\left( \{T_s=t\} \wedge B(\varepsilon ,p) \right) + 2Np\times \mathrm{exp}\left( -\frac{(\varepsilon b - \mu _1)^2}{2\sigma _1^2}\right) , \end{aligned}$$

where the last inequality is from Hoeffding Inequality and assumptions in Sect. 3. The value of $\varepsilon$ is to be specified later.

Denote $\bar{b} = N(1- \frac{ \sqrt{N}\varepsilon \lambda _{2}}{1- \lambda _{2}})b$, then $\bar{b}$ will also tend to infinity as b tends to infinity provided $\varepsilon$ small enough. By Lemma 3, we have

$$\begin{aligned} \sum _{t=1}^{p} \mathbb {P}(T_s=t) \le&\sum _{t=1}^{p} \mathbb {P}\left( \left\{ \sum _{j=1}^{N} y_{j}^{t}> \bar{b}\right\} \wedge B(\varepsilon ,p)\ \right) + 2Np \times \mathrm{exp}\left( -\frac{(\varepsilon b - \mu _1)^2}{2\sigma _1^2}\right) . \end{aligned}$$

(12)

By Lemma 4, we have

$$\begin{aligned} \mathbb {P}\left( \left\{ \sum _{j=1}^{N} y_{j}^{t}> \bar{b}\right\} \wedge B(\varepsilon ,p)\ \right) \le \sum _{C(M,N)} \prod _{j=1}^{N}\mathbb {P}\left( \left\{ y_{j}^{t} > \frac{i_j \bar{b}}{M}\right\} \wedge B_j(\varepsilon ,p) \right) , \end{aligned}$$

(13)

where $B_j(\varepsilon , p) = \{ |L(\mathbf{x}_j^t)|< \varepsilon b, \text{ for } t=1,\ldots ,p \}$ and the value of M is to be specified later. If $y_{j}^{t}>i_j \bar{b}/M$, then there must exist $1\le k\le t$ such that $y_{j}^{t} = \sum _{q=k}^{t} L(\mathbf{x}_j^q) \ge i_j \bar{b}/M$. So, we have

$$\begin{aligned} \mathbb {P}\left( \left\{ y_{j}^{t}> \frac{i_j \bar{b}}{M}\right\} \wedge B_j(\varepsilon ,p) \right) \le \sum _{k=1}^{t} \mathbb {P}\left( \left\{ \sum _{q=k}^{t} L(\mathbf{x}_j^q)> \frac{i_j \bar{b}}{M}\right\} \wedge B_j(\varepsilon ,p) \right) . \end{aligned}$$

(14)

The influence of $B_j(\varepsilon ,p)$ in (14) can be interpreted as truncating the original distribution of $L(\cdot )$. It’s obvious that the new distribution is still sub-Gaussian. Besides, the mean and variance almost keep unchanged provided $\varepsilon b$ large enough.

If $i_j=0$, we just set the upper bound of the probability in (14) to be 1. If $i_j\ne 0$, by Lemma 2, we have

$$\begin{aligned} \sum _{k=1}^{t} \mathbb {P}\left( \left\{ \sum _{q=k}^{t} L(\mathbf{x}_j^q) > \frac{i_j \bar{b}}{M}\right\} \wedge B_j(\varepsilon ,p) \right) < -\frac{2i_j \bar{b}}{M\mu _1} \mathrm{exp}\left( \frac{2i_j \bar{b}\mu _1}{M\sigma _1^{2}}\right) . \end{aligned}$$

(15)

Plugging (15) into (13), we get

$$\begin{aligned}&\sum _{C(M,N)} \prod _{j=1}^{N}\mathbb {P}\left( \left\{ y_{j}^{t} > \frac{i_j \bar{b}}{M}\right\} \wedge B_j(\varepsilon ,p) \right) < \sum _{C(M,N)} \prod _{i_j\ne 0} \frac{2i_j \bar{b}}{-M\mu _1} \mathrm{exp}\left( \frac{2i_j \bar{b}\mu _1}{M\sigma _1^{2}}\right) \\ \nonumber \le&\sum _{C(M,N)} \left( \frac{2\bar{b}}{-\mu _1}\right) ^N \mathrm{exp}\left( \frac{2\bar{b}\mu _1}{\sigma _1^{2}}(1-\frac{N}{M})\right) = |C(M,N)| \left( \frac{2\bar{b}}{-\mu _1}\right) ^N \mathrm{exp}\left( \frac{2\bar{b}\mu _1}{\sigma _1^{2}}(1-\frac{N}{M})\right) . \end{aligned}$$

(16)

Plugging (16) and (13) into (12), we obtain

$$\begin{aligned} \sum _{t=1}^{p} \mathbb {P}(T_s=t) < \sum _{t=1}^{p} |C(M,N)| \left( \frac{2\bar{b}}{-\mu _1}\right) ^N \mathrm{exp}\left( \frac{2\bar{b}\mu _1}{\sigma _1^{2}}(1-\frac{N}{M})\right) + 2Np \times \mathrm{exp}\left( -\frac{(\varepsilon b - \mu _1)^2}{2\sigma _1^2}\right) \end{aligned}$$

(17)

$$\begin{aligned} = p\left| C(M,N)\right| \left( \frac{2\bar{b}}{-\mu _1}\right) ^N \mathrm{exp}\left( \frac{2\bar{b}\mu _1}{\sigma _1^{2}}(1-\frac{N}{M})\right) + 2Np\times \mathrm{exp}\left( -\frac{(\varepsilon b - \mu _1)^2}{2\sigma _1^2}\right) . \end{aligned}$$

(18)

Next, we will show that as b tends to infinity, the second term on the RHS of (17) is a small quantity in comparison with the first term if we choose the value of M and $\varepsilon$ properly. Note that 2Np is a small quantity in comparison with $p\left| C(M,N)\right| \left( -2\bar{b}/\mu _1\right) ^N$, so we only require $\frac{2\bar{b}\mu _1}{\sigma _1^{2}}(1-\frac{N}{M}) \ge -\frac{(\varepsilon b - \mu _1)^2}{2\sigma _1^2}.$ Choose $M=(N+1)^2$. Recall that $\bar{b} = N(1- \frac{ \sqrt{N}\varepsilon \lambda _{2}}{1- \lambda _{2}})b$, the equation above can be rewritten as

$$\begin{aligned} \frac{(\varepsilon b - \mu _1)^2}{2\sigma _1^2} \ge -\frac{2(N^3+N^2+N)\mu _1 b}{(N+1)^2\sigma _1^2}\left( 1- \frac{ \sqrt{N}\varepsilon \lambda _{2}}{1- \lambda _{2}}\right) . \end{aligned}$$

(19)

To ensure that (19) holds as b tends to infinity, $\varepsilon = 2\sqrt{-N\mu _1/b}$ is sufficient. Plugging the value of M and $\varepsilon$ into (17) and neglecting the second term, we get

$$\begin{aligned}&\mathbb {P}\left( T_s \le p \right) \le |C((N+1)^2,N)| \left( \frac{2\bar{b}}{-\mu _1}\right) ^N \\&\cdot \mathrm{exp}\big (\frac{2(N^3+N^2+N)\mu _1 b}{(N+1)^2\sigma _1^{2}}-\frac{4N(N^3+N^2+N) \sqrt{-\mu _1}\mu _1\lambda _{2}\sqrt{b}}{(N+1)^2(1-\lambda _{2})\sigma _1^2}+\ln (p)\big ). \end{aligned}$$

So $\forall l > -\frac{4N(N^3+N^2+N) \sqrt{-\mu _1}\mu _1\lambda _{2}}{(N+1)^2(1-\lambda _{2})\sigma _1^2}$, if we choose $p = \mathrm{exp}\left( -\frac{2(N^3+N^2+N)\mu _1 b}{(N+1)^2\sigma _1^{2}} - l\sqrt{b}\right) ,$

$$\begin{aligned}&\lim _{b\rightarrow +\infty } \mathbb {P}\left( T_s \le p \right) \le \\&\lim _{b\rightarrow +\infty } |C((N+1)^2,N)| \left( \frac{2\bar{b}}{-\mu _1}\right) ^N \mathrm{exp}\left( -\left( l+\frac{4N(N^3+N^2+N) \sqrt{-\mu _1}\mu _1\lambda _{2}}{(N+1)^2(1-\lambda _{2})\sigma _1^2}\right) \sqrt{b}\right) =0, \end{aligned}$$

which together with the definition of $\mathrm{ARL}$ leads to $\mathrm{ARL} \ge p$, $\forall l > -\frac{4N(N^3+N^2+N) \sqrt{-\mu _1}\mu _1\lambda _{2}}{(N+1)^2(1-\lambda _{2})\sigma _1^2}$. This leads to our desired result when b tends to infinity.

1.2 5.2 Proof of Lemma 1

First of all, note that $\mathbb {P}\left( T_s =+\infty \right) =0$, so given $\varepsilon >0$, we have

$$\begin{aligned} \mathrm{EDD} \le \frac{b(1+\varepsilon )}{\mu _2} + \sum _{t=[\frac{b(1+\varepsilon )}{\mu _2}]+1}^{+\infty }\mathbb {P}\left( T_s=t\right) t. \end{aligned}$$

(20)

If $T_s=t$, then we have that $z_j^{t-1}<b$ holds for all j. Since $\sum _{j=1}^{N} z_{j}^{t-1} = \sum _{j=1}^{N} y_{j}^{t-1}$, there must exist some $y_j^{t-1}<b$. Therefore, we have

$$\begin{aligned} \sum _{t=[\frac{b(1+\varepsilon )}{\mu _2}]+1}^{+\infty }\mathbb {P}\left( T_s=t\right) t \le \sum _{t=[\frac{b(1+\varepsilon )}{\mu _2}]+1}^{+\infty } \sum _{j=1}^{N} \mathbb {P}\left( y_j^{t-1}<b \right) t. \end{aligned}$$

(21)

Note that $y_j^{t-1}\ge \sum _{q=1}^{t-1}L(\mathbf{x}_j^q)$, together with Hoeffding Inequality, we get

$$\begin{aligned} \mathbb {P}\left( y_j^{t-1}<b \right) t \le&\mathbb {P}\left( \sum _{q=1}^{t-1}L(\mathbf{x}_j^q)<b \right) = \mathrm{exp}\left( -\frac{1}{2}\left( \frac{b-(t-1)\mu _2}{\sqrt{t-1}\sigma _2}\right) ^{2}\right) t. \end{aligned}$$

(22)

When b is large enough, for any $t>[\frac{b(1+\varepsilon )}{\mu _2}]$, utilizing the similar technique in (9), we get

$$\begin{aligned} \frac{\mathrm{exp}\left( -\frac{1}{2}(\frac{b-t\mu _2}{\sqrt{t}\sigma _2})^{2}\right) }{ \mathrm{exp}\left( -\frac{1}{2}(\frac{b-(t-1)\mu _2}{\sqrt{t-1}\sigma _2})^{2}\right) } \times \frac{t+1}{t} \le \mathrm{exp}\left( -\frac{b}{2}(1-\frac{1}{(1+\varepsilon )^2})\right) . \end{aligned}$$

(23)

Plugging (22) and (23) into (21), utilizing the similar technique in (10), we get

$$\begin{aligned}&\sum _{t=[\frac{b(1+\varepsilon )}{\mu _2}]+1}^{+\infty }\mathbb {P}\left( T_s=t\right) t \le \sum _{t=[\frac{b(1+\varepsilon )}{\mu _2}]+1}^{+\infty } N\times \mathrm{exp}\left( -\frac{1}{2}\left( \frac{b-(t-1)\mu _2}{\sqrt{t-1}\sigma _2}\right) ^{2}\right) t\end{aligned}$$

(24)

$$\begin{aligned} \le&N \left( \frac{b(1+\varepsilon )}{\mu _2}+1\right) \frac{\mathrm{exp}\left( -\frac{\varepsilon ^2 b}{2(1+\varepsilon )\sigma _2^2}\right) }{1-\mathrm{exp}\left( -\frac{b}{2}(1-\frac{1}{(1+\varepsilon )^2})\right) }. \end{aligned}$$

(25)

Note that $\forall \varepsilon >0$, as b tends to infinity, the RHS of (24) would converge to zero. Therefore, by (20), we get $\mathrm{EDD}\le \frac{b\big (1+o(1)\big )}{\mu _2}.$