Running-time analysis of evolutionary programming based on Lebesgue measure of searching space

Abstract

There have been many studies on the runtime analysis of evolutionary algorithms in discrete optimization, and however, relatively few homologous results have been obtained on continuous optimization, such as evolutionary programming (EP). This paper presents an analysis of the running time (as approximated by the mean first hitting time) of two EP algorithms based on Gaussian and Cauchy mutations, using an absorbing Markov process model. Given a constant variation, we analyze the running-time upper bound of special Gaussian mutation EP and Cauchy mutation EP, respectively. Our analysis shows that the upper bounds are impacted by individual number, problem dimension number, searching range, and the Lebesgue measure of the optimal neighborhood. Furthermore, we provide conditions whereby the mean running time of the considered EPs can be no more than a polynomial of n. The condition is that the Lebesgue measure of the optimal neighborhood is larger than a combinatorial computation of an exponential and the given polynomial of n. In the end, we present a case study on sphere function, and the experiment validates the theoretical result in the case study.

Appendix: Proof of the Lemmas, Theorems and Corollaries

Proof of Lemma 1

Recall that $\xi _{t}^{{\text{EP}}}=(\mathbf {v}_{1}^{(t)},\mathbf {v}_{2}^{(t)},\ldots ,\mathbf {v}_{k}^{(t)})$ is a real vector and the status space $\varOmega _{{\text{EP}}}$ of EP is continuous. Furthermore, $\xi _{t}^{{\text{EP}}}$ is only dependent on $\xi _{t-1}^{{\text{EP}}}$ for $t=0,1,\ldots$, as evidenced by steps 3–6 of the EP process. Hence, $\{\xi _{t}^{{\text{EP}}}\}_{t=0}^{+\infty }$ is a Markov process.

Proof of Lemma 2

$\xi _{t}^{{\text{EP}}}\in \varOmega _{{\text{EP}}}^{*}$ when $\exists (\mathbf {x}_{i}^{*(t)},\mathbf {\sigma }_{i}^{(t)})\in \xi _{t}^{{\text{EP}}}$ where $\mathbf {x}_{i}^{*(t)}$ is an optimal solution. Given steps 5 and 6 of EP, $\mathbf {x}_{i}^{*(t)}$ will have the most “wins” as well as the highest fitness, so $\mathbf {x}_{i}^{*(t)}$ will be selected for the next iteration with probability one if there are less than k optimal solutions in the tournament. $\mathbf {x}_{i}^{*(t)}$ might be lost, but other optimal solutions will be selected if there are more than k optimal solutions in the parent and offspring. Thus, $P\{\xi _{t+1}^{{\text{EP}}}\notin \varOmega _{{\text{EP}}}^{*}|\xi _{t}^{{\text{EP}}} \in \varOmega _{{\text{EP}}}^{*}\}=0$, and $\{\xi _{t}^{{\text{EP}}}\}_{t=0}^{+\infty }$ is an absorbing Markov process to $\varOmega _{{\text{EP}}}^{*}$.

Proof of Theorem 1

Observe that $\lambda _{t}^{{\text{EP}}}-\lambda _{t-1}^{{\text{EP}}}=P\{\mu _{{\text{EP}}}\le t\}-P\{\mu _{{\text{EP}}}\le t-1\}=P\{\mu _{{\text{EP}}}=t\}$. Therefore, we have

$$\begin{aligned} {E\mu _{{\text{EP}}}} & {\,=\,\sum \limits _{t=1}^{+\infty }t\cdot (\lambda _{t}^{{\text{EP}}}-\lambda _{t-1}^{{\text{EP}}}) =\lim \limits _{N\rightarrow +\infty }\sum \limits _{t=1}^{N}t\cdot (\lambda _{t}^{{\text{EP}}}-\lambda _{t-1}^{{\text{EP}}})}\\&{\,=\,\lim \limits _{N\rightarrow +\infty }\sum \limits _{i=1}^{N}(\lambda _{N}^{{\text{EP}}}-\lambda _{i-1}^{{\text{EP}}}) {\,=\,\sum \limits _{i=1}^{+\infty }(1-\lambda _{i-1}^{{\text{EP}}})}}\\&{=\sum \limits _{i=0}^{+\infty }(1-\lambda _{i}^{{\text{EP}}})} \end{aligned}$$

Proof of Corollary 1

Let $p_{i}=P\{\xi _{i}^{{\text{EP}}}\in \varOmega _{{\text{EP}}}^{*}|\xi _{i-1}^{{\text{EP}}}\notin \varOmega _{{\text{EP}}}^{*}\}$. Based on the total probability equation, $\lambda _{t}^{{\text{EP}}}=(1-\lambda _{t-1}^{{\text{EP}}})p_{t}+\lambda _{t-1}^{{\text{EP}}}(1-p_{t})$. Substituting it into Theorem 1, we have $E\mu _{{\text{EP}}}=\sum \nolimits _{t=0}^{+\infty }[(1-\lambda _{0}^{{\text{EP}}})\prod _{i=1}^{t}(1-p_{i})]$.

Proof of Corollary 2

According to Corollary 1, $E\mu _{{\text{EP}}}\ge \sum _{t=0}^{+\infty }[(1-\lambda _{0}^{{\text{EP}}}) \prod _{i=1}^{t}(1-\beta _{i})]$. Similarly, we have $E\mu _{{\text{EP}}}\le \sum _{t=0}^{+\infty }[(1-\lambda _{0}^{{\text{EP}}}) \prod _{i=1}^{t}(1-\alpha _{i})]$. Thus, $\beta ^{-1}(1-\lambda _{0}^{{\text{EP}}})\le E\mu _{{\text{EP}}} \le \alpha ^{-1}(1-\lambda _{0}^{{\text{EP}}})$ when $\alpha _{t}=\alpha$ and $\beta _{t}=\beta$.

Proof of Theorem 2

For fixed individual $(\mathbf {x},{\bar{\varvec{\sigma }}})$, ${\bar{\mathbf{x}}}=\mathbf {x}+\mathbf {u}\,\bigotimes\,{\bar{\varvec{\sigma }}}$, where $\mathbf {\bar{\sigma }}=(\bar{\sigma }_1,\bar{\sigma }_2,\ldots ,\bar{\sigma }_n)$ is a n-dimension variation variable and the stochastic disturbance $\mathbf {u}=(u_1,u_2,\ldots ,u_n)$ follows n-dimensional standard Gaussian distribution. Let $\mathbf {z}=(z_1,z_2,\ldots ,z_n)$, that is, $\mathbf {z}=\mathbf {u}\,\bigotimes\,{\bar{\varvec{\sigma }}}=(u_1\bar{\sigma }_1,u_2\bar{\sigma }_2,\ldots ,u_n\bar{\sigma }_n)$. According to Expressions (1) and (2), $z_{j}=\mu _{j}\bar{\sigma }_{j}=\mu _{j}\sigma$. Then, we have $z_{j}\sim N(0,\bar{\sigma }_{j}^{2})$ since $u_{j}\sim N(0,1)$, where $j=1,\ldots ,n$. Therefore, ${\bar{\mathbf{x}}}=\mathbf {x}+\mathbf {u}\,\bigotimes\,{\bar{\varvec{\sigma }}}=\mathbf {x}+\mathbf {z}.$

Given $\mathbf {S}=\prod \nolimits _{j=1}^{n}[-b_{j},b_{j}], \mathbf {S}^{*}(\varepsilon )\subseteq \mathbf {S}.$ Let $\tilde{\mathbf {S}}=\{\mathbf {z}|\mathbf {z}={\bar{\mathbf{x}}}-\mathbf {x},{\bar{\mathbf{x}}}\in \mathbf {S}^{*}(\varepsilon ),\mathbf {x}\in \mathbf {S}\}$. According to the property of measure, $m\big (\mathbf {S}^{*}(\varepsilon )\big )=m(\tilde{\mathbf {S}})$. $P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}(\varepsilon )\}=P\{\mathbf {z}\in \tilde{\mathbf {S}}\}=\int \cdots \int _{\tilde{\mathbf {S}}}\prod \nolimits _{j=1}^{n}\frac{1}{\sqrt{2\pi \bar{\sigma }_{j}}}\exp \{-\frac{z_{j}^{2}}{2\bar{\sigma }_{j}^{2}}\}dz_{1}\ldots dz_{n}.$

Note that ${\bar{\mathbf{x}}},\mathbf {x}\in \mathbf {S}$ and $\mathbf {z}={\bar{\mathbf{x}}}-\mathbf {x}$, so we have $|z_{j}|=|\bar{x}_{j}-x_{j}|\le 2b_{j}=2b$ ($j=1,\ldots ,n$), which leads to the following inequality:

$$\begin{aligned}&\int \cdots \int _{\tilde{\mathbf {S}}}\prod \limits _{j=1}^{n}\frac{1}{\sqrt{2\pi \bar{\sigma }_{j}}}\exp \left\{-\frac{z_{j}^{2}}{2\bar{\sigma }_{j}^{2}}\right\}dz_{1}\ldots dz_{n}\\&\ge \int \cdots \int _{\tilde{\mathbf {S}}}\prod \limits _{j=1}^{n}\frac{1}{\sqrt{2\pi \bar{\sigma }_{j}}}\exp \left\{-\frac{(2b_{j})^{2}}{2\bar{\sigma }_{j}^{2}}\right\}dz_{1}\ldots dz_{n}\\&=m(\tilde{\mathbf {S}})\prod \limits _{j=1}^{n}\frac{1}{\sqrt{2\pi }\sigma }\exp \left\{-\frac{2b_{j}^{2}}{\sigma ^{2}}\right\}\\&=m(\tilde{\mathbf {S}})\left(\frac{1}{\sqrt{2\pi }}\right)^{n} \left(\prod \limits _{j=1}^{n}\frac{1}{\sigma }\right)\exp \left\{-\sum \limits _{j=1}^{n}\frac{2b^{2}}{\sigma ^{2}}\right\} \end{aligned}$$

That is, we obtain a lower bound of $P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}\}$ as follows:

$$\begin{aligned} P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}(\varepsilon )\}\ge m(\tilde{\mathbf {S}})\left(\frac{1}{\sqrt{2\pi }}\right)^{n}\left(\prod \limits _{j=1}^{n}\frac{1}{\sigma }\right)\exp \left\{-\sum \limits _{j=1}^{n}\frac{2b^{2}}{\sigma ^{2}}\right\}. \end{aligned}$$

Noting that $m\big (\mathbf {S}^{*}(\varepsilon )\big )=m(\tilde{\mathbf {S}})$, we also have $P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}(\varepsilon )\}\ge m\big (\mathbf {S}^{*}(\varepsilon )\big )(\frac{1}{\sqrt{2\pi }})^{n} (\prod \nolimits _{j=1}^{n}\frac{1}{\sigma })\exp \{-\sum \nolimits _{j=1}^{n}\frac{2b^{2}}{\sigma ^{2}}\}$ (2) Let $f({\bar{\varvec{\sigma }}})=\log \big (\prod \nolimits _{j=1}^{n}\frac{1}{\sigma }\exp \{-\sum \nolimits _{j=1}^{n}\frac{2b^{2}}{\sigma ^{2}}\}\big ) =-\sum \nolimits _{j=1}^{n}\big (\log \sigma +a_{j}\frac{1}{\sigma ^{2}}\big )$where $a_{j}=2b^{2}$. $\max f({\bar{\varvec{\sigma }}})\Leftrightarrow \min \sum \nolimits _{j=1}^{n}\big (\log \sigma +a_{j}\frac{1}{\sigma ^{2}}\big )$, $\frac{d\big (\log \sigma +a_{j}\frac{1}{\sigma ^{2}}\big )}{d\sigma }=0\Rightarrow \frac{1}{\sigma }-2a_{j}\frac{1}{\sigma ^{3}}=0\Rightarrow \sigma =\sqrt{2a_{j}}.$

Noticing that $a_{j}=2b_{j}^{2}=2b^{2}$, we have $\sigma =2b$, which implies the lower bound of $P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}(\varepsilon )\}$ can be improved into

$$\begin{aligned}&m\big (\mathbf {S}^{*}(\varepsilon )\big )\left(\frac{1}{\sqrt{2\pi }}\right)^{n}\prod \limits _{j=1}^{n}\frac{1}{2b}\exp \left\{-\sum \limits _{j=1}^{n}\frac{2b^{2}}{(2b)^{2}}\right\}\\&=m\big (\mathbf {S}^{*}(\varepsilon )\big )\left(\frac{1}{\sqrt{2\pi }}\right)^{n} e^{-\frac{n}{2}}\prod \limits _{j=1}^{n}\frac{1}{2b}\\&=m\big (\mathbf {S}^{*}(\varepsilon )\big )\left(\frac{1}{\sqrt{2\pi }}\right)^{n} e^{-\frac{n}{2}}(2b)^{-n}\\&=m\big (\mathbf {S}^{*}(\varepsilon )\big )\left(4\sqrt{e}\pi b\right)^{-n} \end{aligned}$$

Therefore, $P\{{\bar{\mathbf{x}}}_{i}^{t}\in \mathbf {S}^{*}(\varepsilon )\}\ge m\big (\mathbf {S}^{*}(\varepsilon )\big )(4\sqrt{e}\pi b)^{-n}$.

(3) By the property of $\varOmega _{{\text{EP}}}^{*}$ and $m\big (\mathbf {S}^{*}(\varepsilon )\big )$ in Definition 4, $\forall \varepsilon >0$ and $t=1,2,\ldots$, $P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C} \notin \varOmega _{{\text{EP}}}^{*}\}=1-\prod _{i=1}^{k}\big (1-P\{{\bar{\mathbf{x}}}_{i}\in \mathbf {S}^{*}(\varepsilon )\}\big )$

Thus, $P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\notin \varOmega _{{\text{EP}}}^{*}\}\ge 1-\big (1-m(\mathbf {S}^{*})(4\sqrt{e}\pi b)^{-n}\big )^{k}\;\text {for}\;\forall \varepsilon >0$.

Proof of Corollary 3

1. Based on the total probability equation, we have

$$\begin{aligned} \lambda _{t}^{C}=(1-\lambda _{t-1}^{C})P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\notin \varOmega _{{\text{EP}}}^{*}\}\\ +\lambda _{t-1}^{C}P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\in \varOmega _{{\text{EP}}}^{*}\} \qquad \quad \end{aligned}$$

Because $\{\xi _{t}^{C}\}_{t=0}^{+\infty }$ can be considered as an absorbing Markov process following Lemma 2, $P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\in \varOmega _{{\text{EP}}}^{*}\}=1$.$\lambda _{t}^{C}=(1-\lambda _{t-1}^{C})P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\notin \varOmega _{{\text{EP}}}^{*}\}+\lambda _{t-1}^{C}$, we have $1-\lambda _{t}^{CEP}=(1-\lambda _{t-1}^{C})(1-P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\notin \varOmega _{{\text{EP}}}^{*}\})$. According to Theorem 2,

$$\begin{aligned} P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\notin \varOmega _{{\text{EP}}}^{*}\} \qquad \quad \\ \ge 1-\Big (1-m\big (\mathbf {S}^{*}(\varepsilon )\big )(\frac{1}{4\sqrt{e}\pi b})^{n}\Big )^{k}. \end{aligned}$$

For $m\big (\mathbf {S}^{*}(\varepsilon )\big )(\frac{1}{4\sqrt{e}\pi b} )^{n}>0$, $d=1-\Big (1-m\big (\mathbf {S}^{*}(\varepsilon )\big )(\frac{1}{4\sqrt{e}\pi b})^{n}\Big )^{k}>0$ and $\lim \nolimits _{t\rightarrow +\infty }(1-d)^{t}=0$. Thus, $1-\lambda _{t}^{C}\le (1-d)(1-\lambda _{t-1}^{C})=(1-\lambda _{0}^{C})(1-d)^{t}$, so $\lim \nolimits _{t\rightarrow +\infty }\lambda _{t}^{C}\ge 1-(1-\lambda _{0}^{C})\lim \nolimits _{t\rightarrow +\infty }(1-d)^{t}=1$. Since $\lambda _{t}^{C}\le 1$, $\lim \nolimits _{t\rightarrow +\infty }\lambda _{t}^{C}=1$.

2. Given $\alpha =1-\Big (1-m\big (\mathbf {S}^{*}(\varepsilon )\big )(\frac{1}{4\sqrt{e}\pi b})^{n}\Big )^{k}>0$, we know that $P\{\xi _{t}^{C}\in \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{C}\notin \varOmega _{{\text{EP}}}^{*}\}\ge \alpha$ based on Theorem 2. According to Corollary 3, $\forall \varepsilon >0$, $E\mu _{C}\le (1-\lambda _{0}^{C})\Big (1-\big (1-m(\mathbf {S}^{*}(\varepsilon ))(\frac{1}{4\sqrt{e}\pi b})^{n}\big )^{k}\Big )^{-1}$.

Proof of Theorem 3

1. For fixed individual $(\mathbf {x},{\bar{\varvec{\sigma }}})$, ${\bar{\mathbf{x}}}=\mathbf {x}+\mathbf {\delta }\,\bigotimes\, {\bar{\varvec{\sigma }}}$, where the stochastic disturbance $\mathbf {\delta }$ follows n-dimensional standard Cauchy distribution. Let $\mathbf {z}=\mathbf {\delta }\,\bigotimes\, {\bar{\varvec{\sigma }}}$, that is, $z_{j}=\sigma \delta _{j}$ ($j=1,\ldots ,n$). Then, we have ${\bar{\mathbf{x}}}=\mathbf {x}+\mathbf {z}$ and $z_{j}\sim \frac{1}{\sigma }C_{\phi =1}(\frac{y}{\sigma })$ since $\delta _{j}\sim C_{\phi =1}(y)$ ($j=1,\ldots ,n$).

In a manner similar to the Proof of Theorem 2, we derive

$$\begin{aligned}&P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}(\varepsilon )\}=P\{\mathbf {z}\in \tilde{\mathbf {S}}\}\\&=\int \cdots \int _{\tilde{\mathbf {S}}}\left(\frac{1}{\sigma \pi }\right)^{n}\frac{1}{\prod \nolimits _{j=1}^{n}\left(1+\frac{z_{j}^{2}}{\sigma ^{2}}\right)}dz_{1}\ldots dz_{n}\\&=\int \cdots \int _{\tilde{\mathbf {S}}}\left(\frac{1}{\sigma \pi }\right)^{n}\frac{1}{\prod \nolimits _{j=1}^{n}\left(1+\frac{z_{j}^{2}}{\sigma ^{2}}\right)}dz_{1}\ldots dz_{n}\\&\ge \int \cdots \int _{\tilde{\mathbf {S}}}\left(\frac{1}{\sigma \pi }\right)^{n}\frac{1}{\prod \nolimits _{j=1}^{n}\left(1+\frac{4b^{2}}{\sigma ^{2}}\right)}dz_{1}\ldots dz_{n}\\&=\left(\frac{1}{\sigma \pi }\right)^{n}\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{\left(1+\frac{4b^{2}}{\sigma ^{2}}\right)^{n}}=\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(\pi )^{n}}\left(\sigma +\frac{4b^{2}}{\sigma }\right)^{-n} \end{aligned}$$

(Note that $|z_{j}|=|\bar{x}_{j}-x_{j}|\le 2b$ ($j=1,\ldots ,n$).)

2. Letting $f({\bar{\varvec{\sigma }}})=-\log (\sigma +\frac{4b^{2}}{\sigma })$ , we have

$$\begin{aligned}&\max f({\bar{\varvec{\sigma }}})\Leftrightarrow \min \big (\sigma +\frac{4b^{2}}{\sigma }\big ),\\&\frac{d\big (\sigma +\frac{4b^{2}}{\sigma }\big )}{d\sigma }=0\Rightarrow 1-\frac{4b^{2}}{\sigma ^{2}}=0\Rightarrow \sigma =2b \end{aligned}$$

It can be improved into $P\{{\bar{\mathbf{x}}}\in \mathbf {S}^{*}\}\ge \frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(4b\pi )^{n}}$ when $\sigma =2b$.

3. Hence, $\forall \varepsilon >0$ and $t=1,2,\ldots$, if $\sigma =2b$, $P\{\xi _{t}^{F}\notin \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{F}\in \varOmega _{{\text{EP}}}^{*}\}\ge 1-\big (1-\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(4b\pi )^{n}}\big )^{k}$.

Proof of Corollary 4

1. According to Theorem 3, $P\{\xi _{t}^{F}\notin \varOmega _{{\text{EP}}}^{*}|\xi _{t-1}^{F}\in \varOmega _{{\text{EP}}}^{*}\}\ge 1-\big (1-\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(4b\pi )^{n}}\big )^{k}$. $0<\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(4b\pi )^{n}}<1$ due to the second assumption in Sect. 2.2. Let $\theta =1-\Big (1-\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(4b\pi )^{n}}\Big )^{k}$ for $t=1,2,\ldots$. Then $0<\theta <1$ and $\lim \nolimits _{t\rightarrow +\infty }(1-\theta )^{t}=0$.

Thus, $\lim \nolimits _{t\rightarrow +\infty }\lambda _{t}^{F}\ge 1- (1-\lambda _{0}^{F})\lim \nolimits _{t\rightarrow +\infty }(1-\theta )^{t}=1$. $\lim \nolimits _{t\rightarrow +\infty }\lambda _{t}^{F}=1$ since $\lambda _{t}^{F}\le 1$.

2. Since $\lim \nolimits _{t\rightarrow +\infty }\lambda _{t}^{F}=1$, following Corollary 2 and Theorem 4, we find that

$$\begin{aligned} E\mu _{F}\le \frac{(1-\lambda _{0}^{F})}{\Big (1-\big (1-\frac{m\big (\mathbf {S}^{*}(\varepsilon )\big )}{(4b\pi )^{n}}\big )^{k}\Big )}, \quad \forall \varepsilon >0. \end{aligned}$$

Proof of Theorem 4

1. According to Eq. (9), for the only individual ${{{\varvec{x}}}}$ per iteration, $\bar{{{\varvec{x}}}} = {{\varvec{x}}} + {{\varvec{u}}}$ , where the stochastic disturbance ${{\varvec{u}}}$ follows n-dimensional standard Gauss distribution. Then, we have ${{\varvec{u}}}=\bar{{{\varvec{x}}}} - {{\varvec{x}}}$ and ${u_j}\sim N(0,1)$ ($j=1, 2, \ldots , n$). Following the Proof of Theorem 2,

$$\begin{aligned}&P\{ \bar{x} \in {{\mathbf{S}}^{\mathbf{*}}}(\varepsilon )\} = P\{ {\mathbf{u}} \in {{\tilde{\mathbf{S}}}}\}\\&= \int \cdots \int _{\tilde{S}} {\prod \limits _{j = 1}^n {\frac{1}{{\sqrt{2\pi } }}} \exp \left\{ - \frac{{u_j^2}}{2}\right\} d} {u_1} \ldots d{u_n}\\&\ge \int \cdots \int _{\tilde{S}} {\prod \limits _{j = 1}^n {\frac{1}{{\sqrt{2\pi } }}} \exp \left\{ - \frac{1}{2}\right\} d} {u_1} \ldots d{u_n}\\&= \int \cdots \int _{\tilde{S}} {{{\left(\frac{1}{{\sqrt{2\pi e} }}\right)}^n}d} {u_1} \ldots d{u_n}\\&= m({{\mathbf{S}}^{\mathbf{*}}}(\varepsilon )){\left(\frac{1}{{\sqrt{2\pi e} }}\right)^n}\\&= {\left(\frac{\varepsilon }{{\sqrt{2\pi e} }}\right)^n} \end{aligned}$$

Based on Corollary 3, $E{\mu _1} \le {(\frac{{\sqrt{2\pi e} }}{\varepsilon })^n}$

2. According to Eq. (10), $\bar{{{\varvec{x}}}} = {{\varvec{x}}} + \varvec{\delta }$ where the stochastic disturbance $\varvec{\delta }$ follows n-dimensional standard Cauchy distribution. Then, we have $\varvec{\delta }=\bar{{{\varvec{x}}}} - {{\varvec{x}}}$ and ${\delta _j}\sim {C_{\phi = 1}}(y)$ ($j=1, 2, \ldots , n$).

Following the Proof of Theorem 3,

$$\begin{aligned}P\{ \bar{x} \in {{\mathbf{S}}^{\mathbf{*}}}(\varepsilon )\} & = P\{ {\varvec{\delta }} \in {{\tilde{\mathbf{S}}}}\}\\&= \int \cdots \int _{\tilde{S}} {\prod \nolimits _{j = 1}^n {{C_{\phi = 1}}({\delta _j})} d} {\delta _1} \ldots d{\delta _n}\\&= \int \cdots \int _{\tilde{S}} {{{\left(\frac{1}{\pi }\right)}^n}\frac{1}{{\prod \nolimits _{j = 1}^n {(1 + \delta _j^2)} }}d} {\delta _1} \ldots d{\delta _n}\\&\ge \int \cdots \int _{\tilde{S}} {{{\left(\frac{1}{\pi }\right)}^n}\frac{1}{{\prod \limits _{j = 1}^n {(1 + 1)} }}d} {\delta _1} \ldots d{\delta _n}\\&= \int \cdots \int _{\tilde{S}} {{{\left(\frac{1}{{2\pi }}\right)}^n}d} {\delta _1} \ldots d{\delta _n}\\&= {\left(\frac{\varepsilon }{{2\pi }}\right)^n} \end{aligned}$$

Hence, $E{\mu _2} \le {(\frac{{2\pi }}{\varepsilon })^n}$ owning to Corollary 4.