## Appendix A Proofs  
**Proposition 1**. : From Equation (2), we have:  
\[D^{(n,m)} = T^{(n)}-T^{(m)}=\sum_{i=1}^{n}T_{ji}-\sum_{j=1}^{m}T_{jj}=\] \[= \sum_{i=1}^{n}T_{ji}+\sum_{j=1}^{m}\left(-T_{jj}\right)\]  
In the case of \(D_{U}^{n,m}\), \(T_{j}\) is a random variable uniformly distributed within \([0,J_{m}]\). Therefore, \(f_{D_{U}^{n,m}}(t)\) can be computed by using the well-known result for the pdf of the sum, \(S_{n}\), of \(n\) independent uniform random variables \(Uniform\left[0,1\right]\)[26]:  
\[f_{S_{n}}(x)=\frac{1}{(n-1)!}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(x-k)_{+}^{n-1} \tag{1}\]  
Note that \(f_{S_{n}}(x)=0\) for \(x>n\) and \(x<0\). Notation \(z_{+}\) denotes the positive part of \(z\), _i.e._, \(z_{+}=z\) if \(z>0\), \(z_{+}=0\) otherwise. Then, the probability of delay inversion between two paths with lengthsand \(m\) (\(n<m\)), becomes \(Pr(D_{U}^{(n,m)}>0)|_{n<m}\). The expression of \(P_{U}\equiv Pr(D_{U}^{(n,m)}>0)\) is computed in Equation (A2):  
\[P_{U} = \frac{1}{(n+m)!}\sum_{k=0}^{n+m}\left[(-1)^{k}{n+m\choose k}\times\right.\] (A2) \[\times\left.\left((n+2m-k)^{n+m}-(m-k)_{+}^{n+m}\right)\right]\]  
\(\square\)  
**Proposition 2**. : \[f_{D}(t) = \left(\bigotimes_{i=1}^{n}f_{T_{j_{i}}^{\prime}}(t)\right)\otimes \left(\bigotimes_{i=1}^{m}f_{(-T_{j_{i}}^{\prime})}(t)\right)\] (A3)  
\[\left\{ \begin{array}{rcl}f_{T_{j_{i}}^{\prime}}(t)&=&\frac{1}{1-\alpha}p \left(\frac{t-\alpha}{1-\alpha}\right)\\ f_{(-T_{j_{i}}^{\prime})}(t)&=&\frac{1}{1-\alpha}p\left(\frac{t-\alpha}{1- \alpha}\right)=\frac{1}{1-\alpha}p\left(\frac{t+1}{1-\alpha}\right)\end{array} \right.\] (A4)  
where:  
\[p(t) = Heaviside(t)-Heaviside(t-1)\]  
Then, Equation (A3) becomes, by using Equation (A4):  
\[(1-\alpha)^{n+m-1}\times\frac{1}{(1-\alpha)^{n+m}}\left(\bigotimes_{i=1}^{n+m }p\right)\left(\frac{t-n\alpha+m}{1-\alpha}\right)\] (A5)  
Note that \(\left(\bigotimes_{i=1}^{n+m}p\right)(x)\) is the pdf of the sum of \(n+m\) uniform independent random variables within \([0,1]\). Therefore, by applying Equation (A1), Equation (A5) becomes:  
\[\frac{(1-\alpha)^{-1}}{(n+m-1)!}\sum_{k=0}^{n+m}(-1)^{k}{n+m\choose k}\left( \frac{t-n\alpha+m}{1-\alpha}-k\right)_{+}^{n+m-1}\] (A6)  
\(\square\)  
**Proposition 3**. : From the definition of probability, \(Pr(D_{W}>0)=\int_{0}^{\infty}f_{D_{W}}(x)dx\), which is:  
\[P_{W} = \int_{0}^{\infty}\frac{dt\sum_{k=0}^{n+m}(-1)^{k}{n+m\choose k} \left(\frac{t-n\alpha+m}{1-\alpha}-k\right)_{+}^{n+m-1}}{(1-\alpha)(n+m-1)!}=\] (A7) \[= \frac{\sum_{k=0}^{n+m}(-1)^{k}{n+m\choose k}\int_{0}^{\infty}dt \left(\frac{t-n\alpha+m}{1-\alpha}-k\right)_{+}^{n+m-1}}{(1-\alpha)(n+m-1)!}\]  
The integrand of Equation (A7) is zero when \(\frac{t-n\alpha+m}{1-\alpha}>n+m\Longleftrightarrow t>n-m\alpha\), and \(n-m\alpha>0\) (see Proposition 2); therefore, Equation (A7) can be rewritten as follows:  
\[P_{W} = \frac{\sum_{k=0}^{n+m}(-1)^{k}{n+m\choose k}\int_{0}^{M}dt\left( \frac{t-n\alpha+m}{1-\alpha}-k\right)_{+}^{n+m-1}}{(1-\alpha)(n+m-1)!}\] (A8)where \(M\equiv\max\{0,n-m\alpha\}\). Consider the integral \(I_{0}\equiv\int_{0}^{M}dt\left(\frac{t-n\alpha+m}{1-\alpha}-k\right)_{+}^{n+m-1}\). The integrand of \(I_{0}\) is non-zero when:  
\[\frac{t-n\alpha+m}{1-\alpha}-k>0\Longrightarrow t>k(1-\alpha)+n\alpha-m \equiv t_{0}\] (A9)  
and zero otherwise. Therefore, integration limits of \(I_{0}\) can be modified as follows:  
\[I_{0} = \int_{\max\{0,t_{0}\}}^{\max\{0,n-m\alpha\}}dt\left(\frac{t-n\alpha +m}{1-\alpha}-k\right)^{n+m-1}\] (A10)  
\(t_{0}\) depends on \(k\). It is worth observing that:  
* \(t_{0}<n-m\alpha\), \(\forall k:0\leq k\leq n+m\). And \(n-m\alpha>0\) if \(\alpha>\frac{n}{m}\). * \(t_{0}>0\Longleftrightarrow k(1-\alpha)+n\alpha-m>0\), for \(k_{\min}=0\) and \(k_{\max}=n+m\). This implies that \(\alpha\leq\frac{n}{m}\) and \(\alpha>\frac{m}{n}\). * For \(\alpha\leq\frac{n}{m}\), \(\alpha\leq\frac{m}{n}\), there is a \(k^{*}=\lfloor\frac{m-n\alpha}{1-\alpha}\rfloor_{+}\geq 0\) such that \(t_{0}>0,\forall k>k^{*}\).