Talk:Random geometric graph

It seems that there is an issue with the expectation value ${\displaystyle E(x)=n(1-\pi r^{2})^{n-1}}${\displaystyle E(x)=n(1-\pi r^{2})^{n-1}}. If one takes the logarithm of this expression and uses a Taylor expansion up to ${\displaystyle {\mathcal {O}}(r^{4}n)}${\displaystyle {\mathcal {O}}(r^{4}n)}, one obtains ${\displaystyle E(x)=n(1-\pi r^{2})^{n-1}=ne^{-\pi r^{2}n-{\mathcal {O}}(r^{4}n)}}${\displaystyle E(x)=n(1-\pi r^{2})^{n-1}=ne^{-\pi r^{2}n-{\mathcal {O}}(r^{4}n)}}. This is also correctly stated in the published version of the cited arXiv article: Díaz, J., Mitsche, D., & Pérez-Giménez, X. (2009). On the probability of the existence of fixed-size components in random geometric graphs. Advances in Applied Probability, 41(2), 344-357. I would suggest to fix it and update the reference. Version 1 of the arXiv article also shows the correct result. — Preceding unsigned comment added by 195.176.113.53 (talk) 17:32, 10 October 2019 (UTC) [reply ]