tends to $0$. Indeed, $D_{v(-\varepsilon)t^{1/(\alpha+d)}}$ has expectation $(1-\varepsilon)^{-(\alpha+d)}t$ and variance $$v(-\varepsilon)^{2\alpha+d}t^{(2\alpha+d)/(\alpha+d)} v_{2\alpha}^{\alpha+d},$$ the latter claim follows from the Chebyshev inequality.

The proof of Theorem \[inEdgeThm\] relies on a rather delicate comparison of $D^{(\alpha)}_{\inn}$ and the random variable $\max_{(\xi,r)\in X}|\xi|^\alpha1_{B_r(o)}(\xi)$.

First, observe that the in-degree of the origin $D^{(0)}_{\inn,n}$ is a Poissonian random variable with mean $$\int_{\T_n}\P(R>|\xi|)\d\xi,$$ which is at least ${d\kappa_d}\int_0^{n/2} r^{d-1}\P(R>r)\d r.$ If $s\le d$, then this lower bound tends to $\infty$ as $n\to\infty$. Using $D^{(\alpha)}_{\inn,n}\le D^{(\alpha)}_{\inn} $, this shows that $\P\big(D^{(\alpha)}_{\inn}=\infty\big)=1$ and that $(D^{(\alpha)}_{\inn,n})_{n\ge1}$ converge in distribution to $D^{(\alpha)}_{\inn}$. In the following, we may therefore assume that $s>d$. To show the assertion on the convergence in distribution, we proceed as in the proof of Theorem \[inEdgeThm\]. Indeed, using the canonical coupling between the $[0,\infty)$-marked Poisson point process $X^{(n)}$ on the torus $\T_n$ and the $[0,\infty)$-marked Poisson point process $X$ on $\R^d$, we deduce that $$\P(D^{(\alpha)}_{\inn,n}\ne D^{(\alpha)}_{\inn})\le \P(X\cap S\setminus ( B_{n/2}(o)\times[0,\infty))\ne\es),$$ where $$S=\{(\xi,r)\in\R^d\times[0,\infty):\, r>|\xi|\}.$$ Note that $\#(X\cap S\setminus (B_{n/2}(o)\times[0,\infty)))$ is Poissonian with mean $$\begin{aligned}
\int_{\R^d\setminus B_{n/2}(o)}\P(R>|\xi|) \d \xi&={d\kappa_d} \int_{n/2}^\infty r^{d-1}\P(R>r) \d r.\end{aligned}$$ The assumption $s>d$ implies that the latter integral tends to $0$ as $n\to\infty$. We also observe that by definition, $D^{(0)}_{\inn}$ is a Poissonian random variable with mean 