obtain that $$\lim_{t\to\infty}t^{(s-d)/\alpha}\P\(\max_{(\xi,r)\in X\cap S}|\xi|\ge(t(1-\varepsilon))^{1/\alpha}\)=d\kappa_d \beta(s-d)^{-1}((1-\varepsilon))^{(d-s)/\alpha}.$$ , $\P(\#(X\cap S)\ge \varepsilon t^{1/4})$ tends to $0$ exponentially fast in $t^{1/4}$ since $\#(X\cap S)$ is a Poissonian random variable with finite mean.

Finally, $\#(X\cap S\setminus (B_{t^{3/(4\alpha)}}(o)\times[0,\infty)))$ is a Poissonian random variable whose mean is at most $2d\kappa_d \beta(s-d)^{-1}t^{3(d-s)/(4\alpha)}$. Hence, the Poisson concentration property yields $$\begin{aligned}
\P(\#(X\cap S\setminus (B_{t^{3/(4\alpha)}}(o)\times[0,\infty)))\ge 2)\in O(t^{-3(s-d)/(2\alpha)}).\end{aligned}$$ Since $\varepsilon>0$ was arbitrary, this completes the proof of Theorem \[inEdgeThm\].

Theorems \[outEdgeThm\] and \[inEdgeThm\] show that on the distributional level, there is a substantial difference between power-weighted sums of outgoing and incoming edge lengths. However, when moving to the level of expectations, these differences disappear. In fact, the equality of expectations is an immediate consequence of Slivnyak’s theorem, and is true for a much more general class of Poisson-based random geometric graphs. Still, for the convenience of the reader, we present some details.

Let $\alpha>0$ be arbitrary. Then $\E D^{(\alpha)}_{\out,n}=\E D^{(\alpha)}_{\inn,n}$.

For $(\xi,r),(\eta,t)\in X^{(n)}$ we write $((\xi,r),(\eta,t))\in G(X^{(n)})$ if there is a directed edge from $(\xi,r)$ to $(\eta,t)$ in the graph $G(X^{(n)})$. Then an application of Slivnyak’s theorem yields $$\begin{aligned}
n^d\E D^{(\alpha)}_{\out,n}&= \E\sum_{(\xi,r)\in X^{(n)}}\sum_{(\eta,t)\in X^{(n)}}\d_{\T_n}(\xi,\eta)^{\alpha}1_{((\xi,r),(\eta,t))\in G(X^{(n)})}\\
&= \E\sum_{(\eta,t)\in X^{(n)}}\sum_{(\xi,r)\in X^{(n)}}\d_{\T_n}(\xi,\eta)^{\alpha}1_{((\xi,r),(\eta,t))\in G(X^{(n)})}\\
&=n^d\E D^{(\alpha)}_{\inn,n}.\end{aligned}$$

Now, we can proceed with the proof of Theorem 