## 4 Theoretical Analysis of Network Congestion Detection Based on Flow Watermarking  
### Analysis of the SRW without Interference  
We define the SRW(success rate of the watermark) as the probability that the two bits of watermark are decoded correctly. That is, when \(f=00\), the SRW is expressed as \(Pr[(\overline{Y^{f}_{r2}}<0)(\overline{Y^{f}_{r3}}>0)]\); when \(f=01\), the SRW is \(Pr[(\overline{Y^{f}_{r2}}\geq 0)(\overline{Y^{f}_{r3}}<0)]\); when \(f=10\), the SRW is \(Pr[(\overline{Y^{f}_{r2}}<0)(\overline{Y^{f}_{r3}}\leq 0)]\); when \(f=11\), the SRW is \(Pr[(\overline{Y^{f}_{r2}}\geq 0)(\overline{Y^{f}_{r3}}\geq 0)]\). \(Y_{k}\) are defined as \(Y^{f}_{k}\) (\(k=1;2,\ldots;r\)) after embedding watermark \(f\). Since \(\overline{Y}_{r}\) is symmetrical about the mean value \(0\), only one case needs to be analyzed. Take \(Pr[(\overline{Y^{f}_{r2}}\geq 0)(\overline{Y^{f}_{r3}}<0)]\) when \(f=01\) as an example.  
\(Y_{k}\) satisfy random sampling, so do \(Y^{f}_{k}\). According to the central limit theorem, in the case of random sampling with a large sample, \(\overline{Y^{f}_{r2}}\) and \(\overline{Y^{f}_{r3}}\) can be considered to satisfy the standard normal distribution; thus, the distribution of probability satisfies:  
\[Pr\left[\frac{\left(\overline{Y^{f}_{r2}}-E\left(\overline{Y^{f}_{r2}}\right) \right)}{\sqrt{V\left(\overline{Y^{f}_{r2}}\right)}}\geq\xi\right]\approx 1- \Phi(\xi)\text{, }Pr\left[\frac{\left(\overline{Y^{f}_{r3}}-E\left(\overline{Y^{f}_{r3}} \right)\right)}{\sqrt{V\left(\overline{Y^{f}_{r3}}\right)}}<\xi\right]\approx \Phi(\xi), \tag{13}\]  
where  
\[\Phi(\xi)=\int_{-\infty}^{\xi}\frac{1}{\sqrt{2\pi}}e^{-p^{2}/2}du \tag{14}\]  
After embedding watermark '01', \(X_{2,k}\) and \(X_{3,k}\) are defined as \(X^{f}_{2,k}\) and \(X^{f}_{3,k}\), namely  
\[X^{f}_{2,k}=X_{1,k}+X_{2,k},X^{f}_{3,k}=0 \tag{15}\]  
From Formula (5) we can obtain:  
\[E\left(X^{f}_{2,k}\right)=2\mu_{x},E\left(X^{f}_{3,k}\right)=0 \tag{16}\]  
\[V\left(X^{f}_{2,k}\right)=\sigma_{x}^{2}+\sigma_{x}^{2}+2 Corr(X_{1,k},X_{2,k} )\sigma_{x}^{2},V\left(X^{f}_{3,k}\right)=0 \tag{17}\]  
The maximum variance occurs when \(Corr(X_{1,k},X_{2,k})=1\).  
Therefore,  
\[E\left(\overline{Y^{f}_{r2}}\right)=\mu_{x},V\left(\overline{Y^{f}_{r2}} \right)\leq 35\sigma_{x}^{2}/8r. \tag{18}\]  
\[E\left(\overline{Y^{f}_{r3}}\right)=-\mu_{x},V\left(\overline{Y^{f}_{r3}} \right)\leq 3\sigma_{x}^{2}/8r \tag{19}\]  
From Formula (13) we can obtain:  
\[P\left[\overline{Y^{f}_{r2}}\geq 0\right]=Pr\left[\frac{\left(\overline{Y^{f}_{r2} }-E\left(\overline{Y^{f}_{r2}}\right)\right)}{\sqrt{V\left(\overline{Y^{f}_{r 2}}\right)}}\geq\frac{-E\left(\overline{Y^{f}_{r2}}\right)}{\sqrt{V(\overline {Y^{f}_{r2}})}}\right]\approx 1-\Phi\left(\frac{-E\left(\overline{Y^{f}_{r2}} \right)}{\sqrt{V\left(\overline{Y^{f}_{r2}}\right)}}\right)\geq 1-\Phi \left(\frac{-\sqrt{8r}\mu_{x}}{\sqrt{35}\sigma_{x}}\right)\!. \tag{20}\]  
Similarly,\[P\left[\overline{\gamma_{r3}^{f}}<0\right]=Pr\left[\frac{\left(\overline{\gamma_{r3 }^{f}}-E\left(\overline{\gamma_{r3}^{f}}\right)\right)}{\sqrt{V\left(\overline{ Y_{r3}^{f}}\right)}}<\frac{-E\left(\overline{Y_{r3}^{f}}\right)}{\sqrt{V \left(\overline{Y_{r3}^{f}}\right)}}\right]\approx\[PERSON]\left(\frac{-E\left( \overline{Y_{r3}^{f}}\right)}{\sqrt{V\left(\overline{Y_{r3}^{f}}\right)}} \right)\geq\[PERSON]\left(\frac{\sqrt{8}\mu_{x}}{\sqrt{3}\sigma_{x}}\right). \tag{21}\]  
The SRW without interference is  
\[P[(\overline{Y_{r2}^{f}}\geq 0)(\overline{Y_{r3}^{f}}<0)]\geq(1-\Phi\left( \frac{-\sqrt{8}\mu_{x}}{\sqrt{35}\sigma_{x}}\right))\cdot\Phi(\frac{\sqrt{8} \mu_{x}}{\sqrt{3}\sigma_{x}}) \tag{22}\]  
It can be concluded that the SRW without external interference increases with the increase of \(u_{x}\), and decreases with the increase of \(\sigma_{x}\).